Polymath Wiki - User contributions [en]

Imo 2010

2010-07-08T17:53:43Z

Mpeake:

This is the wiki page for the mini-polymath2 project, which seeks solutions to Question 5 of the 2010 International Mathematical Olympiad.

The project will start at [http://www.timeanddate.com/worldclock/fixedtime.html?year=2010&month=7&day=8&hour=16&min=0&sec=0&p1=0 16:00 UTC July 8], and is hosted at the [http://polymathprojects.org/ polymath blog]. A discussion thread is hosted at [http://terrytao.wordpress.com Terry Tao's blog].

== Rules ==

This project will follow the [http://polymathprojects.org/general-polymath-rules/ usual polymath rules]. In particular:

* Everyone is welcome to participate, though people who have already seen an external solution to the problem should probably refrain from giving spoilers throughout the experiment.
* This is a team effort, not a race between individuals. Rather than work for extended periods of time in isolation from the rest of the project, the idea is to come up with short observations (or to carry an observation of another participant further) and then report back what one gets to the rest of the team. Partial results or even failures can be worth reporting.
* Participants are encouraged to update the wiki, or to summarise progress within threads, for the benefit of others.

== Threads ==

Discussion and planning:

* [http://terrytao.wordpress.com/2010/06/12/future-mini-polymath-project-2010-imo-q6/ Future mini-polymath project: 2010 IMO Q6?] June 12, 2010.
* [http://terrytao.wordpress.com/2010/06/21/organising-mini-polymath2/ Organising mini-polymath2] June 21, 2010.
* [http://terrytao.wordpress.com/2010/06/27/mini-polymath2-start-time/ Mini-polymath2 start time], June 27, 2010.
* [http://terrytao.wordpress.com/2010/07/08/mini-polymath2-discussion-thread/ Mini-polymath2 discussion thread], July 8 2010.

Research:

* [http://polymathprojects.org/2010/07/08/minipolymath2-project-imo-2010-q5/ Minipolymath2 project: IMO 2010 Q5], July 8 2010.

== The question ==

The question to be solved is Question 5 of the [http://www.imo-official.org/problems.aspx 2010 International Mathematical Olympiad]:

: '''Problem''' In each of six boxes <math>B_1, B_2, B_3, B_4, B_5, B_6</math> there is initially one coin. There are two types of operation allowed:
:
: ''Type 1:'' Choose a nonempty box <math>B_j</math> with <math>1 \leq j \leq 5</math>. Remove one coin from <math>B_j</math> and add two coins to <math>B_{j+1}</math>.
:
: ''Type 2:'' Choose a nonempty box <math>B_k</math> with <math>1 \leq k \leq 4</math>. Remove one coin from <math>B_k</math> and exchange the contents of (possibly empty) boxes <math>B_{k+1}</math> and <math>B_{k+2}</math>.
:
: Determine whether there is a finite sequence of such operations that results in boxes <math>B_1, B_2, B_3, B_4, B_5</math> being empty and box <math>B_6</math> containing exactly <math>2010^{2010^{2010}}</math> coins. (Note that <math>a^{b^c} := a^{(b^c)}</math>.)

== Observations and partial results ==

* If the left-most box <math>B_1</math> becomes empty, then it cannot ever become non-empty again. Furthermore, the left-most box can never have more than one coin; it can be touched exactly once.
* Define the ''worth'' W of a state to be <math>W = B_6 + 2 B_5 + 4 B_4 + 8 B_3 + 16 B_2 + 32 B_1</math>. Then the initial worth is 63, the final desired worth is <math>2010^{2010^{2010}}</math>, and the Type 1 move does not affect the worth. On the other hand, the Type 2 move increases the worth when <math>B_{j+2} - B_{j+1} \geq 4</math>.
* Once one has a large number of coins in one of the first four boxes, say <math>B_k</math>, one can apply the Type 2 move repeatedly to remove coins from <math>B_k</math> while swapping <math>B_{k+1}</math> and <math>B_{k+2}</math> repeatedly. This suggests that it is relatively easy to remove coins from the system; the difficulty is in adding coins to the system.

== Possible strategies ==

* Split the problem into two pieces. Part I: try to show the weaker result that the number of coins in the system can eventually be as large as <math>2010^{2010^{2010}}</math>. Part II: Show that once one has a lot of coins, one can move to the final state where <math>B_1=\ldots=B_5=0</math> and <math>B_6 = 2010^{2010^{2010}}</math>.
* Try to show that a quantity such as the worth increases or decreases in a controlled manner as one applies the Type 1 and Type 2 moves.
* We know that the first box can never contain more than one coin. What can we say about the second box, third box, etc.?
** There may be a recursive formula for the maximal size of box <math>B_j</math>, possibly requiring one to solve the five-box, four-box, etc. problems first.
* Work backwards?
* Try to completely solve the three-box problem (say) first: starting from <math>[X,Y,Z]</math>, what is the most number of coins one can generate?

== Compound moves ==

Here we use Type 1 move <math>[X,Y] \mapsto [X-1,Y+2]</math> and the Type 2 move <math>[N,X,Y] \mapsto [N-1,Y,X]</math> to create more advanced moves.

# We can create the move <math>[X,Y] \mapsto [0,Y+2X]</math> from repeated application of Type 1.
# We have <math>[1,X,Y] \mapsto [0,0,X+2Y]</math> by applying Type 2 once and then Type 1 Y times.
#* Or, by using advanced move 1 first, the move <math>[1, X, Y] \to [1, 0, Y+2X] \to [0, Y+2X, 0] \to [0, 0, 2Y+4X]</math>.
# More generally, we have <math>[N,0,0] \to [N-1,2,0] \to [N-1,0,4] \to [N-2,4,0] \to [N-2,0,8] \to \ldots \to [0,0,2^{N+1}]</math>.
# <math>[a,b,0,0] \to [a,0,0,2^b] \to [a-1,1,0,2^{b+1}] \to [a-2,2^{b+1},0,0] \to [a-2, 0,0,2^{2^{b+1}}] \to \cdots</math>. So let <math> x_0=b, x_{n+1}=2^{x_n}</math> then we obtain <math>[0,x_a,0,0]</math>, which is huge enough.

== World records ==

To make the second box as big as possible:

* <math>[1,1,1] \mapsto [1,0,3] \mapsto [0,3,0]</math> places 3 coins in box 2.

To make the third box as big as possible:

* <math>[1,1,1] \mapsto [0,3,1] \mapsto [0,0,7]</math> places 7 coins in box 3. (Here we use advanced move 2).

To make the fourth box as big as possible

* <math>[1,1,1,1] \to [0,3,0,3] \to [0,2,2,3] \to [0,2,0,7] \to [0,1,7,0] \to [0,1,0,14]</math> <math> \to [0,0,14,0] \to [0,0,0,28]</math> gives 28 coins in box 4.

To make the sixth box as big as possible

* Using the fourth box move we have <math>[0,0,0,28,1,1]</math>. We can move <math>[28,1,1]</math> to <math>[27,0,7]</math> through Type 1; then using a variant of advanced move 3 we get <math>[0,0,7 * 2^{27}]</math>, leading to <math>7 \times 2^{27}</math> coins in the last box.

== Completed solutions ==

=== First solution ===

=== Second solution ===

Integer.tex

2009-12-12T23:11:42Z

Mpeake:

\section{Integer programming}\label{integer-sec}

Integer programming is linear programming with solutions restricted to the integers. A number of packages, for example Maple.12, can solve these problems. It was used in several places during this project.

The first place it was used was to find $c_5$, the size of the maximal subset of $[3]^5$ with no combinatorial lines. One 0/1-variable was used for each point in the cube, and one linear inequality for each combinatorial line. The inequalities were $x_i \geq 0$, $x_i\leq 1$, and $x_i+x_j+x_k\leq 2$ whenever the three points $i,j,k$ form a line. The aim is then to maximize $\sum x_i$. The computer found twelve solutions with 150 points, and found that there were no solutions with 151 points, so $c_5 = 150$. It follows easily that $c_6 = 450$, but the same program was used as a check to show that any set of 451 points in $[3]^6$ has a combinatorial line.

The second application of integer programming was the following routine, used for the 5 dimensional Moser problem to show that $c'_5\ge 124$.
Let G be the 3-hypergraph with vertex set equal to $[3]^5$ as before and set $(I,J,K)\in E_3$ to be a 3-edge if and only if it is a geometric line. $c'_5$ is then the size of a maximal independent set.
Let $$ f(x) = \sum_I x_I - \sum_{(I,J,K)\in E_3} x_Ix_Jx_K. $$

Let $A$ be the solid cube $[0,1]^{3^n}$, and $B$ its set of vertices $\{0,1\}^{3^n}$. The following then holds,
$$c'_n = \max_{x\in B} f(x) = \max_{x\in A} f(x).$$

To check the first equality, if $V$ is a line-free set in $G$ of size $k$, set $x_J=1$ for $J\in V$, and zero otherwise. Then $f(x) = k$ because $V$ is line-free, so all the cubic terms are zero. So $c'_n \le max_{x\in\{0,1\}^{3^n}} f(x)$. Conversely, if for some $x\in\{0,1\}^{3^n}$, $f(x)=k$ and if there are triplets $x_I=x_J=x_K=1$ with $(I,J,K)\in E_3$, we change $x_I$ to 0. This reduces $\sum_I x_I$ by 1 but also reduce the second term of $f(x)$ by at least one. So repeating we can find an $x\in\{0,1\}^{3^n}$ with $f(x)\ge k$ and $x_Ix_Jx_K=0$ if $(I,J,K)\in E_3$, i.e. an independent set of size $\ge k$. So $c'_n \ge max_{x\in\{0,1\}^{3^n}} f(x)$. The second equality follows from the maximum principle since $f$ is a square free polynomial.
$-f$ was minimized using Matlab's built-in minimization routine, and found a 124-point subset of $[3]^5$ with no geometric lines.

A third application of integer programming is expanded in some detail in Section \ref{moser-upper-sec}. Each point was classified by how many of its coordinates were 2. All the possible histograms or `statistics' of points in a low-dimensional line-free set were computed by brute-force. Since a low-dimensional space fits into a higher-dimensional space in many ways, the low-dimensional statistics have consequences for higher-dimensional sets. Many dozens of inequalities were found for six-dimensional and higher spaces, and Maple's integer programming routine was used to find upper bounds on $c'_n$.

A fourth application of integer programming found many of our best lower bounds for both the DHJ problem and the Moser problem. Assume that a solution is made up of complete $\Gamma(a,b,c)$ cells. This drastically reduces the number of variables, from $3^n$ to $(n+2)(n+1)/2$ binary variables, one for each Gamma set. In the DHJ problem, points in a combinatorial line will lie in cells of the form $\Gamma(a+r,b,c)$, $\Gamma(a,b+r,c)$ and $\Gamma(a,b,c+r)$ for some positive $r$. Thus the line-free condition is met if we prevent equilateral triangles of this sort. The Fujimura problem was to maximize the number of Gamma sets under this restriction, searching over the $(n+2)(n+1)/2$ binary variables. To maximize the total number of points, each Gamma set $\Gamma(a,b,c)$ is weighted by its size $(a+b+c)!/(a!b!c!)$, but the restrictions are the same.

Conversely, an attempt was made to disprove the `hyper-optimistic conjecture' mentioned in the Introduction. Each point of $[3]^n$ was given a weight of $1/(a!b!c!)$ when the point's coordinates included $a$ 1s, $b$ 2s and $c$ 3s. This gave each $\Gamma_{a,b,c}$ cell an equal weight of 1. In fact, since an integer programming routine was being used, each point was given the weight $(a+b+c)!/(a!b!c!)$. These numbers quickly grew too large for the routine.

In the Moser-Fujimura problem, we look for collections of $\Gamma(a,b,c)$ that do not contain geometric lines. This is satisfied if there are no isosceles triangles of the form $(a+r,b,c+s)$, $(a,b+r+s,c)$, $(a+s,b,c+r)$. Again, we run a search over the $(n+2)(n+1)/2$ binary variables, with $\Gamma(a,b,c)$ weighted by $(a+b+c)!/(a!b!c!)$, to maximize the total number of points.

Outline of second paper

2009-11-20T06:15:50Z

Mpeake: /* Introduction */

Here is a proposed outline of the second paper, which will focus on the new bounds on DHJ(3) and Moser numbers, and related quantities.

== Metadata ==

* '''Author''': D.H.J. Polymath
* '''Address''': http://michaelnielsen.org/polymath1/index.php (Do we need a more stable address?)
* '''Email''': ???
* '''Title''': Density Hales-Jewett and Moser numbers
* '''AMS Subject classification''': ???

== Abstract ==

(A draft proposal - please edit)

For any <math>n \geq 0</math> and <math>k \geq 1</math>, the density Hales-Jewett number <math>c_{n,k}</math> is defined as the size of the largest subset of the cube <math>[k]^n := \{1,\ldots,k\}^n</math> which contains no combinatorial line; similarly, the Moser number <math>c'_{n,k}</math> is the largest subset of the cube <math>[k]^n</math> which contains no geometric line. A deep theorem of Furstenberg and Katznelson [cite] shows that <math>c_{n,k} = o(k^n)</math> as <math>n \to \infty</math> (which implies a similar claim for <math>c'_{n,k}</math>; this is already non-trivial for <math>k=3</math>. Several new proofs of this result have also been recently established [cite Polymath], [cite Austin].

Using both human and computer-assisted arguments, we compute several values of <math>c_{n,k}</math> and <math>c'_{n,k}</math> for small <math>n,k</math>. For instance the sequence <math>c_{n,3}</math> for <math>n=0,\ldots,6</math> is <math>1,2,6,18,52,150,450</math>, while the sequence <math>c'_{n,3}</math> for <math>n=0,\ldots,6</math> is <math>1,2,6,16,43,124,353</math>. We also establish some results for higher <math>k</math>, showing for instance that an analogue of the LYM inequality (which relates to the <math>k=2</math> case) does not hold for higher <math>k</math>.

== Sections ==

=== Introduction ===

Basic definitions. Definitions and notational conventions include

* [k] = {1, 2, ..., k}
* Subsets of [k]^n are called A
* definition of combinatorial line, geometric line
* Hales-Jewett numbers, Moser numbers

History of and motivation for the problem:

* Sperner's theorem
* Density Hales-Jewett theorem, including new proofs
* Review literature on Moser problem

New results

* Computation of several values of <math>c_{n,3}</math>
* Computation of several values of <math>c'_{n,3}</math>
* Asymptotic lower bounds for <math>c_{n,k}</math>
* Genetic algorithm lower bounds
* Some bounds for <math>c_{n,k}</math> for low n and large k
* Connection between Moser(2k) and DHJ(k)
* Hyper-optimistic conjecture, and its failure
* New bounds for colouring Hales-Jewett numbers
* Kakeya problem for <math>Z_3^n</math>

=== Lower bounds for density Hales-Jewett ===

Fujimura implies DHJ lower bounds; some selected numerics (e.g. lower bounds up to 10 dimensions, plus a few dimensions afterwards).

The precise asymptotic bound of <math>c_{n,k} > C k^{n - \alpha(k)\sqrt[\ell]{\log n}+\beta(k) \log \log n}</math>

Discussion of genetic algorithm

=== Low-dimensional density Hales-Jewett numbers ===

==== Very small n ====

<math>n=0,1,2</math> are trivial. But the six-point examples will get mentioned a lot.

For <math>n=3</math>, one needs to classify the 17-point and 18-point examples.

==== n=4 ====

One needs to classify the 50-point, 51-point, and 52-point examples.

==== n=5 ====

This is the big section, showing there are no 151-point examples.

==== n=6 ====

Easy corollary of n=5 theory

=== Higher k DHJ numbers ===

Exact computations of <math>c_{2,k}, c_{3,k}</math>

Connection between Moser<math>(n,2k)</math> and DHJ<math>(n,k)</math>

Numerics

Failure of hyper-optimistic conjecture

=== Lower bounds for Moser ===

Using Gamma sets to get lower bounds

Adding extra points from degenerate triangles

Higher k; Implications between Moser and DHJ

=== Moser in low dimensions ===

There is some general slicing lemma that needs to be proved here that allows inequalities for low-dim Moser to imply inequalities for higher dim.

For n=0,1,2 the theory is trivial.

For n=3 we need the classification of Pareto optimal configurations etc. So far this is only done by computer brute force search; we may have to find a human version.

n=4 theory: include both computer results and human results

n=5: we have a proof using the n=4 computer data; we should keep looking for a purely human proof.

n=6: we can give the partial results we have.

=== Fujimura's problem ===

=== Coloring DHJ ===

== Files ==

* [[polymath.tex]]
* [[introduction.tex]]
* [[dhj-lown.tex]]
* [[dhj-lown-lower.tex]]
* [[moser.tex]]
* [[moser-lower.tex]]
* [[fujimura.tex]]
* [[higherk.tex]]
* [[genetic.tex]]
* [[integer.tex]]
* [[coloring.tex]]
* [http://terrytao.wordpress.com/files/2009/07/alllinesin3n.pdf A figure depicting combinatorial lines]
* [http://terrytao.wordpress.com/files/2009/07/allgeomlinesin3n.pdf A figure depicting geometric lines]
* [http://thomas1111.files.wordpress.com/2009/06/moser353new.png A figure depicting a 353-point 6D Moser set]

The above are the master copies of the LaTeX files. Below are various compiled versions of the source:

* [http://terrytao.wordpress.com/files/2009/05/polymath.pdf May 24 version]
* [http://terrytao.wordpress.com/files/2009/05/polymath1.pdf May 25 version]
* [http://terrytao.files.wordpress.com/2009/05/polymath2.pdf May 27 version]
* [http://terrytao.wordpress.com/files/2009/05/polymath3.pdf Jun 1 version]
* [http://terrytao.wordpress.com/files/2009/06/polymath.pdf Jun 3 version]
* [http://terrytao.files.wordpress.com/2009/06/polymath2.pdf Jun 18 version]
* [http://terrytao.wordpress.com/files/2009/07/polymath.pdf Jul 9 version]
* [http://terrytao.wordpress.com/files/2009/07/polymath1.pdf Jul 25 version]

Integer.tex

2009-08-05T04:54:40Z

Mpeake:

\section{Integer programming}\label{integer-sec}

Integer programming is linear programming with solutions restricted to the integers. A number of packages, for example Maple.12, can solve these problems. It was used in several places during this project.

The first place it was used was to find $c_5$, the size of the maximal subset of $[3]^5$ with no combinatorial lines. One 0/1-variable was used for each point in the cube, and one linear inequality for each combinatorial line. The inequalities were $x_i \geq 0$, $x_i\leq 1$, and $x_i+x_j+x_k\leq 2$ whenever the three points $i,j,k$ form a line. The aim is then to maximize $\sum x_i$. The computer found twelve solutions with 150 points, and found that there were no solutions with 151 points, so $c_5 = 150$. It follows easily that $c_6 = 450$, but the same program was used as a check to show that any set of 451 points in $[3]^6$ has a combinatorial line.

The following routine was used for the 5 dimensional Moser problem to show that $c'_5\ge 124$.
Let G be the 3-hypergraph with vertex set equal to $[3]^5$ as before and set $(I,J,K)\in E_3$ to be a 3-edge if and only if it is a geometric line. $c'_5$ is then the size of a maximal independent set.
Let $$ f(x) = \sum_I x_I - \sum_{(I,J,K)\in E_3} x_Ix_Jx_K. $$

Let $A$ be the solid cube $[0,1]^{3^n}$, and $B$ its set of vertices $\{0,1\}^{3^n}$. The following then holds,
$$c'_n = \max_{x\in B} f(x) = \max_{x\in A} f(x).$$

To check the first equality, if $V$ is a line-free set in $G$ of size $k$, set $x_J=1$ for $J\in V$, and zero otherwise. Then $f(x) = k$ because $V$ is line-free, so all the cubic terms are zero. So $c'_n \le max_{x\in\{0,1\}^{3^n}} f(x)$. Conversely, if for some $x\in\{0,1\}^{3^n}$, $f(x)=k$ and if there are triplets $x_I=x_J=x_K=1$ with $(I,J,K)\in E_3$, we change $x_I$ to 0. This reduces $\sum_I x_I$ by 1 but also reduce the second term of $f(x)$ by at least one. So repeating we can find an $x\in\{0,1\}^{3^n}$ with $f(x)\ge k$ and $x_Ix_Jx_K=0$ if $(I,J,K)\in E_3$, i.e. an independent set of size $\ge k$. So $c'_n \ge max_{x\in\{0,1\}^{3^n}} f(x)$. The second equality follows from the maximum principle since $f$ is a square free polynomial.
$-f$ was minimized using Matlab's built-in minimization routine, and found a 124-point subset of $[3]^5$ with no geometric lines.

A third application of integer programming is expanded in some detail in Section \ref{moser-upper-sec}. Each point was classified by how many of its coordinates were 2. All the possible histograms or `statistics' of points in a low-dimensional line-free set were computed by brute-force. Since a low-dimensional space fits into a higher-dimensional space in many ways, the low-dimensional statistics have consequences for higher-dimensional sets. Many dozens of inequalities were found for six-dimensional and higher spaces, and Maple's integer programming routine was used to find upper bounds on $c'_n$.

A fourth application of integer programming found many of our best lower bounds for both the DHJ problem and the Moser problem. Assume that a solution is made up of complete $\Gamma(a,b,c)$ cells. This drastically reduces the number of variables, from $3^n$ to $(n+2)(n+1)/2$ binary variables, one for each Gamma set. In the DHJ problem, points in a combinatorial line will lie in cells of the form $\Gamma(a+r,b,c)$, $\Gamma(a,b+r,c)$ and $\Gamma(a,b,c+r)$ for some positive $r$. Thus the line-free condition is met if we prevent equilateral triangles of this sort. The Fujimura problem was to maximize the number of Gamma sets under this restriction, searching over the $(n+2)(n+1)/2$ binary variables. To maximize the total number of points, each Gamma set $\Gamma(a,b,c)$ is weighted by its size $(a+b+c)!/(a!b!c!)$, but the restrictions are the same.

Conversely, an attempt was made to disprove the `hyper-optimistic conjecture' mentioned in the Introduction. Each point of $[3]^n$ was given a weight of $1/(a!b!c!)$ when the point's coordinates included $a$ 1s, $b$ 2s and $c$ 3s. This gave each $\Gamma_{a,b,c}$ cell an equal weight of 1. In fact, since an integer programming routine was being used, each point was given the weight $(a+b+c)!/(a!b!c!)$. These numbers quickly grew too large for the routine.

In the Moser-Fujimura problem, we look for collections of $\Gamma(a,b,c)$ that do not contain geometric lines. This is satisfied if there are no isosceles triangles of the form $(a+r,b,c+s)$, $(a,b+r+s,c)$, $(a+s,b,c+r)$. Again, we run a search over the $(n+2)(n+1)/2$ binary variables, with $\Gamma(a,b,c)$ weighted by $(a+b+c)!/(a!b!c!)$, to maximize the total number of points.

Integer.tex

2009-07-27T14:06:55Z

Mpeake:

\section{Integer programming}\label{integer-sec}

Integer programming is linear programming with solutions restricted to the integers. A number of packages, for example Maple.12, can solve these problems. It was used in several places during this project.

The first place it was used was to find $c_5$, the size of the maximal subset of $[3]^5$ with no combinatorial lines. One 0/1-variable was used for each point in the cube, and one linear inequality for each combinatorial line. The computer found twelve solutions with 150 points, and found that there were no solutions with 151 points, so $c_5 = 150$. It follows easily that $c_6 = 450$, but the same program was used as a check to show that any set of 451 points in $[3]^6$ has a combinatorial line.

The following routine was used for the 5 dimensional Moser problem to show that $c'_5\ge 124$.
Let G be the 3-hypergraph with vertex set equal to $[3]^5$ as before and set $(I,J,K)\in E_3$ to be a 3-edge if and only if it is a geometric line. $c'_5$ is then the size of a maximal independent set.
Let $$ f(x) = \sum_I x_I - \sum_{(I,J,K)\in E_3} x_Ix_Jx_K. $$

The following then holds,
$$c'_n = \max_{x\in\{0,1\}^{3^n}} f(x) = \max_{x\in [0,1]^{3^n}} f(x).$$

To check the first equality, if $V$ is a line-free set in $G$ of size $k$, set $x_J=1$ for $J\in V$, and zero otherwise. Then $f(x) = k$ because $V$ is line-free, so all the cubic terms are zero. So $c'_n \le max_{x\in\{0,1\}^{3^n}} f(x)$. Conversely, if for some $x\in\{0,1\}^{3^n}$, $f(x)=k$ and if there are triplets $x_I=x_J=x_K=1$ with $(I,J,K)\in E_3$, we change $x_I$ to 0. This reduces $\sum_I x_I$ by 1 but also reduce the second term of $f(x)$ by at least one. So repeating we can find an $x\in\{0,1\}^{3^n}$ with $f(x)\ge k$ and $x_Ix_Jx_K=0$ if $(I,J,K)\in E_3$, i.e. an independent set of size $\ge k$. So $c'_n \ge max_{x\in\{0,1\}^{3^n}} f(x)$. The second equality follows from the maximum principle since $f$ is a square free polynomial.
$-f$ was then minimized using Matlab's built-in minimization routine, and found a 124-point subset of $[3]^5$ with no geometric lines.

A third application of integer programming is expanded in some detail in Section \ref{moser-upper-sec}. Each point was classified by how many of its coordinates were 2. All the possible histograms or `statistics' of points in a low-dimensional line-free set were computed by brute-force. Since a low-dimensional space fits into a higher-dimensional space in many ways, the low-dimensional statistics have consequences for higher-dimensional sets. Many dozens of inequalities were found for six-dimensional and higher spaces, and Maple's integer programming routine was used to find upper bounds on $c'_n$.

A fourth application of integer programming found many of our best lower bounds for both the DHJ problem and the Moser problem. Assume that a solution is made up of complete $\Gamma(a,b,c)$ cells. This drastically reduces the number of variables, from $3^n$ to $(n+2)(n+1)/2$ binary variables, one for each Gamma set. In the DHJ problem, points in a combinatorial line will lie in cells of the form $\Gamma(a+r,b,c)$, $\Gamma(a,b+r,c)$ and $\Gamma(a,b,c+r)$ for some positive $r$. Thus the line-free condition is met if we prevent equilateral triangles of this sort. The Fujimura problem was to maximize the number of Gamma sets under this restriction, searching over the $(n+2)(n+1)/2$ binary variables. To maximize the total number of points, each Gamma set $\Gamma(a,b,c)$ is weighted by its size $(a+b+c)!/(a!b!c!)$, but the restrictions are the same.

Conversely, an attempt was made to disprove the `hyper-optimistic conjecture' mentioned in the Introduction. Each point of $[3]^n$ was given a weight of $1/(a!b!c!)$ when the point's coordinates included $a$ 1s, $b$ 2s and $c$ 3s. This gave each $\Gamma_{a,b,c}$ cell an equal weight of 1. In fact, since an integer programming routine was being used, each point was given the weight $(a+b+c)!/(a!b!c!)$. These numbers quickly grew too large for the routine.

In the Moser-Fujimura problem, we look for collections of $\Gamma(a,b,c)$ that do not contain geometric lines. This is satisfied if there are no isosceles triangles of the form $(a+r,b,c+s)$, $(a,b+r+s,c)$, $(a+s,b,c+r)$. Again, we run a search over the $(n+2)(n+1)/2$ binary variables, with $\Gamma(a,b,c)$ weighted by $(a+b+c)!/(a!b!c!)$, to maximize the total number of points.

Higherk.tex

2009-07-27T13:48:59Z

Mpeake:

\section{Higher-k DHJ numbers}\label{higherk-sec}

For any $n$, $k$ let $c_{n,k}$ denote the cardinality of the largest subset of $[k]{}^n$ that does not contain a combinatorial line. When $k=3$, the quantity $c_{n,k} = c_n$ is studied in Sections \ref{dhj-lower-sec}, \ref{dhj-upper-sec}. The density Hales-Jewett theorem asserts that for any fixed $k$, $\lim_{n\rightarrow\infty} c_n/k^n = 0$ .
We trivially have $c_{n,1} = 0$ for $n > 0$, and Sperner's theorem tells us that
$$c_{n,2} = \binom{n}{\lfloor n/2 \rfloor}$$.
Now we look at the opposite regime, in which $n$ is small and $k$ is large. We easily have $c_{1,k} = k-1$ together with the trivial bound $c_{n+1,k} \leq kc_{n,k}$. This implies that
$$c_{n,k}\leq (k-1)k^{n-1}$$
for any $n\geq 1$. Let us call a pair $(n,k)$ with $n > 0$ \emph{saturated} if $c_{n,k}=(k-1)k^{n-1}$, thus there exists a line-free set with exactly one point omitted from every row and column.
The question naturally arises, Which pairs $(n,k)$ are saturated? From the above discussion we see that $(1,k)$ is saturated for all $k \geq 1$, and $(n,1)$ is (rather trivially) saturated for all $n$. Sperner's theorem tells us that $(n,2)$ is saturated only for $n= 1, 2$. Note that if $(n,k)$ is unsaturated then $(n',k)$ will be unsaturated for all $n' > n$.
A computer search has found the following $c_{n,k}$ values for different values of dimension $n$ and edgelength $k$. Several of these values reach the upper bound of $(k - 1)k^{n-1}$.

\begin{tabular}{|l|cccccc}
$n\backslash k$ & 2& 3 & 4 & 5 & 6 & 7 \\
\hline
2 & 2 & 6 & 12 & 20 & 30 & 42\\
3 & 3 & 18 & 48 & 100 & 180 & 294\\
4 & 6 & 52 & 183 & 500 & 1051-1079 & 2058\\
5 & 10 & 150 & 712-732 & 2500 & 6325-6480 & 14406
\end{tabular}

$(2,k)$ is saturated when $k$ is at least $1$. In dimension two the maximal set size is $k(k-1)$. This can be done by removing the diagonal values $11, 22, 33, \ldots, kk$. Since they are in disjoint lines this removal is minimal.
The $k$ missing points are one per line and one per column. So their $y$-coordinates are a shuffle of their $x$-coordinates. There are $k!$ rearrangements of the numbers $1$ to $k$. The $k$ points include a point on the diagonal, so this shuffle is not a derangement. There are $k!/e$ derangements of the numbers $1$ to $k$, so $k!(1-1/e)$ optimal solutions. This number of optimal solutions is sequence A002467 from the Online Encyclopedia of Integer Sequences.

$(3,k)$ is saturated when $k>2$. Let $S$ be a latin square of side $k$ on the symbols $1…k$, with colour $i$ in position $(i,i)$ (This is not possible for $k=2$)
Let axis one in $S$ correspond to coordinate $1$ in $[k]{}^3$, axis two to coordinate $2$ and interpret the colour in position $(i,j)$ as the third coordinate. Delete the points so defined.
The line with three wild cards has now been removed. A line with two wildcards will be missing the point corresponding to the diagonal in $S$. A line with a single wildcard will be missing a point corresponding to an off diagonal point in $S$.

$(n,k)$ is saturated when all prime divisors of $k$ are at least $n$.
First consider the case when $k$ is prime and at least $n$: Delete those points whose coordinates add up to a multiple of $k$. Every combinatorial line has one point deleted, except for the major diagonal of $n=k$, which has all points deleted.
Now consider for instance the case $(n,k) = (4,35)$. Select one value modulo $35$ and eliminate it. Combinatorial lines with one, two, three or four moving coordinates will realize all values modulo $35$ as one, two, three, or four are units modulo $35$, thus $(4,35)$ is saturated.
The same argument tells us that $(n,k)$ is saturated when all prime divisors of $k$ are at least $n$.
On the other hand, computer data shows that $(4,4)$ and $(4,6)$ are not saturated.

\subsection{Failure of Hyper-optimistic conjecture}\label{HOCfailure}
Let $\overline{c}^\mu_{n,4}$ be the largest subset of the tetrahedral grid:
$$\{(a,b,c,d)\in\mathbb{Z}^4_+ : a+b+c+d = n\}$$
which contains no tetrahedrons $(a + r,b,c,d)$,$(a,b + r,c,d)$,$(a,b,c + r,d)$,$(a,b,c,d + r)$ with $r > 0$; call such sets tetrahedron-free.
The first few values, for $n$ from 0 to 7, were found by an integer programming routine, and are given in Figure \ref{Fujimura4}.

\begin{figure}[tb]\centerline{
\begin{tabular}{lllllllll}
n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
$\overline{c}^\mu_{n,4}$ & 1 & 3 & 7 & 14 & 24 & 37 & 55 & 78
\end{tabular}
}
\label{Fujimura4}
\caption{Fujimura numbers $\overline{c}^\mu_{n,4}$}
\end{figure}

Recall the 'hyper-optimistic' Conjecture \ref{hoc} that the weighted sum of the points in any combinatorial line-free set is at most $\overline{c}^\mu_{n,4}$. The following example with $n=2$ and $k=4$ has a weighted sum of $7.5$, and therefore disproves the conjecture for these values of $n$ and $k$. (Note that for all odd numbers including the case $k=3$, which is at the centre of most of Polymath's interest, is still open.

$$\{(1, 1), (1, 2), (1, 3), (2, 1), (2, 3), (2, 4), (3, 2), (3, 3), (3,
4), (4, 1), (4, 2), (4, 4)\}$$

The above counterexample can be extended to all even numbers greater than 2 as follows. One takes one element on the diagonal say $(n,n)$ then one can take $(a,a+1)$ for all $n-1$ values 0 through $n-2$ and $(n-1,0)$ this will block all combinatorial lines and it has weight (n+1)/2. The complement of this thus line free. Its weight is greater than any set of slices. For each such set must contain at least one diagonal blocking point and $n-1$ other blocking points if it has two or more diagonal blocking points the complement will be less than the above set(there must be at least $n$ blocking points so 2 or more diagonal points increase the weight by $1/2$ and pushes it over the limit constructed above. If it has one then because of parity reasons it must have an additional point (there are an odd number of coordinates to take care of and since if $(a,b)$ is in the blocking set $(b,a)$ must be as well, coordinates must be blocked by pairs there is one left over which requires an addition point or an addition point of the form $(a,a)$ which makes the line free set lower than the above construction. The HOC is known to be true for 2 and for even numbers greater than 2 by the above it is false.

Fujimura.tex

2009-07-27T13:46:05Z

Mpeake:

\section{Fujimura's problem}\label{fujimura-sec}

Let $\overline{c}^\mu_n$ be the size of the largest subset of the trianglular grid
$$\Delta_n := \{(a,b,c)\in {\mathbb Z}^3_+ : a+b+c = n\}$$
which contains no equilateral triangles $(a+r,b,c), (a,b+r,c), (a,b,c+r)$ with $r>0$. These are upward-pointing equilateral triangles. We shall refer to such sets as 'triangle-free'.
(Kobon Fujimura is a prolific inventor of puzzles, and in [http://www.puzzles.com/PuzzlePlayground/CoinsAndTriangles/CoinsAndTriangles.htm this] puzzle asked the related question of eliminating all equilateral triangles.)

The table in Figure \ref{lowFujimura} was formed mostly by computer searches for optimal solutions. We also found human proofs for most of them (see {\tt http://michaelnielsen.org/polymath1/index.php?title=Fujimura's\_problem}).

\begin{figure}
\centerline{
\begin{tabular}{l|llllllllllllll}
$n$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13\\
\hline
$\overline{c}^\mu_n$ & 1 & 2 & 4 & 6 & 9 & 12 & 15 & 18 & 22 & 26 & 31 & 35 & 40 & 46
\end{tabular}
}
\label{lowFujimura}
\caption{Fujimura numbers}
\end{figure}

For any equilateral triangle $(a+r,b,c)$,$(a,b+r,c)$ and $(a,b,c+r)$, the value $y+2z$ forms an arithmetic progression of length 3. A Behrend set is a finite set of integers with no arithmetic progression of length 3 (see {\tt http://arxiv.org/PS\_cache/arxiv/pdf/0811/0811.3057v2.pdf}). By looking at those triples $(a,b,c)$ with $a+2b$ inside a Behrend set, one can obtain the lower bound of $\overline{c}^\mu_n \geq n^2 exp(-O(\sqrt{\log n}))$.

It can be shown by a `corners theorem' of Ajtai and Szemeredi \cite{ajtai} that $\overline{c}^\mu_n = o(n^2)$ as $n \rightarrow \infty$.

An explicit lower bound is $3(n-1)$, made of all points in $\Delta_n$ with exactly one coordinate equal to zero.

An explicit upper bound comes from counting the triangles. There are $\binom{n+2}{3}$ triangles, and each point belongs to $n$ of them. So you must remove at least $(n+2)(n+1)/6$ points to remove all triangles, leaving $(n+2)(n+1)/3$ points as an upper bound for $\overline{c}^\mu_n$.

Moser-lower.tex

2009-07-27T13:43:20Z

Mpeake:

\section{Lower bounds for the Moser problem}\label{moser-lower-sec}

Just as for the DHJ problem, we found that Gamma sets $\Gamma_{a,b,c}$ were useful in providing large lower bounds for the Moser problem. This is despite the fact that the symmetries of the cube do not respect Gamma sets.

Observe that if $B \subset \Delta_n$, then the set $A_B := \bigcup_{\vec a \in B} \Gamma_{a,b,c}$ is a Moser set as long as $B$ does not contain any ``isosceles triangles'' $(a+r,b,c+s), (a+s,b,c+r), (a,b+r+s,c)$ for any $r,s \geq 0$ not both zero; in particular, $B$ cannot contain any ``vertical line segments'' $(a+r,b,c+r), (a,b+2r,c)$. An example of such a set is provided by selecting $0 \leq i \leq n-3$ and letting $B$ consist of the triples $(a, n-i, i-a)$ when $a \neq 3 \mod 3$, $(a,n-i-1,i+1-a)$ when $a \neq 1 \mod 3$, $(a,n-i-2,i+2-a)$ when $a=0 \mod 3$, and $(a,n-i-3,i+3-a)$ when $a=2 \mod 3$. Asymptotically, this set includes about two thirds of the spheres $S_{n,i}$, $S_{n,i+1}$ and one third of the spheres $S_{n,i+2}, S_{n,i+3}$ and (setting $i$ close to $n/3$) gives a lower bound \eqref{cpn3} with $C = 2 \times \sqrt{\frac{9}{4\pi}}$.

An integer program was run to obtain the optimal lower bounds achievable by the $A_B$ construction (using \eqref{cn3}, of course). The results for
$1 \leq n \leq 20$ are displayed in Figure \ref{nlow-moser}:

\begin{figure}[tb]
\centerline{
\begin{tabular}{|ll|ll|}
\hline
$n$ & lower bound & $n$ & lower bound \\
\hline
1 & 2 &11& 71766\\
2 & 6 & 12& 212423\\
3 & 16 & 13& 614875\\
4 & 43 & 14& 1794212\\
5 & 122& 15& 5321796\\
6 & 353& 16& 15455256\\
7 & 1017& 17& 45345052\\
8 & 2902&18& 134438520\\
9 & 8622&19& 391796798\\
10& 24786& 20& 1153402148\\
\hline
\end{tabular}}
\caption{Lower bounds for $c'_n$ obtained by the $A_B$ construction.}
\label{nlow-moser}
\end{figure}

More complete data, including the list of optimisers, can be found at
\centerline{{\tt http://abel.math.umu.se/~klasm/Data/HJ/}.}

Note that the lower bound $c'_{6,3} \geq 353$ was first located by a genetic algorithm: see Appendix \ref{genetic-alg}.

\begin{figure}[tb]
\centerline{\includegraphics{moser353new.png}}
\caption{One of the examples of $353$-point sets in $[3]^6$ (elements of the set being indicated by white squares).}
\label{moser353-fig}
\end{figure}

However, we have been unable to locate a lower bound which is asymptotically better than \eqref{cpn3}. Indeed, any method based purely on the $A_B$ construction cannot do asymptotically better than the previous constructions:

\begin{proposition} Let $B \subset \Delta_n$ be such that $A_B$ is a Moser set. Then $|A_B| \leq (2 \sqrt{\frac{9}{4\pi}} + o(1)) \frac{3^n}{\sqrt{n}}$.
\end{proposition}

\begin{proof} By the previous discussion, $B$ cannot contain any pair of the form $(a,b+2r,c), (a+r,b,c+r)$ with $r>0$. In other words, for any $-n \leq h \leq n$, $B$ can contain at most one triple $(a,b,c)$ with $c-a=h$. From this and \eqref{cn3}, we see that
$$ |A_B| \leq \sum_{h=-n}^n \max_{(a,b,c) \in \Delta_n: c-a=h} \frac{n!}{a! b! c!}.$$
From the Chernoff inequality (or the Stirling formula computation below) we see that $\frac{n!}{a! b! c!} \leq \frac{1}{n^{10}} 3^n$ unless $a,b,c = n/3 + O( n^{1/2} \log^{1/2} n )$, so we may restrict to this regime, which also forces $h = O( n^{1/2}\log^{1/2} n)$. If we write $a = n/3 + \alpha$, $b = n/3 + \beta$, $c = n/3+\gamma$ and apply Stirling's formula $n! = (1+o(1)) \sqrt{2\pi n} n^n e^{-n}$, we obtain
$$ \frac{n!}{a! b! c!} = (1+o(1)) \frac{3^{3/2}}{2\pi n} 3^n \exp( - (\frac{n}{3}+\alpha) \log (1 + \frac{3\alpha}{n} ) - (\frac{n}{3}+\beta) \log (1 + \frac{3\beta}{n} ) - (\frac{n}{3}+\gamma) \log (1 + \frac{3\gamma}{n} ) ).$$
From Taylor expansion one has
$$ -(\frac{n}{3}+\alpha) \log (1 + \frac{3\alpha}{n} ) = -\alpha - \frac{3}{2} \frac{\alpha^2}{n} + o(1)$$
and similarly for $\beta,\gamma$; since $\alpha+\beta+\gamma=0$, we conclude that
$$ \frac{n!}{a! b! c!} = (1+o(1)) \frac{3^{3/2}}{2\pi n} 3^n \exp( - \frac{3}{2n} (\alpha^2+\beta^2+\gamma^2) ).$$
If $c-a=h$, then $\alpha^2+\beta^2+\gamma^2 = \frac{3\beta^2}{2} + \frac{h^2}{2}$. Thus we see that
$$ \max_{(a,b,c) \in \Delta_n: c-a=h} \frac{n!}{a! b! c!} \leq (1+o(1)) \frac{3^{3/2}}{2\pi n} 3^n \exp( - \frac{3}{4n} h^2 ).$$
Using the integral test, we thus have
$$ |A_B| \leq (1+o(1)) \frac{3^{3/2}}{2\pi n} 3^n \int_\R \exp( - \frac{3}{4n} x^2 )\ dx.$$
Since $\int_\R \exp( - \frac{3}{4n} x^2 )\ dx = \sqrt{\frac{4\pi n}{3}}$, we obtain the claim.
\end{proof}

Actually it is possible to improve upon these bounds by a slight amount. Observe that if $B$ is a maximiser for the right-hand side of \eqref{cn3} (subject to $B$ not containing isosceles triangles), then any triple $(a,b,c)$ not in $B$ must be the vertex of a (possibly degenerate) isosceles triangle with the other vertices in $B$. If this triangle is non-degenerate, or if $(a,b,c)$ is the upper vertex of a degenerate isosceles triangle, then no point from $\Gamma_{a,b,c}$ can be added to $A_B$ without creating a geometric line. However, if $(a,b,c) = (a'+r,b',c'+r)$ is only the lower vertex of a degenerate isosceles triangle $(a'+r,b',c'+r), (a',b'+2r,c')$, then one can add any subset of $\Gamma_{a,b,c}$ to $A_B$ and still have a Moser set as long as no pair of elements in that subset is separated by Hamming distance $2r$. For instance, in the $n=5$ case, we can start with the 122-point set built from
$$B = \{ (0 0 5),(0 2 3),(1 1 3 ),(1 2 2 ),(2 2 1 ),(3 1 1 ),(3 2 0 ),(5 0 0))\}$$, and add a point each from (1 0 4) and (4 0 1).
This gives an example of the maximal, 124-point solution. Again, in the $n=10$ case, the set
\begin{align*}
B &= \{(0 0 10),(0 2 8 ),(0 3 7 ),(0 4 6 ),(1 4 5 ),(2 1 7 ),(2 3 5 ),
(3 2 5 ),(3 3 4 ),(3 4 3 ),(4 4 2 ),(5 1 4 ), \\
&\quad (5 3 2 ),(6 2 2 ),
(6 3 1 ),(6 4 0 ),(8 1 1 ),(9 0 1 ),(9 1 0 ) \}
\end{align*}

generates the lower bound $c'_{10,3} \geq 24786$ given above (and, up to reflection $a \leftrightarrow c$, is the only such set that does so); but by adding the following twelve elements from $\Gamma_{5,0,5}$ one can increase the lower bound slightly to $24798$: $1111133333$, $1111313333$,
$1113113333$, $1133331113$, $1133331131$, $1133331311$, $3311333111$,
$3313133111$, $3313313111$, $3331111133$, $3331111313$, $3331111331$

A more general form goes with the $B$ set described at the start of this section. Include points from $\Gamma_{(a,n-i-4,i+4-a)}$ when $a=1 (\operatorname{mod}\ 3)$, subject to no two points being included if they differ by the interchange of a $1$ and a $3$. Each of these Gamma sets is the feet of a degenerate isosceles triangle with vertex $\Gamma_{(a-1,n-i-2,a+3-a)}$.

\begin{lemma} If $A$ is a subset of $\Gamma_{(a,b,c)}$ such that no two points of $A$ differ by the interchange of a $1$ and a $3$, then $|A| \leq |\Gamma_{a,b,c}|/(1+\max(a,c))$.
\end{lemma}
\begin{proof}
Say that two points of $\Gamma_{a,b,c}$ are neighbours if they differ by the exchange of a $1$ and a $3$. Each point of $A$ has $ac$ neighbours, none of which are in $A$. Each point of $\Gamma_{(a,b,c)}\backslash A$ has $ac$ neighbours, but only $min(a,c)$ of them may be in $A$. So for each point of $A$ there are on average $ac/\min(a,c) = \max(a,c)$ points not in $A$. So the proportion of points of $\Gamma_{(a,b,c)}$ that are in $A$ is at most one in $1+\max(a,c)$.
\end{proof}

The proportion of extra points for each of the cells $\Gamma_{(a,n-i-4,i+4-a)}$ is no more than $2/(i+6)$. Only one cell in three is included from the $b=n-i-4$ layer, so we expect no more than $\binom{n}{i+4}2^{i+5}/(3i+18)$ new points, all from $S_{n,i+4}$. One can also find extra points from $S_{n,i+5}$ and higher spheres.

Earlier solutions may also give insight into the problem. In this section we discuss lower bounds for $c'_{n,3}$. Clearly we have
$c'_{0,3}=1$ and $c'_{1,3}=2$, so we focus on the case $n \ge 2$.
The first lower bounds may be due to Koml\'{o}s \cite{komlos}, who observed
that the sphere $S_{i,n}$ of elements with exactly $n-i$ 2 entries
(see Section \ref{notation-sec} for definition),
is a Moser set, so that
\begin{equation}\label{cin}
c'_{n,3}\geq \vert S_{i,n}\vert
\end{equation}
holds for all $i$. Choosing $i=\lfloor \frac{2n}{3}\rfloor$ and
applying Stirling's formula, we see that this lower bound takes the form
\begin{equation}\label{cpn3}
c'_{n,3} \geq (C-o(1)) 3^n / \sqrt{n}
\end{equation}
for some absolute constant $C>0$; in fact \eqref{cin} gives \eqref{cpn3} with $C := \sqrt{\frac{9}{4\pi}}$.
In particular $c'_{3,3} \geq 12, c'_{4,3}\geq 24, c'_{5,3}\geq 80,
c'_{6,3}\geq 240$.

These values can be improved by studying combinations of several spheres or
semispheres or applying elementary results from coding theory.

Observe that if $\{w(1),w(2),w(3)\}$ is a geometric line in $[3]^n$,
then $w(1), w(3)$ both lie in the same sphere $S_{i,n}$,
and that $w(2)$ lies in a lower
sphere $S_{i-r,n}$ for some $1 \leq r \leq i \leq n$. Furthermore,
$w(1)$ and $w(3)$ are separated by Hamming distance $r$.

As a consequence, we see that $S_{i-1,n} \cup S_{i,n}^e$ (or $S_{i-1,n}
\cup S_{i,n}^o$) is a Moser set for any $1 \leq i \leq n$, since any two
distinct elements $S_{i,n}^e$ are separated by a Hamming distance of at
least two.
(Recall Section \ref{notation-sec} for definitions),
This leads to the lower bound
\begin{equation}\label{cn3-low}
c'_{n,3} \geq \binom{n}{i-1}
2^{i-1} + \binom{n}{i} 2^{i-1} = \binom{n+1}{i} 2^{i-1}.
\end{equation}
It is not
hard to see that $\binom{n+1}{i+1} 2^{i} > \binom{n+1}{i} 2^{i-1}$ if
and only if $3i < 2n+1$, and so this lower bound is maximised when $i =
\lfloor \frac{2n+1}{3} \rfloor$ for $n \geq 2$, giving the formula
\eqref{binom}. This leads to the lower bounds $$ c'_{2,3} \geq 6;
c'_{3,3} \geq 16; c'_{4,3} \geq 40; c'_{5,3} \geq 120; c'_{6,3} \geq
336$$ which gives the right lower bounds for $n=2,3$, but is slightly
off for $n=4,5$. Asymptotically, Stirling's formula and \eqref{cn3-low} then
give the lower bound \eqref{cpn3} with $C = \frac{3}{2} \times \sqrt{\frac{9}{4\pi}}$, which is asymptotically $50\%$ better than the bound \eqref{cin}.

The work of Chv\'{a}tal \cite{chvatal1}
already contained a refinement of this idea
which we here translate into the usual notation of coding theory:
Let $A(n,d)$ denote the size of the largest binary code of length $n$
and minimal distance $d$.

Then
\begin{equation}\label{cnchvatal}
c'_{n,3}\geq \max_k \left( \sum_{j=0}^k \binom{n}{j} A(n-j, k-j+1)\right).
\end{equation}

With the following values for $A(n,d)$:
{\tiny{
\[
\begin{array}{llllllll}
A(1,1)=2&&&&&&&\\
A(2,1)=4& A(2,2)=2&&&&&&\\
A(3,1)=8&A(3,2)=4&A(3,3)=2&&&&&\\
A(4,1)=16&A(4,2)=8& A(4,3)=2& A(4,4)=2&&&&\\
A(5,1)=32&A(5,2)=16& A(5,3)=4& A(5,4)=2&A(5,5)=2&&&\\
A(6,1)=64&A(6,2)=32& A(6,3)=8& A(6,4)=4&A(6,5)=2&A(6,6)=2&&\\
A(7,1)=128&A(7,2)=64& A(7,3)=16& A(7,4)=8&A(7,5)=2&A(7,6)=2&A(7,7)=2&\\
A(8,1)=256&A(8,2)=128& A(8,3)=20& A(8,4)=16&A(8,5)=4&A(8,6)=2
&A(8,7)=2&A(8,8)=2\\
A(9,1)=512&A(9,2)=256& A(9,3)=40& A(9,4)=20&A(9,5)=6&A(9,6)=4
&A(9,7)=2&A(9,8)=2\\
A(10,1)=1024&A(10,2)=512& A(10,3)=72& A(10,4)=40&A(10,5)=12&A(10,6)=6
&A(10,7)=2&A(10,8)=2\\
A(11,1)=2048&A(11,2)=1024& A(11,3)=144& A(11,4)=72&A(11,5)=24&A(11,6)=12
&A(11,7)=2&A(11,8)=2\\
A(12,1)=4096&A(12,2)=2048& A(12,3)=256& A(12,4)=144&A(12,5)=32&A(12,6)=24
&A(12,7)=4&A(12,8)=2\\
A(13,1)=8192&A(13,2)=4096& A(13,3)=512& A(13,4)=256&A(13,5)=64&A(12,6)=32
&A(13,7)=8&A(13,8)=4\\
\end{array}
\]
}}

Generally, $A(n,1)=2^n, A(n,2)=2^{n-1}, A(n-1,2e-1)=A(n,2e), A(n,d)=2$,
if $d>\frac{2n}{3}$.
The values were taken or derived from Andries Brower's table at
\centerline{\tt http://www.win.tue.nl/$\sim$aeb/codes/binary-1.html}
\textbf{include to references? or other book with explicit values of $A(n,d)$ }

For $c'_{n,3}$ we obtain the following lower bounds:
with $k=2$
\[
\begin{array}{llll}
c'_{4,3}&\geq &\binom{4}{0}A(4,3)+\binom{4}{1}A(3,2)+\binom{4}{2}A(2,1)
=1\cdot 2+4 \cdot 4+6\cdot 4&=42.\\
c'_{5,3}&\geq &\binom{5}{0}A(5,3)+\binom{5}{1}A(4,2)+\binom{5}{2}A(3,1)
=1\cdot 4+5 \cdot 8+10\cdot 8&=124.\\
c'_{6,3}&\geq &\binom{6}{0}A(6,3)+\binom{6}{1}A(5,2)+\binom{6}{2}A(4,1)
=1\cdot 8+6 \cdot 16+15\cdot 16&=344.
\end{array}
\]
With k=3
\[
\begin{array}{llll}
c'_{7,3}&\geq& \binom{7}{0}A(7,4)+\binom{7}{1}A(6,3)+\binom{7}{2}A(5,2)
+ \binom{7}{3}A(4,1)&=960.\\
c'_{8,3}&\geq &\binom{8}{0}A(8,4)+\binom{8}{1}A(7,3)+\binom{8}{2}A(6,2)
+ \binom{8}{3}A(5,1)&=2832.\\
c'_{9,3}&\geq & \binom{9}{0}A(9,4)+\binom{9}{1}A(8,3)+\binom{9}{2}A(7,2)
+ \binom{9}{3}A(6,1)&=7880.
\end{array}\]
With k=4
\[
\begin{array}{llll}
c'_{10,3}&\geq &\binom{10}{0}A(10,5)+\binom{10}{1}A(9,4)+\binom{10}{2}A(8,3)
+ \binom{10}{3}A(7,2)+\binom{10}{4}A(6,1)&=22232.\\
c'_{11,3}&\geq &\binom{11}{0}A(11,5)+\binom{11}{1}A(10,4)+\binom{11}{2}A(9,3)
+ \binom{11}{3}A(8,2)+\binom{11}{4}A(7,1)&=66024.\\
c'_{12,3}&\geq &\binom{12}{0}A(12,5)+\binom{12}{1}A(11,4)+\binom{12}{2}A(10,3)
+ \binom{12}{3}A(9,2)+\binom{12}{4}A(8,1)&=188688.\\
\end{array}\]
With $k=5$
\[ c'_{13,3}\geq 539168.\]

It should be pointed out that these bounds are even numbers, so that
$c'_{4,3}=43$ shows that one cannot generally expect this lower bound
gives the optimum.

The maximum value appears to occur for $k=\lfloor\frac{n+2}{3}\rfloor$,
so that using
Stirling's formula and explicit bounds on $A(n,d)$ the
best possible value known to date
of the constant $C$ in
equation \eqref{cpn3} can be worked out, but we refrain from doing this here.
Using the Singleton bound $A(n,d)\leq 2^{n-d+1}$ Chv\'{a}tal
\cite{chvatal1} proved that the expression on the right hand side of
\eqref{cnchvatal} is also
$O\left( \frac{3^n}{\sqrt{n}}\right)$, so that the refinement described
above gains a constant factor over the initial construction only.

For $n=4$ the above does not yet give the exact value.
The value $c'_{4,3}=43$ was first proven by Chandra \cite{chandra}.
A uniform way of describing examples for the optimum values of
$c'_{4,3}=43$ and $c'_{5,3}=124$ is the following:

Let us consider the sets $$ A := S_{i-1,n}
\cup S_{i,n}^e \cup A'$$ where $A' \subset S_{i+1,n}$ has the property
that any two elements in $A'$ are separated by a Hamming distance of at
least three, or have a Hamming distance of exactly one but their
midpoint lies in $S_{i,n}^o$. By the previous discussion we see that
this is a Moser set, and we have the lower bound
\begin{equation}\label{cnn}
c'_{n,3} \geq \binom{n+1}{i} 2^{i-1} + |A'|.
\end{equation} This gives some improved lower bounds for $c'_{n,3}$:

\begin{itemize} \item By taking $n=4$, $i=3$, and $A' = \{ 1111, 3331,
3333\}$, we obtain $c'_{4,3} \geq 43$; \item By taking $n=5$, $i=4$, and
$A' = \{ 11111, 11333, 33311, 33331 \}$, we obtain $c'_{5,3} \geq 124$.
\item By taking $n=6$, $i=5$, and $A' = \{ 111111, 111113, 111331,
111333, 331111, 331113\}$, we obtain $c'_{6,3} \geq 342$. \end{itemize}

This gives the lower bounds in Theorem \ref{moser} up to $n=5$, but
the bound for $n=6$ is inferior to the lower bound $c'_{6,3}\geq 344$ given above.

\subsection{Higher $k$ values}
One can consider subsets of $[k]{}^n$ that contain no geometric lines. Section \ref{moser-lower-sec} has considered the case $k=3$. Let $c'_{n,k}$ be the greatest number of points in $[k]{}^n$ with no geometric line. For example, $c'_{n,3} = c'_n$. We have the following lower bounds:

$c'_{n,4} \ge \binom{n}{n/2}2^n$.
The set of points with $a$ $1$s,$b$ $2$s,$c$ $3$s and $d$ $4$s, where $a+d$ has the constant value $n/2$, does not form geometric lines because points at the ends of a geometric line have more $a$ or $d$ values than points in the middle of the line.

The following lower bound is asymptotically twice as large. Take all points with $a$ $1$s, $b$ $2$s, $c$ $3$s and $d$ $4$s, for which:

\begin{itemize}
\item Either $a+d = q$ or $q-1$, $a$ and $b$ have the same parity; or
\item $a+d = q-2$ or $q-3$, $a$ and $b$ have opposite parity.
\end{itemize}

This includes half the points of four adjacent layers, and therefore may include $(1+o(1))\binom{n}{n/2}2^{n+1}$ points.

We also have a DHJ(3)-like lower bound for $c'_{n,5}$, namely $c'_{n,5} = 5^{n-O(\sqrt{\log n})}$.
Consider points with $a$ $1$s, $b$ $2$s, $c$ $3$s, $d$ $4$s and $e$ $5$s. For each point, take the value $a+e+2(b+d)+3c$. The first three points in any geometric line give values that form an arithmetic progression of length three.

Select a set of integers with no arithmetic progression of length 3. Select all points whose value belongs to that sequence; there will be no geometric line among those points. By Behrend theory, it is possible to choose these points with density $\exp{-O(\sqrt{\log n})}$.

The $k=6$ version of Moser implies DHJ(3). Indeed, any $k=3$ combinatorial line-free set can be "doubled up" into a $k=6$ geometric line-free set of the same density by pulling back the set from the map that maps 1, 2, 3, 4, 5, 6 to 1, 2, 3, 3, 2, 1 respectively; note that this map sends $k=6$ geometric lines to $k=3$ combinatorial lines. So $c'_{k,6} \geq 2^k c_k$, and more generally, $c'_{k,2n} \geq 2^k c_{k,n}$.

Dhj-lown-lower.tex

2009-07-27T13:40:47Z

Mpeake:

\section{Lower bounds for the density Hales-Jewett problem}\label{dhj-lower-sec}

The purpose of this section is to establish various lower bounds for $c_{n,3}$, in particular establishing Theorem \ref{dhj-lower} and the lower bound component of Theorem \ref{dhj-upper}.

As observed in the introduction, if $B \subset \Delta_{3,n}$ is a Fujimura set (i.e. a subset of $\Delta_{3,n} = \{ (a,b,c) \in \N^3: a+b+c=n\}$ which contains no upward equilateral triangles $(a+r,b,c), (a,b+r,c), (a,b,c+r)$), then the set $A_B := \bigcup_{\vec a \in B} \Gamma_{a,b,c}$ is a line-free subset of $[3]^n$, which gives the lower bound
\begin{equation}\label{cn3}
c_{n,3} \geq |A_B| = \sum_{(a,b,c) \in B} \frac{n!}{a! b! c!}.
\end{equation}
All of the lower bounds for $c_{n,3}$ in this paper will be constructed via this device.

In order to use \eqref{cn3}, one of course needs to build Fujimura sets $B$ which are ``large'' in the sense that the right-hand side of \eqref{cn3} is large. A fruitful starting point for this goal is the sets
$$B_{j,n} := \{ (a,b,c) \in \Delta_{3,n}: a + 2b \neq j \hbox{ mod } 3 \}$$
for $j=0,1,2$. Observe that in order for a triangle $(a+r,b,c), (a,b+r,c), (a,b,c+r)$ to lie in $B_{j,n}$, the length $r$ of the triangle must be a multiple of $3$. This already makes $B_{j,n}$ a Fujimura set for $n < 3$ (and $B_{0,n}$ a Fujimura set for $n = 3$).

When $n$ is not a multiple of $3$, the $B_{j,n}$ are all rotations of each other and give equivalent sets (of size $2 \times 3^{n-1}$). When $n$ is a multiple of $3$, the sets $B_{1,n}$ and $B_{2,n}$ are reflections of each other, but $B_{0,n}$ is not equivalent to the other two sets (in particular, it omits all three corners of $\Delta_{3,n}$); the associated set $A_{B_{0,n}}$ is slightly larger than $A_{B_{1,n}}$ and $A_{B_{2,n}}$ and thus is slightly better for constructing line-free sets.

As mentioned already, $B_{0,n}$ is a Fujimura set for $n \leq 3$, and hence $A_{B_{0,n}}$ is line-free for $n \leq 3$. Applying \eqref{cn3} one obtains the lower bounds
$$ c_{0,3} \geq 1; c_{1,3} \geq 2; c_{2,3} \geq 6; c_{3,3} \geq 18.$$

For $n>3$, $B_{0,n}$ contains some triangles $(a+r,b,c), (a,b+r,c), (a,b,c+r)$ and so is not a Fujimura set, but one can remove points from this set to recover the Fujimura property. For instance, for $n \leq 6$, the only triangles in $B_{0,n}$ have side length $r=3$. One can ``delete'' these triangles by removing one vertex from each; in order to optimise the bound \eqref{cn3} it is preferable to delete vertices near the corners of $\Delta_{3,n}$ rather than near the centre. These considerations lead to the Fujimura sets
\begin{align*}
B_{0,4} &\backslash \{ (0,0,4), (0,4,0), (4,0,0) \}\\
B_{0,5} &\backslash \{ (0,4,1), (0,5,0), (4,0,1), (5,0,0) \}\\
B_{0,6} &\backslash \{ (0,1,5), (0,5,1), (1,0,5), (0,1,5), (1,5,0), (5,1,0) \}
\end{align*}
which by \eqref{cn3} gives the lower bounds
$$ c_{4,3} \geq 52; c_{5,3} \geq 150; c_{6,3} \geq 450.$$
Thus we have established all the lower bounds needed for Theorem \ref{dhj-upper}.

One can of course continue this process by hand, for instance the set
$$ B_{0,7} \backslash \{(0,1,6),(1,0,6),(0,5,2),(5,0,2),(1,5,1),(5,1,1),(1,6,0),(6,1,0) \}$$
gives the lower bound $c_{7,3} \geq 1302$, which we tentatively conjecture to be the correct bound.

A simplification was found when $n$ is a multiple of $3$. Observe that for $n=6$, the sets excluded from $B_{0,6}$ are all permutations of $(0,1,5)$. So the remaining sets are all the permutations of $(1,2,3)$ and $(0,2,4)$. In the same way, sets for $n=9$, $12$ and $15$ can be described as:
\begin{itemize}
\item $n=9$: $(2,3,4),(1,3,5),(0,4,5)$ and permutations;
\item $n=12$: $(3,4,5),(2,4,6),(1,5,6),(0,2,10),(0,5,7)$ and permutations;
\item $n=15$: $(4,5,6),(3,5,7),(2,6,7),(1,3,11),(1,6,8),(0,4,11),(0,7,8)$ and permutations.
\end{itemize}

When $n$ is not a multiple of $3$, say $n=3m-1$ or $n=3m-2$, one first finds a solution for $n=3m$. Then for $n=3m-1$, one restricts the first digit of the $3m$ sequence to equal $1$. This leaves exactly one-third as many points for $3m-1$ as for $3m$. For $n=3m-1$, one restricts the first two digits of the $3m$ sequence to be $12$. This leaves roughly one-ninth as many points for $3m-2$ as for $3m$.

The following is an effective method to find good, though not optimal, solutions for any $n=3m$. (For $n<21$, ignore any triple with a negative entry.)

Start with thirteen groups of points in the centre, formed from adding one of the following points, or its permutation, to $M := (m,m,m)$, when $n=3m$:
$$(-7,-3,+10), (-7, 0,+7),(-7,+3,+4),(-6,-4,+10),(-6,-1,+7),(-6,+2,+4)$$
$$(-5,-1,+6),(-5,+2,+3),(-4,-2,+6),(-4,+1,+3),(-3,+1,+2),(-2,0,+2),(-1,0,+1)$$
Then include eight strings of points, stretching to the edges of the triangle $\Delta_n$;
\begin{itemize}
\item $M+(-8-2x,-6-2x,14+4x)$, $M+(-8-2x,-3-2x,11+4x)$, $M+(-8-2x,x,8+x)$, $M+(-8-2x,3+x,5+x)$ and permutations ($0\le 2x \le M-8$);
\item $M+(-9-2x,-5-2x,14+4x)$, $M+(-9-2x,-2-2x,11+4x)$, $M+(-9-2x,1+x,8+x)$, $M+(-9-2x,4+x,5+x)$ and permutations ($0\le 2x \le M-9$).
\end{itemize}
This solution gives $O(2.7 \sqrt(\log (n)/n)3^n$ points for values of $n$ up to around $1000$. This is asymptotically smaller than the known optimum of $3^{n-O(\sqrt(\log(n)))}$. When $n=99$, it gives more than $3^n/3$ points.

The following solution gives more points for $n>1000$, but not for moderate $n$:

\begin{itemize}
\item Define a sequence, of all positive numbers which, in base 3, do not contain a 1. Add 1 to all multiples of 3 in this sequence. This sequence does not contain a length-3 arithmetic progression. It starts $1,2,7,8,19,20,25,26,55, \ldots$;
\item List all the $(abc)$ triples that sum to $n$, for which the larger two differ by a number from the sequence;
\item Exclude the case when the smaller two differ by 1;
\item Include the case when $(a,b,c)$ is a permutation of $n/3+(-1,0,1)$.
\end{itemize}

An integer program was run to obtain the maximum lower bound one could establish from \eqref{cn3} (see Appendix \ref{integer-sec}). The results for $1 \leq n \leq 20$ are displayed in Figure \ref{nlow}:

\begin{figure}[tb]
\centerline{
\begin{tabular}{|ll|ll|}
\hline
$n$ & lower bound & $n$ & lower bound \\
\hline
1 & 2 &11& 96338\\
2 & 6 & 12& 287892\\
3 & 18 & 13& 854139\\
4 & 52 & 14& 2537821\\
5 & 150& 15& 7528835\\
6 & 450& 16& 22517082\\
7 & 1302& 17& 66944301\\
8 & 3780&18& 198629224\\
9 & 11340&19& 593911730\\
10& 32864& 20& 1766894722\\
\hline
\end{tabular}}
\caption{Lower bounds for $c_n$ obtained by the $A_B$ construction.}
\label{nlow}
\end{figure}

More complete data, including the list of optimisers, can be found at \cite{markstrom}

If $B$ maximises the right-hand side of \eqref{cn3}, it is easy to see that $A_B$ is a line-free set which is maximal in the sense that the addition of any further point to $A_B$ will create a combinatorial line. Thus one might conjecture that the maximal value of the right-hand side \eqref{cn3} is in fact equal to $c_{n,3}$ for all $n$; Theorem \ref{dhj-upper} asserts that this conjecture is true for $n \leq 6$.

Now we prove Theorem \ref{dhj-lower}.

\begin{proof}[Proof of Theorem \ref {dhj-lower}]
Let $M$ be the circulant matrix with first row $(1,2,\ldots,k-1)$, second row $(k-1,1,2,\dots,k-2)$, and so on. Note that $M$ has nonzero determinant by well-known properties of circulant matrices.

Let $S$ be a subset of the interval $[-\sqrt {n}/2, \sqrt {n}/2)$ that contains no nonconstant arithmetic progressions of length k, and let $B\subset\Delta_{n, k}$ be the set
\[ B := \{(n-\sum_{i=1}^{k-1} a_i ,a_1,a_2,\dots, a_{k-1}) :
(a_1,\dots,a_{k-1})= n/k + \det(M) M^{-1}\vec{s} , \vec{s}\in S^{k-1}\}.\]
The map $(m,a_1,\dots,a_{k-1}) \mapsto M (a_1,\dots,a_{k-1})$ takes simplices to nonconstant arithmetic progressions in ${\mathbb Z}^{k-1}$, and takes $B$ to $\{c+det(M) \, \vec{s} \colon \vec{s} \in S^{k-1}\}$, which is a set containing no nonconstant arithmetic progressions. Thus, $B$ is a Fujimura set and so does not contain any combinatorial lines.

If all of $a_1,\ldots,a_k$ are within $C_1\sqrt{n}$ of $n/k$, then $|\Gamma_{\vec{a}}| \geq C k^n/n^{(k-1)/2}$ (where $C$ depends on $C_1$) by the central limit theorem. By our choice of $S$ and applying~\eqref{cn3}, we obtain
$$ c_ {n, k}\geq C k^n/n^{(k-1)/2} |S|^{k-1} = C k^n \left( \frac{|S|}{\sqrt{n}} \right)^{k-1}. $$
One can take $S$ to have cardinality $r_ k (\sqrt {n}) $, which from the results of O'Bryant~\cite {obryant}) satisfies (for all sufficiently large $n$, some $C>0$, and $\ell$ the largest integer satisfying $k> 2^{\ell-1}$)
$$ \frac{r_k (\sqrt{n})}{\sqrt{n}} \geq C (\log n)^{1/(2\ell)}\exp_2 (-\ell 2^{(\ell-1)/2-1/\ell} \sqrt[\ell]{\log_2 n}),$$
which completes the proof.
\end {proof}

Introduction.tex

2009-07-27T13:38:19Z

Mpeake:

\section{Introduction}

For any integers $k \geq 1$ and $n \geq 0$, let $[k] := \{1,\ldots,k\}$, and define $[k]^n$ to be the cube of words of length $n$ with alphabet in $[k]$. Thus for instance $[3]^2 = \{11,12,13,21,22,23,31,32,33\}$.

We define a \emph{combinatorial line} in $[k]^n$ to be a set of the form $\{ w(i): i = 1,\ldots,k\} \subset [k]^n$, where $w \in ([k] \cup \{x\})^n \backslash [k]^n$ is a word of length $n$ with alphabet in $[k]$ together with a ``wildcard'' letter $x$ which appears at least once, and $w(i) \in [k]^n$ is the word obtained from $w$ by replacing $x$ by $i$; we often abuse notation and identify $w$ with the combinatorial line $\{ w(i): i = 1,\ldots,k\}$ it generates. Thus for instance, in $[3]^2$ we have $x2 = \{12,22,32\}$ and $xx = \{11,22,33\}$ as typical examples of combinatorial lines. In general, $[k]^n$ has $k^n$ words and $(k+1)^n-k^n$ lines.

\begin{figure}[tb]
\centerline{\includegraphics{AllLinesIn3n.pdf}}
\caption{Combinatorial lines in $[3]^2$.}
\label{fig-line}
\end{figure}

A set $A \subset [k]^n$ is said to be \emph{line-free} if it contains no combinatorial lines. Define the \emph{$(n,k)$ density Hales-Jewett number} $c_{n,k}$ to be the maximum cardinality $|A|$ of a line-free subset of $[k]^n$. Clearly, one has the trivial bound $c_{n,k} \leq k^n$. A deep theorem of Furstenberg and Katznelson~\cite{fk1}, \cite{fk2} asserts that this bound can be asymptotically improved:

\begin{theorem}[Density Hales-Jewett theorem]\label{dhj} For any fixed $k \geq 2$, one has $\lim_{n \to \infty} c_{n,k}/k^n = 0$.
\end{theorem}

\begin{remark} The difficulty of this theorem increases with $k$. For $k=1$, one clearly has $c_{n,1}=1$. For $k=2$, a classical theorem of Sperner~\cite{sperner} asserts, in our language, that $c_{n,2} = \binom{n}{\lfloor n/2\rfloor}$. The case $k=3$ is already non-trivial (for instance, it implies Roth's theorem~\cite{roth} on arithmetic progressions of length three) and was first established in \cite{fk1} (see also \cite{mcc}). The case of general $k$ was first established in~\cite{fk2} and has a number of implications, in particular implying Szemer\'edi's theorem~\cite{szem} on arithmetic progressions of arbitrary length.
\end{remark}

The Furstenberg-Katznelson proof of Theorem~\ref{dhj} relied on ergodic-theory techniques and did not give an explicit decay rate for $c_{n,k}$. Recently, two further proofs of this theorem have appeared, by Austin~\cite{austin} and by the sister Polymath project to this one~\cite{poly}. The proof of~\cite{austin} also used ergodic theory, but the proof in~\cite{poly} was combinatorial and gave effective bounds for $c_{n,k}$ in the limit $n \to \infty$. For example, if $n$ can be written as an exponential tower $2 \uparrow 2 \uparrow 2 \uparrow \ldots \uparrow 2$ with $m$ 2s, then $c_{n,3} \le 3^n m^{-1/3}$. However, these bounds are not believed to be sharp, and in any case are only non-trivial in the asymptotic regime when $n$ is sufficiently large depending on $k$.

Our first result is the following asymptotic lower bound. The construction is based on the recent refinements \cite{elkin,greenwolf,obryant} of a well-known construction of Behrend~\cite{behrend} and Rankin~\cite{rankin}. The proof of Theorem~\ref{dhj-lower} is in Section~\ref{dhj-lower-sec}. Let $r_k(n)$ be the maximum size of a subset of $[n]$ that does not contain a $k$-term arithmetic progression.

\begin{theorem}[Asymptotic lower bound for $c_{n,k}$]\label{dhj-lower} For each $k\geq 3$, there is an absolute constant $C>0$ such that
\[ c_{n,k} \geq C k^n \left(\frac{r_k(\sqrt{n})}{\sqrt{n}}\right)^{k-1} = k^n \exp\left( - O(\sqrt[\ell]{\log n}) \right), \]
where $\ell$ is the largest integer satisfying $2k>2^{\ell}$. Specifically,
\[ c_{n,k} \geq C k^{n-\alpha(k) \sqrt[\ell]{\log n} + \beta(k) \log\log n},\]
where all logarithms are base-$k$, and $\alpha(k) = (\log 2)^{1-1/\ell} \ell 2^{(\ell-1)/2-1/\ell}$ and $\beta(k)=(k-1)/(2\ell)$.
\end{theorem}

In the case of small $n$, we focus primarily on the first non-trivial case $k=3$. We have computed the following explicit values of $c_{n,3}$:

\begin{theorem}[Explicit values of $c_{n,3}$]\label{dhj-upper} We have $c_{0,3} = 1$, $c_{1,3} = 2$, $c_{2,3} = 6$, $c_{3,3} = 18$, $c_{4,3} = 52$, $c_{5,3}=150$, and $c_{6,3}=450$. (This has been entered in the OEIS~\cite{oeis} as A156762.)
\end{theorem}

This result is established in Sections~\ref{dhj-lower-sec},~\ref{dhj-upper-sec}. Initially these results were established by an integer program (see Appendix~\ref{integer-sec}, but we provide completely computer-free proofs here. The constructions used in Section~\ref{dhj-lower-sec} give reasonably efficient constructions for larger values of $n$; for instance, they show that $3^{99} \leq c_{100,3} \leq 2 \times 3^{99}$. See Section~\ref{dhj-lower-sec} for further discussion.

We also have several partial results for higher values of $k$: see Section~\ref{higherk-sec}. For results on the closely related \emph{Hales-Jewett numbers} $HJ(k,r)$, see Section~\ref{coloring-sec}.

A variant of the density Hales-Jewett theorem has also been studied in the literature. Define a \emph{geometric line} in $[k]^n$ to be any set of the form $\{ a+ir: i=1,\ldots,k\}$ in $[k]^n$, where we identify $[k]^n$ with a subset of $\Z^n$, and $a, r \in \Z^n$ with $r \neq 0$. Equivalently, a geometric line takes the form $\{ w( i, k+1-i ): i =1,\ldots,k \}$, where $w \in ([k] \cup \{x,\overline{x}\})^n \backslash [k]^n$ is a word of length $n$ using the numbers in $[k]$ and two wildcards $x, \overline{x}$ as the alphabet, with at least one wildcard occuring in $w$, and $w(i,j) \in [k]^n$ is the word formed by substituting $i,j$ for $x,\overline{x}$ respectively. Figure~\ref{fig-geomline} shows the eight geometric lines in $[3]^2$. Clearly every combinatorial line is a geometric line, but not conversely. In general, $[k]^n$ has $((k+2)^n-k^n)/2$ geometric lines.

\begin{figure}[tb]
\centerline{\includegraphics{AllGeomLinesIn3n.pdf}}
\caption{Geometric lines in $[3]^2$.}
\label{fig-geomline}
\end{figure}

Define a \emph{Moser set} in $[k]^n$ to be a subset of $[k]^n$ that contains no geometric lines, and let $c'_{n,k}$ be the maximum cardinality $|A|$ of a Moser set in $[k]^n$. Clearly one has $c'_{n,k} \leq c_{n,k}$, so in particular from Theorem~\ref{dhj} one has $c'_{n,k}/k^n \to 0$ as $n \to \infty$. (Interestingly, there is no known proof of this fact that does not go through Theorem~\ref{dhj}, even for $k=3$.) Again, $k=3$ is the first non-trivial case: it is clear that $c'_{n,1}=0$ and $c'_{n,2}=1$ for all $n$.

The question of computing $c'_{n,3}$ was first posed by Moser~\cite{moser}. Prior to our work, the values
$$ c'_{0,3}=1; c'_{1,3}=2; c'_{2,3}=6; c'_{3,3}=16; c'_{4,3}=43$$
were known~\cite{chvatal2},~\cite{chandra} (this is Sequence A003142 in the OEIS~\cite{oeis}). We extend this sequence slightly:

\begin{theorem}[Values of $c'_{n,3}$ for small $n$]\label{moser} We have $c'_{0,3} = 1$, $c'_{1,3} = 2$, $c'_{2,3} = 6$, $c'_{3,3} = 16$, $c'_{4,3} = 43$, $c'_{5,3} = 124$, and $c'_{6,3} = 353$.
\end{theorem}

This result is established in Sections \ref{moser-lower-sec}, \ref{moser-upper-sec}. The arguments given here are computer-assisted; however, we have found alternate (but lengthier) computer-free proofs for the above claims with the the exception of the proof of $c'_{6,3}=353$, which requires one non-trivial computation (Lemma \ref{paretop-4}). These alternate proofs are not given in this paper to save space, but can be found on the wiki for this project at

\centerline{{\tt http://michaelnielsen.org/polymath1/index.php?title=Main\_Page}.}

We establish a lower bound for this problem of $(2+o(1))\binom{n}{i}2^i\leq c'_{n,3}$, which is maximized for $i$ near $2n/3$. This bound is around one-third better than the literature. We also give methods to improve on this construction.

Earlier lower bounds were known:
let $A(n,d)$ denote the size of the largest binary code of length $n$
and minimal distance $d$. Then
\begin{equation}\label{cnchvatalintro}
c'_{n,3}\geq \max_k \left( \sum_{j=0}^k \binom{n}{j} A(n-j, k-j+1)\right).
\end{equation}
which, with $A(n,1)=2^n$ and $A(n,2)=2^{n-1}$, implies in particular that
\begin{equation}\label{binom}
c'_{n,3} \geq \binom{n+1}{\lfloor \frac{2n+1}{3} \rfloor} 2^{\lfloor \frac{2n+1}{3} \rfloor - 1}
\end{equation}
for $n \geq 2$. This bound is not quite optimal; for instance, it gives a lower bound of $c'_{6,3}=344$.

\begin{remark} Let $c''_{n,3}$ be the size of the largest subset of ${\mathbb F}_3^n$ which contains no lines $x, x+r, x+2r$ with $x,r \in {\mathbb F}_3^n$ and $r \neq 0$, where ${\mathbb F}_3$ is the field of three elements. Clearly one has $c''_{n,3} \leq c'_{n,3} \leq c_{n,3}$. It is known that
$$ c''_{0,3}=1; c''_{1,3}=2; c''_{2,3}=4; c'_{3,3}=9; c'_{4,3}=20; c''_{5,3}=45; c''_{6,3} = 112;$$
see \cite{potenchin}.
\end{remark}

As mentioned earlier, the sharp bound on $c_{n,2}$ comes from Sperner's theorem. It is known that Sperner's theorem can be refined to the \emph{Lubell-Meshalkin-Yamamoto (LMY) inequality}, which in our language asserts that
$$
\sum_{a_1,a_2 \geq 0; a_1+a_2 = n} \frac{|A \cap \Gamma_{a_1,a_2}|}{|\Gamma_{a_1,a_2}|} \leq 1
$$
for any line-free subset $A \subset [2]^n$, where the \emph{cell} $\Gamma_{a_1,\ldots,a_k} \subset [k]^n$ is the set of words in $[k]^n$ which contain exactly $a_i$ $i$'s for each $i=1,\ldots,k$. It is natural to ask whether this inequality can be extended to higher $k$. Let $\Delta_{k,n}$ denote the set of all tuples $(a_1,\ldots,a_k)$ of non-negative integers summing to $n$, define a \emph{simplex} to be a set of $k$ points in $\Delta_{k,n}$ of the form
$(a_1+r,a_2,\ldots,a_k), (a_1,a_2+r,\ldots,a_k),\ldots,(a_1,a_2,\ldots,a_k+r)$ for some $0 < r \leq n$ and $a_1,\ldots,a_k$ summing to $n-r$, and define a \emph{Fujimura set} to be a subset $B \subset \Delta_{k,n}$ which contains no simplices. Observe that if $w$ is a combinatorial line in $[k]^n$, then
$$ w(1) \in \Gamma_{a_1+r,a_2,\ldots,a_k}, w(2) \in \Gamma_{a_1,a_2+r,\ldots,a_k}, \ldots, w(k) = \Gamma_{a_1,a_2,\ldots,a_k+r}$$
for some simplex $(a_1+r,a_2,\ldots,a_k), (a_1,a_2+r,\ldots,a_k),\ldots,(a_1,a_2,\ldots,a_k+r)$. Thus, if $B$ is a Fujimura set, then $A := \bigcup_{\vec a \in B} \Gamma_{\vec a}$ is line-free. Note also that
$$
\sum_{\vec a \in \Delta_{k,n}} \frac{|A \cap \Gamma_{\vec a}|}{|\Gamma_{\vec a}|} = |B|.
$$
This motivates a ``hyper-optimistic'' conjecture:

{\bf a picture showing a Fujimura set?}

\begin{conjecture}\label{hoc} For any $k \geq 1$ and $n \geq 0$, and any line-free subset $A$ of $[k]^n$, one has
$$
\sum_{\vec a \in \Delta_{k,n}} \frac{|A \cap \Gamma_{\vec a}|}{|\Gamma_{\vec a}|} \leq c^\mu_{k,n},$$
where $c^\mu_{k,n}$ is the maximal size of a Fujimura set in $\Delta_{k,n}$.
\end{conjecture}

One can show that this conjecture for a fixed value of $k$ would imply Theorem \ref{dhj} for the same value of $k$, in much the same way that the LYM inequality is known to imply Sperner's theorem. The LYM inequality asserts that Conjecture \ref{hoc} is true for $k \leq 2$. As far as we know this conjecture could hold in $k=3$. However, we know that it fails for all even $k\ge 4$, see Section~\ref{higherk-sec}.

We are interested in removing all upward-pointing equilateral triangles from a triangular grid. Fujimura actually proposed the similar problem of removing all equilateral triangles at this website: www.puzzles.com/PuzzlePlayground/CoinsAndTriangles/CoinsAndTriangles.htm

In Section \ref{fujimura-sec} we give some bounds on these numbers. Explicit values are given for $\overline{c}^\mu_n$ up to $n=13$, and general upper and lower bounds are given for all $n$.

\subsection{Notation}\label{notation-sec}

There are several subsets of $[k]^n$ which will be useful in our analysis. We have already introduced combinatorial lines, geometric lines, and cells. One can generalise the notion of a combinatorial line to that of a \emph{combinatorial subspace} in $[k]^n$ of dimension $d$, which is indexed by a word $w$ in $([k] \cup \{x_1,\ldots,x_d\})^n$ containing at least one of each wildcard $x_1,\ldots,x_d$, and which forms the set $\{ w(i_1,\ldots,i_d): i_1,\ldots,i_d \in [k]\}$, where $w(i_1,\ldots,i_d) \in [k]^d$ is the word formed by replacing $x_1,\ldots,x_d$ with $i_1,\ldots,i_d$ respectively. Thus for instance, in $[3]^3$, we have the two-dimensional combinatorial subspace $xxy = \{111,112,113,221,222,223,331,332,333\}$. We similarly have the notion of a \emph{geometric subspace} in $[k]^n$ of dimension $d$, which is defined similarly but with $d$ wildcards $x_1,\ldots,x_d,\overline{x_1},\ldots,\overline{x_d}$, with at least one of either $x_i$ or $\overline{x_i}$ appearing in the word $w$ for each $1 \leq i \leq d$, and the space taking the form $\{ w(i_1,\ldots,i_d,k+1-i_1,\ldots,k+1-i_d): i_1,\ldots,i_d \in [k] \}$. Thus for instance $[3]^3$ contains the two-dimensional geometric subspace $x\overline{x}y = \{ 131, 132, 133, 221, 222, 223, 311, 312, 313\}$.

An important class of combinatorial subspaces in $[k]^n$ will be the \emph{slices} consisting of $n-1$ distinct wildcards and one fixed coordinate. We will denote the distinct wildcards here by asterisks, thus for instance in $[3]^3$ we have $2** = \{ 211, 212, 213, 221, 222, 223, 231, 232, 233\}$. Two slices are \emph{parallel} if their fixed coordinate are in the same position, thus for instance $1**$ and $2**$ are parallel, and one can subdivide $[k]^n$ into $k$ parallel slices, each of which is isomorphic to $[k]^{n-1}$. In the analysis of Moser slices with $k=3$, we will make a distinction between \emph{centre slices}, whose fixed coordinate is equal to $2$, and \emph{side slices}, in which the fixed coordinate is either $1$ or $3$, thus $[3]^n$ can be partitioned into one centre slice and two side slices.

Another important set in the study of $k=3$ Moser sets are the \emph{spheres} $S_{i,n} \subset [3]^n$, defined as those words in $[3]^n$ with exactly $n-i$ $2$'s (and hence $i$ letters that are $1$ or $3$). Thus for instance $S_{1,3} = \{ 122, 322, 212, 232, 221, 223\}$. Observe that $[3]^n = \bigcup_{i=0}^n S_{i,n}$, and each $S_{i,n}$ has cardinality $|S_{i,n}| = \binom{n}{i} 2^{i}$.

It is also convenient to subdivide each sphere $S_{i,n}$ into two components $S_{i,n} = S_{i,n}^o \cup S_{i,n}^e$, where $S_{i,n}^o$ are the words in $S_{i,n}$ with an odd number of $1$'s, and $S_{i,n}^e$ are the words with an even number of $1$'s. Thus for instance $S_{1,3}^o = \{122,212,221\}$ and $S_{1,3}^e = \{322,232,223\}$. Observe that for $i>0$, $S_{i,n}^o$ and $S_{i,n}^e$ both have cardinality $\binom{n}{i} 2^{i-1}$.

The \emph{Hamming distance} between two words $w,w'$ is the number of coordinates in which $w, w'$ differ, e.g. the Hamming distance between $123$ and $321$ is two. Note that $S_{i,n}$ is nothing more than the set of words whose Hamming distance from $2\ldots2$ is $i$, which justifies the terminology ``sphere''.

In the density Hales-Jewett problem, there are two types of symmetries on $[k]^n$ which map combinatorial lines to combinatorial lines (and hence line-free sets to line-free sets). The first is a permutation of the alphabet $[k]$; the second is a permutation of the $n$ coordinates. Together, this gives a symmetry group of order $k!n!$ on the cube $[k]^n$, which we refer to as the \emph{combinatorial symmetry group} of the cube $[k]^n$. Two sets which are related by an element of this symmetry group will be called (combinatorially) \emph{equivalent}, thus for instance any two slices are combinatorially equivalent.

For the analysis of Moser sets in $[k]^n$, the symmetries are a bit different. One can still permute the $n$ coordinates, but one is no longer free to permute the alphabet $[k]$. Instead, one can \emph{reflect} an individual coordinate, for instance sending each word $x_1 \ldots x_n$ to its reflection $x_1 \ldots x_{i-1} (k+1-x_i) x_{i+1} \ldots x_n$. Together, this gives a symmetry group of order $2^n n!$ on the cube $[k]^n$, which we refer to as the \emph{geometric symmetry group} of the cube $[k]^n$; this group maps geometric lines to geometric lines, and thus maps Moser sets to Moser sets. Two Moser sets which are related by an element of this symmetry group will be called (geometrically) \emph{equivalent}. For instance, a sphere $S_{i,n}$ is equivalent only to itself, and $S_{i,n}^o$, $S_{i,n}^e$ are equivalent only to each other.

Polymath.tex

2009-07-26T02:20:10Z

Mpeake:

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\begin{document}

\title{Density Hales-Jewett and Moser numbers}

\author{D.H.J. Polymath}
\address{http://michaelnielsen.org/polymath1/index.php}
\email{???}

\subjclass{???}

\begin{abstract}
For any $n \geq 0$ and $k \geq 1$, the density Hales-Jewett number $c_{n,k}$ is defined as the size of the largest subset of the cube $[k]^n$ := $\{1,\ldots,k\}^n$ which contains no combinatorial line; similarly, the Moser number $c'_{n,k}$ is the largest subset of the cube $[k]^n$ which contains no geometric line. A deep theorem of Furstenberg and Katznelson \cite{fk1}, \cite{fk2}, \cite{mcc} shows that $c_{n,k}$ = $o(k^n)$ as $n \to \infty$ (which implies a similar claim for $c'_{n,k}$; this is already non-trivial for $k = 3$. Several new proofs of this result have also been recently established \cite{poly}, \cite{austin}.

Using both human and computer-assisted arguments, we compute several values of $c_{n,k}$ and $c'_{n,k}$ for small $n,k$. For instance the sequence $c_{n,3}$ for $n=0,\ldots,6$ is $1,2,6,18,52,150,450$, while the sequence $c'_{n,3}$ for $n=0,\ldots,6$ is $1,2,6,16,43,124,353$. We also establish some results for higher $k$, showing for instance that an analogue of the LYM inequality (which relates to the $k = 2$ case) does not hold for higher $k$.
\end{abstract}

\maketitle
%\today

\setcounter{tocdepth}{1}
\tableofcontents

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\include{introduction}
\include{dhj-lown-lower}
\include{dhj-lown}
\include{moser-lower}
\include{moser}
\include{fujimura}
\include{higherk}
\include{coloring}

\appendix

\include{genetic}
\include{integer}

\begin{thebibliography}{10}

\bibitem{ajtai} Ajtai M, Szemerédi E, \emph{Sets of lattice points that form no squares}, Studia Scientiarum Mathematicarum Hungarica, 9, 9-11 (1974), (1975)

\bibitem{austin} T. Austin, \emph{Deducing the density Hales-Jewett theorem from an infinitary removal lemma}, preprint.

\bibitem{behrend}
F. Behrend, \emph{On the sets of integers which contain no three in arithmetic progression}, Proceedings of the National Academy of Sciences \textbf{23} (1946), 331–-332.

\bibitem{chandra}
A. Chandra, \emph{On the solution of Moser's problem in four dimensions}, Canad. Math. Bull. \textbf{16} (1973), 507--511.

\bibitem{chvatal1} V. Chv\'{a}tal, \emph{Remarks on a problem of Moser}, Canadian Math Bulletin, Vol 15, 1972, 19--21.

\bibitem{chvatal2} V. Chv\'{a}tal, \emph{Edmonds polytopes and a hierarchy of combinatorial problems}, Discrete Math. 4 (1973) 305-337.

\bibitem{elkin}
M. Elkin, \emph{An Improved Construction of Progression-Free Sets}, preprint.

\bibitem{fk1} H. Furstenberg, Y. Katznelson, \emph{A density version of the Hales-Jewett theorem for $k = 3$}, Graph Theory and Combinatorics (Cambridge, 1988). Discrete Math. 75 (1989), no. 1-3, 227–-241.

\bibitem{fk2} H. Furstenberg, Y. Katznelson, \emph{A density version of the Hales-Jewett theorem}, J. Anal. Math. 57 (1991), 64–-119. MR1191743

\bibitem{greenwolf}
B. Green, J. Wolf, \emph{A note on Elkin's improvement of Behrend's construction}, preprint.

\bibitem{heule} Marijn Heule, presentation at {\tt http://www.st.ewi.tudelft.nl/sat/slides/waerden.pdf}

\bibitem{komlos}
J. Koml\'{o}s, solution to problem P.170 by Leo Moser, Canad. Math.. Bull. vol {\bf (??check)} (1972), 312--313, 1970.

\bibitem{Krisha} K. Krishna, M. Narasimha Murty, "Genetic K-means algorithm," Systems, Man, and Cybernetics, Part B: Cybernetics, IEEE Transactions on , vol.29, no.3, pp.433-439, Jun 1999

\bibitem{markstrom} {{\tt http://abel.math.umu.se/~klasm/Data/HJ/}}

\bibitem{moser} L. Moser, Problem P.170 in Canad. Math. Bull. 13 (1970), 268.

\bibitem{mcc} R. McCutcheon, \emph{The conclusion of the proof of the density Hales-Jewett theorem for k=3}, unpublished.

\bibitem{obryant}
K. O'Bryant, \emph{Sets of integers that do not contain long arithmetic progressions}, preprint.

\bibitem{oeis}
N. J. A. Sloane, Ed. (2008), The On-Line Encyclopedia of Integer Sequences, {\tt www.research.att.com/~njas/sequences/}

\bibitem{potenchin}
A. Potechin, \emph{Maximal caps in $AG(6, 3)$}, Journal Designs, Codes and Cryptography, Volume 46, Number 3 / March, 2008.

\bibitem{poly} D.H.J. Polymath, ???, preprint. {\bf need title}

\bibitem{rankin}
R. A. Rankin, Sets of integers containing not more than a given number of terms in arithmetical progression, Proc. Roy. Soc. Edinburgh Sect. A 65 (1960/1961), 332–344 (1960/61). MR 0142526 (26 \#95)

\bibitem{roth}
K. Roth, \emph{On certain sets of integers, I}, Journal of the London Mathematical Society \textbf{28} (1953), 104-–109.

\bibitem{Rothlauf} F. Rothlauf, D. E. Goldberg, Representations for Genetic and Evolutionary Algorithms. Physica-Verlag, 2002.

\bibitem{sperner}
E. Sperner, \emph{Ein Satz \"uber Untermengen einer endlichen Menge}, Mathematische Zeitschrift \textbf{27} (1928), 544-–548.

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E. Szemer\'edi, \emph{On sets of integers containing no $k$ elements in arithmetic progression}, Acta Arithmetica \textbf{27} (1975), 199-–245.

\bibitem{waerden} {\tt http://en.wikipedia.org/wiki/Van\_der\_Waerden\_number}

\end{thebibliography}

\end{document}

Dhj-lown-lower.tex

2009-07-26T02:18:38Z

Mpeake:

\section{Lower bounds for the density Hales-Jewett problem}\label{dhj-lower-sec}

The purpose of this section is to establish various lower bounds for $c_{n,3}$, in particular establishing Theorem \ref{dhj-lower} and the lower bound component of Theorem \ref{dhj-upper}.

As observed in the introduction, if $B \subset \Delta_{3,n}$ is a Fujimura set (i.e. a subset of $\Delta_{3,n} = \{ (a,b,c) \in \N^3: a+b+c=n\}$ which contains no upward equilateral triangles $(a+r,b,c), (a,b+r,c), (a,b,c+r)$), then the set $A_B := \bigcup_{\vec a \in B} \Gamma_{a,b,c}$ is a line-free subset of $[3]^n$, which gives the lower bound
\begin{equation}\label{cn3}
c_{n,3} \geq |A_B| = \sum_{(a,b,c) \in B} \frac{n!}{a! b! c!}.
\end{equation}
All of the lower bounds for $c_{n,3}$ in this paper will be constructed via this device.

In order to use \eqref{cn3}, one of course needs to build Fujimura sets $B$ which are ``large'' in the sense that the right-hand side of \eqref{cn3} is large. A fruitful starting point for this goal is the sets
$$B_{j,n} := \{ (a,b,c) \in \Delta_{3,n}: a + 2b \neq j \hbox{ mod } 3 \}$$
for $j=0,1,2$. Observe that in order for a triangle $(a+r,b,c), (a,b+r,c), (a,b,c+r)$ to lie in $B_{j,n}$, the length $r$ of the triangle must be a multiple of $3$. This already makes $B_{j,n}$ a Fujimura set for $n < 3$ (and $B_{0,n}$ a Fujimura set for $n = 3$).

When $n$ is not a multiple of $3$, the $B_{j,n}$ are all rotations of each other and give equivalent sets (of size $2 \times 3^{n-1}$). When $n$ is a multiple of $3$, the sets $B_{1,n}$ and $B_{2,n}$ are reflections of each other, but $B_{0,n}$ is not equivalent to the other two sets (in particular, it omits all three corners of $\Delta_{3,n}$); the associated set $A_{B_{0,n}}$ is slightly larger than $A_{B_{1,n}}$ and $A_{B_{2,n}}$ and thus is slightly better for constructing line-free sets.

As mentioned already, $B_{0,n}$ is a Fujimura set for $n \leq 3$, and hence $A_{B_{0,n}}$ is line-free for $n \leq 3$. Applying \eqref{cn3} one obtains the lower bounds
$$ c_{0,3} \geq 1; c_{1,3} \geq 2; c_{2,3} \geq 6; c_{3,3} \geq 18.$$

For $n>3$, $B_{0,n}$ contains some triangles $(a+r,b,c), (a,b+r,c), (a,b,c+r)$ and so is not a Fujimura set, but one can remove points from this set to recover the Fujimura property. For instance, for $n \leq 6$, the only triangles in $B_{0,n}$ have side length $r=3$. One can ``delete'' these triangles by removing one vertex from each; in order to optimise the bound \eqref{cn3} it is preferable to delete vertices near the corners of $\Delta_{3,n}$ rather than near the centre. These considerations lead to the Fujimura sets
\begin{align*}
B_{0,4} &\backslash \{ (0,0,4), (0,4,0), (4,0,0) \}\\
B_{0,5} &\backslash \{ (0,4,1), (0,5,0), (4,0,1), (5,0,0) \}\\
B_{0,6} &\backslash \{ (0,1,5), (0,5,1), (1,0,5), (0,1,5), (1,5,0), (5,1,0) \}
\end{align*}
which by \eqref{cn3} gives the lower bounds
$$ c_{4,3} \geq 52; c_{5,3} \geq 150; c_{6,3} \geq 450.$$
Thus we have established all the lower bounds needed for Theorem \ref{dhj-upper}.

One can of course continue this process by hand, for instance the set
$$ B_{0,7} \backslash \{(0,1,6),(1,0,6),(0,5,2),(5,0,2),(1,5,1),(5,1,1),(1,6,0),(6,1,0) \}$$
gives the lower bound $c_{7,3} \geq 1302$, which we tentatively conjecture to be the correct bound.

A simplification was found when $n$ is a multiple of $3$. Observe that for $n=6$, the sets excluded from $B_{0,6}$ are all permutations of $(0,1,5)$. So the remaining sets are all the permutations of $(1,2,3)$ and $(0,2,4)$. In the same way, sets for $n=9$, $12$ and $15$ can be described as:
\begin{itemize}
\item $n=9$: $(2,3,4),(1,3,5),(0,4,5)$ and permutations;
\item $n=12$: $(3,4,5),(2,4,6),(1,5,6),(0,2,10),(0,5,7)$ and permutations;
\item $n=15$: $(4,5,6),(3,5,7),(2,6,7),(1,3,11),(1,6,8),(0,4,11),(0,7,8)$ and permutations.
\end{itemize}

When $n$ is not a multiple of $3$, say $n=3m-1$ or $n=3m-2$, one first finds a solution for $n=3m$. Then for $n=3m-1$, one restricts the first digit of the $3m$ sequence to equal $1$. This leaves exactly one-third as many points for $3m-1$ as for $3m$. For $n=3m-1$, one restricts the first two digits of the $3m$ sequence to be $12$. This leaves roughly one-ninth as many points for $3m-2$ as for $3m$.

The following is an effective method to find good, though not optimal, solutions for any $n=3m$. (For $n<21$, ignore any triple with a negative entry.)

Start with thirteen groups of points in the centre, formed from adding one of the following points, or its permutation, to $M := (m,m,m)$, when $n=3m$:
$$(-7,-3,+10), (-7, 0,+7),(-7,+3,+4),(-6,-4,+10),(-6,-1,+7),(-6,+2,+4)$$
$$(-5,-1,+6),(-5,+2,+3),(-4,-2,+6),(-4,+1,+3),(-3,+1,+2),(-2,0,+2),(-1,0,+1)$$
Then include eights string of points, stretching to the edges of the triangle $\Delta_n$;
\begin{itemize}
\item $M+(-8-2x,-6-2x,14+4x)$, $M+(-8-2x,-3-2x,11+4x)$, $M+(-8-2x,x,8+x)$, $M+(-8-2x,3+x,5+x)$ and permutations ($0\le 2x \le M-8$);
\item $M+(-9-2x,-5-2x,14+4x)$, $M+(-9-2x,-2-2x,11+4x)$, $M+(-9-2x,1+x,8+x)$, $M+(-9-2x,4+x,5+x)$ and permutations ($0\le 2x \le M-9$).
\end{itemize}
This solution gives $O(2.7 \sqrt(\log (n)/n)3^n$ points for values of $n$ up to around $1000$. This is asymptotically smaller than the known optimum of $3^{n-O(\sqrt(\log(n)))}$. When $n=99$, it gives more than $3^n/3$ points.

The following solution gives more points for $n>1000$, but not for moderate $n$:

\begin{itemize}
\item Define a sequence, of all positive numbers which, in base 3, do not contain a 1. Add 1 to all multiples of 3 in this sequence. This sequence does not contain a length-3 arithmetic progression. It starts $1,2,7,8,19,20,25,26,55, \ldots$;
\item List all the $(abc)$ triples that sum to $n$, for which the larger two differ by a number from the sequence;
\item Exclude the case when the smaller two differ by 1;
\item Include the case when $(a,b,c)$ is a permutation of $n/3+(-1,0,1)$.
\end{itemize}

An integer program was run obtain the maximum lower bound one could establish from \eqref{cn3} (see Appendix \ref{integer-sec}). The results for $1 \leq n \leq 20$ are displayed in Figure \ref{nlow}:

\begin{figure}[tb]
\centerline{
\begin{tabular}{|ll|ll|}
\hline
$n$ & lower bound & $n$ & lower bound \\
\hline
1 & 2 &11& 96338\\
2 & 6 & 12& 287892\\
3 & 18 & 13& 854139\\
4 & 52 & 14& 2537821\\
5 & 150& 15& 7528835\\
6 & 450& 16& 22517082\\
7 & 1302& 17& 66944301\\
8 & 3780&18& 198629224\\
9 & 11340&19& 593911730\\
10& 32864& 20& 1766894722\\
\hline
\end{tabular}}
\caption{Lower bounds for $c_n$ obtained by the $A_B$ construction.}
\label{nlow}
\end{figure}

More complete data, including the list of optimisers, can be found at \cite{markstrom}

If $B$ maximises the right-hand side of \eqref{cn3}, it is easy to see that $A_B$ is a line-free set which is maximal in the sense that the addition of any further point to $A_B$ will create a combinatorial line. Thus one might conjecture that the maximal value of the right-hand side \eqref{cn3} is in fact equal to $c_{n,3}$ for all $n$; Theorem \ref{dhj-upper} asserts that this conjecture is true for $n \leq 6$.

Now we prove Theorem \ref{dhj-lower}.

\begin{proof}[Proof of Theorem \ref {dhj-lower}]
Let $M$ be the circulant matrix with first row $(1,2,\ldots,k-1)$, second row $(k-1,1,2,\dots,k-2)$, and so on. Note that $M$ has nonzero determinant by well-known properties of circulant matrices.

Let $S$ be a subset of the interval $[-\sqrt {n}/2, \sqrt {n}/2)$ that contains no nonconstant arithmetic progressions of length k, and let $B\subset\Delta_{n, k}$ be the set
\[ B := \{(n-\sum_{i=1}^{k-1} a_i ,a_1,a_2,\dots, a_{k-1}) :
(a_1,\dots,a_{k-1})= n/k + \det(M) M^{-1}\vec{s} , \vec{s}\in S^{k-1}\}.\]
The map $(m,a_1,\dots,a_{k-1}) \mapsto M (a_1,\dots,a_{k-1})$ takes simplices to nonconstant arithmetic progressions in ${\mathbb Z}^{k-1}$, and takes $B$ to $\{c+det(M) \, \vec{s} \colon \vec{s} \in S^{k-1}\}$, which is a set containing no nonconstant arithmetic progressions. Thus, $B$ is a Fujimora set and so does not contain any combinatorial lines.

If all of $a_1,\ldots,a_k$ are within $C_1\sqrt{n}$ of $n/k$, then $|\Gamma_{\vec{a}}| \geq C k^n/n^{(k-1)/2}$ (where $C$ depends on $C_1$) by the central limit theorem. By our choice of $S$ and applying~\eqref{cn3}, we obtain
$$ c_ {n, k}\geq C k^n/n^{(k-1)/2} |S|^{k-1} = C k^n \left( \frac{|S|}{\sqrt{n}} \right)^{k-1}. $$
One can take $S$ to have cardinality $r_ k (\sqrt {n}) $, which from the results of O'Bryant~\cite {obryant}) satisfies (for all sufficiently large $n$, some $C>0$, and $\ell$ the largest integer satisfying $k> 2^{\ell-1}$)
$$ \frac{r_k (\sqrt{n})}{\sqrt{n}} \geq C (\log n)^{1/(2\ell)}\exp_2 (-\ell 2^{(\ell-1)/2-1/\ell} \sqrt[\ell]{\log_2 n}),$$
which completes the proof.
\end {proof}

Dhj-lown-lower.tex

2009-07-26T02:13:29Z

Mpeake:

\section{Lower bounds for the density Hales-Jewett problem}\label{dhj-lower-sec}

The purpose of this section is to establish various lower bounds for $c_{n,3}$, in particular establishing Theorem \ref{dhj-lower} and the lower bound component of Theorem \ref{dhj-upper}.

As observed in the introduction, if $B \subset \Delta_{3,n}$ is a Fujimura set (i.e. a subset of $\Delta_{3,n} = \{ (a,b,c) \in \N^3: a+b+c=n\}$ which contains no upward equilateral triangles $(a+r,b,c), (a,b+r,c), (a,b,c+r)$), then the set $A_B := \bigcup_{\vec a \in B} \Gamma_{a,b,c}$ is a line-free subset of $[3]^n$, which gives the lower bound
\begin{equation}\label{cn3}
c_{n,3} \geq |A_B| = \sum_{(a,b,c) \in B} \frac{n!}{a! b! c!}.
\end{equation}
All of the lower bounds for $c_{n,3}$ in this paper will be constructed via this device.

In order to use \eqref{cn3}, one of course needs to build Fujimura sets $B$ which are ``large'' in the sense that the right-hand side of \eqref{cn3} is large. A fruitful starting point for this goal is the sets
$$B_{j,n} := \{ (a,b,c) \in \Delta_{3,n}: a + 2b \neq j \hbox{ mod } 3 \}$$
for $j=0,1,2$. Observe that in order for a triangle $(a+r,b,c), (a,b+r,c), (a,b,c+r)$ to lie in $B_{j,n}$, the length $r$ of the triangle must be a multiple of $3$. This already makes $B_{j,n}$ a Fujimura set for $n < 3$ (and $B_{0,n}$ a Fujimura set for $n = 3$).

When $n$ is not a multiple of $3$, the $B_{j,n}$ are all rotations of each other and give equivalent sets (of size $2 \times 3^{n-1}$). When $n$ is a multiple of $3$, the sets $B_{1,n}$ and $B_{2,n}$ are reflections of each other, but $B_{0,n}$ is not equivalent to the other two sets (in particular, it omits all three corners of $\Delta_{3,n}$); the associated set $A_{B_{0,n}}$ is slightly larger than $A_{B_{1,n}}$ and $A_{B_{2,n}}$ and thus is slightly better for constructing line-free sets.

As mentioned already, $B_{0,n}$ is a Fujimura set for $n \leq 3$, and hence $A_{B_{0,n}}$ is line-free for $n \leq 3$. Applying \eqref{cn3} one obtains the lower bounds
$$ c_{0,3} \geq 1; c_{1,3} \geq 2; c_{2,3} \geq 6; c_{3,3} \geq 18.$$

For $n>3$, $B_{0,n}$ contains some triangles $(a+r,b,c), (a,b+r,c), (a,b,c+r)$ and so is not a Fujimura set, but one can remove points from this set to recover the Fujimura property. For instance, for $n \leq 6$, the only triangles in $B_{0,n}$ have side length $r=3$. One can ``delete'' these triangles by removing one vertex from each; in order to optimise the bound \eqref{cn3} it is preferable to delete vertices near the corners of $\Delta_{3,n}$ rather than near the centre. These considerations lead to the Fujimura sets
\begin{align*}
B_{0,4} &\backslash \{ (0,0,4), (0,4,0), (4,0,0) \}\\
B_{0,5} &\backslash \{ (0,4,1), (0,5,0), (4,0,1), (5,0,0) \}\\
B_{0,6} &\backslash \{ (0,1,5), (0,5,1), (1,0,5), (0,1,5), (1,5,0), (5,1,0) \}
\end{align*}
which by \eqref{cn3} gives the lower bounds
$$ c_{4,3} \geq 52; c_{5,3} \geq 150; c_{6,3} \geq 450.$$
Thus we have established all the lower bounds needed for Theorem \ref{dhj-upper}.

One can of course continue this process by hand, for instance the set
$$ B_{0,7} \backslash \{(0,1,6),(1,0,6),(0,5,2),(5,0,2),(1,5,1),(5,1,1),(1,6,0),(6,1,0) \}$$
gives the lower bound $c_{7,3} \geq 1302$, which we tentatively conjecture to be the correct bound.

A simplification was found when $n$ is a multiple of $3$. Observe that for $n=6$, the sets excluded from $B_{0,6}$ are all permutations of $(0,1,5)$. So the remaining sets are all the permutations of $(1,2,3)$ and $(0,2,4)$. In the same way, sets for $n=9$, $12$ and $15$ can be described as:
\begin{itemize}
\item $n=9$: $(2,3,4),(1,3,5),(0,4,5)$ and permutations;
\item $n=12$: $(3,4,5),(2,4,6),(1,5,6),(0,2,10),(0,5,7)$ and permutations;
\item $n=15$: $(4,5,6),(3,5,7),(2,6,7),(1,3,11),(1,6,8),(0,4,11),(0,7,8)$ and permutations.
\end{itemize}

When $n$ is not a multiple of $3$, say $n=3m-1$ or $n=3m-2$, one first finds a solution for $n=3m$. Then for $n=3m-1$, one restricts the first digit of the $3m$ sequence to equal $1$. This leaves exactly one-third as many points for $3m-1$ as for $3m$. For $n=3m-1$, one restricts the first two digits of the $3m$ sequence to be $12$. This leaves roughly one-ninth as many points for $3m-2$ as for $3m$.

The following is an effective method to find good, though not optimal, solutions for any $n=3m$. (For $n<21$, ignore any triple with a negative entry.)

Start with thirteen groups of points in the centre, formed from adding one of the following points, or its permutation, to $M := (m,m,m)$, when $n=3m$:
$$(-7,-3,+10), (-7, 0,+7),(-7,+3,+4),(-6,-4,+10),(-6,-1,+7),(-6,+2,+4)$$
$$(-5,-1,+6),(-5,+2,+3),(-4,-2,+6),(-4,+1,+3),(-3,+1,+2),(-2,0,+2),(-1,0,+1)$$
Then include eights string of points, stretching to the edges of the triangle $\Delta_n$;
\begin{itemize}
\item $M+(-8-2x,-6-2x,14+4x)$, $M+(-8-2x,-3-2x,11+4x)$, $M+(-8-2x,x,8+x)$, $M+(-8-2x,3+x,5+x)$ and permutations ($0\le 2x \le M-8$);
\item $M+(-9-2x,-5-2x,14+4x)$, $M+(-9-2x,-2-2x,11+4x)$, $M+(-9-2x,1+x,8+x)$, $M+(-9-2x,4+x,5+x)$ and permutations ($0\le 2x \le M-9$).
\end{itemize}
This solution gives $O(2.7 \sqrt(\log (n)/n)3^n$ points for values of $n$ up to around $1000$. This is asymptotically smaller than the known optimum of $3^{n-O(\sqrt(\log(n)))}$. When $n=99$, it gives more than $3^n/3$ points.

The following solution gives more points for $n>1000$, but not for moderate $n$:

\begin{itemize}
\item Define a sequence, of all positive numbers which, in base 3, do not contain a 1. Add 1 to all multiples of 3 in this sequence. This sequence does not contain a length-3 arithmetic progression. It starts $1,2,7,8,19,20,25,26,55, \ldots$;
\item List all the $(abc)$ triples that sum to $n$, for which the larger two differ by a number from the sequence;
\item Exclude the case when the smaller two differ by 1;
\item Include the case when $(a,b,c)$ is a permutation of $n/3+(-1,0,1)$.
\end{itemize}

An integer program was run obtain the maximum lower bound one could establish from \eqref{cn3} (see Appendix \ref{integer-sec}). The results for $1 \leq n \leq 20$ are displayed in Figure \ref{nlow}:

\begin{figure}[tb]
\centerline{
\begin{tabular}{|ll|ll|}
\hline
$n$ & lower bound & $n$ & lower bound \\
\hline
1 & 2 &11& 96338\\
2 & 6 & 12& 287892\\
3 & 18 & 13& 854139\\
4 & 52 & 14& 2537821\\
5 & 150& 15& 7528835\\
6 & 450& 16& 22517082\\
7 & 1302& 17& 66944301\\
8 & 3780&18& 198629224\\
9 & 11340&19& 593911730\\
10& 32864& 20& 1766894722\\
\hline
\end{tabular}}
\caption{Lower bounds for $c_n$ obtained by the $A_B$ construction.}
\label{nlow}
\end{figure}

More complete data, including the list of optimisers, can be found at
\centerline{{\tt http://abel.math.umu.se/~klasm/Data/HJ/}.}

If $B$ maximises the right-hand side of \eqref{cn3}, it is easy to see that $A_B$ is a line-free set which is maximal in the sense that the addition of any further point to $A_B$ will create a combinatorial line. Thus one might conjecture that the maximal value of the right-hand side \eqref{cn3} is in fact equal to $c_{n,3}$ for all $n$; Theorem \ref{dhj-upper} asserts that this conjecture is true for $n \leq 6$.

Now we prove Theorem \ref{dhj-lower}.

\begin{proof}[Proof of Theorem \ref {dhj-lower}]
Let $M$ be the circulant matrix with first row $(1,2,\ldots,k-1)$, second row $(k-1,1,2,\dots,k-2)$, and so on. Note that $M$ has nonzero determinant by well-known properties of circulant matrices.

Let $S$ be a subset of the interval $[-\sqrt {n}/2, \sqrt {n}/2)$ that contains no nonconstant arithmetic progressions of length k, and let $B\subset\Delta_{n, k}$ be the set
\[ B := \{(n-\sum_{i=1}^{k-1} a_i ,a_1,a_2,\dots, a_{k-1}) :
(a_1,\dots,a_{k-1})= n/k + \det(M) M^{-1}\vec{s} , \vec{s}\in S^{k-1}\}.\]
The map $(m,a_1,\dots,a_{k-1}) \mapsto M (a_1,\dots,a_{k-1})$ takes simplices to nonconstant arithmetic progressions in ${\mathbb Z}^{k-1}$, and takes $B$ to $\{c+det(M) \, \vec{s} \colon \vec{s} \in S^{k-1}\}$, which is a set containing no nonconstant arithmetic progressions. Thus, $B$ is a Fujimora set and so does not contain any combinatorial lines.

If all of $a_1,\ldots,a_k$ are within $C_1\sqrt{n}$ of $n/k$, then $|\Gamma_{\vec{a}}| \geq C k^n/n^{(k-1)/2}$ (where $C$ depends on $C_1$) by the central limit theorem. By our choice of $S$ and applying~\eqref{cn3}, we obtain
$$ c_ {n, k}\geq C k^n/n^{(k-1)/2} |S|^{k-1} = C k^n \left( \frac{|S|}{\sqrt{n}} \right)^{k-1}. $$
One can take $S$ to have cardinality $r_ k (\sqrt {n}) $, which from the results of O'Bryant~\cite {obryant}) satisfies (for all sufficiently large $n$, some $C>0$, and $\ell$ the largest integer satisfying $k> 2^{\ell-1}$)
$$ \frac{r_k (\sqrt{n})}{\sqrt{n}} \geq C (\log n)^{1/(2\ell)}\exp_2 (-\ell 2^{(\ell-1)/2-1/\ell} \sqrt[\ell]{\log_2 n}),$$
which completes the proof.
\end {proof}

Fujimura.tex

2009-07-26T01:38:59Z

Mpeake:

\section{Fujimura's problem}\label{fujimura-sec}

Let $\overline{c}^\mu_n$ be the size of the largest subset of the trianglular grid
$$\Delta_n := \{(a,b,c)\in {\mathbb Z}^3_+ : a+b+c = n\}$$
which contains no equilateral triangles $(a+r,b,c), (a,b+r,c), (a,b,c+r)$ with $r>0$. These are upward-pointing equilateral triangles. We shall refer to such sets as 'triangle-free'.
(Kobon Fujimura is a prolific inventor of puzzles, and in [http://www.puzzles.com/PuzzlePlayground/CoinsAndTriangles/CoinsAndTriangles.htm this] puzzle asked the related question of eliminating all equilateral triangles.)

The following table was formed mostly by computer searches for optimal solutions. We also found human proofs for most of them (see {\tt http://michaelnielsen.org/polymath1/index.php?title=Fujimura's\_problem}).

\begin{figure}
\centerline{
\begin{tabular}{l|llllllllllllll}
$n$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13\\
\hline
$\overline{c}^\mu_n$ & 1 & 2 & 4 & 6 & 9 & 12 & 15 & 18 & 22 & 26 & 31 & 35 & 40 & 46
\end{tabular}
}
\label{lowFujimura}
\caption{Fujimura numbers}
\end{figure}

For any equilateral triangle $(a+r,b,c)$,$(a,b+r,c)$ and $(a,b,c+r)$, the value $y+2z$ forms an arithmetic progression of length 3. A Behrend set is a finite set of integers with no arithmetic progression of length 3 (see {\tt http://arxiv.org/PS\_cache/arxiv/pdf/0811/0811.3057v2.pdf}). By looking at those triples $(a,b,c)$ with $a+2b$ inside a Behrend set, one can obtain the lower bound of $\overline{c}^\mu_n \geq n^2 exp(-O(\sqrt{\log n}))$.

It can be shown by a 'corners theorem' of Ajtai and Szemeredi \cite{ajtai} that $\overline{c}^\mu_n = o(n^2)$ as $n \rightarrow \infty$.

An explicit lower bound is $3(n-1)$, made of all points in $\Delta_n$ with exactly one coordinate equal to zero.

An explicit upper bound comes from counting the triangles. There are $\binom{n+2}{3}$ triangles, and each point belongs to $n$ of them. So you must remove at least $(n+2)(n+1)/6$ points to remove all triangles, leaving $(n+2)(n+1)/3$ points as an upper bound for $\overline{c}^\mu_n$.

Polymath.tex

2009-07-26T01:38:16Z

Mpeake:

\documentclass[12pt,a4paper,reqno]{amsart}
\usepackage{amssymb}
\usepackage{amscd}
%\usepackage{psfig}
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% uncomment this when editing cross-references
\numberwithin{equation}{section}

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\theoremstyle{plain}

\newtheorem{theorem}{Theorem}[section]
%\newtheorem{theorem}[theorem]{Theorem}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{principle}[theorem]{Principle}
\newtheorem{claim}[theorem]{Claim}

\theoremstyle{definition}

%\newtheorem{roughdef}[subsection]{Rough Definition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{remarks}[theorem]{Remarks}
\newtheorem{example}[theorem]{Example}
\newtheorem{examples}[theorem]{Examples}
%\newtheorem{problem}[subsection]{Problem}
%\newtheorem{question}[subsection]{Question}

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\parindent 0mm
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\begin{document}

\title{Density Hales-Jewett and Moser numbers}

\author{D.H.J. Polymath}
\address{http://michaelnielsen.org/polymath1/index.php}
\email{???}

\subjclass{???}

\begin{abstract}
For any $n \geq 0$ and $k \geq 1$, the density Hales-Jewett number $c_{n,k}$ is defined as the size of the largest subset of the cube $[k]^n$ := $\{1,\ldots,k\}^n$ which contains no combinatorial line; similarly, the Moser number $c'_{n,k}$ is the largest subset of the cube $[k]^n$ which contains no geometric line. A deep theorem of Furstenberg and Katznelson \cite{fk1}, \cite{fk2}, \cite{mcc} shows that $c_{n,k}$ = $o(k^n)$ as $n \to \infty$ (which implies a similar claim for $c'_{n,k}$; this is already non-trivial for $k = 3$. Several new proofs of this result have also been recently established \cite{poly}, \cite{austin}.

Using both human and computer-assisted arguments, we compute several values of $c_{n,k}$ and $c'_{n,k}$ for small $n,k$. For instance the sequence $c_{n,3}$ for $n=0,\ldots,6$ is $1,2,6,18,52,150,450$, while the sequence $c'_{n,3}$ for $n=0,\ldots,6$ is $1,2,6,16,43,124,353$. We also establish some results for higher $k$, showing for instance that an analogue of the LYM inequality (which relates to the $k = 2$ case) does not hold for higher $k$.
\end{abstract}

\maketitle
%\today

\setcounter{tocdepth}{1}
\tableofcontents

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\include{introduction}
\include{dhj-lown-lower}
\include{dhj-lown}
\include{moser-lower}
\include{moser}
\include{fujimura}
\include{higherk}
\include{coloring}

\appendix

\include{genetic}
\include{integer}

\begin{thebibliography}{10}

\bibitem{ajtai} Ajtai M, Szemerédi E, \emph{Sets of lattice points that form no squares}, Studia Scientiarum Mathematicarum Hungarica, 9, 9-11 (1974), (1975)

\bibitem{austin} T. Austin, \emph{Deducing the density Hales-Jewett theorem from an infinitary removal lemma}, preprint.

\bibitem{behrend}
F. Behrend, \emph{On the sets of integers which contain no three in arithmetic progression}, Proceedings of the National Academy of Sciences \textbf{23} (1946), 331–-332.

\bibitem{chandra}
A. Chandra, \emph{On the solution of Moser's problem in four dimensions}, Canad. Math. Bull. \textbf{16} (1973), 507--511.

\bibitem{chvatal1} V. Chv\'{a}tal, \emph{Remarks on a problem of Moser}, Canadian Math Bulletin, Vol 15, 1972, 19--21.

\bibitem{chvatal2} V. Chv\'{a}tal, \emph{Edmonds polytopes and a hierarchy of combinatorial problems}, Discrete Math. 4 (1973) 305-337.

\bibitem{elkin}
M. Elkin, \emph{An Improved Construction of Progression-Free Sets}, preprint.

\bibitem{fk1} H. Furstenberg, Y. Katznelson, \emph{A density version of the Hales-Jewett theorem for $k = 3$}, Graph Theory and Combinatorics (Cambridge, 1988). Discrete Math. 75 (1989), no. 1-3, 227–-241.

\bibitem{fk2} H. Furstenberg, Y. Katznelson, \emph{A density version of the Hales-Jewett theorem}, J. Anal. Math. 57 (1991), 64–-119. MR1191743

\bibitem{greenwolf}
B. Green, J. Wolf, \emph{A note on Elkin's improvement of Behrend's construction}, preprint.

\bibitem{heule} Marijn Heule, presentation at {\tt http://www.st.ewi.tudelft.nl/sat/slides/waerden.pdf}

\bibitem{komlos}
J. Koml\'{o}s, solution to problem P.170 by Leo Moser, Canad. Math.. Bull. vol {\bf (??check)} (1972), 312--313, 1970.

\bibitem{Krisha} K. Krishna, M. Narasimha Murty, "Genetic K-means algorithm," Systems, Man, and Cybernetics, Part B: Cybernetics, IEEE Transactions on , vol.29, no.3, pp.433-439, Jun 1999

\bibitem{moser} L. Moser, Problem P.170 in Canad. Math. Bull. 13 (1970), 268.

\bibitem{mcc} R. McCutcheon, \emph{The conclusion of the proof of the density Hales-Jewett theorem for k=3}, unpublished.

\bibitem{obryant}
K. O'Bryant, \emph{Sets of integers that do not contain long arithmetic progressions}, preprint.

\bibitem{oeis}
N. J. A. Sloane, Ed. (2008), The On-Line Encyclopedia of Integer Sequences, {\tt www.research.att.com/~njas/sequences/}

\bibitem{potenchin}
A. Potechin, \emph{Maximal caps in $AG(6, 3)$}, Journal Designs, Codes and Cryptography, Volume 46, Number 3 / March, 2008.

\bibitem{poly} D.H.J. Polymath, ???, preprint. {\bf need title}

\bibitem{rankin}
R. A. Rankin, Sets of integers containing not more than a given number of terms in arithmetical progression, Proc. Roy. Soc. Edinburgh Sect. A 65 (1960/1961), 332–344 (1960/61). MR 0142526 (26 \#95)

\bibitem{roth}
K. Roth, \emph{On certain sets of integers, I}, Journal of the London Mathematical Society \textbf{28} (1953), 104-–109.

\bibitem{Rothlauf} F. Rothlauf, D. E. Goldberg, Representations for Genetic and Evolutionary Algorithms. Physica-Verlag, 2002.

\bibitem{sperner}
E. Sperner, \emph{Ein Satz \"uber Untermengen einer endlichen Menge}, Mathematische Zeitschrift \textbf{27} (1928), 544-–548.

\bibitem{szem}
E. Szemer\'edi, \emph{On sets of integers containing no $k$ elements in arithmetic progression}, Acta Arithmetica \textbf{27} (1975), 199-–245.

\bibitem{waerden} {\tt http://en.wikipedia.org/wiki/Van\_der\_Waerden\_number}

\end{thebibliography}

\end{document}

Integer.tex

2009-07-25T02:57:42Z

Mpeake:

\section{Integer programming}\label{integer-sec}

Integer programming is linear programming with solutions restricted to the integers. A number of packages, for example Maple.12, can solve these problems. It was used in several places during this project.

The first place it was used was to find $c_5$, the maximal subset of $[3]^5$ with no combinatorial lines. One 0/1-variable was used for each point in the cube, and one linear inequality for each combinatorial line. The computer found twelve solutions with 150 points, and found that there were no solutions with 151 points, so $c_5 = 150$. It follows easily that $c_6 = 450$, but the same program was used as a check to show that any set of 451 points in $[3]^6$ has a combinatorial line.

The following routine was used for the 5 dimensional Moser problem to show that $c'_5\ge 124$.
Let G be the 3-hypergraph with vertex set equal to $[3]^5$ as before and set $(I,J,K)\in E_3$ to be a 3-edge if and only if it is a geometric line. $c'_5$ is then the size of a maximal independent set.
Let

$$ f(x) = \Sum_I x_I - \Sum_{(I,J,K)\in E_3} x_Ix_Jx_K. $$

The following then holds,

$$c'_n = max_{x\in\{0,1\}^{3^n}} f(x) = max_{x\in [0,1]^{3^n}} f(x).$$

To check the first equality, if $V$ is a line-free set in $G$ of size $k$, set $x_J=1$ for $J\in V$, and zero otherwise. Then $f(x) = $k$ because $V$ is line-free, so all the cubic terms are zero. So $c'_n \le max_{x\in\{0,1\}^{3^n}} f(x)$. Conversely, if for some $x\in\{0,1\}^{3^n}$, $f(x)=k$ and if there are triplets $x_I=x_J=x_K=1$ with $(I,J,K)\in E_3$, we change $x_I$ to 0. This reduces $\sum_I x_I$ by 1 but also reduce the second term of $f(x)$ by at least one. So repeating we can find an $x\in\{0,1\}^{3^n}$ with $f(x)\ge k$ and $x_Ix_Jx_K=0$ if $(I,J,K)\in E_3$, i.e. an independent set of size $\ge k$. So $c'_n \ge max_{x\in\{0,1\}^{3^n}} f(x)$. The second equality follows from the maximum principle since $f$ is a square free polynomial.
$-f$ was then minimized using Matlab's built-in minimization routine, and found a 124-point subset of $[3]^5$ with no geometric lines.

A third application of integer programming is expanded in some detail in Section \ref{moser-upper-sec}. Each point is classified by how many of its coordinates are 2. Then the possible histograms or 'statistics' of points in a low-dimensional line-free set were computed by brute-force. Since a low-dimensional space fits into a higher-dimensional space in many ways, the low-dimensional statistics had consequences for higher-dimensional sets. Many dozens of inequalities were found for six-dimensional and higher spaces, and Maple's integer programming routine was used to find limits on $c'_n$.

A fourth application of integer programming found many of our best lower bounds for both the DHJ problem and the Moser problem. Assume that a solution is made up of complete $\Gamma(a,b,c)$ cells. This drastically reduces the number of variables, from $3^n$ to $(n+2)(n+1)/2$ binary variables, one for each Gamma set. In the DHJ problem, points in a combinatorial line will lie in cells of the form $\Gamma(a+r,b,c)$, $\Gamma(a,b+r,c)$ and $\Gamma(a,b,c+r)$ for some positive $r$. Thus the line-free condition is met if we prevent equilateral triangles of this sort. The Fujimora problem was to maximize the number of Gamma sets under this restriction, searching over the $(n+2)(n+1)/2$ binary variables. The weighted Fujimora problem was to maximize the total number of points. Thus each Gamma set $\Gamma(a,b,c)$ is weighted by its size $(a+b+c)!/(a!b!c!)$, but the restrictions are the same.

In the Moser-Fujimura problem, we look for collections of $\Gamma(a,b,c)$ that do not contain geometric lines. This is satisfied if there are no isosceles triangles of the form $(a+r,b,c+s)$, $(a,b+r+s,c)$, $(a+s,b,c+r)$. Again, we run a search over the $(n+2)(n+1)/2$ binary variables, with $\Gamma(a,b,c)$ weighted by $(a+b+c)!/(a!b!c!)$, to maximize the total number of points.

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Moser.tex

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\section{Upper bounds for the $k=3$ Moser problem in small dimensions}\label{moser-upper-sec}

In this section we finish the proof of Theorem \ref{moser} by obtaining the upper bounds on $c'_{n,3}$ for $n \leq 6$.

\subsection{Statistics, densities and slices}

Our analysis will revolve around various \emph{statistics} of Moser sets $A \subset [3]^n$, their associated \emph{densities}, and the behavior of such statistics and densities with respect to the operation of passing from the cube $[3]^n$ to various \emph{slices} of that cube.

\begin{definition}[Statistics and densities] Let $A \subset [3]^n$ be a set. For any $0 \leq i \leq n$, set $a_i(A) := |A \cap S_{n-i,n}|$; thus we have
$$ 0 \leq a_i(A) \leq |S_{n-i,n}| = \binom{n}{i} 2^{n-i}$$
for $0 \leq i \leq n$ and
$$ a_0(A) + \ldots + a_n(A) = |A|.$$
We refer to the vector $(a_0(A),\ldots,a_n(A))$ as the \emph{statistics} of $A$. We define the $i^{th}$ \emph{density} $\alpha_i(A)$ to be the quantity
$$ \alpha_i(A) := \frac{a_i(A) }{\binom{n}{i} 2^{n-i}},$$
thus $0 \leq \alpha_i(A) \leq 1$ and
$$ |A| = \sum_{i=0}^n \binom{n}{i} 2^{n-i} a_i(A).$$
\end{definition}

\begin{example}\label{2mos} Let $n=2$ and $A$ be the Moser set $A := \{ 12, 13, 21, 23, 31, 32 \}$. Then the statistics $(a_0(A), a_1(A), a_2(A))$ of $A$ are $(2,4,0)$, and the densities $(\alpha_0(A), \alpha_1(A), \alpha_2(A))$ are $(\frac{1}{2}, 1, 0)$. \textbf{Include picture here? with colours?}
\end{example}

When working with small values of $n$, it will be convenient to write $a(A)$, $b(A)$, $c(A)$, etc. for $a_0(A)$, $a_1(A)$, $a_2(A)$, etc., and similarly write $\alpha(A), \beta(A), \gamma(A)$, etc. for $\alpha_0(A)$, $\alpha_1(A)$, $\alpha_2(A)$, etc. Thus for instance in Example \ref{2mos} we have $b(A) = 4$ and $\alpha(A) = \frac{1}{2}$.

\begin{definition}[Subspace statistics and densities] If $V$ is a $k$-dimensional geometric subspace of $[3]^n$, then we have a map $\phi_V: [3]^k \to [3]^n$ from the $k$-dimensional cube to the $n$-dimensional cube. If $A \subset [3]^n$ is a set and $0 \leq i \leq k$, we write $a_i(V,A)$ for $a_i(\phi_V^{-1}(A))$ and $\alpha_i(V,A)$ for $\alpha_i(\phi_V^{-1}(A))$. If the set $A$ is clear from context, we abbreviate $a_i(V,A)$ as $a_i(V)$ and $\alpha_i(V,A)$ as $\alpha_i(V)$.
\end{definition}

For our problem, a particularly important type of subspace of $[3]^n$ will be the \emph{slices} formed by fixing one coordinate and letting the other $n-1$ coordinates vary. We will denote this by a single string in which the $n-1$ varying coordinates are denoted by asterisks. For instance, in $[3]^2$, $1*$ denotes the slice $1*=\{11,12,13\}$, $*2$ denotes the slice $*2=\{12,22,32\}$, etc.; similarly, in $[3]^3$, $1**$ is the slice $\{111, 112, 113, 121, 122, 123, 131, 132, 133\}$, etc. We call a slice a \emph{centre slice} if the fixed coordinate is $2$ and a \emph{side slice} if it is $1$ or $3$.

\begin{example} We continue Example \ref{2mos}. Then the statistics of the side slice $1*$ are $(a(1*),b(1*)) = (1,1)$, while the statistics of the centre slice $2*$ are $(a(2*),b(2*))=(2,0)$. The corresponding densities are $(\alpha(1*),\beta(1*)) = (1/2,1)$ and $(\alpha(2*),\beta(2*))=(1,0)$.
\end{example}

A simple double counting argument gives the following useful identity:

\begin{lemma}[Double counting identity]\label{dci} Let $A \subset [3]^n$ and $0 \leq i \leq n-1$. Then we have
$$ \frac{1}{n-i-1} \sum_{V \hbox{ a side slice}} a_{i+1}(V) = \frac{1}{i+1} \sum_{W \hbox{ a centre slice}} a_i(W) = a_{i+1}(A)$$
where $V$ ranges over the $2n$ side slices of $[3]^n$, and $W$ ranges over the $n$ centre slices. In other words, the average value of $\alpha_{i+1}(V)$ for side slices $V$ equals the average value of $\alpha_i(W)$ for centre slices $W$, which is in turn equal to $\alpha_{i+1}(A)$.
\end{lemma}

Indeed, this lemma follows from the observation that every string in $A \cap S_{n-i-1,n}$ belongs to $i+1$ centre slices $W$ (and contributes to $a_i(W)$) and to $n-i-1$ side slices $V$ (and contributes to $a_{i+1}(V)$). One can also view this lemma probabilistically, as the assertion that there are three equivalent ways to generate a random string of length $n$:

\begin{itemize}
\item Pick a side slice $V$ at random, and randomly fill in the wildcards in such a way that $i+1$ of the wildcards are $2$'s (i.e. using an element of $S_{n-i-2,n-1}$).
\item Pick a centre slice $V$ at random, and randomly fill in the wildcards in such a way that $i$ of the wildcards are $2$'s (i.e. using an element of $S_{n-i-1,n-1}$).
\item Randomly choose an element of $S_{n-i-1,n}$.
\end{itemize}

\begin{example} We continue Example \ref{2mos}. The average value of $\beta$ for side slices is equal to the average value of $\alpha$ for centre slices, which is equal to $\beta(A) = 1$.
\end{example}

Another very useful fact (essentially due to \cite{chvatal2}) is that linear inequalities for statistics of Moser sets at one dimension propagate to linear inequalities in higher dimensions:

\begin{lemma}[Propagation lemma]\label{prop} Let $n \geq 1$ be an integer. Suppose one has a linear inequality of the form
\begin{equation}\label{alphav}
\sum_{i=0}^n v_i \alpha_i(A) \leq s
\end{equation}
for all Moser sets $A \subset [3]^n$ and some real numbers $v_0,\ldots,v_n,s$. Then we also have the linear inequality
$$ \sum_{i=0}^n v_i \alpha_{qi+r}(A) \leq s$$
whenever $q \geq 1$, $r \geq 0$, $N \geq nq+r$ are integers and $A \subset [3]^N$ is a Moser set.
\end{lemma}

\begin{proof} We run a probabilistic argument (one could of course also use a double counting argument instead). Let $n,v_0,\ldots,v_n,s,q,r,N,A$ be as in the lemma. Let $V$ be a random $n$-dimensional geometric subspace of $[3]^N$, created in the following fashion:
\begin{itemize}
\item Pick $n$ wildcards $x_1,\ldots,x_n$ to run independently from $1$ to $3$. We also introduce dual wildcards $\overline{x_1},\ldots,\overline{x_n}$; each $\overline{x_j}$ will take the value $4-x_j$.
\item We randomly subdivide the $N$ coordinates into $n$ groups of $q$ coordinates, plus a remaining group of $N-nq$ ``fixed'' coordinates.
\item For each coordinate in the $j^{th}$ group of $q$ coordinates for $1 \leq j \leq n$, we randomly assign either a $x_j$ or $\overline{x_j}$.
\item For each coordinate in the $N-nq$ fixed coordinates, we randomly assign a digit $1,2,3$, but condition on the event that exactly $r$ of the digits are equal to $2$ (i.e. we use a random element of $S_{N-nq-r,N-nq}$).
\item Let $V$ be the subspace created by allowing $x_1,\ldots,x_n$ to run independently from $1$ to $3$, and $\overline{x_j}$ to take the value $4-x_j$.
\end{itemize}

For instance, if $n=2, q=2, r=1, N=6$, then a typical subspace $V$ generated in this fashion is
$$ 2x_1\overline{x_2}3x_2x_1 = \{ 213311, 212321, 211331, 223312, 222322, 221332, 233313, 232323, 231333\}.$$
Observe from that the following two ways to generate a random element of $[3]^N$ are equivalent:

\begin{itemize}
\item Pick $V$ randomly as above, and then assign $(x_1,\ldots,x_n)$ randomly from $S_{n-i,n}$. Assign $4-x_j$ to $\overline{x_j}$ for all $1 \leq j \leq n$.
\item Pick a random string in $S_{N-qi-r,N}$.
\end{itemize}

Indeed, both random variables are invariant under the symmetries of the cube, and both random variables always pick out strings in $S_{N-qi-r,N}$, and the claim follows. As a consequence, we see that the expectation of $\alpha_i(V)$ (as $V$ ranges over the recipe described above) is equal to $\alpha_{qi+r}(A)$. On the other hand, from \eqref{alphav} we have
$$ \sum_{i=0}^n v_i \alpha_i(V) \leq s$$
for all such $V$; taking expectations over $V$, we obtain the claim.
\end{proof}

In view of Lemma \ref{prop}, it is of interest to locate linear inequalities relating the densities $\alpha_i(A)$, or (equivalently) the statistics $a_i(A)$. For this, it is convenient to introduce the following notation.

\begin{definition} Let $n \geq 1$ be an integer.
\begin{itemize}
\item A vector $(a_0,\ldots,a_n)$ of non-negative integers is \emph{feasible} if it is the statistics of some Moser set $A$.
\item A feasible vector $(a_0,\ldots,a_n)$ is \emph{Pareto-optimal} if there is no other feasible vector $(b_0,\ldots,b_n) \neq (a_0,\ldots,a_n)$ such that $b_i \geq a_i$ for all $0 \leq i \leq n$.
\item A Pareto-optimal vector $(a_0,\ldots,a_n)$ is \emph{extremal} if it is not a non-trivial convex linear combination of other Pareto-optimal vectors.
\end{itemize}
\end{definition}

To establish a linear inequality of the form \eqref{alphav} with the $v_i$ non-negative, it suffices to test the inequality against densities associated to extremal vectors of statistics. (There is no point considering linear inequalities with negative coefficients $v_i$, since one always has the freedom to reduce a density $\alpha_i(A)$ of a Moser set $A$ to zero, simply by removing all elements of $A$ with exactly $i$ $2$'s.)

We will classify exactly the Pareto-optimal and extremal vectors for $n \leq 3$, which by Lemma \ref{prop} will lead to useful linear inequalities for $n \geq 4$. Using a computer, we have also located a partial list of Pareto-optimal and extremal vectors for $n=4$, which are also useful for the $n=5$ and $n=6$ theory.

\subsection{Up to three dimensions}

We now establish Theorem \ref{moser} for $n \leq 3$, and establish some auxiliary inequalities which will be of use in higher dimensions.

The case $n=0$ is trivial. When $n=1$, it is clear that $c'_{1,3} = 2$, and furthermore that the Pareto-optimal statistics are $(2,0)$ and $(1,1)$, which are both extremal. This leads to the linear inequality
$$ 2\alpha(A) + \beta(A) \leq 2$$
for all Moser sets $A \subset [3]^1$, which by Lemma \ref{prop} implies that
\begin{equation}\label{alpha-1}
2\alpha_r(A) + \alpha_{r+q}(A) \leq 2
\end{equation}
whenever $r \geq 0$, $q \geq 1$, $n \geq q+r$, and $A \subset [3]^n$ is a Moser set.

For $n=2$, we see by partitioning $[3]^2$ into three slices that $c'_{2,3} \leq 3 c'_{1,3} = 6$, and so (by the lower bounds in the previous section) $c'_{2,3} = 6$. Writing $(a,b,c) = (a(A),b(A),c(A)) = (4\alpha(A), 4\beta(A), \gamma(A))$, the inequalities \eqref{alpha-1} become
\begin{equation}\label{abc}
a + 2c \leq 4; b+2c \leq 4; 2a+b <= 8.
\end{equation}

\begin{lemma} When $n=2$, the Pareto-optimal statistics are $(4,0,0), (3,2,0), (2,4,0), (2,2,1)$. In particular, the extremal statistics are $(4,0,0), (2,4,0), (2,2,1)$.
\end{lemma}

\begin{proof} One easily checks that all the statistics listed above are feasible.
Consider the statistics $(a,b,c)$ of a Moser set $A \subset [3]^2$. $c$ is either equal to $0$ or $1$. If $c=1$, then \eqref{abc} implies that $a,b \leq 2$, so the only Pareto-optimal statistic here is $(2,2,1)$. When instead $c=0$, the inequalities \eqref{abc} can easily imply the Pareto-optimality of $(4,0,0), (3,2,0), (2,4,0)$.
\end{proof}

From this lemma we see that we obtain a new inequality $2a+b+2c \leq 8$. Converting this back to densities and using Lemma \ref{prop}, we conclude that
\begin{equation}\label{alpha-2}
4\alpha_r(A) + 2\alpha_{r+q}(A) + \alpha_{r+2q} \leq 4
\end{equation}
whenever $r \geq 0$, $q \geq 1$, $n \geq q+2r$, and $A \subset [3]^n$ is a Moser set.

One can also check by computer that there are exactly $230$ line-free subsets of $[3]^2$.

Now we look at three dimensions. Writing $(a,b,c,d)$ for the statistics of a Moser set $A \subset [3]^n$ (which thus range between $(0,0,0,0)$ and $(8,12,6,1)$), the inequalities \eqref{alpha-1} imply in particular that
\begin{equation}\label{abc-3d}
a+4d \leq 8; b+6d \leq 12; c+3d \leq 6; 3a+2c \leq 24; b+c \leq 12
\end{equation}
while \eqref{alpha-2} implies that
\begin{equation}\label{abcd-3d}
3a+b+c \leq 24; b+c+3d \leq 12.
\end{equation}
Summing the inequalities $b+c \leq 12, 3a+b+c \leq 24, b+c+3d \leq 12$ yields
$$ 3(a+b+c+d) \leq 48$$
and hence $|A| = a+b+c+d \leq 16$; comparing this with the lower bounds of the preceding section we obtain $c'_{3,3} = 16$ as required. (This argument is essentially identical to the one in \cite{chvatal2}).

We have the following useful computation:

\begin{lemma}[3D Pareto-optimals]\label{paretop} When $n=3$, the Pareto-optimal statistics are $$(3,6,3,1),(4,4,3,1),(4,6,2,1),(2,6,6,0),(3,6,5,0),(4,4,5,0),(3,7,4,0),(4,6,4,0),$$
$$ (3,9,3,0),(4,7,3,0),(5,4,3,0),(4,9,2,0),(5,6,2,0),(6,3,2,0),(3,10,1,0),(5,7,1,0),$$
$$ (6,4,1,0),(4,12,0,0),(5,9,0,0),(6,6,0,0),(7,3,0,0),(8,0,0,0).$$
In particular, the extremal statistics are
$$(3,6,3,1),(4,4,3,1),(4,6,2,1),(2,6,6,0),(4,4,5,0),(4,6,4,0),(4,12,0,0),(8,0,0,0).$$
\end{lemma}

\begin{proof} This can be established by a brute-force search over the $2^{27} \approx 1.3 \times 10^8$ different subsets of $[3]^3$. Actually, one can perform a much faster search than this. Firstly, as noted earlier, there are only $230$ line-free subsets of $[3]^2$, so one could search over $230^3 \approx 1.2 \times 10^7$ configurations instead. Secondly, by symmetry we may assume (after enumerating the $230$ sets in a suitable fashion) that the first slice $A \cap 1**$ has an index less than or equal to the third $A \cap 3**$, leading to $\binom{231}{2} \times 230 \approx 6 \times 10^6$ configurations instead. Finally, using the first and third slice one can quickly determine which elements of the second slice $2**$ are prohibited from $A$. There are $2^9 = 512$ possible choices for the prohibited set in $2**$. By crosschecking these against the list of $230$ line-free sets one can compute the Pareto-optimal statistics for the second slices inside the prohibited set (the lists of such statistics turns out to length at most $23$). Storing these statistics in a lookup table, and then running over all choices of the first and third slice (using symmetry), one now has to perform $O( 512 \times 230 ) + O( \binom{231}{2} \times 23) \approx O( 10^6 )$ computations, which is quite a feasible computation.

One could in principle reduce the computations even further, by a factor of up to $8$, by using the symmetry group $D_4$ of the square $[3]^2$ to reduce the number of cases one needs to consider, but we did not implement this.

A computer-free proof of this lemma can be found at
\centerline{{\tt http://michaelnielsen.org/polymath1/index.php?title=Human\_proof\_of\_the\_3D\_Pareto-optimal\_Moser\_statistics}.}
\end{proof}

\begin{remark} A similar computation revealed that the total number of line-free subsets of $[3]^3$ was $3813884$. With respect to the $2^3 \times 3!=48$-element group of geometric symmetries of $[3]^3$, these sets partitioned into $83158$ equivalence classes:
$$
3813884 = 76066 \times 48+6527 \times 24+51 \times 16+338 \times 12 +109 \times 8+41 \times 6+13 \times 4 +5 \times 3+3 \times 2+5 \times 1.
$$
\end{remark}

Lemma \ref{paretop} yields the following new inequalities:
\begin{align*}
2a+b+2c+4d &\leq 22 \\
3a+2b+3c+6d &\leq 36 \\
7a+2b+4c+8d &\leq 56 \\
6a+2b+3c+6d &\leq 48 \\
a+2c+4d &\leq 14 \\
5a+4c+8d &\leq 40.
\end{align*}

Applying Lemma \ref{prop}, we obtain new inequalities:
\begin{align}
8\alpha_r(A)+ 6\alpha_{r+q}(A) + 6\alpha_{r+2q}(A) + 2\alpha_{r+3q}(A) &\leq 11 \label{eleven}\\
4\alpha_r(A)+4\alpha_{r+q}(A)+3\alpha_{r+2q}(A)+\alpha_{r+3q}(A) &\leq 6\label{six}\\
7\alpha_r(A)+3\alpha_{r+q}(A)+3\alpha_{r+2q}(A)+\alpha_{r+3q}(A) &\leq 7\nonumber\\
8\alpha_r(A)+3\alpha_{r+q}(A)+3\alpha_{r+2q}(A)+\alpha_{r+3q}(A) &\leq 8\label{eight}\\
4\alpha_{r+q}(A)+2\alpha_{r+2q}(A)+\alpha_{r+3q}(A) &\leq 4\nonumber\\
4\alpha_r(A)+6\alpha_{r+2q}(A)+2\alpha_{r+3q}(A) &\leq 7\nonumber\\
5\alpha_r(A)+3\alpha_{r+2q}(A)+\alpha_{r+3q}(A) &\leq 5\nonumber
\end{align}
whenever $r \geq 0$, $q \geq 1$, $n \geq r+3q$, and Moser sets $A \subset [3]^n$.

We also note some further corollaries of Lemma \ref{paretop}:

\begin{corollary}[Statistics of large 3D Moser sets]\label{paretop2} Let $(a,b,c,d)$ be the statistics of a Moser set $A$ in $[3]^3$. Then $|A| = a+b+c+d \leq 16$. Furthermore:
\begin{itemize}
\item If $|A|=16$, then $(a,b,c,d) = (4,12,0,0)$.
\item If $|A|=15$, then $(a,b,c,d) = (4,11,0,0)$ or $(3,12,0,0)$.
\item If $|A| \geq 14$, then $b \geq 6$ and $d=0$.
\item If $|A| = 13$ and $d=1$, then $(a,b,c,d) = (4,6,2,1)$ or $(3,6,3,1)$.
\end{itemize}
\end{corollary}

\subsection{Four dimensions}

Now we establish the bound $c'_{4,3}=43$. Let $A$ be a Moser set in $[3]^4$, with attendant statistics $(a,b,c,d,e)$, which range between $(0,0,0,0,0)$ and $(16,32,24,8,1)$. In view of the lower bounds, our task here is to establish the upper bound $a+b+c+d+e \leq 43$.

The linear inequalities already established just barely fail to achieve this bound, but we can obtain the upper bound $a+b+c+d+e \leq 44$ as follows.
First suppose that $e=1$; then from the inequalities \eqref{alpha-1} (or by considering lines passing through $2222$) we see that $a \leq 8, b \leq 16, c \leq 12, d \leq 4$ and hence $a+b+c+d+e \leq 41$, so we may assume that $e=0$.

From Lemma \ref{dci}, we see that $a+b+c+d+e$ is now equal to the sum of $a(V)/4+b(V)/3+c(V)/2+d(V)$, where $V$ ranges over all side slices of $[3]^4$. But from Lemma \ref{paretop} we see that $a(V)/4+b(V)/3+c(V)/2+d(V)$ is at most $\frac{11}{4}$, with equality occuring only when $(a(V),b(V),c(V),d(V))=(2,6,6,0)$. This gives the upper bound $a+b+c+d+e \leq 44$.

The above argument shows that $a+b+c+d+e=44$ can only occur if $e=0$ and if $(a(V),b(V),c(V),d(V))=(2,6,6,0)$ for all side slices $V$. Applying Lemma \ref{paretop} again this implies $(a,b,c,d,e)=(4,16,24,0,0)$. But then $A$ contains all of the sphere $S_{2,4}$, which implies that the four-element set $A \cap S_{4,4}$ cannot contain a pair of strings which differ in exactly two positions (as their midpoint would then lie in $S_{2,4}$, contradicting the hypothesis that $A$ is a Moser set).

Recall that we may partition $S_{4,4} = S_{4,4}^e \cup S_{4,4}^o$, where
$$S_{4,4}^e := \{ 1111, 1133, 1313, 3113, 1331, 3131, 3311, 3333\}$$
is the strings in $S_{4,4}$ with an even number of $1$'s, and
$$S_{4,4}^o := \{ 1113, 1131, 1311, 3111, 1333, 3133, 3313, 3331\}$$
are the strings in $S_{4,4}$ with an odd number. Observe that any two distinct elements in $S_{4,4}^e$ differ in exactly two positions unless they are antipodal. Thus $A \cap S_{4,4}^e$ has size at most two, with equality only when $A \cap S_{4,4}^e$ consists of an antipodal pair. Similarly for $A \cap S_{4,4}^o$. Thus $A$ must consist of two antipodal pairs, one from $S_{4,4}^e$ and one from $S_{4,4}^o$.

By the symmetries of the cube we may assume without loss of generality that these pairs are $\{ 1111, 3333\}$ and $\{1113,3331\}$ respectively. But as $A$ is a Moser set, $A$ must now exclude the strings $1112$ and $3332$. These two strings form two corners of the eight-element set
$$ ***2 \cap S_{3,4} = \{ 1112, 1132, 1312, 3112, 1332, 3132, 3312, 3332 \}.$$
Any pair of points in this set which are ``adjacent'' in the sense that they differ by exactly one entry cannot both lie in $A$, as their midpoint would then lie in $S_{3,4}$, and so $A$ can contain at most four elements from this set, with equality only if $A$ contains all the points in $***2 \cap S_{3,4}$ of the same parity (either all the elements with an even number of $3$s, or all the elements with an odd number of $3$s). But because the two corners removed from this set have the opposite parity (one has an even number of $1$s and one has an odd number), we see in fact that $A$ can contain at most $3$ points from this set. Meanwhile, the same arguments give that $A$ contains at most four points from $**2* \cap S_{3,4}$, $*2** \cap S_{3,4}$, and $2*** \cap S_{3,4}$. Summing we see that $b = |A \cap S_{3,4}| \leq 3+4+4+4=15$, a contradiction. Thus we have $c'_{4,3}=43$ as claimed.

We have the following four-dimensional version of Lemma \ref{paretop}:

\begin{lemma}[4D Pareto-optimals]\label{paretop-4} When $n=4$, the Pareto-optimal statistics listed on Table \ref{table4}.
\end{lemma}

\begin{figure}[tb]
\centered{\tiny
$(3, 16, 24, 0, 0)$,
$(4, 14, 19, 2, 0)$,
$(4, 15, 24, 0, 0)$,
$(4, 16, 8, 4, 1)$,
$(4, 16, 14, 4, 0)$,
$(4, 16, 23, 0, 0)$,
$(4, 17, 21, 0, 0)$,
$(4, 18, 19, 0, 0)$,
$(5, 12, 12, 4, 1)$,
$(5, 12, 13, 6, 0)$,
$(5, 12, 15, 5, 0)$,
$(5, 12, 19, 2, 0)$,
$(5, 13, 10, 4, 1)$,
$(5, 13, 14, 5, 0)$,
$(5, 13, 21, 1, 0)$,
$(5, 15, 9, 4, 1)$,
$(5, 15, 12, 3, 1)$,
$(5, 15, 13, 5, 0)$,
$(5, 15, 18, 3, 0)$,
$(5, 15, 20, 1, 0)$,
$(5, 15, 22, 0, 0)$,
$(5, 16, 7, 4, 1)$,
$(5, 16, 10, 3, 1)$,
$(5, 16, 11, 5, 0)$,
$(5, 16, 12, 2, 1)$,
$(5, 16, 16, 3, 0)$,
$(5, 16, 19, 1, 0)$,
$(5, 16, 21, 0, 0)$,
$(5, 17, 12, 4, 0)$,
$(5, 17, 14, 3, 0)$,
$(5, 17, 16, 2, 0)$,
$(5, 17, 18, 1, 0)$,
$(5, 17, 20, 0, 0)$,
$(5, 18, 13, 3, 0)$,
$(5, 18, 14, 2, 0)$,
$(5, 20, 8, 4, 0)$,
$(5, 20, 10, 3, 0)$,
$(5, 20, 13, 2, 0)$,
$(5, 20, 14, 1, 0)$,
$(5, 20, 18, 0, 0)$,
$(5, 21, 10, 2, 0)$,
$(5, 21, 15, 0, 0)$,
$(5, 22, 13, 0, 0)$,
$(6, 8, 12, 8, 0)$,
$(6, 10, 11, 4, 1)$,
$(6, 11, 12, 7, 0)$,
$(6, 12, 10, 7, 0)$,
$(6, 12, 13, 5, 0)$,
$(6, 12, 18, 4, 0)$,
$(6, 13, 16, 4, 0)$,
$(6, 14, 9, 4, 1)$,
$(6, 14, 9, 7, 0)$,
$(6, 14, 12, 6, 0)$,
$(6, 14, 16, 3, 0)$,
$(6, 14, 19, 1, 0)$,
$(6, 14, 21, 0, 0)$,
$(6, 15, 7, 4, 1)$,
$(6, 15, 10, 3, 1)$,
$(6, 15, 10, 6, 0)$,
$(6, 15, 11, 2, 1)$,
$(6, 15, 12, 5, 0)$,
$(6, 15, 15, 4, 0)$,
$(6, 15, 20, 0, 0)$,
$(6, 16, 7, 3, 1)$,
$(6, 16, 8, 6, 0)$,
$(6, 16, 9, 2, 1)$,
$(6, 16, 10, 5, 0)$,
$(6, 16, 12, 1, 1)$,
$(6, 16, 13, 4, 0)$,
$(6, 16, 14, 3, 0)$,
$(6, 16, 18, 2, 0)$,
$(6, 16, 19, 0, 0)$,
$(6, 17, 9, 5, 0)$,
$(6, 17, 10, 4, 0)$,
$(6, 17, 13, 3, 0)$,
$(6, 17, 15, 2, 0)$,
$(6, 17, 17, 1, 0)$,
$(6, 17, 18, 0, 0)$,
$(6, 18, 13, 2, 0)$,
$(6, 18, 16, 1, 0)$,
$(6, 18, 17, 0, 0)$,
$(6, 19, 9, 4, 0)$,
$(6, 19, 12, 3, 0)$,
$(6, 19, 15, 1, 0)$,
$(6, 20, 7, 4, 0)$,
$(6, 20, 9, 3, 0)$,
$(6, 20, 12, 2, 0)$,
$(6, 20, 13, 1, 0)$,
$(6, 20, 15, 0, 0)$,
$(6, 21, 8, 3, 0)$,
$(6, 21, 9, 2, 0)$,
$(6, 21, 12, 1, 0)$,
$(6, 21, 14, 0, 0)$,
$(6, 22, 7, 3, 0)$,
$(6, 22, 8, 2, 0)$,
$(6, 22, 10, 1, 0)$,
$(6, 23, 9, 1, 0)$,
$(6, 24, 7, 2, 0)$,
$(6, 24, 8, 1, 0)$,
$(6, 24, 12, 0, 0)$,
$(6, 25, 9, 0, 0)$,
$(6, 26, 7, 0, 0)$,
$(7, 8, 6, 8, 0)$,
$(7, 11, 9, 4, 1)$,
$(7, 11, 12, 6, 0)$,
$(7, 12, 8, 4, 1)$,
$(7, 12, 8, 6, 0)$,
$(7, 12, 12, 3, 1)$,
$(7, 12, 12, 5, 0)$,
$(7, 12, 13, 4, 0)$,
$(7, 12, 15, 3, 0)$,
$(7, 12, 17, 2, 0)$,
$(7, 13, 7, 4, 1)$,
$(7, 13, 10, 3, 1)$,
$(7, 13, 11, 5, 0)$,
$(7, 13, 12, 2, 1)$,
$(7, 13, 12, 4, 0)$,
$(7, 13, 14, 3, 0)$,
$(7, 13, 16, 2, 0)$,
$(7, 14, 6, 4, 1)$,
$(7, 14, 6, 7, 0)$,
$(7, 14, 9, 5, 0)$,
$(7, 14, 10, 2, 1)$,
$(7, 14, 12, 1, 1)$,
$(7, 14, 17, 1, 0)$,
$(7, 14, 19, 0, 0)$,
$(7, 15, 7, 5, 0)$,
$(7, 15, 8, 3, 1)$,
$(7, 15, 9, 2, 1)$,
$(7, 15, 11, 1, 1)$,
$(7, 15, 11, 4, 0)$,
$(7, 15, 13, 3, 0)$,
$(7, 15, 16, 1, 0)$,
$(7, 16, 6, 3, 1)$,
$(7, 16, 6, 6, 0)$,
$(7, 16, 8, 2, 1)$,
$(7, 16, 10, 1, 1)$,
$(7, 16, 10, 4, 0)$,
$(7, 16, 12, 0, 1)$,
$(7, 16, 12, 3, 0)$,
$(7, 16, 15, 2, 0)$,
$(7, 16, 17, 0, 0)$,
$(7, 17, 6, 5, 0)$,
$(7, 17, 7, 4, 0)$,
$(7, 17, 11, 3, 0)$,
$(7, 17, 13, 2, 0)$,
$(7, 17, 14, 1, 0)$,
$(7, 17, 16, 0, 0)$,
$(7, 18, 10, 3, 0)$,
$(7, 18, 13, 1, 0)$,
$(7, 18, 15, 0, 0)$,
$(7, 19, 9, 3, 0)$,
$(7, 20, 6, 4, 0)$,
$(7, 20, 11, 2, 0)$,
$(7, 20, 12, 1, 0)$,
$(7, 20, 14, 0, 0)$,
$(7, 21, 8, 2, 0)$,
$(7, 21, 10, 1, 0)$,
$(7, 21, 12, 0, 0)$,
$(7, 22, 9, 1, 0)$,
$(7, 22, 11, 0, 0)$,
$(7, 23, 6, 3, 0)$,
$(7, 23, 7, 1, 0)$,
$(7, 23, 10, 0, 0)$,
$(7, 24, 6, 2, 0)$,
$(7, 24, 9, 0, 0)$,
$(7, 25, 6, 1, 0)$,
$(7, 25, 8, 0, 0)$,
$(7, 26, 3, 1, 0)$,
$(7, 28, 6, 0, 0)$,
$(7, 29, 3, 0, 0)$,
$(7, 30, 1, 0, 0)$,
$(8, 8, 0, 8, 0)$,
$(8, 8, 9, 7, 0)$,
$(8, 8, 12, 6, 0)$,
$(8, 9, 9, 4, 1)$,
$(8, 9, 10, 6, 0)$,
$(8, 9, 12, 3, 1)$,
$(8, 9, 12, 5, 0)$,
$(8, 9, 13, 4, 0)$,
$(8, 9, 15, 3, 0)$,
$(8, 10, 7, 4, 1)$,
$(8, 10, 10, 3, 1)$,
$(8, 10, 10, 5, 0)$,
$(8, 10, 12, 2, 1)$,
$(8, 10, 12, 4, 0)$,
$(8, 10, 13, 3, 0)$,
$(8, 10, 15, 2, 0)$,
$(8, 11, 6, 4, 1)$,
$(8, 11, 9, 6, 0)$,
$(8, 11, 10, 2, 1)$,
$(8, 11, 11, 4, 0)$,
$(8, 12, 7, 6, 0)$,
$(8, 12, 9, 3, 1)$,
$(8, 12, 9, 5, 0)$,
$(8, 12, 10, 4, 0)$,
$(8, 12, 12, 1, 1)$,
$(8, 12, 14, 2, 0)$,
$(8, 12, 16, 1, 0)$,
$(8, 12, 18, 0, 0)$,
$(8, 13, 7, 3, 1)$,
$(8, 13, 7, 5, 0)$,
$(8, 13, 9, 2, 1)$,
$(8, 13, 12, 0, 1)$,
$(8, 13, 12, 3, 0)$,
$(8, 14, 0, 7, 0)$,
$(8, 14, 6, 6, 0)$,
$(8, 14, 7, 2, 1)$,
$(8, 14, 8, 1, 1)$,
$(8, 14, 9, 4, 0)$,
$(8, 14, 11, 0, 1)$,
$(8, 14, 11, 3, 0)$,
$(8, 14, 13, 2, 0)$,
$(8, 14, 15, 1, 0)$,
$(8, 14, 17, 0, 0)$,
$(8, 15, 6, 3, 1)$,
$(8, 15, 6, 5, 0)$,
$(8, 15, 7, 1, 1)$,
$(8, 16, 0, 6, 0)$,
$(8, 16, 4, 3, 1)$,
$(8, 16, 4, 5, 0)$,
$(8, 16, 6, 2, 1)$,
$(8, 16, 8, 4, 0)$,
$(8, 16, 9, 0, 1)$,
$(8, 16, 10, 3, 0)$,
$(8, 16, 12, 2, 0)$,
$(8, 16, 14, 1, 0)$,
$(8, 16, 16, 0, 0)$,
$(8, 17, 0, 5, 0)$,
$(8, 17, 3, 4, 0)$,
$(8, 17, 8, 3, 0)$,
$(8, 17, 10, 2, 0)$,
$(8, 17, 12, 1, 0)$,
$(8, 17, 14, 0, 0)$,
$(8, 18, 9, 2, 0)$,
$(8, 18, 11, 1, 0)$,
$(8, 18, 12, 0, 0)$,
$(8, 19, 6, 3, 0)$,
$(8, 19, 8, 2, 0)$,
$(8, 20, 0, 4, 0)$,
$(8, 20, 4, 3, 0)$,
$(8, 20, 7, 2, 0)$,
$(8, 20, 9, 1, 0)$,
$(8, 20, 11, 0, 0)$,
$(8, 21, 4, 2, 0)$,
$(8, 21, 7, 1, 0)$,
$(8, 22, 3, 2, 0)$,
$(8, 22, 6, 1, 0)$,
$(8, 22, 9, 0, 0)$,
$(8, 23, 0, 3, 0)$,
$(8, 23, 4, 1, 0)$,
$(8, 24, 0, 2, 0)$,
$(8, 24, 3, 1, 0)$,
$(8, 24, 8, 0, 0)$,
$(8, 25, 1, 1, 0)$,
$(8, 25, 6, 0, 0)$,
$(8, 26, 0, 1, 0)$,
$(8, 26, 4, 0, 0)$,
$(8, 28, 3, 0, 0)$,
$(8, 32, 0, 0, 0)$,
$(9, 8, 10, 4, 0)$,
$(9, 9, 9, 4, 0)$,
$(9, 9, 12, 3, 0)$,
$(9, 10, 8, 4, 0)$,
$(9, 10, 10, 3, 0)$,
$(9, 10, 12, 2, 0)$,
$(9, 10, 13, 1, 0)$,
$(9, 10, 15, 0, 0)$,
$(9, 11, 11, 2, 0)$,
$(9, 12, 7, 4, 0)$,
$(9, 12, 9, 3, 0)$,
$(9, 12, 12, 1, 0)$,
$(9, 12, 14, 0, 0)$,
$(9, 13, 7, 3, 0)$,
$(9, 13, 10, 2, 0)$,
$(9, 14, 9, 2, 0)$,
$(9, 14, 11, 1, 0)$,
$(9, 14, 13, 0, 0)$,
$(9, 15, 6, 3, 0)$,
$(9, 16, 0, 4, 0)$,
$(9, 16, 4, 3, 0)$,
$(9, 16, 8, 2, 0)$,
$(9, 16, 10, 1, 0)$,
$(9, 16, 12, 0, 0)$,
$(9, 17, 3, 3, 0)$,
$(9, 17, 6, 2, 0)$,
$(9, 17, 8, 1, 0)$,
$(9, 17, 10, 0, 0)$,
$(9, 18, 2, 3, 0)$,
$(9, 18, 4, 2, 0)$,
$(9, 18, 7, 1, 0)$,
$(9, 18, 9, 0, 0)$,
$(9, 19, 0, 3, 0)$,
$(9, 19, 3, 2, 0)$,
$(9, 19, 6, 1, 0)$,
$(9, 20, 1, 2, 0)$,
$(9, 20, 5, 1, 0)$,
$(9, 20, 8, 0, 0)$,
$(9, 21, 4, 1, 0)$,
$(9, 21, 6, 0, 0)$,
$(9, 22, 1, 1, 0)$,
$(9, 22, 5, 0, 0)$,
$(9, 24, 4, 0, 0)$,
$(9, 25, 2, 0, 0)$,
$(9, 28, 0, 0, 0)$,
$(10, 8, 6, 4, 0)$,
$(10, 8, 8, 3, 0)$,
$(10, 9, 7, 3, 0)$,
$(10, 9, 10, 2, 0)$,
$(10, 9, 11, 1, 0)$,
$(10, 9, 13, 0, 0)$,
$(10, 10, 5, 4, 0)$,
$(10, 10, 9, 2, 0)$,
$(10, 10, 12, 0, 0)$,
$(10, 11, 6, 3, 0)$,
$(10, 12, 4, 4, 0)$,
$(10, 12, 5, 3, 0)$,
$(10, 12, 7, 2, 0)$,
$(10, 12, 10, 1, 0)$,
$(10, 12, 11, 0, 0)$,
$(10, 13, 6, 2, 0)$,
$(10, 13, 8, 1, 0)$,
$(10, 13, 10, 0, 0)$,
$(10, 14, 3, 3, 0)$,
$(10, 14, 5, 2, 0)$,
$(10, 14, 9, 0, 0)$,
$(10, 15, 2, 3, 0)$,
$(10, 15, 7, 1, 0)$,
$(10, 16, 4, 2, 0)$,
$(10, 16, 6, 1, 0)$,
$(10, 16, 8, 0, 0)$,
$(10, 17, 4, 1, 0)$,
$(10, 17, 6, 0, 0)$,
$(10, 18, 2, 1, 0)$,
$(10, 18, 5, 0, 0)$,
$(10, 20, 4, 0, 0)$,
$(10, 21, 2, 0, 0)$,
$(10, 22, 1, 0, 0)$,
$(10, 24, 0, 0, 0)$,
$(11, 4, 6, 4, 0)$,
$(11, 6, 5, 4, 0)$,
$(11, 7, 6, 3, 0)$,
$(11, 8, 4, 4, 0)$,
$(11, 8, 5, 3, 0)$,
$(11, 9, 6, 2, 0)$,
$(11, 9, 8, 1, 0)$,
$(11, 9, 10, 0, 0)$,
$(11, 10, 3, 3, 0)$,
$(11, 10, 5, 2, 0)$,
$(11, 10, 9, 0, 0)$,
$(11, 11, 2, 3, 0)$,
$(11, 11, 7, 1, 0)$,
$(11, 12, 4, 2, 0)$,
$(11, 12, 6, 1, 0)$,
$(11, 12, 8, 0, 0)$,
$(11, 13, 4, 1, 0)$,
$(11, 13, 6, 0, 0)$,
$(11, 14, 2, 1, 0)$,
$(11, 14, 5, 0, 0)$,
$(11, 16, 4, 0, 0)$,
$(11, 17, 2, 0, 0)$,
$(11, 18, 1, 0, 0)$,
$(11, 20, 0, 0, 0)$,
$(12, 4, 3, 3, 0)$,
$(12, 6, 2, 3, 0)$,
$(12, 6, 5, 2, 0)$,
$(12, 6, 7, 1, 0)$,
$(12, 6, 9, 0, 0)$,
$(12, 8, 4, 2, 0)$,
$(12, 8, 6, 1, 0)$,
$(12, 8, 8, 0, 0)$,
$(12, 9, 4, 1, 0)$,
$(12, 9, 6, 0, 0)$,
$(12, 10, 2, 1, 0)$,
$(12, 10, 5, 0, 0)$,
$(12, 12, 4, 0, 0)$,
$(12, 13, 2, 0, 0)$,
$(12, 14, 1, 0, 0)$,
$(12, 16, 0, 0, 0)$,
$(13, 6, 5, 0, 0)$,
$(13, 8, 4, 0, 0)$,
$(13, 9, 2, 0, 0)$,
$(13, 10, 1, 0, 0)$,
$(13, 12, 0, 0, 0)$,
$(14, 4, 3, 0, 0)$,
$(14, 5, 2, 0, 0)$,
$(14, 6, 1, 0, 0)$,
$(14, 8, 0, 0, 0)$,
$(15, 4, 0, 0, 0)$,
$(16, 0, 0, 0, 0)$}}
\caption{The Pareto-optimal statistics of Moser sets in $[3]^4$. This table can also be found at {\tt http://spreadsheets.google.com/ccc?key=rwXB\_Rn3Q1Zf5yaeMQL-RDw}}
\label{table4}
\end{figure}

\begin{proof} This was computed by computer search as follows. First, one observed that if $(a,b,c,d,e)$ was Pareto-optimal, then $a\geq 3$. To see this, it suffices to show that for any Moser set $A \subset [3]^4$ with $a(A)=0$, it is possible to add three points from $S_{4,4}$ to $A$ and still have a Moser set. To show this, suppose first that $A$ contains a point from $S_{1,4}$, such as $2221$. Then $A$ must omit either $2211$ or $2231$; without loss of generality we may assume that it omits $2211$. Similarly we may assume it omits $2121$ and $1221$. Then we can add $1131$, $1311$, $3111$ to $A$, as required. Thus we may assume that $A$ contains no points from $S_{1,4}$. Now suppose that $A$ omits a point from $S_{2,4}$, such as $2211$. Then one can add $3333$, $3111$, $1311$ to $A$, as required. Thus we may assume that A contains all of $S_{2,4}$, which forces $A$ to omit $2222$, as well as at least one point from $S_{3,4}$, such as $2111$. But then $3111$, $1111$, $3333$ can be added to the set, a contradiction.

Thus we only need to search through sets $A \subset [3]^4$ for which $|A \cap S_{4,4}| \geq 3$. A straightforward computer search shows that up to the symmetries of the cube, there are $391$ possible choices for $A \cap S_{4,4}$. For each such choice, we looped through all the possible values of the slices $A \cap 1***$ and $A \cap 3***$, i.e. all three-dimensional Moser sets which had the indicated intersection with $S_{3,3}$. (For fixed $A \cap S_{4,4}$, the number of possibilities for $A \cap 1***$ ranges from $1$ to $87123$, and similarly for $A \cap 3***$). For each pair of slices $A \cap 1***$ and $A \cap 3***$, we computed the lines connecting these two sets to see what subset of $2***$ was excluded from $A$; there are $2^{27}$ possible such exclusion sets. We precomputed a lookup table that gave the Pareto-optimal statistics for $A \cap 2***$ for each such choice of exclusion set; using this lookup table for each choice of $A \cap 1***$ and $A \cap 3***$ and collating the results, we obtained the above list. On a linux cluster, the lookup table took 22 minutes to create, and the loop over the $A \cap 1***$ and $A \cap 3***$ slices took two hours, spread out over $391$ machines (one for each choice of $A \cap S_{4,4}$). Further details (including source code) can be found at
\centerline{{\tt http://michaelnielsen.org/polymath1/index.php?title=4D\_Moser\_brute\_force\_search}.}
\end{proof}

As a consequence of this data, we have the following facts about the statistics of large Moser sets:

\begin{proposition}\label{stat} Let $A \subset [3]^4$ be a Moser set with statistics $(a,b,c,d,e)$.
\begin{itemize}
\item[(i)] If $|A| \geq 40$, then $e=0$.
\item[(ii)] If $|A| \geq 43$, then $d=0$.
\item[(iii)] If $|A| \geq 42$, then $d \leq 2$.
\item[(iv)] If $|A| \geq 41$, then $d \leq 3$.
\item[(v)] If $|A| \geq 40$, then $d \leq 6$.
\item[(vi)] If $|A| \geq 43$, then $c \geq 18$.
\item[(vii)] If $|A| \geq 42$, then $c \geq 12$.
\item[(viii)] If $|A| \geq 43$, then $b \geq 15$.
\end{itemize}
\end{proposition}

\begin{remark} This proposition was first established by an integer program, see Appendix \ref{integer-sec}. A computer-free proof can be found at \centerline{{\tt http://terrytao.files.wordpress.com/2009/06/polymath2.pdf}.}
\end{remark}

\subsection{Five dimensions}

Now we establish the bound $c'_{5,3}=124$. In view of the lower bounds, it suffices to show that there does not exist a Moser set $A \subset [3]^5$ with $|A|=125$.

We argue by contradiction. Let $A$ be as above, and let $(a(A),\ldots,f(A))$ be the statistics of $A$.

\begin{lemma}\label{fvan} $f(A)=0$.
\end{lemma}

\begin{proof} If $f(A)$ is non-zero, then $A$ contains $22222$, then each of the $\frac{3^5-1}{2} = 121$ antipodal pairs in $[3]^5$ can have at most one point in $A$, leading to only $122$ points.
\end{proof}

Let us slice $[3]^5$ into three parallel slices, e.g. $1****, 2****, 3****$. The intersection of $A$ with each of these slices has size at most $43$. In particular, this implies that
\begin{equation}\label{boo}
|A \cap 1****| + |A \cap 3****| = 125 - |A \cap 2****| \geq 82.
\end{equation}
Thus at least one of $A \cap 1****$, $A \cap 3****$ has cardinality at least $41$; by Proposition \ref{stat}(iv) we conclude that
\begin{equation}\label{d13}
\min( d(1****), d(3****) ) \leq 3.
\end{equation}
Furthermore, equality can only hold in \eqref{d13} if $A \cap 1****$, $A \cap 3****$ both have cardinality exactly $41$, in which case from Proposition \ref{stat}(iv) again we must have
\begin{equation}\label{d13a}
d(1****)=d(3****)=3.
\end{equation}
Of course, we have a similar result for permutations.

Now we improve the bound $|A \cap 2****| \leq 43$:

\begin{lemma} $|A \cap 2****| \leq 41$.
\end{lemma}

\begin{proof} Suppose first that $|A \cap 2****|=43$. Let $A' \subset [3]^4$ be the subset of $[3]^4$ corresponding to $A \cap 2****$, thus $A'$ is a Moser set of cardinality $43$. By Proposition \ref{stat}(vi), $c(A') \geq 18$. By Lemma \ref{dci}, the sum of the $c(V)$, where $V$ ranges over the eight side slices of $[3]^4$, is therefore at least $36$. By the pigeonhole principle, we may thus find two opposing side slices, say $1***$ and $3***$, with $c(1***)+c(3****) \geq 9$. Since $c(1***), c(3***)$ cannot exceed $6$, we thus have $c(1***), c(3***) \geq 3$, with at least one of $c(1***), c(3***)$ being at least $5$. Passing back to $A$, this implies that $d(*1***), d(*3***) \geq 3$, with at least one of $d(*1***), d(*3***)$ being at least $5$. But this contradicts \eqref{d13} together with the refinement \eqref{d13a}.

We have just shown that $|A \cap 2****| \leq 42$; we can thus improve \eqref{boo} to
$$ |A \cap 1****| + |A \cap 3****| \geq 83.$$
Combining this with Proposition \ref{stat}(ii)-(v) we see that
\begin{equation}\label{d13-6}
d(1****)+d(3****) \leq 6
\end{equation}
with equality only if $|A \cap 2****|=42$, and similarly for permutations.

Now let $A'$ be defined as before. Then we have
$$ c(1***) + c(3***) \leq 6$$
and similarly for permutations. Applying Lemma \ref{dci}, this implies that $c(2****) = c(A') \leq 12$.

Now suppose for contradiction that $|A'|=|A \cap 2****|=42$. Then by Proposition \ref{stat}(vii) we have
\begin{equation}\label{coo-1}
c(2****) = 12;
\end{equation}
applying Lemma \ref{dci} again, this forces $c(1***)+c(3***)=6$ and similarly for permutations, which then implies that
\begin{equation}\label{doo}
d(*1***)+d(*3***) = d(**1**)+d(**3**) = d(***1*)+d(***3*) = d(****1)+d(****3) = 6
\end{equation}
and hence
$$ |A \cap *2***| = |A \cap **2**| = |A \cap ***2*| = |A \cap ****2| = 42$$
and thus
\begin{equation}\label{coo-2}
c(*2***) = c(**2**) = c(***2*) = c(****2) = 12.
\end{equation}
Combining \eqref{coo-1}, \eqref{doo}, \eqref{coo-2} we conclude that
$$ d(1****)+d(3****) = 16,$$
contradicting \eqref{d13-6}.
\end{proof}

With this proposition, the bound \eqref{boo} now improves to
\begin{equation}\label{84}
|A \cap 1****| + |A \cap 3****| \geq 84
\end{equation}
and in particular
\begin{equation}\label{41}
|A \cap 1****|, |A \cap 3****| \geq 41.
\end{equation}
from this and Proposition \ref{stat}(ii)-(iv) we now have
\begin{equation}\label{d13-improv}
d(1****)+d(3****) \leq 4
\end{equation}
and similarly for permutations.

\begin{lemma}\label{evan} $e(A)=0$.
\end{lemma}

\begin{proof} From \eqref{84}, the intersection of $A$ with any side slice has cardinality at least $41$, and thus by Proposition \ref{stat}(i) such a side slice has an $e$-statistic of zero. The claim then follows from Lemma \ref{dci}.
\end{proof}

We need a technical lemma:

\begin{lemma}\label{tech} Let $B \subset S_{5,5}$. Then there exist at least $|B|-4$ pairs of strings in $B$ which differ in exactly two positions.
\end{lemma}

\begin{proof} The first non-vacuous case is $|B|=5$. It suffices to establish this case, as the higher cases then follow by induction (locating a pair of the desired form, then deleting one element of that pair from $B$).

Suppose for contradiction that one can find a $5$-element set $B \subset S_{5,5}$ such that no two strings in $B$ differ in exactly two positions. Recall that we may split $S_{5,5}=S_{5,5}^e \cup S_{5,5}^o$, where $S_{5,5}^e$ are those strings with an even number of $1$'s, and $S_{5,5}^o$ are those strings with an odd number of $1$'s. By the pigeonhole principle and symmetry we may assume $B$ has at least three elements in $S_{5,5}^o$. Without loss of generality, we can take one of them to be $11111$, thus excluding all elements in $S_{5,5}^o$ with exactly two $3$s, leaving only the elements with exactly four $3$s. But any two of them differ in exactly two positions, a contradiction.
\end{proof}

We can now improve the trivial bound $c(A) \leq 80$:

\begin{corollary}[Non-maximal $c$]\label{cmax} $c(A) \leq 79$. If $a(A) \geq 7$, then $c(A) \leq 78$.
\end{corollary}

\begin{proof} If $c(A)=80$, then $A$ contains all of $S_{3,5}$, which then implies that no two elements in $A \cap S_{5,5}$ can differ in exactly two places. It also implies (from \eqref{alpha-1}) that $d(A)$ must vanish, and that $b(A)$ is at most $40$. By Lemma \ref{tech}, we also have that $a(A) = |A \cap S_{5,5}|$ is at most $4$. Thus $|A| \leq 4 + 40 + 80 + 0 + 0 = 124$, a contradiction.

Now suppose that $a(A) \geq 7$. Then by Lemma \ref{tech} there are at least three pairs in $A \cap S_{5,5}$ that differ in exactly two places. Each such pair eliminates one point from $A \cap S_{3,5}$; but each point in $S_{3,5}$ can be eliminated by at most two such pairs, and so we have at least two points eliminated from $A \cap S_{3,5}$, i.e. $c(A) \leq 78$ as required.
\end{proof}

Next, we rewrite the quantity $125=|A|$ in terms of side slices. From Lemmas \ref{fvan}, \ref{evan} we have
$$ a(A) + b(A) + c(A) + d(A) = 125$$
and hence by Lemma \ref{dci}, the quantity
$$ s(V) := a(V) + \frac{5}{4} b(V) + \frac{5}{3} c(V) + \frac{5}{2} d(V) - \frac{125}{2},$$
where $V$ ranges over side slices, has an average value of zero.

\begin{proposition}[Large values of $s(V)$]\label{suv} For all side slices, we have $s(V) \leq 1/2$. Furthermore, we have $s(V) < -1/2$ unless the statistics $(a(V), b(V), c(V), d(V), e(V))$ are of one of the following four cases:
\begin{itemize}
\item (Type 1) $(a(V),b(V),c(V),d(V),e(V)) = (2,16,24,0,0)$ (and $s(V) = -1/2$ and $|A \cap V| = 42$);
\item (Type 2) $(a(V),b(V),c(V),d(V),e(V)) = (4,16,23,0,0)$ (and $s(V) = -1/6$ and $|A \cap V| = 43$);
\item (Type 3) $(a(V),b(V),c(V),d(V),e(V)) = (4,15,24,0,0)$ (and $s(V) = 1/4$ and $|A \cap V| = 43$);
\item (Type 4) $(a(V),b(V),c(V),d(V),e(V)) = (3,16,24,0,0)$ (and $s(V) = 1/2$ and $|A \cap V| = 43$);
\end{itemize}
\end{proposition}

\begin{proof}
Let $V$ be a side slice. From \eqref{41} we have
$$ 41 \leq a(V)+b(V)+c(V)+d(V) = |A \cap V| \leq 43.$$
First suppose that $|A \cap V| = 43$, then from Proposition \ref{stat}(ii), (viii), $d(V)=0$ and $b(V) \geq 15$.
Also, we have the trivial bound $c(V) \leq 24$, together with the inequality
$$ 3b(V) + 2c(V) \leq 96$$
from \eqref{alpha-1}. To exploit these facts, we rewrite $s(V)$ as
$$ s(V) = \frac{1}{2} - \frac{1}{2}( 24 - c(V) ) - \frac{1}{12} (96-3b(V)-2c(V)).$$
Thus $s(V) \leq 1/2$ in this case. If $s(V) \geq -1/2$, then
$$ 6 (24-c(V)) + (96-3b(V)-2c(V)) \leq 12,$$
which together with the inequalities $b(V) \leq 15$, $c(V) \leq 24$, $3b(V)+2c(V) \leq 96$ we conclude that $(b(V),c(V))$ must be one of $(16,24)$, $(15, 24)$, $(16, 23)$, $(15, 23)$. The first three possibilities lead to Types 4,3,2 respectively. The fourth type would lead to $(a(V),b(V),c(V),d(V),e(V)) = (5,15,23,0,0)$, but this contradicts \eqref{eleven}.

Next, suppose $|A \cap V| = 42$, so by Proposition \ref{stat}(iii) we have $d(V) \leq 2$. From \eqref{alpha-1} we have
\begin{equation}\label{2cd}
2c(V) + 3d(V) \leq 48
\end{equation}
while from \eqref{alpha-2} we have
\begin{equation}\label{3cd}
3b(V)+2c(V)+3d(V) \leq 96
\end{equation}
and so we can rewrite $s(V)$ as
\begin{equation}\label{sv2}
s(V) = -\frac{1}{2} - \frac{1}{4}( 48 - 2c(V) - 3d(V) ) - \frac{1}{12} (96-3b(V)-2c(V)-3d(V)) + \frac{1}{2} d(V).
\end{equation}
This already gives $s(V) \leq 1/2$. If $d(V)=0$, then $s(V) \leq -1/2$, with equality only in Type 1. If $d(V)=1$, then the set $A' \subset [3]^4$ corresponding to $A \cap V$ contains a point in $S_{3,4}$, which without loss of generality we can take to be $2221$. Considering the three lines $*221$, $2*21$, $22*1$, we see that at least three points in $S_{2,4}$ must be missing from $A'$, thus $c(V) \leq 21$. This forces $48-2c(V)-3d(V) \geq 3$, and so $s(V) < -3/4$. Finally, if $d(V)=2$, then $A'$ contains two points in $S_{3,4}$. If they are antipodal (e.g. $2221$ and $2223$), the same argument as above shows that at least six points in $S_{2,4}$ are missing from $A'$; if they are not antipodal (e.g. $2221$ and $2212$) then by considering the lines $*221$, $2*21$, $22*1$, $*212$, $2*12$ we see that five points are missing. Thus we have $c(V) \leq 19$, which forces $48-2c(V)-3d(V) \geq 4$. This forces $s(V) \leq -1/2$, with equality only when $c(V)=19$ and $3b(V)+2c(V)+3d(V)=96$, but this forces $b(V)$ to be the non-integer $52/3$, a contradiction, which concludes the treatment of the $|A \cap V|=42$ case.

Finally, suppose $|A \cap V| = 41$. Using \eqref{2cd}, \eqref{3cd} as before we have
\begin{equation}\label{sv3}
s(V) = -\frac{3}{2} - \frac{1}{4}( 48 - 2c(V) - 3d(V) ) - \frac{1}{12} (96-3b(V)-2c(V)-3d(V)) + \frac{1}{2} d(V),
\end{equation}
while from Proposition \ref{stat}(vi) we have $d(V) \leq 3$. This already gives $s(V) \leq 0$, and $s(V) \leq -1$ when $d(V)=1$. In order to have $s(V) \geq -1/2$, we must then have $d(V)=2$ or $d(V)=3$. But then the arguments of the preceding paragraph give $48-2c(V)-3d(V) \geq 4$, and so $s(V) \leq -1$ in this case.
\end{proof}

Since the $s(V)$ average to zero, by the pigeonhole principle we may find two opposing side slices (e.g. $1****$ and $3****$), whose total $s$-value is non-negative. Actually we can do a little better:

\begin{lemma}\label{side-off} There exists two opposing side slices whose total $s$-value is strictly positive.
\end{lemma}

\begin{proof} If this is not the case, then we must have $s(1****)+s(3****)=0$ and similarly for permutations. Using Proposition \ref{suv} we thus see that for every opposing pair of side slices, one is Type 1 and one is Type 4. In particular $c(V)=24$ for all side slices $V$. But then by Lemma \ref{dci} we have $c(A)=80$, contradicting Lemma \ref{cmax}.
\end{proof}

Let $V, V'$ be the side slices in Lemma \ref{side-off}
By Proposition \ref{suv}, the $V, V'$ slices must then be either Type 2, Type 3, or Type 4, and they cannot both be Type 2. Since $a(A) = a(V)+a(V')$, we conclude
\begin{equation}\label{amix}
6 \leq a(A) \leq 8.
\end{equation}
In a similar spirit, we have
$$ c(V) + c(V') \leq 23+24.$$
On the other hand, by considering the $24$ lines connecting $c$-points of $V, V'$ to $c$-points of the centre slice $W$ between $V$ and $V'$, each of which contains at most two points in $A$, we have
$$ c(V) + c(W) + c(V') \leq 24 \times 2.$$
Thus $c(W) \leq 1$; since
$$ d(A) = d(V) + d(V') + c(W)$$
we conclude from Proposition \ref{suv} that $d(A) \leq 1$. Actually we can do better:

\begin{lemma} $d(A)=0$.
\end{lemma}

\begin{proof} Suppose for contradiction that $d(A)=1$; without loss of generality we may take $11222 \in A$. This implies that $d(1****)=d(*1***)=1$. Also, by the above discussion, $c(**1**)$ and $c(**3**)$ cannot both be $24$, so by Proposition \ref{suv}, $s(**1**)+s(**3**) \leq 1/3$; similarly
$s(***1*)+s(***3*) \leq 1/3$ and $s(****1)+s(****3) \leq 1/3$. Since the $s$ average to zero, we see from the pigeonhole principle that either $s(1****)+s(3****) \geq -1/2$ or $s(*1***)+s(*3***) \geq -1/2$. We may assume by symmetry that
\begin{equation}\label{star-2}
s(1****)+s(3****) \geq -1/2.
\end{equation}
Since $s(3****) \leq 1/2$ by Proposition \ref{suv}, we conclude that
\begin{equation}\label{star}
s(1****) \geq -1.
\end{equation}
If $|A \cap 1****|=41$, then by \eqref{sv3} we have
$$ s(1****) = -1 - \frac{1}{4}( 48 - 2c(1****) - 3d(1****) ) - \frac{1}{12} (96-3b(1****)-2c(1****)-3d(1****))$$
but the arguments in Proposition \ref{suv} give $48 - 2c(1****) - 3d(1****) \geq 3$ and $96-3b(1****)-2c(1****)-3d(1****) \geq 0$, a contradiction. So we must have $|A \cap 1****|=42$ (by Proposition \ref{stat}(ii) and \eqref{41}). In that case, from \eqref{sv2} we have
$$ s(1****) = \frac{1}{4}( 48 - 2c(1****) - 3d(1****) ) - \frac{1}{12} (96-3b(1****)-2c(1****)-3d(1****))$$
while also having $48 - 2c(1****) - 3d(1****) \geq 3$ and $96-3b(1****)-2c(1****)-3d(1****) \geq 0$. Since $s(1****) \geq -1$ and $d(1****)=1$, we soon see that we must have $48 - 2c(1****) - 3d(1****) = 3$ and $96-3b(1****)-2c(1****)-3d(1****) \leq 3$, which forces $c(1****)=21$ and $b(1****)=16$ or $b(1****)=17$; thus the statistics of $1****$ are either $(4,16,21,1,0)$ or $(3,17,21,1,0)$.

We first eliminate the $(3,17,21,1,0)$ case. In this case $s(1****)$ is exactly $-1$. Inspecting the proof of \eqref{star}, we conclude that $s(3****)$ must be $+1/2$ and that $s(**1**)+s(**3**)=1/3$. From the former fact and Proposition \ref{suv} we see that $a(A) = a(1****)+a(3****)=3+3=6$; on the other hand, from the latter fact and Proposition \ref{suv} we have $a(A) = a(**1**)+a(**3**) = 4+3=7$, a contradiction.

So $1****$ has statistics $(4,16,21,1,0)$, which implies that $s(1****)=-3/4$ and $|A \cap 1****|=42$. By \eqref{star-2} we conclude
\begin{equation}\label{s3}
s(3****) \geq 1/4,
\end{equation}
which by Proposition \ref{suv} implies that $|A \cap 3****|=43$, and hence $|A \cap 2****|=40$. On the other hand, since $e(A)=f(A)=0$ and $d(A)=1$, with the latter being caused by $11222$, we see that $c(2****)=d(2****)=e(2****)=0$. From \eqref{alpha-1} we have $4a(2****)+b(2****) \leq 64$, and we also have the trivial inequality $b(2****) \leq 32$; these inequalities are only compatible if $2****$ has statistics $(8,32,0,0,0)$, thus $A \cap 2****$ contains $S_{2,5} \cap 2****$.

If $a(3****)=4$, then $a(A)=a(1****)+a(3****)=8$, which by Proposition \ref{suv} implies that $s(**1**)+s(**3**)$ cannot exceed $1/12$, and similarly for permutations. On the other hand, from Proposition \ref{suv} $s(**1**)+s(**3**)$ cannot exceed $-3/4 + 1/4 = -1/2$, and so the average value of $s$ cannot be zero, a contradiction. Thus $a(3****) \neq 4$, which by \eqref{s3} and Proposition \ref{suv} implies that $**3**$ has statistics $(3,16,24,0,0)$.

In particular, $A$ contains $16$ points from $3**** \cap S_{1,5}$ and all of $3**** \cap S_{2,5}$. As a consequence, no pair of the $16$ points in $A \cap 3**** \cap S_{1,5}$ can differ in only one coordinate; partitioning the $32$-point set $3**** \cap S_{1,5}$ into $16$ such pairs, we conclude that every such pair contains exactly one element of $A$. We conclude that $A \cap 3**** \cap S_{1,5}$ is equal to either $3**** \cap S_{1,5}^e$ or $3**** \cap S_{1,5}^o$.

On the other hand, $A$ contains all of $2**** \cap S_{2,5}$, and exactly sixteen points from $1**** \cap S_{1,5}$. Considering the vertical lines $*xyzw$ where $xyzw \in S_{1,4}$, we conclude that $A \cap 1**** \cap S_{1,5}$ is either equal to $1**** \cap S_{1,5}^o$ or $1**** \cap S_{1,5}^e$.
But either case is incompatible with the fact that $A$ contains $11222$ (consider either the line $11xx2$ or $11x\overline{x}2$, where $x=1,2,3$ and $\overline{x}=4-x$), obtaining the required contradiction.
\end{proof}

We can now eliminate all but three cases for the statistics of $A$:

\begin{proposition}[Statistics of $A$] The statistics $(a(A),b(A),c(A),d(A),e(A),f(A))$ of $A$ must be one of the following three tuples:
\begin{itemize}
\item (Case 1) $(6,40,79,0,0)$;
\item (Case 2) $(7,40,78,0,0)$;
\item (Case 3) $(8,39,78,0,0)$.
\end{itemize}
\end{proposition}

\begin{proof}
Since $d(A)=e(A)=f(A)=0$, we have
$$ c(2****)=d(2****)=e(2****)=0.$$
On the other hand, from \eqref{alpha-1} we have $4a(2****)+b(2****) \leq 64$ as well as the trivial inequality $b(2****) \leq 24$, and also we have
$$ |A \cap 2****| = 125 - |A \cap 1****| - |A \cap 3****| \geq 125 - 43 - 43 = 39.$$
Putting all this together, we see that the only possible statistics for $2****$ are $(8,32,0,0,0)$, $(7,32,0,0,0)$, or $(8,31,0,0,0)$. In particular, $7 \leq a(2****) \leq 8$ and $31 \leq b(2****) \leq 32$, and similarly for permutations. Applying Lemma \ref{dci} we conclude that
$$ 35 \leq b(A) \leq 40$$
and
$$ 77.5 \leq c(A) \leq 80.$$
Combining this with the first part of Corollary \ref{cmax} we conclude that $c(A)$ is either $78$ or $79$. From this and \eqref{amix} we see that the only cases that remain to be eliminated are $(7,39,79,0,0)$ and $(8,38,79,0,0)$, but these cases are incompatible with the second part of Corollary \ref{cmax}.
\end{proof}

We now eliminate each of the three remaining cases in turn.

\subsection{Elimination of $(6,40,79,0,0)$}

Here $A \cap S_{5,5}$ has six points. By Lemma \ref{tech}, there are at least two pairs in this set which differ in two positions. Their midpoints are eliminated from $A \cap S_{3,5}$. But $A$ omits exactly one point from $S_{3,5}$, so these midpoints must be the same. By symmetry, we may then assume that these two pairs are $(11111,11133)$ and $(11113,11131)$. Thus the eliminated point in $S_{3,5}$ is $11122$, i.e. $A$ contains $S_{3,5} \backslash \{11122\}$. Also, $A$ contains $\{11111,11133,11113,11131\}$ and thus must omit $\{11121, 11123, 11112, 11132\}$.

Since $11322 \in A$, at most one of $11312, 11332$ lie in $A$. By symmetry we may assume $11312 \not \in A$, thus there is a pair $(xy1z2, xy3z2)$ with $x,y,z = 1,3$ that is totally omitted from $A$, namely $(11112,11312)$. On the other hand, every other pair of this form can have at most one point in the $A$, thus there are at most seven points in $A$ of the form $xyzw2$ with $x,y,z,w = 1,3$. Similarly there are at most 8 points of the form $xyz2w$, or of $xy2zw$, $x2yzw$, $2xyzw$, leading to $b(A) \leq 7+8+8+8+8=39$, contradicting the statistic $b(A)=40$.

\subsection{Elimination of $(7,40,78,0,0)$}

Here $A \cap S_{5,5}$ has seven points. By Lemma \ref{tech}, there are at least three pairs in this set which differ in two positions. As we can only eliminate two points from $S_{3,5}$, two of the midpoints of these pairs must be the same; thus, as in the previous section, we may assume that $A$ contains $\{11111,11133,11113,11131\}$ and omits $\{11121, 11123, 11112, 11132\}$ and $11122$.

Now consider the $160$ lines $\ell$ connecting two points in $S_{4,5}$ to one point in $S_{3,5}$ (i.e. $*2xyz$ and permutations, where $x,y,z=1,3$). By double counting, the total sum of $|\ell \cap A|$ over all $160$ lines is $4b(A)+2c(A) = 316 = 158 \times 2$. On the other hand, each of these lines contain at most two points in $A$, but two of them (namely $1112*$ and $1112*$) contain no points. Thus we must have $|\ell \cap A|=2$ for the remaining $158$ lines $\ell$.

Since $A$ omits $1112x$ and $111x2$ for $x=1,3$, we thus conclude (by considering the lines $11*2x$ and $11*x2$) that $A$ must contain $1132x$, $113x2$, $1312x$, and $131x2$. Taking midpoints, we conclude that $A$ omits $11322$ and $13122$. But together with $11122$ this implies that at least three points are missing from $A \cap S_{3,5}$, contradicting the hypothesis $c(A)=78$.

\subsection{Elimination of $(8,39,78,0,0)$}

Now $A \cap S_{5,5}$ has eight points. By Lemma \ref{tech}, there are at least three pairs in this set which differ in two positions. As we can only eliminate two points from $S_{3,5}$, two of these pairs $(a,b), (c,d)$ must have the same midpoint $p$, and two other pairs $(a',b'), (c',d')$ must have the same midpoint $p'$, and $A$ contains $S_{3,5} \backslash \{p,p'\}$. As $p,p'$ are distinct, the plane containing $a,b,c,d$ is distinct from the plane containing $a',b',c',d'$.

Again consider the $160$ lines $\ell$ from the previous section. This time, the sum of the $|\ell \cap A|$ is $4b(A)+2c(A) = 312 = 156 \times 2$.
But the two lines in the plane of $a,b,c,d$ passing through $p$, and the two lines in the plane of $a',b',c',d'$ passing through $p'$, have no points; thus we must have $|\ell \cap A|=2$ for the remaining $156$ lines $\ell$.

Without loss of generality we have $(a,b)=(11111,11133)$, $(c,d) = (11113,11131)$, thus $p = 11122$. By permuting the first three indices, we may assume that $p'$ is not of the form $x2y2z, x2yz2, xy22z, xy2z2$ for any $x,y,z=1,3$. Then we have $1112x \not \in A$ and $1122x \in A$ for every $x=1,3$, so by the preceding paragraph we have $1132x \in A$; similarly for $113x2, 1312x, 131x2$. Taking midpoints, this implies that $13122, 11322 \not \in A$, but this (together with 11122) shows that at least three points aremissing from $A \cap S_{3,5}$, contradicting the hypothesis $c(A)=78$.

\subsection{Six dimensions}

Now we establish the bound $c'_{6,3}=353$. In view of the lower bounds, it suffices to show that there does not exist a Moser set $A \subset [3]^5$ with $|A|=354$.

We argue by contradiction. Let $A$ be as above, and let $(a(A),\ldots,g(A))$ be the statistics of $A$.

\begin{lemma}\label{g6} $g(A)=0$.
\end{lemma}

\begin{proof} For any four-dimensional slice $V$ of $A$, define
$$S(V) := 15 a(V) + 5 b(V) + 5 c(V)/2 + 3d(V)/2 + e(V).$$
From Lemma \ref{dci} we see that $|A|$ is equal to $a(A)+b(A)$ plus the average of $S(V)$ where $V$ ranges over the twenty slices which are some permutation of the center slice $22****$.

If $g(A)=1$, then $a(A) \leq 32$ and $b(A) \leq 96$ by \eqref{alpha-1}. Meanwhile, $e(V)=g(A)=1$ for every center slice $V$, so from Lemma \ref{paretop-4}, one can show that $S(V) \leq 223.5$ for every such slice. We conclude that $|A| \leq 351.5$, a contradiction.
\end{proof}

For any four-dimensional slice $V$ of $A$, define the \emph{defects} to be
$$ D(V) := 356 - [4a(V)+6b(V)+10c(V)+20d(V)+60e(V)].$$
Define a \emph{corner slice} to be one of the permutations or reflections of $11****$, thus there are $60$ corner slices. From Lemma \ref{dci} we see that $356-|A|+f(A)=2+f(A)$ is the average of the defects of all the $60$ corner slices. On the other hand, from Lemma \ref{paretop-4} and a straightforward computation, one concludes

\begin{lemma}\label{defects} Let $A$ be a four-dimensional Moser set. Then $D(A) \geq 0$. Furthermore:
\begin{itemize}
\item If $A$ has statistics $(6,12,18,4,0)$, then $D(A)=0$.
\item If $A$ has statistics $(5,12,18,4,0)$, $(5,12,12,4,1)$, or $(6,8,12,8,0)$, then $D(A)=4$.
\item For all other $A$, $D(A) \geq 6$.
\item If $a(A) = 4$, then $D(A) \geq 8$.
\item If $a(A) \geq 7$, then $D(A) \geq 16$.
\item If $a(A) \geq 8$, then $D(A) \geq 30$.
\item If $a(A) \geq 9$, then $D(A) \geq 86$.
\end{itemize}
\end{lemma}

Define a \emph{family} to be a set of four parallel corner slices, thus there are $15$ families, which are all a permutation of $\{11****, 13****, 31****, 33**** \}$. We refer to the family $\{11****, 13****, 31****, 33**** \}$ as $ab****$, and similarly define the family $a*b***$, etc.

\begin{lemma}\label{f6} $f(A)=0$.
\end{lemma}

\begin{proof} For any four-dimensional slice $V$ of $A$, define
$$ s(V) := 12 a(V)+15 b(V)/2+20 c(V)/3+15 d(V)/2 + 12 e(V),$$
and define an \emph{edge slice} to be one of the $30$ permutations or reflections of $12****$. From double counting we see that $|A|-a(A)$ is equal to the average of the $30$ values of $s(V)$ as $V$ ranges over edge slices.

From Lemma \ref{paretop-4} one can verify that $s(V) \leq 336$, and that $s(V) \leq 296 = 336-40$ if $e(V)=1$. The number of edge slices $V$ for which $e(V)=1$ is equal to $5f(A)$, and so the average value of the $s(V)$ is at most $336 - \frac{40 \times 5}{30} f(A)$, and so
$$ |A| - a(A) \leq 336 - \frac{40 \times 5}{30} f(A)$$
which we can rearrange (using $|A|=354$) as
$$ a(A) \geq 18 + \frac{20}{3} f(A).$$
Suppose first that $f(A)=1$; then $a(A) \geq 25$. This means that in any given family, one of the four corner slices has an $a$ value of at least $7$, and thus by Lemma \ref{defects} has a defect of at least $16$. Thus the average defect is at least $4$; on the other hand, the average defect is $2+f(A)=3$, a contradiction.

Now suppose that $f(A) \geq 2$; then $a(A) \geq 32$. Then in any given family, there is a corner slice with an $a$ value at least $9$, or four slices with $a$ value at least $8$, leading to a total defect of at least $86$ by Lemma \ref{defects}. Thus the average defect is at least $21.5$; on the other hand, the average defect is $2+f(A) \leq 2+12$, a contradiction.
\end{proof}

As a consequence of the above lemma, we see that the average defect of all corner slices is $2$, or equivalently that the total defect of these slices is $120$.

Call a corner slice \emph{good} if it has statistics $(6,12,18,4,0)$, and \emph{bad} otherwise. Thus good slices have zero defect, and bad slices have defect at least four. Since the average defect of the $60$ corner slices is $2$, there are at least $30$ good slices.

One can describe the structure of the good slices completely:

\begin{lemma}\label{sixt} The subset of $[3]^4$ consisting of the strings $1111, 1113, 3333, 1332, 1322, 1222, 3322$ and permutations is a Moser set with statistics $(6,12,18,4,0)$. Conversely, every Moser set with statistics $(6,12,18,4,0)$ is of this form up to the symmetries of the cube $[3]^4$.
\end{lemma}

\begin{proof} This can be verified by computer. By symmetry, one assumes 1222,2122,2212 and 2221 are in the set. Then 18 of the 24 'c' points with two 2s must be included; it is quick to check that 1122 and permutations must be the six excluded. Next, one checks that the only possible set of six 'a' points with no 2s is 1111,1113,3333 and permutations. Lastly, in a rather longer computation, one finds there is only possible set of twelve 'b' points, that is points with one 2. A computer-free proof can be found at
\centerline{{\tt http://michaelnielsen.org/polymath1/index.php?title=Classification\_of\_\%286\%2C12\%2C18\%2C4\%2C0\%29\_sets}.}
\end{proof}

As a consequence of this lemma, given any $x,y,z,w \in \{1,3\}$, there is a unique good Moser set in $[3]^4$ set whose intersection with $S_{1,4}$ is $\{x222, 2y22, 22z2, 222w\}$, and these are the only 16 possibilities. Call this set the \emph{good set of type $xyzw$}. It consists of
\begin{itemize}
\item The four points $x222, 2y22, 22z2, 222w$ in $S_{1,4}$;
\item All $24$ elements of $S_{2,4}$ except for $xy22, x2z2, x22w, 2yz2, 2y2w, 22zw$;
\item The twelve points $xYZ2$, $xY2W$, $x2ZW$, $XyZ2$, $Xy2W$, $2yZW$, $XYz2$, $X2zW$, $2YzW$, $XY2w$, $X2Zw$, $2YZw$ in $S_{3,4}$, where $X=4-x$, $Y=4-y$, $Z=4-z$, $W=4-w$;
\item The six points $xyzw, xyzW, xyZw, xYzw, Xyzw, XYZW$ in $S_{4,4}$.
\end{itemize}

We can use this to constrain the types of two intersecting good slices:

\begin{lemma}\label{pqs} Suppose that the $pq****$ slice is of type $xyzw$, and the $p*r***$ slice is of type $x'y'z'w'$, where $p,q,r,x,y,z,w,x',y',z',w'$ are in $\{1,3\}$. Then $x'=x$ iff $q=r$, and $y'z'w'$ is equal to either $yzw$ or $YZW$. If $x=r$ (or equivalently if $x'=q$), then $y'z'w'=yzw$.
\end{lemma}

\begin{proof} By reflection symmetry we can take $p=q=r=1$. Observe that the $11****$ slice contains $111222$ iff $x=1$, and the $1*1***$ slice similarly contains $111222$ iff $x'=1$. This shows that $x=x'$.

Suppose now that $x=x'=1$. Then the $111***$ slice contains the three elements $111y22, 1112z2, 11122w$, and excludes $111Y22, 1112Z2, 11122W$, and similarly with the primes, which forces $yzw=y'z'w'$ as claimed.

Now suppose that $x=x'=3$. Then the $111***$ slice contains the two elements $111yzw, 111YZW$, but does not contain any of the other six points in $S_{6,6} \cap 111***$, and similarly for the primes. Thus $y'z'w'$ is equal to either $yzw$ or $YZW$ as claimed.
\end{proof}

Now we look at two adjacent parallel good slices, such as $11****$ and $13****$. The following lemma asserts that such slices either have opposite type, or else will create a huge amount of defect in other slices:

\begin{lemma}\label{l18} Suppose that the $11****$ and $13****$ slices are good with types $xyzw$ and $x'y'z'w'$ respectively. If $x=x'$, then the $1*x***$ slice has defect at least $30$, and the $1*X***$ slice has defect at least $8$. Also, the $1**1**$, $1**3**$, $1***1*$, $1***3*$, $1****1$, $1****3$ slices have defect at least $6$. In particular, the total defect of slices beginning with $1*$ is at least $74$.
\end{lemma}

\begin{proof} Observe from the $11****, 13****$ hypotheses that $a(1*x***)=9$ and $a(1*X***)=4$, which gives the first two claims by Lemma \ref{defects}. For the other claims, one sees from Lemma \ref{pqs} that the other six slices cannot be good; also, they have an $a$-value of $6$ and a $d$-value of at most $7$, and the claims then follow from Lemma \ref{defects}.
\end{proof}

Now we look at two diagonally opposite parallel good slices, such as $11****$ and $33****$.

\begin{lemma}\label{l14} The $11****$ and $33****$ slices cannot both be good and of the same type.
\end{lemma}

\begin{proof} Suppose not. By symmetry we may assume that $11****$ and $33****$ are of type $1111$. This excludes a lot of points from $22****$. Indeed, by connecting lines between the $11****$ and $33****$ slices, we see that the only points that can still survive in $22****$ are $221133, 221333, 221132, 223332$, and permutations of the last four indices. Double counting the lines $22133*$ and permutations we see that there are at most $12$ points one can place in the permutations of $221133, 221333, 221132$, and so the $22****$ slice has at most $16$ points. Meanwhile, the two five-dimensional slices $1*****, 3*****$ have at most $c'_{5,3} = 124$ points, and the other two four-dimensional slices $21****, 23****$ have at most $c'_{4,3} = 43$ points, leading to at most $16 + 124 * 2 + 43 * 2 = 350$ points in all, a contradiction.
\end{proof}

\begin{lemma}\label{l19} It is not possible for all four slices in a family to be good.
\end{lemma}

\begin{proof} Suppose not. By symmetry we may assume that $11****, 13****, 31****, 33****$ are good. By Lemma \ref{l14}, the $11****$ and $33****$ slices cannot be of the same type, and so they cannot both be of the opposite type to either $13****$ or $31****$. If $13****$ is not of the opposite type to $11****$, then by (a permutation of) Lemma \ref{l18}, the total defect of slices beginning with $1*$ is at least $74$; otherwise, if $13****$ is not of the opposite type to $33****$, then by (a permutation and reflection of) Lemma \ref{l18}, the total defect of slices beginning with $*3$ is at least $74$. Similarly, the total defect of slices beginning with $3*$ or $*1$ is at least $74$, leading to a total defect of at least $148$. But the total defect of all the corner slices is $2 \times 60 = 120$, a contradiction.
\end{proof}

\begin{corollary}\label{l20} At most one family can have a total defect of at least $38$.
\end{corollary}

\begin{proof} Suppose there are two families with defect at least $38$. The remaining thirteen family have defect at least $4$ by Lemma \ref{l19} and Lemma \ref{defects}, leading to a total defect of at least $38*2+13*4=128$. But the total defect is $2 \times 60 = 120$, a contradiction.
\end{proof}

Actually we can refine this:

\begin{proposition} No family can have a total defect of at least $38$.
\end{proposition}

\begin{proof} Suppose for contradiction that the $ab****$ family (say) had a total defect of at least $38$, then by Corollary \ref{l20} no other families have total defect at least $38$.

We claim that the $**ab**$ family can have at most two good slices. Indeed, suppose the $**ab**$ has three good slices, say $**11**, **13**, **33**$. By Lemma \ref{l14}, the $**11**$ and $**33**$ slices cannot be of the same type, and so cannot both be of opposite type to $**13**$. Suppose $**11**$ and $**13**$ are not of opposite type. Then by (a permutation of) Lemma \ref{l18}, one of the families $a*b***, *ab***, **b*a*, **b**a$ has a net defect of at least $38$, contradicting the normalisation.

Thus each of the six families $**ab**, **a*b*, **a**b, ***ab*, ***a*b, ****ab$ have at least two bad slices. Meanwhile, the eight families $a*b***, a**b**, a***b*, a****b, *ab***, *a*b**, *a**b*, *a***b$ have at least one bad slice by Corollary \ref{l19}, leading to twenty bad slices in addition to the defect of at least $38$ arising from the $ab****$ slice. To add up to a total defect of $120$, we conclude from Lemma \ref{defects} that all bad slices outside of the $ab****$ family have a defect of four, with at most one exception; but then by Lemma \ref{l18} this shows that (for instance) the $1*1***$ and $1*3***$ slices cannot be good unless they are of opposite type. The previous argument then shows that the a*b*** slice cannot have three good slices, which increases the number of bad slices outside of $ab****$ to at least twenty-one, and now there is no way to add up to $120$, a contradiction.
\end{proof}

\begin{corollary} Every family can have at most two good slices.
\end{corollary}

\begin{proof} If for instance $11****, 13****, 33****$ are all good, then by Lemma \ref{l14} at least one of $11****, 33****$ is not of the opposite type to $13****$, which by Lemma \ref{l18} implies that there is a family with a total defect of at least $38$, contradicting the previous proposition.
\end{proof}

From this corollary and Lemma \ref{defects}, we see that every family has a defect of at least $8$. Since there are $15$ families, and $8 \times 15$ is exactly equal to $120$, we conclude

\begin{corollary}\label{coda} Every family has \emph{exactly} two good slices, and the remaining two slices have defect $4$. In particular, by Lemma \ref{defects}, the bad slices must have statistics $(5,12,18,4,0)$, $(5,12,12,4,1)$, or $(6,8,12,8,0)$.
\end{corollary}

We now limit how these slices can interact with good slices.

\begin{lemma}\label{goodgood} Suppose that $1*1***$ is a good slice.
\begin{itemize}
\item[(i)] The $11****$ slice cannot have statistics $(6,8,12,8,0)$.
\item[(ii)] The $11****$ slice cannot have statistics $(5,12,12,4,1)$.
\item[(iii)] If the $11****$ slice has statistics $(5,12,18,4,0)$, then the $112***$ slice has statistics $(3,9,3,0)$.
\end{itemize}
\end{lemma}

\begin{proof} This can be verified through computer search; there are $16$ possible configurations for the good slices, and one can calculate that there are $27520$ configurations for the $(5,12,12,4,1)$ slices, $4368$ configurations for the $(5,12,18,4,0)$ slices, and $80000$ configurations for the $(6,8,12,8,0)$ slices. {\bf we could put human proofs for all this somewhere, presumably.}

%We first prove (i). Suppose for contradiction that the $11****$ slice has statistics $(6,8,12,8,0)$, then $A$ contains $111222$, and so the $1*1***$ slice is of type $1xyz$ for some $x,y,z$. By symmetry we may assume it is of type $1111$, thus the $111***$ slice consists of
%$111111, 111113, 111332, 111322, 111222$ and permutations of the last three indices. On the other hand, the $11****$ slice has all eight of the points in $11**** \cap S_{2,6}$. Drawing lines between these points and $111111, 111113$ and permutations, we see that the $113***$ slice cannot contain $113111, 113113, 113133$, or permutations, leaving $113333$ as the only possible element of $113*** \cap S_{6,6}$. This makes $a(11****)=5$, a contradiction.
\end{proof}

\begin{corollary}\label{slic} The $111***$ slice has statistics $(4,3,3,1)$, $(2,6,6,0)$, $(3,3,3,1)$, or $(1,6,6,0)$.
\end{corollary}

\begin{proof} From Corollary \ref{coda}, we know that at least one of the slices $13****, 31****, 11****$ are good. If $11****$ or $1*1***$ is good, then the slice $111***$ has statistics $(4,3,3,1)$ or $(2,6,6,0)$, by Lemma \ref{sixt}. By symmetry we may thus reduce to the case where $13****$ is good and $1*1***$ is bad. Then by Lemma \ref{goodgood}, the $1*1***$ slice has statistics $(5,12,18,4,0)$ and the $121***$ slice has statistics $(3,9,3,0)$. Since the $131***$ slice, as a side slice of the good $13****$ slice, has statistics $(4,3,3,1)$ or $(2,6,6,0)$, we conclude that the $111***$ slice has statistics $(1,6,6,0)$ or $(3,3,3,1)$, and the claim follows.
\end{proof}

\begin{corollary}\label{slic2} All corner slices have statistics $(6,12,18,4,0)$ or $(5,12,18,4,0)$.
\end{corollary}

\begin{proof} Suppose first that a corner slice, say $11****$ has statistic $(6,8,12,8,0)$. Then $111***$ and $113***$ contain one ``d'' point each, and have six ``a'' points between them, so by Corollary \ref{slic}, they both have statistic $(3,3,3,1)$. This forces the $1*1***$, $1*3***$ slices to be bad, which by Corollary \ref{coda} forces the $3*1***,3*3***$ slices to be good. This forces the $311***, 313***$ slices to have statistics either $(2,6,6,0)$ or $(4,3,3,1)$. But the $311***$ slice (say) cannot have statistic $(4,3,3,1)$, since when combined with the $(3,3,3,1)$ statistics of $111***$ would give $a(*11***)=7$, which contradicts Corollary \ref{coda}; thus the $311***$ slice has statistic $(2,6,6,0)$, and similarly for $331***$. But then $a(3*1***)=4$, which again contradicts Corollary \ref{coda}.

Thus no corner slice has statistic $(6,8,12,8,0)$. Now suppose that a corner slice, say $11****$ has statistic $(5,12,12,4,1)$. By Lemma \ref{goodgood}, the $1*1***, 1*3***$ slices are bad, so by repeating the preceding arguments we conclude that the $311***, 313***$ slices have statistics $(2,6,6,0)$ or $(4,3,3,1)$; in particular, their $a$-value is even. However, the $*11***$ and $*13***$ slices are bad by Lemma \ref{goodgood}, and thus have an $a$-value of $5$; thus the $111***$ and $113***$ slices have an odd $a$-value. Thus forces $a(11****)$ to be even; but it is equal to $5$, a contradiction.
\end{proof}

From this and Lemma \ref{dci}, we see that $A$ has statistics $(22,72,180,80,0,0,0)$. In particular, we have $2\alpha_2(A)+\alpha_3(A) = 2$, which by double counting (cf. \eqref{alpha-1}) shows that for every line of the form $11122*$ (or a reflection or permutation thereof) intersects $A$ in exactly two points. Note that such lines connect a ``$d$'' point to two ``$c$'' points.

Also, we observe that two adjacent ``$d$'' points, such as $111222$ and $113222$, cannot both lie in $A$; for this would force the $*13***$ and $*11***$ slices to have statistics $(4,3,3,1)$ or $(3,3,3,1)$ by Corollary \ref{slic}, which forces $a(*1****)=6$, and thus $*1****$ must be good by Corollary \ref{slic2}; but this contradicts Lemma \ref{sixt}. Since $\alpha_3(A)=1/2$, we conclude that given any two adjacent ``$d$'' points, exactly one of them lies in $A$. In particular, the d points of the form $***222$ consist either of those strings with an even number of $1$s, or those with an odd number of $1$s.

Let's say it's the former, thus the set contains $111222, 133222$, and permutations of the first three coordinates, but omits $113222, 333222$ and permutations of the first three coordinates. Since the ``$d$'' points $113222, 333222$ are omitted, we conclude that the ``$c$'' points $113122, 113322, 333122, 333322$ must lie in the set, and similarly for permutations of the first three and last three coordinates. But this gives at least $15$ of the $16$ ``$c$'' points ending in $22$; by symmetry this leads to $225$ $c$-points in all; but $c(A)=180$, contradiction. This (finally!) completes the proof that $c'_{6,3}=353$.

Moser.tex

2009-07-22T12:22:47Z

Mpeake:

\section{Upper bounds for the $k=3$ Moser problem in small dimensions}\label{moser-upper-sec}

In this section we finish the proof of Theorem \ref{moser} by obtaining the upper bounds on $c'_{n,3}$ for $n \leq 6$.

\subsection{Statistics, densities and slices}

Our analysis will revolve around various \emph{statistics} of Moser sets $A \subset [3]^n$, their associated \emph{densities}, and the behavior of such statistics and densities with respect to the operation of passing from the cube $[3]^n$ to various \emph{slices} of that cube.

\begin{definition}[Statistics and densities] Let $A \subset [3]^n$ be a set. For any $0 \leq i \leq n$, set $a_i(A) := |A \cap S_{n-i,n}|$; thus we have
$$ 0 \leq a_i(A) \leq |S_{n-i,n}| = \binom{n}{i} 2^{n-i}$$
for $0 \leq i \leq n$ and
$$ a_0(A) + \ldots + a_n(A) = |A|.$$
We refer to the vector $(a_0(A),\ldots,a_n(A))$ as the \emph{statistics} of $A$. We define the $i^{th}$ \emph{density} $\alpha_i(A)$ to be the quantity
$$ \alpha_i(A) := \frac{a_i(A) }{\binom{n}{i} 2^{n-i}},$$
thus $0 \leq \alpha_i(A) \leq 1$ and
$$ |A| = \sum_{i=0}^n \binom{n}{i} 2^{n-i} a_i(A).$$
\end{definition}

\begin{example}\label{2mos} Let $n=2$ and $A$ be the Moser set $A := \{ 12, 13, 21, 23, 31, 32 \}$. Then the statistics $(a_0(A), a_1(A), a_2(A))$ of $A$ are $(2,4,0)$, and the densities $(\alpha_0(A), \alpha_1(A), \alpha_2(A))$ are $(\frac{1}{2}, 1, 0)$. \textbf{Include picture here? with colours?}
\end{example}

When working with small values of $n$, it will be convenient to write $a(A)$, $b(A)$, $c(A)$, etc. for $a_0(A)$, $a_1(A)$, $a_2(A)$, etc., and similarly write $\alpha(A), \beta(A), \gamma(A)$, etc. for $\alpha_0(A)$, $\alpha_1(A)$, $\alpha_2(A)$, etc. Thus for instance in Example \ref{2mos} we have $b(A) = 4$ and $\alpha(A) = \frac{1}{2}$.

\begin{definition}[Subspace statistics and densities] If $V$ is a $k$-dimensional geometric subspace of $[3]^n$, then we have a map $\phi_V: [3]^k \to [3]^n$ from the $k$-dimensional cube to the $n$-dimensional cube. If $A \subset [3]^n$ is a set and $0 \leq i \leq k$, we write $a_i(V,A)$ for $a_i(\phi_V^{-1}(A))$ and $\alpha_i(V,A)$ for $\alpha_i(\phi_V^{-1}(A))$. If the set $A$ is clear from context, we abbreviate $a_i(V,A)$ as $a_i(V)$ and $\alpha_i(V,A)$ as $\alpha_i(V)$.
\end{definition}

For our problem, a particularly important type of subspace of $[3]^n$ will be the \emph{slices} formed by fixing one coordinate and letting the other $n-1$ coordinates vary. We will denote this by a single string in which the $n-1$ varying coordinates are denoted by asterisks. For instance, in $[3]^2$, $1*$ denotes the slice $1*=\{11,12,13\}$, $*2$ denotes the slice $*2=\{12,22,32\}$, etc.; similarly, in $[3]^3$, $1**$ is the slice $\{111, 112, 113, 121, 122, 123, 131, 132, 133\}$, etc. We call a slice a \emph{centre slice} if the fixed coordinate is $2$ and a \emph{side slice} if it is $1$ or $3$.

\begin{example} We continue Example \ref{2mos}. Then the statistics of the side slice $1*$ are $(a(1*),b(1*)) = (1,1)$, while the statistics of the centre slice $2*$ are $(a(2*),b(2*))=(2,0)$. The corresponding densities are $(\alpha(1*),\beta(1*)) = (1/2,1)$ and $(\alpha(2*),\beta(2*))=(1,0)$.
\end{example}

A simple double counting argument gives the following useful identity:

\begin{lemma}[Double counting identity]\label{dci} Let $A \subset [3]^n$ and $0 \leq i \leq n-1$. Then we have
$$ \frac{1}{n-i-1} \sum_{V \hbox{ a side slice}} a_{i+1}(V) = \frac{1}{i+1} \sum_{W \hbox{ a centre slice}} a_i(W) = a_{i+1}(A)$$
where $V$ ranges over the $2n$ side slices of $[3]^n$, and $W$ ranges over the $n$ centre slices. In other words, the average value of $\alpha_{i+1}(V)$ for side slices $V$ equals the average value of $\alpha_i(W)$ for centre slices $W$, which is in turn equal to $\alpha_{i+1}(A)$.
\end{lemma}

Indeed, this lemma follows from the observation that every string in $A \cap S_{n-i-1,n}$ belongs to $i+1$ centre slices $W$ (and contributes to $a_i(W)$) and to $n-i-1$ side slices $V$ (and contributes to $a_{i+1}(V)$). One can also view this lemma probabilistically, as the assertion that there are three equivalent ways to generate a random string of length $n$:

\begin{itemize}
\item Pick a side slice $V$ at random, and randomly fill in the wildcards in such a way that $i+1$ of the wildcards are $2$'s (i.e. using an element of $S_{n-i-2,n-1}$).
\item Pick a centre slice $V$ at random, and randomly fill in the wildcards in such a way that $i$ of the wildcards are $2$'s (i.e. using an element of $S_{n-i-1,n-1}$).
\item Randomly choose an element of $S_{n-i-1,n}$.
\end{itemize}

\begin{example} We continue Example \ref{2mos}. The average value of $\beta$ for side slices is equal to the average value of $\alpha$ for centre slices, which is equal to $\beta(A) = 1$.
\end{example}

Another very useful fact (essentially due to \cite{chvatal2}) is that linear inequalities for statistics of Moser sets at one dimension propagate to linear inequalities in higher dimensions:

\begin{lemma}[Propagation lemma]\label{prop} Let $n \geq 1$ be an integer. Suppose one has a linear inequality of the form
\begin{equation}\label{alphav}
\sum_{i=0}^n v_i \alpha_i(A) \leq s
\end{equation}
for all Moser sets $A \subset [3]^n$ and some real numbers $v_0,\ldots,v_n,s$. Then we also have the linear inequality
$$ \sum_{i=0}^n v_i \alpha_{qi+r}(A) \leq s$$
whenever $q \geq 1$, $r \geq 0$, $N \geq nq+r$ are integers and $A \subset [3]^N$ is a Moser set.
\end{lemma}

\begin{proof} We run a probabilistic argument (one could of course also use a double counting argument instead). Let $n,v_0,\ldots,v_n,s,q,r,N,A$ be as in the lemma. Let $V$ be a random $n$-dimensional geometric subspace of $[3]^N$, created in the following fashion:
\begin{itemize}
\item Pick $n$ wildcards $x_1,\ldots,x_n$ to run independently from $1$ to $3$. We also introduce dual wildcards $\overline{x_1},\ldots,\overline{x_n}$; each $\overline{x_j}$ will take the value $4-x_j$.
\item We randomly subdivide the $N$ coordinates into $n$ groups of $q$ coordinates, plus a remaining group of $N-nq$ ``fixed'' coordinates.
\item For each coordinate in the $j^{th}$ group of $q$ coordinates for $1 \leq j \leq n$, we randomly assign either a $x_j$ or $\overline{x_j}$.
\item For each coordinate in the $N-nq$ fixed coordinates, we randomly assign a digit $1,2,3$, but condition on the event that exactly $r$ of the digits are equal to $2$ (i.e. we use a random element of $S_{N-nq-r,N-nq}$).
\item Let $V$ be the subspace created by allowing $x_1,\ldots,x_n$ to run independently from $1$ to $3$, and $\overline{x_j}$ to take the value $4-x_j$.
\end{itemize}

For instance, if $n=2, q=2, r=1, N=6$, then a typical subspace $V$ generated in this fashion is
$$ 2x_1\overline{x_2}3x_2x_1 = \{ 213311, 212321, 211331, 223312, 222322, 221332, 233313, 232323, 231333\}.$$
Observe from that the following two ways to generate a random element of $[3]^N$ are equivalent:

\begin{itemize}
\item Pick $V$ randomly as above, and then assign $(x_1,\ldots,x_n)$ randomly from $S_{n-i,n}$. Assign $4-x_j$ to $\overline{x_j}$ for all $1 \leq j \leq n$.
\item Pick a random string in $S_{N-qi-r,N}$.
\end{itemize}

Indeed, both random variables are invariant under the symmetries of the cube, and both random variables always pick out strings in $S_{N-qi-r,N}$, and the claim follows. As a consequence, we see that the expectation of $\alpha_i(V)$ (as $V$ ranges over the recipe described above) is equal to $\alpha_{qi+r}(A)$. On the other hand, from \eqref{alphav} we have
$$ \sum_{i=0}^n v_i \alpha_i(V) \leq s$$
for all such $V$; taking expectations over $V$, we obtain the claim.
\end{proof}

In view of Lemma \ref{prop}, it is of interest to locate linear inequalities relating the densities $\alpha_i(A)$, or (equivalently) the statistics $a_i(A)$. For this, it is convenient to introduce the following notation.

\begin{definition} Let $n \geq 1$ be an integer.
\begin{itemize}
\item A vector $(a_0,\ldots,a_n)$ of non-negative integers is \emph{feasible} if it is the statistics of some Moser set $A$.
\item A feasible vector $(a_0,\ldots,a_n)$ is \emph{Pareto-optimal} if there is no other feasible vector $(b_0,\ldots,b_n) \neq (a_0,\ldots,a_n)$ such that $b_i \geq a_i$ for all $0 \leq i \leq n$.
\item A Pareto-optimal vector $(a_0,\ldots,a_n)$ is \emph{extremal} if it is not a non-trivial convex linear combination of other Pareto-optimal vectors.
\end{itemize}
\end{definition}

To establish a linear inequality of the form \eqref{alphav} with the $v_i$ non-negative, it suffices to test the inequality against densities associated to extremal vectors of statistics. (There is no point considering linear inequalities with negative coefficients $v_i$, since one always has the freedom to reduce a density $\alpha_i(A)$ of a Moser set $A$ to zero, simply by removing all elements of $A$ with exactly $i$ $2$'s.)

We will classify exactly the Pareto-optimal and extremal vectors for $n \leq 3$, which by Lemma \ref{prop} will lead to useful linear inequalities for $n \geq 4$. Using a computer, we have also located a partial list of Pareto-optimal and extremal vectors for $n=4$, which are also useful for the $n=5$ and $n=6$ theory.

\subsection{Up to three dimensions}

We now establish Theorem \ref{moser} for $n \leq 3$, and establish some auxiliary inequalities which will be of use in higher dimensions.

The case $n=0$ is trivial. When $n=1$, it is clear that $c'_{1,3} = 2$, and furthermore that the Pareto-optimal statistics are $(2,0)$ and $(1,1)$, which are both extremal. This leads to the linear inequality
$$ 2\alpha(A) + \beta(A) \leq 2$$
for all Moser sets $A \subset [3]^1$, which by Lemma \ref{prop} implies that
\begin{equation}\label{alpha-1}
2\alpha_r(A) + \alpha_{r+q}(A) \leq 2
\end{equation}
whenever $r \geq 0$, $q \geq 1$, $n \geq q+r$, and $A \subset [3]^n$ is a Moser set.

For $n=2$, we see by partitioning $[3]^2$ into three slices that $c'_{2,3} \leq 3 c'_{1,3} = 6$, and so (by the lower bounds in the previous section) $c'_{2,3} = 6$. Writing $(a,b,c) = (a(A),b(A),c(A)) = (4\alpha(A), 4\beta(A), \gamma(A))$, the inequalities \eqref{alpha-1} become
\begin{equation}\label{abc}
a + 2c \leq 4; b+2c \leq 4; 2a+b <= 8.
\end{equation}

\begin{lemma} When $n=2$, the Pareto-optimal statistics are $(4,0,0), (3,2,0), (2,4,0), (2,2,1)$. In particular, the extremal statistics are $(4,0,0), (2,4,0), (2,2,1)$.
\end{lemma}

\begin{proof} One easily checks that all the statistics listed above are feasible.
Consider the statistics $(a,b,c)$ of a Moser set $A \subset [3]^2$. $c$ is either equal to $0$ or $1$. If $c=1$, then \eqref{abc} implies that $a,b \leq 2$, so the only Pareto-optimal statistic here is $(2,2,1)$. When instead $c=0$, the inequalities \eqref{abc} can easily imply the Pareto-optimality of $(4,0,0), (3,2,0), (2,4,0)$.
\end{proof}

From this lemma we see that we obtain a new inequality $2a+b+2c \leq 8$. Converting this back to densities and using Lemma \ref{prop}, we conclude that
\begin{equation}\label{alpha-2}
4\alpha_r(A) + 2\alpha_{r+q}(A) + \alpha_{r+2q} \leq 4
\end{equation}
whenever $r \geq 0$, $q \geq 1$, $n \geq q+2r$, and $A \subset [3]^n$ is a Moser set.

One can also check by computer that there are exactly $230$ line-free subsets of $[3]^2$.

Now we look at three dimensions. Writing $(a,b,c,d)$ for the statistics of a Moser set $A \subset [3]^n$ (which thus range between $(0,0,0,0)$ and $(8,12,6,1)$), the inequalities \eqref{alpha-1} imply in particular that
\begin{equation}\label{abc-3d}
a+4d \leq 8; b+6d \leq 12; c+3d \leq 6; 3a+2c \leq 24; b+c \leq 12
\end{equation}
while \eqref{alpha-2} implies that
\begin{equation}\label{abcd-3d}
3a+b+c \leq 24; b+c+3d \leq 12.
\end{equation}
Summing the inequalities $b+c \leq 12, 3a+b+c \leq 24, b+c+3d \leq 12$ yields
$$ 3(a+b+c+d) \leq 48$$
and hence $|A| = a+b+c+d \leq 16$; comparing this with the lower bounds of the preceding section we obtain $c'_{3,3} = 16$ as required. (This argument is essentially identical to the one in \cite{chvatal2}).

We have the following useful computation:

\begin{lemma}[3D Pareto-optimals]\label{paretop} When $n=3$, the Pareto-optimal statistics are $$(3,6,3,1),(4,4,3,1),(4,6,2,1),(2,6,6,0),(3,6,5,0),(4,4,5,0),(3,7,4,0),(4,6,4,0),$$
$$ (3,9,3,0),(4,7,3,0),(5,4,3,0),(4,9,2,0),(5,6,2,0),(6,3,2,0),(3,10,1,0),(5,7,1,0),$$
$$ (6,4,1,0),(4,12,0,0),(5,9,0,0),(6,6,0,0),(7,3,0,0),(8,0,0,0).$$
In particular, the extremal statistics are
$$(3,6,3,1),(4,4,3,1),(4,6,2,1),(2,6,6,0),(4,4,5,0),(4,6,4,0),(4,12,0,0),(8,0,0,0).$$
\end{lemma}

\begin{proof} This can be established by a brute-force search over the $2^{27} \approx 1.3 \times 10^8$ different subsets of $[3]^3$. Actually, one can perform a much faster search than this. Firstly, as noted earlier, there are only $230$ line-free subsets of $[3]^2$, so one could search over $230^3 \approx 1.2 \times 10^7$ configurations instead. Secondly, by symmetry we may assume (after enumerating the $230$ sets in a suitable fashion) that the first slice $A \cap 1**$ has an index less than or equal to the third $A \cap 3**$, leading to $\binom{231}{2} \times 230 \approx 6 \times 10^6$ configurations instead. Finally, using the first and third slice one can quickly determine which elements of the second slice $2**$ are prohibited from $A$. There are $2^9 = 512$ possible choices for the prohibited set in $2**$. By crosschecking these against the list of $230$ line-free sets one can compute the Pareto-optimal statistics for the second slices inside the prohibited set (the lists of such statistics turns out to length at most $23$). Storing these statistics in a lookup table, and then running over all choices of the first and third slice (using symmetry), one now has to perform $O( 512 \times 230 ) + O( \binom{231}{2} \times 23) \approx O( 10^6 )$ computations, which is quite a feasible computation.

One could in principle reduce the computations even further, by a factor of up to $8$, by using the symmetry group $D_4$ of the square $[3]^2$ to reduce the number of cases one needs to consider, but we did not implement this.

A computer-free proof of this lemma can be found at
\centerline{{\tt http://michaelnielsen.org/polymath1/index.php?title=Human\_proof\_of\_the\_3D\_Pareto-optimal\_Moser\_statistics}.}
\end{proof}

\begin{remark} A similar computation revealed that the total number of line-free subsets of $[3]^3$ was $3813884$. With respect to the $2^3 \times 3!=48$-element group of geometric symmetries of $[3]^3$, these sets partitioned into $83158$ equivalence classes:
$$
3813884 = 76066 \times 48+6527 \times 24+51 \times 16+338 \times 12 +109 \times 8+41 \times 6+13 \times 4 +5 \times 3+3 \times 2+5 \times 1.
$$
\end{remark}

Lemma \ref{paretop} yields the following new inequalities:
\begin{align*}
2a+b+2c+4d &\leq 22 \\
3a+2b+3c+6d &\leq 36 \\
7a+2b+4c+8d &\leq 56 \\
6a+2b+3c+6d &\leq 48 \\
a+2c+4d &\leq 14 \\
5a+4c+8d &\leq 40.
\end{align*}

Applying Lemma \ref{prop}, we obtain new inequalities:
\begin{align}
8\alpha_r(A)+ 6\alpha_{r+q}(A) + 6\alpha_{r+2q}(A) + 2\alpha_{r+3q}(A) &\leq 11 \label{eleven}\\
4\alpha_r(A)+4\alpha_{r+q}(A)+3\alpha_{r+2q}(A)+\alpha_{r+3q}(A) &\leq 6\label{six}\\
7\alpha_r(A)+3\alpha_{r+q}(A)+3\alpha_{r+2q}(A)+\alpha_{r+3q}(A) &\leq 7\nonumber\\
8\alpha_r(A)+3\alpha_{r+q}(A)+3\alpha_{r+2q}(A)+\alpha_{r+3q}(A) &\leq 8\label{eight}\\
4\alpha_{r+q}(A)+2\alpha_{r+2q}(A)+\alpha_{r+3q}(A) &\leq 4\nonumber\\
4\alpha_r(A)+6\alpha_{r+2q}(A)+2\alpha_{r+3q}(A) &\leq 7\nonumber\\
5\alpha_r(A)+3\alpha_{r+2q}(A)+\alpha_{r+3q}(A) &\leq 5\nonumber
\end{align}
whenever $r \geq 0$, $q \geq 1$, $n \geq r+3q$, and Moser sets $A \subset [3]^n$.

We also note some further corollaries of Lemma \ref{paretop}:

\begin{corollary}[Statistics of large 3D Moser sets]\label{paretop2} Let $(a,b,c,d)$ be the statistics of a Moser set $A$ in $[3]^3$. Then $|A| = a+b+c+d \leq 16$. Furthermore:
\begin{itemize}
\item If $|A|=16$, then $(a,b,c,d) = (4,12,0,0)$.
\item If $|A|=15$, then $(a,b,c,d) = (4,11,0,0)$ or $(3,12,0,0)$.
\item If $|A| \geq 14$, then $b \geq 6$ and $d=0$.
\item If $|A| = 13$ and $d=1$, then $(a,b,c,d) = (4,6,2,1)$ or $(3,6,3,1)$.
\end{itemize}
\end{corollary}

\subsection{Four dimensions}

Now we establish the bound $c'_{4,3}=43$. Let $A$ be a Moser set in $[3]^4$, with attendant statistics $(a,b,c,d,e)$, which range between $(0,0,0,0,0)$ and $(16,32,24,8,1)$. In view of the lower bounds, our task here is to establish the upper bound $a+b+c+d+e \leq 43$.

The linear inequalities already established just barely fail to achieve this bound, but we can obtain the upper bound $a+b+c+d+e \leq 44$ as follows.
First suppose that $e=1$; then from the inequalities \eqref{alpha-1} (or by considering lines passing through $2222$) we see that $a \leq 8, b \leq 16, c \leq 12, d \leq 4$ and hence $a+b+c+d+e \leq 41$, so we may assume that $e=0$.

From Lemma \ref{dci}, we see that $a+b+c+d+e$ is now equal to the sum of $a(V)/4+b(V)/3+c(V)/2+d(V)$, where $V$ ranges over all side slices of $[3]^4$. But from Lemma \ref{paretop} we see that $a(V)/4+b(V)/3+c(V)/2+d(V)$ is at most $\frac{11}{4}$, with equality occuring only when $(a(V),b(V),c(V),d(V))=(2,6,6,0)$. This gives the upper bound $a+b+c+d+e \leq 44$.

The above argument shows that $a+b+c+d+e=44$ can only occur if $e=0$ and if $(a(V),b(V),c(V),d(V))=(2,6,6,0)$ for all side slices $V$. Applying Lemma \ref{paretop} again this implies $(a,b,c,d,e)=(4,16,24,0,0)$. But then $A$ contains all of the sphere $S_{2,4}$, which implies that the four-element set $A \cap S_{4,4}$ cannot contain a pair of strings which differ in exactly two positions (as their midpoint would then lie in $S_{2,4}$, contradicting the hypothesis that $A$ is a Moser set).

Recall that we may partition $S_{4,4} = S_{4,4}^e \cup S_{4,4}^o$, where
$$S_{4,4}^e := \{ 1111, 1133, 1313, 3113, 1331, 3131, 3311, 3333\}$$
is the strings in $S_{4,4}$ with an even number of $1$'s, and
$$S_{4,4}^o := \{ 1113, 1131, 1311, 3111, 1333, 3133, 3313, 3331\}$$
are the strings in $S_{4,4}$ with an odd number. Observe that any two distinct elements in $S_{4,4}^e$ differ in exactly two positions unless they are antipodal. Thus $A \cap S_{4,4}^e$ has size at most two, with equality only when $A \cap S_{4,4}^e$ consists of an antipodal pair. Similarly for $A \cap S_{4,4}^o$. Thus $A$ must consist of two antipodal pairs, one from $S_{4,4}^e$ and one from $S_{4,4}^o$.

By the symmetries of the cube we may assume without loss of generality that these pairs are $\{ 1111, 3333\}$ and $\{1113,3331\}$ respectively. But as $A$ is a Moser set, $A$ must now exclude the strings $1112$ and $3332$. These two strings form two corners of the eight-element set
$$ ***2 \cap S_{3,4} = \{ 1112, 1132, 1312, 3112, 1332, 3132, 3312, 3332 \}.$$
Any pair of points in this set which are ``adjacent'' in the sense that they differ by exactly one entry cannot both lie in $A$, as their midpoint would then lie in $S_{3,4}$, and so $A$ can contain at most four elements from this set, with equality only if $A$ contains all the points in $***2 \cap S_{3,4}$ of the same parity (either all the elements with an even number of $3$s, or all the elements with an odd number of $3$s). But because the two corners removed from this set have the opposite parity (one has an even number of $1$s and one has an odd number), we see in fact that $A$ can contain at most $3$ points from this set. Meanwhile, the same arguments give that $A$ contains at most four points from $**2* \cap S_{3,4}$, $*2** \cap S_{3,4}$, and $2*** \cap S_{3,4}$. Summing we see that $b = |A \cap S_{3,4}| \leq 3+4+4+4=15$, a contradiction. Thus we have $c'_{4,3}=43$ as claimed.

We have the following four-dimensional version of Lemma \ref{paretop}:

\begin{lemma}[4D Pareto-optimals]\label{paretop-4} When $n=4$, the Pareto-optimal statistics listed on Table \ref{table4}.
\end{lemma}

\begin{figure}[tb]
\centered{\tiny
$(3, 16, 24, 0, 0)$,
$(4, 14, 19, 2, 0)$,
$(4, 15, 24, 0, 0)$,
$(4, 16, 8, 4, 1)$,
$(4, 16, 14, 4, 0)$,
$(4, 16, 23, 0, 0)$,
$(4, 17, 21, 0, 0)$,
$(4, 18, 19, 0, 0)$,
$(5, 12, 12, 4, 1)$,
$(5, 12, 13, 6, 0)$,
$(5, 12, 15, 5, 0)$,
$(5, 12, 19, 2, 0)$,
$(5, 13, 10, 4, 1)$,
$(5, 13, 14, 5, 0)$,
$(5, 13, 21, 1, 0)$,
$(5, 15, 9, 4, 1)$,
$(5, 15, 12, 3, 1)$,
$(5, 15, 13, 5, 0)$,
$(5, 15, 18, 3, 0)$,
$(5, 15, 20, 1, 0)$,
$(5, 15, 22, 0, 0)$,
$(5, 16, 7, 4, 1)$,
$(5, 16, 10, 3, 1)$,
$(5, 16, 11, 5, 0)$,
$(5, 16, 12, 2, 1)$,
$(5, 16, 16, 3, 0)$,
$(5, 16, 19, 1, 0)$,
$(5, 16, 21, 0, 0)$,
$(5, 17, 12, 4, 0)$,
$(5, 17, 14, 3, 0)$,
$(5, 17, 16, 2, 0)$,
$(5, 17, 18, 1, 0)$,
$(5, 17, 20, 0, 0)$,
$(5, 18, 13, 3, 0)$,
$(5, 18, 14, 2, 0)$,
$(5, 20, 8, 4, 0)$,
$(5, 20, 10, 3, 0)$,
$(5, 20, 13, 2, 0)$,
$(5, 20, 14, 1, 0)$,
$(5, 20, 18, 0, 0)$,
$(5, 21, 10, 2, 0)$,
$(5, 21, 15, 0, 0)$,
$(5, 22, 13, 0, 0)$,
$(6, 8, 12, 8, 0)$,
$(6, 10, 11, 4, 1)$,
$(6, 11, 12, 7, 0)$,
$(6, 12, 10, 7, 0)$,
$(6, 12, 13, 5, 0)$,
$(6, 12, 18, 4, 0)$,
$(6, 13, 16, 4, 0)$,
$(6, 14, 9, 4, 1)$,
$(6, 14, 9, 7, 0)$,
$(6, 14, 12, 6, 0)$,
$(6, 14, 16, 3, 0)$,
$(6, 14, 19, 1, 0)$,
$(6, 14, 21, 0, 0)$,
$(6, 15, 7, 4, 1)$,
$(6, 15, 10, 3, 1)$,
$(6, 15, 10, 6, 0)$,
$(6, 15, 11, 2, 1)$,
$(6, 15, 12, 5, 0)$,
$(6, 15, 15, 4, 0)$,
$(6, 15, 20, 0, 0)$,
$(6, 16, 7, 3, 1)$,
$(6, 16, 8, 6, 0)$,
$(6, 16, 9, 2, 1)$,
$(6, 16, 10, 5, 0)$,
$(6, 16, 12, 1, 1)$,
$(6, 16, 13, 4, 0)$,
$(6, 16, 14, 3, 0)$,
$(6, 16, 18, 2, 0)$,
$(6, 16, 19, 0, 0)$,
$(6, 17, 9, 5, 0)$,
$(6, 17, 10, 4, 0)$,
$(6, 17, 13, 3, 0)$,
$(6, 17, 15, 2, 0)$,
$(6, 17, 17, 1, 0)$,
$(6, 17, 18, 0, 0)$,
$(6, 18, 13, 2, 0)$,
$(6, 18, 16, 1, 0)$,
$(6, 18, 17, 0, 0)$,
$(6, 19, 9, 4, 0)$,
$(6, 19, 12, 3, 0)$,
$(6, 19, 15, 1, 0)$,
$(6, 20, 7, 4, 0)$,
$(6, 20, 9, 3, 0)$,
$(6, 20, 12, 2, 0)$,
$(6, 20, 13, 1, 0)$,
$(6, 20, 15, 0, 0)$,
$(6, 21, 8, 3, 0)$,
$(6, 21, 9, 2, 0)$,
$(6, 21, 12, 1, 0)$,
$(6, 21, 14, 0, 0)$,
$(6, 22, 7, 3, 0)$,
$(6, 22, 8, 2, 0)$,
$(6, 22, 10, 1, 0)$,
$(6, 23, 9, 1, 0)$,
$(6, 24, 7, 2, 0)$,
$(6, 24, 8, 1, 0)$,
$(6, 24, 12, 0, 0)$,
$(6, 25, 9, 0, 0)$,
$(6, 26, 7, 0, 0)$,
$(7, 8, 6, 8, 0)$,
$(7, 11, 9, 4, 1)$,
$(7, 11, 12, 6, 0)$,
$(7, 12, 8, 4, 1)$,
$(7, 12, 8, 6, 0)$,
$(7, 12, 12, 3, 1)$,
$(7, 12, 12, 5, 0)$,
$(7, 12, 13, 4, 0)$,
$(7, 12, 15, 3, 0)$,
$(7, 12, 17, 2, 0)$,
$(7, 13, 7, 4, 1)$,
$(7, 13, 10, 3, 1)$,
$(7, 13, 11, 5, 0)$,
$(7, 13, 12, 2, 1)$,
$(7, 13, 12, 4, 0)$,
$(7, 13, 14, 3, 0)$,
$(7, 13, 16, 2, 0)$,
$(7, 14, 6, 4, 1)$,
$(7, 14, 6, 7, 0)$,
$(7, 14, 9, 5, 0)$,
$(7, 14, 10, 2, 1)$,
$(7, 14, 12, 1, 1)$,
$(7, 14, 17, 1, 0)$,
$(7, 14, 19, 0, 0)$,
$(7, 15, 7, 5, 0)$,
$(7, 15, 8, 3, 1)$,
$(7, 15, 9, 2, 1)$,
$(7, 15, 11, 1, 1)$,
$(7, 15, 11, 4, 0)$,
$(7, 15, 13, 3, 0)$,
$(7, 15, 16, 1, 0)$,
$(7, 16, 6, 3, 1)$,
$(7, 16, 6, 6, 0)$,
$(7, 16, 8, 2, 1)$,
$(7, 16, 10, 1, 1)$,
$(7, 16, 10, 4, 0)$,
$(7, 16, 12, 0, 1)$,
$(7, 16, 12, 3, 0)$,
$(7, 16, 15, 2, 0)$,
$(7, 16, 17, 0, 0)$,
$(7, 17, 6, 5, 0)$,
$(7, 17, 7, 4, 0)$,
$(7, 17, 11, 3, 0)$,
$(7, 17, 13, 2, 0)$,
$(7, 17, 14, 1, 0)$,
$(7, 17, 16, 0, 0)$,
$(7, 18, 10, 3, 0)$,
$(7, 18, 13, 1, 0)$,
$(7, 18, 15, 0, 0)$,
$(7, 19, 9, 3, 0)$,
$(7, 20, 6, 4, 0)$,
$(7, 20, 11, 2, 0)$,
$(7, 20, 12, 1, 0)$,
$(7, 20, 14, 0, 0)$,
$(7, 21, 8, 2, 0)$,
$(7, 21, 10, 1, 0)$,
$(7, 21, 12, 0, 0)$,
$(7, 22, 9, 1, 0)$,
$(7, 22, 11, 0, 0)$,
$(7, 23, 6, 3, 0)$,
$(7, 23, 7, 1, 0)$,
$(7, 23, 10, 0, 0)$,
$(7, 24, 6, 2, 0)$,
$(7, 24, 9, 0, 0)$,
$(7, 25, 6, 1, 0)$,
$(7, 25, 8, 0, 0)$,
$(7, 26, 3, 1, 0)$,
$(7, 28, 6, 0, 0)$,
$(7, 29, 3, 0, 0)$,
$(7, 30, 1, 0, 0)$,
$(8, 8, 0, 8, 0)$,
$(8, 8, 9, 7, 0)$,
$(8, 8, 12, 6, 0)$,
$(8, 9, 9, 4, 1)$,
$(8, 9, 10, 6, 0)$,
$(8, 9, 12, 3, 1)$,
$(8, 9, 12, 5, 0)$,
$(8, 9, 13, 4, 0)$,
$(8, 9, 15, 3, 0)$,
$(8, 10, 7, 4, 1)$,
$(8, 10, 10, 3, 1)$,
$(8, 10, 10, 5, 0)$,
$(8, 10, 12, 2, 1)$,
$(8, 10, 12, 4, 0)$,
$(8, 10, 13, 3, 0)$,
$(8, 10, 15, 2, 0)$,
$(8, 11, 6, 4, 1)$,
$(8, 11, 9, 6, 0)$,
$(8, 11, 10, 2, 1)$,
$(8, 11, 11, 4, 0)$,
$(8, 12, 7, 6, 0)$,
$(8, 12, 9, 3, 1)$,
$(8, 12, 9, 5, 0)$,
$(8, 12, 10, 4, 0)$,
$(8, 12, 12, 1, 1)$,
$(8, 12, 14, 2, 0)$,
$(8, 12, 16, 1, 0)$,
$(8, 12, 18, 0, 0)$,
$(8, 13, 7, 3, 1)$,
$(8, 13, 7, 5, 0)$,
$(8, 13, 9, 2, 1)$,
$(8, 13, 12, 0, 1)$,
$(8, 13, 12, 3, 0)$,
$(8, 14, 0, 7, 0)$,
$(8, 14, 6, 6, 0)$,
$(8, 14, 7, 2, 1)$,
$(8, 14, 8, 1, 1)$,
$(8, 14, 9, 4, 0)$,
$(8, 14, 11, 0, 1)$,
$(8, 14, 11, 3, 0)$,
$(8, 14, 13, 2, 0)$,
$(8, 14, 15, 1, 0)$,
$(8, 14, 17, 0, 0)$,
$(8, 15, 6, 3, 1)$,
$(8, 15, 6, 5, 0)$,
$(8, 15, 7, 1, 1)$,
$(8, 16, 0, 6, 0)$,
$(8, 16, 4, 3, 1)$,
$(8, 16, 4, 5, 0)$,
$(8, 16, 6, 2, 1)$,
$(8, 16, 8, 4, 0)$,
$(8, 16, 9, 0, 1)$,
$(8, 16, 10, 3, 0)$,
$(8, 16, 12, 2, 0)$,
$(8, 16, 14, 1, 0)$,
$(8, 16, 16, 0, 0)$,
$(8, 17, 0, 5, 0)$,
$(8, 17, 3, 4, 0)$,
$(8, 17, 8, 3, 0)$,
$(8, 17, 10, 2, 0)$,
$(8, 17, 12, 1, 0)$,
$(8, 17, 14, 0, 0)$,
$(8, 18, 9, 2, 0)$,
$(8, 18, 11, 1, 0)$,
$(8, 18, 12, 0, 0)$,
$(8, 19, 6, 3, 0)$,
$(8, 19, 8, 2, 0)$,
$(8, 20, 0, 4, 0)$,
$(8, 20, 4, 3, 0)$,
$(8, 20, 7, 2, 0)$,
$(8, 20, 9, 1, 0)$,
$(8, 20, 11, 0, 0)$,
$(8, 21, 4, 2, 0)$,
$(8, 21, 7, 1, 0)$,
$(8, 22, 3, 2, 0)$,
$(8, 22, 6, 1, 0)$,
$(8, 22, 9, 0, 0)$,
$(8, 23, 0, 3, 0)$,
$(8, 23, 4, 1, 0)$,
$(8, 24, 0, 2, 0)$,
$(8, 24, 3, 1, 0)$,
$(8, 24, 8, 0, 0)$,
$(8, 25, 1, 1, 0)$,
$(8, 25, 6, 0, 0)$,
$(8, 26, 0, 1, 0)$,
$(8, 26, 4, 0, 0)$,
$(8, 28, 3, 0, 0)$,
$(8, 32, 0, 0, 0)$,
$(9, 8, 10, 4, 0)$,
$(9, 9, 9, 4, 0)$,
$(9, 9, 12, 3, 0)$,
$(9, 10, 8, 4, 0)$,
$(9, 10, 10, 3, 0)$,
$(9, 10, 12, 2, 0)$,
$(9, 10, 13, 1, 0)$,
$(9, 10, 15, 0, 0)$,
$(9, 11, 11, 2, 0)$,
$(9, 12, 7, 4, 0)$,
$(9, 12, 9, 3, 0)$,
$(9, 12, 12, 1, 0)$,
$(9, 12, 14, 0, 0)$,
$(9, 13, 7, 3, 0)$,
$(9, 13, 10, 2, 0)$,
$(9, 14, 9, 2, 0)$,
$(9, 14, 11, 1, 0)$,
$(9, 14, 13, 0, 0)$,
$(9, 15, 6, 3, 0)$,
$(9, 16, 0, 4, 0)$,
$(9, 16, 4, 3, 0)$,
$(9, 16, 8, 2, 0)$,
$(9, 16, 10, 1, 0)$,
$(9, 16, 12, 0, 0)$,
$(9, 17, 3, 3, 0)$,
$(9, 17, 6, 2, 0)$,
$(9, 17, 8, 1, 0)$,
$(9, 17, 10, 0, 0)$,
$(9, 18, 2, 3, 0)$,
$(9, 18, 4, 2, 0)$,
$(9, 18, 7, 1, 0)$,
$(9, 18, 9, 0, 0)$,
$(9, 19, 0, 3, 0)$,
$(9, 19, 3, 2, 0)$,
$(9, 19, 6, 1, 0)$,
$(9, 20, 1, 2, 0)$,
$(9, 20, 5, 1, 0)$,
$(9, 20, 8, 0, 0)$,
$(9, 21, 4, 1, 0)$,
$(9, 21, 6, 0, 0)$,
$(9, 22, 1, 1, 0)$,
$(9, 22, 5, 0, 0)$,
$(9, 24, 4, 0, 0)$,
$(9, 25, 2, 0, 0)$,
$(9, 28, 0, 0, 0)$,
$(10, 8, 6, 4, 0)$,
$(10, 8, 8, 3, 0)$,
$(10, 9, 7, 3, 0)$,
$(10, 9, 10, 2, 0)$,
$(10, 9, 11, 1, 0)$,
$(10, 9, 13, 0, 0)$,
$(10, 10, 5, 4, 0)$,
$(10, 10, 9, 2, 0)$,
$(10, 10, 12, 0, 0)$,
$(10, 11, 6, 3, 0)$,
$(10, 12, 4, 4, 0)$,
$(10, 12, 5, 3, 0)$,
$(10, 12, 7, 2, 0)$,
$(10, 12, 10, 1, 0)$,
$(10, 12, 11, 0, 0)$,
$(10, 13, 6, 2, 0)$,
$(10, 13, 8, 1, 0)$,
$(10, 13, 10, 0, 0)$,
$(10, 14, 3, 3, 0)$,
$(10, 14, 5, 2, 0)$,
$(10, 14, 9, 0, 0)$,
$(10, 15, 2, 3, 0)$,
$(10, 15, 7, 1, 0)$,
$(10, 16, 4, 2, 0)$,
$(10, 16, 6, 1, 0)$,
$(10, 16, 8, 0, 0)$,
$(10, 17, 4, 1, 0)$,
$(10, 17, 6, 0, 0)$,
$(10, 18, 2, 1, 0)$,
$(10, 18, 5, 0, 0)$,
$(10, 20, 4, 0, 0)$,
$(10, 21, 2, 0, 0)$,
$(10, 22, 1, 0, 0)$,
$(10, 24, 0, 0, 0)$,
$(11, 4, 6, 4, 0)$,
$(11, 6, 5, 4, 0)$,
$(11, 7, 6, 3, 0)$,
$(11, 8, 4, 4, 0)$,
$(11, 8, 5, 3, 0)$,
$(11, 9, 6, 2, 0)$,
$(11, 9, 8, 1, 0)$,
$(11, 9, 10, 0, 0)$,
$(11, 10, 3, 3, 0)$,
$(11, 10, 5, 2, 0)$,
$(11, 10, 9, 0, 0)$,
$(11, 11, 2, 3, 0)$,
$(11, 11, 7, 1, 0)$,
$(11, 12, 4, 2, 0)$,
$(11, 12, 6, 1, 0)$,
$(11, 12, 8, 0, 0)$,
$(11, 13, 4, 1, 0)$,
$(11, 13, 6, 0, 0)$,
$(11, 14, 2, 1, 0)$,
$(11, 14, 5, 0, 0)$,
$(11, 16, 4, 0, 0)$,
$(11, 17, 2, 0, 0)$,
$(11, 18, 1, 0, 0)$,
$(11, 20, 0, 0, 0)$,
$(12, 4, 3, 3, 0)$,
$(12, 6, 2, 3, 0)$,
$(12, 6, 5, 2, 0)$,
$(12, 6, 7, 1, 0)$,
$(12, 6, 9, 0, 0)$,
$(12, 8, 4, 2, 0)$,
$(12, 8, 6, 1, 0)$,
$(12, 8, 8, 0, 0)$,
$(12, 9, 4, 1, 0)$,
$(12, 9, 6, 0, 0)$,
$(12, 10, 2, 1, 0)$,
$(12, 10, 5, 0, 0)$,
$(12, 12, 4, 0, 0)$,
$(12, 13, 2, 0, 0)$,
$(12, 14, 1, 0, 0)$,
$(12, 16, 0, 0, 0)$,
$(13, 6, 5, 0, 0)$,
$(13, 8, 4, 0, 0)$,
$(13, 9, 2, 0, 0)$,
$(13, 10, 1, 0, 0)$,
$(13, 12, 0, 0, 0)$,
$(14, 4, 3, 0, 0)$,
$(14, 5, 2, 0, 0)$,
$(14, 6, 1, 0, 0)$,
$(14, 8, 0, 0, 0)$,
$(15, 4, 0, 0, 0)$,
$(16, 0, 0, 0, 0)$}}
\caption{The Pareto-optimal statistics of Moser sets in $[3]^4$. This table can also be found at {\tt http://spreadsheets.google.com/ccc?key=rwXB\_Rn3Q1Zf5yaeMQL-RDw}}
\label{table4}
\end{figure}

\begin{proof} This was computed by computer search as follows. First, one observed that if $(a,b,c,d,e)$ was Pareto-optimal, then $a\geq 3$. To see this, it suffices to show that for any Moser set $A \subset [3]^4$ with $a(A)=0$, it is possible to add three points from $S_{4,4}$ to $A$ and still have a Moser set. To show this, suppose first that $A$ contains a point from $S_{1,4}$, such as $2221$. Then $A$ must omit either $2211$ or $2231$; without loss of generality we may assume that it omits $2211$. Similarly we may assume it omits $2121$ and $1221$. Then we can add $1131$, $1311$, $3111$ to $A$, as required. Thus we may assume that $A$ contains no points from $S_{1,4}$. Now suppose that $A$ omits a point from $S_{2,4}$, such as $2211$. Then one can add $3333$, $3111$, $1311$ to $A$, as required. Thus we may assume that A contains all of $S_{2,4}$, which forces $A$ to omit $2222$, as well as at least one point from $S_{3,4}$, such as $2111$. But then $3111$, $1111$, $3333$ can be added to the set, a contradiction.

Thus we only need to search through sets $A \subset [3]^4$ for which $|A \cap S_{4,4}| \geq 3$. A straightforward computer search shows that up to the symmetries of the cube, there are $391$ possible choices for $A \cap S_{4,4}$. For each such choice, we looped through all the possible values of the slices $A \cap 1***$ and $A \cap 3***$, i.e. all three-dimensional Moser sets which had the indicated intersection with $S_{3,3}$. (For fixed $A \cap S_{4,4}$, the number of possibilities for $A \cap 1***$ ranges from $1$ to $87123$, and similarly for $A \cap 3***$). For each pair of slices $A \cap 1***$ and $A \cap 3***$, we computed the lines connecting these two sets to see what subset of $2***$ was excluded from $A$; there are $2^{27}$ possible such exclusion sets. We precomputed a lookup table that gave the Pareto-optimal statistics for $A \cap 2***$ for each such choice of exclusion set; using this lookup table for each choice of $A \cap 1***$ and $A \cap 3***$ and collating the results, we obtained the above list. On a linux cluster, the lookup table took 22 minutes to create, and the loop over the $A \cap 1***$ and $A \cap 3***$ slices took two hours, spread out over $391$ machines (one for each choice of $A \cap S_{4,4}$). Further details (including source code) can be found at
\centerline{{\tt http://michaelnielsen.org/polymath1/index.php?title=4D\_Moser\_brute\_force\_search}.}
\end{proof}

As a consequence of this data, we have the following facts about the statistics of large Moser sets:

\begin{proposition}\label{stat} Let $A \subset [3]^4$ be a Moser set with statistics $(a,b,c,d,e)$.
\begin{itemize}
\item[(i)] If $|A| \geq 40$, then $e=0$.
\item[(ii)] If $|A| \geq 43$, then $d=0$.
\item[(iii)] If $|A| \geq 42$, then $d \leq 2$.
\item[(iv)] If $|A| \geq 41$, then $d \leq 3$.
\item[(v)] If $|A| \geq 40$, then $d \leq 6$.
\item[(vi)] If $|A| \geq 43$, then $c \geq 18$.
\item[(vii)] If $|A| \geq 42$, then $c \geq 12$.
\item[(viii)] If $|A| \geq 43$, then $b \geq 15$.
\end{itemize}
\end{proposition}

\begin{remark} This proposition was first established by an integer program, see Appendix \ref{integer-sec}. A computer-free proof can be found at \centerline{{\tt http://terrytao.files.wordpress.com/2009/06/polymath2.pdf}.}
\end{remark}

\subsection{Five dimensions}

Now we establish the bound $c'_{5,3}=124$. In view of the lower bounds, it suffices to show that there does not exist a Moser set $A \subset [3]^5$ with $|A|=125$.

We argue by contradiction. Let $A$ be as above, and let $(a(A),\ldots,f(A))$ be the statistics of $A$.

\begin{lemma}\label{fvan} $f(A)=0$.
\end{lemma}

\begin{proof} If $f(A)$ is non-zero, then $A$ contains $22222$, then each of the $\frac{3^5-1}{2} = 121$ antipodal pairs in $[3]^5$ can have at most one point in $A$, leading to only $122$ points.
\end{proof}

Let us slice $[3]^5$ into three parallel slices, e.g. $1****, 2****, 3****$. The intersection of $A$ with each of these slices has size at most $43$. In particular, this implies that
\begin{equation}\label{boo}
|A \cap 1****| + |A \cap 3****| = 125 - |A \cap 2****| \geq 82.
\end{equation}
Thus at least one of $A \cap 1****$, $A \cap 3****$ has cardinality at least $41$; by Proposition \ref{stat}(iv) we conclude that
\begin{equation}\label{d13}
\min( d(1****), d(3****) ) \leq 3.
\end{equation}
Furthermore, equality can only hold in \eqref{d13} if $A \cap 1****$, $A \cap 3****$ both have cardinality exactly $41$, in which case from Proposition \ref{stat}(iv) again we must have
\begin{equation}\label{d13a}
d(1****)=d(3****)=3.
\end{equation}
Of course, we have a similar result for permutations.

Now we improve the bound $|A \cap 2****| \leq 43$:

\begin{lemma} $|A \cap 2****| \leq 41$.
\end{lemma}

\begin{proof} Suppose first that $|A \cap 2****|=43$. Let $A' \subset [3]^4$ be the subset of $[3]^4$ corresponding to $A \cap 2****$, thus $A'$ is a Moser set of cardinality $43$. By Proposition \ref{stat}(vi), $c(A') \geq 18$. By Lemma \ref{dci}, the sum of the $c(V)$, where $V$ ranges over the eight side slices of $[3]^4$, is therefore at least $36$. By the pigeonhole principle, we may thus find two opposing side slices, say $1***$ and $3***$, with $c(1***)+c(3****) \geq 9$. Since $c(1***), c(3***)$ cannot exceed $6$, we thus have $c(1***), c(3***) \geq 3$, with at least one of $c(1***), c(3***)$ being at least $5$. Passing back to $A$, this implies that $d(*1***), d(*3***) \geq 3$, with at least one of $d(*1***), d(*3***)$ being at least $5$. But this contradicts \eqref{d13} together with the refinement \eqref{d13a}.

We have just shown that $|A \cap 2****| \leq 42$; we can thus improve \eqref{boo} to
$$ |A \cap 1****| + |A \cap 3****| \geq 83.$$
Combining this with Proposition \ref{stat}(ii)-(v) we see that
\begin{equation}\label{d13-6}
d(1****)+d(3****) \leq 6
\end{equation}
with equality only if $|A \cap 2****|=42$, and similarly for permutations.

Now let $A'$ be defined as before. Then we have
$$ c(1***) + c(3***) \leq 6$$
and similarly for permutations. Applying Lemma \ref{dci}, this implies that $c(2****) = c(A') \leq 12$.

Now suppose for contradiction that $|A'|=|A \cap 2****|=42$. Then by Proposition \ref{stat}(vii) we have
\begin{equation}\label{coo-1}
c(2****) = 12;
\end{equation}
applying Lemma \ref{dci} again, this forces $c(1***)+c(3***)=6$ and similarly for permutations, which then implies that
\begin{equation}\label{doo}
d(*1***)+d(*3***) = d(**1**)+d(**3**) = d(***1*)+d(***3*) = d(****1)+d(****3) = 6
\end{equation}
and hence
$$ |A \cap *2***| = |A \cap **2**| = |A \cap ***2*| = |A \cap ****2| = 42$$
and thus
\begin{equation}\label{coo-2}
c(*2***) = c(**2**) = c(***2*) = c(****2) = 12.
\end{equation}
Combining \eqref{coo-1}, \eqref{doo}, \eqref{coo-2} we conclude that
$$ d(1****)+d(3****) = 16,$$
contradicting \eqref{d13-6}.
\end{proof}

With this proposition, the bound \eqref{boo} now improves to
\begin{equation}\label{84}
|A \cap 1****| + |A \cap 3****| \geq 84
\end{equation}
and in particular
\begin{equation}\label{41}
|A \cap 1****|, |A \cap 3****| \geq 41.
\end{equation}
from this and Proposition \ref{stat}(ii)-(iv) we now have
\begin{equation}\label{d13-improv}
d(1****)+d(3****) \leq 4
\end{equation}
and similarly for permutations.

\begin{lemma}\label{evan} $e(A)=0$.
\end{lemma}

\begin{proof} From \eqref{84}, the intersection of $A$ with any side slice has cardinality at least $41$, and thus by Proposition \ref{stat}(i) such a side slice has an $e$-statistic of zero. The claim then follows from Lemma \ref{dci}.
\end{proof}

We need a technical lemma:

\begin{lemma}\label{tech} Let $B \subset S_{5,5}$. Then there exist at least $|B|-4$ pairs of strings in $B$ which differ in exactly two positions.
\end{lemma}

\begin{proof} The first non-vacuous case is $|B|=5$. It suffices to establish this case, as the higher cases then follow by induction (locating a pair of the desired form, then deleting one element of that pair from $B$).

Suppose for contradiction that one can find a $5$-element set $B \subset S_{5,5}$ such that no two strings in $B$ differ in exactly two positions. Recall that we may split $S_{5,5}=S_{5,5}^e \cup S_{5,5}^o$, where $S_{5,5}^e$ are those strings with an even number of $1$'s, and $S_{5,5}^o$ are those strings with an odd number of $1$'s. By the pigeonhole principle and symmetry we may assume $B$ has at least three elements in $S_{5,5}^o$. Without loss of generality, we can take one of them to be $11111$, thus excluding all elements in $S_{5,5}^o$ with exactly two $3$s, leaving only the elements with exactly four $3$s. But any two of them differ in exactly two positions, a contradiction.
\end{proof}

We can now improve the trivial bound $c(A) \leq 80$:

\begin{corollary}[Non-maximal $c$]\label{cmax} $c(A) \leq 79$. If $a(A) \geq 7$, then $c(A) \leq 78$.
\end{corollary}

\begin{proof} If $c(A)=80$, then $A$ contains all of $S_{3,5}$, which then implies that no two elements in $A \cap S_{5,5}$ can differ in exactly two places. It also implies (from \eqref{alpha-1}) that $d(A)$ must vanish, and that $b(A)$ is at most $40$. By Lemma \ref{tech}, we also have that $a(A) = |A \cap S_{5,5}|$ is at most $4$. Thus $|A| \leq 4 + 40 + 80 + 0 + 0 = 124$, a contradiction.

Now suppose that $a(A) \geq 7$. Then by Lemma \ref{tech} there are at least three pairs in $A \cap S_{5,5}$ that differ in exactly two places. Each such pair eliminates one point from $A \cap S_{3,5}$; but each point in $S_{3,5}$ can be eliminated by at most two such pairs, and so we have at least two points eliminated from $A \cap S_{3,5}$, i.e. $c(A) \leq 78$ as required.
\end{proof}

Next, we rewrite the quantity $125=|A|$ in terms of side slices. From Lemmas \ref{fvan}, \ref{evan} we have
$$ a(A) + b(A) + c(A) + d(A) = 125$$
and hence by Lemma \ref{dci}, the quantity
$$ s(V) := a(V) + \frac{5}{4} b(V) + \frac{5}{3} c(V) + \frac{5}{2} d(V) - \frac{125}{2},$$
where $V$ ranges over side slices, has an average value of zero.

\begin{proposition}[Large values of $s(V)$]\label{suv} For all side slices, we have $s(V) \leq 1/2$. Furthermore, we have $s(V) < -1/2$ unless the statistics $(a(V), b(V), c(V), d(V), e(V))$ are of one of the following four cases:
\begin{itemize}
\item (Type 1) $(a(V),b(V),c(V),d(V),e(V)) = (2,16,24,0,0)$ (and $s(V) = -1/2$ and $|A \cap V| = 42$);
\item (Type 2) $(a(V),b(V),c(V),d(V),e(V)) = (4,16,23,0,0)$ (and $s(V) = -1/6$ and $|A \cap V| = 43$);
\item (Type 3) $(a(V),b(V),c(V),d(V),e(V)) = (4,15,24,0,0)$ (and $s(V) = 1/4$ and $|A \cap V| = 43$);
\item (Type 4) $(a(V),b(V),c(V),d(V),e(V)) = (3,16,24,0,0)$ (and $s(V) = 1/2$ and $|A \cap V| = 43$);
\end{itemize}
\end{proposition}

\begin{proof}
Let $V$ be a side slice. From \eqref{41} we have
$$ 41 \leq a(V)+b(V)+c(V)+d(V) = |A \cap V| \leq 43.$$
First suppose that $|A \cap V| = 43$, then from Proposition \ref{stat}(ii), (viii), $d(V)=0$ and $b(V) \geq 15$.
Also, we have the trivial bound $c(V) \leq 24$, together with the inequality
$$ 3b(V) + 2c(V) \leq 96$$
from \eqref{alpha-1}. To exploit these facts, we rewrite $s(V)$ as
$$ s(V) = \frac{1}{2} - \frac{1}{2}( 24 - c(V) ) - \frac{1}{12} (96-3b(V)-2c(V)).$$
Thus $s(V) \leq 1/2$ in this case. If $s(V) \geq -1/2$, then
$$ 6 (24-c(V)) + (96-3b(V)-2c(V)) \leq 12,$$
which together with the inequalities $b(V) \leq 15$, $c(V) \leq 24$, $3b(V)+2c(V) \leq 96$ we conclude that $(b(V),c(V))$ must be one of $(16,24)$, $(15, 24)$, $(16, 23)$, $(15, 23)$. The first three possibilities lead to Types 4,3,2 respectively. The fourth type would lead to $(a(V),b(V),c(V),d(V),e(V)) = (5,15,23,0,0)$, but this contradicts \eqref{eleven}.

Next, suppose $|A \cap V| = 42$, so by Proposition \ref{stat}(iii) we have $d(V) \leq 2$. From \eqref{alpha-1} we have
\begin{equation}\label{2cd}
2c(V) + 3d(V) \leq 48
\end{equation}
while from \eqref{alpha-2} we have
\begin{equation}\label{3cd}
3b(V)+2c(V)+3d(V) \leq 96
\end{equation}
and so we can rewrite $s(V)$ as
\begin{equation}\label{sv2}
s(V) = -\frac{1}{2} - \frac{1}{4}( 48 - 2c(V) - 3d(V) ) - \frac{1}{12} (96-3b(V)-2c(V)-3d(V)) + \frac{1}{2} d(V).
\end{equation}
This already gives $s(V) \leq 1/2$. If $d(V)=0$, then $s(V) \leq -1/2$, with equality only in Type 1. If $d(V)=1$, then the set $A' \subset [3]^4$ corresponding to $A \cap V$ contains a point in $S_{3,4}$, which without loss of generality we can take to be $2221$. Considering the three lines $*221$, $2*21$, $22*1$, we see that at least three points in $S_{2,4}$ must be missing from $A'$, thus $c(V) \leq 21$. This forces $48-2c(V)-3d(V) \geq 3$, and so $s(V) < -3/4$. Finally, if $d(V)=2$, then $A'$ contains two points in $S_{3,4}$. If they are antipodal (e.g. $2221$ and $2223$), the same argument as above shows that at least six points in $S_{2,4}$ are missing from $A'$; if they are not antipodal (e.g. $2221$ and $2212$) then by considering the lines $*221$, $2*21$, $22*1$, $*212$, $2*12$ we see that five points are missing. Thus we have $c(V) \leq 19$, which forces $48-2c(V)-3d(V) \geq 4$. This forces $s(V) \leq -1/2$, with equality only when $c(V)=19$ and $3b(V)+2c(V)+3d(V)=96$, but this forces $b(V)$ to be the non-integer $52/3$, a contradiction, which concludes the treatment of the $|A \cap V|=42$ case.

Finally, suppose $|A \cap V| = 41$. Using \eqref{2cd}, \eqref{3cd} as before we have
\begin{equation}\label{sv3}
s(V) = -\frac{3}{2} - \frac{1}{4}( 48 - 2c(V) - 3d(V) ) - \frac{1}{12} (96-3b(V)-2c(V)-3d(V)) + \frac{1}{2} d(V),
\end{equation}
while from Proposition \ref{stat}(vi) we have $d(V) \leq 3$. This already gives $s(V) \leq 0$, and $s(V) \leq -1$ when $d(V)=1$. In order to have $s(V) \geq -1/2$, we must then have $d(V)=2$ or $d(V)=3$. But then the arguments of the preceding paragraph give $48-2c(V)-3d(V) \geq 4$, and so $s(V) \leq -1$ in this case.
\end{proof}

Since the $s(V)$ average to zero, by the pigeonhole principle we may find two opposing side slices (e.g. $1****$ and $3****$), whose total $s$-value is non-negative. Actually we can do a little better:

\begin{lemma}\label{side-off} There exists two opposing side slices whose total $s$-value is strictly positive.
\end{lemma}

\begin{proof} If this is not the case, then we must have $s(1****)+s(3****)=0$ and similarly for permutations. Using Proposition \ref{suv} we thus see that for every opposing pair of side slices, one is Type 1 and one is Type 4. In particular $c(V)=24$ for all side slices $V$. But then by Lemma \ref{dci} we have $c(A)=80$, contradicting Lemma \ref{cmax}.
\end{proof}

Let $V, V'$ be the side slices in Lemma \ref{side-off}
By Proposition \ref{suv}, the $V, V'$ slices must then be either Type 2, Type 3, or Type 4, and they cannot both be Type 2. Since $a(A) = a(V)+a(V')$, we conclude
\begin{equation}\label{amix}
6 \leq a(A) \leq 8.
\end{equation}
In a similar spirit, we have
$$ c(V) + c(V') \leq 23+24.$$
On the other hand, by considering the $24$ lines connecting $c$-points of $V, V'$ to $c$-points of the centre slice $W$ between $V$ and $V'$, each of which contains at most two points in $A$, we have
$$ c(V) + c(W) + c(V') \leq 24 \times 2.$$
Thus $c(W) \leq 1$; since
$$ d(A) = d(V) + d(V') + c(W)$$
we conclude from Proposition \ref{suv} that $d(A) \leq 1$. Actually we can do better:

\begin{lemma} $d(A)=0$.
\end{lemma}

\begin{proof} Suppose for contradiction that $d(A)=1$; without loss of generality we may take $11222 \in A$. This implies that $d(1****)=d(*1***)=1$. Also, by the above discussion, $c(**1**)$ and $c(**3**)$ cannot both be $24$, so by Proposition \ref{suv}, $s(**1**)+s(**3**) \leq 1/3$; similarly
$s(***1*)+s(***3*) \leq 1/3$ and $s(****1)+s(****3) \leq 1/3$. Since the $s$ average to zero, we see from the pigeonhole principle that either $s(1****)+s(3****) \geq -1/2$ or $s(*1***)+s(*3***) \geq -1/2$. We may assume by symmetry that
\begin{equation}\label{star-2}
s(1****)+s(3****) \geq -1/2.
\end{equation}
Since $s(3****) \leq 1/2$ by Proposition \ref{suv}, we conclude that
\begin{equation}\label{star}
s(1****) \geq -1.
\end{equation}
If $|A \cap 1****|=41$, then by \eqref{sv3} we have
$$ s(1****) = -1 - \frac{1}{4}( 48 - 2c(1****) - 3d(1****) ) - \frac{1}{12} (96-3b(1****)-2c(1****)-3d(1****))$$
but the arguments in Proposition \ref{suv} give $48 - 2c(1****) - 3d(1****) \geq 3$ and $96-3b(1****)-2c(1****)-3d(1****) \geq 0$, a contradiction. So we must have $|A \cap 1****|=42$ (by Proposition \ref{stat}(ii) and \eqref{41}). In that case, from \eqref{sv2} we have
$$ s(1****) = \frac{1}{4}( 48 - 2c(1****) - 3d(1****) ) - \frac{1}{12} (96-3b(1****)-2c(1****)-3d(1****))$$
while also having $48 - 2c(1****) - 3d(1****) \geq 3$ and $96-3b(1****)-2c(1****)-3d(1****) \geq 0$. Since $s(1****) \geq -1$ and $d(1****)=1$, we soon see that we must have $48 - 2c(1****) - 3d(1****) = 3$ and $96-3b(1****)-2c(1****)-3d(1****) \leq 3$, which forces $c(1****)=21$ and $b(1****)=16$ or $b(1****)=17$; thus the statistics of $1****$ are either $(4,16,21,1,0)$ or $(3,17,21,1,0)$.

We first eliminate the $(3,17,21,1,0)$ case. In this case $s(1****)$ is exactly $-1$. Inspecting the proof of \eqref{star}, we conclude that $s(3****)$ must be $+1/2$ and that $s(**1**)+s(**3**)=1/3$. From the former fact and Proposition \ref{suv} we see that $a(A) = a(1****)+a(3****)=3+3=6$; on the other hand, from the latter fact and Proposition \ref{suv} we have $a(A) = a(**1**)+a(**3**) = 4+3=7$, a contradiction.

So $1****$ has statistics $(4,16,21,1,0)$, which implies that $s(1****)=-3/4$ and $|A \cap 1****|=42$. By \eqref{star-2} we conclude
\begin{equation}\label{s3}
s(3****) \geq 1/4,
\end{equation}
which by Proposition \ref{suv} implies that $|A \cap 3****|=43$, and hence $|A \cap 2****|=40$. On the other hand, since $e(A)=f(A)=0$ and $d(A)=1$, with the latter being caused by $11222$, we see that $c(2****)=d(2****)=e(2****)=0$. From \eqref{alpha-1} we have $4a(2****)+b(2****) \leq 64$, and we also have the trivial inequality $b(2****) \leq 32$; these inequalities are only compatible if $2****$ has statistics $(8,32,0,0,0)$, thus $A \cap 2****$ contains $S_{2,5} \cap 2****$.

If $a(3****)=4$, then $a(A)=a(1****)+a(3****)=8$, which by Proposition \ref{suv} implies that $s(**1**)+s(**3**)$ cannot exceed $1/12$, and similarly for permutations. On the other hand, from Proposition \ref{suv} $s(**1**)+s(**3**)$ cannot exceed $-3/4 + 1/4 = -1/2$, and so the average value of $s$ cannot be zero, a contradiction. Thus $a(3****) \neq 4$, which by \eqref{s3} and Proposition \ref{suv} implies that $**3**$ has statistics $(3,16,24,0,0)$.

In particular, $A$ contains $16$ points from $3**** \cap S_{1,5}$ and all of $3**** \cap S_{2,5}$. As a consequence, no pair of the $16$ points in $A \cap 3**** \cap S_{1,5}$ can differ in only one coordinate; partitioning the $32$-point set $3**** \cap S_{1,5}$ into $16$ such pairs, we conclude that every such pair contains exactly one element of $A$. We conclude that $A \cap 3**** \cap S_{1,5}$ is equal to either $3**** \cap S_{1,5}^e$ or $3**** \cap S_{1,5}^o$.

On the other hand, $A$ contains all of $2**** \cap S_{2,5}$, and exactly sixteen points from $1**** \cap S_{1,5}$. Considering the vertical lines $*xyzw$ where $xyzw \in S_{1,4}$, we conclude that $A \cap 1**** \cap S_{1,5}$ is either equal to $1**** \cap S_{1,5}^o$ or $1**** \cap S_{1,5}^e$.
But either case is incompatible with the fact that $A$ contains $11222$ (consider either the line $11xx2$ or $11x\overline{x}2$, where $x=1,2,3$ and $\overline{x}=4-x$), obtaining the required contradiction.
\end{proof}

We can now eliminate all but three cases for the statistics of $A$:

\begin{proposition}[Statistics of $A$] The statistics $(a(A),b(A),c(A),d(A),e(A),f(A))$ of $A$ must be one of the following three tuples:
\begin{itemize}
\item (Case 1) $(6,40,79,0,0)$;
\item (Case 2) $(7,40,78,0,0)$;
\item (Case 3) $(8,39,78,0,0)$.
\end{itemize}
\end{proposition}

\begin{proof}
Since $d(A)=e(A)=f(A)=0$, we have
$$ c(2****)=d(2****)=e(2****)=0.$$
On the other hand, from \eqref{alpha-1} we have $4a(2****)+b(2****) \leq 64$ as well as the trivial inequality $b(2****) \leq 24$, and also we have
$$ |A \cap 2****| = 125 - |A \cap 1****| - |A \cap 3****| \geq 125 - 43 - 43 = 39.$$
Putting all this together, we see that the only possible statistics for $2****$ are $(8,32,0,0,0)$, $(7,32,0,0,0)$, or $(8,31,0,0,0)$. In particular, $7 \leq a(2****) \leq 8$ and $31 \leq b(2****) \leq 32$, and similarly for permutations. Applying Lemma \ref{dci} we conclude that
$$ 35 \leq b(A) \leq 40$$
and
$$ 77.5 \leq c(A) \leq 80.$$
Combining this with the first part of Corollary \ref{cmax} we conclude that $c(A)$ is either $78$ or $79$. From this and \eqref{amix} we see that the only cases that remain to be eliminated are $(7,39,79,0,0)$ and $(8,38,79,0,0)$, but these cases are incompatible with the second part of Corollary \ref{cmax}.
\end{proof}

We now eliminate each of the three remaining cases in turn.

\subsection{Elimination of $(6,40,79,0,0)$}

Here $A \cap S_{5,5}$ has six points. By Lemma \ref{tech}, there are at least two pairs in this set which differ in two positions. Their midpoints are eliminated from $A \cap S_{3,5}$. But $A$ omits exactly one point from $S_{3,5}$, so these midpoints must be the same. By symmetry, we may then assume that these two pairs are $(11111,11133)$ and $(11113,11131)$. Thus the eliminated point in $S_{3,5}$ is $11122$, i.e. $A$ contains $S_{3,5} \backslash \{11122\}$. Also, $A$ contains $\{11111,11133,11113,11131\}$ and thus must omit $\{11121, 11123, 11112, 11132\}$.

Since $11322 \in A$, at most one of $11312, 11332$ lie in $A$. By symmetry we may assume $11312 \not \in A$, thus there is a pair $(xy1z2, xy3z2)$ with $x,y,z = 1,3$ that is totally omitted from $A$, namely $(11112,11312)$. On the other hand, every other pair of this form can have at most one point in the $A$, thus there are at most seven points in $A$ of the form $xyzw2$ with $x,y,z,w = 1,3$. Similarly there are at most 8 points of the form $xyz2w$, or of $xy2zw$, $x2yzw$, $2xyzw$, leading to $b(A) \leq 7+8+8+8+8=39$, contradicting the statistic $b(A)=40$.

\subsection{Elimination of $(7,40,78,0,0)$}

Here $A \cap S_{5,5}$ has seven points. By Lemma \ref{tech}, there are at least three pairs in this set which differ in two positions. As we can only eliminate two points from $S_{3,5}$, two of the midpoints of these pairs must be the same; thus, as in the previous section, we may assume that $A$ contains $\{11111,11133,11113,11131\}$ and omits $\{11121, 11123, 11112, 11132\}$ and $11122$.

Now consider the $160$ lines $\ell$ connecting two points in $S_{4,5}$ to one point in $S_{3,5}$ (i.e. $*2xyz$ and permutations, where $x,y,z=1,3$). By double counting, the total sum of $|\ell \cap A|$ over all $160$ lines is $4b(A)+2c(A) = 316 = 158 \times 2$. On the other hand, each of these lines contain at most two points in $A$, but two of them (namely $1112*$ and $1112*$) contain no points. Thus we must have $|\ell \cap A|=2$ for the remaining $158$ lines $\ell$.

Since $A$ omits $1112x$ and $111x2$ for $x=1,3$, we thus conclude (by considering the lines $11*2x$ and $11*x2$) that $A$ must contain $1132x$, $113x2$, $1312x$, and $131x2$. Taking midpoints, we conclude that $A$ omits $11322$ and $13122$. But together with $11122$ this implies that at least three points are missing from $A \cap S_{3,5}$, contradicting the hypothesis $c(A)=78$.

\subsection{Elimination of $(8,39,78,0,0)$}

Now $A \cap S_{5,5}$ has eight points. By Lemma \ref{tech}, there are at least three pairs in this set which differ in two positions. As we can only eliminate two points from $S_{3,5}$, two of these pairs $(a,b), (c,d)$ must have the same midpoint $p$, and two other pairs $(a',b'), (c',d')$ must have the same midpoint $p'$, and $A$ contains $S_{3,5} \backslash \{p,p'\}$. As $p,p'$ are distinct, the plane containing $a,b,c,d$ is distinct from the plane containing $a',b',c',d'$.

Again consider the $160$ lines $\ell$ from the previous section. This time, the sum of the $|\ell \cap A|$ is $4b(A)+2c(A) = 312 = 156 \times 2$.
But the two lines in the plane of $a,b,c,d$ passing through $p$, and the two lines in the plane of $a',b',c',d'$ passing through $p'$, have no points; thus we must have $|\ell \cap A|=2$ for the remaining $156$ lines $\ell$.

Without loss of generality we have $(a,b)=(11111,11133)$, $(c,d) = (11113,11131)$, thus $p = 11122$. By permuting the first three indices, we may assume that $p'$ is not of the form $x2y2z, x2yz2, xy22z, xy2z2$ for any $x,y,z=1,3$. Then we have $1112x \not \in A$ and $1122x \in A$ for every $x=1,3$, so by the preceding paragraph we have $1132x \in A$; similarly for $113x2, 1312x, 131x2$. Taking midpoints, this implies that $13122, 11322 \not \in A$, but this (together with 11122) shows that at least three points aremissing from $A \cap S_{3,5}$, contradicting the hypothesis $c(A)=78$.

\subsection{Six dimensions}

Now we establish the bound $c'_{6,3}=353$. In view of the lower bounds, it suffices to show that there does not exist a Moser set $A \subset [3]^5$ with $|A|=354$.

We argue by contradiction. Let $A$ be as above, and let $(a(A),\ldots,g(A))$ be the statistics of $A$.

\begin{lemma}\label{g6} $g(A)=0$.
\end{lemma}

\begin{proof} For any four-dimensional slice $V$ of $A$, define
$$S(V) := 15 a(V) + 5 b(V) + 5 c(V)/2 + 3d(V)/2 + e(V).$$
From Lemma \ref{dci} we see that $|A|$ is equal to $a(A)+b(A)$ plus the average of $S(V)$ where $V$ ranges over the twenty slices which are some permutation of the center slice $22****$.

If $g(A)=1$, then $a(A) \leq 32$ and $b(A) \leq 96$ by \eqref{alpha-1}. Meanwhile, $e(V)=g(A)=1$ for every center slice $V$, so from Lemma \ref{paretop-4}, one can show that $S(V) \leq 223.5$ for every such slice. We conclude that $|A| \leq 351.5$, a contradiction.
\end{proof}

For any four-dimensional slice $V$ of $A$, define the \emph{defects} to be
$$ D(V) := 356 - [4a(V)+6b(V)+10c(V)+20d(V)+60e(V)].$$
Define a \emph{corner slice} to be one of the permutations or reflections of $11****$, thus there are $60$ corner slices. From Lemma \ref{dci} we see that $356-|A|+f(A)=2+f(A)$ is the average of the defects of all the $60$ corner slices. On the other hand, from Lemma \ref{paretop-4} and a straightforward computation, one concludes

\begin{lemma}\label{defects} Let $A$ be a four-dimensional Moser set. Then $D(A) \geq 0$. Furthermore:
\begin{itemize}
\item If $A$ has statistics $(6,12,18,4,0)$, then $D(A)=0$.
\item If $A$ has statistics $(5,12,18,4,0)$, $(5,12,12,4,1)$, or $(6,8,12,8,0)$, then $D(A)=4$.
\item For all other $A$, $D(A) \geq 6$.
\item If $a(A) = 4$, then $D(A) \geq 8$.
\item If $a(A) \geq 7$, then $D(A) \geq 16$.
\item If $a(A) \geq 8$, then $D(A) \geq 30$.
\item If $a(A) \geq 9$, then $D(A) \geq 86$.
\end{itemize}
\end{lemma}

Define a \emph{family} to be a set of four parallel corner slices, thus there are $15$ families, which are all a permutation of $\{11****, 13****, 31****, 33**** \}$. We refer to the family $\{11****, 13****, 31****, 33**** \}$ as $ab****$, and similarly define the family $a*b***$, etc.

\begin{lemma}\label{f6} $f(A)=0$.
\end{lemma}

\begin{proof} For any four-dimensional slice $V$ of $A$, define
$$ s(V) := 12 a(V)+15 b(V)/2+20 c(V)/3+15 d(V)/2 + 12 e(V),$$
and define an \emph{edge slice} to be one of the $30$ permutations or reflections of $12****$. From double counting we see that $|A|-a(A)$ is equal to the average of the $30$ values of $s(V)$ as $V$ ranges over edge slices.

From Lemma \ref{paretop-4} one can verify that $s(V) \leq 336$, and that $s(V) \leq 296 = 336-40$ if $e(V)=1$. The number of edge slices $V$ for which $e(V)=1$ is equal to $5f(A)$, and so the average value of the $s(V)$ is at most $336 - \frac{40 \times 5}{30} f(A)$, and so
$$ |A| - a(A) \leq 336 - \frac{40 \times 5}{30} f(A)$$
which we can rearrange (using $|A|=354$) as
$$ a(A) \geq 18 + \frac{20}{3} f(A).$$
Suppose first that $f(A)=1$; then $a(A) \geq 25$. This means that in any given family, one of the four corner slices has an $a$ value of at least $7$, and thus by Lemma \ref{defects} has a defect of at least $16$. Thus the average defect is at least $4$; on the other hand, the average defect is $2+f(A)=3$, a contradiction.

Now suppose that $f(A) \geq 2$; then $a(A) \geq 32$. Then in any given family, there is a corner slice with an $a$ value at least $9$, or four slices with $a$ value at least $8$, leading to a total defect of at least $86$ by Lemma \ref{defects}. Thus the average defect is at least $21.5$; on the other hand, the average defect is $2+f(A) \leq 2+12$, a contradiction.
\end{proof}

As a consequence of the above lemma, we see that the average defect of all corner slices is $2$, or equivalently that the total defect of these slices is $120$.

Call a corner slice \emph{good} if it has statistics $(6,12,18,4,0)$, and \emph{bad} otherwise. Thus good slices have zero defect, and bad slices have defect at least four. Since the average defect of the $60$ corner slices is $2$, there are at least $30$ good slices.

One can describe the structure of the good slices completely:

\begin{lemma}\label{sixt} The subset of $[3]^4$ consisting of the strings $1111, 1113, 3333, 1332, 1322, 1222, 3322$ and permutations is a Moser set with statistics $(6,12,18,4,0)$. Conversely, every Moser set with statistics $(6,12,18,4,0)$ is of this form up to the symmetries of the cube $[3]^4$.
\end{lemma}

\begin{proof} This can be verified by computer. By symmetry, one assumes 1222,2122,2212 and 2221 are in the set. Then 18 of the 24 'c' points with two 2s must be included; it is quick to check that 1122 and permutations must be the six excluded. Next, one checks that the only possible set of six 'a' points with no 2s is 1111,1113,3333 and permutations. Lastly, in a rather longer computation, one finds there is only possible set of twelve 'b' points, that is points with one 2. A computer-free proof can be found at
\centerline{{\tt http://michaelnielsen.org/polymath1/index.php?title=Classification\_of\_\%286\%2C12\%2C18\%2C4\%2C0\%29\_sets}.}
\end{proof}

As a consequence of this lemma, given any $x,y,z,w \in \{1,3\}$, there is a unique good Moser set in $[3]^4$ set whose intersection with $S_{1,4}$ is $\{x222, 2y22, 22z2, 222w\}$, and these are the only 16 possibilities. Call this set the \emph{good set of type $xyzw$}. It consists of
\begin{itemize}
\item The four points $x222, 2y22, 22z2, 222w$ in $S_{1,4}$;
\item All $24$ elements of $S_{2,4}$ except for $xy22, x2z2, x22w, 2yz2, 2y2w, 22zw$;
\item The twelve points $xYZ2$, $xY2W$, $x2ZW$, $XyZ2$, $Xy2W$, $2yZW$, $XYz2$, $X2zW$, $2YzW$, $XY2w$, $X2Zw$, $2YZw$ in $S_{3,4}$, where $X=4-x$, $Y=4-y$, $Z=4-z$, $W=4-w$;
\item The six points $xyzw, xyzW, xyZw, xYzw, Xyzw, XYZW$ in $S_{4,4}$.
\end{itemize}

We can use this to constrain the types of two intersecting good slices:

\begin{lemma}\label{pqs} Suppose that the $pq****$ slice is of type $xyzw$, and the $p*r***$ slice is of type $x'y'z'w'$, where $p,q,r,x,y,z,w,x',y',z',w'$ are in $\{1,3\}$. Then $x'=x$ iff $q=r$, and $y'z'w'$ is equal to either $yzw$ or $YZW$. If $x=r$ (or equivalently if $x'=q$), then $y'z'w'=yzw$.
\end{lemma}

\begin{proof} By reflection symmetry we can take $p=q=r=1$. Observe that the $11****$ slice contains $111222$ iff $x=1$, and the $1*1***$ slice similarly contains $111222$ iff $x'=1$. This shows that $x=x'$.

Suppose now that $x=x'=1$. Then the $111***$ slice contains the three elements $111y22, 1112z2, 11122w$, and excludes $111Y22, 1112Z2, 11122W$, and similarly with the primes, which forces $yzw=y'z'w'$ as claimed.

Now suppose that $x=x'=3$. Then the $111***$ slice contains the two elements $111yzw, 111YZW$, but does not contain any of the other six points in $S_{6,6} \cap 111***$, and similarly for the primes. Thus $y'z'w'$ is equal to either $yzw$ or $YZW$ as claimed.
\end{proof}

Now we look at two adjacent parallel good slices, such as $11****$ and $13****$. The following lemma asserts that such slices either have opposite type, or else will create a huge amount of defect in other slices:

\begin{lemma}\label{l18} Suppose that the $11****$ and $13****$ slices are good with types $xyzw$ and $x'y'z'w'$ respectively. If $x=x'$, then the $1*x***$ slice has defect at least $30$, and the $1*X***$ slice has defect at least $8$. Also, the $1**1**$, $1**3**$, $1***1*$, $1***3*$, $1****1$, $1****3$ slices have defect at least $6$. In particular, the total defect of slices beginning with $1*$ is at least $74$.
\end{lemma}

\begin{proof} Observe from the $11****, 13****$ hypotheses that $a(1*x***)=9$ and $a(1*X***)=4$, which gives the first two claims by Lemma \ref{defects}. For the other claims, one sees from Lemma \ref{pqs} that the other six slices cannot be good; also, they have an $a$-value of $6$ and a $d$-value of at most $7$, and the claims then follow from Lemma \ref{defects}.
\end{proof}

Now we look at two diagonally opposite parallel good slices, such as $11****$ and $33****$.

\begin{lemma}\label{l14} The $11****$ and $33****$ slices cannot both be good and of the same type.
\end{lemma}

\begin{proof} Suppose not. By symmetry we may assume that $11****$ and $33****$ are of type $1111$. This excludes a lot of points from $22****$. Indeed, by connecting lines between the $11****$ and $33****$ slices, we see that the only points that can still survive in $22****$ are $221133, 221333, 221132, 223332$, and permutations of the last four indices. Double counting the lines $22133*$ and permutations we see that there are at most $12$ points one can place in the permutations of $221133, 221333, 221132$, and so the $22****$ slice has at most $16$ points. Meanwhile, the two five-dimensional slices $1*****, 3*****$ have at most $c'_{5,3} = 124$ points, and the other two four-dimensional slices $21****, 23****$ have at most $c'_{4,3} = 43$ points, leading to at most $16 + 124 * 2 + 43 * 2 = 350$ points in all, a contradiction.
\end{proof}

\begin{lemma}\label{l19} It is not possible for all four slices in a family to be good.
\end{lemma}

\begin{proof} Suppose not. By symmetry we may assume that $11****, 13****, 31****, 33****$ are good. By Lemma \ref{l14}, the $11****$ and $33****$ slices cannot be of the same type, and so they cannot both be of the opposite type to either $13****$ or $31****$. If $13****$ is not of the opposite type to $11****$, then by (a permutation of) Lemma \ref{l18}, the total defect of slices beginning with $1*$ is at least $74$; otherwise, if $13****$ is not of the opposite type to $33****$, then by (a permutation and reflection of) Lemma \ref{l18}, the total defect of slices beginning with $*3$ is at least $74$. Similarly, the total defect of slices beginning with $3*$ or $*1$ is at least $74$, leading to a total defect of at least $148$. But the total defect of all the corner slices is $2 \times 60 = 120$, a contradiction.
\end{proof}

\begin{corollary}\label{l20} At most one family can have a total defect of at least $38$.
\end{corollary}

\begin{proof} Suppose there are two families with defect at least $38$. The remaining thirteen family have defect at least $4$ by Lemma \ref{l19} and Lemma \ref{defects}, leading to a total defect of at least $38*2+13*4=128$. But the total defect is $2 \times 60 = 120$, a contradiction.
\end{proof}

Actually we can refine this:

\begin{proposition} No family can have a total defect of at least $38$.
\end{proposition}

\begin{proof} Suppose for contradiction that the $ab****$ family (say) had a total defect of at least $38$, then by Corollary \ref{l20} no other families have total defect at least $38$.

We claim that the $**ab**$ family can have at most two good slices. Indeed, suppose the $**ab**$ has three good slices, say $**11**, **13**, **33**$. By Lemma \ref{l14}, the $**11**$ and $**33**$ slices cannot be of the same type, and so cannot both be of opposite type to $**13**$. Suppose $**11**$ and $**13**$ are not of opposite type. Then by (a permutation of) Lemma \ref{l18}, one of the families $a*b***, *ab***, **b*a*, **b**a$ has a net defect of at least $38$, contradicting the normalisation.

Thus each of the six families $**ab**, **a*b*, **a**b, ***ab*, ***a*b, ****ab$ have at least two bad slices. Meanwhile, the eight families $a*b***, a**b**, a***b*, a****b, *ab***, *a*b**, *a**b*, *a***b$ have at least one bad slice by Corollary \ref{l19}, leading to twenty bad slices in addition to the defect of at least $38$ arising from the $ab****$ slice. To add up to a total defect of $120$, we conclude from Lemma \ref{defects} that all bad slices outside of the $ab****$ family have a defect of four, with at most one exception; but then by Lemma \ref{l18} this shows that (for instance) the $1*1***$ and $1*3***$ slices cannot be good unless they are of opposite type. The previous argument then shows that the a*b*** slice cannot have three good slices, which increases the number of bad slices outside of $ab****$ to at least twenty-one, and now there is no way to add up to $120$, a contradiction.
\end{proof}

\begin{corollary} Every family can have at most two good slices.
\end{corollary}

\begin{proof} If for instance $11****, 13****, 33****$ are all good, then by Lemma \ref{l14} at least one of $11****, 33****$ is not of the opposite type to $13****$, which by Lemma \ref{l18} implies that there is a family with a total defect of at least $38$, contradicting the previous proposition.
\end{proof}

From this corollary and Lemma \ref{defects}, we see that every family has a defect of at least $8$. Since there are $15$ families, and $8 \times 15$ is exactly equal to $120$, we conclude

\begin{corollary}\label{coda} Every family has \emph{exactly} two good slices, and the remaining two slices have defect $4$. In particular, by Lemma \ref{defects}, the bad slices must have statistics $(5,12,18,4,0)$, $(5,12,12,4,1)$, or $(6,8,12,8,0)$.
\end{corollary}

We now limit how these slices can interact with good slices.

\begin{lemma}\label{goodgood} Suppose that $1*1***$ is a good slice.
\begin{itemize}
\item[(i)] The $11****$ slice cannot have statistics $(6,8,12,8,0)$.
\item[(ii)] The $11****$ slice cannot have statistics $(5,12,12,4,1)$.
\item[(iii)] If the $11****$ slice has statistics $(5,12,18,4,0)$, then the $112***$ slice has statistics $(3,9,3,0)$.
\end{itemize}
\end{lemma}

\begin{proof} This can be verified through computer search; there are $16$ possible configurations for the good slices, and one can calculate that there are $27520$ configurations for the $(5,12,12,4,1)$ slices, $4368$ configurations for the $(5,12,18,4,0)$ slices, and $80000$ configurations for the $(6,8,12,8,0)$ slices. {\bf we could put human proofs for all this somewhere, presumably.}

%We first prove (i). Suppose for contradiction that the $11****$ slice has statistics $(6,8,12,8,0)$, then $A$ contains $111222$, and so the $1*1***$ slice is of type $1xyz$ for some $x,y,z$. By symmetry we may assume it is of type $1111$, thus the $111***$ slice consists of
%$111111, 111113, 111332, 111322, 111222$ and permutations of the last three indices. On the other hand, the $11****$ slice has all eight of the points in $11**** \cap S_{2,6}$. Drawing lines between these points and $111111, 111113$ and permutations, we see that the $113***$ slice cannot contain $113111, 113113, 113133$, or permutations, leaving $113333$ as the only possible element of $113*** \cap S_{6,6}$. This makes $a(11****)=5$, a contradiction.
\end{proof}

\begin{corollary}\label{slic} The $111***$ slice has statistics $(4,3,3,1)$, $(2,6,6,0)$, $(3,3,3,1)$, or $(1,6,6,0)$.
\end{corollary}

\begin{proof} From Corollary \ref{coda}, we know that at least one of the slices $13****, 31****, 11****$ are good. If $11****$ or $1*1***$ is good, then the slice $111***$ has statistics $(4,3,3,1)$ or $(2,6,6,0)$, by Lemma \ref{sixt}. By symmetry we may thus reduce to the case where $13****$ is good and $1*1***$ is bad. Then by Lemma \ref{goodgood}, the $1*1***$ slice has statistics $(5,12,18,4,0)$ and the $121***$ slice has statistics $(3,9,3,0)$. Since the $131***$ slice, as a side slice of the good $13****$ slice, has statistics $(4,3,3,1)$ or $(2,6,6,0)$, we conclude that the $111***$ slice has statistics $(1,6,6,0)$ or $(3,3,3,1)$, and the claim follows.
\end{proof}

\begin{corollary}\label{slic2} All corner slices have statistics $(6,12,18,4,0)$ or $(5,12,18,4,0)$.
\end{corollary}

\begin{proof} Suppose first that a corner slice, say $11****$ has statistic $(6,8,12,8,0)$. Then $111***$ and $113***$ contain one ``d'' point each, and have six ``a'' points between them, so by Corollary \ref{slic}, they both have statistic $(3,3,3,1)$. This forces the $1*1***$, $1*3***$ slices to be bad, which by Corollary \ref{coda} forces the $3*1***,3*3***$ slices to be good. This forces the $311***, 313***$ slices to have statistics either $(2,6,6,0)$ or $(4,3,3,1)$. But the $311***$ slice (say) cannot have statistic $(4,3,3,1)$, since when combined with the $(3,3,3,1)$ statistics of $111***$ would give $a(*11***)=7$, which contradicts Corollary \ref{coda}; thus the $311***$ slice has statistic $(2,6,6,0)$, and similarly for $331***$. But then $a(3*1***)=4$, which again contradicts Corollary \ref{coda}.

Thus no corner slice has statistic $(6,8,12,8,0)$. Now suppose that a corner slice, say $11****$ has statistic $(5,12,12,4,1)$. By Lemma \ref{goodgood}, the $1*1***, 1*3***$ slices are bad, so by repeating the preceding arguments we conclude that the $311***, 313***$ slices have statistics $(2,6,6,0)$ or $(4,3,3,1)$; in particular, their $a$-value is even. However, the $*11***$ and $*13***$ slices are bad by Lemma \ref{goodgood}, and thus have an $a$-value of $5$; thus the $111***$ and $113***$ slices have an odd $a$-value. Thus forces $a(11****)$ to be even; but it is equal to $5$, a contradiction.
\end{proof}

From this and Lemma \ref{dci}, we see that $A$ has statistics $(22,72,180,80,0,0,0)$. In particular, we have $2\alpha_2(A)+\alpha_3(A) = 2$, which by double counting (cf. \eqref{alpha-1}) shows that for every line of the form $12222*$ (or a reflection or permutation thereof) intersects $A$ in exactly two points. Note that such lines connect a ``$d$'' point to two ``$c$'' points.

Also, we observe that two adjacent ``$d$'' points, such as $111222$ and $113222$, cannot both lie in $A$; for this would force the $*13***$ and $*11***$ slices to have statistics $(4,3,3,1)$ or $(3,3,3,1)$ by Corollary \ref{slic}, which forces $a(*1****)=6$, and thus $*1****$ must be good by Corollary \ref{slic2}; but this contradicts Lemma \ref{sixt}. Since $\alpha_3(A)=1/2$, we conclude that given any two adjacent ``$d$'' points, exactly one of them lies in $A$. In particular, the d points of the form $***222$ consist either of those strings with an even number of $1$s, or those with an odd number of $1$s.

Let's say it's the former, thus the set contains $111222, 133222$, and permutations of the first three coordinates, but omits $113222, 333222$ and permutations of the first three coordinates. Since the ``$d$'' points $113222, 333222$ are omitted, we conclude that the ``$c$'' points $113122, 113322, 333122, 333322$ must lie in the set, and similarly for permutations of the first three and last three coordinates. But this gives at least $15$ of the $16$ ``$c$'' points ending in $22$; by symmetry this leads to $225$ $c$-points in all; but $c(A)=180$, contradiction. This (finally!) completes the proof that $c'_{6,3}=353$.

Moser.tex

2009-07-22T12:04:54Z

Mpeake:

\section{Upper bounds for the $k=3$ Moser problem in small dimensions}\label{moser-upper-sec}

In this section we finish the proof of Theorem \ref{moser} by obtaining the upper bounds on $c'_{n,3}$ for $n \leq 6$.

\subsection{Statistics, densities and slices}

Our analysis will revolve around various \emph{statistics} of Moser sets $A \subset [3]^n$, their associated \emph{densities}, and the behavior of such statistics and densities with respect to the operation of passing from the cube $[3]^n$ to various \emph{slices} of that cube.

\begin{definition}[Statistics and densities] Let $A \subset [3]^n$ be a set. For any $0 \leq i \leq n$, set $a_i(A) := |A \cap S_{n-i,n}|$; thus we have
$$ 0 \leq a_i(A) \leq |S_{n-i,n}| = \binom{n}{i} 2^{n-i}$$
for $0 \leq i \leq n$ and
$$ a_0(A) + \ldots + a_n(A) = |A|.$$
We refer to the vector $(a_0(A),\ldots,a_n(A))$ as the \emph{statistics} of $A$. We define the $i^{th}$ \emph{density} $\alpha_i(A)$ to be the quantity
$$ \alpha_i(A) := \frac{a_i(A) }{\binom{n}{i} 2^{n-i}},$$
thus $0 \leq \alpha_i(A) \leq 1$ and
$$ |A| = \sum_{i=0}^n \binom{n}{i} 2^{n-i} a_i(A).$$
\end{definition}

\begin{example}\label{2mos} Let $n=2$ and $A$ be the Moser set $A := \{ 12, 13, 21, 23, 31, 32 \}$. Then the statistics $(a_0(A), a_1(A), a_2(A))$ of $A$ are $(2,4,0)$, and the densities $(\alpha_0(A), \alpha_1(A), \alpha_2(A))$ are $(\frac{1}{2}, 1, 0)$. \textbf{Include picture here? with colours?}
\end{example}

When working with small values of $n$, it will be convenient to write $a(A)$, $b(A)$, $c(A)$, etc. for $a_0(A)$, $a_1(A)$, $a_2(A)$, etc., and similarly write $\alpha(A), \beta(A), \gamma(A)$, etc. for $\alpha_0(A)$, $\alpha_1(A)$, $\alpha_2(A)$, etc. Thus for instance in Example \ref{2mos} we have $b(A) = 4$ and $\alpha(A) = \frac{1}{2}$.

\begin{definition}[Subspace statistics and densities] If $V$ is a $k$-dimensional geometric subspace of $[3]^n$, then we have a map $\phi_V: [3]^k \to [3]^n$ from the $k$-dimensional cube to the $n$-dimensional cube. If $A \subset [3]^n$ is a set and $0 \leq i \leq k$, we write $a_i(V,A)$ for $a_i(\phi_V^{-1}(A))$ and $\alpha_i(V,A)$ for $\alpha_i(\phi_V^{-1}(A))$. If the set $A$ is clear from context, we abbreviate $a_i(V,A)$ as $a_i(V)$ and $\alpha_i(V,A)$ as $\alpha_i(V)$.
\end{definition}

For our problem, a particularly important type of subspace of $[3]^n$ will be the \emph{slices} formed by fixing one coordinate and letting the other $n-1$ coordinates vary. We will denote this by a single string in which the $n-1$ varying coordinates are denoted by asterisks. For instance, in $[3]^2$, $1*$ denotes the slice $1*=\{11,12,13\}$, $*2$ denotes the slice $*2=\{12,22,32\}$, etc.; similarly, in $[3]^3$, $1**$ is the slice $\{111, 112, 113, 121, 122, 123, 131, 132, 133\}$, etc. We call a slice a \emph{centre slice} if the fixed coordinate is $2$ and a \emph{side slice} if it is $1$ or $3$.

\begin{example} We continue Example \ref{2mos}. Then the statistics of the side slice $1*$ are $(a(1*),b(1*)) = (1,1)$, while the statistics of the centre slice $2*$ are $(a(2*),b(2*))=(2,0)$. The corresponding densities are $(\alpha(1*),\beta(1*)) = (1/2,1)$ and $(\alpha(2*),\beta(2*))=(1,0)$.
\end{example}

A simple double counting argument gives the following useful identity:

\begin{lemma}[Double counting identity]\label{dci} Let $A \subset [3]^n$ and $0 \leq i \leq n-1$. Then we have
$$ \frac{1}{n-i-1} \sum_{V \hbox{ a side slice}} a_{i+1}(V) = \frac{1}{i+1} \sum_{W \hbox{ a centre slice}} a_i(W) = a_{i+1}(A)$$
where $V$ ranges over the $2n$ side slices of $[3]^n$, and $W$ ranges over the $n$ centre slices. In other words, the average value of $\alpha_{i+1}(V)$ for side slices $V$ equals the average value of $\alpha_i(W)$ for centre slices $W$, which is in turn equal to $\alpha_{i+1}(A)$.
\end{lemma}

Indeed, this lemma follows from the observation that every string in $A \cap S_{n-i-1,n}$ belongs to $i+1$ centre slices $W$ (and contributes to $a_i(W)$) and to $n-i-1$ side slices $V$ (and contributes to $a_{i+1}(V)$). One can also view this lemma probabilistically, as the assertion that there are three equivalent ways to generate a random string of length $n$:

\begin{itemize}
\item Pick a side slice $V$ at random, and randomly fill in the wildcards in such a way that $i+1$ of the wildcards are $2$'s (i.e. using an element of $S_{n-i-2,n-1}$).
\item Pick a centre slice $V$ at random, and randomly fill in the wildcards in such a way that $i$ of the wildcards are $2$'s (i.e. using an element of $S_{n-i-1,n-1}$).
\item Randomly choose an element of $S_{n-i-1,n}$.
\end{itemize}

\begin{example} We continue Example \ref{2mos}. The average value of $\beta$ for side slices is equal to the average value of $\alpha$ for centre slices, which is equal to $\beta(A) = 1$.
\end{example}

Another very useful fact (essentially due to \cite{chvatal2}) is that linear inequalities for statistics of Moser sets at one dimension propagate to linear inequalities in higher dimensions:

\begin{lemma}[Propagation lemma]\label{prop} Let $n \geq 1$ be an integer. Suppose one has a linear inequality of the form
\begin{equation}\label{alphav}
\sum_{i=0}^n v_i \alpha_i(A) \leq s
\end{equation}
for all Moser sets $A \subset [3]^n$ and some real numbers $v_0,\ldots,v_n,s$. Then we also have the linear inequality
$$ \sum_{i=0}^n v_i \alpha_{qi+r}(A) \leq s$$
whenever $q \geq 1$, $r \geq 0$, $N \geq nq+r$ are integers and $A \subset [3]^N$ is a Moser set.
\end{lemma}

\begin{proof} We run a probabilistic argument (one could of course also use a double counting argument instead). Let $n,v_0,\ldots,v_n,s,q,r,N,A$ be as in the lemma. Let $V$ be a random $n$-dimensional geometric subspace of $[3]^N$, created in the following fashion:
\begin{itemize}
\item Pick $n$ wildcards $x_1,\ldots,x_n$ to run independently from $1$ to $3$. We also introduce dual wildcards $\overline{x_1},\ldots,\overline{x_n}$; each $\overline{x_j}$ will take the value $4-x_j$.
\item We randomly subdivide the $N$ coordinates into $n$ groups of $q$ coordinates, plus a remaining group of $N-nq$ ``fixed'' coordinates.
\item For each coordinate in the $j^{th}$ group of $q$ coordinates for $1 \leq j \leq n$, we randomly assign either a $x_j$ or $\overline{x_j}$.
\item For each coordinate in the $N-nq$ fixed coordinates, we randomly assign a digit $1,2,3$, but condition on the event that exactly $r$ of the digits are equal to $2$ (i.e. we use a random element of $S_{N-nq-r,N-nq}$).
\item Let $V$ be the subspace created by allowing $x_1,\ldots,x_n$ to run independently from $1$ to $3$, and $\overline{x_j}$ to take the value $4-x_j$.
\end{itemize}

For instance, if $n=2, q=2, r=1, N=6$, then a typical subspace $V$ generated in this fashion is
$$ 2x_1\overline{x_2}3x_2x_1 = \{ 213311, 212321, 211331, 223312, 222322, 221332, 233313, 232323, 231333\}.$$
Observe from that the following two ways to generate a random element of $[3]^N$ are equivalent:

\begin{itemize}
\item Pick $V$ randomly as above, and then assign $(x_1,\ldots,x_n)$ randomly from $S_{n-i,n}$. Assign $4-x_j$ to $\overline{x_j}$ for all $1 \leq j \leq n$.
\item Pick a random string in $S_{N-qi-r,N}$.
\end{itemize}

Indeed, both random variables are invariant under the symmetries of the cube, and both random variables always pick out strings in $S_{N-qi-r,N}$, and the claim follows. As a consequence, we see that the expectation of $\alpha_i(V)$ (as $V$ ranges over the recipe described above) is equal to $\alpha_{qi+r}(A)$. On the other hand, from \eqref{alphav} we have
$$ \sum_{i=0}^n v_i \alpha_i(V) \leq s$$
for all such $V$; taking expectations over $V$, we obtain the claim.
\end{proof}

In view of Lemma \ref{prop}, it is of interest to locate linear inequalities relating the densities $\alpha_i(A)$, or (equivalently) the statistics $a_i(A)$. For this, it is convenient to introduce the following notation.

\begin{definition} Let $n \geq 1$ be an integer.
\begin{itemize}
\item A vector $(a_0,\ldots,a_n)$ of non-negative integers is \emph{feasible} if it is the statistics of some Moser set $A$.
\item A feasible vector $(a_0,\ldots,a_n)$ is \emph{Pareto-optimal} if there is no other feasible vector $(b_0,\ldots,b_n) \neq (a_0,\ldots,a_n)$ such that $b_i \geq a_i$ for all $0 \leq i \leq n$.
\item A Pareto-optimal vector $(a_0,\ldots,a_n)$ is \emph{extremal} if it is not a non-trivial convex linear combination of other Pareto-optimal vectors.
\end{itemize}
\end{definition}

To establish a linear inequality of the form \eqref{alphav} with the $v_i$ non-negative, it suffices to test the inequality against densities associated to extremal vectors of statistics. (There is no point considering linear inequalities with negative coefficients $v_i$, since one always has the freedom to reduce a density $\alpha_i(A)$ of a Moser set $A$ to zero, simply by removing all elements of $A$ with exactly $i$ $2$'s.)

We will classify exactly the Pareto-optimal and extremal vectors for $n \leq 3$, which by Lemma \ref{prop} will lead to useful linear inequalities for $n \geq 4$. Using a computer, we have also located a partial list of Pareto-optimal and extremal vectors for $n=4$, which are also useful for the $n=5$ and $n=6$ theory.

\subsection{Up to three dimensions}

We now establish Theorem \ref{moser} for $n \leq 3$, and establish some auxiliary inequalities which will be of use in higher dimensions.

The case $n=0$ is trivial. When $n=1$, it is clear that $c'_{1,3} = 2$, and furthermore that the Pareto-optimal statistics are $(2,0)$ and $(1,1)$, which are both extremal. This leads to the linear inequality
$$ 2\alpha(A) + \beta(A) \leq 2$$
for all Moser sets $A \subset [3]^1$, which by Lemma \ref{prop} implies that
\begin{equation}\label{alpha-1}
2\alpha_r(A) + \alpha_{r+q}(A) \leq 2
\end{equation}
whenever $r \geq 0$, $q \geq 1$, $n \geq q+r$, and $A \subset [3]^n$ is a Moser set.

For $n=2$, we see by partitioning $[3]^2$ into three slices that $c'_{2,3} \leq 3 c'_{1,3} = 6$, and so (by the lower bounds in the previous section) $c'_{2,3} = 6$. Writing $(a,b,c) = (a(A),b(A),c(A)) = (4\alpha(A), 4\beta(A), \gamma(A))$, the inequalities \eqref{alpha-1} become
\begin{equation}\label{abc}
a + 2c \leq 4; b+2c \leq 4; 2a+b <= 8.
\end{equation}

\begin{lemma} When $n=2$, the Pareto-optimal statistics are $(4,0,0), (3,2,0), (2,4,0), (2,2,1)$. In particular, the extremal statistics are $(4,0,0), (2,4,0), (2,2,1)$.
\end{lemma}

\begin{proof} One easily checks that all the statistics listed above are feasible.
Consider the statistics $(a,b,c)$ of a Moser set $A \subset [3]^2$. $c$ is either equal to $0$ or $1$. If $c=1$, then \eqref{abc} implies that $a,b \leq 2$, so the only Pareto-optimal statistic here is $(2,2,1)$. When instead $c=0$, the inequalities \eqref{abc} can easily imply the Pareto-optimality of $(4,0,0), (3,2,0), (2,4,0)$.
\end{proof}

From this lemma we see that we obtain a new inequality $2a+b+2c \leq 8$. Converting this back to densities and using Lemma \ref{prop}, we conclude that
\begin{equation}\label{alpha-2}
4\alpha_r(A) + 2\alpha_{r+q}(A) + \alpha_{r+2q} \leq 4
\end{equation}
whenever $r \geq 0$, $q \geq 1$, $n \geq q+2r$, and $A \subset [3]^n$ is a Moser set.

One can also check by computer that there are exactly $230$ line-free subsets of $[3]^2$.

Now we look at three dimensions. Writing $(a,b,c,d)$ for the statistics of a Moser set $A \subset [3]^n$ (which thus range between $(0,0,0,0)$ and $(8,12,6,1)$), the inequalities \eqref{alpha-1} imply in particular that
\begin{equation}\label{abc-3d}
a+4d \leq 8; b+6d \leq 12; c+3d \leq 6; 3a+2c \leq 24; b+c \leq 12
\end{equation}
while \eqref{alpha-2} implies that
\begin{equation}\label{abcd-3d}
3a+b+c \leq 24; b+c+3d \leq 12.
\end{equation}
Summing the inequalities $b+c \leq 12, 3a+b+c \leq 24, b+c+3d \leq 12$ yields
$$ 3(a+b+c+d) \leq 48$$
and hence $|A| = a+b+c+d \leq 16$; comparing this with the lower bounds of the preceding section we obtain $c'_{3,3} = 16$ as required. (This argument is essentially identical to the one in \cite{chvatal2}).

We have the following useful computation:

\begin{lemma}[3D Pareto-optimals]\label{paretop} When $n=3$, the Pareto-optimal statistics are $$(3,6,3,1),(4,4,3,1),(4,6,2,1),(2,6,6,0),(3,6,5,0),(4,4,5,0),(3,7,4,0),(4,6,4,0),$$
$$ (3,9,3,0),(4,7,3,0),(5,4,3,0),(4,9,2,0),(5,6,2,0),(6,3,2,0),(3,10,1,0),(5,7,1,0),$$
$$ (6,4,1,0),(4,12,0,0),(5,9,0,0),(6,6,0,0),(7,3,0,0),(8,0,0,0).$$
In particular, the extremal statistics are
$$(3,6,3,1),(4,4,3,1),(4,6,2,1),(2,6,6,0),(4,4,5,0),(4,6,4,0),(4,12,0,0),(8,0,0,0).$$
\end{lemma}

\begin{proof} This can be established by a brute-force search over the $2^{27} \approx 1.3 \times 10^8$ different subsets of $[3]^3$. Actually, one can perform a much faster search than this. Firstly, as noted earlier, there are only $230$ line-free subsets of $[3]^2$, so one could search over $230^3 \approx 1.2 \times 10^7$ configurations instead. Secondly, by symmetry we may assume (after enumerating the $230$ sets in a suitable fashion) that the first slice $A \cap 1**$ has an index less than or equal to the third $A \cap 3**$, leading to $\binom{231}{2} \times 230 \approx 6 \times 10^6$ configurations instead. Finally, using the first and third slice one can quickly determine which elements of the second slice $2**$ are prohibited from $A$. There are $2^9 = 512$ possible choices for the prohibited set in $2**$. By crosschecking these against the list of $230$ line-free sets one can compute the Pareto-optimal statistics for the second slices inside the prohibited set (the lists of such statistics turns out to length at most $23$). Storing these statistics in a lookup table, and then running over all choices of the first and third slice (using symmetry), one now has to perform $O( 512 \times 230 ) + O( \binom{231}{2} \times 23) \approx O( 10^6 )$ computations, which is quite a feasible computation.

One could in principle reduce the computations even further, by a factor of up to $8$, by using the symmetry group $D_4$ of the square $[3]^2$ to reduce the number of cases one needs to consider, but we did not implement this.

A computer-free proof of this lemma can be found at
\centerline{{\tt http://michaelnielsen.org/polymath1/index.php?title=Human\_proof\_of\_the\_3D\_Pareto-optimal\_Moser\_statistics}.}
\end{proof}

\begin{remark} A similar computation revealed that the total number of line-free subsets of $[3]^3$ was $3813884$. With respect to the $2^3 \times 3!=48$-element group of geometric symmetries of $[3]^3$, these sets partitioned into $83158$ equivalence classes:
$$
3813884 = 76066 \times 48+6527 \times 24+51 \times 16+338 \times 12 +109 \times 8+41 \times 6+13 \times 4 +5 \times 3+3 \times 2+5 \times 1.
$$
\end{remark}

Lemma \ref{paretop} yields the following new inequalities:
\begin{align*}
2a+b+2c+4d &\leq 22 \\
3a+2b+3c+6d &\leq 36 \\
7a+2b+4c+8d &\leq 56 \\
6a+2b+3c+6d &\leq 48 \\
a+2c+4d &\leq 14 \\
5a+4c+8d &\leq 40.
\end{align*}

Applying Lemma \ref{prop}, we obtain new inequalities:
\begin{align}
8\alpha_r(A)+ 6\alpha_{r+q}(A) + 6\alpha_{r+2q}(A) + 2\alpha_{r+3q}(A) &\leq 11 \label{eleven}\\
4\alpha_r(A)+4\alpha_{r+q}(A)+3\alpha_{r+2q}(A)+\alpha_{r+3q}(A) &\leq 6\label{six}\\
7\alpha_r(A)+3\alpha_{r+q}(A)+3\alpha_{r+2q}(A)+\alpha_{r+3q}(A) &\leq 7\nonumber\\
8\alpha_r(A)+3\alpha_{r+q}(A)+3\alpha_{r+2q}(A)+\alpha_{r+3q}(A) &\leq 8\label{eight}\\
4\alpha_{r+q}(A)+2\alpha_{r+2q}(A)+\alpha_{r+3q}(A) &\leq 4\nonumber\\
4\alpha_r(A)+6\alpha_{r+2q}(A)+2\alpha_{r+3q}(A) &\leq 7\nonumber\\
5\alpha_r(A)+3\alpha_{r+2q}(A)+\alpha_{r+3q}(A) &\leq 5\nonumber
\end{align}
whenever $r \geq 0$, $q \geq 1$, $n \geq r+3q$, and Moser sets $A \subset [3]^n$.

We also note some further corollaries of Lemma \ref{paretop}:

\begin{corollary}[Statistics of large 3D Moser sets]\label{paretop2} Let $(a,b,c,d)$ be the statistics of a Moser set $A$ in $[3]^3$. Then $|A| = a+b+c+d \leq 16$. Furthermore:
\begin{itemize}
\item If $|A|=16$, then $(a,b,c,d) = (4,12,0,0)$.
\item If $|A|=15$, then $(a,b,c,d) = (4,11,0,0)$ or $(3,12,0,0)$.
\item If $|A| \geq 14$, then $b \geq 6$ and $d=0$.
\item If $|A| = 13$ and $d=1$, then $(a,b,c,d) = (4,6,2,1)$ or $(3,6,3,1)$.
\end{itemize}
\end{corollary}

\subsection{Four dimensions}

Now we establish the bound $c'_{4,3}=43$. Let $A$ be a Moser set in $[3]^4$, with attendant statistics $(a,b,c,d,e)$, which range between $(0,0,0,0,0)$ and $(16,32,24,8,1)$. In view of the lower bounds, our task here is to establish the upper bound $a+b+c+d+e \leq 43$.

The linear inequalities already established just barely fail to achieve this bound, but we can obtain the upper bound $a+b+c+d+e \leq 44$ as follows.
First suppose that $e=1$; then from the inequalities \eqref{alpha-1} (or by considering lines passing through $2222$) we see that $a \leq 8, b \leq 16, c \leq 12, d \leq 4$ and hence $a+b+c+d+e \leq 41$, so we may assume that $e=0$.

From Lemma \ref{dci}, we see that $a+b+c+d+e$ is now equal to the sum of $a(V)/4+b(V)/3+c(V)/2+d(V)$, where $V$ ranges over all side slices of $[3]^4$. But from Lemma \ref{paretop} we see that $a(V)/4+b(V)/3+c(V)/2+d(V)$ is at most $\frac{11}{4}$, with equality occuring only when $(a(V),b(V),c(V),d(V))=(2,6,6,0)$. This gives the upper bound $a+b+c+d+e \leq 44$.

The above argument shows that $a+b+c+d+e=44$ can only occur if $e=0$ and if $(a(V),b(V),c(V),d(V))=(2,6,6,0)$ for all side slices $V$. Applying Lemma \ref{paretop} again this implies $(a,b,c,d,e)=(4,16,24,0,0)$. But then $A$ contains all of the sphere $S_{2,4}$, which implies that the four-element set $A \cap S_{4,4}$ cannot contain a pair of strings which differ in exactly two positions (as their midpoint would then lie in $S_{2,4}$, contradicting the hypothesis that $A$ is a Moser set).

Recall that we may partition $S_{4,4} = S_{4,4}^e \cup S_{4,4}^o$, where
$$S_{4,4}^e := \{ 1111, 1133, 1313, 3113, 1331, 3131, 3311, 3333\}$$
is the strings in $S_{4,4}$ with an even number of $1$'s, and
$$S_{4,4}^o := \{ 1113, 1131, 1311, 3111, 1333, 3133, 3313, 3331\}$$
are the strings in $S_{4,4}$ with an odd number. Observe that any two distinct elements in $S_{4,4}^e$ differ in exactly two positions unless they are antipodal. Thus $A \cap S_{4,4}^e$ has size at most two, with equality only when $A \cap S_{4,4}^e$ consists of an antipodal pair. Similarly for $A \cap S_{4,4}^o$. Thus $A$ must consist of two antipodal pairs, one from $S_{4,4}^e$ and one from $S_{4,4}^o$.

By the symmetries of the cube we may assume without loss of generality that these pairs are $\{ 1111, 3333\}$ and $\{1113,3331\}$ respectively. But as $A$ is a Moser set, $A$ must now exclude the strings $1112$ and $3332$. These two strings form two corners of the eight-element set
$$ ***2 \cap S_{3,4} = \{ 1112, 1132, 1312, 3112, 1332, 3132, 3312, 3332 \}.$$
Any pair of points in this set which are ``adjacent'' in the sense that they differ by exactly one entry cannot both lie in $A$, as their midpoint would then lie in $S_{3,4}$, and so $A$ can contain at most four elements from this set, with equality only if $A$ contains all the points in $***2 \cap S_{3,4}$ of the same parity (either all the elements with an even number of $3$s, or all the elements with an odd number of $3$s). But because the two corners removed from this set have the opposite parity (one has an even number of $1$s and one has an odd number), we see in fact that $A$ can contain at most $3$ points from this set. Meanwhile, the same arguments give that $A$ contains at most four points from $**2* \cap S_{3,4}$, $*2** \cap S_{3,4}$, and $2*** \cap S_{3,4}$. Summing we see that $b = |A \cap S_{3,4}| \leq 3+4+4+4=15$, a contradiction. Thus we have $c'_{4,3}=43$ as claimed.

We have the following four-dimensional version of Lemma \ref{paretop}:

\begin{lemma}[4D Pareto-optimals]\label{paretop-4} When $n=4$, the Pareto-optimal statistics listed on Table \ref{table4}.
\end{lemma}

\begin{figure}[tb]
\centered{\tiny
$(3, 16, 24, 0, 0)$,
$(4, 14, 19, 2, 0)$,
$(4, 15, 24, 0, 0)$,
$(4, 16, 8, 4, 1)$,
$(4, 16, 14, 4, 0)$,
$(4, 16, 23, 0, 0)$,
$(4, 17, 21, 0, 0)$,
$(4, 18, 19, 0, 0)$,
$(5, 12, 12, 4, 1)$,
$(5, 12, 13, 6, 0)$,
$(5, 12, 15, 5, 0)$,
$(5, 12, 19, 2, 0)$,
$(5, 13, 10, 4, 1)$,
$(5, 13, 14, 5, 0)$,
$(5, 13, 21, 1, 0)$,
$(5, 15, 9, 4, 1)$,
$(5, 15, 12, 3, 1)$,
$(5, 15, 13, 5, 0)$,
$(5, 15, 18, 3, 0)$,
$(5, 15, 20, 1, 0)$,
$(5, 15, 22, 0, 0)$,
$(5, 16, 7, 4, 1)$,
$(5, 16, 10, 3, 1)$,
$(5, 16, 11, 5, 0)$,
$(5, 16, 12, 2, 1)$,
$(5, 16, 16, 3, 0)$,
$(5, 16, 19, 1, 0)$,
$(5, 16, 21, 0, 0)$,
$(5, 17, 12, 4, 0)$,
$(5, 17, 14, 3, 0)$,
$(5, 17, 16, 2, 0)$,
$(5, 17, 18, 1, 0)$,
$(5, 17, 20, 0, 0)$,
$(5, 18, 13, 3, 0)$,
$(5, 18, 14, 2, 0)$,
$(5, 20, 8, 4, 0)$,
$(5, 20, 10, 3, 0)$,
$(5, 20, 13, 2, 0)$,
$(5, 20, 14, 1, 0)$,
$(5, 20, 18, 0, 0)$,
$(5, 21, 10, 2, 0)$,
$(5, 21, 15, 0, 0)$,
$(5, 22, 13, 0, 0)$,
$(6, 8, 12, 8, 0)$,
$(6, 10, 11, 4, 1)$,
$(6, 11, 12, 7, 0)$,
$(6, 12, 10, 7, 0)$,
$(6, 12, 13, 5, 0)$,
$(6, 12, 18, 4, 0)$,
$(6, 13, 16, 4, 0)$,
$(6, 14, 9, 4, 1)$,
$(6, 14, 9, 7, 0)$,
$(6, 14, 12, 6, 0)$,
$(6, 14, 16, 3, 0)$,
$(6, 14, 19, 1, 0)$,
$(6, 14, 21, 0, 0)$,
$(6, 15, 7, 4, 1)$,
$(6, 15, 10, 3, 1)$,
$(6, 15, 10, 6, 0)$,
$(6, 15, 11, 2, 1)$,
$(6, 15, 12, 5, 0)$,
$(6, 15, 15, 4, 0)$,
$(6, 15, 20, 0, 0)$,
$(6, 16, 7, 3, 1)$,
$(6, 16, 8, 6, 0)$,
$(6, 16, 9, 2, 1)$,
$(6, 16, 10, 5, 0)$,
$(6, 16, 12, 1, 1)$,
$(6, 16, 13, 4, 0)$,
$(6, 16, 14, 3, 0)$,
$(6, 16, 18, 2, 0)$,
$(6, 16, 19, 0, 0)$,
$(6, 17, 9, 5, 0)$,
$(6, 17, 10, 4, 0)$,
$(6, 17, 13, 3, 0)$,
$(6, 17, 15, 2, 0)$,
$(6, 17, 17, 1, 0)$,
$(6, 17, 18, 0, 0)$,
$(6, 18, 13, 2, 0)$,
$(6, 18, 16, 1, 0)$,
$(6, 18, 17, 0, 0)$,
$(6, 19, 9, 4, 0)$,
$(6, 19, 12, 3, 0)$,
$(6, 19, 15, 1, 0)$,
$(6, 20, 7, 4, 0)$,
$(6, 20, 9, 3, 0)$,
$(6, 20, 12, 2, 0)$,
$(6, 20, 13, 1, 0)$,
$(6, 20, 15, 0, 0)$,
$(6, 21, 8, 3, 0)$,
$(6, 21, 9, 2, 0)$,
$(6, 21, 12, 1, 0)$,
$(6, 21, 14, 0, 0)$,
$(6, 22, 7, 3, 0)$,
$(6, 22, 8, 2, 0)$,
$(6, 22, 10, 1, 0)$,
$(6, 23, 9, 1, 0)$,
$(6, 24, 7, 2, 0)$,
$(6, 24, 8, 1, 0)$,
$(6, 24, 12, 0, 0)$,
$(6, 25, 9, 0, 0)$,
$(6, 26, 7, 0, 0)$,
$(7, 8, 6, 8, 0)$,
$(7, 11, 9, 4, 1)$,
$(7, 11, 12, 6, 0)$,
$(7, 12, 8, 4, 1)$,
$(7, 12, 8, 6, 0)$,
$(7, 12, 12, 3, 1)$,
$(7, 12, 12, 5, 0)$,
$(7, 12, 13, 4, 0)$,
$(7, 12, 15, 3, 0)$,
$(7, 12, 17, 2, 0)$,
$(7, 13, 7, 4, 1)$,
$(7, 13, 10, 3, 1)$,
$(7, 13, 11, 5, 0)$,
$(7, 13, 12, 2, 1)$,
$(7, 13, 12, 4, 0)$,
$(7, 13, 14, 3, 0)$,
$(7, 13, 16, 2, 0)$,
$(7, 14, 6, 4, 1)$,
$(7, 14, 6, 7, 0)$,
$(7, 14, 9, 5, 0)$,
$(7, 14, 10, 2, 1)$,
$(7, 14, 12, 1, 1)$,
$(7, 14, 17, 1, 0)$,
$(7, 14, 19, 0, 0)$,
$(7, 15, 7, 5, 0)$,
$(7, 15, 8, 3, 1)$,
$(7, 15, 9, 2, 1)$,
$(7, 15, 11, 1, 1)$,
$(7, 15, 11, 4, 0)$,
$(7, 15, 13, 3, 0)$,
$(7, 15, 16, 1, 0)$,
$(7, 16, 6, 3, 1)$,
$(7, 16, 6, 6, 0)$,
$(7, 16, 8, 2, 1)$,
$(7, 16, 10, 1, 1)$,
$(7, 16, 10, 4, 0)$,
$(7, 16, 12, 0, 1)$,
$(7, 16, 12, 3, 0)$,
$(7, 16, 15, 2, 0)$,
$(7, 16, 17, 0, 0)$,
$(7, 17, 6, 5, 0)$,
$(7, 17, 7, 4, 0)$,
$(7, 17, 11, 3, 0)$,
$(7, 17, 13, 2, 0)$,
$(7, 17, 14, 1, 0)$,
$(7, 17, 16, 0, 0)$,
$(7, 18, 10, 3, 0)$,
$(7, 18, 13, 1, 0)$,
$(7, 18, 15, 0, 0)$,
$(7, 19, 9, 3, 0)$,
$(7, 20, 6, 4, 0)$,
$(7, 20, 11, 2, 0)$,
$(7, 20, 12, 1, 0)$,
$(7, 20, 14, 0, 0)$,
$(7, 21, 8, 2, 0)$,
$(7, 21, 10, 1, 0)$,
$(7, 21, 12, 0, 0)$,
$(7, 22, 9, 1, 0)$,
$(7, 22, 11, 0, 0)$,
$(7, 23, 6, 3, 0)$,
$(7, 23, 7, 1, 0)$,
$(7, 23, 10, 0, 0)$,
$(7, 24, 6, 2, 0)$,
$(7, 24, 9, 0, 0)$,
$(7, 25, 6, 1, 0)$,
$(7, 25, 8, 0, 0)$,
$(7, 26, 3, 1, 0)$,
$(7, 28, 6, 0, 0)$,
$(7, 29, 3, 0, 0)$,
$(7, 30, 1, 0, 0)$,
$(8, 8, 0, 8, 0)$,
$(8, 8, 9, 7, 0)$,
$(8, 8, 12, 6, 0)$,
$(8, 9, 9, 4, 1)$,
$(8, 9, 10, 6, 0)$,
$(8, 9, 12, 3, 1)$,
$(8, 9, 12, 5, 0)$,
$(8, 9, 13, 4, 0)$,
$(8, 9, 15, 3, 0)$,
$(8, 10, 7, 4, 1)$,
$(8, 10, 10, 3, 1)$,
$(8, 10, 10, 5, 0)$,
$(8, 10, 12, 2, 1)$,
$(8, 10, 12, 4, 0)$,
$(8, 10, 13, 3, 0)$,
$(8, 10, 15, 2, 0)$,
$(8, 11, 6, 4, 1)$,
$(8, 11, 9, 6, 0)$,
$(8, 11, 10, 2, 1)$,
$(8, 11, 11, 4, 0)$,
$(8, 12, 7, 6, 0)$,
$(8, 12, 9, 3, 1)$,
$(8, 12, 9, 5, 0)$,
$(8, 12, 10, 4, 0)$,
$(8, 12, 12, 1, 1)$,
$(8, 12, 14, 2, 0)$,
$(8, 12, 16, 1, 0)$,
$(8, 12, 18, 0, 0)$,
$(8, 13, 7, 3, 1)$,
$(8, 13, 7, 5, 0)$,
$(8, 13, 9, 2, 1)$,
$(8, 13, 12, 0, 1)$,
$(8, 13, 12, 3, 0)$,
$(8, 14, 0, 7, 0)$,
$(8, 14, 6, 6, 0)$,
$(8, 14, 7, 2, 1)$,
$(8, 14, 8, 1, 1)$,
$(8, 14, 9, 4, 0)$,
$(8, 14, 11, 0, 1)$,
$(8, 14, 11, 3, 0)$,
$(8, 14, 13, 2, 0)$,
$(8, 14, 15, 1, 0)$,
$(8, 14, 17, 0, 0)$,
$(8, 15, 6, 3, 1)$,
$(8, 15, 6, 5, 0)$,
$(8, 15, 7, 1, 1)$,
$(8, 16, 0, 6, 0)$,
$(8, 16, 4, 3, 1)$,
$(8, 16, 4, 5, 0)$,
$(8, 16, 6, 2, 1)$,
$(8, 16, 8, 4, 0)$,
$(8, 16, 9, 0, 1)$,
$(8, 16, 10, 3, 0)$,
$(8, 16, 12, 2, 0)$,
$(8, 16, 14, 1, 0)$,
$(8, 16, 16, 0, 0)$,
$(8, 17, 0, 5, 0)$,
$(8, 17, 3, 4, 0)$,
$(8, 17, 8, 3, 0)$,
$(8, 17, 10, 2, 0)$,
$(8, 17, 12, 1, 0)$,
$(8, 17, 14, 0, 0)$,
$(8, 18, 9, 2, 0)$,
$(8, 18, 11, 1, 0)$,
$(8, 18, 12, 0, 0)$,
$(8, 19, 6, 3, 0)$,
$(8, 19, 8, 2, 0)$,
$(8, 20, 0, 4, 0)$,
$(8, 20, 4, 3, 0)$,
$(8, 20, 7, 2, 0)$,
$(8, 20, 9, 1, 0)$,
$(8, 20, 11, 0, 0)$,
$(8, 21, 4, 2, 0)$,
$(8, 21, 7, 1, 0)$,
$(8, 22, 3, 2, 0)$,
$(8, 22, 6, 1, 0)$,
$(8, 22, 9, 0, 0)$,
$(8, 23, 0, 3, 0)$,
$(8, 23, 4, 1, 0)$,
$(8, 24, 0, 2, 0)$,
$(8, 24, 3, 1, 0)$,
$(8, 24, 8, 0, 0)$,
$(8, 25, 1, 1, 0)$,
$(8, 25, 6, 0, 0)$,
$(8, 26, 0, 1, 0)$,
$(8, 26, 4, 0, 0)$,
$(8, 28, 3, 0, 0)$,
$(8, 32, 0, 0, 0)$,
$(9, 8, 10, 4, 0)$,
$(9, 9, 9, 4, 0)$,
$(9, 9, 12, 3, 0)$,
$(9, 10, 8, 4, 0)$,
$(9, 10, 10, 3, 0)$,
$(9, 10, 12, 2, 0)$,
$(9, 10, 13, 1, 0)$,
$(9, 10, 15, 0, 0)$,
$(9, 11, 11, 2, 0)$,
$(9, 12, 7, 4, 0)$,
$(9, 12, 9, 3, 0)$,
$(9, 12, 12, 1, 0)$,
$(9, 12, 14, 0, 0)$,
$(9, 13, 7, 3, 0)$,
$(9, 13, 10, 2, 0)$,
$(9, 14, 9, 2, 0)$,
$(9, 14, 11, 1, 0)$,
$(9, 14, 13, 0, 0)$,
$(9, 15, 6, 3, 0)$,
$(9, 16, 0, 4, 0)$,
$(9, 16, 4, 3, 0)$,
$(9, 16, 8, 2, 0)$,
$(9, 16, 10, 1, 0)$,
$(9, 16, 12, 0, 0)$,
$(9, 17, 3, 3, 0)$,
$(9, 17, 6, 2, 0)$,
$(9, 17, 8, 1, 0)$,
$(9, 17, 10, 0, 0)$,
$(9, 18, 2, 3, 0)$,
$(9, 18, 4, 2, 0)$,
$(9, 18, 7, 1, 0)$,
$(9, 18, 9, 0, 0)$,
$(9, 19, 0, 3, 0)$,
$(9, 19, 3, 2, 0)$,
$(9, 19, 6, 1, 0)$,
$(9, 20, 1, 2, 0)$,
$(9, 20, 5, 1, 0)$,
$(9, 20, 8, 0, 0)$,
$(9, 21, 4, 1, 0)$,
$(9, 21, 6, 0, 0)$,
$(9, 22, 1, 1, 0)$,
$(9, 22, 5, 0, 0)$,
$(9, 24, 4, 0, 0)$,
$(9, 25, 2, 0, 0)$,
$(9, 28, 0, 0, 0)$,
$(10, 8, 6, 4, 0)$,
$(10, 8, 8, 3, 0)$,
$(10, 9, 7, 3, 0)$,
$(10, 9, 10, 2, 0)$,
$(10, 9, 11, 1, 0)$,
$(10, 9, 13, 0, 0)$,
$(10, 10, 5, 4, 0)$,
$(10, 10, 9, 2, 0)$,
$(10, 10, 12, 0, 0)$,
$(10, 11, 6, 3, 0)$,
$(10, 12, 4, 4, 0)$,
$(10, 12, 5, 3, 0)$,
$(10, 12, 7, 2, 0)$,
$(10, 12, 10, 1, 0)$,
$(10, 12, 11, 0, 0)$,
$(10, 13, 6, 2, 0)$,
$(10, 13, 8, 1, 0)$,
$(10, 13, 10, 0, 0)$,
$(10, 14, 3, 3, 0)$,
$(10, 14, 5, 2, 0)$,
$(10, 14, 9, 0, 0)$,
$(10, 15, 2, 3, 0)$,
$(10, 15, 7, 1, 0)$,
$(10, 16, 4, 2, 0)$,
$(10, 16, 6, 1, 0)$,
$(10, 16, 8, 0, 0)$,
$(10, 17, 4, 1, 0)$,
$(10, 17, 6, 0, 0)$,
$(10, 18, 2, 1, 0)$,
$(10, 18, 5, 0, 0)$,
$(10, 20, 4, 0, 0)$,
$(10, 21, 2, 0, 0)$,
$(10, 22, 1, 0, 0)$,
$(10, 24, 0, 0, 0)$,
$(11, 4, 6, 4, 0)$,
$(11, 6, 5, 4, 0)$,
$(11, 7, 6, 3, 0)$,
$(11, 8, 4, 4, 0)$,
$(11, 8, 5, 3, 0)$,
$(11, 9, 6, 2, 0)$,
$(11, 9, 8, 1, 0)$,
$(11, 9, 10, 0, 0)$,
$(11, 10, 3, 3, 0)$,
$(11, 10, 5, 2, 0)$,
$(11, 10, 9, 0, 0)$,
$(11, 11, 2, 3, 0)$,
$(11, 11, 7, 1, 0)$,
$(11, 12, 4, 2, 0)$,
$(11, 12, 6, 1, 0)$,
$(11, 12, 8, 0, 0)$,
$(11, 13, 4, 1, 0)$,
$(11, 13, 6, 0, 0)$,
$(11, 14, 2, 1, 0)$,
$(11, 14, 5, 0, 0)$,
$(11, 16, 4, 0, 0)$,
$(11, 17, 2, 0, 0)$,
$(11, 18, 1, 0, 0)$,
$(11, 20, 0, 0, 0)$,
$(12, 4, 3, 3, 0)$,
$(12, 6, 2, 3, 0)$,
$(12, 6, 5, 2, 0)$,
$(12, 6, 7, 1, 0)$,
$(12, 6, 9, 0, 0)$,
$(12, 8, 4, 2, 0)$,
$(12, 8, 6, 1, 0)$,
$(12, 8, 8, 0, 0)$,
$(12, 9, 4, 1, 0)$,
$(12, 9, 6, 0, 0)$,
$(12, 10, 2, 1, 0)$,
$(12, 10, 5, 0, 0)$,
$(12, 12, 4, 0, 0)$,
$(12, 13, 2, 0, 0)$,
$(12, 14, 1, 0, 0)$,
$(12, 16, 0, 0, 0)$,
$(13, 6, 5, 0, 0)$,
$(13, 8, 4, 0, 0)$,
$(13, 9, 2, 0, 0)$,
$(13, 10, 1, 0, 0)$,
$(13, 12, 0, 0, 0)$,
$(14, 4, 3, 0, 0)$,
$(14, 5, 2, 0, 0)$,
$(14, 6, 1, 0, 0)$,
$(14, 8, 0, 0, 0)$,
$(15, 4, 0, 0, 0)$,
$(16, 0, 0, 0, 0)$}}
\caption{The Pareto-optimal statistics of Moser sets in $[3]^4$. This table can also be found at {\tt http://spreadsheets.google.com/ccc?key=rwXB\_Rn3Q1Zf5yaeMQL-RDw}}
\label{table4}
\end{figure}

\begin{proof} This was computed by computer search as follows. First, one observed that if $(a,b,c,d,e)$ was Pareto-optimal, then $a\geq 3$. To see this, it suffices to show that for any Moser set $A \subset [3]^4$ with $a(A)=0$, it is possible to add three points from $S_{4,4}$ to $A$ and still have a Moser set. To show this, suppose first that $A$ contains a point from $S_{1,4}$, such as $2221$. Then $A$ must omit either $2211$ or $2231$; without loss of generality we may assume that it omits $2211$. Similarly we may assume it omits $2121$ and $1221$. Then we can add $1131$, $1311$, $3111$ to $A$, as required. Thus we may assume that $A$ contains no points from $S_{1,4}$. Now suppose that $A$ omits a point from $S_{2,4}$, such as $2211$. Then one can add $3333$, $3111$, $1311$ to $A$, as required. Thus we may assume that A contains all of $S_{2,4}$, which forces $A$ to omit $2222$, as well as at least one point from $S_{3,4}$, such as $2111$. But then $3111$, $1111$, $3333$ can be added to the set, a contradiction.

Thus we only need to search through sets $A \subset [3]^4$ for which $|A \cap S_{4,4}| \geq 3$. A straightforward computer search shows that up to the symmetries of the cube, there are $391$ possible choices for $A \cap S_{4,4}$. For each such choice, we looped through all the possible values of the slices $A \cap 1***$ and $A \cap 3***$, i.e. all three-dimensional Moser sets which had the indicated intersection with $S_{3,3}$. (For fixed $A \cap S_{4,4}$, the number of possibilities for $A \cap 1***$ ranges from $1$ to $87123$, and similarly for $A \cap 3***$). For each pair of slices $A \cap 1***$ and $A \cap 3***$, we computed the lines connecting these two sets to see what subset of $2***$ was excluded from $A$; there are $2^{27}$ possible such exclusion sets. We precomputed a lookup table that gave the Pareto-optimal statistics for $A \cap 2***$ for each such choice of exclusion set; using this lookup table for each choice of $A \cap 1***$ and $A \cap 3***$ and collating the results, we obtained the above list. On a linux cluster, the lookup table took 22 minutes to create, and the loop over the $A \cap 1***$ and $A \cap 3***$ slices took two hours, spread out over $391$ machines (one for each choice of $A \cap S_{4,4}$). Further details (including source code) can be found at
\centerline{{\tt http://michaelnielsen.org/polymath1/index.php?title=4D\_Moser\_brute\_force\_search}.}
\end{proof}

As a consequence of this data, we have the following facts about the statistics of large Moser sets:

\begin{proposition}\label{stat} Let $A \subset [3]^4$ be a Moser set with statistics $(a,b,c,d,e)$.
\begin{itemize}
\item[(i)] If $|A| \geq 40$, then $e=0$.
\item[(ii)] If $|A| \geq 43$, then $d=0$.
\item[(iii)] If $|A| \geq 42$, then $d \leq 2$.
\item[(iv)] If $|A| \geq 41$, then $d \leq 3$.
\item[(v)] If $|A| \geq 40$, then $d \leq 6$.
\item[(vi)] If $|A| \geq 43$, then $c \geq 18$.
\item[(vii)] If $|A| \geq 42$, then $c \geq 12$.
\item[(viii)] If $|A| \geq 43$, then $b \geq 15$.
\end{itemize}
\end{proposition}

\begin{remark} This proposition was first established by an integer program, see Appendix \ref{integer-sec}. A computer-free proof can be found at \centerline{{\tt http://terrytao.files.wordpress.com/2009/06/polymath2.pdf}.}
\end{remark}

\subsection{Five dimensions}

Now we establish the bound $c'_{5,3}=124$. In view of the lower bounds, it suffices to show that there does not exist a Moser set $A \subset [3]^5$ with $|A|=125$.

We argue by contradiction. Let $A$ be as above, and let $(a(A),\ldots,f(A))$ be the statistics of $A$.

\begin{lemma}\label{fvan} $f(A)=0$.
\end{lemma}

\begin{proof} If $f(A)$ is non-zero, then $A$ contains $22222$, then each of the $\frac{3^5-1}{2} = 121$ antipodal pairs in $[3]^5$ can have at most one point in $A$, leading to only $122$ points.
\end{proof}

Let us slice $[3]^5$ into three parallel slices, e.g. $1****, 2****, 3****$. The intersection of $A$ with each of these slices has size at most $43$. In particular, this implies that
\begin{equation}\label{boo}
|A \cap 1****| + |A \cap 3****| = 125 - |A \cap 2****| \geq 82.
\end{equation}
Thus at least one of $A \cap 1****$, $A \cap 3****$ has cardinality at least $41$; by Proposition \ref{stat}(iv) we conclude that
\begin{equation}\label{d13}
\min( d(1****), d(3****) ) \leq 3.
\end{equation}
Furthermore, equality can only hold in \eqref{d13} if $A \cap 1****$, $A \cap 3****$ both have cardinality exactly $41$, in which case from Proposition \ref{stat}(iv) again we must have
\begin{equation}\label{d13a}
d(1****)=d(3****)=3.
\end{equation}
Of course, we have a similar result for permutations.

Now we improve the bound $|A \cap 2****| \leq 43$:

\begin{lemma} $|A \cap 2****| \leq 41$.
\end{lemma}

\begin{proof} Suppose first that $|A \cap 2****|=43$. Let $A' \subset [3]^4$ be the subset of $[3]^4$ corresponding to $A \cap 2****$, thus $A'$ is a Moser set of cardinality $43$. By Proposition \ref{stat}(vi), $c(A') \geq 18$. By Lemma \ref{dci}, the sum of the $c(V)$, where $V$ ranges over the eight side slices of $[3]^4$, is therefore at least $36$. By the pigeonhole principle, we may thus find two opposing side slices, say $1***$ and $3***$, with $c(1***)+c(3****) \geq 9$. Since $c(1***), c(3***)$ cannot exceed $6$, we thus have $c(1***), c(3***) \geq 3$, with at least one of $c(1***), c(3***)$ being at least $5$. Passing back to $A$, this implies that $d(*1***), d(*3***) \geq 3$, with at least one of $d(*1***), d(*3***)$ being at least $5$. But this contradicts \eqref{d13} together with the refinement \eqref{d13a}.

We have just shown that $|A \cap 2****| \leq 42$; we can thus improve \eqref{boo} to
$$ |A \cap 1****| + |A \cap 3****| \geq 83.$$
Combining this with Proposition \ref{stat}(ii)-(v) we see that
\begin{equation}\label{d13-6}
d(1****)+d(3****) \leq 6
\end{equation}
with equality only if $|A \cap 2****|=42$, and similarly for permutations.

Now let $A'$ be defined as before. Then we have
$$ c(1***) + c(3***) \leq 6$$
and similarly for permutations. Applying Lemma \ref{dci}, this implies that $c(2****) = c(A') \leq 12$.

Now suppose for contradiction that $|A'|=|A \cap 2****|=42$. Then by Proposition \ref{stat}(vii) we have
\begin{equation}\label{coo-1}
c(2****) = 12;
\end{equation}
applying Lemma \ref{dci} again, this forces $c(1***)+c(3***)=6$ and similarly for permutations, which then implies that
\begin{equation}\label{doo}
d(*1***)+d(*3***) = d(**1**)+d(**3**) = d(***1*)+d(***3*) = d(****1)+d(****3) = 6
\end{equation}
and hence
$$ |A \cap *2***| = |A \cap **2**| = |A \cap ***2*| = |A \cap ****2| = 42$$
and thus
\begin{equation}\label{coo-2}
c(*2***) = c(**2**) = c(***2*) = c(****2) = 12.
\end{equation}
Combining \eqref{coo-1}, \eqref{doo}, \eqref{coo-2} we conclude that
$$ d(1****)+d(3****) = 16,$$
contradicting \eqref{d13-6}.
\end{proof}

With this proposition, the bound \eqref{boo} now improves to
\begin{equation}\label{84}
|A \cap 1****| + |A \cap 3****| \geq 84
\end{equation}
and in particular
\begin{equation}\label{41}
|A \cap 1****|, |A \cap 3****| \geq 41.
\end{equation}
from this and Proposition \ref{stat}(ii)-(iv) we now have
\begin{equation}\label{d13-improv}
d(1****)+d(3****) \leq 4
\end{equation}
and similarly for permutations.

\begin{lemma}\label{evan} $e(A)=0$.
\end{lemma}

\begin{proof} From \eqref{84}, the intersection of $A$ with any side slice has cardinality at least $41$, and thus by Proposition \ref{stat}(i) such a side slice has an $e$-statistic of zero. The claim then follows from Lemma \ref{dci}.
\end{proof}

We need a technical lemma:

\begin{lemma}\label{tech} Let $B \subset S_{5,5}$. Then there exist at least $|B|-4$ pairs of strings in $B$ which differ in exactly two positions.
\end{lemma}

\begin{proof} The first non-vacuous case is $|B|=5$. It suffices to establish this case, as the higher cases then follow by induction (locating a pair of the desired form, then deleting one element of that pair from $B$).

Suppose for contradiction that one can find a $5$-element set $B \subset S_{5,5}$ such that no two strings in $B$ differ in exactly two positions. Recall that we may split $S_{5,5}=S_{5,5}^e \cup S_{5,5}^o$, where $S_{5,5}^e$ are those strings with an even number of $1$'s, and $S_{5,5}^o$ are those strings with an odd number of $1$'s. By the pigeonhole principle and symmetry we may assume $B$ has at least three elements in $S_{5,5}^o$. Without loss of generality, we can take one of them to be $11111$, thus excluding all elements in $S_{5,5}^o$ with exactly two $3$s, leaving only the elements with exactly four $3$s. But any two of them differ in exactly two positions, a contradiction.
\end{proof}

We can now improve the trivial bound $c(A) \leq 80$:

\begin{corollary}[Non-maximal $c$]\label{cmax} $c(A) \leq 79$. If $a(A) \geq 7$, then $c(A) \leq 78$.
\end{corollary}

\begin{proof} If $c(A)=80$, then $A$ contains all of $S_{3,5}$, which then implies that no two elements in $A \cap S_{5,5}$ can differ in exactly two places. It also implies (from \eqref{alpha-1}) that $d(A)$ must vanish, and that $b(A)$ is at most $40$. By Lemma \ref{tech}, we also have that $a(A) = |A \cap S_{5,5}|$ is at most $4$. Thus $|A| \leq 4 + 40 + 80 + 0 + 0 = 124$, a contradiction.

Now suppose that $a(A) \geq 7$. Then by Lemma \ref{tech} there are at least three pairs in $A \cap S_{5,5}$ that differ in exactly two places. Each such pair eliminates one point from $A \cap S_{3,5}$; but each point in $S_{3,5}$ can be eliminated by at most two such pairs, and so we have at least two points eliminated from $A \cap S_{3,5}$, i.e. $c(A) \leq 78$ as required.
\end{proof}

Next, we rewrite the quantity $125=|A|$ in terms of side slices. From Lemmas \ref{fvan}, \ref{evan} we have
$$ a(A) + b(A) + c(A) + d(A) = 125$$
and hence by Lemma \ref{dci}, the quantity
$$ s(V) := a(V) + \frac{5}{4} b(V) + \frac{5}{3} c(V) + \frac{5}{2} d(V) - \frac{125}{2},$$
where $V$ ranges over side slices, has an average value of zero.

\begin{proposition}[Large values of $s(V)$]\label{suv} For all side slices, we have $s(V) \leq 1/2$. Furthermore, we have $s(V) < -1/2$ unless the statistics $(a(V), b(V), c(V), d(V), e(V))$ are of one of the following four cases:
\begin{itemize}
\item (Type 1) $(a(V),b(V),c(V),d(V),e(V)) = (2,16,24,0,0)$ (and $s(V) = -1/2$ and $|A \cap V| = 42$);
\item (Type 2) $(a(V),b(V),c(V),d(V),e(V)) = (4,16,23,0,0)$ (and $s(V) = -1/6$ and $|A \cap V| = 43$);
\item (Type 3) $(a(V),b(V),c(V),d(V),e(V)) = (4,15,24,0,0)$ (and $s(V) = 1/4$ and $|A \cap V| = 43$);
\item (Type 4) $(a(V),b(V),c(V),d(V),e(V)) = (3,16,24,0,0)$ (and $s(V) = 1/2$ and $|A \cap V| = 43$);
\end{itemize}
\end{proposition}

\begin{proof}
Let $V$ be a side slice. From \eqref{41} we have
$$ 41 \leq a(V)+b(V)+c(V)+d(V) = |A \cap V| \leq 43.$$
First suppose that $|A \cap V| = 43$, then from Proposition \ref{stat}(ii), (viii), $d(V)=0$ and $b(V) \geq 15$.
Also, we have the trivial bound $c(V) \leq 24$, together with the inequality
$$ 3b(V) + 2c(V) \leq 96$$
from \eqref{alpha-1}. To exploit these facts, we rewrite $s(V)$ as
$$ s(V) = \frac{1}{2} - \frac{1}{2}( 24 - c(V) ) - \frac{1}{12} (96-3b(V)-2c(V)).$$
Thus $s(V) \leq 1/2$ in this case. If $s(V) \geq -1/2$, then
$$ 6 (24-c(V)) + (96-3b(V)-2c(V)) \leq 12,$$
which together with the inequalities $b(V) \leq 15$, $c(V) \leq 24$, $3b(V)+2c(V) \leq 96$ we conclude that $(b(V),c(V))$ must be one of $(16,24)$, $(15, 24)$, $(16, 23)$, $(15, 23)$. The first three possibilities lead to Types 4,3,2 respectively. The fourth type would lead to $(a(V),b(V),c(V),d(V),e(V)) = (5,15,23,0,0)$, but this contradicts \eqref{eleven}.

Next, suppose $|A \cap V| = 42$, so by Proposition \ref{stat}(iii) we have $d(V) \leq 2$. From \eqref{alpha-1} we have
\begin{equation}\label{2cd}
2c(V) + 3d(V) \leq 48
\end{equation}
while from \eqref{alpha-2} we have
\begin{equation}\label{3cd}
3b(V)+2c(V)+3d(V) \leq 96
\end{equation}
and so we can rewrite $s(V)$ as
\begin{equation}\label{sv2}
s(V) = -\frac{1}{2} - \frac{1}{4}( 48 - 2c(V) - 3d(V) ) - \frac{1}{12} (96-3b(V)-2c(V)-3d(V)) + \frac{1}{2} d(V).
\end{equation}
This already gives $s(V) \leq 1/2$. If $d(V)=0$, then $s(V) \leq -1/2$, with equality only in Type 1. If $d(V)=1$, then the set $A' \subset [3]^4$ corresponding to $A \cap V$ contains a point in $S_{3,4}$, which without loss of generality we can take to be $2221$. Considering the three lines $*221$, $2*21$, $22*1$, we see that at least three points in $S_{2,4}$ must be missing from $A'$, thus $c(V) \leq 21$. This forces $48-2c(V)-3d(V) \geq 3$, and so $s(V) < -3/4$. Finally, if $d(V)=2$, then $A'$ contains two points in $S_{3,4}$. If they are antipodal (e.g. $2221$ and $2223$), the same argument as above shows that at least six points in $S_{2,4}$ are missing from $A'$; if they are not antipodal (e.g. $2221$ and $2212$) then by considering the lines $*221$, $2*21$, $22*1$, $*212$, $2*12$ we see that five points are missing. Thus we have $c(V) \leq 19$, which forces $48-2c(V)-3d(V) \geq 4$. This forces $s(V) \leq -1/2$, with equality only when $c(V)=19$ and $3b(V)+2c(V)+3d(V)=96$, but this forces $b(V)$ to be the non-integer $52/3$, a contradiction, which concludes the treatment of the $|A \cap V|=42$ case.

Finally, suppose $|A \cap V| = 41$. Using \eqref{2cd}, \eqref{3cd} as before we have
\begin{equation}\label{sv3}
s(V) = -\frac{3}{2} - \frac{1}{4}( 48 - 2c(V) - 3d(V) ) - \frac{1}{12} (96-3b(V)-2c(V)-3d(V)) + \frac{1}{2} d(V),
\end{equation}
while from Proposition \ref{stat}(vi) we have $d(V) \leq 3$. This already gives $s(V) \leq 0$, and $s(V) \leq -1$ when $d(V)=1$. In order to have $s(V) \geq -1/2$, we must then have $d(V)=2$ or $d(V)=3$. But then the arguments of the preceding paragraph give $48-2c(V)-3d(V) \geq 4$, and so $s(V) \leq -1$ in this case.
\end{proof}

Since the $s(V)$ average to zero, by the pigeonhole principle we may find two opposing side slices (e.g. $1****$ and $3****$), whose total $s$-value is non-negative. Actually we can do a little better:

\begin{lemma}\label{side-off} There exists two opposing side slices whose total $s$-value is strictly positive.
\end{lemma}

\begin{proof} If this is not the case, then we must have $s(1****)+s(3****)=0$ and similarly for permutations. Using Proposition \ref{suv} we thus see that for every opposing pair of side slices, one is Type 1 and one is Type 4. In particular $c(V)=24$ for all side slices $V$. But then by Lemma \ref{dci} we have $c(A)=80$, contradicting Lemma \ref{cmax}.
\end{proof}

Let $V, V'$ be the side slices in Lemma \ref{side-off}
By Proposition \ref{suv}, the $V, V'$ slices must then be either Type 2, Type 3, or Type 4, and they cannot both be Type 2. Since $a(A) = a(V)+a(V')$, we conclude
\begin{equation}\label{amix}
6 \leq a(A) \leq 8.
\end{equation}
In a similar spirit, we have
$$ c(V) + c(V') \leq 23+24.$$
On the other hand, by considering the $24$ lines connecting $c$-points of $V, V'$ to $c$-points of the centre slice $W$ between $V$ and $V'$, each of which contains at most two points in $A$, we have
$$ c(V) + c(W) + c(V') \leq 24 \times 2.$$
Thus $c(W) \leq 1$; since
$$ d(A) = d(V) + d(V') + c(W)$$
we conclude from Proposition \ref{suv} that $d(A) \leq 1$. Actually we can do better:

\begin{lemma} $d(A)=0$.
\end{lemma}

\begin{proof} Suppose for contradiction that $d(A)=1$; without loss of generality we may take $11222 \in A$. This implies that $d(1****)=d(*1***)=1$. Also, by the above discussion, $c(**1**)$ and $c(**3**)$ cannot both be $24$, so by Proposition \ref{suv}, $s(**1**)+s(**3**) \leq 1/3$; similarly
$s(***1*)+s(***3*) \leq 1/3$ and $s(****1)+s(****3) \leq 1/3$. Since the $s$ average to zero, we see from the pigeonhole principle that either $s(1****)+s(3****) \geq -1/2$ or $s(*1***)+s(*3***) \geq -1/2$. We may assume by symmetry that
\begin{equation}\label{star-2}
s(1****)+s(3****) \geq -1/2.
\end{equation}
Since $s(3****) \leq 1/2$ by Proposition \ref{suv}, we conclude that
\begin{equation}\label{star}
s(1****) \geq -1.
\end{equation}
If $|A \cap 1****|=41$, then by \eqref{sv3} we have
$$ s(1****) = -1 - \frac{1}{4}( 48 - 2c(1****) - 3d(1****) ) - \frac{1}{12} (96-3b(1****)-2c(1****)-3d(1****))$$
but the arguments in Proposition \ref{suv} give $48 - 2c(1****) - 3d(1****) \geq 3$ and $96-3b(1****)-2c(1****)-3d(1****) \geq 0$, a contradiction. So we must have $|A \cap 1****|=42$ (by Proposition \ref{stat}(ii) and \eqref{41}). In that case, from \eqref{sv2} we have
$$ s(1****) = \frac{1}{4}( 48 - 2c(1****) - 3d(1****) ) - \frac{1}{12} (96-3b(1****)-2c(1****)-3d(1****))$$
while also having $48 - 2c(1****) - 3d(1****) \geq 3$ and $96-3b(1****)-2c(1****)-3d(1****) \geq 0$. Since $s(1****) \geq -1$ and $d(1****)=1$, we soon see that we must have $48 - 2c(1****) - 3d(1****) = 3$ and $96-3b(1****)-2c(1****)-3d(1****) \leq 3$, which forces $c(1****)=21$ and $b(1****)=16$ or $b(1****)=17$; thus the statistics of $1****$ are either $(4,16,21,1,0)$ or $(3,17,21,1,0)$.

We first eliminate the $(3,17,21,1,0)$ case. In this case $s(1****)$ is exactly $-1$. Inspecting the proof of \eqref{star}, we conclude that $s(3****)$ must be $+1/2$ and that $s(**1**)+s(**3**)=1/3$. From the former fact and Proposition \ref{suv} we see that $a(A) = a(1****)+a(3****)=3+3=6$; on the other hand, from the latter fact and Proposition \ref{suv} we have $a(A) = a(**1**)+a(**3**) = 4+3=7$, a contradiction.

So $1****$ has statistics $(4,16,21,1,0)$, which implies that $s(1****)=-3/4$ and $|A \cap 1****|=42$. By \eqref{star-2} we conclude
\begin{equation}\label{s3}
s(3****) \geq 1/4,
\end{equation}
which by Proposition \ref{suv} implies that $|A \cap 3****|=43$, and hence $|A \cap 2****|=40$. On the other hand, since $e(A)=f(A)=0$ and $d(A)=1$, with the latter being caused by $11222$, we see that $c(2****)=d(2****)=e(2****)=0$. From \eqref{alpha-1} we have $4a(2****)+b(2****) \leq 64$, and we also have the trivial inequality $b(2****) \leq 32$; these inequalities are only compatible if $2****$ has statistics $(8,32,0,0,0)$, thus $A \cap 2****$ contains $S_{2,5} \cap 2****$.

If $a(3****)=4$, then $a(A)=a(1****)+a(3****)=8$, which by Proposition \ref{suv} implies that $s(**1**)+s(**3**)$ cannot exceed $1/12$, and similarly for permutations. On the other hand, from Proposition \ref{suv} $s(**1**)+s(**3**)$ cannot exceed $-3/4 + 1/4 = -1/2$, and so the average value of $s$ cannot be zero, a contradiction. Thus $a(3****) \neq 4$, which by \eqref{s3} and Proposition \ref{suv} implies that $**3**$ has statistics $(3,16,24,0,0)$.

In particular, $A$ contains $16$ points from $3**** \cap S_{1,5}$ and all of $3**** \cap S_{2,5}$. As a consequence, no pair of the $16$ points in $A \cap 3**** \cap S_{1,5}$ can differ in only one coordinate; partitioning the $32$-point set $3**** \cap S_{1,5}$ into $16$ such pairs, we conclude that every such pair contains exactly one element of $A$. We conclude that $A \cap 3**** \cap S_{1,5}$ is equal to either $3**** \cap S_{1,5}^e$ or $3**** \cap S_{1,5}^o$.

On the other hand, $A$ contains all of $2**** \cap S_{2,5}$, and exactly sixteen points from $1**** \cap S_{1,5}$. Considering the vertical lines $*xyzw$ where $xyzw \in S_{1,4}$, we conclude that $A \cap 1**** \cap S_{1,5}$ is either equal to $1**** \cap S_{1,5}^o$ or $1**** \cap S_{1,5}^e$.
But either case is incompatible with the fact that $A$ contains $11222$ (consider either the line $11xx2$ or $11x\overline{x}2$, where $x=1,2,3$ and $\overline{x}=4-x$), obtaining the required contradiction.
\end{proof}

We can now eliminate all but three cases for the statistics of $A$:

\begin{proposition}[Statistics of $A$] The statistics $(a(A),b(A),c(A),d(A),e(A),f(A))$ of $A$ must be one of the following three tuples:
\begin{itemize}
\item (Case 1) $(6,40,79,0,0)$;
\item (Case 2) $(7,40,78,0,0)$;
\item (Case 3) $(8,39,78,0,0)$.
\end{itemize}
\end{proposition}

\begin{proof}
Since $d(A)=e(A)=f(A)=0$, we have
$$ c(2****)=d(2****)=e(2****)=0.$$
On the other hand, from \eqref{alpha-1} we have $4a(2****)+b(2****) \leq 64$ as well as the trivial inequality $b(2****) \leq 24$, and also we have
$$ |A \cap 2****| = 125 - |A \cap 1****| - |A \cap 3****| \geq 125 - 43 - 43 = 39.$$
Putting all this together, we see that the only possible statistics for $2****$ are $(8,32,0,0,0)$, $(7,32,0,0,0)$, or $(8,31,0,0,0)$. In particular, $7 \leq a(2****) \leq 8$ and $31 \leq b(2****) \leq 32$, and similarly for permutations. Applying Lemma \ref{dci} we conclude that
$$ 35 \leq b(A) \leq 40$$
and
$$ 77.5 \leq c(A) \leq 80.$$
Combining this with the first part of Corollary \ref{cmax} we conclude that $c(A)$ is either $78$ or $79$. From this and \eqref{amix} we see that the only cases that remain to be eliminated are $(7,39,79,0,0)$ and $(8,38,79,0,0)$, but these cases are incompatible with the second part of Corollary \ref{cmax}.
\end{proof}

We now eliminate each of the three remaining cases in turn.

\subsection{Elimination of $(6,40,79,0,0)$}

Here $A \cap S_{5,5}$ has six points. By Lemma \ref{tech}, there are at least two pairs in this set which differ in two positions. Their midpoints are eliminated from $A \cap S_{3,5}$. But $A$ omits exactly one point from $S_{3,5}$, so these midpoints must be the same. By symmetry, we may then assume that these two pairs are $(11111,11133)$ and $(11113,11131)$. Thus the eliminated point in $S_{3,5}$ is $11122$, i.e. $A$ contains $S_{3,5} \backslash \{11122\}$. Also, $A$ contains $\{11111,11133,11113,11131\}$ and thus must omit $\{11121, 11123, 11112, 11132\}$.

Since $11322 \in A$, at most one of $11312, 11332$ lie in $A$. By symmetry we may assume $11312 \not \in A$, thus there is a pair $(xy1z2, xy3z2)$ with $x,y,z = 1,3$ that is totally omitted from $A$, namely $(11112,11312)$. On the other hand, every other pair of this form can have at most one point in the $A$, thus there are at most seven points in $A$ of the form $xyzw2$ with $x,y,z,w = 1,3$. Similarly there are at most 8 points of the form $xyz2w$, or of $xy2zw$, $x2yzw$, $2xyzw$, leading to $b(A) \leq 7+8+8+8+8=39$, contradicting the statistic $b(A)=40$.

\subsection{Elimination of $(7,40,78,0,0)$}

Here $A \cap S_{5,5}$ has seven points. By Lemma \ref{tech}, there are at least three pairs in this set which differ in two positions. As we can only eliminate two points from $S_{3,5}$, two of the midpoints of these pairs must be the same; thus, as in the previous section, we may assume that $A$ contains $\{11111,11133,11113,11131\}$ and omits $\{11121, 11123, 11112, 11132\}$ and $11122$.

Now consider the $160$ lines $\ell$ connecting two points in $S_{4,5}$ to one point in $S_{3,5}$ (i.e. $*2xyz$ and permutations, where $x,y,z=1,3$). By double counting, the total sum of $|\ell \cap A|$ over all $160$ lines is $4b(A)+2c(A) = 316 = 158 \times 2$. On the other hand, each of these lines contain at most two points in $A$, but two of them (namely $1112*$ and $1112*$) contain no points. Thus we must have $|\ell \cap A|=2$ for the remaining $158$ lines $\ell$.

Since $A$ omits $1112x$ and $111x2$ for $x=1,3$, we thus conclude (by considering the lines $11*2x$ and $11*x2$) that $A$ must contain $1132x$, $113x2$, $1312x$, and $131x2$. Taking midpoints, we conclude that $A$ omits $11322$ and $13122$. But together with $11122$ this implies that at least three points are missing from $A \cap S_{3,5}$, contradicting the hypothesis $c(A)=78$.

\subsection{Elimination of $(8,39,78,0,0)$}

Now $A \cap S_{5,5}$ has eight points. By Lemma \ref{tech}, there are at least three pairs in this set which differ in two positions. As we can only eliminate two points from $S_{3,5}$, two of these pairs $(a,b), (c,d)$ must have the same midpoint $p$, and two other pairs $(a',b'), (c',d')$ must have the same midpoint $p'$, and $A$ contains $S_{3,5} \backslash \{p,p'\}$. As $p,p'$ are distinct, the plane containing $a,b,c,d$ is distinct from the plane containing $a',b',c',d'$.

Again consider the $160$ lines $\ell$ from the previous section. This time, the sum of the $|\ell \cap A|$ is $4b(A)+2c(A) = 312 = 156 \times 2$.
But the two lines in the plane of $a,b,c,d$ passing through $p$, and the two lines in the plane of $a',b',c',d'$ passing through $p'$, have no points; thus we must have $|\ell \cap A|=2$ for the remaining $156$ lines $\ell$.

Without loss of generality we have $(a,b)=(11111,11133)$, $(c,d) = (11113,11131)$, thus $p = 11122$. By permuting the first three indices, we may assume that $p'$ is not of the form $x2y2z, x2yz2, xy22z, xy2z2$ for any $x,y,z=1,3$. Then we have $1112x \not \in A$ and $1122x \in A$ for every $x=1,3$, so by the preceding paragraph we have $1132x \in A$; similarly for $113x2, 1312x, 131x2$. Taking midpoints, this implies that $13122, 11322 \not \in A$, but this (together with 11122) shows that at least three points aremissing from $A \cap S_{3,5}$, contradicting the hypothesis $c(A)=78$.

\subsection{Six dimensions}

Now we establish the bound $c'_{6,3}=353$. In view of the lower bounds, it suffices to show that there does not exist a Moser set $A \subset [3]^5$ with $|A|=354$.

We argue by contradiction. Let $A$ be as above, and let $(a(A),\ldots,g(A))$ be the statistics of $A$.

\begin{lemma}\label{g6} $g(A)=0$.
\end{lemma}

\begin{proof} For any four-dimensional slice $V$ of $A$, define
$$S(V) := 15 a(V) + 5 b(V) + 5 c(V)/2 + 3d(V)/2 + e(V).$$
From Lemma \ref{dci} we see that $|A|$ is equal to $a(A)+b(A)$ plus the average of $S(V)$ where $V$ ranges over the twenty slices which are some permutation of the center slice $22****$.

If $g(A)=1$, then $a(A) \leq 32$ and $b(A) \leq 96$ by \eqref{alpha-1}. Meanwhile, $e(V)=g(A)=1$ for every center slice $V$, so from Lemma \ref{paretop-4}, one can show that $S(V) \leq 223.5$ for every such slice. We conclude that $|A| \leq 351.5$, a contradiction.
\end{proof}

For any four-dimensional slice $V$ of $A$, define the \emph{defects} to be
$$ D(V) := 356 - [4a(V)+6b(V)+10c(V)+20d(V)+60e(V)].$$
Define a \emph{corner slice} to be one of the permutations or reflections of $11****$, thus there are $60$ corner slices. From Lemma \ref{dci} we see that $356-|A|+f(A)=2+f(A)$ is the average of the defects of all the $60$ corner slices. On the other hand, from Lemma \ref{paretop-4} and a straightforward computation, one concludes

\begin{lemma}\label{defects} Let $A$ be a four-dimensional Moser set. Then $D(A) \geq 0$. Furthermore:
\begin{itemize}
\item If $A$ has statistics $(6,12,18,4,0)$, then $D(A)=0$.
\item If $A$ has statistics $(5,12,18,4,0)$, $(5,12,12,4,1)$, or $(6,8,12,8,0)$, then $D(A)=4$.
\item For all other $A$, $D(A) \geq 6$.
\item If $a(A) = 4$, then $D(A) \geq 8$.
\item If $a(A) \geq 7$, then $D(A) \geq 16$.
\item If $a(A) \geq 8$, then $D(A) \geq 30$.
\item If $a(A) \geq 9$, then $D(A) \geq 86$.
\end{itemize}
\end{lemma}

Define a \emph{family} to be a set of four parallel corner slices, thus there are $15$ families, which are all a permutation of $\{11****, 13****, 31****, 33**** \}$. We refer to the family $\{11****, 13****, 31****, 33**** \}$ as $ab****$, and similarly define the family $a*b***$, etc.

\begin{lemma}\label{f6} $f(A)=0$.
\end{lemma}

\begin{proof} For any four-dimensional slice $V$ of $A$, define
$$ s(V) := 12 a(V)+15 b(V)/2+20 c(V)/3+15 d(V)/2 + 12 e(V),$$
and define an \emph{edge slice} to be one of the $30$ permutations or reflections of $12****$. From double counting we see that $|A|-a(A)$ is equal to the average of the $30$ values of $s(V)$ as $V$ ranges over edge slices.

From Lemma \ref{paretop-4} one can verify that $s(V) \leq 336$, and that $s(V) \leq 296 = 336-40$ if $e(V)=1$. The number of edge slices $V$ for which $e(V)=1$ is equal to $5f(A)$, and so the average value of the $s(V)$ is at most $336 - \frac{40 \times 5}{30} f(A)$, and so
$$ |A| - a(A) \leq 336 - \frac{40 \times 5}{30} f(A)$$
which we can rearrange (using $|A|=354$) as
$$ a(A) \geq 18 + \frac{20}{3} f(A).$$
Suppose first that $f(A)=1$; then $a(A) \geq 25$. This means that in any given family, one of the four corner slices has an $a$ value of at least $7$, and thus by Lemma \ref{defects} has a defect of at least $16$. Thus the average defect is at least $4$; on the other hand, the average defect is $2+f(A)=3$, a contradiction.

Now suppose that $f(A) \geq 2$; then $a(A) \geq 32$. Then in any given family, there is a corner slice with an $a$ value at least $9$, or four slices with $a$ value at least $8$, leading to a total defect of at least $86$ by Lemma \ref{defects}. Thus the average defect is at least $21.5$; on the other hand, the average defect is $2+f(A) \leq 2+12$, a contradiction.
\end{proof}

As a consequence of the above lemma, we see that the average defect of all corner slices is $2$, or equivalently that the total defect of these slices is $120$.

Call a corner slice \emph{good} if it has statistics $(6,12,18,4,0)$, and \emph{bad} otherwise. Thus good slices have zero defect, and bad slices have defect at least four. Since the average defect of the $60$ corner slices is $2$, there are at least $30$ good slices.

One can describe the structure of the good slices completely:

\begin{lemma}\label{sixt} The subset of $[3]^4$ consisting of the strings $1111, 1113, 3333, 1332, 1322, 1222, 3322$ and permutations is a Moser set with statistics $(6,12,18,4,0)$. Conversely, every Moser set with statistics $(6,12,18,4,0)$ is of this form up to the symmetries of the cube $[3]^4$.
\end{lemma}

\begin{proof} This can be verified by computer. By symmetry, one assumes 1222,2122,2212 and 2221 are in the set. Then 18 of the 24 'c' points with two 2s must be included; it is quick to check that 1122 and permutations must be the six excluded. Next, one checks that the only possible set of six 'a' points with no 2s is 1111,1113,3333 and permutations. Lastly, in a rather longer computation, one finds there is only possible set of twelve 'b' points, that is points with one 2. A computer-free proof can be found at
\centerline{{\tt http://michaelnielsen.org/polymath1/index.php?title=Classification\_of\_\%286\%2C12\%2C18\%2C4\%2C0\%29\_sets}.}
\end{proof}

As a consequence of this lemma, given any $x,y,z,w \in \{1,3\}$, there is a unique good Moser set in $[3]^4$ set whose intersection with $S_{1,4}$ is $\{x222, 2y22, 22z2, 222w\}$, and these are the only 16 possibilities. Call this set the \emph{good set of type $xyzw$}. It consists of
\begin{itemize}
\item The four points $x222, 2y22, 22z2, 222w$ in $S_{1,4}$;
\item All $24$ elements of $S_{2,4}$ except for $xy22, x2z2, x22w, 2yz2, 2y2w, 22zw$;
\item The twelve points $xYZ2$, $xY2W$, $x2ZW$, $XyZ2$, $Xy2W$, $2yZW$, $XYz2$, $X2zW$, $2YzW$, $XY2w$, $X2Zw$, $2YZw$ in $S_{3,4}$, where $X=4-x$, $Y=4-y$, $Z=4-z$, $W=4-w$;
\item The six points $xyzw, xyzW, xyZw, xYzw, Xyzw, XYZW$ in $S_{4,4}$.
\end{itemize}

We can use this to constrain the types of two intersecting good slices:

\begin{lemma}\label{pqs} Suppose that the $pq****$ slice is of type $xyzw$, and the $p*r***$ slice is of type $x'y'z'w'$, where $p,q,r,x,y,z,w,x',y',z',w'$ are in $\{1,3\}$. Then $x'=x$ iff $q=r$, and $y'z'w'$ is equal to either $yzw$ or $YZW$. If $x=r$ (or equivalently if $x'=q$), then $y'z'w'=yzw$.
\end{lemma}

\begin{proof} By reflection symmetry we can take $p=q=r=1$. Observe that the $11****$ slice contains $111222$ iff $x=1$, and the $1*1***$ slice similarly contains $111222$ iff $x'=1$. This shows that $x=x'$.

Suppose now that $x=x'=1$. Then the $111***$ slice contains the three elements $111y22, 1112z2, 11122w$, and excludes $111Y22, 1112Z2, 11122W$, and similarly with the primes, which forces $yzw=y'z'w'$ as claimed.

Now suppose that $x=x'=3$. Then the $111***$ slice contains the two elements $111yzw, 111YZW$, but does not contain any of the other six points in $S_{6,6} \cap 111***$, and similarly for the primes. Thus $y'z'w'$ is equal to either $yzw$ or $YZW$ as claimed.
\end{proof}

Now we look at two adjacent parallel good slices, such as $11****$ and $13****$. The following lemma asserts that such slices either have opposite type, or else will create a huge amount of defect in other slices:

\begin{lemma}\label{l18} Suppose that the $11****$ and $13****$ slices are good with types $xyzw$ and $x'y'z'w'$ respectively. If $x=x'$, then the $1*x***$ slice has defect at least $30$, and the $1*X***$ slice has defect at least $8$. Also, the $1**1**$, $1**3**$, $1***1*$, $1***3*$, $1****1$, $1****3$ slices have defect at least $6$. In particular, the total defect of slices beginning with $1*$ is at least $74$.
\end{lemma}

\begin{proof} Observe from the $11****, 13****$ hypotheses that $a(1*x***)=9$ and $a(1*X***)=4$, which gives the first two claims by Lemma \ref{defects}. For the other claims, one sees from Lemma \ref{pqs} that the other six slices cannot be good; also, they have an $a$-value of $6$ and a $d$-value of at most $7$, and the claims then follow from Lemma \ref{defects}.
\end{proof}

Now we look at two diagonally opposite parallel good slices, such as $11****$ and $33****$.

\begin{lemma}\label{l14} The $11****$ and $33****$ slices cannot both be good and of the same type.
\end{lemma}

\begin{proof} Suppose not. By symmetry we may assume that $11****$ and $33****$ are of type $1111$. This excludes a lot of points from $22****$. Indeed, by connecting lines between the $11****$ and $33****$ slices, we see that the only points that can still survive in $22****$ are $221133, 221333, 221132, 223332$, and permutations of the last four indices. Double counting the lines $22133*$ and permutations we see that there are at most $12$ points one can place in the permutations of $221133, 221333, 221132$, and so the $22****$ slice has at most $16$ points. Meanwhile, the two five-dimensional slices $1*****, 3*****$ have at most $c'_{5,3} = 124$ points, and the other two four-dimensional slices $21****, 23****$ have at most $c'_{4,3} = 43$ points, leading to at most $16 + 124 * 2 + 43 * 2 = 350$ points in all, a contradiction.
\end{proof}

\begin{lemma}\label{l19} It is not possible for all four slices in a family to be good.
\end{lemma}

\begin{proof} Suppose not. By symmetry we may assume that $11****, 13****, 31****, 33****$ are good. By Lemma \ref{l14}, the $11****$ and $33****$ slices cannot be of the same type, and so they cannot both be of the opposite type to either $13****$ or $31****$. If $13****$ is not of the opposite type to $11****$, then by (a permutation of) Lemma \ref{l18}, the total defect of slices beginning with $1*$ is at least $74$; otherwise, if $13****$ is not of the opposite type to $33****$, then by (a permutation and reflection of) Lemma \ref{l18}, the total defect of slices beginning with $*3$ is at least $74$. Similarly, the total defect of slices beginning with $3*$ or $*1$ is at least $74$, leading to a total defect of at least $148$. But the total defect of all the corner slices is $2 \times 60 = 120$, a contradiction.
\end{proof}

\begin{corollary}\label{l20} At most one family can have a total defect of at least $38$.
\end{corollary}

\begin{proof} Suppose there are two families with defect at least $38$. The remaining thirteen family have defect at least $4$ by Lemma \ref{l19} and Lemma \ref{defects}, leading to a total defect of at least $38*2+13*4=128$. But the total defect is $2 \times 60 = 120$, a contradiction.
\end{proof}

Actually we can refine this:

\begin{proposition} No family can have a total defect of at least $38$.
\end{proposition}

\begin{proof} Suppose for contradiction that the $ab****$ family (say) had a total defect of at least $38$, then by Corollary \ref{l20} no other families have total defect at least $38$.

We claim that the $**ab**$ family can have at most two good slices. Indeed, suppose the $**ab**$ has three good slices, say $**11**, **13**, **33**$. By Lemma \ref{l14}, the $**11**$ and $**33**$ slices cannot be of the same type, and so cannot both be of opposite type to $**13**$. Suppose $**11**$ and $**13**$ are not of opposite type. Then by (a permutation of) Lemma \ref{l18}, one of the families $a*b***, *ab***, **b*a*, **b**a$ has a net defect of at least $38$, contradicting the normalisation.

Thus each of the six families $**ab**, **a*b*, **a**b, ***ab*, ***a*b$ have at least two bad slices. Meanwhile, the eight families $a*b***, a**b**, a***b*, a****b, *ab***, *a*b**, *a**b*, *a***b$ have at least one bad slice by Corollary \ref{l19}, leading to twenty bad slices in addition to the defect of at least $38$ arising from the $ab****$ slice. To add up to a total defect of $120$, we conclude from Lemma \ref{defects} that all bad slices outside of the $ab****$ family have a defect of four, with at most one exception; but then by Lemma \ref{l18} this shows that (for instance) the $1*1***$ and $1*3***$ slices cannot be good unless they are of opposite type. The previous argument then shows that the a*b*** slice cannot have three good slices, which increases the number of bad slices outside of $ab****$ to at least twenty-one, and now there is no way to add up to $120$, a contradiction.
\end{proof}

\begin{corollary} Every family can have at most two good slices.
\end{corollary}

\begin{proof} If for instance $11****, 13****, 33****$ are both good, then by Lemma \ref{l14} at least one of $11****, 33****$ is not of the opposite type to $13****$, which by Lemma \ref{l18} implies that there is a family with a total defect of at least $38$, contradicting the previous proposition.
\end{proof}

From this corollary and Lemma \ref{defects}, we see that every family has a defect of at least $8$. Since there are $15$ families, and $8 \times 15$ is exactly equal to $120$, we conclude

\begin{corollary}\label{coda} Every family has \emph{exactly} two good slices, and the remaining two slices have defect $4$. In particular, by Lemma \ref{defects}, the bad slices must have statistics $(5,12,18,4,0)$, $(5,12,12,4,1)$, or $(6,8,12,8,0)$.
\end{corollary}

We now limit how these slices can interact with good slices.

\begin{lemma}\label{goodgood} Suppose that $1*1***$ is a good slice.
\begin{itemize}
\item[(i)] The $11****$ slice cannot have statistics $(6,8,12,8,0)$.
\item[(ii)] The $11****$ slice cannot have statistics $(5,12,12,4,1)$.
\item[(iii)] If the $11****$ slice has statistics $(5,12,18,4,0)$, then the $112***$ slice has statistics $(3,9,3,0)$.
\end{itemize}
\end{lemma}

\begin{proof} This can be verified through computer search; there are $16$ possible configurations for the good slices, and one can calculate that there are $27520$ configurations for the $(5,12,12,4,1)$ slices, $4368$ configurations for the $(5,12,18,4,0)$ slices, and $80000$ configurations for the $(6,8,12,8,0)$ slices. {\bf we could put human proofs for all this somewhere, presumably.}

%We first prove (i). Suppose for contradiction that the $11****$ slice has statistics $(6,8,12,8,0)$, then $A$ contains $111222$, and so the $1*1***$ slice is of type $1xyz$ for some $x,y,z$. By symmetry we may assume it is of type $1111$, thus the $111***$ slice consists of
%$111111, 111113, 111332, 111322, 111222$ and permutations of the last three indices. On the other hand, the $11****$ slice has all eight of the points in $11**** \cap S_{2,6}$. Drawing lines between these points and $111111, 111113$ and permutations, we see that the $113***$ slice cannot contain $113111, 113113, 113133$, or permutations, leaving $113333$ as the only possible element of $113*** \cap S_{6,6}$. This makes $a(11****)=5$, a contradiction.
\end{proof}

\begin{corollary}\label{slic} The $111***$ slice has statistics $(4,3,3,1)$, $(2,6,6,0)$, $(3,3,3,1)$, or $(1,6,6,0)$.
\end{corollary}

\begin{proof} From Corollary \ref{coda}, we know that at least one of the slices $13****, 31****, 11****$ are good. If $11****$ or $1*1***$ is good, then the slice $111***$ has statistics $(4,3,3,1)$ or $(2,6,6,0)$, by Lemma \ref{sixt}. By symmetry we may thus reduce to the case where $13****$ is good and $1*1***$ is bad. Then by Lemma \ref{goodgood}, the $1*1***$ slice has statistics $(5,12,18,4,0)$ and the $121***$ slice has statistics $(3,9,3,0)$. Since the $131***$ slice, as a side slice of the good $13****$ slice, has statistics $(4,3,3,1)$ or $(2,6,6,0)$, we conclude that the $111***$ slice has statistics $(1,6,6,0)$ or $(3,3,3,1)$, and the claim follows.
\end{proof}

\begin{corollary}\label{slic2} All corner slices have statistics $(6,12,18,4,0)$ or $(5,12,18,4,0)$.
\end{corollary}

\begin{proof} Suppose first that a corner slice, say $11****$ has statistic $(6,8,12,8,0)$. Then $111***$ and $113***$ contain one ``d'' point each, and have six ``a'' points between them, so by Corollary \ref{slic}, they both have statistic $(3,3,3,1)$. This forces the $1*1***$, $1*3***$ slices to be bad, which by Corollary \ref{coda} forces the $3*1***,3*3***$ slices to be good. This forces the $311***, 313***$ slices to have statistics either $(2,6,6,0)$ or $(4,3,3,1)$. But the $311***$ slice (say) cannot have statistic $(4,3,3,1)$, since when combined with the $(3,3,3,1)$ statistics of $111***$ would give $a(*11***)=7$, which contradicts Corollary \ref{coda}; thus the $311***$ slice has statistic $(2,6,6,0)$, and similarly for $331***$. But then $a(3*1***)=4$, which again contradicts Corollary \ref{coda}.

Thus no corner slice has statistic $(6,8,12,8,0)$. Now suppose that a corner slice, say $11****$ has statistic $(5,12,12,4,1)$. By Lemma \ref{goodgood}, the $1*1***, 1*3***$ slices are bad, so by repeating the preceding arguments we conclude that the $311***, 313***$ slices have statistics $(2,6,6,0)$ or $(4,3,3,1)$; in particular, their $a$-value is even. However, the $*11***$ and $*13***$ slices are bad by Lemma \ref{goodgood}, and thus have an $a$-value of $5$; thus the $111***$ and $113***$ slices have an odd $a$-value. Thus forces $a(11****)$ to be even; but it is equal to $5$, a contradiction.
\end{proof}

From this and Lemma \ref{dci}, we see that $A$ has statistics $(22,72,180,80,0,0,0)$. In particular, we have $2\alpha_2(A)+\alpha_3(A) = 2$, which by double counting (cf. \eqref{alpha-1}) shows that for every line of the form $12222*$ (or a reflection or permutation thereof) intersects $A$ in exactly two points. Note that such lines connect a ``$d$'' point to two ``$c$'' points.

Also, we observe that two adjacent ``$d$'' points, such as $111222$ and $113222$, cannot both lie in $A$; for this would force the $*13***$ and $*11***$ slices to have statistics $(4,3,3,1)$ or $(3,3,3,1)$ by Corollary \ref{slic}, which forces $a(*1****)=6$, and thus $*1****$ must be good by Corollary \ref{slic2}; but this contradicts Lemma \ref{sixt}. Since $\alpha_3(A)=1/2$, we conclude that given any two adjacent ``$d$'' points, exactly one of them lies in $A$. In particular, the d points of the form $***222$ consist either of those strings with an even number of $1$s, or those with an odd number of $1$s.

Let's say it's the former, thus the set contains $111222, 133222$, and permutations of the first three coordinates, but omits $113222, 333222$ and permutations of the first three coordinates. Since the ``$d$'' points $113222, 333222$ are omitted, we conclude that the ``$c$'' points $113122, 113322, 333122, 333322$ must lie in the set, and similarly for permutations of the first three and last three coordinates. But this gives at least $15$ of the $16$ ``$c$'' points ending in $22$; by symmetry this leads to $225$ $c$-points in all; but $c(A)=180$, contradiction. This (finally!) completes the proof that $c'_{6,3}=353$.

Moser.tex

2009-07-20T13:29:03Z

Mpeake:

\section{Upper bounds for the $k=3$ Moser problem in small dimensions}\label{moser-upper-sec}

In this section we finish the proof of Theorem \ref{moser} by obtaining the upper bounds on $c'_{n,3}$ for $n \leq 6$.

\subsection{Statistics, densities and slices}

Our analysis will revolve around various \emph{statistics} of Moser sets $A \subset [3]^n$, their associated \emph{densities}, and the behavior of such statistics and densities with respect to the operation of passing from the cube $[3]^n$ to various \emph{slices} of that cube.

\begin{definition}[Statistics and densities] Let $A \subset [3]^n$ be a set. For any $0 \leq i \leq n$, set $a_i(A) := |A \cap S_{n-i,n}|$; thus we have
$$ 0 \leq a_i(A) \leq |S_{n-i,n}| = \binom{n}{i} 2^{n-i}$$
for $0 \leq i \leq n$ and
$$ a_0(A) + \ldots + a_n(A) = |A|.$$
We refer to the vector $(a_0(A),\ldots,a_n(A))$ as the \emph{statistics} of $A$. We define the $i^{th}$ \emph{density} $\alpha_i(A)$ to be the quantity
$$ \alpha_i(A) := \frac{a_i(A) }{\binom{n}{i} 2^{n-i}},$$
thus $0 \leq \alpha_i(A) \leq 1$ and
$$ |A| = \sum_{i=0}^n \binom{n}{i} 2^{n-i} a_i(A).$$
\end{definition}

\begin{example}\label{2mos} Let $n=2$ and $A$ be the Moser set $A := \{ 12, 13, 21, 23, 31, 32 \}$. Then the statistics $(a_0(A), a_1(A), a_2(A))$ of $A$ are $(2,4,0)$, and the densities $(\alpha_0(A), \alpha_1(A), \alpha_2(A))$ are $(\frac{1}{2}, 1, 0)$. \textbf{Include picture here? with colours?}
\end{example}

When working with small values of $n$, it will be convenient to write $a(A)$, $b(A)$, $c(A)$, etc. for $a_0(A)$, $a_1(A)$, $a_2(A)$, etc., and similarly write $\alpha(A), \beta(A), \gamma(A)$, etc. for $\alpha_0(A)$, $\alpha_1(A)$, $\alpha_2(A)$, etc. Thus for instance in Example \ref{2mos} we have $b(A) = 4$ and $\alpha(A) = \frac{1}{2}$.

\begin{definition}[Subspace statistics and densities] If $V$ is a $k$-dimensional geometric subspace of $[3]^n$, then we have a map $\phi_V: [3]^k \to [3]^n$ from the $k$-dimensional cube to the $n$-dimensional cube. If $A \subset [3]^n$ is a set and $0 \leq i \leq k$, we write $a_i(V,A)$ for $a_i(\phi_V^{-1}(A))$ and $\alpha_i(V,A)$ for $\alpha_i(\phi_V^{-1}(A))$. If the set $A$ is clear from context, we abbreviate $a_i(V,A)$ as $a_i(V)$ and $\alpha_i(V,A)$ as $\alpha_i(V)$.
\end{definition}

For our problem, a particularly important type of subspace of $[3]^n$ will be the \emph{slices} formed by fixing one coordinate and letting the other $n-1$ coordinates vary. We will denote this by a single string in which the $n-1$ varying coordinates are denoted by asterisks. For instance, in $[3]^2$, $1*$ denotes the slice $1*=\{11,12,13\}$, $*2$ denotes the slice $*2=\{12,22,32\}$, etc.; similarly, in $[3]^3$, $1**$ is the slice $\{111, 112, 113, 121, 122, 123, 131, 132, 133\}$, etc. We call a slice a \emph{centre slice} if the fixed coordinate is $2$ and a \emph{side slice} if it is $1$ or $3$.

\begin{example} We continue Example \ref{2mos}. Then the statistics of the side slice $1*$ are $(a(1*),b(1*)) = (1,1)$, while the statistics of the centre slice $2*$ are $(a(2*),b(2*))=(2,0)$. The corresponding densities are $(\alpha(1*),\beta(1*)) = (1/2,1)$ and $(\alpha(2*),\beta(2*))=(1,0)$.
\end{example}

A simple double counting argument gives the following useful identity:

\begin{lemma}[Double counting identity]\label{dci} Let $A \subset [3]^n$ and $0 \leq i \leq n-1$. Then we have
$$ \frac{1}{n-i-1} \sum_{V \hbox{ a side slice}} a_{i+1}(V) = \frac{1}{i+1} \sum_{W \hbox{ a centre slice}} a_i(W) = a_{i+1}(A)$$
where $V$ ranges over the $2n$ side slices of $[3]^n$, and $W$ ranges over the $n$ centre slices. In other words, the average value of $\alpha_{i+1}(V)$ for side slices $V$ equals the average value of $\alpha_i(W)$ for centre slices $W$, which is in turn equal to $\alpha_{i+1}(A)$.
\end{lemma}

Indeed, this lemma follows from the observation that every string in $A \cap S_{n-i-1,n}$ belongs to $i+1$ centre slices $W$ (and contributes to $a_i(W)$) and to $n-i-1$ side slices $V$ (and contributes to $a_{i+1}(V)$). One can also view this lemma probabilistically, as the assertion that there are three equivalent ways to generate a random string of length $n$:

\begin{itemize}
\item Pick a side slice $V$ at random, and randomly fill in the wildcards in such a way that $i+1$ of the wildcards are $2$'s (i.e. using an element of $S_{n-i-2,n-1}$).
\item Pick a centre slice $V$ at random, and randomly fill in the wildcards in such a way that $i$ of the wildcards are $2$'s (i.e. using an element of $S_{n-i-1,n-1}$).
\item Randomly choose an element of $S_{n-i-1,n}$.
\end{itemize}

\begin{example} We continue Example \ref{2mos}. The average value of $\beta$ for side slices is equal to the average value of $\alpha$ for centre slices, which is equal to $\beta(A) = 1$.
\end{example}

Another very useful fact (essentially due to \cite{chvatal2}) is that linear inequalities for statistics of Moser sets at one dimension propagate to linear inequalities in higher dimensions:

\begin{lemma}[Propagation lemma]\label{prop} Let $n \geq 1$ be an integer. Suppose one has a linear inequality of the form
\begin{equation}\label{alphav}
\sum_{i=0}^n v_i \alpha_i(A) \leq s
\end{equation}
for all Moser sets $A \subset [3]^n$ and some real numbers $v_0,\ldots,v_n,s$. Then we also have the linear inequality
$$ \sum_{i=0}^n v_i \alpha_{qi+r}(A) \leq s$$
whenever $q \geq 1$, $r \geq 0$, $N \geq nq+r$ are integers and $A \subset [3]^N$ is a Moser set.
\end{lemma}

\begin{proof} We run a probabilistic argument (one could of course also use a double counting argument instead). Let $n,v_0,\ldots,v_n,s,q,r,N,A$ be as in the lemma. Let $V$ be a random $n$-dimensional geometric subspace of $[3]^N$, created in the following fashion:
\begin{itemize}
\item Pick $n$ wildcards $x_1,\ldots,x_n$ to run independently from $1$ to $3$. We also introduce dual wildcards $\overline{x_1},\ldots,\overline{x_n}$; each $\overline{x_j}$ will take the value $4-x_j$.
\item We randomly subdivide the $N$ coordinates into $n$ groups of $q$ coordinates, plus a remaining group of $N-nq$ ``fixed'' coordinates.
\item For each coordinate in the $j^{th}$ group of $q$ coordinates for $1 \leq j \leq n$, we randomly assign either a $x_j$ or $\overline{x_j}$.
\item For each coordinate in the $N-nq$ fixed coordinates, we randomly assign a digit $1,2,3$, but condition on the event that exactly $r$ of the digits are equal to $2$ (i.e. we use a random element of $S_{N-nq-r,N-nq}$).
\item Let $V$ be the subspace created by allowing $x_1,\ldots,x_n$ to run independently from $1$ to $3$, and $\overline{x_j}$ to take the value $4-x_j$.
\end{itemize}

For instance, if $n=2, q=2, r=1, N=6$, then a typical subspace $V$ generated in this fashion is
$$ 2x_1\overline{x_2}3x_2x_1 = \{ 213311, 212321, 211331, 223312, 222322, 221332, 233313, 232323, 231333\}.$$
Observe from that the following two ways to generate a random element of $[3]^N$ are equivalent:

\begin{itemize}
\item Pick $V$ randomly as above, and then assign $(x_1,\ldots,x_n)$ randomly from $S_{n-i,n}$. Assign $4-x_j$ to $\overline{x_j}$ for all $1 \leq j \leq n$.
\item Pick a random string in $S_{N-qi-r,N}$.
\end{itemize}

Indeed, both random variables are invariant under the symmetries of the cube, and both random variables always pick out strings in $S_{N-qi-r,N}$, and the claim follows. As a consequence, we see that the expectation of $\alpha_i(V)$ (as $V$ ranges over the recipe described above) is equal to $\alpha_{qi+r}(A)$. On the other hand, from \eqref{alphav} we have
$$ \sum_{i=0}^n v_i \alpha_i(V) \leq s$$
for all such $V$; taking expectations over $V$, we obtain the claim.
\end{proof}

In view of Lemma \ref{prop}, it is of interest to locate linear inequalities relating the densities $\alpha_i(A)$, or (equivalently) the statistics $a_i(A)$. For this, it is convenient to introduce the following notation.

\begin{definition} Let $n \geq 1$ be an integer.
\begin{itemize}
\item A vector $(a_0,\ldots,a_n)$ of non-negative integers is \emph{feasible} if it is the statistics of some Moser set $A$.
\item A feasible vector $(a_0,\ldots,a_n)$ is \emph{Pareto-optimal} if there is no other feasible vector $(b_0,\ldots,b_n) \neq (a_0,\ldots,a_n)$ such that $b_i \geq a_i$ for all $0 \leq i \leq n$.
\item A Pareto-optimal vector $(a_0,\ldots,a_n)$ is \emph{extremal} if it is not a non-trivial convex linear combination of other Pareto-optimal vectors.
\end{itemize}
\end{definition}

To establish a linear inequality of the form \eqref{alphav} with the $v_i$ non-negative, it suffices to test the inequality against densities associated to extremal vectors of statistics. (There is no point considering linear inequalities with negative coefficients $v_i$, since one always has the freedom to reduce a density $\alpha_i(A)$ of a Moser set $A$ to zero, simply by removing all elements of $A$ with exactly $i$ $2$'s.)

We will classify exactly the Pareto-optimal and extremal vectors for $n \leq 3$, which by Lemma \ref{prop} will lead to useful linear inequalities for $n \geq 4$. Using a computer, we have also located a partial list of Pareto-optimal and extremal vectors for $n=4$, which are also useful for the $n=5$ and $n=6$ theory.

\subsection{Up to three dimensions}

We now establish Theorem \ref{moser} for $n \leq 3$, and establish some auxiliary inequalities which will be of use in higher dimensions.

The case $n=0$ is trivial. When $n=1$, it is clear that $c'_{1,3} = 2$, and furthermore that the Pareto-optimal statistics are $(2,0)$ and $(1,1)$, which are both extremal. This leads to the linear inequality
$$ 2\alpha(A) + \beta(A) \leq 2$$
for all Moser sets $A \subset [3]^1$, which by Lemma \ref{prop} implies that
\begin{equation}\label{alpha-1}
2\alpha_r(A) + \alpha_{r+q}(A) \leq 2
\end{equation}
whenever $r \geq 0$, $q \geq 1$, $n \geq q+r$, and $A \subset [3]^n$ is a Moser set.

For $n=2$, we see by partitioning $[3]^2$ into three slices that $c'_{2,3} \leq 3 c'_{1,3} = 6$, and so (by the lower bounds in the previous section) $c'_{2,3} = 6$. Writing $(a,b,c) = (a(A),b(A),c(A)) = (4\alpha(A), 4\beta(A), \gamma(A))$, the inequalities \eqref{alpha-1} become
\begin{equation}\label{abc}
a + 2c \leq 4; b+2c \leq 4; 2a+b <= 8.
\end{equation}

\begin{lemma} When $n=2$, the Pareto-optimal statistics are $(4,0,0), (3,2,0), (2,4,0), (2,2,1)$. In particular, the extremal statistics are $(4,0,0), (2,4,0), (2,2,1)$.
\end{lemma}

\begin{proof} One easily checks that all the statistics listed above are feasible.
Consider the statistics $(a,b,c)$ of a Moser set $A \subset [3]^2$. $c$ is either equal to $0$ or $1$. If $c=1$, then \eqref{abc} implies that $a,b \leq 2$, so the only Pareto-optimal statistic here is $(2,2,1)$. When instead $c=0$, the inequalities \eqref{abc} can easily imply the Pareto-optimality of $(4,0,0), (3,2,0), (2,4,0)$.
\end{proof}

From this lemma we see that we obtain a new inequality $2a+b+2c \leq 8$. Converting this back to densities and using Lemma \ref{prop}, we conclude that
\begin{equation}\label{alpha-2}
4\alpha_r(A) + 2\alpha_{r+q}(A) + \alpha_{r+2q} \leq 4
\end{equation}
whenever $r \geq 0$, $q \geq 1$, $n \geq q+2r$, and $A \subset [3]^n$ is a Moser set.

One can also check by computer that there are exactly $230$ line-free subsets of $[3]^2$.

Now we look at three dimensions. Writing $(a,b,c,d)$ for the statistics of a Moser set $A \subset [3]^n$ (which thus range between $(0,0,0,0)$ and $(8,12,6,1)$), the inequalities \eqref{alpha-1} imply in particular that
\begin{equation}\label{abc-3d}
a+4d \leq 8; b+6d \leq 12; c+3d \leq 6; 3a+2c \leq 24; b+c \leq 12
\end{equation}
while \eqref{alpha-2} implies that
\begin{equation}\label{abcd-3d}
3a+b+c \leq 24; b+c+3d \leq 12.
\end{equation}
Summing the inequalities $b+c \leq 12, 3a+b+c \leq 24, b+c+3d \leq 12$ yields
$$ 3(a+b+c+d) \leq 48$$
and hence $|A| = a+b+c+d \leq 16$; comparing this with the lower bounds of the preceding section we obtain $c'_{3,3} = 16$ as required. (This argument is essentially identical to the one in \cite{chvatal2}).

We have the following useful computation:

\begin{lemma}[3D Pareto-optimals]\label{paretop} When $n=3$, the Pareto-optimal statistics are $$(3,6,3,1),(4,4,3,1),(4,6,2,1),(2,6,6,0),(3,6,5,0),(4,4,5,0),(3,7,4,0),(4,6,4,0),$$
$$ (3,9,3,0),(4,7,3,0),(5,4,3,0),(4,9,2,0),(5,6,2,0),(6,3,2,0),(3,10,1,0),(5,7,1,0),$$
$$ (6,4,1,0),(4,12,0,0),(5,9,0,0),(6,6,0,0),(7,3,0,0),(8,0,0,0).$$
In particular, the extremal statistics are
$$(3,6,3,1),(4,4,3,1),(4,6,2,1),(2,6,6,0),(4,4,5,0),(4,6,4,0),(4,12,0,0),(8,0,0,0).$$
\end{lemma}

\begin{proof} This can be established by a brute-force search over the $2^{27} \approx 1.3 \times 10^8$ different subsets of $[3]^3$. Actually, one can perform a much faster search than this. Firstly, as noted earlier, there are only $230$ line-free subsets of $[3]^2$, so one could search over $230^3 \approx 1.2 \times 10^7$ configurations instead. Secondly, by symmetry we may assume (after enumerating the $230$ sets in a suitable fashion) that the first slice $A \cap 1**$ has an index less than or equal to the third $A \cap 3**$, leading to $\binom{231}{2} \times 230 \approx 6 \times 10^6$ configurations instead. Finally, using the first and third slice one can quickly determine which elements of the second slice $2**$ are prohibited from $A$. There are $2^9 = 512$ possible choices for the prohibited set in $2**$. By crosschecking these against the list of $230$ line-free sets one can compute the Pareto-optimal statistics for the second slices inside the prohibited set (the lists of such statistics turns out to length at most $23$). Storing these statistics in a lookup table, and then running over all choices of the first and third slice (using symmetry), one now has to perform $O( 512 \times 230 ) + O( \binom{231}{2} \times 23) \approx O( 10^6 )$ computations, which is quite a feasible computation.

One could in principle reduce the computations even further, by a factor of up to $8$, by using the symmetry group $D_4$ of the square $[3]^2$ to reduce the number of cases one needs to consider, but we did not implement this.

A computer-free proof of this lemma can be found at
\centerline{{\tt http://michaelnielsen.org/polymath1/index.php?title=Human\_proof\_of\_the\_3D\_Pareto-optimal\_Moser\_statistics}.}
\end{proof}

\begin{remark} A similar computation revealed that the total number of line-free subsets of $[3]^3$ was $3813884$. With respect to the $2^3 \times 3!=48$-element group of geometric symmetries of $[3]^3$, these sets partitioned into $83158$ equivalence classes:
$$
3813884 = 76066 \times 48+6527 \times 24+51 \times 16+338 \times 12 +109 \times 8+41 \times 6+13 \times 4 +5 \times 3+3 \times 2+5 \times 1.
$$
\end{remark}

Lemma \ref{paretop} yields the following new inequalities:
\begin{align*}
2a+b+2c+4d &\leq 22 \\
3a+2b+3c+6d &\leq 36 \\
7a+2b+4c+8d &\leq 56 \\
6a+2b+3c+6d &\leq 48 \\
a+2c+4d &\leq 14 \\
5a+4c+8d &\leq 40.
\end{align*}

Applying Lemma \ref{prop}, we obtain new inequalities:
\begin{align}
8\alpha_r(A)+ 6\alpha_{r+q}(A) + 6\alpha_{r+2q}(A) + 2\alpha_{r+3q}(A) &\leq 11 \label{eleven}\\
4\alpha_r(A)+4\alpha_{r+q}(A)+3\alpha_{r+2q}(A)+\alpha_{r+3q}(A) &\leq 6\label{six}\\
7\alpha_r(A)+3\alpha_{r+q}(A)+3\alpha_{r+2q}(A)+\alpha_{r+3q}(A) &\leq 7\nonumber\\
8\alpha_r(A)+3\alpha_{r+q}(A)+3\alpha_{r+2q}(A)+\alpha_{r+3q}(A) &\leq 8\label{eight}\\
4\alpha_{r+q}(A)+2\alpha_{r+2q}(A)+\alpha_{r+3q}(A) &\leq 4\nonumber\\
4\alpha_r(A)+6\alpha_{r+2q}(A)+2\alpha_{r+3q}(A) &\leq 7\nonumber\\
5\alpha_r(A)+3\alpha_{r+2q}(A)+\alpha_{r+3q}(A) &\leq 5\nonumber
\end{align}
whenever $r \geq 0$, $q \geq 1$, $n \geq r+3q$, and Moser sets $A \subset [3]^n$.

We also note some further corollaries of Lemma \ref{paretop}:

\begin{corollary}[Statistics of large 3D Moser sets]\label{paretop2} Let $(a,b,c,d)$ be the statistics of a Moser set $A$ in $[3]^3$. Then $|A| = a+b+c+d \leq 16$. Furthermore:
\begin{itemize}
\item If $|A|=16$, then $(a,b,c,d) = (4,12,0,0)$.
\item If $|A|=15$, then $(a,b,c,d) = (4,11,0,0)$ or $(3,12,0,0)$.
\item If $|A| \geq 14$, then $b \geq 6$ and $d=0$.
\item If $|A| = 13$ and $d=1$, then $(a,b,c,d) = (4,6,2,1)$ or $(3,6,3,1)$.
\end{itemize}
\end{corollary}

\subsection{Four dimensions}

Now we establish the bound $c'_{4,3}=43$. Let $A$ be a Moser set in $[3]^4$, with attendant statistics $(a,b,c,d,e)$, which range between $(0,0,0,0,0)$ and $(16,32,24,8,1)$. In view of the lower bounds, our task here is to establish the upper bound $a+b+c+d+e \leq 43$.

The linear inequalities already established just barely fail to achieve this bound, but we can obtain the upper bound $a+b+c+d+e \leq 44$ as follows.
First suppose that $e=1$; then from the inequalities \eqref{alpha-1} (or by considering lines passing through $2222$) we see that $a \leq 8, b \leq 16, c \leq 12, d \leq 4$ and hence $a+b+c+d+e \leq 41$, so we may assume that $e=0$.

From Lemma \ref{dci}, we see that $a+b+c+d+e$ is now equal to the sum of $a(V)/4+b(V)/3+c(V)/2+d(V)$, where $V$ ranges over all side slices of $[3]^4$. But from Lemma \ref{paretop} we see that $a(V)/4+b(V)/3+c(V)/2+d(V)$ is at most $\frac{11}{4}$, with equality occuring only when $(a(V),b(V),c(V),d(V))=(2,6,6,0)$. This gives the upper bound $a+b+c+d+e \leq 44$.

The above argument shows that $a+b+c+d+e=44$ can only occur if $e=0$ and if $(a(V),b(V),c(V),d(V))=(2,6,6,0)$ for all side slices $V$. Applying Lemma \ref{paretop} again this implies $(a,b,c,d,e)=(4,16,24,0,0)$. But then $A$ contains all of the sphere $S_{2,4}$, which implies that the four-element set $A \cap S_{4,4}$ cannot contain a pair of strings which differ in exactly two positions (as their midpoint would then lie in $S_{2,4}$, contradicting the hypothesis that $A$ is a Moser set).

Recall that we may partition $S_{4,4} = S_{4,4}^e \cup S_{4,4}^o$, where
$$S_{4,4}^e := \{ 1111, 1133, 1313, 3113, 1331, 3131, 3311, 3333\}$$
is the strings in $S_{4,4}$ with an even number of $1$'s, and
$$S_{4,4}^o := \{ 1113, 1131, 1311, 3111, 1333, 3133, 3313, 3331\}$$
are the strings in $S_{4,4}$ with an odd number. Observe that any two distinct elements in $S_{4,4}^e$ differ in exactly two positions unless they are antipodal. Thus $A \cap S_{4,4}^e$ has size at most two, with equality only when $A \cap S_{4,4}^e$ consists of an antipodal pair. Similarly for $A \cap S_{4,4}^o$. Thus $A$ must consist of two antipodal pairs, one from $S_{4,4}^e$ and one from $S_{4,4}^o$.

By the symmetries of the cube we may assume without loss of generality that these pairs are $\{ 1111, 3333\}$ and $\{1113,3331\}$ respectively. But as $A$ is a Moser set, $A$ must now exclude the strings $1112$ and $3332$. These two strings form two corners of the eight-element set
$$ ***2 \cap S_{3,4} = \{ 1112, 1132, 1312, 3112, 1332, 3132, 3312, 3332 \}.$$
Any pair of points in this set which are ``adjacent'' in the sense that they differ by exactly one entry cannot both lie in $A$, as their midpoint would then lie in $S_{3,4}$, and so $A$ can contain at most four elements from this set, with equality only if $A$ contains all the points in $***2 \cap S_{3,4}$ of the same parity (either all the elements with an even number of $3$s, or all the elements with an odd number of $3$s). But because the two corners removed from this set have the opposite parity (one has an even number of $1$s and one has an odd number), we see in fact that $A$ can contain at most $3$ points from this set. Meanwhile, the same arguments give that $A$ contains at most four points from $**2* \cap S_{3,4}$, $*2** \cap S_{3,4}$, and $2*** \cap S_{3,4}$. Summing we see that $b = |A \cap S_{3,4}| \leq 3+4+4+4=15$, a contradiction. Thus we have $c'_{4,3}=43$ as claimed.

We have the following four-dimensional version of Lemma \ref{paretop}:

\begin{lemma}[4D Pareto-optimals]\label{paretop-4} When $n=4$, the Pareto-optimal statistics listed on Table \ref{table4}.
\end{lemma}

\begin{figure}[tb]
\centered{\tiny
$(3, 16, 24, 0, 0)$,
$(4, 14, 19, 2, 0)$,
$(4, 15, 24, 0, 0)$,
$(4, 16, 8, 4, 1)$,
$(4, 16, 14, 4, 0)$,
$(4, 16, 23, 0, 0)$,
$(4, 17, 21, 0, 0)$,
$(4, 18, 19, 0, 0)$,
$(5, 12, 12, 4, 1)$,
$(5, 12, 13, 6, 0)$,
$(5, 12, 15, 5, 0)$,
$(5, 12, 19, 2, 0)$,
$(5, 13, 10, 4, 1)$,
$(5, 13, 14, 5, 0)$,
$(5, 13, 21, 1, 0)$,
$(5, 15, 9, 4, 1)$,
$(5, 15, 12, 3, 1)$,
$(5, 15, 13, 5, 0)$,
$(5, 15, 18, 3, 0)$,
$(5, 15, 20, 1, 0)$,
$(5, 15, 22, 0, 0)$,
$(5, 16, 7, 4, 1)$,
$(5, 16, 10, 3, 1)$,
$(5, 16, 11, 5, 0)$,
$(5, 16, 12, 2, 1)$,
$(5, 16, 16, 3, 0)$,
$(5, 16, 19, 1, 0)$,
$(5, 16, 21, 0, 0)$,
$(5, 17, 12, 4, 0)$,
$(5, 17, 14, 3, 0)$,
$(5, 17, 16, 2, 0)$,
$(5, 17, 18, 1, 0)$,
$(5, 17, 20, 0, 0)$,
$(5, 18, 13, 3, 0)$,
$(5, 18, 14, 2, 0)$,
$(5, 20, 8, 4, 0)$,
$(5, 20, 10, 3, 0)$,
$(5, 20, 13, 2, 0)$,
$(5, 20, 14, 1, 0)$,
$(5, 20, 18, 0, 0)$,
$(5, 21, 10, 2, 0)$,
$(5, 21, 15, 0, 0)$,
$(5, 22, 13, 0, 0)$,
$(6, 8, 12, 8, 0)$,
$(6, 10, 11, 4, 1)$,
$(6, 11, 12, 7, 0)$,
$(6, 12, 10, 7, 0)$,
$(6, 12, 13, 5, 0)$,
$(6, 12, 18, 4, 0)$,
$(6, 13, 16, 4, 0)$,
$(6, 14, 9, 4, 1)$,
$(6, 14, 9, 7, 0)$,
$(6, 14, 12, 6, 0)$,
$(6, 14, 16, 3, 0)$,
$(6, 14, 19, 1, 0)$,
$(6, 14, 21, 0, 0)$,
$(6, 15, 7, 4, 1)$,
$(6, 15, 10, 3, 1)$,
$(6, 15, 10, 6, 0)$,
$(6, 15, 11, 2, 1)$,
$(6, 15, 12, 5, 0)$,
$(6, 15, 15, 4, 0)$,
$(6, 15, 20, 0, 0)$,
$(6, 16, 7, 3, 1)$,
$(6, 16, 8, 6, 0)$,
$(6, 16, 9, 2, 1)$,
$(6, 16, 10, 5, 0)$,
$(6, 16, 12, 1, 1)$,
$(6, 16, 13, 4, 0)$,
$(6, 16, 14, 3, 0)$,
$(6, 16, 18, 2, 0)$,
$(6, 16, 19, 0, 0)$,
$(6, 17, 9, 5, 0)$,
$(6, 17, 10, 4, 0)$,
$(6, 17, 13, 3, 0)$,
$(6, 17, 15, 2, 0)$,
$(6, 17, 17, 1, 0)$,
$(6, 17, 18, 0, 0)$,
$(6, 18, 13, 2, 0)$,
$(6, 18, 16, 1, 0)$,
$(6, 18, 17, 0, 0)$,
$(6, 19, 9, 4, 0)$,
$(6, 19, 12, 3, 0)$,
$(6, 19, 15, 1, 0)$,
$(6, 20, 7, 4, 0)$,
$(6, 20, 9, 3, 0)$,
$(6, 20, 12, 2, 0)$,
$(6, 20, 13, 1, 0)$,
$(6, 20, 15, 0, 0)$,
$(6, 21, 8, 3, 0)$,
$(6, 21, 9, 2, 0)$,
$(6, 21, 12, 1, 0)$,
$(6, 21, 14, 0, 0)$,
$(6, 22, 7, 3, 0)$,
$(6, 22, 8, 2, 0)$,
$(6, 22, 10, 1, 0)$,
$(6, 23, 9, 1, 0)$,
$(6, 24, 7, 2, 0)$,
$(6, 24, 8, 1, 0)$,
$(6, 24, 12, 0, 0)$,
$(6, 25, 9, 0, 0)$,
$(6, 26, 7, 0, 0)$,
$(7, 8, 6, 8, 0)$,
$(7, 11, 9, 4, 1)$,
$(7, 11, 12, 6, 0)$,
$(7, 12, 8, 4, 1)$,
$(7, 12, 8, 6, 0)$,
$(7, 12, 12, 3, 1)$,
$(7, 12, 12, 5, 0)$,
$(7, 12, 13, 4, 0)$,
$(7, 12, 15, 3, 0)$,
$(7, 12, 17, 2, 0)$,
$(7, 13, 7, 4, 1)$,
$(7, 13, 10, 3, 1)$,
$(7, 13, 11, 5, 0)$,
$(7, 13, 12, 2, 1)$,
$(7, 13, 12, 4, 0)$,
$(7, 13, 14, 3, 0)$,
$(7, 13, 16, 2, 0)$,
$(7, 14, 6, 4, 1)$,
$(7, 14, 6, 7, 0)$,
$(7, 14, 9, 5, 0)$,
$(7, 14, 10, 2, 1)$,
$(7, 14, 12, 1, 1)$,
$(7, 14, 17, 1, 0)$,
$(7, 14, 19, 0, 0)$,
$(7, 15, 7, 5, 0)$,
$(7, 15, 8, 3, 1)$,
$(7, 15, 9, 2, 1)$,
$(7, 15, 11, 1, 1)$,
$(7, 15, 11, 4, 0)$,
$(7, 15, 13, 3, 0)$,
$(7, 15, 16, 1, 0)$,
$(7, 16, 6, 3, 1)$,
$(7, 16, 6, 6, 0)$,
$(7, 16, 8, 2, 1)$,
$(7, 16, 10, 1, 1)$,
$(7, 16, 10, 4, 0)$,
$(7, 16, 12, 0, 1)$,
$(7, 16, 12, 3, 0)$,
$(7, 16, 15, 2, 0)$,
$(7, 16, 17, 0, 0)$,
$(7, 17, 6, 5, 0)$,
$(7, 17, 7, 4, 0)$,
$(7, 17, 11, 3, 0)$,
$(7, 17, 13, 2, 0)$,
$(7, 17, 14, 1, 0)$,
$(7, 17, 16, 0, 0)$,
$(7, 18, 10, 3, 0)$,
$(7, 18, 13, 1, 0)$,
$(7, 18, 15, 0, 0)$,
$(7, 19, 9, 3, 0)$,
$(7, 20, 6, 4, 0)$,
$(7, 20, 11, 2, 0)$,
$(7, 20, 12, 1, 0)$,
$(7, 20, 14, 0, 0)$,
$(7, 21, 8, 2, 0)$,
$(7, 21, 10, 1, 0)$,
$(7, 21, 12, 0, 0)$,
$(7, 22, 9, 1, 0)$,
$(7, 22, 11, 0, 0)$,
$(7, 23, 6, 3, 0)$,
$(7, 23, 7, 1, 0)$,
$(7, 23, 10, 0, 0)$,
$(7, 24, 6, 2, 0)$,
$(7, 24, 9, 0, 0)$,
$(7, 25, 6, 1, 0)$,
$(7, 25, 8, 0, 0)$,
$(7, 26, 3, 1, 0)$,
$(7, 28, 6, 0, 0)$,
$(7, 29, 3, 0, 0)$,
$(7, 30, 1, 0, 0)$,
$(8, 8, 0, 8, 0)$,
$(8, 8, 9, 7, 0)$,
$(8, 8, 12, 6, 0)$,
$(8, 9, 9, 4, 1)$,
$(8, 9, 10, 6, 0)$,
$(8, 9, 12, 3, 1)$,
$(8, 9, 12, 5, 0)$,
$(8, 9, 13, 4, 0)$,
$(8, 9, 15, 3, 0)$,
$(8, 10, 7, 4, 1)$,
$(8, 10, 10, 3, 1)$,
$(8, 10, 10, 5, 0)$,
$(8, 10, 12, 2, 1)$,
$(8, 10, 12, 4, 0)$,
$(8, 10, 13, 3, 0)$,
$(8, 10, 15, 2, 0)$,
$(8, 11, 6, 4, 1)$,
$(8, 11, 9, 6, 0)$,
$(8, 11, 10, 2, 1)$,
$(8, 11, 11, 4, 0)$,
$(8, 12, 7, 6, 0)$,
$(8, 12, 9, 3, 1)$,
$(8, 12, 9, 5, 0)$,
$(8, 12, 10, 4, 0)$,
$(8, 12, 12, 1, 1)$,
$(8, 12, 14, 2, 0)$,
$(8, 12, 16, 1, 0)$,
$(8, 12, 18, 0, 0)$,
$(8, 13, 7, 3, 1)$,
$(8, 13, 7, 5, 0)$,
$(8, 13, 9, 2, 1)$,
$(8, 13, 12, 0, 1)$,
$(8, 13, 12, 3, 0)$,
$(8, 14, 0, 7, 0)$,
$(8, 14, 6, 6, 0)$,
$(8, 14, 7, 2, 1)$,
$(8, 14, 8, 1, 1)$,
$(8, 14, 9, 4, 0)$,
$(8, 14, 11, 0, 1)$,
$(8, 14, 11, 3, 0)$,
$(8, 14, 13, 2, 0)$,
$(8, 14, 15, 1, 0)$,
$(8, 14, 17, 0, 0)$,
$(8, 15, 6, 3, 1)$,
$(8, 15, 6, 5, 0)$,
$(8, 15, 7, 1, 1)$,
$(8, 16, 0, 6, 0)$,
$(8, 16, 4, 3, 1)$,
$(8, 16, 4, 5, 0)$,
$(8, 16, 6, 2, 1)$,
$(8, 16, 8, 4, 0)$,
$(8, 16, 9, 0, 1)$,
$(8, 16, 10, 3, 0)$,
$(8, 16, 12, 2, 0)$,
$(8, 16, 14, 1, 0)$,
$(8, 16, 16, 0, 0)$,
$(8, 17, 0, 5, 0)$,
$(8, 17, 3, 4, 0)$,
$(8, 17, 8, 3, 0)$,
$(8, 17, 10, 2, 0)$,
$(8, 17, 12, 1, 0)$,
$(8, 17, 14, 0, 0)$,
$(8, 18, 9, 2, 0)$,
$(8, 18, 11, 1, 0)$,
$(8, 18, 12, 0, 0)$,
$(8, 19, 6, 3, 0)$,
$(8, 19, 8, 2, 0)$,
$(8, 20, 0, 4, 0)$,
$(8, 20, 4, 3, 0)$,
$(8, 20, 7, 2, 0)$,
$(8, 20, 9, 1, 0)$,
$(8, 20, 11, 0, 0)$,
$(8, 21, 4, 2, 0)$,
$(8, 21, 7, 1, 0)$,
$(8, 22, 3, 2, 0)$,
$(8, 22, 6, 1, 0)$,
$(8, 22, 9, 0, 0)$,
$(8, 23, 0, 3, 0)$,
$(8, 23, 4, 1, 0)$,
$(8, 24, 0, 2, 0)$,
$(8, 24, 3, 1, 0)$,
$(8, 24, 8, 0, 0)$,
$(8, 25, 1, 1, 0)$,
$(8, 25, 6, 0, 0)$,
$(8, 26, 0, 1, 0)$,
$(8, 26, 4, 0, 0)$,
$(8, 28, 3, 0, 0)$,
$(8, 32, 0, 0, 0)$,
$(9, 8, 10, 4, 0)$,
$(9, 9, 9, 4, 0)$,
$(9, 9, 12, 3, 0)$,
$(9, 10, 8, 4, 0)$,
$(9, 10, 10, 3, 0)$,
$(9, 10, 12, 2, 0)$,
$(9, 10, 13, 1, 0)$,
$(9, 10, 15, 0, 0)$,
$(9, 11, 11, 2, 0)$,
$(9, 12, 7, 4, 0)$,
$(9, 12, 9, 3, 0)$,
$(9, 12, 12, 1, 0)$,
$(9, 12, 14, 0, 0)$,
$(9, 13, 7, 3, 0)$,
$(9, 13, 10, 2, 0)$,
$(9, 14, 9, 2, 0)$,
$(9, 14, 11, 1, 0)$,
$(9, 14, 13, 0, 0)$,
$(9, 15, 6, 3, 0)$,
$(9, 16, 0, 4, 0)$,
$(9, 16, 4, 3, 0)$,
$(9, 16, 8, 2, 0)$,
$(9, 16, 10, 1, 0)$,
$(9, 16, 12, 0, 0)$,
$(9, 17, 3, 3, 0)$,
$(9, 17, 6, 2, 0)$,
$(9, 17, 8, 1, 0)$,
$(9, 17, 10, 0, 0)$,
$(9, 18, 2, 3, 0)$,
$(9, 18, 4, 2, 0)$,
$(9, 18, 7, 1, 0)$,
$(9, 18, 9, 0, 0)$,
$(9, 19, 0, 3, 0)$,
$(9, 19, 3, 2, 0)$,
$(9, 19, 6, 1, 0)$,
$(9, 20, 1, 2, 0)$,
$(9, 20, 5, 1, 0)$,
$(9, 20, 8, 0, 0)$,
$(9, 21, 4, 1, 0)$,
$(9, 21, 6, 0, 0)$,
$(9, 22, 1, 1, 0)$,
$(9, 22, 5, 0, 0)$,
$(9, 24, 4, 0, 0)$,
$(9, 25, 2, 0, 0)$,
$(9, 28, 0, 0, 0)$,
$(10, 8, 6, 4, 0)$,
$(10, 8, 8, 3, 0)$,
$(10, 9, 7, 3, 0)$,
$(10, 9, 10, 2, 0)$,
$(10, 9, 11, 1, 0)$,
$(10, 9, 13, 0, 0)$,
$(10, 10, 5, 4, 0)$,
$(10, 10, 9, 2, 0)$,
$(10, 10, 12, 0, 0)$,
$(10, 11, 6, 3, 0)$,
$(10, 12, 4, 4, 0)$,
$(10, 12, 5, 3, 0)$,
$(10, 12, 7, 2, 0)$,
$(10, 12, 10, 1, 0)$,
$(10, 12, 11, 0, 0)$,
$(10, 13, 6, 2, 0)$,
$(10, 13, 8, 1, 0)$,
$(10, 13, 10, 0, 0)$,
$(10, 14, 3, 3, 0)$,
$(10, 14, 5, 2, 0)$,
$(10, 14, 9, 0, 0)$,
$(10, 15, 2, 3, 0)$,
$(10, 15, 7, 1, 0)$,
$(10, 16, 4, 2, 0)$,
$(10, 16, 6, 1, 0)$,
$(10, 16, 8, 0, 0)$,
$(10, 17, 4, 1, 0)$,
$(10, 17, 6, 0, 0)$,
$(10, 18, 2, 1, 0)$,
$(10, 18, 5, 0, 0)$,
$(10, 20, 4, 0, 0)$,
$(10, 21, 2, 0, 0)$,
$(10, 22, 1, 0, 0)$,
$(10, 24, 0, 0, 0)$,
$(11, 4, 6, 4, 0)$,
$(11, 6, 5, 4, 0)$,
$(11, 7, 6, 3, 0)$,
$(11, 8, 4, 4, 0)$,
$(11, 8, 5, 3, 0)$,
$(11, 9, 6, 2, 0)$,
$(11, 9, 8, 1, 0)$,
$(11, 9, 10, 0, 0)$,
$(11, 10, 3, 3, 0)$,
$(11, 10, 5, 2, 0)$,
$(11, 10, 9, 0, 0)$,
$(11, 11, 2, 3, 0)$,
$(11, 11, 7, 1, 0)$,
$(11, 12, 4, 2, 0)$,
$(11, 12, 6, 1, 0)$,
$(11, 12, 8, 0, 0)$,
$(11, 13, 4, 1, 0)$,
$(11, 13, 6, 0, 0)$,
$(11, 14, 2, 1, 0)$,
$(11, 14, 5, 0, 0)$,
$(11, 16, 4, 0, 0)$,
$(11, 17, 2, 0, 0)$,
$(11, 18, 1, 0, 0)$,
$(11, 20, 0, 0, 0)$,
$(12, 4, 3, 3, 0)$,
$(12, 6, 2, 3, 0)$,
$(12, 6, 5, 2, 0)$,
$(12, 6, 7, 1, 0)$,
$(12, 6, 9, 0, 0)$,
$(12, 8, 4, 2, 0)$,
$(12, 8, 6, 1, 0)$,
$(12, 8, 8, 0, 0)$,
$(12, 9, 4, 1, 0)$,
$(12, 9, 6, 0, 0)$,
$(12, 10, 2, 1, 0)$,
$(12, 10, 5, 0, 0)$,
$(12, 12, 4, 0, 0)$,
$(12, 13, 2, 0, 0)$,
$(12, 14, 1, 0, 0)$,
$(12, 16, 0, 0, 0)$,
$(13, 6, 5, 0, 0)$,
$(13, 8, 4, 0, 0)$,
$(13, 9, 2, 0, 0)$,
$(13, 10, 1, 0, 0)$,
$(13, 12, 0, 0, 0)$,
$(14, 4, 3, 0, 0)$,
$(14, 5, 2, 0, 0)$,
$(14, 6, 1, 0, 0)$,
$(14, 8, 0, 0, 0)$,
$(15, 4, 0, 0, 0)$,
$(16, 0, 0, 0, 0)$}}
\caption{The Pareto-optimal statistics of Moser sets in $[3]^4$. This table can also be found at {\tt http://spreadsheets.google.com/ccc?key=rwXB\_Rn3Q1Zf5yaeMQL-RDw}}
\label{table4}
\end{figure}

\begin{proof} This was computed by computer search as follows. First, one observed that if $(a,b,c,d,e)$ was Pareto-optimal, then $a\geq 3$. To see this, it suffices to show that for any Moser set $A \subset [3]^4$ with $a(A)=0$, it is possible to add three points from $S_{4,4}$ to $A$ and still have a Moser set. To show this, suppose first that $A$ contains a point from $S_{1,4}$, such as $2221$. Then $A$ must omit either $2211$ or $2231$; without loss of generality we may assume that it omits $2211$. Similarly we may assume it omits $2121$ and $1221$. Then we can add $1131$, $1311$, $3111$ to $A$, as required. Thus we may assume that $A$ contains no points from $S_{1,4}$. Now suppose that $A$ omits a point from $S_{2,4}$, such as $2211$. Then one can add $3333$, $3111$, $1311$ to $A$, as required. Thus we may assume that A contains all of $S_{2,4}$, which forces $A$ to omit $2222$, as well as at least one point from $S_{3,4}$, such as $2111$. But then $3111$, $1111$, $3333$ can be added to the set, a contradiction.

Thus we only need to search through sets $A \subset [3]^4$ for which $|A \cap S_{4,4}| \geq 3$. A straightforward computer search shows that up to the symmetries of the cube, there are $391$ possible choices for $A \cap S_{4,4}$. For each such choice, we looped through all the possible values of the slices $A \cap 1***$ and $A \cap 3***$, i.e. all three-dimensional Moser sets which had the indicated intersection with $S_{3,3}$. (For fixed $A \cap S_{4,4}$, the number of possibilities for $A \cap 1***$ ranges from $1$ to $87123$, and similarly for $A \cap 3***$). For each pair of slices $A \cap 1***$ and $A \cap 3***$, we computed the lines connecting these two sets to see what subset of $2***$ was excluded from $A$; there are $2^{27}$ possible such exclusion sets. We precomputed a lookup table that gave the Pareto-optimal statistics for $A \cap 2***$ for each such choice of exclusion set; using this lookup table for each choice of $A \cap 1***$ and $A \cap 3***$ and collating the results, we obtained the above list. On a linux cluster, the lookup table took 22 minutes to create, and the loop over the $A \cap 1***$ and $A \cap 3***$ slices took two hours, spread out over $391$ machines (one for each choice of $A \cap S_{4,4}$). Further details (including source code) can be found at
\centerline{{\tt http://michaelnielsen.org/polymath1/index.php?title=4D\_Moser\_brute\_force\_search}.}
\end{proof}

As a consequence of this data, we have the following facts about the statistics of large Moser sets:

\begin{proposition}\label{stat} Let $A \subset [3]^4$ be a Moser set with statistics $(a,b,c,d,e)$.
\begin{itemize}
\item[(i)] If $|A| \geq 40$, then $e=0$.
\item[(ii)] If $|A| \geq 43$, then $d=0$.
\item[(iii)] If $|A| \geq 42$, then $d \leq 2$.
\item[(iv)] If $|A| \geq 41$, then $d \leq 3$.
\item[(v)] If $|A| \geq 40$, then $d \leq 6$.
\item[(vi)] If $|A| \geq 43$, then $c \geq 18$.
\item[(vii)] If $|A| \geq 42$, then $c \geq 12$.
\item[(viii)] If $|A| \geq 43$, then $b \geq 15$.
\end{itemize}
\end{proposition}

\begin{remark} This proposition was first established by an integer program, see Appendix \ref{integer-sec}. A computer-free proof can be found at \centerline{{\tt http://terrytao.files.wordpress.com/2009/06/polymath2.pdf}.}
\end{remark}

\subsection{Five dimensions}

Now we establish the bound $c'_{5,3}=124$. In view of the lower bounds, it suffices to show that there does not exist a Moser set $A \subset [3]^5$ with $|A|=125$.

We argue by contradiction. Let $A$ be as above, and let $(a(A),\ldots,f(A))$ be the statistics of $A$.

\begin{lemma}\label{fvan} $f(A)=0$.
\end{lemma}

\begin{proof} If $f(A)$ is non-zero, then $A$ contains $22222$, then each of the $\frac{3^5-1}{2} = 121$ antipodal pairs in $[3]^5$ can have at most one point in $A$, leading to only $122$ points.
\end{proof}

Let us slice $[3]^5$ into three parallel slices, e.g. $1****, 2****, 3****$. The intersection of $A$ with each of these slices has size at most $43$. In particular, this implies that
\begin{equation}\label{boo}
|A \cap 1****| + |A \cap 3****| = 125 - |A \cap 2****| \geq 82.
\end{equation}
Thus at least one of $A \cap 1****$, $A \cap 3****$ has cardinality at least $41$; by Proposition \ref{stat}(iv) we conclude that
\begin{equation}\label{d13}
\min( d(1****), d(3****) ) \leq 3.
\end{equation}
Furthermore, equality can only hold in \eqref{d13} if $A \cap 1****$, $A \cap 3****$ both have cardinality exactly $41$, in which case from Proposition \ref{stat}(iv) again we must have
\begin{equation}\label{d13a}
d(1****)=d(3****)=3.
\end{equation}
Of course, we have a similar result for permutations.

Now we improve the bound $|A \cap 2****| \leq 43$:

\begin{lemma} $|A \cap 2****| \leq 41$.
\end{lemma}

\begin{proof} Suppose first that $|A \cap 2****|=43$. Let $A' \subset [3]^4$ be the subset of $[3]^4$ corresponding to $A \cap 2****$, thus $A'$ is a Moser set of cardinality $43$. By Proposition \ref{stat}(vi), $c(A') \geq 18$. By Lemma \ref{dci}, the sum of the $c(V)$, where $V$ ranges over the eight side slices of $[3]^4$, is therefore at least $36$. By the pigeonhole principle, we may thus find two opposing side slices, say $1***$ and $3***$, with $c(1***)+c(3****) \geq 9$. Since $c(1***), c(3***)$ cannot exceed $6$, we thus have $c(1***), c(3***) \geq 3$, with at least one of $c(1***), c(3***)$ being at least $5$. Passing back to $A$, this implies that $d(*1***), d(*3***) \geq 3$, with at least one of $d(*1***), d(*3***)$ being at least $5$. But this contradicts \eqref{d13} together with the refinement \eqref{d13a}.

We have just shown that $|A \cap 2****| \leq 42$; we can thus improve \eqref{boo} to
$$ |A \cap 1****| + |A \cap 3****| \geq 83.$$
Combining this with Proposition \ref{stat}(ii)-(v) we see that
\begin{equation}\label{d13-6}
d(1****)+d(3****) \leq 6
\end{equation}
with equality only if $|A \cap 2****|=42$, and similarly for permutations.

Now let $A'$ be defined as before. Then we have
$$ c(1***) + c(3***) \leq 6$$
and similarly for permutations. Applying Lemma \ref{dci}, this implies that $c(2****) = c(A') \leq 12$.

Now suppose for contradiction that $|A'|=|A \cap 2****|=42$. Then by Proposition \ref{stat}(vii) we have
\begin{equation}\label{coo-1}
c(2****) = 12;
\end{equation}
applying Lemma \ref{dci} again, this forces $c(1***)+c(3***)=6$ and similarly for permutations, which then implies that
\begin{equation}\label{doo}
d(*1***)+d(*3***) = d(**1**)+d(**3**) = d(***1*)+d(***3*) = d(****1)+d(****3) = 6
\end{equation}
and hence
$$ |A \cap *2***| = |A \cap **2**| = |A \cap ***2*| = |A \cap ****2| = 42$$
and thus
\begin{equation}\label{coo-2}
c(*2***) = c(**2**) = c(***2*) = c(****2) = 12.
\end{equation}
Combining \eqref{coo-1}, \eqref{doo}, \eqref{coo-2} we conclude that
$$ d(1****)+d(3****) = 16,$$
contradicting \eqref{d13-6}.
\end{proof}

With this proposition, the bound \eqref{boo} now improves to
\begin{equation}\label{84}
|A \cap 1****| + |A \cap 3****| \geq 84
\end{equation}
and in particular
\begin{equation}\label{41}
|A \cap 1****|, |A \cap 3****| \geq 41.
\end{equation}
from this and Proposition \ref{stat}(ii)-(iv) we now have
\begin{equation}\label{d13-improv}
d(1****)+d(3****) \leq 4
\end{equation}
and similarly for permutations.

\begin{lemma}\label{evan} $e(A)=0$.
\end{lemma}

\begin{proof} From \eqref{84}, the intersection of $A$ with any side slice has cardinality at least $41$, and thus by Proposition \ref{stat}(i) such a side slice has an $e$-statistic of zero. The claim then follows from Lemma \ref{dci}.
\end{proof}

We need a technical lemma:

\begin{lemma}\label{tech} Let $B \subset S_{5,5}$. Then there exist at least $|B|-4$ pairs of strings in $B$ which differ in exactly two positions.
\end{lemma}

\begin{proof} The first non-vacuous case is $|B|=5$. It suffices to establish this case, as the higher cases then follow by induction (locating a pair of the desired form, then deleting one element of that pair from $B$).

Suppose for contradiction that one can find a $5$-element set $B \subset S_{5,5}$ such that no two strings in $B$ differ in exactly two positions. Recall that we may split $S_{5,5}=S_{5,5}^e \cup S_{5,5}^o$, where $S_{5,5}^e$ are those strings with an even number of $1$'s, and $S_{5,5}^o$ are those strings with an odd number of $1$'s. By the pigeonhole principle and symmetry we may assume $B$ has at least three elements in $S_{5,5}^o$. Without loss of generality, we can take one of them to be $11111$, thus excluding all elements in $S_{5,5}^o$ with exactly two $3$s, leaving only the elements with exactly four $3$s. But any two of them differ in exactly two positions, a contradiction.
\end{proof}

We can now improve the trivial bound $c(A) \leq 80$:

\begin{corollary}[Non-maximal $c$]\label{cmax} $c(A) \leq 79$. If $a(A) \geq 7$, then $c(A) \leq 78$.
\end{corollary}

\begin{proof} If $c(A)=80$, then $A$ contains all of $S_{3,5}$, which then implies that no two elements in $A \cap S_{5,5}$ can differ in exactly two places. It also implies (from \eqref{alpha-1}) that $d(A)$ must vanish, and that $b(A)$ is at most $40$. By Lemma \ref{tech}, we also have that $a(A) = |A \cap S_{5,5}|$ is at most $4$. Thus $|A| \leq 4 + 40 + 80 + 0 + 0 = 124$, a contradiction.

Now suppose that $a(A) \geq 7$. Then by Lemma \ref{tech} there are at least three pairs in $A \cap S_{5,5}$ that differ in exactly two places. Each such pair eliminates one point from $A \cap S_{3,5}$; but each point in $S_{3,5}$ can be eliminated by at most two such pairs, and so we have at least two points eliminated from $A \cap S_{3,5}$, i.e. $c(A) \leq 78$ as required.
\end{proof}

Next, we rewrite the quantity $125=|A|$ in terms of side slices. From Lemmas \ref{fvan}, \ref{evan} we have
$$ a(A) + b(A) + c(A) + d(A) = 125$$
and hence by Lemma \ref{dci}, the quantity
$$ s(V) := a(V) + \frac{5}{4} b(V) + \frac{5}{3} c(V) + \frac{5}{2} d(V) - \frac{125}{2},$$
where $V$ ranges over side slices, has an average value of zero.

\begin{proposition}[Large values of $s(V)$]\label{suv} For all side slices, we have $s(V) \leq 1/2$. Furthermore, we have $s(V) < -1/2$ unless the statistics $(a(V), b(V), c(V), d(V), e(V))$ are of one of the following four cases:
\begin{itemize}
\item (Type 1) $(a(V),b(V),c(V),d(V),e(V)) = (2,16,24,0,0)$ (and $s(V) = -1/2$ and $|A \cap V| = 42$);
\item (Type 2) $(a(V),b(V),c(V),d(V),e(V)) = (4,16,23,0,0)$ (and $s(V) = -1/6$ and $|A \cap V| = 43$);
\item (Type 3) $(a(V),b(V),c(V),d(V),e(V)) = (4,15,24,0,0)$ (and $s(V) = 1/4$ and $|A \cap V| = 43$);
\item (Type 4) $(a(V),b(V),c(V),d(V),e(V)) = (3,16,24,0,0)$ (and $s(V) = 1/2$ and $|A \cap V| = 43$);
\end{itemize}
\end{proposition}

\begin{proof}
Let $V$ be a side slice. From \eqref{41} we have
$$ 41 \leq a(V)+b(V)+c(V)+d(V) = |A \cap V| \leq 43.$$
First suppose that $|A \cap V| = 43$, then from Proposition \ref{stat}(ii), (viii), $d(V)=0$ and $b(V) \geq 15$.
Also, we have the trivial bound $c(V) \leq 24$, together with the inequality
$$ 3b(V) + 2c(V) \leq 96$$
from \eqref{alpha-1}. To exploit these facts, we rewrite $s(V)$ as
$$ s(V) = \frac{1}{2} - \frac{1}{2}( 24 - c(V) ) - \frac{1}{12} (96-3b(V)-2c(V)).$$
Thus $s(V) \leq 1/2$ in this case. If $s(V) \geq -1/2$, then
$$ 6 (24-c(V)) + (96-3b(V)-2c(V)) \leq 12,$$
which together with the inequalities $b(V) \leq 15$, $c(V) \leq 24$, $3b(V)+2c(V) \leq 96$ we conclude that $(b(V),c(V))$ must be one of $(16,24)$, $(15, 24)$, $(16, 23)$, $(15, 23)$. The first three possibilities lead to Types 4,3,2 respectively. The fourth type would lead to $(a(V),b(V),c(V),d(V),e(V)) = (5,15,23,0,0)$, but this contradicts \eqref{eleven}.

Next, suppose $|A \cap V| = 42$, so by Proposition \ref{stat}(iii) we have $d(V) \leq 2$. From \eqref{alpha-1} we have
\begin{equation}\label{2cd}
2c(V) + 3d(V) \leq 48
\end{equation}
while from \eqref{alpha-2} we have
\begin{equation}\label{3cd}
3b(V)+2c(V)+3d(V) \leq 96
\end{equation}
and so we can rewrite $s(V)$ as
\begin{equation}\label{sv2}
s(V) = -\frac{1}{2} - \frac{1}{4}( 48 - 2c(V) - 3d(V) ) - \frac{1}{12} (96-3b(V)-2c(V)-3d(V)) + \frac{1}{2} d(V).
\end{equation}
This already gives $s(V) \leq 1/2$. If $d(V)=0$, then $s(V) \leq -1/2$, with equality only in Type 1. If $d(V)=1$, then the set $A' \subset [3]^4$ corresponding to $A \cap V$ contains a point in $S_{3,4}$, which without loss of generality we can take to be $2221$. Considering the three lines $*221$, $2*21$, $22*1$, we see that at least three points in $S_{2,4}$ must be missing from $A'$, thus $c(V) \leq 21$. This forces $48-2c(V)-3d(V) \geq 3$, and so $s(V) < -3/4$. Finally, if $d(V)=2$, then $A'$ contains two points in $S_{3,4}$. If they are antipodal (e.g. $2221$ and $2223$), the same argument as above shows that at least six points in $S_{2,4}$ are missing from $A'$; if they are not antipodal (e.g. $2221$ and $2212$) then by considering the lines $*221$, $2*21$, $22*1$, $*212$, $2*12$ we see that five points are missing. Thus we have $c(V) \leq 19$, which forces $48-2c(V)-3d(V) \geq 4$. This forces $s(V) \leq -1/2$, with equality only when $c(V)=19$ and $3b(V)+2c(V)+3d(V)=96$, but this forces $b(V)$ to be the non-integer $52/3$, a contradiction, which concludes the treatment of the $|A \cap V|=42$ case.

Finally, suppose $|A \cap V| = 41$. Using \eqref{2cd}, \eqref{3cd} as before we have
\begin{equation}\label{sv3}
s(V) = -\frac{3}{2} - \frac{1}{4}( 48 - 2c(V) - 3d(V) ) - \frac{1}{12} (96-3b(V)-2c(V)-3d(V)) + \frac{1}{2} d(V),
\end{equation}
while from Proposition \ref{stat}(vi) we have $d(V) \leq 3$. This already gives $s(V) \leq 0$, and $s(V) \leq -1$ when $d(V)=1$. In order to have $s(V) \geq -1/2$, we must then have $d(V)=2$ or $d(V)=3$. But then the arguments of the preceding paragraph give $48-2c(V)-3d(V) \geq 4$, and so $s(V) \leq -1$ in this case.
\end{proof}

Since the $s(V)$ average to zero, by the pigeonhole principle we may find two opposing side slices (e.g. $1****$ and $3****$), whose total $s$-value is non-negative. Actually we can do a little better:

\begin{lemma}\label{side-off} There exists two opposing side slices whose total $s$-value is strictly positive.
\end{lemma}

\begin{proof} If this is not the case, then we must have $s(1****)+s(3****)=0$ and similarly for permutations. Using Proposition \ref{suv} we thus see that for every opposing pair of side slices, one is Type 1 and one is Type 4. In particular $c(V)=24$ for all side slices $V$. But then by Lemma \ref{dci} we have $c(A)=80$, contradicting Lemma \ref{cmax}.
\end{proof}

Let $V, V'$ be the side slices in Lemma \ref{side-off}
By Proposition \ref{suv}, the $V, V'$ slices must then be either Type 2, Type 3, or Type 4, and they cannot both be Type 2. Since $a(A) = a(V)+a(V')$, we conclude
\begin{equation}\label{amix}
6 \leq a(A) \leq 8.
\end{equation}
In a similar spirit, we have
$$ c(V) + c(V') \leq 23+24.$$
On the other hand, by considering the $24$ lines connecting $c$-points of $V, V'$ to $c$-points of the centre slice $W$ between $V$ and $V'$, each of which contains at most two points in $A$, we have
$$ c(V) + c(W) + c(V') \leq 24 \times 2.$$
Thus $c(W) \leq 1$; since
$$ d(A) = d(V) + d(V') + c(W)$$
we conclude from Proposition \ref{suv} that $d(A) \leq 1$. Actually we can do better:

\begin{lemma} $d(A)=0$.
\end{lemma}

\begin{proof} Suppose for contradiction that $d(A)=1$; without loss of generality we may take $11222 \in A$. This implies that $d(1****)=d(*1***)=1$. Also, by the above discussion, $c(**1**)$ and $c(**3**)$ cannot both be $24$, so by Proposition \ref{suv}, $s(**1**)+s(**3**) \leq 1/3$; similarly
$s(***1*)+s(***3*) \leq 1/3$ and $s(****1)+s(****3) \leq 1/3$. Since the $s$ average to zero, we see from the pigeonhole principle that either $s(1****)+s(3****) \geq -1/2$ or $s(*1***)+s(*3***) \geq -1/2$. We may assume by symmetry that
\begin{equation}\label{star-2}
s(1****)+s(3****) \geq -1/2.
\end{equation}
Since $s(3****) \leq 1/2$ by Proposition \ref{suv}, we conclude that
\begin{equation}\label{star}
s(1****) \geq -1.
\end{equation}
If $|A \cap 1****|=41$, then by \eqref{sv3} we have
$$ s(1****) = -1 - \frac{1}{4}( 48 - 2c(1****) - 3d(1****) ) - \frac{1}{12} (96-3b(1****)-2c(1****)-3d(1****))$$
but the arguments in Proposition \ref{suv} give $48 - 2c(1****) - 3d(1****) \geq 3$ and $96-3b(1****)-2c(1****)-3d(1****) \geq 0$, a contradiction. So we must have $|A \cap 1****|=42$ (by Proposition \ref{stat}(ii) and \eqref{41}). In that case, from \eqref{sv2} we have
$$ s(1****) = \frac{1}{4}( 48 - 2c(1****) - 3d(1****) ) - \frac{1}{12} (96-3b(1****)-2c(1****)-3d(1****))$$
while also having $48 - 2c(1****) - 3d(1****) \geq 3$ and $96-3b(1****)-2c(1****)-3d(1****) \geq 0$. Since $s(1****) \geq -1$ and $d(1****)=1$, we soon see that we must have $48 - 2c(1****) - 3d(1****) = 3$ and $96-3b(1****)-2c(1****)-3d(1****) \leq 3$, which forces $c(1****)=21$ and $b(1****)=16$ or $b(1****)=17$; thus the statistics of $1****$ are either $(4,16,21,1,0)$ or $(3,17,21,1,0)$.

We first eliminate the $(3,17,21,1,0)$ case. In this case $s(1****)$ is exactly $-1$. Inspecting the proof of \eqref{star}, we conclude that $s(3****)$ must be $+1/2$ and that $s(**1**)+s(**3**)=1/3$. From the former fact and Proposition \ref{suv} we see that $a(A) = a(1****)+a(3****)=3+3=6$; on the other hand, from the latter fact and Proposition \ref{suv} we have $a(A) = a(**1**)+a(**3**) = 4+3=7$, a contradiction.

So $1****$ has statistics $(4,16,21,1,0)$, which implies that $s(1****)=-3/4$ and $|A \cap 1****|=42$. By \eqref{star-2} we conclude
\begin{equation}\label{s3}
s(3****) \geq 1/4,
\end{equation}
which by Proposition \ref{suv} implies that $|A \cap 3****|=43$, and hence $|A \cap 2****|=40$. On the other hand, since $e(A)=f(A)=0$ and $d(A)=1$, with the latter being caused by $11222$, we see that $c(2****)=d(2****)=e(2****)=0$. From \eqref{alpha-1} we have $4a(2****)+b(2****) \leq 64$, and we also have the trivial inequality $b(2****) \leq 32$; these inequalities are only compatible if $2****$ has statistics $(8,32,0,0,0)$, thus $A \cap 2****$ contains $S_{2,5} \cap 2****$.

If $a(3****)=4$, then $a(A)=a(1****)+a(3****)=8$, which by Proposition \ref{suv} implies that $s(**1**)+s(**3**)$ cannot exceed $1/12$, and similarly for permutations. On the other hand, from Proposition \ref{suv} $s(**1**)+s(**3**)$ cannot exceed $-3/4 + 1/4 = -1/2$, and so the average value of $s$ cannot be zero, a contradiction. Thus $a(3****) \neq 4$, which by \eqref{s3} and Proposition \ref{suv} implies that $**3**$ has statistics $(3,16,24,0,0)$.

In particular, $A$ contains $16$ points from $3**** \cap S_{1,5}$ and all of $3**** \cap S_{2,5}$. As a consequence, no pair of the $16$ points in $A \cap 3**** \cap S_{1,5}$ can differ in only one coordinate; partitioning the $32$-point set $3**** \cap S_{1,5}$ into $16$ such pairs, we conclude that every such pair contains exactly one element of $A$. We conclude that $A \cap 3**** \cap S_{1,5}$ is equal to either $3**** \cap S_{1,5}^e$ or $3**** \cap S_{1,5}^o$.

On the other hand, $A$ contains all of $2**** \cap S_{2,5}$, and exactly sixteen points from $1**** \cap S_{1,5}$. Considering the vertical lines $*xyzw$ where $xyzw \in S_{1,4}$, we conclude that $A \cap 1**** \cap S_{1,5}$ is either equal to $1**** \cap S_{1,5}^o$ or $1**** \cap S_{1,5}^e$.
But either case is incompatible with the fact that $A$ contains $11222$ (consider either the line $11xx2$ or $11x\overline{x}2$, where $x=1,2,3$ and $\overline{x}=4-x$), obtaining the required contradiction.
\end{proof}

We can now eliminate all but three cases for the statistics of $A$:

\begin{proposition}[Statistics of $A$] The statistics $(a(A),b(A),c(A),d(A),e(A),f(A))$ of $A$ must be one of the following three tuples:
\begin{itemize}
\item (Case 1) $(6,40,79,0,0)$;
\item (Case 2) $(7,40,78,0,0)$;
\item (Case 3) $(8,39,78,0,0)$.
\end{itemize}
\end{proposition}

\begin{proof}
Since $d(A)=e(A)=f(A)=0$, we have
$$ c(2****)=d(2****)=e(2****)=0.$$
On the other hand, from \eqref{alpha-1} we have $4a(2****)+b(2****) \leq 64$ as well as the trivial inequality $b(2****) \leq 24$, and also we have
$$ |A \cap 2****| = 125 - |A \cap 1****| - |A \cap 3****| \geq 125 - 43 - 43 = 39.$$
Putting all this together, we see that the only possible statistics for $2****$ are $(8,32,0,0,0)$, $(7,32,0,0,0)$, or $(8,31,0,0,0)$. In particular, $7 \leq a(2****) \leq 8$ and $31 \leq b(2****) \leq 32$, and similarly for permutations. Applying Lemma \ref{dci} we conclude that
$$ 35 \leq b(A) \leq 40$$
and
$$ 77.5 \leq c(A) \leq 80.$$
Combining this with the first part of Corollary \ref{cmax} we conclude that $c(A)$ is either $78$ or $79$. From this and \eqref{amix} we see that the only cases that remain to be eliminated are $(7,39,79,0,0)$ and $(8,38,79,0,0)$, but these cases are incompatible with the second part of Corollary \ref{cmax}.
\end{proof}

We now eliminate each of the three remaining cases in turn.

\subsection{Elimination of $(6,40,79,0,0)$}

Here $A \cap S_{5,5}$ has six points. By Lemma \ref{tech}, there are at least two pairs in this set which differ in two positions. Their midpoints are eliminated from $A \cap S_{3,5}$. But $A$ omits exactly one point from $S_{3,5}$, so these midpoints must be the same. By symmetry, we may then assume that these two pairs are $(11111,11133)$ and $(11113,11131)$. Thus the eliminated point in $S_{3,5}$ is $11122$, i.e. $A$ contains $S_{3,5} \backslash \{11122\}$. Also, $A$ contains $\{11111,11133,11113,11131\}$ and thus must omit $\{11121, 11123, 11112, 11132\}$.

Since $11322 \in A$, at most one of $11312, 11332$ lie in $A$. By symmetry we may assume $11312 \not \in A$, thus there is a pair $(xy1z2, xy3z2)$ with $x,y,z = 1,3$ that is totally omitted from $A$, namely $(11112,11312)$. On the other hand, every other pair of this form can have at most one point in the $A$, thus there are at most seven points in $A$ of the form $xyzw2$ with $x,y,z,w = 1,3$. Similarly there are at most 8 points of the form $xyz2w$, or of $xy2zw$, $x2yzw$, $2xyzw$, leading to $b(A) \leq 7+8+8+8+8=39$, contradicting the statistic $b(A)=40$.

\subsection{Elimination of $(7,40,78,0,0)$}

Here $A \cap S_{5,5}$ has seven points. By Lemma \ref{tech}, there are at least three pairs in this set which differ in two positions. As we can only eliminate two points from $S_{3,5}$, two of the midpoints of these pairs must be the same; thus, as in the previous section, we may assume that $A$ contains $\{11111,11133,11113,11131\}$ and omits $\{11121, 11123, 11112, 11132\}$ and $11122$.

Now consider the $160$ lines $\ell$ connecting two points in $S_{4,5}$ to one point in $S_{3,5}$ (i.e. $*2xyz$ and permutations, where $x,y,z=1,3$). By double counting, the total sum of $|\ell \cap A|$ over all $160$ lines is $4b(A)+2c(A) = 316 = 158 \times 2$. On the other hand, each of these lines contain at most two points in $A$, but two of them (namely $1112*$ and $1112*$) contain no points. Thus we must have $|\ell \cap A|=2$ for the remaining $158$ lines $\ell$.

Since $A$ omits $1112x$ and $111x2$ for $x=1,3$, we thus conclude (by considering the lines $11*2x$ and $11*x2$) that $A$ must contain $1132x$, $113x2$, $1312x$, and $131x2$. Taking midpoints, we conclude that $A$ omits $11322$ and $13122$. But together with $11122$ this implies that at least three points are missing from $A \cap S_{3,5}$, contradicting the hypothesis $c(A)=78$.

\subsection{Elimination of $(8,39,78,0,0)$}

Now $A \cap S_{5,5}$ has eight points. By Lemma \ref{tech}, there are at least three pairs in this set which differ in two positions. As we can only eliminate two points from $S_{3,5}$, two of these pairs $(a,b), (c,d)$ must have the same midpoint $p$, and two other pairs $(a',b'), (c',d')$ must have the same midpoint $p'$, and $A$ contains $S_{3,5} \backslash \{p,p'\}$. As $p,p'$ are distinct, the plane containing $a,b,c,d$ is distinct from the plane containing $a',b',c',d'$.

Again consider the $160$ lines $\ell$ from the previous section. This time, the sum of the $|\ell \cap A|$ is $4b(A)+2c(A) = 312 = 156 \times 2$.
But the two lines in the plane of $a,b,c,d$ passing through $p$, and the two lines in the plane of $a',b',c',d'$ passing through $p'$, have no points; thus we must have $|\ell \cap A|=2$ for the remaining $156$ lines $\ell$.

Without loss of generality we have $(a,b)=(11111,11133)$, $(c,d) = (11113,11131)$, thus $p = 11122$. By permuting the first three indices, we may assume that $p'$ is not of the form $x2y2z, x2yz2, xy22z, xy2z2$ for any $x,y,z=1,3$. Then we have $1112x \not \in A$ and $1122x \in A$ for every $x=1,3$, so by the preceding paragraph we have $1132x \in A$; similarly for $113x2, 1312x, 131x2$. Taking midpoints, this implies that $13122, 11322 \not \in A$, but this (together with 11122) shows that at least three points aremissing from $A \cap S_{3,5}$, contradicting the hypothesis $c(A)=78$.

\subsection{Six dimensions}

Now we establish the bound $c'_{6,3}=353$. In view of the lower bounds, it suffices to show that there does not exist a Moser set $A \subset [3]^5$ with $|A|=354$.

We argue by contradiction. Let $A$ be as above, and let $(a(A),\ldots,g(A))$ be the statistics of $A$.

\begin{lemma}\label{g6} $g(A)=0$.
\end{lemma}

\begin{proof} For any four-dimensional slice $V$ of $A$, define
$$S(V) := 15 a(V) + 5 b(V) + 5 c(V)/2 + 3d(V)/2 + e(V).$$
From Lemma \ref{dci} we see that $|A|$ is equal to $a(A)+b(A)$ plus the average of $S(V)$ where $V$ ranges over the twenty slices which are some permutation of the center slice $22****$.

If $g(A)=1$, then $a(A) \leq 32$ and $b(A) \leq 96$ by \eqref{alpha-1}. Meanwhile, $e(V)=g(A)=1$ for every center slice $V$, so from Lemma \ref{paretop-4}, one can show that $S(V) \leq 223.5$ for every such slice. We conclude that $|A| \leq 351.5$, a contradiction.
\end{proof}

For any four-dimensional slice $V$ of $A$, define the \emph{defects} to be
$$ D(V) := 356 - [4a(V)+6b(V)+10c(V)+20d(V)+60e(V)].$$
Define a \emph{corner slice} to be one of the permutations or reflections of $11****$, thus there are $60$ corner slices. From Lemma \ref{dci} we see that $356-|A|+f(A)=2+f(A)$ is the average of the defects of all the $60$ corner slices. On the other hand, from Lemma \ref{paretop-4} and a straightforward computation, one concludes

\begin{lemma}\label{defects} Let $A$ be a four-dimensional Moser set. Then $D(A) \geq 0$. Furthermore:
\begin{itemize}
\item If $A$ has statistics $(6,12,18,4,0)$, then $D(A)=0$.
\item If $A$ has statistics $(5,12,18,4,0)$, $(5,12,12,4,1)$, or $(6,8,12,8,0)$, then $D(A)=4$.
\item For all other $A$, $D(A) \geq 6$.
\item If $a(A) = 4$, then $D(A) \geq 8$.
\item If $a(A) \geq 7$, then $D(A) \geq 16$.
\item If $a(A) \geq 8$, then $D(A) \geq 30$.
\item If $a(A) \geq 9$, then $D(A) \geq 86$.
\end{itemize}
\end{lemma}

Define a \emph{family} to be a set of four parallel corner slices, thus there are $15$ families, which are all a permutation of $\{11****, 13****, 31****, 33**** \}$. We refer to the family $\{11****, 13****, 31****, 33**** \}$ as $ab****$, and similarly define the family $a*b***$, etc.

\begin{lemma}\label{f6} $f(A)=0$.
\end{lemma}

\begin{proof} For any four-dimensional slice $V$ of $A$, define
$$ s(V) := 12 a(V)+15 b(V)/2+20 c(V)/3+15 d(V)/2 + 12 e(V),$$
and define an \emph{edge slice} to be one of the $30$ permutations or reflections of $12****$. From double counting we see that $|A|-a(A)$ is equal to the average of the $30$ values of $s(V)$ as $V$ ranges over edge slices.

From Lemma \ref{paretop-4} one can verify that $s(V) \leq 336$, and that $s(V) \leq 296 = 336-40$ if $e(V)=1$. The number of edge slices $V$ for which $e(V)=1$ is equal to $5f(A)$, and so the average value of the $s(V)$ is at most $336 - \frac{40 \times 5}{30} f(A)$, and so
$$ |A| - a(A) \leq 336 - \frac{40 \times 5}{30} f(A)$$
which we can rearrange (using $|A|=354$) as
$$ a(A) \geq 18 + \frac{20}{3} f(A).$$
Suppose first that $f(A)=1$; then $a(A) \geq 25$. This means that in any given family, one of the four corner slices has an $a$ value of at least $7$, and thus by Lemma \ref{defects} has a defect of at least $16$. Thus the average defect is at least $4$; on the other hand, the average defect is $2+f(A)=3$, a contradiction.

Now suppose that $f(A) \geq 2$; then $a(A) \geq 32$. Then in any given family, there is a corner slice with an $a$ value at least $9$, or four slices with $a$ value at least $8$, leading to a total defect of at least $86$ by Lemma \ref{defects}. Thus the average defect is at least $21.5$; on the other hand, the average defect is $2+f(A) \leq 2+12$, a contradiction.
\end{proof}

As a consequence of the above lemma, we see that the average defect of all corner slices is $2$, or equivalently that the total defect of these slices is $120$.

Call a corner slice \emph{good} if it has statistics $(6,12,18,4,0)$, and \emph{bad} otherwise. Thus good slices have zero defect, and bad slices have defect at least four. Since the average defect of the $60$ corner slices is $2$, there are at least $30$ good slices.

One can describe the structure of the good slices completely:

\begin{lemma}\label{sixt} The subset of $[3]^4$ consisting of the strings $1111, 1113, 3333, 1332, 1322, 1222, 3322$ and permutations is a Moser set with statistics $(6,12,18,4,0)$. Conversely, every Moser set with statistics $(6,12,18,4,0)$ is of this form up to the symmetries of the cube $[3]^4$.
\end{lemma}

\begin{proof} This can be verified by computer. By symmetry, one assumes 1222,2122,2212 and 2221 are in the set. Then 18 of the 24 'c' points with two 2s must be included; it is quick to check that 1122 and permutations must be the six excluded. Next, one checks that the only possible set of six 'a' points with no 2s is 1111,1113,3333 and permutations. Lastly, in a rather longer computation, one finds there is only possible set of twelve 'b' points, that is points with one 2. A computer-free proof can be found at
\centerline{{\tt http://michaelnielsen.org/polymath1/index.php?title=Classification\_of\_\%286\%2C12\%2C18\%2C4\%2C0\%29\_sets}.}
\end{proof}

As a consequence of this lemma, given any $x,y,z,w \in \{1,3\}$, there is a unique good Moser set in $[3]^4$ set whose intersection with $S_{1,4}$ is $\{x222, 2y22, 22z2, 222w\}$, and these are the only 16 possibilities. Call this set the \emph{good set of type $xyzw$}. It consists of
\begin{itemize}
\item The four points $x222, 2y22, 22z2, 222w$ in $S_{1,4}$;
\item All $24$ elements of $S_{2,4}$ except for $xy22, x2z2, x22w, 2yz2, 2y2w, 22zw$;
\item The twelve points $xYZ2$, $xY2W$, $x2ZW$, $XyZ2$, $Xy2W$, $2yZW$, $XYz2$, $X2zW$, $2YzW$, $XY2w$, $X2Zw$, $2YZw$ in $S_{3,4}$, where $X=4-x$, $Y=4-y$, $Z=4-z$, $W=4-w$;
\item The six points $xyzw, xyzW, xyZw, xYzw, Xyzw, XYZW$ in $S_{4,4}$.
\end{itemize}

We can use this to constrain the types of two intersecting good slices:

\begin{lemma}\label{pqs} Suppose that the $pq****$ slice is of type $xyzw$, and the $p*r***$ slice is of type $x'y'z'w'$, where $p,q,r,x,y,z,w,x',y',z',w'$ are in $\{1,3\}$. Then $x'=x$ iff $q=r$, and $y'z'w'$ is equal to either $yzw$ or $YZW$. If $x=r$ (or equivalently if $x'=q$), then $y'z'w'=yzw$.
\end{lemma}

\begin{proof} By reflection symmetry we can take $p=q=r=1$. Observe that the $11****$ slice contains $111222$ iff $x=1$, and the $1*1***$ slice similarly contains $111222$ iff $x'=1$. This shows that $x=x'$.

Suppose now that $x=x'=1$. Then the $111***$ slice contains the three elements $111y22, 1112z2, 11122w$, and excludes $111Y22, 1112Z2, 11122W$, and similarly with the primes, which forces $yzw=y'z'w'$ as claimed.

Now suppose that $x=x'=3$. Then the $111***$ slice contains the two elements $111yzw, 111YZW$, but does not contain any of the other six points in $S_{6,6} \cap 111***$, and similarly for the primes. Thus $y'z'w'$ is equal to either $yzw$ or $YZW$ as claimed.
\end{proof}

Now we look at two adjacent parallel good slices, such as $11****$ and $13****$. The following lemma asserts that such slices either have opposite type, or else will create a huge amount of defect in other slices:

\begin{lemma}\label{l18} Suppose that the $11****$ and $13****$ slices are good with types $xyzw$ and $x'y'z'w'$ respectively. If $x=x'$, then the $1*x***$ slice has defect at least $30$, and the $1*X***$ slice has defect at least $8$. Also, the $1**1**$, $1**3**$, $1***1*$, $1***3*$, $1****1$, $1****3$ slices have defect at least $6$. In particular, the total defect of slices beginning with $1*$ is at least $74$.
\end{lemma}

\begin{proof} Observe from the $11****, 13****$ hypotheses that $a(1*x***)=9$ and $a(1*X***)=4$, which gives the first two claims by Lemma \ref{defects}. For the other claims, one sees from Lemma \ref{pqs} that the other six slices cannot be good; also, they have an $a$-value of $6$ and a $d$-value of at most $7$, and the claims then follow from Lemma \ref{defects}.
\end{proof}

Now we look at two diagonally opposite parallel good slices, such as $11****$ and $33****$.

\begin{lemma}\label{l14} The $11****$ and $33****$ slices cannot both be good and of the same type.
\end{lemma}

\begin{proof} Suppose not. By symmetry we may assume that $11****$ and $33****$ are of type $1111$. This excludes a lot of points from $22****$. Indeed, by connecting lines between the $11****$ and $33****$ slices, we see that the only points that can still survive in $22****$ are $221133, 221333, 221132, 223332$, and permutations of the last four indices. Double counting the lines $22133*$ and permutations we see that there are at most $12$ points one can place in the permutations of $221133, 221333, 221132$, and so the $22****$ slice has at most $16$ points. Meanwhile, the two five-dimensional slices $1*****, 3*****$ have at most $c'_{5,3} = 124$ points, and the other two four-dimensional slices $21****, 23****$ have at most $c'_{4,3} = 43$ points, leading to at most $16 + 124 * 2 + 43 * 2 = 350$ points in all, a contradiction.
\end{proof}

\begin{lemma}\label{l19} It is not possible for all four slices in a family to be good.
\end{lemma}

\begin{proof} Suppose not. By symmetry we may assume that $11****, 13****, 31****, 33****$ are good. By Lemma \ref{l14}, the $11****$ and $33****$ slices cannot be of the same type, and so they cannot both be of the opposite type to either $13****$ or $31****$. If $13****$ is not of the opposite type to $11****$, then by (a permutation of) Lemma \ref{l18}, the total defect of slices beginning with $1*$ is at least $74$; otherwise, if $13****$ is not of the opposite type to $33****$, then by (a permutation and reflection of) Lemma \ref{l18}, the total defect of slices beginning with $*3$ is at least $74$. Similarly, the total defect of slices beginning with $3*$ or $*1$ is at least $74$, leading to a total defect of at least $148$. But the total defect of all the corner slices is $2 \times 60 = 120$, a contradiction.
\end{proof}

\begin{corollary}\label{l20} At most one family can have a total defect of at least $38$.
\end{corollary}

\begin{proof} Suppose there are two families with defect at least $38$. The remaining thirteen family have defect at least $4$ by Lemma \ref{l19} and Lemma \ref{defects}, leading to a total defect of at least $38*2+13*4=128$. But the total defect is $2 \times 60 = 120$, a contradiction.
\end{proof}

Actually we can refine this:

\begin{proposition} No family can have a total defect of at least $38$.
\end{proposition}

\begin{proof} Suppose for contradiction that the $ab****$ family (say) had a total defect of at least $38$, then by Corollary \ref{l20} no other families have total defect at least $38$.

We claim that the $**ab**$ family can have at most two good slices. Indeed, suppose the $**ab**$ has three good slices, say $**11**, **13**, **33**$. By Lemma \ref{l14}, the $**11**$ and $**33**$ slices cannot be of the same type, and so cannot both be of opposite type to $**13**$. Suppose $**11**$ and $**13**$ are not of opposite type. Then by (a permutation of) Lemma \ref{l18}, one of the families $a*b***, *ab***, **b*a*, **b**a$ has a net defect of at least $38$, contradicting the normalisation.

Thus each of the six families $**ab**, **a*b*, **a**b, ***ab*, ***a*b$ have at least two bad slices. Meanwhile, the eight families $a*b***, a**b**, a***b*, a****b, *ab***, *a*b**, *a**b*, *a***b$ have at least one bad slice by Corollary \ref{l19}, leading to twenty bad slices in addition to the defect of at least $38$ arising from the $ab****$ slice. To add up to a total defect of $120$, we conclude from Lemma \ref{defects} that all bad slices outside of the $ab****$ family have a defect of four, with at most one exception; but then by Lemma \ref{l18} this shows that (for instance) the $1*1***$ and $1*3***$ slices cannot be good unless they are of opposite type. The previous argument then shows that the a*b*** slice cannot have three good slices, which increases the number of bad slices outside of $ab****$ to at least twenty-one, and now there is no way to add up to $120$, a contradiction.
\end{proof}

\begin{corollary} Every family can have at most two good slices.
\end{corollary}

\begin{proof} If for instance $11****, 13****, 33****$ are both good, then by Lemma \ref{l14} at least one of $11****, 33****$ is not of the opposite type to $13****$, which by Lemma \ref{l18} implies that there is a family with a total defect of at least $38$, contradicting the previous proposition.
\end{proof}

From this corollary and Lemma \ref{defects}, we see that every family has a defect of at least $8$. Since there are $15$ families, and $8 \times 15$ is exactly equal to $120$, we conclude

\begin{corollary}\label{coda} Every family has \emph{exactly} two good slices, and the remaining two slices have defect $4$. In particular, by Lemma \ref{defects}, the bad slices must have statistics $(5,12,18,4,0)$, $(5,12,12,4,1)$, or $(6,8,12,8,0)$.
\end{corollary}

We now limit how these slices can interact with good slices.

\begin{lemma}\label{goodgood} Suppose that $1*1***$ is a good slice.
\begin{itemize}
\item[(i)] The $11****$ slice cannot have statistics $(6,8,12,8,0)$.
\item[(ii)] The $11****$ slice cannot have statistics $(5,12,12,4,1)$.
\item[(iii)] If the $11****$ slice has statistics $(5,12,18,4,0)$, then the $112***$ slice has statistics $(3,9,3,0)$.
\end{itemize}
\end{lemma}

\begin{proof} This can be verified through computer search; there are $16$ possible configurations for the good slices, and one can calculate that there are $2750$ configurations for the $(5,12,12,4,1)$ slices, $4368$ configurations for the $(5,12,18,4,0)$ slices, and $10000$ configurations for the $(6,8,12,8,0)$ slices. {\bf we could put human proofs for all this somewhere, presumably.}

%We first prove (i). Suppose for contradiction that the $11****$ slice has statistics $(6,8,12,8,0)$, then $A$ contains $111222$, and so the $1*1***$ slice is of type $1xyz$ for some $x,y,z$. By symmetry we may assume it is of type $1111$, thus the $111***$ slice consists of
%$111111, 111113, 111332, 111322, 111222$ and permutations of the last three indices. On the other hand, the $11****$ slice has all eight of the points in $11**** \cap S_{2,6}$. Drawing lines between these points and $111111, 111113$ and permutations, we see that the $113***$ slice cannot contain $113111, 113113, 113133$, or permutations, leaving $113333$ as the only possible element of $113*** \cap S_{6,6}$. This makes $a(11****)=5$, a contradiction.
\end{proof}

\begin{corollary}\label{slic} The $111***$ slice has statistics $(4,3,3,1)$, $(2,6,6,0)$, $(3,3,3,1)$, or $(1,6,6,0)$.
\end{corollary}

\begin{proof} From Corollary \ref{coda}, we know that at least one of the slices $13****, 31****, 11****$ are good. If $11****$ or $1*1***$ is good, then the slice $111***$ has statistics $(4,3,3,1)$ or $(2,6,6,0)$, by Lemma \ref{sixt}. By symmetry we may thus reduce to the case where $13****$ is good and $1*1***$ is bad. Then by Lemma \ref{goodgood}, the $1*1***$ slice has statistics $(5,12,18,4,0)$ and the $121***$ slice has statistics $(3,9,3,0)$. Since the $131***$ slice, as a side slice of the good $13****$ slice, has statistics $(4,3,3,1)$ or $(2,6,6,0)$, we conclude that the $111***$ slice has statistics $(1,6,6,0)$ or $(3,3,3,1)$, and the claim follows.
\end{proof}

\begin{corollary}\label{slic2} All corner slices have statistics $(6,12,18,4,0)$ or $(5,12,18,4,0)$.
\end{corollary}

\begin{proof} Suppose first that a corner slice, say $11****$ has statistic $(6,8,12,8,0)$. Then $111***$ and $113***$ contain one ``d'' point each, and have six ``a'' points between them, so by Corollary \ref{slic}, they both have statistic $(3,3,3,1)$. This forces the $1*1***$, $1*3***$ slices to be bad, which by Corollary \ref{coda} forces the $3*1***,3*3***$ slices to be good. This forces the $311***, 313***$ slices to have statistics either $(2,6,6,0)$ or $(4,3,3,1)$. But the $311***$ slice (say) cannot have statistic $(4,3,3,1)$, since when combined with the $(3,3,3,1)$ statistics of $111***$ would give $a(*11***)=7$, which contradicts Corollary \ref{coda}; thus the $311***$ slice has statistic $(2,6,6,0)$, and similarly for $331***$. But then $a(3*1***)=4$, which again contradicts Corollary \ref{coda}.

Thus no corner slice has statistic $(6,8,12,8,0)$. Now suppose that a corner slice, say $11****$ has statistic $(5,12,12,4,1)$. By Lemma \ref{goodgood}, the $1*1***, 1*3***$ slices are bad, so by repeating the preceding arguments we conclude that the $311***, 313***$ slices have statistics $(2,6,6,0)$ or $(4,3,3,1)$; in particular, their $a$-value is even. However, the $*11***$ and $*13***$ slices are bad by Lemma \ref{goodgood}, and thus have an $a$-value of $5$; thus the $111***$ and $113***$ slices have an odd $a$-value. Thus forces $a(11****)$ to be even; but it is equal to $5$, a contradiction.
\end{proof}

From this and Lemma \ref{dci}, we see that $A$ has statistics $(22,72,180,80,0,0,0)$. In particular, we have $2\alpha_2(A)+\alpha_3(A) = 2$, which by double counting (cf. \eqref{alpha-1}) shows that for every line of the form $12222*$ (or a reflection or permutation thereof) intersects $A$ in exactly two points. Note that such lines connect a ``$d$'' point to two ``$c$'' points.

Also, we observe that two adjacent ``$d$'' points, such as $111222$ and $113222$, cannot both lie in $A$; for this would force the $*13***$ and $*11***$ slices to have statistics $(4,3,3,1)$ or $(3,3,3,1)$ by Corollary \ref{slic}, which forces $a(*1****)=6$, and thus $*1****$ must be good by Corollary \ref{slic2}; but this contradicts Lemma \ref{sixt}. Since $\alpha_3(A)=1/2$, we conclude that given any two adjacent ``$d$'' points, exactly one of them lies in $A$. In particular, the d points of the form $***222$ consist either of those strings with an even number of $1$s, or those with an odd number of $1$s.

Let's say it's the former, thus the set contains $111222, 133222$, and permutations of the first three coordinates, but omits $113222, 333222$ and permutations of the first three coordinates. Since the ``$d$'' points $113222, 333222$ are omitted, we conclude that the ``$c$'' points $113122, 113322, 333122, 333322$ must lie in the set, and similarly for permutations of the first three and last three coordinates. But this gives at least $15$ of the $16$ ``$c$'' points ending in $22$; by symmetry this leads to $225$ $c$-points in all; but $c(A)=180$, contradiction. This (finally!) completes the proof that $c'_{6,3}=353$.

Introduction.tex

2009-07-14T10:05:05Z

Mpeake:

\section{Introduction}

For any integers $k \geq 1$ and $n \geq 0$, let $[k] := \{1,\ldots,k\}$, and define $[k]^n$ to be the cube of words of length $n$ with alphabet in $[k]$. Thus for instance $[3]^2 = \{11,12,13,21,22,23,31,32,33\}$.

We define a \emph{combinatorial line} in $[k]^n$ to be a set of the form $\{ w(i): i = 1,\ldots,k\} \subset [k]^n$, where $w \in ([k] \cup \{x\})^n \backslash [k]^n$ is a word of length $n$ with alphabet in $[k]$ together with a ``wildcard'' letter $x$ which appears at least once, and $w(i) \in [k]^n$ is the word obtained from $w$ by replacing $x$ by $i$; we often abuse notation and identify $w$ with the combinatorial line $\{ w(i): i = 1,\ldots,k\}$ it generates. Thus for instance, in $[3]^2$ we have $x2 = \{12,22,32\}$ and $xx = \{11,22,33\}$ as typical examples of combinatorial lines. In general, $[k]^n$ has $k^n$ words and $(k+1)^n-k^n$ lines.

\begin{figure}[tb]
\centerline{\includegraphics{AllLinesIn3n.pdf}}
\caption{Combinatorial lines in $[3]^2$.}
\label{fig-line}
\end{figure}

A set $A \subset [k]^n$ is said to be \emph{line-free} if it contains no combinatorial lines. Define the \emph{$(n,k)$ density Hales-Jewett number} $c_{n,k}$ to be the maximum cardinality $|A|$ of a line-free subset of $[k]^n$. Clearly, one has the trivial bound $c_{n,k} \leq k^n$. A deep theorem of Furstenberg and Katznelson~\cite{fk1}, \cite{fk2} asserts that this bound can be asymptotically improved:

\begin{theorem}[Density Hales-Jewett theorem]\label{dhj} For any fixed $k \geq 2$, one has $\lim_{n \to \infty} c_{n,k}/k^n = 0$.
\end{theorem}

\begin{remark} The difficulty of this theorem increases with $k$. For $k=1$, one clearly has $c_{n,1}=1$. For $k=2$, a classical theorem of Sperner~\cite{sperner} asserts, in our language, that $c_{n,2} = \binom{n}{\lfloor n/2\rfloor}$. The case $k=3$ is already non-trivial (for instance, it implies Roth's theorem~\cite{roth} on arithmetic progressions of length three) and was first established in \cite{fk1} (see also \cite{mcc}). The case of general $k$ was first established in~\cite{fk2} and has a number of implications, in particular implying Szemer\'edi's theorem~\cite{szem} on arithmetic progressions of arbitrary length.
\end{remark}

The Furstenberg-Katznelson proof of Theorem~\ref{dhj} relied on ergodic-theory techniques and did not give an explicit decay rate for $c_{n,k}$. Recently, two further proofs of this theorem have appeared, by Austin~\cite{austin} and by the sister Polymath project to this one~\cite{poly}. The proof of~\cite{austin} also used ergodic theory, but the proof in~\cite{poly} was combinatorial and gave effective bounds for $c_{n,k}$ in the limit $n \to \infty$. For example, if $n$ can be written as an exponential tower 2^2^2^...^2 with $m$ 2s, then $c_{n,3} \le 3^n m^{-1/3}$. However, these bounds are not believed to be sharp, and in any case are only non-trivial in the asymptotic regime when $n$ is sufficiently large depending on $k$.

Our first result is the following asymptotic lower bound. The construction is based on the recent refinements \cite{elkin,greenwolf,obryant} of a well-known construction of Behrend~\cite{behrend} and Rankin~\cite{rankin}. The proof of Theorem~\ref{dhj-lower} is in Section~\ref{dhj-lower-sec}. Let $r_k(n)$ be the maximum size of a subset of $[n]$ that does not contain a $k$-term arithmetic progression.

\begin{theorem}[Asymptotic lower bound for $c_{n,k}$]\label{dhj-lower} For each $k\geq 3$, there is an absolute constant $C>0$ such that
\[ c_{n,k} \geq C k^n \left(\frac{r_k(\sqrt{n})}{\sqrt{n}}\right)^{k-1} = k^n \exp\left( - O(\sqrt[\ell]{\log n}) \right), \]
where $\ell$ is the largest integer satisfying $2k>2^{\ell}$. Specifically,
\[ c_{n,k} \geq C k^{n-\alpha(k) \sqrt[\ell]{\log n} + \beta(k) \log\log n},\]
where all logarithms are base-$k$, and $\alpha(k) = (\log 2)^{1-1/\ell} \ell 2^{(\ell-1)/2-1/\ell}$ and $\beta(k)=(k-1)/(2\ell)$.
\end{theorem}

In the case of small $n$, we focus primarily on the first non-trivial case $k=3$. We have computed the following explicit values of $c_{n,3}$:

\begin{theorem}[Explicit values of $c_{n,3}$]\label{dhj-upper} We have $c_{0,3} = 1$, $c_{1,3} = 2$, $c_{2,3} = 6$, $c_{3,3} = 18$, $c_{4,3} = 52$, $c_{5,3}=150$, and $c_{6,3}=450$. (This has been entered in the OEIS~\cite{oeis} as A156762.)
\end{theorem}

This result is established in Sections~\ref{dhj-lower-sec},~\ref{dhj-upper-sec}. Initially these results were established by an integer program (see Appendix~\ref{integer-sec}, but we provide completely computer-free proofs here. The constructions used in Section~\ref{dhj-lower-sec} give reasonably efficient constructions for larger values of $n$; for instance, they show that $3^{99} \leq c_{100,3} \leq 2 \times 3^{99}$. See Section~\ref{dhj-lower-sec} for further discussion.

We also have several partial results for higher values of $k$: see Section~\ref{higherk-sec}. For results on the closely related \emph{Hales-Jewett numbers} $HJ(k,r)$, see Section~\ref{coloring-sec}.

A variant of the density Hales-Jewett theorem has also been studied in the literature. Define a \emph{geometric line} in $[k]^n$ to be any set of the form $\{ a+ir: i=1,\ldots,k\}$ in $[k]^n$, where we identify $[k]^n$ with a subset of $\Z^n$, and $a, r \in \Z^n$ with $r \neq 0$. Equivalently, a geometric line takes the form $\{ w( i, k+1-i ): i =1,\ldots,k \}$, where $w \in ([k] \cup \{x,\overline{x}\})^n \backslash [k]^n$ is a word of length $n$ using the numbers in $[k]$ and two wildcards $x, \overline{x}$ as the alphabet, with at least one wildcard occuring in $w$, and $w(i,j) \in [k]^n$ is the word formed by substituting $i,j$ for $x,\overline{x}$ respectively. Figure~\ref{fig-geomline} shows the eight geometric lines in $[3]^2$. Clearly every combinatorial line is a geometric line, but not conversely. In general, $[k]^n$ has $((k+2)^n-k^n)/2$ geometric lines.

\begin{figure}[tb]
\centerline{\includegraphics{AllGeomLinesIn3n.pdf}}
\caption{Geometric lines in $[3]^2$.}
\label{fig-geomline}
\end{figure}

Define a \emph{Moser set} in $[k]^n$ to be a subset of $[k]^n$ that contains no geometric lines, and let $c'_{n,k}$ be the maximum cardinality $|A|$ of a Moser set in $[k]^n$. Clearly one has $c'_{n,k} \leq c_{n,k}$, so in particular from Theorem~\ref{dhj} one has $c'_{n,k}/k^n \to 0$ as $n \to \infty$. (Interestingly, there is no known proof of this fact that does not go through Theorem~\ref{dhj}, even for $k=3$.) Again, $k=3$ is the first non-trivial case: it is clear that $c'_{n,1}=0$ and $c'_{n,2}=1$ for all $n$.

The question of computing $c'_{n,3}$ was first posed by Moser~\cite{moser}. Prior to our work, the values
$$ c'_{0,3}=1; c'_{1,3}=2; c'_{2,3}=6; c'_{3,3}=16; c'_{4,3}=43$$
were known~\cite{chvatal2},~\cite{chandra} (this is Sequence A003142 in the OEIS~\cite{oeis}). We extend this sequence slightly:

\begin{theorem}[Values of $c'_{n,3}$ for small $n$]\label{moser} We have $c'_{0,3} = 1$, $c'_{1,3} = 2$, $c'_{2,3} = 6$, $c'_{3,3} = 16$, $c'_{4,3} = 43$, $c'_{5,3} = 124$, and $c'_{6,3} = 353$.
\end{theorem}

This result is established in Sections \ref{moser-lower-sec}, \ref{moser-upper-sec}. The arguments given here are computer-assisted; however, we have found alternate (but lengthier) computer-free proofs for the above claims with the the exception of the proof of $c'_{6,3}=353$, which requires one non-trivial computation (Lemma \ref{paretop-4}). These alternate proofs are not given in this paper to save space, but can be found on the wiki for this project at

\centerline{{\tt http://michaelnielsen.org/polymath1/index.php?title=Main\_Page}.}

We establish a lower bound for this problem of $(2+o(1))\binom{n}{i}2^i\leq c'_{n,3}$, which is maximized for $i$ near $2n/3$. This bound is around one-third better than the literature. We also give methods to improve on this construction.

Earlier lower bounds were known:
let $A(n,d)$ denote the size of the largest binary code of length $n$
and minimal distance $d$. Then
\begin{equation}\label{cnchvatalintro}
c'_{n,3}\geq \max_k \left( \sum_{j=0}^k \binom{n}{j} A(n-j, k-j+1)\right).
\end{equation}
which, with $A(n,1)=2^n$ and $A(n,2)=2^{n-1}$, implies in particular that
\begin{equation}\label{binom}
c'_{n,3} \geq \binom{n+1}{\lfloor \frac{2n+1}{3} \rfloor} 2^{\lfloor \frac{2n+1}{3} \rfloor - 1}
\end{equation}
for $n \geq 2$. This bound is not quite optimal; for instance, it gives a lower bound of $c'_{6,3}=344$.

\begin{remark} Let $c''_{n,3}$ be the size of the largest subset of ${\Bbb F}_3^n$ which contains no lines $x, x+r, x+2r$ with $x,r \in {\mathbb F}_3^n$ and $r \neq 0$, where ${\mathbb F}_3$ is the field of three elements. Clearly one has $c''_{n,3} \leq c'_{n,3} \leq c_{n,3}$. It is known that
$$ c''_{0,3}=1; c''_{1,3}=2; c''_{2,3}=4; c'_{3,3}=9; c'_{4,3}=20; c''_{5,3}=45; c''_{6,3} = 112;$$
see \cite{potenchin}.
\end{remark}

As mentioned earlier, the sharp bound on $c_{n,2}$ comes from Sperner's theorem. It is known that Sperner's theorem can be refined to the \emph{Lubell-Meshalkin-Yamamoto (LMY) inequality}, which in our language asserts that
$$
\sum_{a_1,a_2 \geq 0; a_1+a_2 = n} \frac{|A \cap \Gamma_{a_1,a_2}|}{|\Gamma_{a_1,a_2}|} \leq 1
$$
for any line-free subset $A \subset [2]^n$, where the \emph{cell} $\Gamma_{a_1,\ldots,a_k} \subset [k]^n$ is the set of words in $[k]^n$ which contain exactly $a_i$ $i$'s for each $i=1,\ldots,k$. It is natural to ask whether this inequality can be extended to higher $k$. Let $\Delta_{k,n}$ denote the set of all tuples $(a_1,\ldots,a_k)$ of non-negative integers summing to $n$, define a \emph{simplex} to be a set of $k$ points in $\Delta_{k,n}$ of the form
$(a_1+r,a_2,\ldots,a_k), (a_1,a_2+r,\ldots,a_k),\ldots,(a_1,a_2,\ldots,a_k+r)$ for some $0 < r \leq n$ and $a_1,\ldots,a_k$ summing to $n-r$, and define a \emph{Fujimura set} to be a subset $B \subset \Delta_{k,n}$ which contains no simplices. Observe that if $w$ is a combinatorial line in $[k]^n$, then
$$ w(1) \in \Gamma_{a_1+r,a_2,\ldots,a_k}, w(2) \in \Gamma_{a_1,a_2+r,\ldots,a_k}, \ldots, w(k) = \Gamma_{a_1,a_2,\ldots,a_k+r}$$
for some simplex $(a_1+r,a_2,\ldots,a_k), (a_1,a_2+r,\ldots,a_k),\ldots,(a_1,a_2,\ldots,a_k+r)$. Thus, if $B$ is a Fujimura set, then $A := \bigcup_{\vec a \in B} \Gamma_{\vec a}$ is line-free. Note also that
$$
\sum_{\vec a \in \Delta_{k,n}} \frac{|A \cap \Gamma_{\vec a}|}{|\Gamma_{\vec a}|} = |B|.
$$
This motivates a ``hyper-optimistic'' conjecture:

{\bf a picture showing a Fujimura set?}

\begin{conjecture}\label{hoc} For any $k \geq 1$ and $n \geq 0$, and any line-free subset $A$ of $[k]^n$, one has
$$
\sum_{\vec a \in \Delta_{k,n}} \frac{|A \cap \Gamma_{\vec a}|}{|\Gamma_{\vec a}|} \leq c^\mu_{k,n},$$
where $c^\mu_{k,n}$ is the maximal size of a Fujimura set in $\Delta_{k,n}$.
\end{conjecture}

One can show that this conjecture for a fixed value of $k$ would imply Theorem \ref{dhj} for the same value of $k$, in much the same way that the LYM inequality is known to imply Sperner's theorem. The LYM inequality asserts that Conjecture \ref{hoc} is true for $k \leq 2$. As far as we know this conjecture could hold in $k=3$. However, we know that it fails for all even $k\ge 4$, see Section~\ref{higherk-sec}.

We are interested in removing all upward-pointing equilateral triangles from a triangular grid. Fujimura actually proposed the similar problem of removing all equilateral triangles at this website: www.puzzles.com/PuzzlePlayground/CoinsAndTriangles/CoinsAndTriangles.htm

In Section \ref{fujimura-sec} we give some bounds on these numbers. Explicit values are given for $\overline{c}^\mu_n$ up to $n=13$, and general upper and lower bounds are given for all $n$.

\subsection{Notation}\label{notation-sec}

There are several subsets of $[k]^n$ which will be useful in our analysis. We have already introduced combinatorial lines, geometric lines, and cells. One can generalise the notion of a combinatorial line to that of a \emph{combinatorial subspace} in $[k]^n$ of dimension $d$, which is indexed by a word $w$ in $([k] \cup \{x_1,\ldots,x_d\})^n$ containing at least one of each wildcard $x_1,\ldots,x_d$, and which forms the set $\{ w(i_1,\ldots,i_d): i_1,\ldots,i_d \in [k]\}$, where $w(i_1,\ldots,i_d) \in [k]^d$ is the word formed by replacing $x_1,\ldots,x_d$ with $i_1,\ldots,i_d$ respectively. Thus for instance, in $[3]^3$, we have the two-dimensional combinatorial subspace $xxy = \{111,112,113,221,222,223,331,332,333\}$. We similarly have the notion of a \emph{geometric subspace} in $[k]^n$ of dimension $d$, which is defined similarly but with $d$ wildcards $x_1,\ldots,x_d,\overline{x_1},\ldots,\overline{x_d}$, with at least one of either $x_i$ or $\overline{x_i}$ appearing in the word $w$ for each $1 \leq i \leq d$, and the space taking the form $\{ w(i_1,\ldots,i_d,k+1-i_1,\ldots,k+1-i_d): i_1,\ldots,i_d \in [k] \}$. Thus for instance $[3]^3$ contains the two-dimensional geometric subspace $x\overline{x}y = \{ 131, 132, 133, 221, 222, 223, 311, 312, 313\}$.

An important class of combinatorial subspaces in $[k]^n$ will be the \emph{slices} consisting of $n-1$ distinct wildcards and one fixed coordinate. We will denote the distinct wildcards here by asterisks, thus for instance in $[3]^3$ we have $2** = \{ 211, 212, 213, 221, 222, 223, 231, 232, 233\}$. Two slices are \emph{parallel} if their fixed coordinate are in the same position, thus for instance $1**$ and $2**$ are parallel, and one can subdivide $[k]^n$ into $k$ parallel slices, each of which is isomorphic to $[k]^{n-1}$. In the analysis of Moser slices with $k=3$, we will make a distinction between \emph{centre slices}, whose fixed coordinate is equal to $2$, and \emph{side slices}, in which the fixed coordinate is either $1$ or $3$, thus $[3]^n$ can be partitioned into one centre slice and two side slices.

Another important set in the study of $k=3$ Moser sets are the \emph{spheres} $S_{i,n} \subset [3]^n$, defined as those words in $[3]^n$ with exactly $n-i$ $2$'s (and hence $i$ letters that are $1$ or $3$). Thus for instance $S_{1,3} = \{ 122, 322, 212, 232, 221, 223\}$. Observe that $[3]^n = \bigcup_{i=0}^n S_{i,n}$, and each $S_{i,n}$ has cardinality $|S_{i,n}| = \binom{n}{i} 2^{i}$.

It is also convenient to subdivide each sphere $S_{i,n}$ into two components $S_{i,n} = S_{i,n}^o \cup S_{i,n}^e$, where $S_{i,n}^o$ are the words in $S_{i,n}$ with an odd number of $1$'s, and $S_{i,n}^e$ are the words with an even number of $1$'s. Thus for instance $S_{1,3}^o = \{122,212,221\}$ and $S_{1,3}^e = \{322,232,223\}$. Observe that for $i>0$, $S_{i,n}^o$ and $S_{i,n}^e$ both have cardinality $\binom{n}{i} 2^{i-1}$.

The \emph{Hamming distance} between two words $w,w'$ is the number of coordinates in which $w, w'$ differ, e.g. the Hamming distance between $123$ and $321$ is two. Note that $S_{i,n}$ is nothing more than the set of words whose Hamming distance from $2\ldots2$ is $i$, which justifies the terminology ``sphere''.

In the density Hales-Jewett problem, there are two types of symmetries on $[k]^n$ which map combinatorial lines to combinatorial lines (and hence line-free sets to line-free sets). The first is a permutation of the alphabet $[k]$; the second is a permutation of the $n$ coordinates. Together, this gives a symmetry group of order $k!n!$ on the cube $[k]^n$, which we refer to as the \emph{combinatorial symmetry group} of the cube $[k]^n$. Two sets which are related by an element of this symmetry group will be called (combinatorially) \emph{equivalent}, thus for instance any two slices are combinatorially equivalent.

For the analysis of Moser sets in $[k]^n$, the symmetries are a bit different. One can still permute the $n$ coordinates, but one is no longer free to permute the alphabet $[k]$. Instead, one can \emph{reflect} an individual coordinate, for instance sending each word $x_1 \ldots x_n$ to its reflection $x_1 \ldots x_{i-1} (k+1-x_i) x_{i+1} \ldots x_n$. Together, this gives a symmetry group of order $2^k n!$ on the cube $[k]^n$, which we refer to as the \emph{geometric symmetry group} of the cube $[k]^n$; this group maps geometric lines to geometric lines, and thus maps Moser sets to Moser sets. Two Moser sets which are related by an element of this symmetry group will be called (geometrically) \emph{equivalent}. For instance, a sphere $S_{i,n}$ is equivalent only to itself, and $S_{i,n}^o$, $S_{i,n}^e$ are equivalent only to each other.

Moser.tex

2009-07-14T06:45:44Z

Mpeake:

\section{Upper bounds for the $k=3$ Moser problem in small dimensions}\label{moser-upper-sec}

In this section we finish the proof of Theorem \ref{moser} by obtaining the upper bounds on $c'_{n,3}$ for $n \leq 6$.

\subsection{Statistics, densities and slices}

Our analysis will revolve around various \emph{statistics} of Moser sets $A \subset [3]^n$, their associated \emph{densities}, and the behavior of such statistics and densities with respect to the operation of passing from the cube $[3]^n$ to various \emph{slices} of that cube.

\begin{definition}[Statistics and densities] Let $A \subset [3]^n$ be a set. For any $0 \leq i \leq n$, set $a_i(A) := |A \cap S_{n-i,n}|$; thus we have
$$ 0 \leq a_i(A) \leq |S_{n-i,n}| = \binom{n}{i} 2^{n-i}$$
for $0 \leq i \leq n$ and
$$ a_0(A) + \ldots + a_n(A) = |A|.$$
We refer to the vector $(a_0(A),\ldots,a_n(A))$ as the \emph{statistics} of $A$. We define the $i^{th}$ \emph{density} $\alpha_i(A)$ to be the quantity
$$ \alpha_i(A) := \frac{a_i(A) }{\binom{n}{i} 2^{n-i}},$$
thus $0 \leq \alpha_i(A) \leq 1$ and
$$ |A| = \sum_{i=0}^n \binom{n}{i} 2^{n-i} a_i(A).$$
\end{definition}

\begin{example}\label{2mos} Let $n=2$ and $A$ be the Moser set $A := \{ 12, 13, 21, 23, 31, 32 \}$. Then the statistics $(a_0(A), a_1(A), a_2(A))$ of $A$ are $(2,4,0)$, and the densities $(\alpha_0(A), \alpha_1(A), \alpha_2(A))$ are $(\frac{1}{2}, 1, 0)$. \textbf{Include picture here? with colours?}
\end{example}

When working with small values of $n$, it will be convenient to write $a(A)$, $b(A)$, $c(A)$, etc. for $a_0(A)$, $a_1(A)$, $a_2(A)$, etc., and similarly write $\alpha(A), \beta(A), \gamma(A)$, etc. for $\alpha_0(A)$, $\alpha_1(A)$, $\alpha_2(A)$, etc. Thus for instance in Example \ref{2mos} we have $b(A) = 4$ and $\alpha(A) = \frac{1}{2}$.

\begin{definition}[Subspace statistics and densities] If $V$ is a $k$-dimensional geometric subspace of $[3]^n$, then we have a map $\phi_V: [3]^k \to [3]^n$ from the $k$-dimensional cube to the $n$-dimensional cube. If $A \subset [3]^n$ is a set and $0 \leq i \leq k$, we write $a_i(V,A)$ for $a_i(\phi_V^{-1}(A))$ and $\alpha_i(V,A)$ for $\alpha_i(\phi_V^{-1}(A))$. If the set $A$ is clear from context, we abbreviate $a_i(V,A)$ as $a_i(V)$ and $\alpha_i(V,A)$ as $\alpha_i(V)$.
\end{definition}

For our problem, a particularly important type of subspace of $[3]^n$ will be the \emph{slices} formed by fixing one coordinate and letting the other $n-1$ coordinates vary. We will denote this by a single string in which the $n-1$ varying coordinates are denoted by asterisks. For instance, in $[3]^2$, $1*$ denotes the slice $1*=\{11,12,13\}$, $*2$ denotes the slice $*2=\{12,22,32\}$, etc.; similarly, in $[3]^3$, $1**$ is the slice $\{111, 112, 113, 121, 122, 123, 131, 132, 133\}$, etc. We call a slice a \emph{centre slice} if the fixed coordinate is $2$ and a \emph{side slice} if it is $1$ or $3$.

\begin{example} We continue Example \ref{2mos}. Then the statistics of the side slice $1*$ are $(a(1*),b(1*)) = (1,1)$, while the statistics of the centre slice $2*$ are $(a(2*),b(2*))=(2,0)$. The corresponding densities are $(\alpha(1*),\beta(1*)) = (1/2,1)$ and $(\alpha(2*),\beta(2*))=(1,0)$.
\end{example}

A simple double counting argument gives the following useful identity:

\begin{lemma}[Double counting identity]\label{dci} Let $A \subset [3]^n$ and $0 \leq i \leq n-1$. Then we have
$$ \frac{1}{n-i-1} \sum_{V \hbox{ a side slice}} a_{i+1}(V) = \frac{1}{i+1} \sum_{W \hbox{ a centre slice}} a_i(W) = a_{i+1}(A)$$
where $V$ ranges over the $2n$ side slices of $[3]^n$, and $W$ ranges over the $n$ centre slices. In other words, the average value of $\alpha_{i+1}(V)$ for side slices $V$ equals the average value of $\alpha_i(W)$ for centre slices $W$, which is in turn equal to $\alpha_{i+1}(A)$.
\end{lemma}

Indeed, this lemma follows from the observation that every string in $A \cap S_{n-i-1,n}$ belongs to $i+1$ centre slices $W$ (and contributes to $a_i(W)$) and to $n-i-1$ side slices $V$ (and contributes to $a_{i+1}(V)$). One can also view this lemma probabilistically, as the assertion that there are three equivalent ways to generate a random string of length $n$:

\begin{itemize}
\item Pick a side slice $V$ at random, and randomly fill in the wildcards in such a way that $i+1$ of the wildcards are $2$'s (i.e. using an element of $S_{n-i-2,n-1}$).
\item Pick a centre slice $V$ at random, and randomly fill in the wildcards in such a way that $i$ of the wildcards are $2$'s (i.e. using an element of $S_{n-i-1,n-1}$).
\item Randomly choose an element of $S_{n-i-1,n}$.
\end{itemize}

\begin{example} We continue Example \ref{2mos}. The average value of $\beta$ for side slices is equal to the average value of $\alpha$ for centre slices, which is equal to $\beta(A) = 1$.
\end{example}

Another very useful fact (essentially due to \cite{chvatal2}) is that linear inequalities for statistics of Moser sets at one dimension propagate to linear inequalities in higher dimensions:

\begin{lemma}[Propagation lemma]\label{prop} Let $n \geq 1$ be an integer. Suppose one has a linear inequality of the form
\begin{equation}\label{alphav}
\sum_{i=0}^n v_i \alpha_i(A) \leq s
\end{equation}
for all Moser sets $A \subset [3]^n$ and some real numbers $v_0,\ldots,v_n,s$. Then we also have the linear inequality
$$ \sum_{i=0}^n v_i \alpha_{qi+r}(A) \leq s$$
whenever $q \geq 1$, $r \geq 0$, $N \geq nq+r$ are integers and $A \subset [3]^N$ is a Moser set.
\end{lemma}

\begin{proof} We run a probabilistic argument (one could of course also use a double counting argument instead). Let $n,v_0,\ldots,v_n,s,q,r,N,A$ be as in the lemma. Let $V$ be a random $n$-dimensional geometric subspace of $[3]^N$, created in the following fashion:
\begin{itemize}
\item Pick $n$ wildcards $x_1,\ldots,x_n$ to run independently from $1$ to $3$. We also introduce dual wildcards $\overline{x_1},\ldots,\overline{x_n}$; each $\overline{x_j}$ will take the value $4-x_j$.
\item We randomly subdivide the $N$ coordinates into $n$ groups of $q$ coordinates, plus a remaining group of $N-nq$ ``fixed'' coordinates.
\item For each coordinate in the $j^{th}$ group of $q$ coordinates for $1 \leq j \leq n$, we randomly assign either a $x_j$ or $\overline{x_j}$.
\item For each coordinate in the $N-nq$ fixed coordinates, we randomly assign a digit $1,2,3$, but condition on the event that exactly $r$ of the digits are equal to $2$ (i.e. we use a random element of $S_{N-nq-r,N-nq}$).
\item Let $V$ be the subspace created by allowing $x_1,\ldots,x_n$ to run independently from $1$ to $3$, and $\overline{x_j}$ to take the value $4-x_j$.
\end{itemize}

For instance, if $n=2, q=2, r=1, N=6$, then a typical subspace $V$ generated in this fashion is
$$ 2x_1\overline{x_2}3x_2x_1 = \{ 213311, 212321, 211331, 223312, 222322, 221332, 233313, 232323, 231333\}.$$
Observe from that the following two ways to generate a random element of $[3]^N$ are equivalent:

\begin{itemize}
\item Pick $V$ randomly as above, and then assign $(x_1,\ldots,x_n)$ randomly from $S_{n-i,n}$. Assign $4-x_j$ to $\overline{x_j}$ for all $1 \leq j \leq n$.
\item Pick a random string in $S_{N-qi-r,N}$.
\end{itemize}

Indeed, both random variables are invariant under the symmetries of the cube, and both random variables always pick out strings in $S_{N-qi-r,N}$, and the claim follows. As a consequence, we see that the expectation of $\alpha_i(V)$ (as $V$ ranges over the recipe described above) is equal to $\alpha_{qi+r}(A)$. On the other hand, from \eqref{alphav} we have
$$ \sum_{i=0}^n v_i \alpha_i(V) \leq s$$
for all such $V$; taking expectations over $V$, we obtain the claim.
\end{proof}

In view of Lemma \ref{prop}, it is of interest to locate linear inequalities relating the densities $\alpha_i(A)$, or (equivalently) the statistics $a_i(A)$. For this, it is convenient to introduce the following notation.

\begin{definition} Let $n \geq 1$ be an integer.
\begin{itemize}
\item A vector $(a_0,\ldots,a_n)$ of non-negative integers is \emph{feasible} if it is the statistics of some Moser set $A$.
\item A feasible vector $(a_0,\ldots,a_n)$ is \emph{Pareto-optimal} if there is no other feasible vector $(b_0,\ldots,b_n) \neq (a_0,\ldots,a_n)$ such that $b_i \geq a_i$ for all $0 \leq i \leq n$.
\item A Pareto-optimal vector $(a_0,\ldots,a_n)$ is \emph{extremal} if it is not a non-trivial convex linear combination of other Pareto-optimal vectors.
\end{itemize}
\end{definition}

To establish a linear inequality of the form \eqref{alphav} with the $v_i$ non-negative, it suffices to test the inequality against densities associated to extremal vectors of statistics. (There is no point considering linear inequalities with negative coefficients $v_i$, since one always has the freedom to reduce a density $\alpha_i(A)$ of a Moser set $A$ to zero, simply by removing all elements of $A$ with exactly $i$ $2$'s.)

We will classify exactly the Pareto-optimal and extremal vectors for $n \leq 3$, which by Lemma \ref{prop} will lead to useful linear inequalities for $n \geq 4$. Using a computer, we have also located a partial list of Pareto-optimal and extremal vectors for $n=4$, which are also useful for the $n=5$ and $n=6$ theory.

\subsection{Up to three dimensions}

We now establish Theorem \ref{moser} for $n \leq 3$, and establish some auxiliary inequalities which will be of use in higher dimensions.

The case $n=0$ is trivial. When $n=1$, it is clear that $c'_{1,3} = 2$, and furthermore that the Pareto-optimal statistics are $(2,0)$ and $(1,1)$, which are both extremal. This leads to the linear inequality
$$ 2\alpha(A) + \beta(A) \leq 2$$
for all Moser sets $A \subset [3]^1$, which by Lemma \ref{prop} implies that
\begin{equation}\label{alpha-1}
2\alpha_r(A) + \alpha_{r+q}(A) \leq 2
\end{equation}
whenever $r \geq 0$, $q \geq 1$, $n \geq q+r$, and $A \subset [3]^n$ is a Moser set.

For $n=2$, we see by partitioning $[3]^2$ into three slices that $c'_{2,3} \leq 3 c'_{1,3} = 6$, and so (by the lower bounds in the previous section) $c'_{2,3} = 6$. Writing $(a,b,c) = (a(A),b(A),c(A)) = (4\alpha(A), 4\beta(A), \gamma(A))$, the inequalities \eqref{alpha-1} become
\begin{equation}\label{abc}
a + 2c \leq 4; b+2c \leq 4; 2a+b <= 8.
\end{equation}

\begin{lemma} When $n=2$, the Pareto-optimal statistics are $(4,0,0), (3,2,0), (2,4,0), (2,2,1)$. In particular, the extremal statistics are $(4,0,0), (2,4,0), (2,2,1)$.
\end{lemma}

\begin{proof} One easily checks that all the statistics listed above are feasible.
Consider the statistics $(a,b,c)$ of a Moser set $A \subset [3]^2$. $c$ is either equal to $0$ or $1$. If $c=1$, then \eqref{abc} implies that $a,b \leq 2$, so the only Pareto-optimal statistic here is $(2,2,1)$. When instead $c=0$, the inequalities \eqref{abc} can easily imply the Pareto-optimality of $(4,0,0), (3,2,0), (2,4,0)$.
\end{proof}

From this lemma we see that we obtain a new inequality $2a+b+2c \leq 8$. Converting this back to densities and using Lemma \ref{prop}, we conclude that
\begin{equation}\label{alpha-2}
4\alpha_r(A) + 2\alpha_{r+q}(A) + \alpha_{r+2q} \leq 4
\end{equation}
whenever $r \geq 0$, $q \geq 1$, $n \geq q+2r$, and $A \subset [3]^n$ is a Moser set.

One can also check by computer that there are exactly $230$ line-free subsets of $[3]^2$.

Now we look at three dimensions. Writing $(a,b,c,d)$ for the statistics of a Moser set $A \subset [3]^n$ (which thus range between $(0,0,0,0)$ and $(8,12,6,1)$), the inequalities \eqref{alpha-1} imply in particular that
\begin{equation}\label{abc-3d}
a+4d \leq 8; b+6d \leq 12; c+3d \leq 6; 3a+2c \leq 24; b+c \leq 12
\end{equation}
while \eqref{alpha-2} implies that
\begin{equation}\label{abcd-3d}
3a+b+c \leq 24; b+c+3d \leq 12.
\end{equation}
Summing the inequalities $b+c \leq 12, 3a+b+c \leq 24, b+c+3d \leq 12$ yields
$$ 3(a+b+c+d) \leq 48$$
and hence $|A| = a+b+c+d \leq 16$; comparing this with the lower bounds of the preceding section we obtain $c'_{3,3} = 16$ as required. (This argument is essentially identical to the one in \cite{chvatal2}).

We have the following useful computation:

\begin{lemma}[3D Pareto-optimals]\label{paretop} When $n=3$, the Pareto-optimal statistics are $$(3,6,3,1),(4,4,3,1),(4,6,2,1),(2,6,6,0),(3,6,5,0),(4,4,5,0),(3,7,4,0),(4,6,4,0),$$
$$ (3,9,3,0),(4,7,3,0),(5,4,3,0),(4,9,2,0),(5,6,2,0),(6,3,2,0),(3,10,1,0),(5,7,1,0),$$
$$ (6,4,1,0),(4,12,0,0),(5,9,0,0),(6,6,0,0),(7,3,0,0),(8,0,0,0).$$
In particular, the extremal statistics are
$$(3,6,3,1),(4,4,3,1),(4,6,2,1),(2,6,6,0),(4,4,5,0),(4,6,4,0),(4,12,0,0),(8,0,0,0).$$
\end{lemma}

\begin{proof} This can be established by a brute-force search over the $2^{27} \approx 1.3 \times 10^8$ different subsets of $[3]^3$. Actually, one can perform a much faster search than this. Firstly, as noted earlier, there are only $230$ line-free subsets of $[3]^2$, so one could search over $230^3 \approx 1.2 \times 10^7$ configurations instead. Secondly, by symmetry we may assume (after enumerating the $230$ sets in a suitable fashion) that the first slice $A \cap 1**$ has an index less than or equal to the third $A \cap 3**$, leading to $\binom{231}{2} \times 230 \approx 6 \times 10^6$ configurations instead. Finally, using the first and third slice one can quickly determine which elements of the second slice $2**$ are prohibited from $A$. There are $2^9 = 512$ possible choices for the prohibited set in $2**$. By crosschecking these against the list of $230$ line-free sets one can compute the Pareto-optimal statistics for the second slices inside the prohibited set (the lists of such statistics turns out to length at most $23$). Storing these statistics in a lookup table, and then running over all choices of the first and third slice (using symmetry), one now has to perform $O( 512 \times 230 ) + O( \binom{231}{2} \times 23) \approx O( 10^6 )$ computations, which is quite a feasible computation.

One could in principle reduce the computations even further, by a factor of up to $8$, by using the symmetry group $D_4$ of the square $[3]^2$ to reduce the number of cases one needs to consider, but we did not implement this.

A computer-free proof of this lemma can be found at
\centerline{{\tt http://michaelnielsen.org/polymath1/index.php?title=Human\_proof\_of\_the\_3D\_Pareto-optimal\_Moser\_statistics}.}
\end{proof}

\begin{remark} A similar computation revealed that the total number of line-free subsets of $[3]^3$ was $3813884$. With respect to the $2^3 \times 3!=48$-element group of geometric symmetries of $[3]^3$, these sets partitioned into $83158$ equivalence classes:
$$
3813884 = 76066 \times 48+6527 \times 24+51 \times 16+338 \times 12 +109 \times 8+41 \times 6+13 \times 4 +5 \times 3+3 \times 2+5 \times 1.
$$
\end{remark}

Lemma \ref{paretop} yields the following new inequalities:
\begin{align*}
2a+b+2c+4d &\leq 22 \\
3a+2b+3c+6d &\leq 36 \\
7a+2b+4c+8d &\leq 56 \\
6a+2b+3c+6d &\leq 48 \\
a+2c+4d &\leq 14 \\
5a+4c+8d &\leq 40.
\end{align*}

Applying Lemma \ref{prop}, we obtain new inequalities:
\begin{align}
8\alpha_r(A)+ 6\alpha_{r+q}(A) + 6\alpha_{r+2q}(A) + 2\alpha_{r+3q}(A) &\leq 11 \label{eleven}\\
4\alpha_r(A)+4\alpha_{r+q}(A)+3\alpha_{r+2q}(A)+\alpha_{r+3q}(A) &\leq 6\label{six}\\
7\alpha_r(A)+3\alpha_{r+q}(A)+3\alpha_{r+2q}(A)+\alpha_{r+3q}(A) &\leq 7\nonumber\\
8\alpha_r(A)+3\alpha_{r+q}(A)+3\alpha_{r+2q}(A)+\alpha_{r+3q}(A) &\leq 8\label{eight}\\
4\alpha_{r+q}(A)+2\alpha_{r+2q}(A)+\alpha_{r+3q}(A) &\leq 4\nonumber\\
4\alpha_r(A)+6\alpha_{r+2q}(A)+2\alpha_{r+3q}(A) &\leq 7\nonumber\\
5\alpha_r(A)+3\alpha_{r+2q}(A)+\alpha_{r+3q}(A) &\leq 5\nonumber
\end{align}
whenever $r \geq 0$, $q \geq 1$, $n \geq r+3q$, and Moser sets $A \subset [3]^n$.

We also note some further corollaries of Lemma \ref{paretop}:

\begin{corollary}[Statistics of large 3D Moser sets]\label{paretop2} Let $(a,b,c,d)$ be the statistics of a Moser set $A$ in $[3]^3$. Then $|A| = a+b+c+d \leq 16$. Furthermore:
\begin{itemize}
\item If $|A|=16$, then $(a,b,c,d) = (4,12,0,0)$.
\item If $|A|=15$, then $(a,b,c,d) = (4,11,0,0)$ or $(3,12,0,0)$.
\item If $|A| \geq 14$, then $b \geq 6$ and $d=0$.
\item If $|A| = 13$ and $d=1$, then $(a,b,c,d) = (4,6,2,1)$ or $(3,6,3,1)$.
\end{itemize}
\end{corollary}

\subsection{Four dimensions}

Now we establish the bound $c'_{4,3}=43$. Let $A$ be a Moser set in $[3]^4$, with attendant statistics $(a,b,c,d,e)$, which range between $(0,0,0,0,0)$ and $(16,32,24,8,1)$. In view of the lower bounds, our task here is to establish the upper bound $a+b+c+d+e \leq 43$.

The linear inequalities already established just barely fail to achieve this bound, but we can obtain the upper bound $a+b+c+d+e \leq 44$ as follows.
First suppose that $e=1$; then from the inequalities \eqref{alpha-1} (or by considering lines passing through $2222$) we see that $a \leq 8, b \leq 16, c \leq 12, d \leq 4$ and hence $a+b+c+d+e \leq 41$, so we may assume that $e=0$.

From Lemma \ref{dci}, we see that $a+b+c+d+e$ is now equal to the sum of $a(V)/4+b(V)/3+c(V)/2+d(V)$, where $V$ ranges over all side slices of $[3]^4$. But from Lemma \ref{paretop} we see that $a(V)/4+b(V)/3+c(V)/2+d(V)$ is at most $\frac{11}{4}$, with equality occuring only when $(a(V),b(V),c(V),d(V))=(2,6,6,0)$. This gives the upper bound $a+b+c+d+e \leq 44$.

The above argument shows that $a+b+c+d+e=44$ can only occur if $e=0$ and if $(a(V),b(V),c(V),d(V))=(2,6,6,0)$ for all side slices $V$. Applying Lemma \ref{paretop} again this implies $(a,b,c,d,e)=(4,16,24,0,0)$. But then $A$ contains all of the sphere $S_{2,4}$, which implies that the four-element set $A \cap S_{4,4}$ cannot contain a pair of strings which differ in exactly two positions (as their midpoint would then lie in $S_{2,4}$, contradicting the hypothesis that $A$ is a Moser set).

Recall that we may partition $S_{4,4} = S_{4,4}^e \cup S_{4,4}^o$, where
$$S_{4,4}^e := \{ 1111, 1133, 1313, 3113, 1331, 3131, 3311, 3333\}$$
is the strings in $S_{4,4}$ with an even number of $1$'s, and
$$S_{4,4}^o := \{ 1113, 1131, 1311, 3111, 1333, 3133, 3313, 3331\}$$
are the strings in $S_{4,4}$ with an odd number. Observe that any two distinct elements in $S_{4,4}^e$ differ in exactly two positions unless they are antipodal. Thus $A \cap S_{4,4}^e$ has size at most two, with equality only when $A \cap S_{4,4}^e$ consists of an antipodal pair. Similarly for $A \cap S_{4,4}^o$. Thus $A$ must consist of two antipodal pairs, one from $S_{4,4}^e$ and one from $S_{4,4}^o$.

By the symmetries of the cube we may assume without loss of generality that these pairs are $\{ 1111, 3333\}$ and $\{1113,3331\}$ respectively. But as $A$ is a Moser set, $A$ must now exclude the strings $1112$ and $3332$. These two strings form two corners of the eight-element set
$$ ***2 \cap S_{3,4} = \{ 1112, 1132, 1312, 3112, 1332, 3132, 3312, 3332 \}.$$
Any pair of points in this set which are ``adjacent'' in the sense that they differ by exactly one entry cannot both lie in $A$, as their midpoint would then lie in $S_{3,4}$, and so $A$ can contain at most four elements from this set, with equality only if $A$ contains all the points in $***2 \cap S_{3,4}$ of the same parity (either all the elements with an even number of $3$s, or all the elements with an odd number of $3$s). But because the two corners removed from this set have the opposite parity (one has an even number of $1$s and one has an odd number), we see in fact that $A$ can contain at most $3$ points from this set. Meanwhile, the same arguments give that $A$ contains at most four points from $**2* \cap S_{3,4}$, $*2** \cap S_{3,4}$, and $2*** \cap S_{3,4}$. Summing we see that $b = |A \cap S_{3,4}| \leq 3+4+4+4=15$, a contradiction. Thus we have $c'_{4,3}=43$ as claimed.

We have the following four-dimensional version of Lemma \ref{paretop}:

\begin{lemma}[4D Pareto-optimals]\label{paretop-4} When $n=4$, the Pareto-optimal statistics listed on Table \ref{table4}.
\end{lemma}

\begin{figure}[tb]
\centered{\tiny
$(3, 16, 24, 0, 0)$,
$(4, 14, 19, 2, 0)$,
$(4, 15, 24, 0, 0)$,
$(4, 16, 8, 4, 1)$,
$(4, 16, 14, 4, 0)$,
$(4, 16, 23, 0, 0)$,
$(4, 17, 21, 0, 0)$,
$(4, 18, 19, 0, 0)$,
$(5, 12, 12, 4, 1)$,
$(5, 12, 13, 6, 0)$,
$(5, 12, 15, 5, 0)$,
$(5, 12, 19, 2, 0)$,
$(5, 13, 10, 4, 1)$,
$(5, 13, 14, 5, 0)$,
$(5, 13, 21, 1, 0)$,
$(5, 15, 9, 4, 1)$,
$(5, 15, 12, 3, 1)$,
$(5, 15, 13, 5, 0)$,
$(5, 15, 18, 3, 0)$,
$(5, 15, 20, 1, 0)$,
$(5, 15, 22, 0, 0)$,
$(5, 16, 7, 4, 1)$,
$(5, 16, 10, 3, 1)$,
$(5, 16, 11, 5, 0)$,
$(5, 16, 12, 2, 1)$,
$(5, 16, 16, 3, 0)$,
$(5, 16, 19, 1, 0)$,
$(5, 16, 21, 0, 0)$,
$(5, 17, 12, 4, 0)$,
$(5, 17, 14, 3, 0)$,
$(5, 17, 16, 2, 0)$,
$(5, 17, 18, 1, 0)$,
$(5, 17, 20, 0, 0)$,
$(5, 18, 13, 3, 0)$,
$(5, 18, 14, 2, 0)$,
$(5, 20, 8, 4, 0)$,
$(5, 20, 10, 3, 0)$,
$(5, 20, 13, 2, 0)$,
$(5, 20, 14, 1, 0)$,
$(5, 20, 18, 0, 0)$,
$(5, 21, 10, 2, 0)$,
$(5, 21, 15, 0, 0)$,
$(5, 22, 13, 0, 0)$,
$(6, 8, 12, 8, 0)$,
$(6, 10, 11, 4, 1)$,
$(6, 11, 12, 7, 0)$,
$(6, 12, 10, 7, 0)$,
$(6, 12, 13, 5, 0)$,
$(6, 12, 18, 4, 0)$,
$(6, 13, 16, 4, 0)$,
$(6, 14, 9, 4, 1)$,
$(6, 14, 9, 7, 0)$,
$(6, 14, 12, 6, 0)$,
$(6, 14, 16, 3, 0)$,
$(6, 14, 19, 1, 0)$,
$(6, 14, 21, 0, 0)$,
$(6, 15, 7, 4, 1)$,
$(6, 15, 10, 3, 1)$,
$(6, 15, 10, 6, 0)$,
$(6, 15, 11, 2, 1)$,
$(6, 15, 12, 5, 0)$,
$(6, 15, 15, 4, 0)$,
$(6, 15, 20, 0, 0)$,
$(6, 16, 7, 3, 1)$,
$(6, 16, 8, 6, 0)$,
$(6, 16, 9, 2, 1)$,
$(6, 16, 10, 5, 0)$,
$(6, 16, 12, 1, 1)$,
$(6, 16, 13, 4, 0)$,
$(6, 16, 14, 3, 0)$,
$(6, 16, 18, 2, 0)$,
$(6, 16, 19, 0, 0)$,
$(6, 17, 9, 5, 0)$,
$(6, 17, 10, 4, 0)$,
$(6, 17, 13, 3, 0)$,
$(6, 17, 15, 2, 0)$,
$(6, 17, 17, 1, 0)$,
$(6, 17, 18, 0, 0)$,
$(6, 18, 13, 2, 0)$,
$(6, 18, 16, 1, 0)$,
$(6, 18, 17, 0, 0)$,
$(6, 19, 9, 4, 0)$,
$(6, 19, 12, 3, 0)$,
$(6, 19, 15, 1, 0)$,
$(6, 20, 7, 4, 0)$,
$(6, 20, 9, 3, 0)$,
$(6, 20, 12, 2, 0)$,
$(6, 20, 13, 1, 0)$,
$(6, 20, 15, 0, 0)$,
$(6, 21, 8, 3, 0)$,
$(6, 21, 9, 2, 0)$,
$(6, 21, 12, 1, 0)$,
$(6, 21, 14, 0, 0)$,
$(6, 22, 7, 3, 0)$,
$(6, 22, 8, 2, 0)$,
$(6, 22, 10, 1, 0)$,
$(6, 23, 9, 1, 0)$,
$(6, 24, 7, 2, 0)$,
$(6, 24, 8, 1, 0)$,
$(6, 24, 12, 0, 0)$,
$(6, 25, 9, 0, 0)$,
$(6, 26, 7, 0, 0)$,
$(7, 8, 6, 8, 0)$,
$(7, 11, 9, 4, 1)$,
$(7, 11, 12, 6, 0)$,
$(7, 12, 8, 4, 1)$,
$(7, 12, 8, 6, 0)$,
$(7, 12, 12, 3, 1)$,
$(7, 12, 12, 5, 0)$,
$(7, 12, 13, 4, 0)$,
$(7, 12, 15, 3, 0)$,
$(7, 12, 17, 2, 0)$,
$(7, 13, 7, 4, 1)$,
$(7, 13, 10, 3, 1)$,
$(7, 13, 11, 5, 0)$,
$(7, 13, 12, 2, 1)$,
$(7, 13, 12, 4, 0)$,
$(7, 13, 14, 3, 0)$,
$(7, 13, 16, 2, 0)$,
$(7, 14, 6, 4, 1)$,
$(7, 14, 6, 7, 0)$,
$(7, 14, 9, 5, 0)$,
$(7, 14, 10, 2, 1)$,
$(7, 14, 12, 1, 1)$,
$(7, 14, 17, 1, 0)$,
$(7, 14, 19, 0, 0)$,
$(7, 15, 7, 5, 0)$,
$(7, 15, 8, 3, 1)$,
$(7, 15, 9, 2, 1)$,
$(7, 15, 11, 1, 1)$,
$(7, 15, 11, 4, 0)$,
$(7, 15, 13, 3, 0)$,
$(7, 15, 16, 1, 0)$,
$(7, 16, 6, 3, 1)$,
$(7, 16, 6, 6, 0)$,
$(7, 16, 8, 2, 1)$,
$(7, 16, 10, 1, 1)$,
$(7, 16, 10, 4, 0)$,
$(7, 16, 12, 0, 1)$,
$(7, 16, 12, 3, 0)$,
$(7, 16, 15, 2, 0)$,
$(7, 16, 17, 0, 0)$,
$(7, 17, 6, 5, 0)$,
$(7, 17, 7, 4, 0)$,
$(7, 17, 11, 3, 0)$,
$(7, 17, 13, 2, 0)$,
$(7, 17, 14, 1, 0)$,
$(7, 17, 16, 0, 0)$,
$(7, 18, 10, 3, 0)$,
$(7, 18, 13, 1, 0)$,
$(7, 18, 15, 0, 0)$,
$(7, 19, 9, 3, 0)$,
$(7, 20, 6, 4, 0)$,
$(7, 20, 11, 2, 0)$,
$(7, 20, 12, 1, 0)$,
$(7, 20, 14, 0, 0)$,
$(7, 21, 8, 2, 0)$,
$(7, 21, 10, 1, 0)$,
$(7, 21, 12, 0, 0)$,
$(7, 22, 9, 1, 0)$,
$(7, 22, 11, 0, 0)$,
$(7, 23, 6, 3, 0)$,
$(7, 23, 7, 1, 0)$,
$(7, 23, 10, 0, 0)$,
$(7, 24, 6, 2, 0)$,
$(7, 24, 9, 0, 0)$,
$(7, 25, 6, 1, 0)$,
$(7, 25, 8, 0, 0)$,
$(7, 26, 3, 1, 0)$,
$(7, 28, 6, 0, 0)$,
$(7, 29, 3, 0, 0)$,
$(7, 30, 1, 0, 0)$,
$(8, 8, 0, 8, 0)$,
$(8, 8, 9, 7, 0)$,
$(8, 8, 12, 6, 0)$,
$(8, 9, 9, 4, 1)$,
$(8, 9, 10, 6, 0)$,
$(8, 9, 12, 3, 1)$,
$(8, 9, 12, 5, 0)$,
$(8, 9, 13, 4, 0)$,
$(8, 9, 15, 3, 0)$,
$(8, 10, 7, 4, 1)$,
$(8, 10, 10, 3, 1)$,
$(8, 10, 10, 5, 0)$,
$(8, 10, 12, 2, 1)$,
$(8, 10, 12, 4, 0)$,
$(8, 10, 13, 3, 0)$,
$(8, 10, 15, 2, 0)$,
$(8, 11, 6, 4, 1)$,
$(8, 11, 9, 6, 0)$,
$(8, 11, 10, 2, 1)$,
$(8, 11, 11, 4, 0)$,
$(8, 12, 7, 6, 0)$,
$(8, 12, 9, 3, 1)$,
$(8, 12, 9, 5, 0)$,
$(8, 12, 10, 4, 0)$,
$(8, 12, 12, 1, 1)$,
$(8, 12, 14, 2, 0)$,
$(8, 12, 16, 1, 0)$,
$(8, 12, 18, 0, 0)$,
$(8, 13, 7, 3, 1)$,
$(8, 13, 7, 5, 0)$,
$(8, 13, 9, 2, 1)$,
$(8, 13, 12, 0, 1)$,
$(8, 13, 12, 3, 0)$,
$(8, 14, 0, 7, 0)$,
$(8, 14, 6, 6, 0)$,
$(8, 14, 7, 2, 1)$,
$(8, 14, 8, 1, 1)$,
$(8, 14, 9, 4, 0)$,
$(8, 14, 11, 0, 1)$,
$(8, 14, 11, 3, 0)$,
$(8, 14, 13, 2, 0)$,
$(8, 14, 15, 1, 0)$,
$(8, 14, 17, 0, 0)$,
$(8, 15, 6, 3, 1)$,
$(8, 15, 6, 5, 0)$,
$(8, 15, 7, 1, 1)$,
$(8, 16, 0, 6, 0)$,
$(8, 16, 4, 3, 1)$,
$(8, 16, 4, 5, 0)$,
$(8, 16, 6, 2, 1)$,
$(8, 16, 8, 4, 0)$,
$(8, 16, 9, 0, 1)$,
$(8, 16, 10, 3, 0)$,
$(8, 16, 12, 2, 0)$,
$(8, 16, 14, 1, 0)$,
$(8, 16, 16, 0, 0)$,
$(8, 17, 0, 5, 0)$,
$(8, 17, 3, 4, 0)$,
$(8, 17, 8, 3, 0)$,
$(8, 17, 10, 2, 0)$,
$(8, 17, 12, 1, 0)$,
$(8, 17, 14, 0, 0)$,
$(8, 18, 9, 2, 0)$,
$(8, 18, 11, 1, 0)$,
$(8, 18, 12, 0, 0)$,
$(8, 19, 6, 3, 0)$,
$(8, 19, 8, 2, 0)$,
$(8, 20, 0, 4, 0)$,
$(8, 20, 4, 3, 0)$,
$(8, 20, 7, 2, 0)$,
$(8, 20, 9, 1, 0)$,
$(8, 20, 11, 0, 0)$,
$(8, 21, 4, 2, 0)$,
$(8, 21, 7, 1, 0)$,
$(8, 22, 3, 2, 0)$,
$(8, 22, 6, 1, 0)$,
$(8, 22, 9, 0, 0)$,
$(8, 23, 0, 3, 0)$,
$(8, 23, 4, 1, 0)$,
$(8, 24, 0, 2, 0)$,
$(8, 24, 3, 1, 0)$,
$(8, 24, 8, 0, 0)$,
$(8, 25, 1, 1, 0)$,
$(8, 25, 6, 0, 0)$,
$(8, 26, 0, 1, 0)$,
$(8, 26, 4, 0, 0)$,
$(8, 28, 3, 0, 0)$,
$(8, 32, 0, 0, 0)$,
$(9, 8, 10, 4, 0)$,
$(9, 9, 9, 4, 0)$,
$(9, 9, 12, 3, 0)$,
$(9, 10, 8, 4, 0)$,
$(9, 10, 10, 3, 0)$,
$(9, 10, 12, 2, 0)$,
$(9, 10, 13, 1, 0)$,
$(9, 10, 15, 0, 0)$,
$(9, 11, 11, 2, 0)$,
$(9, 12, 7, 4, 0)$,
$(9, 12, 9, 3, 0)$,
$(9, 12, 12, 1, 0)$,
$(9, 12, 14, 0, 0)$,
$(9, 13, 7, 3, 0)$,
$(9, 13, 10, 2, 0)$,
$(9, 14, 9, 2, 0)$,
$(9, 14, 11, 1, 0)$,
$(9, 14, 13, 0, 0)$,
$(9, 15, 6, 3, 0)$,
$(9, 16, 0, 4, 0)$,
$(9, 16, 4, 3, 0)$,
$(9, 16, 8, 2, 0)$,
$(9, 16, 10, 1, 0)$,
$(9, 16, 12, 0, 0)$,
$(9, 17, 3, 3, 0)$,
$(9, 17, 6, 2, 0)$,
$(9, 17, 8, 1, 0)$,
$(9, 17, 10, 0, 0)$,
$(9, 18, 2, 3, 0)$,
$(9, 18, 4, 2, 0)$,
$(9, 18, 7, 1, 0)$,
$(9, 18, 9, 0, 0)$,
$(9, 19, 0, 3, 0)$,
$(9, 19, 3, 2, 0)$,
$(9, 19, 6, 1, 0)$,
$(9, 20, 1, 2, 0)$,
$(9, 20, 5, 1, 0)$,
$(9, 20, 8, 0, 0)$,
$(9, 21, 4, 1, 0)$,
$(9, 21, 6, 0, 0)$,
$(9, 22, 1, 1, 0)$,
$(9, 22, 5, 0, 0)$,
$(9, 24, 4, 0, 0)$,
$(9, 25, 2, 0, 0)$,
$(9, 28, 0, 0, 0)$,
$(10, 8, 6, 4, 0)$,
$(10, 8, 8, 3, 0)$,
$(10, 9, 7, 3, 0)$,
$(10, 9, 10, 2, 0)$,
$(10, 9, 11, 1, 0)$,
$(10, 9, 13, 0, 0)$,
$(10, 10, 5, 4, 0)$,
$(10, 10, 9, 2, 0)$,
$(10, 10, 12, 0, 0)$,
$(10, 11, 6, 3, 0)$,
$(10, 12, 4, 4, 0)$,
$(10, 12, 5, 3, 0)$,
$(10, 12, 7, 2, 0)$,
$(10, 12, 10, 1, 0)$,
$(10, 12, 11, 0, 0)$,
$(10, 13, 6, 2, 0)$,
$(10, 13, 8, 1, 0)$,
$(10, 13, 10, 0, 0)$,
$(10, 14, 3, 3, 0)$,
$(10, 14, 5, 2, 0)$,
$(10, 14, 9, 0, 0)$,
$(10, 15, 2, 3, 0)$,
$(10, 15, 7, 1, 0)$,
$(10, 16, 4, 2, 0)$,
$(10, 16, 6, 1, 0)$,
$(10, 16, 8, 0, 0)$,
$(10, 17, 4, 1, 0)$,
$(10, 17, 6, 0, 0)$,
$(10, 18, 2, 1, 0)$,
$(10, 18, 5, 0, 0)$,
$(10, 20, 4, 0, 0)$,
$(10, 21, 2, 0, 0)$,
$(10, 22, 1, 0, 0)$,
$(10, 24, 0, 0, 0)$,
$(11, 4, 6, 4, 0)$,
$(11, 6, 5, 4, 0)$,
$(11, 7, 6, 3, 0)$,
$(11, 8, 4, 4, 0)$,
$(11, 8, 5, 3, 0)$,
$(11, 9, 6, 2, 0)$,
$(11, 9, 8, 1, 0)$,
$(11, 9, 10, 0, 0)$,
$(11, 10, 3, 3, 0)$,
$(11, 10, 5, 2, 0)$,
$(11, 10, 9, 0, 0)$,
$(11, 11, 2, 3, 0)$,
$(11, 11, 7, 1, 0)$,
$(11, 12, 4, 2, 0)$,
$(11, 12, 6, 1, 0)$,
$(11, 12, 8, 0, 0)$,
$(11, 13, 4, 1, 0)$,
$(11, 13, 6, 0, 0)$,
$(11, 14, 2, 1, 0)$,
$(11, 14, 5, 0, 0)$,
$(11, 16, 4, 0, 0)$,
$(11, 17, 2, 0, 0)$,
$(11, 18, 1, 0, 0)$,
$(11, 20, 0, 0, 0)$,
$(12, 4, 3, 3, 0)$,
$(12, 6, 2, 3, 0)$,
$(12, 6, 5, 2, 0)$,
$(12, 6, 7, 1, 0)$,
$(12, 6, 9, 0, 0)$,
$(12, 8, 4, 2, 0)$,
$(12, 8, 6, 1, 0)$,
$(12, 8, 8, 0, 0)$,
$(12, 9, 4, 1, 0)$,
$(12, 9, 6, 0, 0)$,
$(12, 10, 2, 1, 0)$,
$(12, 10, 5, 0, 0)$,
$(12, 12, 4, 0, 0)$,
$(12, 13, 2, 0, 0)$,
$(12, 14, 1, 0, 0)$,
$(12, 16, 0, 0, 0)$,
$(13, 6, 5, 0, 0)$,
$(13, 8, 4, 0, 0)$,
$(13, 9, 2, 0, 0)$,
$(13, 10, 1, 0, 0)$,
$(13, 12, 0, 0, 0)$,
$(14, 4, 3, 0, 0)$,
$(14, 5, 2, 0, 0)$,
$(14, 6, 1, 0, 0)$,
$(14, 8, 0, 0, 0)$,
$(15, 4, 0, 0, 0)$,
$(16, 0, 0, 0, 0)$}}
\caption{The Pareto-optimal statistics of Moser sets in $[3]^4$. This table can also be found at {\tt http://spreadsheets.google.com/ccc?key=rwXB\_Rn3Q1Zf5yaeMQL-RDw}}
\label{table4}
\end{figure}

\begin{proof} This was computed by computer search as follows. First, one observed that if $(a,b,c,d,e)$ was Pareto-optimal, then $a\geq 3$. To see this, it suffices to show that for any Moser set $A \subset [3]^4$ with $a(A)=0$, it is possible to add three points from $S_{4,4}$ to $A$ and still have a Moser set. To show this, suppose first that $A$ contains a point from $S_{1,4}$, such as $2221$. Then $A$ must omit either $2211$ or $2231$; without loss of generality we may assume that it omits $2211$. Similarly we may assume it omits $2121$ and $1221$. Then we can add $1131$, $1311$, $3111$ to $A$, as required. Thus we may assume that $A$ contains no points from $S_{1,4}$. Now suppose that $A$ omits a point from $S_{2,4}$, such as $2211$. Then one can add $3333$, $3111$, $1311$ to $A$, as required. Thus we may assume that A contains all of $S_{2,4}$, which forces $A$ to omit $2222$, as well as at least one point from $S_{3,4}$, such as $2111$. But then $3111$, $1111$, $3333$ can be added to the set, a contradiction.

Thus we only need to search through sets $A \subset [3]^4$ for which $|A \cap S_{4,4}| \geq 3$. A straightforward computer search shows that up to the symmetries of the cube, there are $391$ possible choices for $A \cap S_{4,4}$. For each such choice, we looped through all the possible values of the slices $A \cap 1***$ and $A \cap 3***$, i.e. all three-dimensional Moser sets which had the indicated intersection with $S_{3,3}$. (For fixed $A \cap S_{4,4}$, the number of possibilities for $A \cap 1***$ ranges from $1$ to $87123$, and similarly for $A \cap 3***$). For each pair of slices $A \cap 1***$ and $A \cap 3***$, we computed the lines connecting these two sets to see what subset of $2***$ was excluded from $A$; there are $2^{27}$ possible such exclusion sets. We precomputed a lookup table that gave the Pareto-optimal statistics for $A \cap 2***$ for each such choice of exclusion set; using this lookup table for each choice of $A \cap 1***$ and $A \cap 3***$ and collating the results, we obtained the above list. On a linux cluster, the lookup table took 22 minutes to create, and the loop over the $A \cap 1***$ and $A \cap 3***$ slices took two hours, spread out over $391$ machines (one for each choice of $A \cap S_{4,4}$). Further details (including source code) can be found at
\centerline{{\tt http://michaelnielsen.org/polymath1/index.php?title=4D\_Moser\_brute\_force\_search}.}
\end{proof}

As a consequence of this data, we have the following facts about the statistics of large Moser sets:

\begin{proposition}\label{stat} Let $A \subset [3]^4$ be a Moser set with statistics $(a,b,c,d,e)$.
\begin{itemize}
\item[(i)] If $|A| \geq 40$, then $e=0$.
\item[(ii)] If $|A| \geq 43$, then $d=0$.
\item[(iii)] If $|A| \geq 42$, then $d \leq 2$.
\item[(iv)] If $|A| \geq 41$, then $d \leq 3$.
\item[(v)] If $|A| \geq 40$, then $d \leq 6$.
\item[(vi)] If $|A| \geq 43$, then $c \geq 18$.
\item[(vii)] If $|A| \geq 42$, then $c \geq 12$.
\item[(viii)] If $|A| \geq 43$, then $b \geq 15$.
\end{itemize}
\end{proposition}

\begin{remark} This proposition was first established by an integer program, see Appendix \ref{integer-sec}. A computer-free proof can be found at \centerline{{\tt http://terrytao.files.wordpress.com/2009/06/polymath2.pdf}.}
\end{remark}

\subsection{Five dimensions}

Now we establish the bound $c'_{5,3}=124$. In view of the lower bounds, it suffices to show that there does not exist a Moser set $A \subset [3]^5$ with $|A|=125$.

We argue by contradiction. Let $A$ be as above, and let $(a(A),\ldots,f(A))$ be the statistics of $A$.

\begin{lemma}\label{fvan} $f(A)=0$.
\end{lemma}

\begin{proof} If $f(A)$ is non-zero, then $A$ contains $22222$, then each of the $\frac{3^5-1}{2} = 121$ antipodal pairs in $[3]^5$ can have at most one point in $A$, leading to only $122$ points.
\end{proof}

Let us slice $[3]^5$ into three parallel slices, e.g. $1****, 2****, 3****$. The intersection of $A$ with each of these slices has size at most $43$. In particular, this implies that
\begin{equation}\label{boo}
|A \cap 1****| + |A \cap 3****| = 125 - |A \cap 2****| \geq 82.
\end{equation}
Thus at least one of $A \cap 1****$, $A \cap 3****$ has cardinality at least $41$; by Proposition \ref{stat}(iv) we conclude that
\begin{equation}\label{d13}
\min( d(1****), d(3****) ) \leq 3.
\end{equation}
Furthermore, equality can only hold in \eqref{d13} if $A \cap 1****$, $A \cap 3****$ both have cardinality exactly $41$, in which case from Proposition \ref{stat}(iv) again we must have
\begin{equation}\label{d13a}
d(1****)=d(3****)=3.
\end{equation}
Of course, we have a similar result for permutations.

Now we improve the bound $|A \cap 2****| \leq 43$:

\begin{lemma} $|A \cap 2****| \leq 41$.
\end{lemma}

\begin{proof} Suppose first that $|A \cap 2****|=43$. Let $A' \subset [3]^4$ be the subset of $[3]^4$ corresponding to $A \cap 2****$, thus $A'$ is a Moser set of cardinality $43$. By Proposition \ref{stat}(vi), $c(A') \geq 18$. By Lemma \ref{dci}, the sum of the $c(V)$, where $V$ ranges over the eight side slices of $[3]^4$, is therefore at least $36$. By the pigeonhole principle, we may thus find two opposing side slices, say $1***$ and $3***$, with $c(1***)+c(3****) \geq 9$. Since $c(1***), c(3***)$ cannot exceed $6$, we thus have $c(1***), c(3***) \geq 3$, with at least one of $c(1***), c(3***)$ being at least $5$. Passing back to $A$, this implies that $d(*1***), d(*3***) \geq 3$, with at least one of $d(*1***), d(*3***)$ being at least $5$. But this contradicts \eqref{d13} together with the refinement \eqref{d13a}.

We have just shown that $|A \cap 2****| \leq 42$; we can thus improve \eqref{boo} to
$$ |A \cap 1****| + |A \cap 3****| \geq 83.$$
Combining this with Proposition \ref{stat}(ii)-(v) we see that
\begin{equation}\label{d13-6}
d(1****)+d(3****) \leq 6
\end{equation}
with equality only if $|A \cap 2****|=42$, and similarly for permutations.

Now let $A'$ be defined as before. Then we have
$$ c(1***) + c(3***) \leq 6$$
and similarly for permutations. Applying Lemma \ref{dci}, this implies that $c(2****) = c(A') \leq 12$.

Now suppose for contradiction that $|A'|=|A \cap 2****|=42$. Then by Proposition \ref{stat}(vii) we have
\begin{equation}\label{coo-1}
c(2****) = 12;
\end{equation}
applying Lemma \ref{dci} again, this forces $c(1***)+c(3***)=6$ and similarly for permutations, which then implies that
\begin{equation}\label{doo}
d(*1***)+d(*3***) = d(**1**)+d(**3**) = d(***1*)+d(***3*) = d(****1)+d(****3) = 6
\end{equation}
and hence
$$ |A \cap *2***| = |A \cap **2**| = |A \cap ***2*| = |A \cap ****2| = 42$$
and thus
\begin{equation}\label{coo-2}
c(*2***) = c(**2**) = c(***2*) = c(****2) = 12.
\end{equation}
Combining \eqref{coo-1}, \eqref{doo}, \eqref{coo-2} we conclude that
$$ d(1****)+d(3****) = 16,$$
contradicting \eqref{d13-6}.
\end{proof}

With this proposition, the bound \eqref{boo} now improves to
\begin{equation}\label{84}
|A \cap 1****| + |A \cap 3****| \geq 84
\end{equation}
and in particular
\begin{equation}\label{41}
|A \cap 1****|, |A \cap 3****| \geq 41.
\end{equation}
from this and Proposition \ref{stat}(ii)-(iv) we now have
\begin{equation}\label{d13-improv}
d(1****)+d(3****) \leq 4
\end{equation}
and similarly for permutations.

\begin{lemma}\label{evan} $e(A)=0$.
\end{lemma}

\begin{proof} From \eqref{84}, the intersection of $A$ with any side slice has cardinality at least $41$, and thus by Proposition \ref{stat}(i) such a side slice has an $e$-statistic of zero. The claim then follows from Lemma \ref{dci}.
\end{proof}

We need a technical lemma:

\begin{lemma}\label{tech} Let $B \subset S_{5,5}$. Then there exist at least $|B|-4$ pairs of strings in $B$ which differ in exactly two positions.
\end{lemma}

\begin{proof} The first non-vacuous case is $|B|=5$. It suffices to establish this case, as the higher cases then follow by induction (locating a pair of the desired form, then deleting one element of that pair from $B$).

Suppose for contradiction that one can find a $5$-element set $B \subset S_{5,5}$ such that no two strings in $B$ differ in exactly two positions. Recall that we may split $S_{5,5}=S_{5,5}^e \cup S_{5,5}^o$, where $S_{5,5}^e$ are those strings with an even number of $1$'s, and $S_{5,5}^o$ are those strings with an odd number of $1$'s. By the pigeonhole principle and symmetry we may assume $B$ has at least three elements in $S_{5,5}^o$. Without loss of generality, we can take one of them to be $11111$, thus excluding all elements in $S_{5,5}^o$ with exactly two $3$s, leaving only the elements with exactly four $3$s. But any two of them differ in exactly two positions, a contradiction.
\end{proof}

We can now improve the trivial bound $c(A) \leq 80$:

\begin{corollary}[Non-maximal $c$]\label{cmax} $c(A) \leq 79$. If $a(A) \geq 7$, then $c(A) \leq 78$.
\end{corollary}

\begin{proof} If $c(A)=80$, then $A$ contains all of $S_{3,5}$, which then implies that no two elements in $A \cap S_{5,5}$ can differ in exactly two places. It also implies (from \eqref{alpha-1}) that $d(A)$ must vanish, and that $b(A)$ is at most $40$. By Lemma \ref{tech}, we also have that $a(A) = |A \cap S_{5,5}|$ is at most $4$. Thus $|A| \leq 4 + 40 + 80 + 0 + 0 = 124$, a contradiction.

Now suppose that $a(A) \geq 7$. Then by Lemma \ref{tech} there are at least three pairs in $A \cap S_{5,5}$ that differ in exactly two places. Each such pair eliminates one point from $A \cap S_{3,5}$; but each point in $S_{3,5}$ can be eliminated by at most two such pairs, and so we have at least two points eliminated from $A \cap S_{3,5}$, i.e. $c(A) \leq 78$ as required.
\end{proof}

Next, we rewrite the quantity $125=|A|$ in terms of side slices. From Lemmas \ref{fvan}, \ref{evan} we have
$$ a(A) + b(A) + c(A) + d(A) = 125$$
and hence by Lemma \ref{dci}, the quantity
$$ s(V) := a(V) + \frac{5}{4} b(V) + \frac{5}{3} c(V) + \frac{5}{2} d(V) - \frac{125}{2},$$
where $V$ ranges over side slices, has an average value of zero.

\begin{proposition}[Large values of $s(V)$]\label{suv} For all side slices, we have $s(V) \leq 1/2$. Furthermore, we have $s(V) < -1/2$ unless the statistics $(a(V), b(V), c(V), d(V), e(V))$ are of one of the following four cases:
\begin{itemize}
\item (Type 1) $(a(V),b(V),c(V),d(V),e(V)) = (2,16,24,0,0)$ (and $s(V) = -1/2$ and $|A \cap V| = 42$);
\item (Type 2) $(a(V),b(V),c(V),d(V),e(V)) = (4,16,23,0,0)$ (and $s(V) = -1/6$ and $|A \cap V| = 43$);
\item (Type 3) $(a(V),b(V),c(V),d(V),e(V)) = (4,15,24,0,0)$ (and $s(V) = 1/4$ and $|A \cap V| = 43$);
\item (Type 4) $(a(V),b(V),c(V),d(V),e(V)) = (3,16,24,0,0)$ (and $s(V) = 1/2$ and $|A \cap V| = 43$);
\end{itemize}
\end{proposition}

\begin{proof}
Let $V$ be a side slice. From \eqref{41} we have
$$ 41 \leq a(V)+b(V)+c(V)+d(V) = |A \cap V| \leq 43.$$
First suppose that $|A \cap V| = 43$, then from Proposition \ref{stat}(ii), (viii), $d(V)=0$ and $b(V) \geq 15$.
Also, we have the trivial bound $c(V) \leq 24$, together with the inequality
$$ 3b(V) + 2c(V) \leq 96$$
from \eqref{alpha-1}. To exploit these facts, we rewrite $s(V)$ as
$$ s(V) = \frac{1}{2} - \frac{1}{2}( 24 - c(V) ) - \frac{1}{12} (96-3b(V)-2c(V)).$$
Thus $s(V) \leq 1/2$ in this case. If $s(V) \geq -1/2$, then
$$ 6 (24-c(V)) + (96-3b(V)-2c(V)) \leq 12,$$
which together with the inequalities $b(V) \leq 15$, $c(V) \leq 24$, $3b(V)+2c(V) \leq 96$ we conclude that $(b(V),c(V))$ must be one of $(16,24)$, $(15, 24)$, $(16, 23)$, $(15, 23)$. The first three possibilities lead to Types 4,3,2 respectively. The fourth type would lead to $(a(V),b(V),c(V),d(V),e(V)) = (5,15,23,0,0)$, but this contradicts \eqref{eleven}.

Next, suppose $|A \cap V| = 42$, so by Proposition \ref{stat}(iii) we have $d(V) \leq 2$. From \eqref{alpha-1} we have
\begin{equation}\label{2cd}
2c(V) + 3d(V) \leq 48
\end{equation}
while from \eqref{alpha-2} we have
\begin{equation}\label{3cd}
3b(V)+2c(V)+3d(V) \leq 96
\end{equation}
and so we can rewrite $s(V)$ as
\begin{equation}\label{sv2}
s(V) = -\frac{1}{2} - \frac{1}{4}( 48 - 2c(V) - 3d(V) ) - \frac{1}{12} (96-3b(V)-2c(V)-3d(V)) + \frac{1}{2} d(V).
\end{equation}
This already gives $s(V) \leq 1/2$. If $d(V)=0$, then $s(V) \leq -1/2$, with equality only in Type 1. If $d(V)=1$, then the set $A' \subset [3]^4$ corresponding to $A \cap V$ contains a point in $S_{3,4}$, which without loss of generality we can take to be $2221$. Considering the three lines $*221$, $2*21$, $22*1$, we see that at least three points in $S_{2,4}$ must be missing from $A'$, thus $c(V) \leq 21$. This forces $48-2c(V)-3d(V) \geq 3$, and so $s(V) < -3/4$. Finally, if $d(V)=2$, then $A'$ contains two points in $S_{3,4}$. If they are antipodal (e.g. $2221$ and $2223$), the same argument as above shows that at least six points in $S_{2,4}$ are missing from $A'$; if they are not antipodal (e.g. $2221$ and $2212$) then by considering the lines $*221$, $2*21$, $22*1$, $*212$, $2*12$ we see that five points are missing. Thus we have $c(V) \leq 19$, which forces $48-2c(V)-3d(V) \geq 4$. This forces $s(V) \leq -1/2$, with equality only when $c(V)=19$ and $3b(V)+2c(V)+3d(V)=96$, but this forces $b(V)$ to be the non-integer $52/3$, a contradiction, which concludes the treatment of the $|A \cap V|=42$ case.

Finally, suppose $|A \cap V| = 41$. Using \eqref{2cd}, \eqref{3cd} as before we have
\begin{equation}\label{sv3}
s(V) = -\frac{3}{2} - \frac{1}{4}( 48 - 2c(V) - 3d(V) ) - \frac{1}{12} (96-3b(V)-2c(V)-3d(V)) + \frac{1}{2} d(V),
\end{equation}
while from Proposition \ref{stat}(vi) we have $d(V) \leq 3$. This already gives $s(V) \leq 0$, and $s(V) \leq -1$ when $d(V)=1$. In order to have $s(V) \geq -1/2$, we must then have $d(V)=2$ or $d(V)=3$. But then the arguments of the preceding paragraph give $48-2c(V)-3d(V) \geq 4$, and so $s(V) \leq -1$ in this case.
\end{proof}

Since the $s(V)$ average to zero, by the pigeonhole principle we may find two opposing side slices (e.g. $1****$ and $3****$), whose total $s$-value is non-negative. Actually we can do a little better:

\begin{lemma}\label{side-off} There exists two opposing side slices whose total $s$-value is strictly positive.
\end{lemma}

\begin{proof} If this is not the case, then we must have $s(1****)+s(3****)=0$ and similarly for permutations. Using Proposition \ref{suv} we thus see that for every opposing pair of side slices, one is Type 1 and one is Type 4. In particular $c(V)=24$ for all side slices $V$. But then by Lemma \ref{dci} we have $c(A)=80$, contradicting Lemma \ref{cmax}.
\end{proof}

Let $V, V'$ be the side slices in Lemma \ref{side-off}
By Proposition \ref{suv}, the $V, V'$ slices must then be either Type 2, Type 3, or Type 4, and they cannot both be Type 2. Since $a(A) = a(V)+a(V')$, we conclude
\begin{equation}\label{amix}
6 \leq a(A) \leq 8.
\end{equation}
In a similar spirit, we have
$$ c(V) + c(V') \leq 23+24.$$
On the other hand, by considering the $24$ lines connecting $c$-points of $V, V'$ to $c$-points of the centre slice $W$ between $V$ and $V'$, each of which contains at most two points in $A$, we have
$$ c(V) + c(W) + c(V') \leq 24 \times 2.$$
Thus $c(W) \leq 1$; since
$$ d(A) = d(V) + d(V') + c(W)$$
we conclude from Proposition \ref{suv} that $d(A) \leq 1$. Actually we can do better:

\begin{lemma} $d(A)=0$.
\end{lemma}

\begin{proof} Suppose for contradiction that $d(A)=1$; without loss of generality we may take $11222 \in A$. This implies that $d(1****)=d(*1***)=1$. Also, by the above discussion, $c(**1**)$ and $c(**3**)$ cannot both be $24$, so by Proposition \ref{suv}, $s(**1**)+s(**3**) \leq 1/3$; similarly
$s(***1*)+s(***3*) \leq 1/3$ and $s(****1)+s(****3) \leq 1/3$. Since the $s$ average to zero, we see from the pigeonhole principle that either $s(1****)+s(3****) \geq -1/2$ or $s(*1***)+s(*3***) \geq -1/2$. We may assume by symmetry that
\begin{equation}\label{star-2}
s(1****)+s(3****) \geq -1/2.
\end{equation}
Since $s(3****) \leq 1/2$ by Proposition \ref{suv}, we conclude that
\begin{equation}\label{star}
s(1****) \geq -1.
\end{equation}
If $|A \cap 1****|=41$, then by \eqref{sv3} we have
$$ s(1****) = -1 - \frac{1}{4}( 48 - 2c(1****) - 3d(1****) ) - \frac{1}{12} (96-3b(1****)-2c(1****)-3d(1****))$$
but the arguments in Proposition \ref{suv} give $48 - 2c(1****) - 3d(1****) \geq 3$ and $96-3b(1****)-2c(1****)-3d(1****) \geq 0$, a contradiction. So we must have $|A \cap 1****|=42$ (by Proposition \ref{stat}(ii) and \eqref{41}). In that case, from \eqref{sv2} we have
$$ s(1****) = \frac{1}{4}( 48 - 2c(1****) - 3d(1****) ) - \frac{1}{12} (96-3b(1****)-2c(1****)-3d(1****))$$
while also having $48 - 2c(1****) - 3d(1****) \geq 3$ and $96-3b(1****)-2c(1****)-3d(1****) \geq 0$. Since $s(1****) \geq -1$ and $d(1****)=1$, we soon see that we must have $48 - 2c(1****) - 3d(1****) = 3$ and $96-3b(1****)-2c(1****)-3d(1****) \leq 3$, which forces $c(1****)=21$ and $b(1****)=16$ or $b(1****)=17$; thus the statistics of $1****$ are either $(4,16,21,1,0)$ or $(3,17,21,1,0)$.

We first eliminate the $(3,17,21,1,0)$ case. In this case $s(1****)$ is exactly $-1$. Inspecting the proof of \eqref{star}, we conclude that $s(3****)$ must be $+1/2$ and that $s(**1**)+s(**3**)=1/3$. From the former fact and Proposition \ref{suv} we see that $a(A) = a(1****)+a(3****)=3+3=6$; on the other hand, from the latter fact and Proposition \ref{suv} we have $a(A) = a(**1**)+a(**3**) = 4+3=7$, a contradiction.

So $1****$ has statistics $(4,16,21,1,0)$, which implies that $s(1****)=-3/4$ and $|A \cap 1****|=42$. By \eqref{star-2} we conclude
\begin{equation}\label{s3}
s(3****) \geq 1/4,
\end{equation}
which by Proposition \ref{suv} implies that $|A \cap 3****|=43$, and hence $|A \cap 2****|=40$. On the other hand, since $e(A)=f(A)=0$ and $d(A)=1$, with the latter being caused by $11222$, we see that $c(2****)=d(2****)=e(2****)=0$. From \eqref{alpha-1} we have $4a(2****)+b(2****) \leq 64$, and we also have the trivial inequality $b(2****) \leq 32$; these inequalities are only compatible if $2****$ has statistics $(8,32,0,0,0)$, thus $A \cap 2****$ contains $S_{2,5} \cap 2****$.

If $a(3****)=4$, then $a(A)=a(1****)+a(3****)=8$, which by Proposition \ref{suv} implies that $s(**1**)+s(**3**)$ cannot exceed $1/12$, and similarly for permutations. On the other hand, from Proposition \ref{suv} $s(**1**)+s(**3**)$ cannot exceed $-3/4 + 1/4 = -1/2$, and so the average value of $s$ cannot be zero, a contradiction. Thus $a(3****) \neq 4$, which by \eqref{s3} and Proposition \ref{suv} implies that $**3**$ has statistics $(3,16,24,0,0)$.

In particular, $A$ contains $16$ points from $3**** \cap S_{1,5}$ and all of $3**** \cap S_{2,5}$. As a consequence, no pair of the $16$ points in $A \cap 3**** \cap S_{1,5}$ can differ in only one coordinate; partitioning the $32$-point set $3**** \cap S_{1,5}$ into $16$ such pairs, we conclude that every such pair contains exactly one element of $A$. We conclude that $A \cap 3**** \cap S_{1,5}$ is equal to either $3**** \cap S_{1,5}^e$ or $3**** \cap S_{1,5}^o$.

On the other hand, $A$ contains all of $2**** \cap S_{2,5}$, and exactly sixteen points from $1**** \cap S_{1,5}$. Considering the vertical lines $*xyzw$ where $xyzw \in S_{1,4}$, we conclude that $A \cap 1**** \cap S_{1,5}$ is either equal to $1**** \cap S_{1,5}^o$ or $1**** \cap S_{1,5}^e$.
But either case is incompatible with the fact that $A$ contains $11222$ (consider either the line $11xx2$ or $11x\overline{x}2$, where $x=1,2,3$ and $\overline{x}=4-x$), obtaining the required contradiction.
\end{proof}

We can now eliminate all but three cases for the statistics of $A$:

\begin{proposition}[Statistics of $A$] The statistics $(a(A),b(A),c(A),d(A),e(A),f(A))$ of $A$ must be one of the following three tuples:
\begin{itemize}
\item (Case 1) $(6,40,79,0,0)$;
\item (Case 2) $(7,40,78,0,0)$;
\item (Case 3) $(8,39,78,0,0)$.
\end{itemize}
\end{proposition}

\begin{proof}
Since $d(A)=e(A)=f(A)=0$, we have
$$ c(2****)=d(2****)=e(2****)=0.$$
On the other hand, from \eqref{alpha-1} we have $4a(2****)+b(2****) \leq 64$ as well as the trivial inequality $b(2****) \leq 24$, and also we have
$$ |A \cap 2****| = 125 - |A \cap 1****| - |A \cap 3****| \geq 125 - 43 - 43 = 39.$$
Putting all this together, we see that the only possible statistics for $2****$ are $(8,32,0,0,0)$, $(7,32,0,0,0)$, or $(8,31,0,0,0)$. In particular, $7 \leq a(2****) \leq 8$ and $31 \leq b(2****) \leq 32$, and similarly for permutations. Applying Lemma \ref{dci} we conclude that
$$ 35 \leq b(A) \leq 40$$
and
$$ 77.5 \leq c(A) \leq 80.$$
Combining this with the first part of Corollary \ref{cmax} we conclude that $c(A)$ is either $78$ or $79$. From this and \eqref{amix} we see that the only cases that remain to be eliminated are $(7,39,79,0,0)$ and $(8,38,79,0,0)$, but these cases are incompatible with the second part of Corollary \ref{cmax}.
\end{proof}

We now eliminate each of the three remaining cases in turn.

\subsection{Elimination of $(6,40,79,0,0)$}

Here $A \cap S_{5,5}$ has six points. By Lemma \ref{tech}, there are at least two pairs in this set which differ in two positions. Their midpoints are eliminated from $A \cap S_{3,5}$. But $A$ omits exactly one point from $S_{3,5}$, so these midpoints must be the same. By symmetry, we may then assume that these two pairs are $(11111,11133)$ and $(11113,11131)$. Thus the eliminated point in $S_{3,5}$ is $11122$, i.e. $A$ contains $S_{3,5} \backslash \{11122\}$. Also, $A$ contains $\{11111,11133,11113,11131\}$ and thus must omit $\{11121, 11123, 11112, 11132\}$.

Since $11322 \in A$, at most one of $11312, 11332$ lie in $A$. By symmetry we may assume $11312 \not \in A$, thus there is a pair $(xy1z2, xy3z2)$ with $x,y,z = 1,3$ that is totally omitted from $A$, namely $(11112,11312)$. On the other hand, every other pair of this form can have at most one point in the $A$, thus there are at most seven points in $A$ of the form $xyzw2$ with $x,y,z,w = 1,3$. Similarly there are at most 8 points of the form $xyz2w$, or of $xy2zw$, $x2yzw$, $2xyzw$, leading to $b(A) \leq 7+8+8+8+8=39$, contradicting the statistic $b(A)=40$.

\subsection{Elimination of $(7,40,78,0,0)$}

Here $A \cap S_{5,5}$ has seven points. By Lemma \ref{tech}, there are at least three pairs in this set which differ in two positions. As we can only eliminate two points from $S_{3,5}$, two of the midpoints of these pairs must be the same; thus, as in the previous section, we may assume that $A$ contains $\{11111,11133,11113,11131\}$ and omits $\{11121, 11123, 11112, 11132\}$ and $11122$.

Now consider the $160$ lines $\ell$ connecting two points in $S_{4,5}$ to one point in $S_{3,5}$ (i.e. $*2xyz$ and permutations, where $x,y,z=1,3$). By double counting, the total sum of $|\ell \cap A|$ over all $160$ lines is $4b(A)+2c(A) = 316 = 158 \times 2$. On the other hand, each of these lines contain at most two points in $A$, but two of them (namely $1112*$ and $1112*$) contain no points. Thus we must have $|\ell \cap A|=2$ for the remaining $158$ lines $\ell$.

Since $A$ omits $1112x$ and $111x2$ for $x=1,3$, we thus conclude (by considering the lines $11*2x$ and $11*x2$) that $A$ must contain $1132x$, $113x2$, $1312x$, and $131x2$. Taking midpoints, we conclude that $A$ omits $11322$ and $13122$. But together with $11122$ this implies that at least three points are missing from $A \cap S_{3,5}$, contradicting the hypothesis $c(A)=78$.

\subsection{Elimination of $(8,39,78,0,0)$}

Now $A \cap S_{5,5}$ has eight points. By Lemma \ref{tech}, there are at least three pairs in this set which differ in two positions. As we can only eliminate two points from $S_{3,5}$, two of these pairs $(a,b), (c,d)$ must have the same midpoint $p$, and two other pairs $(a',b'), (c',d')$ must have the same midpoint $p'$, and $A$ contains $S_{3,5} \backslash \{p,p'\}$. As $p,p'$ are distinct, the plane containing $a,b,c,d$ is distinct from the plane containing $a',b',c',d'$.

Again consider the $160$ lines $\ell$ from the previous section. This time, the sum of the $|\ell \cap A|$ is $4b(A)+2c(A) = 312 = 156 \times 2$.
But the two lines in the plane of $a,b,c,d$ passing through $p$, and the two lines in the plane of $a',b',c',d'$ passing through $p'$, have no points; thus we must have $|\ell \cap A|=2$ for the remaining $156$ lines $\ell$.

Without loss of generality we have $(a,b)=(11111,11133)$, $(c,d) = (11113,11131)$, thus $p = 11122$. By permuting the first three indices, we may assume that $p'$ is not of the form $x2y2z, x2yz2, xy22z, xy2z2$ for any $x,y,z=1,3$. Then we have $1112x \not \in A$ and $1122x \in A$ for every $x=1,3$, so by the preceding paragraph we have $1132x \in A$; similarly for $113x2, 1312x, 131x2$. Taking midpoints, this implies that $13122, 11322 \not \in A$, but this (together with 11122) shows that at least three points aremissing from $A \cap S_{3,5}$, contradicting the hypothesis $c(A)=78$.

\subsection{Six dimensions}

Now we establish the bound $c'_{6,3}=353$. In view of the lower bounds, it suffices to show that there does not exist a Moser set $A \subset [3]^5$ with $|A|=354$.

We argue by contradiction. Let $A$ be as above, and let $(a(A),\ldots,g(A))$ be the statistics of $A$.

\begin{lemma}\label{g6} $g(A)=0$.
\end{lemma}

\begin{proof} For any four-dimensional slice $V$ of $A$, define
$$S(V) := 15 a(V) + 5 b(V) + 5 c(V)/2 + 3d(V)/2 + e(V).$$
From Lemma \ref{dci} we see that $|A|$ is equal to $a(A)+b(A)$ plus the average of $S(V)$ where $V$ ranges over the twenty slices which are some permutation of the center slice $22****$.

If $g(A)=1$, then $a(A) \leq 32$ and $b(A) \leq 96$ by \eqref{alpha-1}. Meanwhile, $e(V)=g(A)=1$ for every center slice $V$, so from Lemma \ref{paretop-4}, one can show that $S(V) \leq 223.5$ for every such slice. We conclude that $|A| \leq 351.5$, a contradiction.
\end{proof}

For any four-dimensional slice $V$ of $A$, define the \emph{defects} to be
$$ D(V) := 356 - [4a(V)+6b(V)+10c(V)+20d(V)+60e(V)].$$
Define a \emph{corner slice} to be one of the permutations or reflections of $11****$, thus there are $60$ corner slices. From Lemma \ref{dci} we see that $356-|A|+f(A)=2+f(A)$ is the average of the defects of all the $60$ corner slices. On the other hand, from Lemma \ref{paretop-4} and a straightforward computation, one concludes

\begin{lemma}\label{defects} Let $A$ be a four-dimensional Moser set. Then $D(A) \geq 0$. Furthermore:
\begin{itemize}
\item If $A$ has statistics $(6,12,18,4,0)$, then $D(A)=0$.
\item If $A$ has statistics $(5,12,18,4,0)$, $(5,12,12,4,1)$, or $(6,8,12,8,0)$, then $D(A)=4$.
\item For all other $A$, $D(A) \geq 6$.
\item If $a(A) = 4$, then $D(A) \geq 8$.
\item If $a(A) \geq 7$, then $D(A) \geq 16$.
\item If $a(A) \geq 8$, then $D(A) \geq 30$.
\item If $a(A) \geq 9$, then $D(A) \geq 86$.
\end{itemize}
\end{lemma}

Define a \emph{family} to be a set of four parallel corner slices, thus there are $15$ families, which are all a permutation of $\{11****, 13****, 31****, 33**** \}$. We refer to the family $\{11****, 13****, 31****, 33**** \}$ as $ab****$, and similarly define the family $a*b***$, etc.

\begin{lemma}\label{f6} $f(A)=0$.
\end{lemma}

\begin{proof} For any four-dimensional slice $V$ of $A$, define
$$ s(V) := 12 a(V)+15 b(V)/2+20 c(V)/3+15 d(V)/2 + 12 e(V),$$
and define an \emph{edge slice} to be one of the $30$ permutations or reflections of $12****$. From double counting we see that $|A|-a(A)$ is equal to the average of the $30$ values of $s(V)$ as $V$ ranges over edge slices.

From Lemma \ref{paretop-4} one can verify that $s(V) \leq 336$, and that $s(V) \leq 296 = 336-40$ if $e(V)=1$. The number of edge slices $V$ for which $e(V)=1$ is equal to $5f(A)$, and so the average value of the $s(V)$ is at most $336 - \frac{40 \times 5}{30} f(A)$, and so
$$ |A| - a(A) \leq 336 - \frac{40 \times 5}{30} f(A)$$
which we can rearrange (using $|A|=354$) as
$$ a(A) \geq 18 + \frac{20}{3} f(A).$$
Suppose first that $f(A)=1$; then $a(A) \geq 25$. This means that in any given family, one of the four corner slices has an $a$ value of at least $7$, and thus by Lemma \ref{defects} has a defect of at least $16$. Thus the average defect is at least $4$; on the other hand, the average defect is $2+f(A)=3$, a contradiction.

Now suppose that $f(A) \geq 2$; then $a(A) \geq 32$. Then in any given family, there is a corner slice with an $a$ value at least $9$, or four slices with $a$ value at least $8$, leading to a total defect of at least $86$ by Lemma \ref{defects}. Thus the average defect is at least $21.5$; on the other hand, the average defect is $2+f(A) \leq 2+12$, a contradiction.
\end{proof}

As a consequence of the above lemma, we see that the average defect of all corner slices is $2$, or equivalently that the total defect of these slices is $120$.

Call a corner slice \emph{good} if it has statistics $(6,12,18,4,0)$, and \emph{bad} otherwise. Thus good slices have zero defect, and bad slices have defect at least four. Since the average defect of the $60$ corner slices is $2$, there are at least $30$ good slices.

One can describe the structure of the good slices completely:

\begin{lemma}\label{sixt} The subset of $[3]^4$ consisting of the strings $1111, 1113, 3333, 1332, 1322, 1222, 3322$ and permutations is a Moser set with statistics $(6,12,18,4,0)$. Conversely, every Moser set with statistics $(6,12,18,4,0)$ is of this form up to the symmetries of the cube $[3]^4$.
\end{lemma}

\begin{proof} This can be verified by computer ({\bf need details}). A computer-free proof can be found at
\centerline{{\tt http://michaelnielsen.org/polymath1/index.php?title=Classification\_of\_\%286\%2C12\%2C18\%2C4\%2C0\%29\_sets}.}
\end{proof}

As a consequence of this lemma, given any $x,y,z,w \in \{1,3\}$, there is a unique good Moser set in $[3]^4$ set whose intersection with $S_{1,4}$ is $\{x222, 2y22, 22z2, 222w\}$, and these are the only 16 possibilities. Call this set the \emph{good set of type $xyzw$}. It consists of
\begin{itemize}
\item The four points $x222, 2y22, 22z2, 222w$ in $S_{1,4}$;
\item All $24$ elements of $S_{2,4}$ except for $xy22, x2z2, x22w, 2yz2, 2y2w, 22zw$;
\item The twelve points $xYZ2$, $xY2W$, $x2ZW$, $XyZ2$, $Xy2W$, $2yZW$, $XYz2$, $X2zW$, $2YzW$, $XY2w$, $X2Zw$, $2YZw$ in $S_{3,4}$, where $X=4-x$, $Y=4-y$, $Z=4-z$, $W=4-w$;
\item The six points $xyzw, xyzW, xyZw, xYzw, Xyzw, XYZW$ in $S_{4,4}$.
\end{itemize}

We can use this to constrain the types of two intersecting good slices:

\begin{lemma}\label{pqs} Suppose that the $pq****$ slice is of type $xyzw$, and the $p*r***$ slice is of type $x'y'z'w'$, where $p,q,r,x,y,z,w,x',y',z',w'$ are in $\{1,3\}$. Then $x'=x$ iff $q=r$, and $y'z'w'$ is equal to either $yzw$ or $YZW$. If $x=r$ (or equivalently if $x'=q$), then $y'z'w'=yzw$.
\end{lemma}

\begin{proof} By reflection symmetry we can take $p=q=r=1$. Observe that the $11****$ slice contains $111222$ iff $x=1$, and the $1*1***$ slice similarly contains $111222$ iff $x'=1$. This shows that $x=x'$.

Suppose now that $x=x'=1$. Then the $111***$ slice contains the three elements $111y22, 1112z2, 11122w$, and excludes $111Y22, 1112Z2, 11122W$, and similarly with the primes, which forces $yzw=y'z'w'$ as claimed.

Now suppose that $x=x'=3$. Then the $111***$ slice contains the two elements $111yzw, 111YZW$, but does not contain any of the other six points in $S_{6,6} \cap 111***$, and similarly for the primes. Thus $y'z'w'$ is equal to either $yzw$ or $YZW$ as claimed.
\end{proof}

Now we look at two adjacent parallel good slices, such as $11****$ and $13****$. The following lemma asserts that such slices either have opposite type, or else will create a huge amount of defect in other slices:

\begin{lemma}\label{l18} Suppose that the $11****$ and $13****$ slices are good with types $xyzw$ and $x'y'z'w'$ respectively. If $x=x'$, then the $1*x***$ slice has defect at least $30$, and the $1*X***$ slice has defect at least $8$. Also, the $1**1**$, $1**3**$, $1***1*$, $1***3*$, $1****1$, $1****3$ slices have defect at least $6$. In particular, the total defect of slices beginning with $1*$ is at least $74$.
\end{lemma}

\begin{proof} Observe from the $11****, 13****$ hypotheses that $a(1*x***)=9$ and $a(1*X***)=4$, which gives the first two claims by Lemma \ref{defects}. For the other claims, one sees from Lemma \ref{pqs} that the other six slices cannot be good; also, they have an $a$-value of $6$ and a $d$-value of at most $7$, and the claims then follow from Lemma \ref{defects}.
\end{proof}

Now we look at two diagonally opposite parallel good slices, such as $11****$ and $33****$.

\begin{lemma}\label{l14} The $11****$ and $33****$ slices cannot both be good and of the same type.
\end{lemma}

\begin{proof} Suppose not. By symmetry we may assume that $11****$ and $33****$ are of type $1111$. This excludes a lot of points from $22****$. Indeed, by connecting lines between the $11****$ and $33****$ slices, we see that the only points that can still survive in $22****$ are $221133, 221333, 221132, 223332$, and permutations of the last four indices. Double counting the lines $22133*$ and permutations we see that there are at most $12$ points one can place in the permutations of $221133, 221333, 221132$, and so the $22****$ slice has at most $16$ points. Meanwhile, the two five-dimensional slices $1*****, 3*****$ have at most $c'_{5,3} = 124$ points, and the other two four-dimensional slices $21****, 23****$ have at most $c'_{4,3} = 43$ points, leading to at most $16 + 124 * 2 + 43 * 2 = 350$ points in all, a contradiction.
\end{proof}

\begin{lemma}\label{l19} It is not possible for all four slices in a family to be good.
\end{lemma}

\begin{proof} Suppose not. By symmetry we may assume that $11****, 13****, 31****, 33****$ are good. By Lemma \ref{l14}, the $11****$ and $33****$ slices cannot be of the same type, and so they cannot both be of the opposite type to either $13****$ or $31****$. If $13****$ is not of the opposite type to $11****$, then by (a permutation of) Lemma \ref{l18}, the total defect of slices beginning with $1*$ is at least $74$; otherwise, if $13****$ is not of the opposite type to $33****$, then by (a permutation and reflection of) Lemma \ref{l18}, the total defect of slices beginning with $*3$ is at least $74$. Similarly, the total defect of slices beginning with $3*$ or $*1$ is at least $74$, leading to a total defect of at least $148$. But the total defect of all the corner slices is $2 \times 60 = 120$, a contradiction.
\end{proof}

\begin{corollary}\label{l20} At most one family can have a total defect of at least $38$.
\end{corollary}

\begin{proof} Suppose there are two families with defect at least $38$. The remaining thirteen family have defect at least $4$ by Lemma \ref{l19} and Lemma \ref{defects}, leading to a total defect of at least $38*2+13*4=128$. But the total defect is $2 \times 60 = 120$, a contradiction.
\end{proof}

Actually we can refine this:

\begin{proposition} No family can have a total defect of at least $38$.
\end{proposition}

\begin{proof} Suppose for contradiction that the $ab****$ family (say) had a total defect of at least $38$, then by Corollary \ref{l20} no other families have total defect at least $38$.

We claim that the $**ab**$ family can have at most two good slices. Indeed, suppose the $**ab**$ has three good slices, say $**11**, **13**, **33**$. By Lemma \ref{l14}, the $**11**$ and $**33**$ slices cannot be of the same type, and so cannot both be of opposite type to $**13**$. Suppose $**11**$ and $**13**$ are not of opposite type. Then by (a permutation of) Lemma \ref{l18}, one of the families $a*b***, *ab***, **b*a*, **b**a$ has a net defect of at least $38$, contradicting the normalisation.

Thus each of the six families $**ab**, **a*b*, **a**b, ***ab*, ***a*b$ have at least two bad slices. Meanwhile, the eight families $a*b***, a**b**, a***b*, a****b, *ab***, *a*b**, *a**b*, *a***b$ have at least one bad slice by Corollary \ref{l19}, leading to twenty bad slices in addition to the defect of at least $38$ arising from the $ab****$ slice. To add up to a total defect of $120$, we conclude from Lemma \ref{defects} that all bad slices outside of the $ab****$ family have a defect of four, with at most one exception; but then by Lemma \ref{l18} this shows that (for instance) the $1*1***$ and $1*3***$ slices cannot be good unless they are of opposite type. The previous argument then shows that the a*b*** slice cannot have three good slices, which increases the number of bad slices outside of $ab****$ to at least twenty-one, and now there is no way to add up to $120$, a contradiction.
\end{proof}

\begin{corollary} Every family can have at most two good slices.
\end{corollary}

\begin{proof} If for instance $11****, 13****, 33****$ are both good, then by Lemma \ref{l14} at least one of $11****, 33****$ is not of the opposite type to $13****$, which by Lemma \ref{l18} implies that there is a family with a total defect of at least $38$, contradicting the previous proposition.
\end{proof}

From this corollary and Lemma \ref{defects}, we see that every family has a defect of at least $8$. Since there are $15$ families, and $8 \times 15$ is exactly equal to $120$, we conclude

\begin{corollary}\label{coda} Every family has \emph{exactly} two good slices, and the remaining two slices have defect $4$. In particular, by Lemma \ref{defects}, the bad slices must have statistics $(5,12,18,4,0)$, $(5,12,12,4,1)$, or $(6,8,12,8,0)$.
\end{corollary}

We now limit how these slices can interact with good slices.

\begin{lemma}\label{goodgood} Suppose that $1*1***$ is a good slice.
\begin{itemize}
\item[(i)] The $11****$ slice cannot have statistics $(6,8,12,8,0)$.
\item[(ii)] The $11****$ slice cannot have statistics $(5,12,12,4,1)$.
\item[(iii)] If the $11****$ slice has statistics $(5,12,18,4,0)$, then the $112***$ slice has statistics $(3,9,3,0)$.
\end{itemize}
\end{lemma}

\begin{proof} This can be verified through computer search; there are $16$ possible configurations for the good slices, and one can calculate that there are $2750$ configurations for the $(5,12,12,4,1)$ slices, $4368$ configurations for the $(5,12,18,4,0)$ slices, and $10000$ configurations for the $(6,8,12,8,0)$ slices. {\bf we could put human proofs for all this somewhere, presumably.}

%We first prove (i). Suppose for contradiction that the $11****$ slice has statistics $(6,8,12,8,0)$, then $A$ contains $111222$, and so the $1*1***$ slice is of type $1xyz$ for some $x,y,z$. By symmetry we may assume it is of type $1111$, thus the $111***$ slice consists of
%$111111, 111113, 111332, 111322, 111222$ and permutations of the last three indices. On the other hand, the $11****$ slice has all eight of the points in $11**** \cap S_{2,6}$. Drawing lines between these points and $111111, 111113$ and permutations, we see that the $113***$ slice cannot contain $113111, 113113, 113133$, or permutations, leaving $113333$ as the only possible element of $113*** \cap S_{6,6}$. This makes $a(11****)=5$, a contradiction.
\end{proof}

\begin{corollary}\label{slic} The $111***$ slice has statistics $(4,3,3,1)$, $(2,6,6,0)$, $(3,3,3,1)$, or $(1,6,6,0)$.
\end{corollary}

\begin{proof} From Corollary \ref{coda}, we know that at least one of the slices $13****, 31****, 11****$ are good. If $11****$ or $1*1***$ is good, then the slice $111***$ has statistics $(4,3,3,1)$ or $(2,6,6,0)$, by Lemma \ref{sixt}. By symmetry we may thus reduce to the case where $13****$ is good and $1*1***$ is bad. Then by Lemma \ref{goodgood}, the $1*1***$ slice has statistics $(5,12,18,4,0)$ and the $121***$ slice has statistics $(3,9,3,0)$. Since the $131***$ slice, as a side slice of the good $13****$ slice, has statistics $(4,3,3,1)$ or $(2,6,6,0)$, we conclude that the $111***$ slice has statistics $(1,6,6,0)$ or $(3,3,3,1)$, and the claim follows.
\end{proof}

\begin{corollary}\label{slic2} All corner slices have statistics $(6,12,18,4,0)$ or $(5,12,18,4,0)$.
\end{corollary}

\begin{proof} Suppose first that a corner slice, say $11****$ has statistic $(6,8,12,8,0)$. Then $111***$ and $113***$ contain one ``d'' point each, and have six ``a'' points between them, so by Corollary \ref{slic}, they both have statistic $(3,3,3,1)$. This forces the $1*1***$, $1*3***$ slices to be bad, which by Corollary \ref{coda} forces the $3*1***,3*3***$ slices to be good. This forces the $311***, 313***$ slices to have statistics either $(2,6,6,0)$ or $(4,3,3,1)$. But the $311***$ slice (say) cannot have statistic $(4,3,3,1)$, since when combined with the $(3,3,3,1)$ statistics of $111***$ would give $a(*11***)=7$, which contradicts Corollary \ref{coda}; thus the $311***$ slice has statistic $(2,6,6,0)$, and similarly for $331***$. But then $a(3*1***)=4$, which again contradicts Corollary \ref{coda}.

Thus no corner slice has statistic $(6,8,12,8,0)$. Now suppose that a corner slice, say $11****$ has statistic $(5,12,12,4,1)$. By Lemma \ref{goodgood}, the $1*1***, 1*3***$ slices are bad, so by repeating the preceding arguments we conclude that the $311***, 313***$ slices have statistics $(2,6,6,0)$ or $(4,3,3,1)$; in particular, their $a$-value is even. However, the $*11***$ and $*13***$ slices are bad by Lemma \ref{goodgood}, and thus have an $a$-value of $5$; thus the $111***$ and $113***$ slices have an odd $a$-value. Thus forces $a(11****)$ to be even; but it is equal to $5$, a contradiction.
\end{proof}

From this and Lemma \ref{dci}, we see that $A$ has statistics $(22,72,180,80,0,0,0)$. In particular, we have $2\alpha_2(A)+\alpha_3(A) = 2$, which by double counting (cf. \eqref{alpha-1}) shows that for every line of the form $12222*$ (or a reflection or permutation thereof) intersects $A$ in exactly two points. Note that such lines connect a ``$d$'' point to two ``$c$'' points.

Also, we observe that two adjacent ``$d$'' points, such as $111222$ and $113222$, cannot both lie in $A$; for this would force the $*13***$ and $*11***$ slices to have statistics $(4,3,3,1)$ or $(3,3,3,1)$ by Corollary \ref{slic}, which forces $a(*1****)=6$, and thus $*1****$ must be good by Corollary \ref{slic2}; but this contradicts Lemma \ref{sixt}. Since $\alpha_3(A)=1/2$, we conclude that given any two adjacent ``$d$'' points, exactly one of them lies in $A$. In particular, the d points of the form $***222$ consist either of those strings with an even number of $1$s, or those with an odd number of $1$s.

Let's say it's the former, thus the set contains $111222, 133222$, and permutations of the first three coordinates, but omits $113222, 333222$ and permutations of the first three coordinates. Since the ``$d$'' points $113222, 333222$ are omitted, we conclude that the ``$c$'' points $113122, 113322, 333122, 333322$ must lie in the set, and similarly for permutations of the first three and last three coordinates. But this gives at least $15$ of the $16$ ``$c$'' points ending in $22$; by symmetry this leads to $225$ $c$-points in all; but $c(A)=180$, contradiction. This (finally!) completes the proof that $c'_{6,3}=353$.

Moser.tex

2009-07-14T06:44:13Z

Mpeake:

\section{Upper bounds for the $k=3$ Moser problem in small dimensions}\label{moser-upper-sec}

In this section we finish the proof of Theorem \ref{moser} by obtaining the upper bounds on $c'_{n,3}$ for $n \leq 6$.

\subsection{Statistics, densities and slices}

Our analysis will revolve around various \emph{statistics} of Moser sets $A \subset [3]^n$, their associated \emph{densities}, and the behavior of such statistics and densities with respect to the operation of passing from the cube $[3]^n$ to various \emph{slices} of that cube.

\begin{definition}[Statistics and densities] Let $A \subset [3]^n$ be a set. For any $0 \leq i \leq n$, set $a_i(A) := |A \cap S_{n-i,n}|$; thus we have
$$ 0 \leq a_i(A) \leq |S_{n-i,n}| = \binom{n}{i} 2^{n-i}$$
for $0 \leq i \leq n$ and
$$ a_0(A) + \ldots + a_n(A) = |A|.$$
We refer to the vector $(a_0(A),\ldots,a_n(A))$ as the \emph{statistics} of $A$. We define the $i^{th}$ \emph{density} $\alpha_i(A)$ to be the quantity
$$ \alpha_i(A) := \frac{a_i(A) }{\binom{n}{i} 2^{n-i}},$$
thus $0 \leq \alpha_i(A) \leq 1$ and
$$ |A| = \sum_{i=0}^n \binom{n}{i} 2^{n-i} a_i(A).$$
\end{definition}

\begin{example}\label{2mos} Let $n=2$ and $A$ be the Moser set $A := \{ 12, 13, 21, 23, 31, 32 \}$. Then the statistics $(a_0(A), a_1(A), a_2(A))$ of $A$ are $(2,4,0)$, and the densities $(\alpha_0(A), \alpha_1(A), \alpha_2(A))$ are $(\frac{1}{2}, 1, 0)$. \textbf{Include picture here? with colours?}
\end{example}

When working with small values of $n$, it will be convenient to write $a(A)$, $b(A)$, $c(A)$, etc. for $a_0(A)$, $a_1(A)$, $a_2(A)$, etc., and similarly write $\alpha(A), \beta(A), \gamma(A)$, etc. for $\alpha_0(A)$, $\alpha_1(A)$, $\alpha_2(A)$, etc. Thus for instance in Example \ref{2mos} we have $b(A) = 4$ and $\alpha(A) = \frac{1}{2}$.

\begin{definition}[Subspace statistics and densities] If $V$ is a $k$-dimensional geometric subspace of $[3]^n$, then we have a map $\phi_V: [3]^k \to [3]^n$ from the $k$-dimensional cube to the $n$-dimensional cube. If $A \subset [3]^n$ is a set and $0 \leq i \leq k$, we write $a_i(V,A)$ for $a_i(\phi_V^{-1}(A))$ and $\alpha_i(V,A)$ for $\alpha_i(\phi_V^{-1}(A))$. If the set $A$ is clear from context, we abbreviate $a_i(V,A)$ as $a_i(V)$ and $\alpha_i(V,A)$ as $\alpha_i(V)$.
\end{definition}

For our problem, a particularly important type of subspace of $[3]^n$ will be the \emph{slices} formed by fixing one coordinate and letting the other $n-1$ coordinates vary. We will denote this by a single string in which the $n-1$ varying coordinates are denoted by asterisks. For instance, in $[3]^2$, $1*$ denotes the slice $1*=\{11,12,13\}$, $*2$ denotes the slice $*2=\{12,22,32\}$, etc.; similarly, in $[3]^3$, $1**$ is the slice $\{111, 112, 113, 121, 122, 123, 131, 132, 133\}$, etc. We call a slice a \emph{centre slice} if the fixed coordinate is $2$ and a \emph{side slice} if it is $1$ or $3$.

\begin{example} We continue Example \ref{2mos}. Then the statistics of the side slice $1*$ are $(a(1*),b(1*)) = (1,1)$, while the statistics of the centre slice $2*$ are $(a(2*),b(2*))=(2,0)$. The corresponding densities are $(\alpha(1*),\beta(1*)) = (1/2,1)$ and $(\alpha(2*),\beta(2*))=(1,0)$.
\end{example}

A simple double counting argument gives the following useful identity:

\begin{lemma}[Double counting identity]\label{dci} Let $A \subset [3]^n$ and $0 \leq i \leq n-1$. Then we have
$$ \frac{1}{n-i-1} \sum_{V \hbox{ a side slice}} a_{i+1}(V) = \frac{1}{i+1} \sum_{W \hbox{ a centre slice}} a_i(W) = a_{i+1}(A)$$
where $V$ ranges over the $2n$ side slices of $[3]^n$, and $W$ ranges over the $n$ centre slices. In other words, the average value of $\alpha_{i+1}(V)$ for side slices $V$ equals the average value of $\alpha_i(W)$ for centre slices $W$, which is in turn equal to $\alpha_{i+1}(A)$.
\end{lemma}

Indeed, this lemma follows from the observation that every string in $A \cap S_{n-i-1,n}$ belongs to $i+1$ centre slices $W$ (and contributes to $a_i(W)$) and to $n-i-1$ side slices $V$ (and contributes to $a_{i+1}(V)$). One can also view this lemma probabilistically, as the assertion that there are three equivalent ways to generate a random string of length $n$:

\begin{itemize}
\item Pick a side slice $V$ at random, and randomly fill in the wildcards in such a way that $i+1$ of the wildcards are $2$'s (i.e. using an element of $S_{n-i-2,n-1}$).
\item Pick a centre slice $V$ at random, and randomly fill in the wildcards in such a way that $i$ of the wildcards are $2$'s (i.e. using an element of $S_{n-i-1,n-1}$).
\item Randomly choose an element of $S_{n-i-1,n}$.
\end{itemize}

\begin{example} We continue Example \ref{2mos}. The average value of $\beta$ for side slices is equal to the average value of $\alpha$ for centre slices, which is equal to $\beta(A) = 1$.
\end{example}

Another very useful fact (essentially due to \cite{chvatal2}) is that linear inequalities for statistics of Moser sets at one dimension propagate to linear inequalities in higher dimensions:

\begin{lemma}[Propagation lemma]\label{prop} Let $n \geq 1$ be an integer. Suppose one has a linear inequality of the form
\begin{equation}\label{alphav}
\sum_{i=0}^n v_i \alpha_i(A) \leq s
\end{equation}
for all Moser sets $A \subset [3]^n$ and some real numbers $v_0,\ldots,v_n,s$. Then we also have the linear inequality
$$ \sum_{i=0}^n v_i \alpha_{qi+r}(A) \leq s$$
whenever $q \geq 1$, $r \geq 0$, $N \geq nq+r$ are integers and $A \subset [3]^N$ is a Moser set.
\end{lemma}

\begin{proof} We run a probabilistic argument (one could of course also use a double counting argument instead). Let $n,v_0,\ldots,v_n,s,q,r,N,A$ be as in the lemma. Let $V$ be a random $n$-dimensional geometric subspace of $[3]^N$, created in the following fashion:
\begin{itemize}
\item Pick $n$ wildcards $x_1,\ldots,x_n$ to run independently from $1$ to $3$. We also introduce dual wildcards $\overline{x_1},\ldots,\overline{x_n}$; each $\overline{x_j}$ will take the value $4-x_j$.
\item We randomly subdivide the $N$ coordinates into $n$ groups of $q$ coordinates, plus a remaining group of $N-nq$ ``fixed'' coordinates.
\item For each coordinate in the $j^{th}$ group of $q$ coordinates for $1 \leq j \leq n$, we randomly assign either a $x_j$ or $\overline{x_j}$.
\item For each coordinate in the $N-nq$ fixed coordinates, we randomly assign a digit $1,2,3$, but condition on the event that exactly $r$ of the digits are equal to $2$ (i.e. we use a random element of $S_{N-nq-r,N-nq}$).
\item Let $V$ be the subspace created by allowing $x_1,\ldots,x_n$ to run independently from $1$ to $3$, and $\overline{x_j}$ to take the value $4-x_j$.
\end{itemize}

For instance, if $n=2, q=2, r=1, N=6$, then a typical subspace $V$ generated in this fashion is
$$ 2x_1\overline{x_2}3x_2x_1 = \{ 213311, 212321, 211331, 223312, 222322, 221332, 233313, 232323, 231333\}.$$
Observe from that the following two ways to generate a random element of $[3]^N$ are equivalent:

\begin{itemize}
\item Pick $V$ randomly as above, and then assign $(x_1,\ldots,x_n)$ randomly from $S_{n-i,n}$. Assign $4-x_j$ to $\overline{x_j}$ for all $1 \leq j \leq n$.
\item Pick a random string in $S_{N-qi-r,N}$.
\end{itemize}

Indeed, both random variables are invariant under the symmetries of the cube, and both random variables always pick out strings in $S_{N-qi-r,N}$, and the claim follows. As a consequence, we see that the expectation of $\alpha_i(V)$ (as $V$ ranges over the recipe described above) is equal to $\alpha_{qi+r}(A)$. On the other hand, from \eqref{alphav} we have
$$ \sum_{i=0}^n v_i \alpha_i(V) \leq s$$
for all such $V$; taking expectations over $V$, we obtain the claim.
\end{proof}

In view of Lemma \ref{prop}, it is of interest to locate linear inequalities relating the densities $\alpha_i(A)$, or (equivalently) the statistics $a_i(A)$. For this, it is convenient to introduce the following notation.

\begin{definition} Let $n \geq 1$ be an integer.
\begin{itemize}
\item A vector $(a_0,\ldots,a_n)$ of non-negative integers is \emph{feasible} if it is the statistics of some Moser set $A$.
\item A feasible vector $(a_0,\ldots,a_n)$ is \emph{Pareto-optimal} if there is no other feasible vector $(b_0,\ldots,b_n) \neq (a_0,\ldots,a_n)$ such that $b_i \geq a_i$ for all $0 \leq i \leq n$.
\item A Pareto-optimal vector $(a_0,\ldots,a_n)$ is \emph{extremal} if it is not a non-trivial convex linear combination of other Pareto-optimal vectors.
\end{itemize}
\end{definition}

To establish a linear inequality of the form \eqref{alphav} with the $v_i$ non-negative, it suffices to test the inequality against densities associated to extremal vectors of statistics. (There is no point considering linear inequalities with negative coefficients $v_i$, since one always has the freedom to reduce a density $\alpha_i(A)$ of a Moser set $A$ to zero, simply by removing all elements of $A$ with exactly $i$ $2$'s.)

We will classify exactly the Pareto-optimal and extremal vectors for $n \leq 3$, which by Lemma \ref{prop} will lead to useful linear inequalities for $n \geq 4$. Using a computer, we have also located a partial list of Pareto-optimal and extremal vectors for $n=4$, which are also useful for the $n=5$ and $n=6$ theory.

\subsection{Up to three dimensions}

We now establish Theorem \ref{moser} for $n \leq 3$, and establish some auxiliary inequalities which will be of use in higher dimensions.

The case $n=0$ is trivial. When $n=1$, it is clear that $c'_{1,3} = 2$, and furthermore that the Pareto-optimal statistics are $(2,0)$ and $(1,1)$, which are both extremal. This leads to the linear inequality
$$ 2\alpha(A) + \beta(A) \leq 2$$
for all Moser sets $A \subset [3]^1$, which by Lemma \ref{prop} implies that
\begin{equation}\label{alpha-1}
2\alpha_r(A) + \alpha_{r+q}(A) \leq 2
\end{equation}
whenever $r \geq 0$, $q \geq 1$, $n \geq q+r$, and $A \subset [3]^n$ is a Moser set.

For $n=2$, we see by partitioning $[3]^2$ into three slices that $c'_{2,3} \leq 3 c'_{1,3} = 6$, and so (by the lower bounds in the previous section) $c'_{2,3} = 6$. Writing $(a,b,c) = (a(A),b(A),c(A)) = (4\alpha(A), 4\beta(A), \gamma(A))$, the inequalities \eqref{alpha-1} become
\begin{equation}\label{abc}
a + 2c \leq 4; b+2c \leq 4; 2a+b <= 8.
\end{equation}

\begin{lemma} When $n=2$, the Pareto-optimal statistics are $(4,0,0), (3,2,0), (2,4,0), (2,2,1)$. In particular, the extremal statistics are $(4,0,0), (2,4,0), (2,2,1)$.
\end{lemma}

\begin{proof} One easily checks that all the statistics listed above are feasible.
Consider the statistics $(a,b,c)$ of a Moser set $A \subset [3]^2$. $c$ is either equal to $0$ or $1$. If $c=1$, then \eqref{abc} implies that $a,b \leq 2$, so the only Pareto-optimal statistic here is $(2,2,1)$. When instead $c=0$, the inequalities \eqref{abc} can easily imply the Pareto-optimality of $(4,0,0), (3,2,0), (2,4,0)$.
\end{proof}

From this lemma we see that we obtain a new inequality $2a+b+2c \leq 8$. Converting this back to densities and using Lemma \ref{prop}, we conclude that
\begin{equation}\label{alpha-2}
4\alpha_r(A) + 2\alpha_{r+q}(A) + \alpha_{r+2q} \leq 4
\end{equation}
whenever $r \geq 0$, $q \geq 1$, $n \geq q+2r$, and $A \subset [3]^n$ is a Moser set.

One can also check by computer that there are exactly $230$ line-free subsets of $[3]^2$.

Now we look at three dimensions. Writing $(a,b,c,d)$ for the statistics of a Moser set $A \subset [3]^n$ (which thus range between $(0,0,0,0)$ and $(8,12,6,1)$), the inequalities \eqref{alpha-1} imply in particular that
\begin{equation}\label{abc-3d}
a+4d \leq 8; b+6d \leq 12; c+3d \leq 6; 3a+2c \leq 24; b+c \leq 12
\end{equation}
while \eqref{alpha-2} implies that
\begin{equation}\label{abcd-3d}
3a+b+c \leq 24; b+c+3d \leq 12.
\end{equation}
Summing the inequalities $b+c \leq 12, 3a+b+c \leq 24, b+c+3d \leq 12$ yields
$$ 3(a+b+c+d) \leq 48$$
and hence $|A| = a+b+c+d \leq 16$; comparing this with the lower bounds of the preceding section we obtain $c'_{3,3} = 16$ as required. (This argument is essentially identical to the one in \cite{chvatal2}).

We have the following useful computation:

\begin{lemma}[3D Pareto-optimals]\label{paretop} When $n=3$, the Pareto-optimal statistics are $$(3,6,3,1),(4,4,3,1),(4,6,2,1),(2,6,6,0),(3,6,5,0),(4,4,5,0),(3,7,4,0),(4,6,4,0),$$
$$ (3,9,3,0),(4,7,3,0),(5,4,3,0),(4,9,2,0),(5,6,2,0),(6,3,2,0),(3,10,1,0),(5,7,1,0),$$
$$ (6,4,1,0),(4,12,0,0),(5,9,0,0),(6,6,0,0),(7,3,0,0),(8,0,0,0).$$
In particular, the extremal statistics are
$$(3,6,3,1),(4,4,3,1),(4,6,2,1),(2,6,6,0),(4,4,5,0),(4,6,4,0),(4,12,0,0),(8,0,0,0).$$
\end{lemma}

\begin{proof} This can be established by a brute-force search over the $2^{27} \approx 1.3 \times 10^8$ different subsets of $[3]^3$. Actually, one can perform a much faster search than this. Firstly, as noted earlier, there are only $230$ line-free subsets of $[3]^2$, so one could search over $230^3 \approx 1.2 \times 10^7$ configurations instead. Secondly, by symmetry we may assume (after enumerating the $230$ sets in a suitable fashion) that the first slice $A \cap 1**$ has an index less than or equal to the third $A \cap 3**$, leading to $\binom{231}{2} \times 230 \approx 6 \times 10^6$ configurations instead. Finally, using the first and third slice one can quickly determine which elements of the second slice $2**$ are prohibited from $A$. There are $2^9 = 512$ possible choices for the prohibited set in $2**$. By crosschecking these against the list of $230$ line-free sets one can compute the Pareto-optimal statistics for the second slices inside the prohibited set (the lists of such statistics turns out to length at most $23$). Storing these statistics in a lookup table, and then running over all choices of the first and third slice (using symmetry), one now has to perform $O( 512 \times 230 ) + O( \binom{231}{2} \times 23) \approx O( 10^6 )$ computations, which is quite a feasible computation.

One could in principle reduce the computations even further, by a factor of up to $8$, by using the symmetry group $D_4$ of the square $[3]^2$ to reduce the number of cases one needs to consider, but we did not implement this.

A computer-free proof of this lemma can be found at
\centerline{{\tt http://michaelnielsen.org/polymath1/index.php?title=Human\_proof\_of\_the\_3D\_Pareto-optimal\_Moser\_statistics}.}
\end{proof}

\begin{remark} A similar computation revealed that the total number of line-free subsets of $[3]^3$ was $3813884$. With respect to the $2^3 \times 3!=48$-element group of geometric symmetries of $[3]^3$, these sets partitioned into $83158$ equivalence classes:
$$
3813884 = 76066 \times 48+6527 \times 24+51 \times 16+338 \times 12 +109 \times 8+41 \times 6+13 \times 4 +5 \times 3+3 \times 2+5 \times 1.
$$
\end{remark}

Lemma \ref{paretop} yields the following new inequalities:
\begin{align*}
2a+b+2c+4d &\leq 22 \\
3a+2b+3c+6d &\leq 36 \\
7a+2b+4c+8d &\leq 56 \\
6a+2b+3c+6d &\leq 48 \\
a+2c+4d &\leq 14 \\
5a+4c+8d &\leq 40.
\end{align*}

Applying Lemma \ref{prop}, we obtain new inequalities:
\begin{align}
8\alpha_r(A)+ 6\alpha_{r+q}(A) + 6\alpha_{r+2q}(A) + 2\alpha_{r+3q}(A) &\leq 11 \label{eleven}\\
4\alpha_r(A)+4\alpha_{r+q}(A)+3\alpha_{r+2q}(A)+\alpha_{r+3q}(A) &\leq 6\label{six}\\
7\alpha_r(A)+3\alpha_{r+q}(A)+3\alpha_{r+2q}(A)+\alpha_{r+3q}(A) &\leq 7\nonumber\\
8\alpha_r(A)+3\alpha_{r+q}(A)+3\alpha_{r+2q}(A)+\alpha_{r+3q}(A) &\leq 8\label{eight}\\
4\alpha_{r+q}(A)+2\alpha_{r+2q}(A)+\alpha_{r+3q}(A) &\leq 4\nonumber\\
4\alpha_r(A)+6\alpha_{r+2q}(A)+2\alpha_{r+3q}(A) &\leq 7\nonumber\\
5\alpha_r(A)+3\alpha_{r+2q}(A)+\alpha_{r+3q}(A) &\leq 5\nonumber
\end{align}
whenever $r \geq 0$, $q \geq 1$, $n \geq r+3q$, and Moser sets $A \subset [3]^n$.

We also note some further corollaries of Lemma \ref{paretop}:

\begin{corollary}[Statistics of large 3D Moser sets]\label{paretop2} Let $(a,b,c,d)$ be the statistics of a Moser set $A$ in $[3]^3$. Then $|A| = a+b+c+d \leq 16$. Furthermore:
\begin{itemize}
\item If $|A|=16$, then $(a,b,c,d) = (4,12,0,0)$.
\item If $|A|=15$, then $(a,b,c,d) = (4,11,0,0)$ or $(3,12,0,0)$.
\item If $|A| \geq 14$, then $b \geq 6$ and $d=0$.
\item If $|A| = 13$ and $d=1$, then $(a,b,c,d) = (4,6,2,1)$ or $(3,6,3,1)$.
\end{itemize}
\end{corollary}

\subsection{Four dimensions}

Now we establish the bound $c'_{4,3}=43$. Let $A$ be a Moser set in $[3]^4$, with attendant statistics $(a,b,c,d,e)$, which range between $(0,0,0,0,0)$ and $(16,32,24,8,1)$. In view of the lower bounds, our task here is to establish the upper bound $a+b+c+d+e \leq 43$.

The linear inequalities already established just barely fail to achieve this bound, but we can obtain the upper bound $a+b+c+d+e \leq 44$ as follows.
First suppose that $e=1$; then from the inequalities \eqref{alpha-1} (or by considering lines passing through $2222$) we see that $a \leq 8, b \leq 16, c \leq 12, d \leq 4$ and hence $a+b+c+d+e \leq 41$, so we may assume that $e=0$.

From Lemma \ref{dci}, we see that $a+b+c+d+e$ is now equal to the sum of $a(V)/4+b(V)/3+c(V)/2+d(V)$, where $V$ ranges over all side slices of $[3]^4$. But from Lemma \ref{paretop} we see that $a(V)/4+b(V)/3+c(V)/2+d(V)$ is at most $\frac{11}{4}$, with equality occuring only when $(a(V),b(V),c(V),d(V))=(2,6,6,0)$. This gives the upper bound $a+b+c+d+e \leq 44$.

The above argument shows that $a+b+c+d+e=44$ can only occur if $e=0$ and if $(a(V),b(V),c(V),d(V))=(2,6,6,0)$ for all side slices $V$. Applying Lemma \ref{paretop} again this implies $(a,b,c,d,e)=(4,16,24,0,0)$. But then $A$ contains all of the sphere $S_{2,4}$, which implies that the four-element set $A \cap S_{4,4}$ cannot contain a pair of strings which differ in exactly two positions (as their midpoint would then lie in $S_{2,4}$, contradicting the hypothesis that $A$ is a Moser set).

Recall that we may partition $S_{4,4} = S_{4,4}^e \cup S_{4,4}^o$, where
$$S_{4,4}^e := \{ 1111, 1133, 1313, 3113, 1331, 3131, 3311, 3333\}$$
is the strings in $S_{4,4}$ with an even number of $1$'s, and
$$S_{4,4}^o := \{ 1113, 1131, 1311, 3111, 1333, 3133, 3313, 3331\}$$
are the strings in $S_{4,4}$ with an odd number. Observe that any two distinct elements in $S_{4,4}^e$ differ in exactly two positions unless they are antipodal. Thus $A \cap S_{4,4}^e$ has size at most two, with equality only when $A \cap S_{4,4}^e$ consists of an antipodal pair. Similarly for $A \cap S_{4,4}^o$. Thus $A$ must consist of two antipodal pairs, one from $S_{4,4}^e$ and one from $S_{4,4}^o$.

By the symmetries of the cube we may assume without loss of generality that these pairs are $\{ 1111, 3333\}$ and $\{1113,3331\}$ respectively. But as $A$ is a Moser set, $A$ must now exclude the strings $1112$ and $3332$. These two strings form two corners of the eight-element set
$$ ***2 \cap S_{3,4} = \{ 1112, 1132, 1312, 3112, 1332, 3132, 3312, 3332 \}.$$
Any pair of points in this set which are ``adjacent'' in the sense that they differ by exactly one entry cannot both lie in $A$, as their midpoint would then lie in $S_{3,4}$, and so $A$ can contain at most four elements from this set, with equality only if $A$ contains all the points in $***2 \cap S_{3,4}$ of the same parity (either all the elements with an even number of $3$s, or all the elements with an odd number of $3$s). But because the two corners removed from this set have the opposite parity (one has an even number of $1$s and one has an odd number), we see in fact that $A$ can contain at most $3$ points from this set. Meanwhile, the same arguments give that $A$ contains at most four points from $**2* \cap S_{3,4}$, $*2** \cap S_{3,4}$, and $2*** \cap S_{3,4}$. Summing we see that $b = |A \cap S_{3,4}| \leq 3+4+4+4=15$, a contradiction. Thus we have $c'_{4,3}=43$ as claimed.

We have the following four-dimensional version of Lemma \ref{paretop}:

\begin{lemma}[4D Pareto-optimals]\label{paretop-4} When $n=4$, the Pareto-optimal statistics listed on Table \ref{table4}.
\end{lemma}

\begin{figure}[tb]
\centered{\tiny
$(3, 16, 24, 0, 0)$,
$(4, 14, 19, 2, 0)$,
$(4, 15, 24, 0, 0)$,
$(4, 16, 8, 4, 1)$,
$(4, 16, 14, 4, 0)$,
$(4, 16, 23, 0, 0)$,
$(4, 17, 21, 0, 0)$,
$(4, 18, 19, 0, 0)$,
$(5, 12, 12, 4, 1)$,
$(5, 12, 13, 6, 0)$,
$(5, 12, 15, 5, 0)$,
$(5, 12, 19, 2, 0)$,
$(5, 13, 10, 4, 1)$,
$(5, 13, 14, 5, 0)$,
$(5, 13, 21, 1, 0)$,
$(5, 15, 9, 4, 1)$,
$(5, 15, 12, 3, 1)$,
$(5, 15, 13, 5, 0)$,
$(5, 15, 18, 3, 0)$,
$(5, 15, 20, 1, 0)$,
$(5, 15, 22, 0, 0)$,
$(5, 16, 7, 4, 1)$,
$(5, 16, 10, 3, 1)$,
$(5, 16, 11, 5, 0)$,
$(5, 16, 12, 2, 1)$,
$(5, 16, 16, 3, 0)$,
$(5, 16, 19, 1, 0)$,
$(5, 16, 21, 0, 0)$,
$(5, 17, 12, 4, 0)$,
$(5, 17, 14, 3, 0)$,
$(5, 17, 16, 2, 0)$,
$(5, 17, 18, 1, 0)$,
$(5, 17, 20, 0, 0)$,
$(5, 18, 13, 3, 0)$,
$(5, 18, 14, 2, 0)$,
$(5, 20, 8, 4, 0)$,
$(5, 20, 10, 3, 0)$,
$(5, 20, 13, 2, 0)$,
$(5, 20, 14, 1, 0)$,
$(5, 20, 18, 0, 0)$,
$(5, 21, 10, 2, 0)$,
$(5, 21, 15, 0, 0)$,
$(5, 22, 13, 0, 0)$,
$(6, 8, 12, 8, 0)$,
$(6, 10, 11, 4, 1)$,
$(6, 11, 12, 7, 0)$,
$(6, 12, 10, 7, 0)$,
$(6, 12, 13, 5, 0)$,
$(6, 12, 18, 4, 0)$,
$(6, 13, 16, 4, 0)$,
$(6, 14, 9, 4, 1)$,
$(6, 14, 9, 7, 0)$,
$(6, 14, 12, 6, 0)$,
$(6, 14, 16, 3, 0)$,
$(6, 14, 19, 1, 0)$,
$(6, 14, 21, 0, 0)$,
$(6, 15, 7, 4, 1)$,
$(6, 15, 10, 3, 1)$,
$(6, 15, 10, 6, 0)$,
$(6, 15, 11, 2, 1)$,
$(6, 15, 12, 5, 0)$,
$(6, 15, 15, 4, 0)$,
$(6, 15, 20, 0, 0)$,
$(6, 16, 7, 3, 1)$,
$(6, 16, 8, 6, 0)$,
$(6, 16, 9, 2, 1)$,
$(6, 16, 10, 5, 0)$,
$(6, 16, 12, 1, 1)$,
$(6, 16, 13, 4, 0)$,
$(6, 16, 14, 3, 0)$,
$(6, 16, 18, 2, 0)$,
$(6, 16, 19, 0, 0)$,
$(6, 17, 9, 5, 0)$,
$(6, 17, 10, 4, 0)$,
$(6, 17, 13, 3, 0)$,
$(6, 17, 15, 2, 0)$,
$(6, 17, 17, 1, 0)$,
$(6, 17, 18, 0, 0)$,
$(6, 18, 13, 2, 0)$,
$(6, 18, 16, 1, 0)$,
$(6, 18, 17, 0, 0)$,
$(6, 19, 9, 4, 0)$,
$(6, 19, 12, 3, 0)$,
$(6, 19, 15, 1, 0)$,
$(6, 20, 7, 4, 0)$,
$(6, 20, 9, 3, 0)$,
$(6, 20, 12, 2, 0)$,
$(6, 20, 13, 1, 0)$,
$(6, 20, 15, 0, 0)$,
$(6, 21, 8, 3, 0)$,
$(6, 21, 9, 2, 0)$,
$(6, 21, 12, 1, 0)$,
$(6, 21, 14, 0, 0)$,
$(6, 22, 7, 3, 0)$,
$(6, 22, 8, 2, 0)$,
$(6, 22, 10, 1, 0)$,
$(6, 23, 9, 1, 0)$,
$(6, 24, 7, 2, 0)$,
$(6, 24, 8, 1, 0)$,
$(6, 24, 12, 0, 0)$,
$(6, 25, 9, 0, 0)$,
$(6, 26, 7, 0, 0)$,
$(7, 8, 6, 8, 0)$,
$(7, 11, 9, 4, 1)$,
$(7, 11, 12, 6, 0)$,
$(7, 12, 8, 4, 1)$,
$(7, 12, 8, 6, 0)$,
$(7, 12, 12, 3, 1)$,
$(7, 12, 12, 5, 0)$,
$(7, 12, 13, 4, 0)$,
$(7, 12, 15, 3, 0)$,
$(7, 12, 17, 2, 0)$,
$(7, 13, 7, 4, 1)$,
$(7, 13, 10, 3, 1)$,
$(7, 13, 11, 5, 0)$,
$(7, 13, 12, 2, 1)$,
$(7, 13, 12, 4, 0)$,
$(7, 13, 14, 3, 0)$,
$(7, 13, 16, 2, 0)$,
$(7, 14, 6, 4, 1)$,
$(7, 14, 6, 7, 0)$,
$(7, 14, 9, 5, 0)$,
$(7, 14, 10, 2, 1)$,
$(7, 14, 12, 1, 1)$,
$(7, 14, 17, 1, 0)$,
$(7, 14, 19, 0, 0)$,
$(7, 15, 7, 5, 0)$,
$(7, 15, 8, 3, 1)$,
$(7, 15, 9, 2, 1)$,
$(7, 15, 11, 1, 1)$,
$(7, 15, 11, 4, 0)$,
$(7, 15, 13, 3, 0)$,
$(7, 15, 16, 1, 0)$,
$(7, 16, 6, 3, 1)$,
$(7, 16, 6, 6, 0)$,
$(7, 16, 8, 2, 1)$,
$(7, 16, 10, 1, 1)$,
$(7, 16, 10, 4, 0)$,
$(7, 16, 12, 0, 1)$,
$(7, 16, 12, 3, 0)$,
$(7, 16, 15, 2, 0)$,
$(7, 16, 17, 0, 0)$,
$(7, 17, 6, 5, 0)$,
$(7, 17, 7, 4, 0)$,
$(7, 17, 11, 3, 0)$,
$(7, 17, 13, 2, 0)$,
$(7, 17, 14, 1, 0)$,
$(7, 17, 16, 0, 0)$,
$(7, 18, 10, 3, 0)$,
$(7, 18, 13, 1, 0)$,
$(7, 18, 15, 0, 0)$,
$(7, 19, 9, 3, 0)$,
$(7, 20, 6, 4, 0)$,
$(7, 20, 11, 2, 0)$,
$(7, 20, 12, 1, 0)$,
$(7, 20, 14, 0, 0)$,
$(7, 21, 8, 2, 0)$,
$(7, 21, 10, 1, 0)$,
$(7, 21, 12, 0, 0)$,
$(7, 22, 9, 1, 0)$,
$(7, 22, 11, 0, 0)$,
$(7, 23, 6, 3, 0)$,
$(7, 23, 7, 1, 0)$,
$(7, 23, 10, 0, 0)$,
$(7, 24, 6, 2, 0)$,
$(7, 24, 9, 0, 0)$,
$(7, 25, 6, 1, 0)$,
$(7, 25, 8, 0, 0)$,
$(7, 26, 3, 1, 0)$,
$(7, 28, 6, 0, 0)$,
$(7, 29, 3, 0, 0)$,
$(7, 30, 1, 0, 0)$,
$(8, 8, 0, 8, 0)$,
$(8, 8, 9, 7, 0)$,
$(8, 8, 12, 6, 0)$,
$(8, 9, 9, 4, 1)$,
$(8, 9, 10, 6, 0)$,
$(8, 9, 12, 3, 1)$,
$(8, 9, 12, 5, 0)$,
$(8, 9, 13, 4, 0)$,
$(8, 9, 15, 3, 0)$,
$(8, 10, 7, 4, 1)$,
$(8, 10, 10, 3, 1)$,
$(8, 10, 10, 5, 0)$,
$(8, 10, 12, 2, 1)$,
$(8, 10, 12, 4, 0)$,
$(8, 10, 13, 3, 0)$,
$(8, 10, 15, 2, 0)$,
$(8, 11, 6, 4, 1)$,
$(8, 11, 9, 6, 0)$,
$(8, 11, 10, 2, 1)$,
$(8, 11, 11, 4, 0)$,
$(8, 12, 7, 6, 0)$,
$(8, 12, 9, 3, 1)$,
$(8, 12, 9, 5, 0)$,
$(8, 12, 10, 4, 0)$,
$(8, 12, 12, 1, 1)$,
$(8, 12, 14, 2, 0)$,
$(8, 12, 16, 1, 0)$,
$(8, 12, 18, 0, 0)$,
$(8, 13, 7, 3, 1)$,
$(8, 13, 7, 5, 0)$,
$(8, 13, 9, 2, 1)$,
$(8, 13, 12, 0, 1)$,
$(8, 13, 12, 3, 0)$,
$(8, 14, 0, 7, 0)$,
$(8, 14, 6, 6, 0)$,
$(8, 14, 7, 2, 1)$,
$(8, 14, 8, 1, 1)$,
$(8, 14, 9, 4, 0)$,
$(8, 14, 11, 0, 1)$,
$(8, 14, 11, 3, 0)$,
$(8, 14, 13, 2, 0)$,
$(8, 14, 15, 1, 0)$,
$(8, 14, 17, 0, 0)$,
$(8, 15, 6, 3, 1)$,
$(8, 15, 6, 5, 0)$,
$(8, 15, 7, 1, 1)$,
$(8, 16, 0, 6, 0)$,
$(8, 16, 4, 3, 1)$,
$(8, 16, 4, 5, 0)$,
$(8, 16, 6, 2, 1)$,
$(8, 16, 8, 4, 0)$,
$(8, 16, 9, 0, 1)$,
$(8, 16, 10, 3, 0)$,
$(8, 16, 12, 2, 0)$,
$(8, 16, 14, 1, 0)$,
$(8, 16, 16, 0, 0)$,
$(8, 17, 0, 5, 0)$,
$(8, 17, 3, 4, 0)$,
$(8, 17, 8, 3, 0)$,
$(8, 17, 10, 2, 0)$,
$(8, 17, 12, 1, 0)$,
$(8, 17, 14, 0, 0)$,
$(8, 18, 9, 2, 0)$,
$(8, 18, 11, 1, 0)$,
$(8, 18, 12, 0, 0)$,
$(8, 19, 6, 3, 0)$,
$(8, 19, 8, 2, 0)$,
$(8, 20, 0, 4, 0)$,
$(8, 20, 4, 3, 0)$,
$(8, 20, 7, 2, 0)$,
$(8, 20, 9, 1, 0)$,
$(8, 20, 11, 0, 0)$,
$(8, 21, 4, 2, 0)$,
$(8, 21, 7, 1, 0)$,
$(8, 22, 3, 2, 0)$,
$(8, 22, 6, 1, 0)$,
$(8, 22, 9, 0, 0)$,
$(8, 23, 0, 3, 0)$,
$(8, 23, 4, 1, 0)$,
$(8, 24, 0, 2, 0)$,
$(8, 24, 3, 1, 0)$,
$(8, 24, 8, 0, 0)$,
$(8, 25, 1, 1, 0)$,
$(8, 25, 6, 0, 0)$,
$(8, 26, 0, 1, 0)$,
$(8, 26, 4, 0, 0)$,
$(8, 28, 3, 0, 0)$,
$(8, 32, 0, 0, 0)$,
$(9, 8, 10, 4, 0)$,
$(9, 9, 9, 4, 0)$,
$(9, 9, 12, 3, 0)$,
$(9, 10, 8, 4, 0)$,
$(9, 10, 10, 3, 0)$,
$(9, 10, 12, 2, 0)$,
$(9, 10, 13, 1, 0)$,
$(9, 10, 15, 0, 0)$,
$(9, 11, 11, 2, 0)$,
$(9, 12, 7, 4, 0)$,
$(9, 12, 9, 3, 0)$,
$(9, 12, 12, 1, 0)$,
$(9, 12, 14, 0, 0)$,
$(9, 13, 7, 3, 0)$,
$(9, 13, 10, 2, 0)$,
$(9, 14, 9, 2, 0)$,
$(9, 14, 11, 1, 0)$,
$(9, 14, 13, 0, 0)$,
$(9, 15, 6, 3, 0)$,
$(9, 16, 0, 4, 0)$,
$(9, 16, 4, 3, 0)$,
$(9, 16, 8, 2, 0)$,
$(9, 16, 10, 1, 0)$,
$(9, 16, 12, 0, 0)$,
$(9, 17, 3, 3, 0)$,
$(9, 17, 6, 2, 0)$,
$(9, 17, 8, 1, 0)$,
$(9, 17, 10, 0, 0)$,
$(9, 18, 2, 3, 0)$,
$(9, 18, 4, 2, 0)$,
$(9, 18, 7, 1, 0)$,
$(9, 18, 9, 0, 0)$,
$(9, 19, 0, 3, 0)$,
$(9, 19, 3, 2, 0)$,
$(9, 19, 6, 1, 0)$,
$(9, 20, 1, 2, 0)$,
$(9, 20, 5, 1, 0)$,
$(9, 20, 8, 0, 0)$,
$(9, 21, 4, 1, 0)$,
$(9, 21, 6, 0, 0)$,
$(9, 22, 1, 1, 0)$,
$(9, 22, 5, 0, 0)$,
$(9, 24, 4, 0, 0)$,
$(9, 25, 2, 0, 0)$,
$(9, 28, 0, 0, 0)$,
$(10, 8, 6, 4, 0)$,
$(10, 8, 8, 3, 0)$,
$(10, 9, 7, 3, 0)$,
$(10, 9, 10, 2, 0)$,
$(10, 9, 11, 1, 0)$,
$(10, 9, 13, 0, 0)$,
$(10, 10, 5, 4, 0)$,
$(10, 10, 9, 2, 0)$,
$(10, 10, 12, 0, 0)$,
$(10, 11, 6, 3, 0)$,
$(10, 12, 4, 4, 0)$,
$(10, 12, 5, 3, 0)$,
$(10, 12, 7, 2, 0)$,
$(10, 12, 10, 1, 0)$,
$(10, 12, 11, 0, 0)$,
$(10, 13, 6, 2, 0)$,
$(10, 13, 8, 1, 0)$,
$(10, 13, 10, 0, 0)$,
$(10, 14, 3, 3, 0)$,
$(10, 14, 5, 2, 0)$,
$(10, 14, 9, 0, 0)$,
$(10, 15, 2, 3, 0)$,
$(10, 15, 7, 1, 0)$,
$(10, 16, 4, 2, 0)$,
$(10, 16, 6, 1, 0)$,
$(10, 16, 8, 0, 0)$,
$(10, 17, 4, 1, 0)$,
$(10, 17, 6, 0, 0)$,
$(10, 18, 2, 1, 0)$,
$(10, 18, 5, 0, 0)$,
$(10, 20, 4, 0, 0)$,
$(10, 21, 2, 0, 0)$,
$(10, 22, 1, 0, 0)$,
$(10, 24, 0, 0, 0)$,
$(11, 4, 6, 4, 0)$,
$(11, 6, 5, 4, 0)$,
$(11, 7, 6, 3, 0)$,
$(11, 8, 4, 4, 0)$,
$(11, 8, 5, 3, 0)$,
$(11, 9, 6, 2, 0)$,
$(11, 9, 8, 1, 0)$,
$(11, 9, 10, 0, 0)$,
$(11, 10, 3, 3, 0)$,
$(11, 10, 5, 2, 0)$,
$(11, 10, 9, 0, 0)$,
$(11, 11, 2, 3, 0)$,
$(11, 11, 7, 1, 0)$,
$(11, 12, 4, 2, 0)$,
$(11, 12, 6, 1, 0)$,
$(11, 12, 8, 0, 0)$,
$(11, 13, 4, 1, 0)$,
$(11, 13, 6, 0, 0)$,
$(11, 14, 2, 1, 0)$,
$(11, 14, 5, 0, 0)$,
$(11, 16, 4, 0, 0)$,
$(11, 17, 2, 0, 0)$,
$(11, 18, 1, 0, 0)$,
$(11, 20, 0, 0, 0)$,
$(12, 4, 3, 3, 0)$,
$(12, 6, 2, 3, 0)$,
$(12, 6, 5, 2, 0)$,
$(12, 6, 7, 1, 0)$,
$(12, 6, 9, 0, 0)$,
$(12, 8, 4, 2, 0)$,
$(12, 8, 6, 1, 0)$,
$(12, 8, 8, 0, 0)$,
$(12, 9, 4, 1, 0)$,
$(12, 9, 6, 0, 0)$,
$(12, 10, 2, 1, 0)$,
$(12, 10, 5, 0, 0)$,
$(12, 12, 4, 0, 0)$,
$(12, 13, 2, 0, 0)$,
$(12, 14, 1, 0, 0)$,
$(12, 16, 0, 0, 0)$,
$(13, 6, 5, 0, 0)$,
$(13, 8, 4, 0, 0)$,
$(13, 9, 2, 0, 0)$,
$(13, 10, 1, 0, 0)$,
$(13, 12, 0, 0, 0)$,
$(14, 4, 3, 0, 0)$,
$(14, 5, 2, 0, 0)$,
$(14, 6, 1, 0, 0)$,
$(14, 8, 0, 0, 0)$,
$(15, 4, 0, 0, 0)$,
$(16, 0, 0, 0, 0)$}}
\caption{The Pareto-optimal statistics of Moser sets in $[3]^4$. This table can also be found at {\tt http://spreadsheets.google.com/ccc?key=rwXB\_Rn3Q1Zf5yaeMQL-RDw}}
\label{table4}
\end{figure}

\begin{proof} This was computed by computer search as follows. First, one observed that if $(a,b,c,d,e)$ was Pareto-optimal, then $a\geq 3$. To see this, it suffices to show that for any Moser set $A \subset [3]^4$ with $a(A)=0$, it is possible to add three points from $S_{4,4}$ to $A$ and still have a Moser set. To show this, suppose first that $A$ contains a point from $S_{1,4}$, such as $2221$. Then $A$ must omit either $2211$ or $2231$; without loss of generality we may assume that it omits $2211$. Similarly we may assume it omits $2121$ and $1221$. Then we can add $1131$, $1311$, $3111$ to $A$, as required. Thus we may assume that $A$ contains no points from $S_{1,4}$. Now suppose that $A$ omits a point from $S_{2,4}$, such as $2211$. Then one can add $3333$, $3111$, $1311$ to $A$, as required. Thus we may assume that A contains all of $S_{2,4}$, which forces $A$ to omit $2222$, as well as at least one point from $S_{3,4}$, such as $2111$. But then $3111$, $1111$, $3333$ can be added to the set, a contradiction.

Thus we only need to search through sets $A \subset [3]^4$ for which $|A \cap S_{4,4}| \geq 3$. A straightforward computer search shows that up to the symmetries of the cube, there are $391$ possible choices for $A \cap S_{4,4}$. For each such choice, we looped through all the possible values of the slices $A \cap 1***$ and $A \cap 3***$, i.e. all three-dimensional Moser sets which had the indicated intersection with $S_{3,3}$. (For fixed $A \cap S_{4,4}$, the number of possibilities for $A \cap 1***$ ranges from $1$ to $87123$, and similarly for $A \cap 3***$). For each pair of slices $A \cap 1***$ and $A \cap 3***$, we computed the lines connecting these two sets to see what subset of $2***$ was excluded from $A$; there are $2^{27}$ possible such exclusion sets. We precomputed a lookup table that gave the Pareto-optimal statistics for $A \cap 2***$ for each such choice of exclusion set; using this lookup table for each choice of $A \cap 1***$ and $A \cap 3***$ and collating the results, we obtained the above list. On a linux cluster, the lookup table took 22 minutes to create, and the loop over the $A \cap 1***$ and $A \cap 3***$ slices took two hours, spread out over $391$ machines (one for each choice of $A \cap S_{4,4}$). Further details (including source code) can be found at
\centerline{{\tt http://michaelnielsen.org/polymath1/index.php?title=4D\_Moser\_brute\_force\_search}.}
\end{proof}

As a consequence of this data, we have the following facts about the statistics of large Moser sets:

\begin{proposition}\label{stat} Let $A \subset [3]^4$ be a Moser set with statistics $(a,b,c,d,e)$.
\begin{itemize}
\item[(i)] If $|A| \geq 40$, then $e=0$.
\item[(ii)] If $|A| \geq 43$, then $d=0$.
\item[(iii)] If $|A| \geq 42$, then $d \leq 2$.
\item[(iv)] If $|A| \geq 41$, then $d \leq 3$.
\item[(v)] If $|A| \geq 40$, then $d \leq 6$.
\item[(vi)] If $|A| \geq 43$, then $c \geq 18$.
\item[(vii)] If $|A| \geq 42$, then $c \geq 12$.
\item[(viii)] If $|A| \geq 43$, then $b \geq 15$.
\end{itemize}
\end{proposition}

\begin{remark} This proposition was first established by an integer program, see Appendix \ref{integer-sec}. A computer-free proof can be found at \centerline{{\tt http://terrytao.files.wordpress.com/2009/06/polymath2.pdf}.}
\end{remark}

\subsection{Five dimensions}

Now we establish the bound $c'_{5,3}=124$. In view of the lower bounds, it suffices to show that there does not exist a Moser set $A \subset [3]^5$ with $|A|=125$.

We argue by contradiction. Let $A$ be as above, and let $(a(A),\ldots,f(A))$ be the statistics of $A$.

\begin{lemma}\label{fvan} $f(A)=0$.
\end{lemma}

\begin{proof} If $f(A)$ is non-zero, then $A$ contains $22222$, then each of the $\frac{3^5-1}{2} = 121$ antipodal pairs in $[3]^5$ can have at most one point in $A$, leading to only $122$ points.
\end{proof}

Let us slice $[3]^5$ into three parallel slices, e.g. $1****, 2****, 3****$. The intersection of $A$ with each of these slices has size at most $43$. In particular, this implies that
\begin{equation}\label{boo}
|A \cap 1****| + |A \cap 3****| = 125 - |A \cap 2****| \geq 82.
\end{equation}
Thus at least one of $A \cap 1****$, $A \cap 3****$ has cardinality at least $41$; by Proposition \ref{stat}(iv) we conclude that
\begin{equation}\label{d13}
\min( d(1****), d(3****) ) \leq 3.
\end{equation}
Furthermore, equality can only hold in \eqref{d13} if $A \cap 1****$, $A \cap 3****$ both have cardinality exactly $41$, in which case from Proposition \ref{stat}(iv) again we must have
\begin{equation}\label{d13a}
d(1****)=d(3****)=3.
\end{equation}
Of course, we have a similar result for permutations.

Now we improve the bound $|A \cap 2****| \leq 43$:

\begin{lemma} $|A \cap 2****| \leq 41$.
\end{lemma}

\begin{proof} Suppose first that $|A \cap 2****|=43$. Let $A' \subset [3]^4$ be the subset of $[3]^4$ corresponding to $A \cap 2****$, thus $A'$ is a Moser set of cardinality $43$. By Proposition \ref{stat}(vi), $c(A') \geq 18$. By Lemma \ref{dci}, the sum of the $c(V)$, where $V$ ranges over the eight side slices of $[3]^4$, is therefore at least $36$. By the pigeonhole principle, we may thus find two opposing side slices, say $1***$ and $3***$, with $c(1***)+c(3****) \geq 9$. Since $c(1***), c(3***)$ cannot exceed $6$, we thus have $c(1***), c(3***) \geq 3$, with at least one of $c(1***), c(3***)$ being at least $5$. Passing back to $A$, this implies that $d(*1***), d(*3***) \geq 3$, with at least one of $d(*1***), d(*3***)$ being at least $5$. But this contradicts \eqref{d13} together with the refinement \eqref{d13a}.

We have just shown that $|A \cap 2****| \leq 42$; we can thus improve \eqref{boo} to
$$ |A \cap 1****| + |A \cap 3****| \geq 83.$$
Combining this with Proposition \ref{stat}(ii)-(v) we see that
\begin{equation}\label{d13-6}
d(1****)+d(3****) \leq 6
\end{equation}
with equality only if $|A \cap 2****|=42$, and similarly for permutations.

Now let $A'$ be defined as before. Then we have
$$ c(1***) + c(3***) \leq 6$$
and similarly for permutations. Applying Lemma \ref{dci}, this implies that $c(2****) = c(A') \leq 12$.

Now suppose for contradiction that $|A'|=|A \cap 2****|=42$. Then by Proposition \ref{stat}(vii) we have
\begin{equation}\label{coo-1}
c(2****) = 12;
\end{equation}
applying Lemma \ref{dci} again, this forces $c(1***)+c(3***)=6$ and similarly for permutations, which then implies that
\begin{equation}\label{doo}
d(*1***)+d(*3***) = d(**1**)+d(**3**) = d(***1*)+d(***3*) = d(****1)+d(****3) = 6
\end{equation}
and hence
$$ |A \cap *2***| = |A \cap **2**| = |A \cap ***2*| = |A \cap ****2| = 42$$
and thus
\begin{equation}\label{coo-2}
c(*2***) = c(**2**) = c(***2*) = c(****2) = 12.
\end{equation}
Combining \eqref{coo-1}, \eqref{doo}, \eqref{coo-2} we conclude that
$$ d(1****)+d(3****) = 16,$$
contradicting \eqref{d13-6}.
\end{proof}

With this proposition, the bound \eqref{boo} now improves to
\begin{equation}\label{84}
|A \cap 1****| + |A \cap 3****| \geq 84
\end{equation}
and in particular
\begin{equation}\label{41}
|A \cap 1****|, |A \cap 3****| \geq 41.
\end{equation}
from this and Proposition \ref{stat}(ii)-(iv) we now have
\begin{equation}\label{d13-improv}
d(1****)+d(3****) \leq 4
\end{equation}
and similarly for permutations.

\begin{lemma}\label{evan} $e(A)=0$.
\end{lemma}

\begin{proof} From \eqref{84}, the intersection of $A$ with any side slice has cardinality at least $41$, and thus by Proposition \ref{stat}(i) such a side slice has an $e$-statistic of zero. The claim then follows from Lemma \ref{dci}.
\end{proof}

We need a technical lemma:

\begin{lemma}\label{tech} Let $B \subset S_{5,5}$. Then there exist at least $|B|-4$ pairs of strings in $B$ which differ in exactly two positions.
\end{lemma}

\begin{proof} The first non-vacuous case is $|B|=5$. It suffices to establish this case, as the higher cases then follow by induction (locating a pair of the desired form, then deleting one element of that pair from $B$).

Suppose for contradiction that one can find a $5$-element set $B \subset S_{5,5}$ such that no two strings in $B$ differ in exactly two positions. Recall that we may split $S_{5,5}=S_{5,5}^e \cup S_{5,5}^o$, where $S_{5,5}^e$ are those strings with an even number of $1$'s, and $S_{5,5}^o$ are those strings with an odd number of $1$'s. By the pigeonhole principle and symmetry we may assume $B$ has at least three elements in $S_{5,5}^o$. Without loss of generality, we can take one of them to be $11111$, thus excluding all elements in $S_{5,5}^o$ with exactly two $3$s, leaving only the elements with exactly four $3$s. But any two of them differ in exactly two positions, a contradiction.
\end{proof}

We can now improve the trivial bound $c(A) \leq 80$:

\begin{corollary}[Non-maximal $c$]\label{cmax} $c(A) \leq 79$. If $a(A) \geq 7$, then $c(A) \leq 78$.
\end{corollary}

\begin{proof} If $c(A)=80$, then $A$ contains all of $S_{3,5}$, which then implies that no two elements in $A \cap S_{5,5}$ can differ in exactly two places. It also implies (from \eqref{alpha-1}) that $d(A)$ must vanish, and that $b(A)$ is at most $40$. By Lemma \ref{tech}, we also have that $a(A) = |A \cap S_{5,5}|$ is at most $4$. Thus $|A| \leq 4 + 40 + 80 + 0 + 0 = 124$, a contradiction.

Now suppose that $a(A) \geq 7$. Then by Lemma \ref{tech} there are at least three pairs in $A \cap S_{5,5}$ that differ in exactly two places. Each such pair eliminates one point from $A \cap S_{3,5}$; but each point in $S_{3,5}$ can be eliminated by at most two such pairs, and so we have at least two points eliminated from $A \cap S_{3,5}$, i.e. $c(A) \leq 78$ as required.
\end{proof}

Next, we rewrite the quantity $125=|A|$ in terms of side slices. From Lemmas \ref{fvan}, \ref{evan} we have
$$ a(A) + b(A) + c(A) + d(A) = 125$$
and hence by Lemma \ref{dci}, the quantity
$$ s(V) := a(V) + \frac{5}{4} b(V) + \frac{5}{3} c(V) + \frac{5}{2} d(V) - \frac{125}{2},$$
where $V$ ranges over side slices, has an average value of zero.

\begin{proposition}[Large values of $s(V)$]\label{suv} For all side slices, we have $s(V) \leq 1/2$. Furthermore, we have $s(V) < -1/2$ unless the statistics $(a(V), b(V), c(V), d(V), e(V))$ are of one of the following four cases:
\begin{itemize}
\item (Type 1) $(a(V),b(V),c(V),d(V),e(V)) = (2,16,24,0,0)$ (and $s(V) = -1/2$ and $|A \cap V| = 42$);
\item (Type 2) $(a(V),b(V),c(V),d(V),e(V)) = (4,16,23,0,0)$ (and $s(V) = -1/6$ and $|A \cap V| = 43$);
\item (Type 3) $(a(V),b(V),c(V),d(V),e(V)) = (4,15,24,0,0)$ (and $s(V) = 1/4$ and $|A \cap V| = 43$);
\item (Type 4) $(a(V),b(V),c(V),d(V),e(V)) = (3,16,24,0,0)$ (and $s(V) = 1/2$ and $|A \cap V| = 43$);
\end{itemize}
\end{proposition}

\begin{proof}
Let $V$ be a side slice. From \eqref{41} we have
$$ 41 \leq a(V)+b(V)+c(V)+d(V) = |A \cap V| \leq 43.$$
First suppose that $|A \cap V| = 43$, then from Proposition \ref{stat}(ii), (viii), $d(V)=0$ and $b(V) \geq 15$.
Also, we have the trivial bound $c(V) \leq 24$, together with the inequality
$$ 3b(V) + 2c(V) \leq 96$$
from \eqref{alpha-1}. To exploit these facts, we rewrite $s(V)$ as
$$ s(V) = \frac{1}{2} - \frac{1}{2}( 24 - c(V) ) - \frac{1}{12} (96-3b(V)-2c(V)).$$
Thus $s(V) \leq 1/2$ in this case. If $s(V) \geq -1/2$, then
$$ 6 (24-c(V)) + (96-3b(V)-2c(V)) \leq 12,$$
which together with the inequalities $b(V) \leq 15$, $c(V) \leq 24$, $3b(V)+2c(V) \leq 96$ we conclude that $(b(V),c(V))$ must be one of $(16,24)$, $(15, 24)$, $(16, 23)$, $(15, 23)$. The first three possibilities lead to Types 4,3,2 respectively. The fourth type would lead to $(a(V),b(V),c(V),d(V),e(V)) = (5,15,23,0,0)$, but this contradicts \eqref{eleven}.

Next, suppose $|A \cap V| = 42$, so by Proposition \ref{stat}(iii) we have $d(V) \leq 2$. From \eqref{alpha-1} we have
\begin{equation}\label{2cd}
2c(V) + 3d(V) \leq 48
\end{equation}
while from \eqref{alpha-2} we have
\begin{equation}\label{3cd}
3b(V)+2c(V)+3d(V) \leq 96
\end{equation}
and so we can rewrite $s(V)$ as
\begin{equation}\label{sv2}
s(V) = -\frac{1}{2} - \frac{1}{4}( 48 - 2c(V) - 3d(V) ) - \frac{1}{12} (96-3b(V)-2c(V)-3d(V)) + \frac{1}{2} d(V).
\end{equation}
This already gives $s(V) \leq 1/2$. If $d(V)=0$, then $s(V) \leq -1/2$, with equality only in Type 1. If $d(V)=1$, then the set $A' \subset [3]^4$ corresponding to $A \cap V$ contains a point in $S_{3,4}$, which without loss of generality we can take to be $2221$. Considering the three lines $*221$, $2*21$, $22*1$, we see that at least three points in $S_{2,4}$ must be missing from $A'$, thus $c(V) \leq 21$. This forces $48-2c(V)-3d(V) \geq 3$, and so $s(V) < -3/4$. Finally, if $d(V)=2$, then $A'$ contains two points in $S_{3,4}$. If they are antipodal (e.g. $2221$ and $2223$), the same argument as above shows that at least six points in $S_{2,4}$ are missing from $A'$; if they are not antipodal (e.g. $2221$ and $2212$) then by considering the lines $*221$, $2*21$, $22*1$, $*212$, $2*12$ we see that five points are missing. Thus we have $c(V) \leq 19$, which forces $48-2c(V)-3d(V) \geq 4$. This forces $s(V) \leq -1/2$, with equality only when $c(V)=19$ and $3b(V)+2c(V)+3d(V)=96$, but this forces $b(V)$ to be the non-integer $52/3$, a contradiction, which concludes the treatment of the $|A \cap V|=42$ case.

Finally, suppose $|A \cap V| = 41$. Using \eqref{2cd}, \eqref{3cd} as before we have
\begin{equation}\label{sv3}
s(V) = -\frac{3}{2} - \frac{1}{4}( 48 - 2c(V) - 3d(V) ) - \frac{1}{12} (96-3b(V)-2c(V)-3d(V)) + \frac{1}{2} d(V),
\end{equation}
while from Proposition \ref{stat}(vi) we have $d(V) \leq 3$. This already gives $s(V) \leq 0$, and $s(V) \leq -1$ when $d(V)=1$. In order to have $s(V) \geq -1/2$, we must then have $d(V)=2$ or $d(V)=3$. But then the arguments of the preceding paragraph give $48-2c(V)-3d(V) \geq 4$, and so $s(V) \leq -1$ in this case.
\end{proof}

Since the $s(V)$ average to zero, by the pigeonhole principle we may find two opposing side slices (e.g. $1****$ and $3****$), whose total $s$-value is non-negative. Actually we can do a little better:

\begin{lemma}\label{side-off} There exists two opposing side slices whose total $s$-value is strictly positive.
\end{lemma}

\begin{proof} If this is not the case, then we must have $s(1****)+s(3****)=0$ and similarly for permutations. Using Proposition \ref{suv} we thus see that for every opposing pair of side slices, one is Type 1 and one is Type 4. In particular $c(V)=24$ for all side slices $V$. But then by Lemma \ref{dci} we have $c(A)=80$, contradicting Lemma \ref{cmax}.
\end{proof}

Let $V, V'$ be the side slices in Lemma \ref{side-off}
By Proposition \ref{suv}, the $V, V'$ slices must then be either Type 2, Type 3, or Type 4, and they cannot both be Type 2. Since $a(A) = a(V)+a(V')$, we conclude
\begin{equation}\label{amix}
6 \leq a(A) \leq 8.
\end{equation}
In a similar spirit, we have
$$ c(V) + c(V') \leq 23+24.$$
On the other hand, by considering the $24$ lines connecting $c$-points of $V, V'$ to $c$-points of the centre slice $W$ between $V$ and $V'$, each of which contains at most two points in $A$, we have
$$ c(V) + c(W) + c(V') \leq 24 \times 2.$$
Thus $c(W) \leq 1$; since
$$ d(A) = d(V) + d(V') + c(W)$$
we conclude from Proposition \ref{suv} that $d(A) \leq 1$. Actually we can do better:

\begin{lemma} $d(A)=0$.
\end{lemma}

\begin{proof} Suppose for contradiction that $d(A)=1$; without loss of generality we may take $11222 \in A$. This implies that $d(1****)=d(*1***)=1$. Also, by the above discussion, $c(**1**)$ and $c(**3**)$ cannot both be $24$, so by Proposition \ref{suv}, $s(**1**)+s(**3**) \leq 1/3$; similarly
$s(***1*)+s(***3*) \leq 1/3$ and $s(****1)+s(****3) \leq 1/3$. Since the $s$ average to zero, we see from the pigeonhole principle that either $s(1****)+s(3****) \geq -1/2$ or $s(*1***)+s(*3***) \geq -1/2$. We may assume by symmetry that
\begin{equation}\label{star-2}
s(1****)+s(3****) \geq -1/2.
\end{equation}
Since $s(3****) \leq 1/2$ by Proposition \ref{suv}, we conclude that
\begin{equation}\label{star}
s(1****) \geq -1.
\end{equation}
If $|A \cap 1****|=41$, then by \eqref{sv3} we have
$$ s(1****) = -1 - \frac{1}{4}( 48 - 2c(1****) - 3d(1****) ) - \frac{1}{12} (96-3b(1****)-2c(1****)-3d(1****))$$
but the arguments in Proposition \ref{suv} give $48 - 2c(1****) - 3d(1****) \geq 3$ and $96-3b(1****)-2c(1****)-3d(1****) \geq 0$, a contradiction. So we must have $|A \cap 1****|=42$ (by Proposition \ref{stat}(ii) and \eqref{41}). In that case, from \eqref{sv2} we have
$$ s(1****) = \frac{1}{4}( 48 - 2c(1****) - 3d(1****) ) - \frac{1}{12} (96-3b(1****)-2c(1****)-3d(1****))$$
while also having $48 - 2c(1****) - 3d(1****) \geq 3$ and $96-3b(1****)-2c(1****)-3d(1****) \geq 0$. Since $s(1****) \geq -1$ and $d(1****)=1$, we soon see that we must have $48 - 2c(1****) - 3d(1****) = 3$ and $96-3b(1****)-2c(1****)-3d(1****) \leq 3$, which forces $c(1****)=21$ and $b(1****)=16$ or $b(1****)=17$; thus the statistics of $1****$ are either $(4,16,21,1,0)$ or $(3,17,21,1,0)$.

We first eliminate the $(3,17,21,1,0)$ case. In this case $s(1****)$ is exactly $-1$. Inspecting the proof of \eqref{star}, we conclude that $s(3****)$ must be $+1/2$ and that $s(**1**)+s(**3**)=1/3$. From the former fact and Proposition \ref{suv} we see that $a(A) = a(1****)+a(3****)=3+3=6$; on the other hand, from the latter fact and Proposition \ref{suv} we have $a(A) = a(**1**)+a(**3**) = 4+3=7$, a contradiction.

So $1****$ has statistics $(4,16,21,1,0)$, which implies that $s(1****)=-3/4$ and $|A \cap 1****|=42$. By \eqref{star-2} we conclude
\begin{equation}\label{s3}
s(3****) \geq 1/4,
\end{equation}
which by Proposition \ref{suv} implies that $|A \cap 3****|=43$, and hence $|A \cap 2****|=40$. On the other hand, since $e(A)=f(A)=0$ and $d(A)=1$, with the latter being caused by $11222$, we see that $c(2****)=d(2****)=e(2****)=0$. From \eqref{alpha-1} we have $4a(2****)+b(2****) \leq 64$, and we also have the trivial inequality $b(2****) \leq 32$; these inequalities are only compatible if $2****$ has statistics $(8,32,0,0,0)$, thus $A \cap 2****$ contains $S_{2,5} \cap 2****$.

If $a(3****)=4$, then $a(A)=a(1****)+a(3****)=8$, which by Proposition \ref{suv} implies that $s(**1**)+s(**3**)$ cannot exceed $1/12$, and similarly for permutations. On the other hand, from Proposition \ref{suv} $s(**1**)+s(**3**)$ cannot exceed $-3/4 + 1/4 = -1/2$, and so the average value of $s$ cannot be zero, a contradiction. Thus $a(3****) \neq 4$, which by \eqref{s3} and Proposition \ref{suv} implies that $**3**$ has statistics $(3,16,24,0,0)$.

In particular, $A$ contains $16$ points from $3**** \cap S_{1,5}$ and all of $3**** \cap S_{2,5}$. As a consequence, no pair of the $16$ points in $A \cap 3**** \cap S_{1,5}$ can differ in only one coordinate; partitioning the $32$-point set $3**** \cap S_{1,5}$ into $16$ such pairs, we conclude that every such pair contains exactly one element of $A$. We conclude that $A \cap 3**** \cap S_{1,5}$ is equal to either $3**** \cap S_{1,5}^e$ or $3**** \cap S_{1,5}^o$.

On the other hand, $A$ contains all of $2**** \cap S_{2,5}$, and exactly sixteen points from $1**** \cap S_{1,5}$. Considering the vertical lines $*xyzw$ where $xyzw \in S_{1,4}$, we conclude that $A \cap 1**** \cap S_{1,5}$ is either equal to $1**** \cap S_{1,5}^o$ or $1**** \cap S_{1,5}^e$.
But either case is incompatible with the fact that $A$ contains $11222$ (consider either the line $11xx2$ or $11x\overline{x}2$, where $x=1,2,3$ and $\overline{x}=4-x$), obtaining the required contradiction.
\end{proof}

We can now eliminate all but three cases for the statistics of $A$:

\begin{proposition}[Statistics of $A$] The statistics $(a(A),b(A),c(A),d(A),e(A),f(A))$ of $A$ must be one of the following three tuples:
\begin{itemize}
\item (Case 1) $(6,40,79,0,0)$;
\item (Case 2) $(7,40,78,0,0)$;
\item (Case 3) $(8,39,78,0,0)$.
\end{itemize}
\end{proposition}

\begin{proof}
Since $d(A)=e(A)=f(A)=0$, we have
$$ c(2****)=d(2****)=e(2****)=0.$$
On the other hand, from \eqref{alpha-1} we have $4a(2****)+b(2****) \leq 64$ as well as the trivial inequality $b(2****) \leq 24$, and also we have
$$ |A \cap 2****| = 125 - |A \cap 1****| - |A \cap 3****| \geq 125 - 43 - 43 = 39.$$
Putting all this together, we see that the only possible statistics for $2****$ are $(8,32,0,0,0)$, $(7,32,0,0,0)$, or $(8,31,0,0,0)$. In particular, $7 \leq a(2****) \leq 8$ and $31 \leq b(2****) \leq 32$, and similarly for permutations. Applying Lemma \ref{dci} we conclude that
$$ 35 \leq b(A) \leq 40$$
and
$$ 77.5 \leq c(A) \leq 80.$$
Combining this with the first part of Corollary \ref{cmax} we conclude that $c(A)$ is either $78$ or $79$. From this and \eqref{amix} we see that the only cases that remain to be eliminated are $(7,39,79,0,0)$ and $(8,38,79,0,0)$, but these cases are incompatible with the second part of Corollary \ref{cmax}.
\end{proof}

We now eliminate each of the three remaining cases in turn.

\subsection{Elimination of $(6,40,79,0,0)$}

Here $A \cap S_{5,5}$ has six points. By Lemma \ref{tech}, there are at least two pairs in this set which differ in two positions. Their midpoints are eliminated from $A \cap S_{3,5}$. But $A$ omits exactly one point from $S_{3,5}$, so these midpoints must be the same. By symmetry, we may then assume that these two pairs are $(11111,11133)$ and $(11113,11131)$. Thus the eliminated point in $S_{3,5}$ is $11122$, i.e. $A$ contains $S_{3,5} \backslash \{11122\}$. Also, $A$ contains $\{11111,11133,11113,11131\}$ and thus must omit $\{11121, 11123, 11112, 11132\}$.

Since $11322 \in A$, at most one of $11312, 11332$ lie in $A$. By symmetry we may assume $11312 \not \in A$, thus there is a pair $(xy1z2, xy3z2)$ with $x,y,z = 1,3$ that is totally omitted from $A$, namely $(11112,11312)$. On the other hand, every other pair of this form can have at most one point in the $A$, thus there are at most seven points in $A$ of the form $xyzw2$ with $x,y,z,w = 1,3$. Similarly there are at most 8 points of the form $xyz2w$, or of $xy2zw$, $x2yzw$, $2xyzw$, leading to $b(A) \leq 7+8+8+8+8=39$, contradicting the statistic $b(A)=40$.

\subsection{Elimination of $(7,40,78,0,0)$}

Here $A \cap S_{5,5}$ has seven points. By Lemma \ref{tech}, there are at least three pairs in this set which differ in two positions. As we can only eliminate two points from $S_{3,5}$, two of the midpoints of these pairs must be the same; thus, as in the previous section, we may assume that $A$ contains $\{11111,11133,11113,11131\}$ and omits $\{11121, 11123, 11112, 11132\}$ and $11122$.

Now consider the $160$ lines $\ell$ connecting two points in $S_{4,5}$ to one point in $S_{3,5}$ (i.e. $*2xyz$ and permutations, where $x,y,z=1,3$). By double counting, the total sum of $|\ell \cap A|$ over all $160$ lines is $4b(A)+2c(A) = 316 = 158 \times 2$. On the other hand, each of these lines contain at most two points in $A$, but two of them (namely $1112*$ and $1112*$) contain no points. Thus we must have $|\ell \cap A|=2$ for the remaining $158$ lines $\ell$.

Since $A$ omits $1112x$ and $111x2$ for $x=1,3$, we thus conclude (by considering the lines $11*2x$ and $11*x2$) that $A$ must contain $1132x$, $113x2$, $1312x$, and $131x2$. Taking midpoints, we conclude that $A$ omits $11322$ and $13122$. But together with $11122$ this implies that at least three points are missing from $A \cap S_{3,5}$, contradicting the hypothesis $c(A)=78$.

\subsection{Elimination of $(8,39,78,0,0)$}

Now $A \cap S_{5,5}$ has eight points. By Lemma \ref{tech}, there are at least three pairs in this set which differ in two positions. As we can only eliminate two points from $S_{3,5}$, two of these pairs $(a,b), (c,d)$ must have the same midpoint $p$, and two other pairs $(a',b'), (c',d')$ must have the same midpoint $p'$, and $A$ contains $S_{3,5} \backslash \{p,p'\}$. As $p,p'$ are distinct, the plane containing $a,b,c,d$ is distinct from the plane containing $a',b',c',d'$.

Again consider the $160$ lines $\ell$ from the previous section. This time, the sum of the $|\ell \cap A|$ is $4b(A)+2c(A) = 312 = 156 \times 2$.
But the two lines in the plane of $a,b,c,d$ passing through $p$, and the two lines in the plane of $a',b',c',d'$ passing through $p'$, have no points; thus we must have $|\ell \cap A|=2$ for the remaining $156$ lines $\ell$.

Without loss of generality we have $(a,b)=(11111,11133)$, $(c,d) = (11113,11131)$, thus $p = 11122$. By permuting the first three indices, we may assume that $p'$ is not of the form $x2y2z, x2yz2, xy22z, xy2z2$ for any $x,y,z=1,3$. Then we have $1112x \not \in A$ and $1122x \in A$ for every $x=1,3$, so by the preceding paragraph we have $1132x \in A$; similarly for $113x2, 1312x, 131x2$. Taking midpoints, this implies that $13122, 11322 \not \in A$, but this (together with 11122) shows that at least three points aremissing from $A \cap S_{3,5}$, contradicting the hypothesis $c(A)=78$.

\subsection{Six dimensions}

Now we establish the bound $c'_{6,3}=353$. In view of the lower bounds, it suffices to show that there does not exist a Moser set $A \subset [3]^5$ with $|A|=354$.

We argue by contradiction. Let $A$ be as above, and let $(a(A),\ldots,g(A))$ be the statistics of $A$.

\begin{lemma}\label{g6} $g(A)=0$.
\end{lemma}

\begin{proof} For any four-dimensional slice $V$ of $A$, define
$$S(V) := 15 a(V) + 5 b(V) + 5 c(V)/2 + 3d(V)/2 + e(V).$$
From Lemma \ref{dci} we see that $|A|$ is equal to $a(A)+b(A)$ plus the average of $S(V)$ where $V$ ranges over the twenty slices which are some permutation of the center slice $22****$.

If $g(A)=1$, then $a(A) \leq 32$ and $b(A) \leq 96$ by \eqref{alpha-1}. Meanwhile, $e(V)=g(A)=1$ for every center slice $V$, so from Lemma \ref{paretop-4}, one can show that $S(V) \leq 223.5$ for every such slice. We conclude that $|A| \leq 351.5$, a contradiction.
\end{proof}

For any four-dimensional slice $V$ of $A$, define the \emph{defects} to be
$$ D(V) := 356 - [4a(V)+6b(V)+10c(V)+20d(V)+60e(V)].$$
Define a \emph{corner slice} to be one of the permutations or reflections of $11****$, thus there are $60$ corner slices. From Lemma \ref{dci} we see that $356-|A|+f(A)=2+f(A)$ is the average of the defects of all the $60$ corner slices. On the other hand, from Lemma \ref{paretop-4} and a straightforward computation, one concludes

\begin{lemma}\label{defects} Let $A$ be a four-dimensional Moser set. Then $D(A) \geq 0$. Furthermore:
\begin{itemize}
\item If $A$ has statistics $(6,12,18,4,0)$, then $D(A)=0$.
\item If $A$ has statistics $(5,12,18,4,0)$, $(5,12,12,4,1)$, or $(6,8,12,8,0)$, then $D(A)=4$.
\item For all other $A$, $D(A) \geq 6$.
\item If $a(A) = 4$, then $D(A) \geq 8$.
\item If $a(A) \geq 7$, then $D(A) \geq 16$.
\item If $a(A) \geq 8$, then $D(A) \geq 30$.
\item If $a(A) \geq 9$, then $D(A) \geq 86$.
\end{itemize}
\end{lemma}

Define a \emph{family} to be a set of four parallel corner slices, thus there are $15$ families, which are all a permutation of $\{11****, 13****, 31****, 33**** \}$. We refer to the family $\{11****, 13****, 31****, 33**** \}$ as $ab****$, and similarly define the family $a*b***$, etc.

\begin{lemma}\label{f6} $f(A)=0$.
\end{lemma}

\begin{proof} For any four-dimensional slice $V$ of $A$, define
$$ s(V) := 12 a(V)+15 b(V)/2+20 c(V)/3+15 d(V)/2 + 12 e(V),$$
and define an \emph{edge slice} to be one of the $30$ permutations or reflections of $12****$. From double counting we see that $|A|-a(A)$ is equal to the average of the $30$ values of $s(V)$ as $V$ ranges over edge slices.

From Lemma \ref{paretop-4} one can verify that $s(V) \leq 336$, and that $s(V) \leq 296 = 336-40$ if $e(V)=1$. The number of edge slices $V$ for which $e(V)=1$ is equal to $5f(A)$, and so the average value of the $s(V)$ is at most $336 - \frac{40 \times 5}{30} f(A)$, and so
$$ |A| - a(A) \leq 336 - \frac{40 \times 5}{30} f(A)$$
which we can rearrange (using $|A|=354$) as
$$ a(A) \geq 18 + \frac{20}{3} f(A).$$
Suppose first that $f(A)=1$; then $a(A) \geq 25$. This means that in any given family, one of the four corner slices has an $a$ value of at least $7$, and thus by Lemma \ref{defects} has a defect of at least $16$. Thus the average defect is at least $4$; on the other hand, the average defect is $2+f(A)=3$, a contradiction.

Now suppose that $f(A) \geq 2$; then $a(A) \geq 32$. Then in any given family, there is a corner slice with an $a$ value at least $9$, or four slices with $a$ value at least $8$, leading to a total defect of at least $86$ by Lemma \ref{defects}. Thus the average defect is at least $21.5$; on the other hand, the average defect is $2+f(A) \leq 2+12$, a contradiction.
\end{proof}

As a consequence of the above lemma, we see that the average defect of all corner slices is $2$, or equivalently that the total defect of these slices is $120$.

Call a corner slice \emph{good} if it has statistics $(6,12,18,4,0)$, and \emph{bad} otherwise. Thus good slices have zero defect, and bad slices have defect at least four. Since the average defect of the $60$ corner slices is $2$, there are at least $30$ good slices.

One can describe the structure of the good slices completely:

\begin{lemma}\label{sixt} The subset of $[3]^4$ consisting of the strings $1111, 1113, 3333, 1332, 1322, 1222, 3322$ and permutations is a Moser set with statistics $(6,12,18,4,0)$. Conversely, every Moser set with statistics $(6,12,18,4,0)$ is of this form up to the symmetries of the cube $[3]^4$.
\end{lemma}

\begin{proof} This can be verified by computer ({\bf need details}). A computer-free proof can be found at
\centerline{{\tt http://michaelnielsen.org/polymath1/index.php?title=Classification\_of\_\%286\%2C12\%2C18\%2C4\%2C0\%29\_sets}.}
\end{proof}

As a consequence of this lemma, given any $x,y,z,w \in \{1,3\}$, there is a unique good Moser set in $[3]^4$ set whose intersection with $S_{1,4}$ is $\{x222, 2y22, 22z2, 222w\}$, and these are the only 16 possibilities. Call this set the \emph{good set of type $xyzw$}. It consists of
\begin{itemize}
\item The four points $x222, 2y22, 22z2, 222w$ in $S_{1,4}$;
\item All $24$ elements of $S_{2,4}$ except for $xy22, x2z2, x22w, 2yz2, 2y2w, 22zw$;
\item The twelve points $xYZ2$, $xY2W$, $x2ZW$, $XyZ2$, $Xy2W$, $2yZW$, $XYz2$, $X2zW$, $2YzW$, $XY2w$, $X2Zw$, $2YZw$ in $S_{3,4}$, where $X=4-x$, $Y=4-y$, $Z=4-z$, $W=4-w$;
\item The six points $xyzw, xyzW, xyZw, xYzw, Xyzw, XYZW$ in $S_{4,4}$.
\end{itemize}

We can use this to constrain the types of two intersecting good slices:

\begin{lemma}\label{pqs} Suppose that the $pq****$ slice is of type $xyzw$, and the $p*r***$ slice is of type $x'y'z'w'$, where $p,q,r,x,y,z,w,x',y',z',w'$ are in $\{1,3\}$. Then $x'=x$ iff $q=r$, and $y'z'w'$ is equal to either $yzw$ or $YZW$. If $x=r$ (or equivalently if $x'=q$), then $y'z'w'=yzw$.
\end{lemma}

\begin{proof} By reflection symmetry we can take $p=q=r=1$. Observe that the $11****$ slice contains $111222$ iff $x=1$, and the $1*1***$ slice similarly contains $111222$ iff $x'=1$. This shows that $x=x'$.

Suppose now that $x=x'=1$. Then the $111***$ slice contains the three elements $111y22, 1112z2, 11122w$, and excludes $111Y22, 1112Z2, 11122W$, and similarly with the primes, which forces $yzw=y'z'w'$ as claimed.

Now suppose that $x=x'=3$. Then the $111***$ slice contains the two elements $111yzw, 111YZW$, but does not contain any of the other six points in $S_{6,6} \cap 111***$, and similarly for the primes. Thus $y'z'w'$ is equal to either $yzw$ or $YZW$ as claimed.
\end{proof}

Now we look at two adjacent parallel good slices, such as $11****$ and $13****$. The following lemma asserts that such slices either have opposite type, or else will create a huge amount of defect in other slices:

\begin{lemma}\label{l18} Suppose that the $11****$ and $13****$ slices are good with types $xyzw$ and $x'y'z'w'$ respectively. If $x=x'$, then the $1*x***$ slice has defect at least $30$, and the $1*X***$ slice has defect at least $8$. Also, the $1**1**$, $1**3**$, $1***1*$, $1***3*$, $1****1$, $1****3$ slices have defect at least $6$. In particular, the total defect of slices beginning with $1*$ is at least $74$.
\end{lemma}

\begin{proof} Observe from the $11****, 13****$ hypotheses that $a(1*x***)=9$ and $a(1*X***)=4$, which gives the first two claims by Lemma \ref{defects}. For the other claims, one sees from Lemma \ref{pqs} that the other six slices cannot be good; also, they have an $a$-value of $6$ and a $d$-value of at most $7$, and the claims then follow from Lemma \ref{defects}.
\end{proof}

Now we look at two diagonally opposite parallel good slices, such as $11****$ and $13****$.

\begin{lemma}\label{l14} The $11****$ and $33****$ slices cannot both be good and of the same type.
\end{lemma}

\begin{proof} Suppose not. By symmetry we may assume that $11****$ and $33****$ are of type $1111$. This excludes a lot of points from $22****$. Indeed, by connecting lines between the $11****$ and $33****$ slices, we see that the only points that can still survive in $22****$ are $221133, 221333, 221132, 223332$, and permutations of the last four indices. Double counting the lines $22133*$ and permutations we see that there are at most $12$ points one can place in the permutations of $221133, 221333, 221132$, and so the $22****$ slice has at most $16$ points. Meanwhile, the two five-dimensional slices $1*****, 3*****$ have at most $c'_{5,3} = 124$ points, and the other two four-dimensional slices $21****, 23****$ have at most $c'_{4,3} = 43$ points, leading to at most $16 + 124 * 2 + 43 * 2 = 350$ points in all, a contradiction.
\end{proof}

\begin{lemma}\label{l19} It is not possible for all four slices in a family to be good.
\end{lemma}

\begin{proof} Suppose not. By symmetry we may assume that $11****, 13****, 31****, 33****$ are good. By Lemma \ref{l14}, the $11****$ and $33****$ slices cannot be of the same type, and so they cannot both be of the opposite type to either $13****$ or $31****$. If $13****$ is not of the opposite type to $11****$, then by (a permutation of) Lemma \ref{l18}, the total defect of slices beginning with $1*$ is at least $74$; otherwise, if $13****$ is not of the opposite type to $33****$, then by (a permutation and reflection of) Lemma \ref{l18}, the total defect of slices beginning with $*3$ is at least $74$. Similarly, the total defect of slices beginning with $3*$ or $*1$ is at least $74$, leading to a total defect of at least $148$. But the total defect of all the corner slices is $2 \times 60 = 120$, a contradiction.
\end{proof}

\begin{corollary}\label{l20} At most one family can have a total defect of at least $38$.
\end{corollary}

\begin{proof} Suppose there are two families with defect at least $38$. The remaining thirteen family have defect at least $4$ by Lemma \ref{l19} and Lemma \ref{defects}, leading to a total defect of at least $38*2+13*4=128$. But the total defect is $2 \times 60 = 120$, a contradiction.
\end{proof}

Actually we can refine this:

\begin{proposition} No family can have a total defect of at least $38$.
\end{proposition}

\begin{proof} Suppose for contradiction that the $ab****$ family (say) had a total defect of at least $38$, then by Corollary \ref{l20} all other families have total defect at least $38$.

We claim that the $**ab**$ family can have at most two good slices. Indeed, suppose the $**ab**$ has three good slices, say $**11**, **13**, **33**$. By Lemma \ref{l14}, the $**11**$ and $**33**$ slices cannot be of the same type, and so cannot both be of opposite type to $**13**$. Suppose $**11**$ and $**13**$ are not of opposite type. Then by (a permutation of) Lemma \ref{l18}, one of the families $a*b***, *ab***, **b*a*, **b**a$ has a net defect of at least $38$, contradicting the normalisation.

Thus each of the six families $**ab**, **a*b*, **a**b, ***ab*, ***a*b$ have at least two bad slices. Meanwhile, the eight families $a*b***, a**b**, a***b*, a****b, *ab***, *a*b**, *a**b*, *a***b$ have at least one bad slice by Corollary \ref{l19}, leading to twenty bad slices in addition to the defect of at least $38$ arising from the $ab****$ slice. To add up to a total defect of $120$, we conclude from Lemma \ref{defects} that all bad slices outside of the $ab****$ family have a defect of four, with at most one exception; but then by Lemma \ref{l18} this shows that (for instance) the $1*1***$ and $1*3***$ slices cannot be good unless they are of opposite type. The previous argument then shows that the a*b*** slice cannot have three good slices, which increases the number of bad slices outside of $ab****$ to at least twenty-one, and now there is no way to add up to $120$, a contradiction.
\end{proof}

\begin{corollary} Every family can have at most two good slices.
\end{corollary}

\begin{proof} If for instance $11****, 13****, 33****$ are both good, then by Lemma \ref{l14} at least one of $11****, 33****$ is not of the opposite type to $13****$, which by Lemma \ref{l18} implies that there is a family with a total defect of at least $38$, contradicting the previous proposition.
\end{proof}

From this corollary and Lemma \ref{defects}, we see that every family has a defect of at least $8$. Since there are $15$ families, and $8 \times 15$ is exactly equal to $120$, we conclude

\begin{corollary}\label{coda} Every family has \emph{exactly} two good slices, and the remaining two slices have defect $4$. In particular, by Lemma \ref{defects}, the bad slices must have statistics $(5,12,18,4,0)$, $(5,12,12,4,1)$, or $(6,8,12,8,0)$.
\end{corollary}

We now limit how these slices can interact with good slices.

\begin{lemma}\label{goodgood} Suppose that $1*1***$ is a good slice.
\begin{itemize}
\item[(i)] The $11****$ slice cannot have statistics $(6,8,12,8,0)$.
\item[(ii)] The $11****$ slice cannot have statistics $(5,12,12,4,1)$.
\item[(iii)] If the $11****$ slice has statistics $(5,12,18,4,0)$, then the $112***$ slice has statistics $(3,9,3,0)$.
\end{itemize}
\end{lemma}

\begin{proof} This can be verified through computer search; there are $16$ possible configurations for the good slices, and one can calculate that there are $2750$ configurations for the $(5,12,12,4,1)$ slices, $4368$ configurations for the $(5,12,18,4,0)$ slices, and $10000$ configurations for the $(6,8,12,8,0)$ slices. {\bf we could put human proofs for all this somewhere, presumably.}

%We first prove (i). Suppose for contradiction that the $11****$ slice has statistics $(6,8,12,8,0)$, then $A$ contains $111222$, and so the $1*1***$ slice is of type $1xyz$ for some $x,y,z$. By symmetry we may assume it is of type $1111$, thus the $111***$ slice consists of
%$111111, 111113, 111332, 111322, 111222$ and permutations of the last three indices. On the other hand, the $11****$ slice has all eight of the points in $11**** \cap S_{2,6}$. Drawing lines between these points and $111111, 111113$ and permutations, we see that the $113***$ slice cannot contain $113111, 113113, 113133$, or permutations, leaving $113333$ as the only possible element of $113*** \cap S_{6,6}$. This makes $a(11****)=5$, a contradiction.
\end{proof}

\begin{corollary}\label{slic} The $111***$ slice has statistics $(4,3,3,1)$, $(2,6,6,0)$, $(3,3,3,1)$, or $(1,6,6,0)$.
\end{corollary}

\begin{proof} From Corollary \ref{coda}, we know that at least one of the slices $13****, 31****, 11****$ are good. If $11****$ or $1*1***$ is good, then the slice $111***$ has statistics $(4,3,3,1)$ or $(2,6,6,0)$, by Lemma \ref{sixt}. By symmetry we may thus reduce to the case where $13****$ is good and $1*1***$ is bad. Then by Lemma \ref{goodgood}, the $1*1***$ slice has statistics $(5,12,18,4,0)$ and the $121***$ slice has statistics $(3,9,3,0)$. Since the $131***$ slice, as a side slice of the good $13****$ slice, has statistics $(4,3,3,1)$ or $(2,6,6,0)$, we conclude that the $111***$ slice has statistics $(1,6,6,0)$ or $(3,3,3,1)$, and the claim follows.
\end{proof}

\begin{corollary}\label{slic2} All corner slices have statistics $(6,12,18,4,0)$ or $(5,12,18,4,0)$.
\end{corollary}

\begin{proof} Suppose first that a corner slice, say $11****$ has statistic $(6,8,12,8,0)$. Then $111***$ and $113***$ contain one ``d'' point each, and have six ``a'' points between them, so by Corollary \ref{slic}, they both have statistic $(3,3,3,1)$. This forces the $1*1***$, $1*3***$ slices to be bad, which by Corollary \ref{coda} forces the $3*1***,3*3***$ slices to be good. This forces the $311***, 313***$ slices to have statistics either $(2,6,6,0)$ or $(4,3,3,1)$. But the $311***$ slice (say) cannot have statistic $(4,3,3,1)$, since when combined with the $(3,3,3,1)$ statistics of $111***$ would give $a(*11***)=7$, which contradicts Corollary \ref{coda}; thus the $311***$ slice has statistic $(2,6,6,0)$, and similarly for $331***$. But then $a(3*1***)=4$, which again contradicts Corollary \ref{coda}.

Thus no corner slice has statistic $(6,8,12,8,0)$. Now suppose that a corner slice, say $11****$ has statistic $(5,12,12,4,1)$. By Lemma \ref{goodgood}, the $1*1***, 1*3***$ slices are bad, so by repeating the preceding arguments we conclude that the $311***, 313***$ slices have statistics $(2,6,6,0)$ or $(4,3,3,1)$; in particular, their $a$-value is even. However, the $*11***$ and $*13***$ slices are bad by Lemma \ref{goodgood}, and thus have an $a$-value of $5$; thus the $111***$ and $113***$ slices have an odd $a$-value. Thus forces $a(11****)$ to be even; but it is equal to $5$, a contradiction.
\end{proof}

From this and Lemma \ref{dci}, we see that $A$ has statistics $(22,72,180,80,0,0,0)$. In particular, we have $2\alpha_2(A)+\alpha_3(A) = 2$, which by double counting (cf. \eqref{alpha-1}) shows that for every line of the form $12222*$ (or a reflection or permutation thereof) intersects $A$ in exactly two points. Note that such lines connect a ``$d$'' point to two ``$c$'' points.

Also, we observe that two adjacent ``$d$'' points, such as $111222$ and $113222$, cannot both lie in $A$; for this would force the $*13***$ and $*11***$ slices to have statistics $(4,3,3,1)$ or $(3,3,3,1)$ by Corollary \ref{slic}, which forces $a(*1****)=6$, and thus $*1****$ must be good by Corollary \ref{slic2}; but this contradicts Lemma \ref{sixt}. Since $\alpha_3(A)=1/2$, we conclude that given any two adjacent ``$d$'' points, exactly one of them lies in $A$. In particular, the d points of the form $***222$ consist either of those strings with an even number of $1$s, or those with an odd number of $1$s.

Let's say it's the former, thus the set contains $111222, 133222$, and permutations of the first three coordinates, but omits $113222, 333222$ and permutations of the first three coordinates. Since the ``$d$'' points $113222, 333222$ are omitted, we conclude that the ``$c$'' points $113122, 113322, 333122, 333322$ must lie in the set, and similarly for permutations of the first three and last three coordinates. But this gives at least $15$ of the $16$ ``$c$'' points ending in $22$; by symmetry this leads to $225$ $c$-points in all; but $c(A)=180$, contradiction. This (finally!) completes the proof that $c'_{6,3}=353$.

Fujimura.tex

2009-07-14T06:39:18Z

Mpeake:

\section{Fujimura's problem}\label{fujimura-sec}
Let $\overline{c}^\mu_n$ be the size of the largest subset of the trianglular grid
$$\Delta_n := \{(a,b,c)\in {\mathbb Z}^3_+ : a+b+c = n\}$$
which contains no equilateral triangles $(a+r,b,c), (a,b+r,c), (a,b,c+r)$ with $r>0$. These are upward-pointing equilateral triangles. We shall refer to such sets as 'triangle-free'.
(Kobon Fujimura is a prolific inventor of puzzles, and in [http://www.puzzles.com/PuzzlePlayground/CoinsAndTriangles/CoinsAndTriangles.htm this] puzzle asked the related question of eliminating all equilateral triangles.)

The following table was formed mostly by computer searches for optimal solutions. We also found human proofs for most of them (see [http://michaelnielsen.org/polymath1/index.php?title=Fujimura's_problem here]).

\begin{figure}\centerline{
\begin{tabular}[l|lllllllllllll]
$n$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13\\
\hline $\overline{c}^\mu_n$ & 1 & 2 & 4 & 6 & 9 & 12 & 15 & 18 & 22 & 26 & 31 & 35 & 40 & 46
\end{tabular}\label{lowFujimura}\caption{Fujimura numbers}}

For any equilateral triangle $(a+r,b,c)$,$(a,b+r,c)$ and $(a,b,c+r)$, the value $y+2z$ forms an arithmetic progression of length 3. A Behrend set is a finite set of integers with no arithmetic progression of length 3 (see [http://arxiv.org/PS_cache/arxiv/pdf/0811/0811.3057v2.pdf this paper]). By looking at those triples (a,b,c) with a+2b inside a Behrend set, one can obtain the lower bound of $\overline{c}^\mu_n \geq n^2 exp(-O(\sqrt{\log n}))$.

It can be shown by a 'corners theorem' of Ajtai and Szemeredi that $\overline{c}^\mu_n = o(n^2)$ as $n \rightarrow \infty$.

An explicit lower bound is $3(n-1)$, made of all points in $\Delta_n$ with exactly one coordinate equal to zero.

An explicit upper bound comes from counting the triangles. There are $\binom{n+2}{3} triangles, and each point belongs to n of them. So you must remove at least (n+2)(n+1)/6 points to remove all triangles, leaving (n+2)(n+1)/3 points as an upper bound for $\overline{c}^mu_n$.

Introduction.tex

2009-07-14T06:32:40Z

Mpeake:

\section{Introduction}

For any integers $k \geq 1$ and $n \geq 0$, let $[k] := \{1,\ldots,k\}$, and define $[k]^n$ to be the cube of words of length $n$ with alphabet in $[k]$. Thus for instance $[3]^2 = \{11,12,13,21,22,23,31,32,33\}$.

We define a \emph{combinatorial line} in $[k]^n$ to be a set of the form $\{ w(i): i = 1,\ldots,k\} \subset [k]^n$, where $w \in ([k] \cup \{x\})^n \backslash [k]^n$ is a word of length $n$ with alphabet in $[k]$ together with a ``wildcard'' letter $x$ which appears at least once, and $w(i) \in [k]^n$ is the word obtained from $w$ by replacing $x$ by $i$; we often abuse notation and identify $w$ with the combinatorial line $\{ w(i): i = 1,\ldots,k\}$ it generates. Thus for instance, in $[3]^2$ we have $x2 = \{12,22,32\}$ and $xx = \{11,22,33\}$ as typical examples of combinatorial lines. In general, $[k]^n$ has $k^n$ words and $(k+1)^n-k^n$ lines.

\begin{figure}[tb]
\centerline{\includegraphics{AllLinesIn3n.pdf}}
\caption{Combinatorial lines in $[3]^2$.}
\label{fig-line}
\end{figure}

A set $A \subset [k]^n$ is said to be \emph{line-free} if it contains no combinatorial lines. Define the \emph{$(n,k)$ density Hales-Jewett number} $c_{n,k}$ to be the maximum cardinality $|A|$ of a line-free subset of $[k]^n$. Clearly, one has the trivial bound $c_{n,k} \leq k^n$. A deep theorem of Furstenberg and Katznelson~\cite{fk1}, \cite{fk2} asserts that this bound can be asymptotically improved:

\begin{theorem}[Density Hales-Jewett theorem]\label{dhj} For any fixed $k \geq 2$, one has $\lim_{n \to \infty} c_{n,k}/k^n = 0$.
\end{theorem}

\begin{remark} The difficulty of this theorem increases with $k$. For $k=1$, one clearly has $c_{n,1}=1$. For $k=2$, a classical theorem of Sperner~\cite{sperner} asserts, in our language, that $c_{n,2} = \binom{n}{\lfloor n/2\rfloor}$. The case $k=3$ is already non-trivial (for instance, it implies Roth's theorem~\cite{roth} on arithmetic progressions of length three) and was first established in \cite{fk1} (see also \cite{mcc}). The case of general $k$ was first established in~\cite{fk2} and has a number of implications, in particular implying Szemer\'edi's theorem~\cite{szem} on arithmetic progressions of arbitrary length.
\end{remark}

The Furstenberg-Katznelson proof of Theorem~\ref{dhj} relied on ergodic-theory techniques and did not give an explicit decay rate for $c_{n,k}$. Recently, two further proofs of this theorem have appeared, by Austin~\cite{austin} and by the sister Polymath project to this one~\cite{poly}. The proof of~\cite{austin} also used ergodic theory, but the proof in~\cite{poly} was combinatorial and gave effective bounds for $c_{n,k}$ in the limit $n \to \infty$. {\bf Note - need to ask what those explicit bounds are!} However, these bounds are not believed to be sharp, and in any case are only non-trivial in the asymptotic regime when $n$ is sufficiently large depending on $k$.

Our first result is the following asymptotic lower bound. The construction is based on the recent refinements \cite{elkin,greenwolf,obryant} of a well-known construction of Behrend~\cite{behrend} and Rankin~\cite{rankin}. The proof of Theorem~\ref{dhj-lower} is in Section~\ref{dhj-lower-sec}. Let $r_k(n)$ be the maximum size of a subset of $[n]$ that does not contain a $k$-term arithmetic progression.

\begin{theorem}[Asymptotic lower bound for $c_{n,k}$]\label{dhj-lower} For each $k\geq 3$, there is an absolute constant $C>0$ such that
\[ c_{n,k} \geq C k^n \left(\frac{r_k(\sqrt{n})}{\sqrt{n}}\right)^{k-1} = k^n \exp\left( - O(\sqrt[\ell]{\log n}) \right), \]
where $\ell$ is the largest integer satisfying $2k>2^{\ell}$. Specifically,
\[ c_{n,k} \geq C k^{n-\alpha(k) \sqrt[\ell]{\log n} + \beta(k) \log\log n},\]
where all logarithms are base-$k$, and $\alpha(k) = (\log 2)^{1-1/\ell} \ell 2^{(\ell-1)/2-1/\ell}$ and $\beta(k)=(k-1)/(2\ell)$.
\end{theorem}

In the case of small $n$, we focus primarily on the first non-trivial case $k=3$. We have computed the following explicit values of $c_{n,3}$:

\begin{theorem}[Explicit values of $c_{n,3}$]\label{dhj-upper} We have $c_{0,3} = 1$, $c_{1,3} = 2$, $c_{2,3} = 6$, $c_{3,3} = 18$, $c_{4,3} = 52$, $c_{5,3}=150$, and $c_{6,3}=450$. (This has been entered in the OEIS~\cite{oeis} as A156762.)
\end{theorem}

This result is established in Sections~\ref{dhj-lower-sec},~\ref{dhj-upper-sec}. Initially these results were established by an integer program (see Appendix~\ref{integer-sec}, but we provide completely computer-free proofs here. The constructions used in Section~\ref{dhj-lower-sec} give reasonably efficient constructions for larger values of $n$; for instance, they show that $3^{99} \leq c_{100,3} \leq 2 \times 3^{99}$. See Section~\ref{dhj-lower-sec} for further discussion.

We also have several partial results for higher values of $k$: see Section~\ref{higherk-sec}. For results on the closely related \emph{Hales-Jewett numbers} $HJ(k,r)$, see Section~\ref{coloring-sec}.

A variant of the density Hales-Jewett theorem has also been studied in the literature. Define a \emph{geometric line} in $[k]^n$ to be any set of the form $\{ a+ir: i=1,\ldots,k\}$ in $[k]^n$, where we identify $[k]^n$ with a subset of $\Z^n$, and $a, r \in \Z^n$ with $r \neq 0$. Equivalently, a geometric line takes the form $\{ w( i, k+1-i ): i =1,\ldots,k \}$, where $w \in ([k] \cup \{x,\overline{x}\})^n \backslash [k]^n$ is a word of length $n$ using the numbers in $[k]$ and two wildcards $x, \overline{x}$ as the alphabet, with at least one wildcard occuring in $w$, and $w(i,j) \in [k]^n$ is the word formed by substituting $i,j$ for $x,\overline{x}$ respectively. Figure~\ref{fig-geomline} shows the eight geometric lines in $[3]^2$. Clearly every combinatorial line is a geometric line, but not conversely. In general, $[k]^n$ has $((k+2)^n-k^n)/2$ geometric lines.

\begin{figure}[tb]
\centerline{\includegraphics{AllGeomLinesIn3n.pdf}}
\caption{Geometric lines in $[3]^2$.}
\label{fig-geomline}
\end{figure}

Define a \emph{Moser set} in $[k]^n$ to be a subset of $[k]^n$ that contains no geometric lines, and let $c'_{n,k}$ be the maximum cardinality $|A|$ of a Moser set in $[k]^n$. Clearly one has $c'_{n,k} \leq c_{n,k}$, so in particular from Theorem~\ref{dhj} one has $c'_{n,k}/k^n \to 0$ as $n \to \infty$. (Interestingly, there is no known proof of this fact that does not go through Theorem~\ref{dhj}, even for $k=3$.) Again, $k=3$ is the first non-trivial case: it is clear that $c'_{n,1}=0$ and $c'_{n,2}=1$ for all $n$.

The question of computing $c'_{n,3}$ was first posed by Moser~\cite{moser}. Prior to our work, the values
$$ c'_{0,3}=1; c'_{1,3}=2; c'_{2,3}=6; c'_{3,3}=16; c'_{4,3}=43$$
were known~\cite{chvatal2},~\cite{chandra} (this is Sequence A003142 in the OEIS~\cite{oeis}). We extend this sequence slightly:

\begin{theorem}[Values of $c'_{n,3}$ for small $n$]\label{moser} We have $c'_{0,3} = 1$, $c'_{1,3} = 2$, $c'_{2,3} = 6$, $c'_{3,3} = 16$, $c'_{4,3} = 43$, $c'_{5,3} = 124$, and $c'_{6,3} = 353$.
\end{theorem}

This result is established in Sections \ref{moser-lower-sec}, \ref{moser-upper-sec}. The arguments given here are computer-assisted; however, we have found alternate (but lengthier) computer-free proofs for the above claims with the the exception of the proof of $c'_{6,3}=353$, which requires one non-trivial computation (Lemma \ref{paretop-4}). These alternate proofs are not given in this paper to save space, but can be found on the wiki for this project at

\centerline{{\tt http://michaelnielsen.org/polymath1/index.php?title=Main\_Page}.}

We establish a lower bound for this problem of $(2+o(1))\binom{n}{i}2^i\leq c'_{n,3}$, which is maximized for $i$ near $2n/3$. This bound is around one-third better than the literature. We also give methods to improve on this construction.

Earlier lower bounds were known:
let $A(n,d)$ denote the size of the largest binary code of length $n$
and minimal distance $d$. Then
\begin{equation}\label{cnchvatalintro}
c'_{n,3}\geq \max_k \left( \sum_{j=0}^k \binom{n}{j} A(n-j, k-j+1)\right).
\end{equation}
which, with $A(n,1)=2^n$ and $A(n,2)=2^{n-1}$, implies in particular that
\begin{equation}\label{binom}
c'_{n,3} \geq \binom{n+1}{\lfloor \frac{2n+1}{3} \rfloor} 2^{\lfloor \frac{2n+1}{3} \rfloor - 1}
\end{equation}
for $n \geq 2$. This bound is not quite optimal; for instance, it gives a lower bound of $c'_{6,3}=344$.

\begin{remark} Let $c''_{n,3}$ be the size of the largest subset of ${\Bbb F}_3^n$ which contains no lines $x, x+r, x+2r$ with $x,r \in {\mathbb F}_3^n$ and $r \neq 0$, where ${\mathbb F}_3$ is the field of three elements. Clearly one has $c''_{n,3} \leq c'_{n,3} \leq c_{n,3}$. It is known that
$$ c''_{0,3}=1; c''_{1,3}=2; c''_{2,3}=4; c'_{3,3}=9; c'_{4,3}=20; c''_{5,3}=45; c''_{6,3} = 112;$$
see \cite{potenchin}.
\end{remark}

As mentioned earlier, the sharp bound on $c_{n,2}$ comes from Sperner's theorem. It is known that Sperner's theorem can be refined to the \emph{Lubell-Meshalkin-Yamamoto (LMY) inequality}, which in our language asserts that
$$
\sum_{a_1,a_2 \geq 0; a_1+a_2 = n} \frac{|A \cap \Gamma_{a_1,a_2}|}{|\Gamma_{a_1,a_2}|} \leq 1
$$
for any line-free subset $A \subset [2]^n$, where the \emph{cell} $\Gamma_{a_1,\ldots,a_k} \subset [k]^n$ is the set of words in $[k]^n$ which contain exactly $a_i$ $i$'s for each $i=1,\ldots,k$. It is natural to ask whether this inequality can be extended to higher $k$. Let $\Delta_{k,n}$ denote the set of all tuples $(a_1,\ldots,a_k)$ of non-negative integers summing to $n$, define a \emph{simplex} to be a set of $k$ points in $\Delta_{k,n}$ of the form
$(a_1+r,a_2,\ldots,a_k), (a_1,a_2+r,\ldots,a_k),\ldots,(a_1,a_2,\ldots,a_k+r)$ for some $0 < r \leq n$ and $a_1,\ldots,a_k$ summing to $n-r$, and define a \emph{Fujimura set} to be a subset $B \subset \Delta_{k,n}$ which contains no simplices. Observe that if $w$ is a combinatorial line in $[k]^n$, then
$$ w(1) \in \Gamma_{a_1+r,a_2,\ldots,a_k}, w(2) \in \Gamma_{a_1,a_2+r,\ldots,a_k}, \ldots, w(k) = \Gamma_{a_1,a_2,\ldots,a_k+r}$$
for some simplex $(a_1+r,a_2,\ldots,a_k), (a_1,a_2+r,\ldots,a_k),\ldots,(a_1,a_2,\ldots,a_k+r)$. Thus, if $B$ is a Fujimura set, then $A := \bigcup_{\vec a \in B} \Gamma_{\vec a}$ is line-free. Note also that
$$
\sum_{\vec a \in \Delta_{k,n}} \frac{|A \cap \Gamma_{\vec a}|}{|\Gamma_{\vec a}|} = |B|.
$$
This motivates a ``hyper-optimistic'' conjecture:

{\bf a picture showing a Fujimura set?}

\begin{conjecture}\label{hoc} For any $k \geq 1$ and $n \geq 0$, and any line-free subset $A$ of $[k]^n$, one has
$$
\sum_{\vec a \in \Delta_{k,n}} \frac{|A \cap \Gamma_{\vec a}|}{|\Gamma_{\vec a}|} \leq c^\mu_{k,n},$$
where $c^\mu_{k,n}$ is the maximal size of a Fujimura set in $\Delta_{k,n}$.
\end{conjecture}

One can show that this conjecture for a fixed value of $k$ would imply Theorem \ref{dhj} for the same value of $k$, in much the same way that the LYM inequality is known to imply Sperner's theorem. The LYM inequality asserts that Conjecture \ref{hoc} is true for $k \leq 2$. As far as we know this conjecture could hold in $k=3$. However, we know that it fails for all even $k\ge 4$, see Section~\ref{higherk-sec}.

We are interested in removing all upward-pointing equilateral triangles from a triangular grid. Fujimura actually proposed the similar problem of removing all equilateral triangles at this website: www.puzzles.com/PuzzlePlayground/CoinsAndTriangles/CoinsAndTriangles.htm

In Section \ref{fujimura-sec} we give some bounds on these numbers. Explicit values are given for $\overline{c}^\mu_n$ up to $n=13$, and general upper and lower bounds are given for all $n$.

\subsection{Notation}\label{notation-sec}

There are several subsets of $[k]^n$ which will be useful in our analysis. We have already introduced combinatorial lines, geometric lines, and cells. One can generalise the notion of a combinatorial line to that of a \emph{combinatorial subspace} in $[k]^n$ of dimension $d$, which is indexed by a word $w$ in $([k] \cup \{x_1,\ldots,x_d\})^n$ containing at least one of each wildcard $x_1,\ldots,x_d$, and which forms the set $\{ w(i_1,\ldots,i_d): i_1,\ldots,i_d \in [k]\}$, where $w(i_1,\ldots,i_d) \in [k]^d$ is the word formed by replacing $x_1,\ldots,x_d$ with $i_1,\ldots,i_d$ respectively. Thus for instance, in $[3]^3$, we have the two-dimensional combinatorial subspace $xxy = \{111,112,113,221,222,223,331,332,333\}$. We similarly have the notion of a \emph{geometric subspace} in $[k]^n$ of dimension $d$, which is defined similarly but with $d$ wildcards $x_1,\ldots,x_d,\overline{x_1},\ldots,\overline{x_d}$, with at least one of either $x_i$ or $\overline{x_i}$ appearing in the word $w$ for each $1 \leq i \leq d$, and the space taking the form $\{ w(i_1,\ldots,i_d,k+1-i_1,\ldots,k+1-i_d): i_1,\ldots,i_d \in [k] \}$. Thus for instance $[3]^3$ contains the two-dimensional geometric subspace $x\overline{x}y = \{ 131, 132, 133, 221, 222, 223, 311, 312, 313\}$.

An important class of combinatorial subspaces in $[k]^n$ will be the \emph{slices} consisting of $n-1$ distinct wildcards and one fixed coordinate. We will denote the distinct wildcards here by asterisks, thus for instance in $[3]^3$ we have $2** = \{ 211, 212, 213, 221, 222, 223, 231, 232, 233\}$. Two slices are \emph{parallel} if their fixed coordinate are in the same position, thus for instance $1**$ and $2**$ are parallel, and one can subdivide $[k]^n$ into $k$ parallel slices, each of which is isomorphic to $[k]^{n-1}$. In the analysis of Moser slices with $k=3$, we will make a distinction between \emph{centre slices}, whose fixed coordinate is equal to $2$, and \emph{side slices}, in which the fixed coordinate is either $1$ or $3$, thus $[3]^n$ can be partitioned into one centre slice and two side slices.

Another important set in the study of $k=3$ Moser sets are the \emph{spheres} $S_{i,n} \subset [3]^n$, defined as those words in $[3]^n$ with exactly $n-i$ $2$'s (and hence $i$ letters that are $1$ or $3$). Thus for instance $S_{1,3} = \{ 122, 322, 212, 232, 221, 223\}$. Observe that $[3]^n = \bigcup_{i=0}^n S_{i,n}$, and each $S_{i,n}$ has cardinality $|S_{i,n}| = \binom{n}{i} 2^{i}$.

It is also convenient to subdivide each sphere $S_{i,n}$ into two components $S_{i,n} = S_{i,n}^o \cup S_{i,n}^e$, where $S_{i,n}^o$ are the words in $S_{i,n}$ with an odd number of $1$'s, and $S_{i,n}^e$ are the words with an even number of $1$'s. Thus for instance $S_{1,3}^o = \{122,212,221\}$ and $S_{1,3}^e = \{322,232,223\}$. Observe that for $i>0$, $S_{i,n}^o$ and $S_{i,n}^e$ both have cardinality $\binom{n}{i} 2^{i-1}$.

The \emph{Hamming distance} between two words $w,w'$ is the number of coordinates in which $w, w'$ differ, e.g. the Hamming distance between $123$ and $321$ is two. Note that $S_{i,n}$ is nothing more than the set of words whose Hamming distance from $2\ldots2$ is $i$, which justifies the terminology ``sphere''.

In the density Hales-Jewett problem, there are two types of symmetries on $[k]^n$ which map combinatorial lines to combinatorial lines (and hence line-free sets to line-free sets). The first is a permutation of the alphabet $[k]$; the second is a permutation of the $n$ coordinates. Together, this gives a symmetry group of order $k!n!$ on the cube $[k]^n$, which we refer to as the \emph{combinatorial symmetry group} of the cube $[k]^n$. Two sets which are related by an element of this symmetry group will be called (combinatorially) \emph{equivalent}, thus for instance any two slices are combinatorially equivalent.

For the analysis of Moser sets in $[k]^n$, the symmetries are a bit different. One can still permute the $n$ coordinates, but one is no longer free to permute the alphabet $[k]$. Instead, one can \emph{reflect} an individual coordinate, for instance sending each word $x_1 \ldots x_n$ to its reflection $x_1 \ldots x_{i-1} (k+1-x_i) x_{i+1} \ldots x_n$. Together, this gives a symmetry group of order $2^k n!$ on the cube $[k]^n$, which we refer to as the \emph{geometric symmetry group} of the cube $[k]^n$; this group maps geometric lines to geometric lines, and thus maps Moser sets to Moser sets. Two Moser sets which are related by an element of this symmetry group will be called (geometrically) \emph{equivalent}. For instance, a sphere $S_{i,n}$ is equivalent only to itself, and $S_{i,n}^o$, $S_{i,n}^e$ are equivalent only to each other.

Introduction.tex

2009-07-14T06:30:42Z

Mpeake:

\section{Introduction}

For any integers $k \geq 1$ and $n \geq 0$, let $[k] := \{1,\ldots,k\}$, and define $[k]^n$ to be the cube of words of length $n$ with alphabet in $[k]$. Thus for instance $[3]^2 = \{11,12,13,21,22,23,31,32,33\}$.

We define a \emph{combinatorial line} in $[k]^n$ to be a set of the form $\{ w(i): i = 1,\ldots,k\} \subset [k]^n$, where $w \in ([k] \cup \{x\})^n \backslash [k]^n$ is a word of length $n$ with alphabet in $[k]$ together with a ``wildcard'' letter $x$ which appears at least once, and $w(i) \in [k]^n$ is the word obtained from $w$ by replacing $x$ by $i$; we often abuse notation and identify $w$ with the combinatorial line $\{ w(i): i = 1,\ldots,k\}$ it generates. Thus for instance, in $[3]^2$ we have $x2 = \{12,22,32\}$ and $xx = \{11,22,33\}$ as typical examples of combinatorial lines. In general, $[k]^n$ has $k^n$ words and $(k+1)^n-k^n$ lines.

\begin{figure}[tb]
\centerline{\includegraphics{AllLinesIn3n.pdf}}
\caption{Combinatorial lines in $[3]^2$.}
\label{fig-line}
\end{figure}

A set $A \subset [k]^n$ is said to be \emph{line-free} if it contains no combinatorial lines. Define the \emph{$(n,k)$ density Hales-Jewett number} $c_{n,k}$ to be the maximum cardinality $|A|$ of a line-free subset of $[k]^n$. Clearly, one has the trivial bound $c_{n,k} \leq k^n$. A deep theorem of Furstenberg and Katznelson~\cite{fk1}, \cite{fk2} asserts that this bound can be asymptotically improved:

\begin{theorem}[Density Hales-Jewett theorem]\label{dhj} For any fixed $k \geq 2$, one has $\lim_{n \to \infty} c_{n,k}/k^n = 0$.
\end{theorem}

\begin{remark} The difficulty of this theorem increases with $k$. For $k=1$, one clearly has $c_{n,1}=1$. For $k=2$, a classical theorem of Sperner~\cite{sperner} asserts, in our language, that $c_{n,2} = \binom{n}{\lfloor n/2\rfloor}$. The case $k=3$ is already non-trivial (for instance, it implies Roth's theorem~\cite{roth} on arithmetic progressions of length three) and was first established in \cite{fk1} (see also \cite{mcc}). The case of general $k$ was first established in~\cite{fk2} and has a number of implications, in particular implying Szemer\'edi's theorem~\cite{szem} on arithmetic progressions of arbitrary length.
\end{remark}

The Furstenberg-Katznelson proof of Theorem~\ref{dhj} relied on ergodic-theory techniques and did not give an explicit decay rate for $c_{n,k}$. Recently, two further proofs of this theorem have appeared, by Austin~\cite{austin} and by the sister Polymath project to this one~\cite{poly}. The proof of~\cite{austin} also used ergodic theory, but the proof in~\cite{poly} was combinatorial and gave effective bounds for $c_{n,k}$ in the limit $n \to \infty$. {\bf Note - need to ask what those explicit bounds are!} However, these bounds are not believed to be sharp, and in any case are only non-trivial in the asymptotic regime when $n$ is sufficiently large depending on $k$.

Our first result is the following asymptotic lower bound. The construction is based on the recent refinements \cite{elkin,greenwolf,obryant} of a well-known construction of Behrend~\cite{behrend} and Rankin~\cite{rankin}. The proof of Theorem~\ref{dhj-lower} is in Section~\ref{dhj-lower-sec}. Let $r_k(n)$ be the maximum size of a subset of $[n]$ that does not contain a $k$-term arithmetic progression.

\begin{theorem}[Asymptotic lower bound for $c_{n,k}$]\label{dhj-lower} For each $k\geq 3$, there is an absolute constant $C>0$ such that
\[ c_{n,k} \geq C k^n \left(\frac{r_k(\sqrt{n})}{\sqrt{n}}\right)^{k-1} = k^n \exp\left( - O(\sqrt[\ell]{\log n}) \right), \]
where $\ell$ is the largest integer satisfying $2k>2^{\ell}$. Specifically,
\[ c_{n,k} \geq C k^{n-\alpha(k) \sqrt[\ell]{\log n} + \beta(k) \log\log n},\]
where all logarithms are base-$k$, and $\alpha(k) = (\log 2)^{1-1/\ell} \ell 2^{(\ell-1)/2-1/\ell}$ and $\beta(k)=(k-1)/(2\ell)$.
\end{theorem}

In the case of small $n$, we focus primarily on the first non-trivial case $k=3$. We have computed the following explicit values of $c_{n,3}$:

\begin{theorem}[Explicit values of $c_{n,3}$]\label{dhj-upper} We have $c_{0,3} = 1$, $c_{1,3} = 2$, $c_{2,3} = 6$, $c_{3,3} = 18$, $c_{4,3} = 52$, $c_{5,3}=150$, and $c_{6,3}=450$. (This has been entered in the OEIS~\cite{oeis} as A156762.)
\end{theorem}

This result is established in Sections~\ref{dhj-lower-sec},~\ref{dhj-upper-sec}. Initially these results were established by an integer program (see Appendix~\ref{integer-sec}, but we provide completely computer-free proofs here. The constructions used in Section~\ref{dhj-lower-sec} give reasonably efficient constructions for larger values of $n$; for instance, they show that $3^{99} \leq c_{100,3} \leq 2 \times 3^{99}$. See Section~\ref{dhj-lower-sec} for further discussion.

We also have several partial results for higher values of $k$: see Section~\ref{higherk-sec}. For results on the closely related \emph{Hales-Jewett numbers} $HJ(k,r)$, see Section~\ref{coloring-sec}.

A variant of the density Hales-Jewett theorem has also been studied in the literature. Define a \emph{geometric line} in $[k]^n$ to be any set of the form $\{ a+ir: i=1,\ldots,k\}$ in $[k]^n$, where we identify $[k]^n$ with a subset of $\Z^n$, and $a, r \in \Z^n$ with $r \neq 0$. Equivalently, a geometric line takes the form $\{ w( i, k+1-i ): i =1,\ldots,k \}$, where $w \in ([k] \cup \{x,\overline{x}\})^n \backslash [k]^n$ is a word of length $n$ using the numbers in $[k]$ and two wildcards $x, \overline{x}$ as the alphabet, with at least one wildcard occuring in $w$, and $w(i,j) \in [k]^n$ is the word formed by substituting $i,j$ for $x,\overline{x}$ respectively. Figure~\ref{fig-geomline} shows the eight geometric lines in $[3]^2$. Clearly every combinatorial line is a geometric line, but not conversely. In general, $[k]^n$ has $((k+2)^n-k^n)/2$ geometric lines.

\begin{figure}[tb]
\centerline{\includegraphics{AllGeomLinesIn3n.pdf}}
\caption{Geometric lines in $[3]^2$.}
\label{fig-geomline}
\end{figure}

Define a \emph{Moser set} in $[k]^n$ to be a subset of $[k]^n$ that contains no geometric lines, and let $c'_{n,k}$ be the maximum cardinality $|A|$ of a Moser set in $[k]^n$. Clearly one has $c'_{n,k} \leq c_{n,k}$, so in particular from Theorem~\ref{dhj} one has $c'_{n,k}/k^n \to 0$ as $n \to \infty$. (Interestingly, there is no known proof of this fact that does not go through Theorem~\ref{dhj}, even for $k=3$.) Again, $k=3$ is the first non-trivial case: it is clear that $c'_{n,1}=0$ and $c'_{n,2}=1$ for all $n$.

The question of computing $c'_{n,3}$ was first posed by Moser~\cite{moser}. Prior to our work, the values
$$ c'_{0,3}=1; c'_{1,3}=2; c'_{2,3}=6; c'_{3,3}=16; c'_{4,3}=43$$
were known~\cite{chvatal2},~\cite{chandra} (this is Sequence A003142 in the OEIS~\cite{oeis}). We extend this sequence slightly:

\begin{theorem}[Values of $c'_{n,3}$ for small $n$]\label{moser} We have $c'_{0,3} = 1$, $c'_{1,3} = 2$, $c'_{2,3} = 6$, $c'_{3,3} = 16$, $c'_{4,3} = 43$, $c'_{5,3} = 124$, and $c'_{6,3} = 353$.
\end{theorem}

This result is established in Sections \ref{moser-lower-sec}, \ref{moser-upper-sec}. The arguments given here are computer-assisted; however, we have found alternate (but lengthier) computer-free proofs for the above claims with the the exception of the proof of $c'_{6,3}=353$, which requires one non-trivial computation (Lemma \ref{paretop-4}). These alternate proofs are not given in this paper to save space, but can be found on the wiki for this project at

\centerline{{\tt http://michaelnielsen.org/polymath1/index.php?title=Main\_Page}.}

We establish a lower bound for this problem of $(2+o(1))\binom{n}{i}2^i\leq c'_{n,3}$, which is maximized for $i$ near $2n/3$. This bound is around one-third better than the literature. We also give methods to improve on this construction.

Earlier lower bounds were known:
let $A(n,d)$ denote the size of the largest binary code of length $n$
and minimal distance $d$. Then
\begin{equation}\label{cnchvatalintro}
c'_{n,3}\geq \max_k \left( \sum_{j=0}^k \binom{n}{j} A(n-j, k-j+1)\right).
\end{equation}
which, with $A(n,1)=2^n$ and $A(n,2)=2^{n-1}$, implies in particular that
\begin{equation}\label{binom}
c'_{n,3} \geq \binom{n+1}{\lfloor \frac{2n+1}{3} \rfloor} 2^{\lfloor \frac{2n+1}{3} \rfloor - 1}
\end{equation}
for $n \geq 2$. This bound is not quite optimal; for instance, it gives a lower bound of $c'_{6,3}=344$.

\begin{remark} Let $c''_{n,3}$ be the size of the largest subset of ${\Bbb F}_3^n$ which contains no lines $x, x+r, x+2r$ with $x,r \in {\mathbb F}_3^n$ and $r \neq 0$, where ${\mathbb F}_3$ is the field of three elements. Clearly one has $c''_{n,3} \leq c'_{n,3} \leq c_{n,3}$. It is known that
$$ c''_{0,3}=1; c''_{1,3}=2; c''_{2,3}=4; c'_{3,3}=9; c'_{4,3}=20; c''_{5,3}=45; c''_{6,3} = 112;$$
see \cite{potenchin}.
\end{remark}

As mentioned earlier, the sharp bound on $c_{n,2}$ comes from Sperner's theorem. It is known that Sperner's theorem can be refined to the \emph{Lubell-Meshalkin-Yamamoto (LMY) inequality}, which in our language asserts that
$$
\sum_{a_1,a_2 \geq 0; a_1+a_2 = n} \frac{|A \cap \Gamma_{a_1,a_2}|}{|\Gamma_{a_1,a_2}|} \leq 1
$$
for any line-free subset $A \subset [2]^n$, where the \emph{cell} $\Gamma_{a_1,\ldots,a_k} \subset [k]^n$ is the set of words in $[k]^n$ which contain exactly $a_i$ $i$'s for each $i=1,\ldots,k$. It is natural to ask whether this inequality can be extended to higher $k$. Let $\Delta_{k,n}$ denote the set of all tuples $(a_1,\ldots,a_k)$ of non-negative integers summing to $n$, define a \emph{simplex} to be a set of $k$ points in $\Delta_{k,n}$ of the form
$(a_1+r,a_2,\ldots,a_k), (a_1,a_2+r,\ldots,a_k),\ldots,(a_1,a_2,\ldots,a_k+r)$ for some $0 < r \leq n$ and $a_1,\ldots,a_k$ summing to $n-r$, and define a \emph{Fujimura set} to be a subset $B \subset \Delta_{k,n}$ which contains no simplices. Observe that if $w$ is a combinatorial line in $[k]^n$, then
$$ w(1) \in \Gamma_{a_1+r,a_2,\ldots,a_k}, w(2) \in \Gamma_{a_1,a_2+r,\ldots,a_k}, \ldots, w(k) = \Gamma_{a_1,a_2,\ldots,a_k+r}$$
for some simplex $(a_1+r,a_2,\ldots,a_k), (a_1,a_2+r,\ldots,a_k),\ldots,(a_1,a_2,\ldots,a_k+r)$. Thus, if $B$ is a Fujimura set, then $A := \bigcup_{\vec a \in B} \Gamma_{\vec a}$ is line-free. Note also that
$$
\sum_{\vec a \in \Delta_{k,n}} \frac{|A \cap \Gamma_{\vec a}|}{|\Gamma_{\vec a}|} = |B|.
$$
This motivates a ``hyper-optimistic'' conjecture:

{\bf a picture showing a Fujimura set?}

\begin{conjecture}\label{hoc} For any $k \geq 1$ and $n \geq 0$, and any line-free subset $A$ of $[k]^n$, one has
$$
\sum_{\vec a \in \Delta_{k,n}} \frac{|A \cap \Gamma_{\vec a}|}{|\Gamma_{\vec a}|} \leq c^\mu_{k,n},$$
where $c^\mu_{k,n}$ is the maximal size of a Fujimura set in $\Delta_{k,n}$.
\end{conjecture}

One can show that this conjecture for a fixed value of $k$ would imply Theorem \ref{dhj} for the same value of $k$, in much the same way that the LYM inequality is known to imply Sperner's theorem. The LYM inequality asserts that Conjecture \ref{hoc} is true for $k \leq 2$. As far as we know this conjecture could hold in $k=3$. However, we know that it fails for all even $k\ge 4$, see Section~\ref{fujimura-sec}.

We are interested in removing all upward-pointing equilateral triangles from a triangular grid. Fujimura actually proposed the similar problem of removing all equilateral triangles at this website: www.puzzles.com/PuzzlePlayground/CoinsAndTriangles/CoinsAndTriangles.htm

In Section \ref{fujimura-sec} we give some bounds on these numbers. Explicit values are given for $\overline{c}^\mu_n$ up to $n=13$, and general upper and lower bounds are given for all $n$.

\subsection{Notation}\label{notation-sec}

There are several subsets of $[k]^n$ which will be useful in our analysis. We have already introduced combinatorial lines, geometric lines, and cells. One can generalise the notion of a combinatorial line to that of a \emph{combinatorial subspace} in $[k]^n$ of dimension $d$, which is indexed by a word $w$ in $([k] \cup \{x_1,\ldots,x_d\})^n$ containing at least one of each wildcard $x_1,\ldots,x_d$, and which forms the set $\{ w(i_1,\ldots,i_d): i_1,\ldots,i_d \in [k]\}$, where $w(i_1,\ldots,i_d) \in [k]^d$ is the word formed by replacing $x_1,\ldots,x_d$ with $i_1,\ldots,i_d$ respectively. Thus for instance, in $[3]^3$, we have the two-dimensional combinatorial subspace $xxy = \{111,112,113,221,222,223,331,332,333\}$. We similarly have the notion of a \emph{geometric subspace} in $[k]^n$ of dimension $d$, which is defined similarly but with $d$ wildcards $x_1,\ldots,x_d,\overline{x_1},\ldots,\overline{x_d}$, with at least one of either $x_i$ or $\overline{x_i}$ appearing in the word $w$ for each $1 \leq i \leq d$, and the space taking the form $\{ w(i_1,\ldots,i_d,k+1-i_1,\ldots,k+1-i_d): i_1,\ldots,i_d \in [k] \}$. Thus for instance $[3]^3$ contains the two-dimensional geometric subspace $x\overline{x}y = \{ 131, 132, 133, 221, 222, 223, 311, 312, 313\}$.

An important class of combinatorial subspaces in $[k]^n$ will be the \emph{slices} consisting of $n-1$ distinct wildcards and one fixed coordinate. We will denote the distinct wildcards here by asterisks, thus for instance in $[3]^3$ we have $2** = \{ 211, 212, 213, 221, 222, 223, 231, 232, 233\}$. Two slices are \emph{parallel} if their fixed coordinate are in the same position, thus for instance $1**$ and $2**$ are parallel, and one can subdivide $[k]^n$ into $k$ parallel slices, each of which is isomorphic to $[k]^{n-1}$. In the analysis of Moser slices with $k=3$, we will make a distinction between \emph{centre slices}, whose fixed coordinate is equal to $2$, and \emph{side slices}, in which the fixed coordinate is either $1$ or $3$, thus $[3]^n$ can be partitioned into one centre slice and two side slices.

Another important set in the study of $k=3$ Moser sets are the \emph{spheres} $S_{i,n} \subset [3]^n$, defined as those words in $[3]^n$ with exactly $n-i$ $2$'s (and hence $i$ letters that are $1$ or $3$). Thus for instance $S_{1,3} = \{ 122, 322, 212, 232, 221, 223\}$. Observe that $[3]^n = \bigcup_{i=0}^n S_{i,n}$, and each $S_{i,n}$ has cardinality $|S_{i,n}| = \binom{n}{i} 2^{i}$.

It is also convenient to subdivide each sphere $S_{i,n}$ into two components $S_{i,n} = S_{i,n}^o \cup S_{i,n}^e$, where $S_{i,n}^o$ are the words in $S_{i,n}$ with an odd number of $1$'s, and $S_{i,n}^e$ are the words with an even number of $1$'s. Thus for instance $S_{1,3}^o = \{122,212,221\}$ and $S_{1,3}^e = \{322,232,223\}$. Observe that for $i>0$, $S_{i,n}^o$ and $S_{i,n}^e$ both have cardinality $\binom{n}{i} 2^{i-1}$.

The \emph{Hamming distance} between two words $w,w'$ is the number of coordinates in which $w, w'$ differ, e.g. the Hamming distance between $123$ and $321$ is two. Note that $S_{i,n}$ is nothing more than the set of words whose Hamming distance from $2\ldots2$ is $i$, which justifies the terminology ``sphere''.

In the density Hales-Jewett problem, there are two types of symmetries on $[k]^n$ which map combinatorial lines to combinatorial lines (and hence line-free sets to line-free sets). The first is a permutation of the alphabet $[k]$; the second is a permutation of the $n$ coordinates. Together, this gives a symmetry group of order $k!n!$ on the cube $[k]^n$, which we refer to as the \emph{combinatorial symmetry group} of the cube $[k]^n$. Two sets which are related by an element of this symmetry group will be called (combinatorially) \emph{equivalent}, thus for instance any two slices are combinatorially equivalent.

For the analysis of Moser sets in $[k]^n$, the symmetries are a bit different. One can still permute the $n$ coordinates, but one is no longer free to permute the alphabet $[k]$. Instead, one can \emph{reflect} an individual coordinate, for instance sending each word $x_1 \ldots x_n$ to its reflection $x_1 \ldots x_{i-1} (k+1-x_i) x_{i+1} \ldots x_n$. Together, this gives a symmetry group of order $2^k n!$ on the cube $[k]^n$, which we refer to as the \emph{geometric symmetry group} of the cube $[k]^n$; this group maps geometric lines to geometric lines, and thus maps Moser sets to Moser sets. Two Moser sets which are related by an element of this symmetry group will be called (geometrically) \emph{equivalent}. For instance, a sphere $S_{i,n}$ is equivalent only to itself, and $S_{i,n}^o$, $S_{i,n}^e$ are equivalent only to each other.

Moser-lower.tex

2009-07-13T12:17:59Z

Mpeake:

\section{Lower bounds for the Moser problem}\label{moser-lower-sec}

Just as for the DHJ problem, we found that Gamma sets $\Gamma_{a,b,c}$ were useful in providing large lower bounds for the Moser problem. This is despite the fact that the symmetries of the cube do not respect Gamma sets.

Observe that if $B \subset \Delta_n$, then the set $A_B := \bigcup_{\vec a \in B} \Gamma_{a,b,c}$ is a Moser set as long as $B$ does not contain any ``isosceles triangles'' $(a+r,b,c+s), (a+s,b,c+r), (a,b+r+s,c)$ for any $r,s \geq 0$ not both zero; in particular, $B$ cannot contain any ``vertical line segments'' $(a+r,b,c+r), (a,b+2r,c)$. An example of such a set is provided by selecting $0 \leq i \leq n-3$ and letting $B$ consist of the triples $(a, n-i, i-a)$ when $a \neq 3 \mod 3$, $(a,n-i-1,i+1-a)$ when $a \neq 1 \mod 3$, $(a,n-i-2,i+2-a)$ when $a=0 \mod 3$, and $(a,n-i-3,i+3-a)$ when $a=2 \mod 3$. Asymptotically, this set includes about two thirds of the spheres $S_{n,i}$, $S_{n,i+1}$ and one third of the spheres $S_{n,i+2}, S_{n,i+3}$ and (setting $i$ close to $n/3$) gives a lower bound \eqref{cpn3} with $C = 2 \times \sqrt{\frac{9}{4\pi}}$.

An integer program was run to obtain the optimal lower bounds achievable by the $A_B$ construction (using \eqref{cn3}, of course). The results for
$1 \leq n \leq 20$ are displayed in Figure \ref{nlow-moser}:

\begin{figure}[tb]
\centerline{
\begin{tabular}{|ll|ll|}
\hline
$n$ & lower bound & $n$ & lower bound \\
\hline
1 & 2 &11& 71766\\
2 & 6 & 12& 212423\\
3 & 16 & 13& 614875\\
4 & 43 & 14& 1794212\\
5 & 122& 15& 5321796\\
6 & 353& 16& 15455256\\
7 & 1017& 17& 45345052\\
8 & 2902&18& 134438520\\
9 & 8622&19& 391796798\\
10& 24786& 20& 1153402148\\
\hline
\end{tabular}}
\caption{Lower bounds for $c'_n$ obtained by the $A_B$ construction.}
\label{nlow-moser}
\end{figure}

More complete data, including the list of optimisers, can be found at
\centerline{{\tt http://abel.math.umu.se/~klasm/Data/HJ/}.}

Note that the lower bound $c'_{6,3} \geq 353$ was first located by a genetic algorithm: see Appendix \ref{genetic-alg}.

\begin{figure}[tb]
\centerline{\includegraphics{moser353new.png}}
\caption{One of the examples of $353$-point sets in $[3]^6$ (elements of the set being indicated by white squares).}
\label{moser353-fig}
\end{figure}

However, we have been unable to locate a lower bound which is asymptotically better than \eqref{cpn3}. Indeed, any method based purely on the $A_B$ construction cannot do asymptotically better than the previous constructions:

\begin{proposition} Let $B \subset \Delta_n$ be such that $A_B$ is a Moser set. Then $|A_B| \leq (2 \sqrt{\frac{9}{4\pi}} + o(1)) \frac{3^n}{\sqrt{n}}$.
\end{proposition}

\begin{proof} By the previous discussion, $B$ cannot contain any pair of the form $(a,b+2r,c), (a+r,b,c+r)$ with $r>0$. In other words, for any $-n \leq h \leq n$, $B$ can contain at most one triple $(a,b,c)$ with $c-a=h$. From this and \eqref{cn3}, we see that
$$ |A_B| \leq \sum_{h=-n}^n \max_{(a,b,c) \in \Delta_n: c-a=h} \frac{n!}{a! b! c!}.$$
From the Chernoff inequality (or the Stirling formula computation below) we see that $\frac{n!}{a! b! c!} \leq \frac{1}{n^{10}} 3^n$ unless $a,b,c = n/3 + O( n^{1/2} \log^{1/2} n )$, so we may restrict to this regime, which also forces $h = O( n^{1/2}\log^{1/2} n)$. If we write $a = n/3 + \alpha$, $b = n/3 + \beta$, $c = n/3+\gamma$ and apply Stirling's formula $n! = (1+o(1)) \sqrt{2\pi n} n^n e^{-n}$, we obtain
$$ \frac{n!}{a! b! c!} = (1+o(1)) \frac{3^{3/2}}{2\pi n} 3^n \exp( - (\frac{n}{3}+\alpha) \log (1 + \frac{3\alpha}{n} ) - (\frac{n}{3}+\beta) \log (1 + \frac{3\beta}{n} ) - (\frac{n}{3}+\gamma) \log (1 + \frac{3\gamma}{n} ) ).$$
From Taylor expansion one has
$$ -(\frac{n}{3}+\alpha) \log (1 + \frac{3\alpha}{n} ) = -\alpha - \frac{3}{2} \frac{\alpha^2}{n} + o(1)$$
and similarly for $\beta,\gamma$; since $\alpha+\beta+\gamma=0$, we conclude that
$$ \frac{n!}{a! b! c!} = (1+o(1)) \frac{3^{3/2}}{2\pi n} 3^n \exp( - \frac{3}{2n} (\alpha^2+\beta^2+\gamma^2) ).$$
If $c-a=h$, then $\alpha^2+\beta^2+\gamma^2 = \frac{3\beta^2}{2} + \frac{h^2}{2}$. Thus we see that
$$ \max_{(a,b,c) \in \Delta_n: c-a=h} \frac{n!}{a! b! c!} \leq (1+o(1)) \frac{3^{3/2}}{2\pi n} 3^n \exp( - \frac{3}{4n} h^2 ).$$
Using the integral test, we thus have
$$ |A_B| \leq (1+o(1)) \frac{3^{3/2}}{2\pi n} 3^n \int_\R \exp( - \frac{3}{4n} x^2 )\ dx.$$
Since $\int_\R \exp( - \frac{3}{4n} x^2 )\ dx = \sqrt{\frac{4\pi n}{3}}$, we obtain the claim.
\end{proof}

Actually it is possible to improve upon these bounds by a slight amount. Observe that if $B$ is a maximiser for the right-hand side of \eqref{cn3} (subject to $B$ not containing isosceles triangles), then any triple $(a,b,c)$ not in $B$ must be the vertex of a (possibly degenerate) isosceles triangle with the other vertices in $B$. If this triangle is non-degenerate, or if $(a,b,c)$ is the upper vertex of a degenerate isosceles triangle, then no point from $\Gamma_{a,b,c}$ can be added to $A_B$ without creating a geometric line. However, if $(a,b,c) = (a'+r,b',c'+r)$ is only the lower vertex of a degenerate isosceles triangle $(a'+r,b',c'+r), (a',b'+2r,c')$, then one can add any subset of $\Gamma_{a,b,c}$ to $A_B$ and still have a Moser set as long as no pair of elements in that subset is separated by Hamming distance $2r$. For instance, in the $n=5$ case, we can start with the 122-point set built from
$$B = \{ (0 0 5),(0 2 3),(1 1 3 ),(1 2 2 ),(2 2 1 ),(3 1 1 ),(3 2 0 ),(5 0 0))\}$$, and add a point each from (1 0 4) and (4 0 1).
This gives an example of the maximal, 124-point solution. Again, in the $n=10$ case, the set
\begin{align*}
B &= \{(0 0 10),(0 2 8 ),(0 3 7 ),(0 4 6 ),(1 4 5 ),(2 1 7 ),(2 3 5 ),
(3 2 5 ),(3 3 4 ),(3 4 3 ),(4 4 2 ),(5 1 4 ), \\
&\quad (5 3 2 ),(6 2 2 ),
(6 3 1 ),(6 4 0 ),(8 1 1 ),(9 0 1 ),(9 1 0 ) \}
\end{align*}

generates the lower bound $c'_{10,3} \geq 24786$ given above (and, up to reflection $a \leftrightarrow c$, is the only such set that does so); but by adding the following twelve elements from $\Gamma_{5,0,5}$ one can increase the lower bound slightly to $24798$: $1111133333$, $1111313333$,
$1113113333$, $1133331113$, $1133331131$, $1133331311$, $3311333111$,
$3313133111$, $3313313111$, $3331111133$, $3331111313$, $3331111331$

A more general form goes with the $B$ set described at the start of this section. Include points from $\Gamma_{(a,n-i-4,i+4-a)}$ when $a=1 (\operatorname{mod}\ 3)$, subject to no two points being included if they differ by the interchange of a $1$ and a $3$. Each of these Gamma sets is the feet of a degenerate isosceles triangle with vertex $\Gamma_{(a-1,n-i-2,a+3-a)}$.

\begin{lemma} If $A$ is a subset of $\Gamma_{(a,b,c)}$ such that no two points of $A$ differ by the interchange of a $1$ and a $3$, then $|A| \leq |\Gamma_{a,b,c}|/(1+\max(a,c))$.
\end{lemma}
\begin{proof}
Say that two points of $\Gamma_{a,b,c}$ are neighbours if they differ by the exchange of a $1$ and a $3$. Each point of $A$ has $ac$ neighbours, none of which are in $A$. Each point of $\Gamma_{(a,b,c)}\backslash A$ has $ac$ neighbours, but only $min(a,c)$ of them may be in $A$. So for each point of $A$ there are on average $ac/\min(a,c) = \max(a,c)$ points not in $A$. So the proportion of points of $\Gamma_{(a,b,c)}$ that are in $A$ is at most one in $1+\max(a,c)$.
\end{proof}

The proportion of extra points for each of the cells $\Gamma_{(a,n-i-4,i+4-a)}$ is no more than $2/(i+6)$. Only one cell in three is included from the $b=n-i-4$ layer, so we expect no more than $\binom{n}{i+4}2^{i+5}/(3i+18)$ new points, all from $S_{n,i+4}$. One can also find extra points from $S_{n,i+5}$ and higher spheres.

Earlier solutions may also give insight into the problem. In this section we discuss lower bounds for $c'_{n,3}$. Clearly we have
$c'_{0,3}=1$ and $c'_{1,3}=2$, so we focus on the case $n \ge 2$.
The first lower bounds may be due to Koml\'{o}s \cite{komlos}, who observed
that the sphere $S_{i,n}$ of elements with exactly $n-i$ 2 entries
(see Section \ref{notation-sec} for definition),
is a Moser set, so that
\begin{equation}\label{cin}
c'_{n,3}\geq \vert S_{i,n}\vert
\end{equation}
holds for all $i$. Choosing $i=\lfloor \frac{2n}{3}\rfloor$ and
applying Stirling's formula, we see that this lower bound takes the form
\begin{equation}\label{cpn3}
c'_{n,3} \geq (C-o(1)) 3^n / \sqrt{n}
\end{equation}
for some absolute constant $C>0$; in fact \eqref{cin} gives \eqref{cpn3} with $C := \sqrt{\frac{9}{4\pi}}$.
In particular $c'_{3,3} \geq 12, c'_{4,3}\geq 24, c'_{5,3}\geq 80,
c'_{6,3}\geq 240$.

These values can be improved by studying combinations of several spheres or
semispheres or applying elementary results from coding theory.

Observe that if $\{w(1),w(2),w(3)\}$ is a geometric line in $[3]^n$,
then $w(1), w(3)$ both lie in the same sphere $S_{i,n}$,
and that $w(2)$ lies in a lower
sphere $S_{i-r,n}$ for some $1 \leq r \leq i \leq n$. Furthermore,
$w(1)$ and $w(3)$ are separated by Hamming distance $r$.

As a consequence, we see that $S_{i-1,n} \cup S_{i,n}^e$ (or $S_{i-1,n}
\cup S_{i,n}^o$) is a Moser set for any $1 \leq i \leq n$, since any two
distinct elements $S_{i,n}^e$ are separated by a Hamming distance of at
least two.
(Recall Section \ref{notation-sec} for definitions),
This leads to the lower bound
\begin{equation}\label{cn3-low}
c'_{n,3} \geq \binom{n}{i-1}
2^{i-1} + \binom{n}{i} 2^{i-1} = \binom{n+1}{i} 2^{i-1}.
\end{equation}
It is not
hard to see that $\binom{n+1}{i+1} 2^{i} > \binom{n+1}{i} 2^{i-1}$ if
and only if $3i < 2n+1$, and so this lower bound is maximised when $i =
\lfloor \frac{2n+1}{3} \rfloor$ for $n \geq 2$, giving the formula
\eqref{binom}. This leads to the lower bounds $$ c'_{2,3} \geq 6;
c'_{3,3} \geq 16; c'_{4,3} \geq 40; c'_{5,3} \geq 120; c'_{6,3} \geq
336$$ which gives the right lower bounds for $n=2,3$, but is slightly
off for $n=4,5$. Asymptotically, Stirling's formula and \eqref{cn3-low} then
give the lower bound \eqref{cpn3} with $C = \frac{3}{2} \times \sqrt{\frac{9}{4\pi}}$, which is asymptotically $50\%$ better than the bound \eqref{cin}.

The work of Chv\'{a}tal \cite{chvatal1}
already contained a refinement of this idea
which we here translate into the usual notation of coding theory:
Let $A(n,d)$ denote the size of the largest binary code of length $n$
and minimal distance $d$.

Then
\begin{equation}\label{cnchvatal}
c'_{n,3}\geq \max_k \left( \sum_{j=0}^k \binom{n}{j} A(n-j, k-j+1)\right).
\end{equation}

With the following values for $A(n,d)$:
{\tiny{
\[
\begin{array}{llllllll}
A(1,1)=2&&&&&&&\\
A(2,1)=4& A(2,2)=2&&&&&&\\
A(3,1)=8&A(3,2)=4&A(3,3)=2&&&&&\\
A(4,1)=16&A(4,2)=8& A(4,3)=2& A(4,4)=2&&&&\\
A(5,1)=32&A(5,2)=16& A(5,3)=4& A(5,4)=2&A(5,5)=2&&&\\
A(6,1)=64&A(6,2)=32& A(6,3)=8& A(6,4)=4&A(6,5)=2&A(6,6)=2&&\\
A(7,1)=128&A(7,2)=64& A(7,3)=16& A(7,4)=8&A(7,5)=2&A(7,6)=2&A(7,7)=2&\\
A(8,1)=256&A(8,2)=128& A(8,3)=20& A(8,4)=16&A(8,5)=4&A(8,6)=2
&A(8,7)=2&A(8,8)=2\\
A(9,1)=512&A(9,2)=256& A(9,3)=40& A(9,4)=20&A(9,5)=6&A(9,6)=4
&A(9,7)=2&A(9,8)=2\\
A(10,1)=1024&A(10,2)=512& A(10,3)=72& A(10,4)=40&A(10,5)=12&A(10,6)=6
&A(10,7)=2&A(10,8)=2\\
A(11,1)=2048&A(11,2)=1024& A(11,3)=144& A(11,4)=72&A(11,5)=24&A(11,6)=12
&A(11,7)=2&A(11,8)=2\\
A(12,1)=4096&A(12,2)=2048& A(12,3)=256& A(12,4)=144&A(12,5)=32&A(12,6)=24
&A(12,7)=4&A(12,8)=2\\
A(13,1)=8192&A(13,2)=4096& A(13,3)=512& A(13,4)=256&A(13,5)=64&A(12,6)=32
&A(13,7)=8&A(13,8)=4\\
\end{array}
\]
}}

Generally, $A(n,1)=2^n, A(n,2)=2^{n-1}, A(n-1,2e-1)=A(n,2e), A(n,d)=2$,
if $d>\frac{2n}{3}$.
The values were taken or derived from Andries Brower's table at
\centerline{\tt http://www.win.tue.nl/$\sim$aeb/codes/binary-1.html}
\textbf{include to references? or other book with explicit values of $A(n,d)$ }

For $c'_{n,3}$ we obtain the following lower bounds:
with $k=2$
\[
\begin{array}{llll}
c'_{4,3}&\geq &\binom{4}{0}A(4,3)+\binom{4}{1}A(3,2)+\binom{4}{2}A(2,1)
=1\cdot 2+4 \cdot 4+6\cdot 4&=42.\\
c'_{5,3}&\geq &\binom{5}{0}A(5,3)+\binom{5}{1}A(4,2)+\binom{5}{2}A(3,1)
=1\cdot 4+5 \cdot 8+10\cdot 8&=124.\\
c'_{6,3}&\geq &\binom{6}{0}A(6,3)+\binom{6}{1}A(5,2)+\binom{6}{2}A(4,1)
=1\cdot 8+6 \cdot 16+15\cdot 16&=344.
\end{array}
\]
With k=3
\[
\begin{array}{llll}
c'_{7,3}&\geq& \binom{7}{0}A(7,4)+\binom{7}{1}A(6,3)+\binom{7}{2}A(5,2)
+ \binom{7}{3}A(4,1)&=960.\\
c'_{8,3}&\geq &\binom{8}{0}A(8,4)+\binom{8}{1}A(7,3)+\binom{8}{2}A(6,2)
+ \binom{8}{3}A(5,1)&=2832.\\
c'_{9,3}&\geq & \binom{9}{0}A(9,4)+\binom{9}{1}A(8,3)+\binom{9}{2}A(7,2)
+ \binom{9}{3}A(6,1)&=7880.
\end{array}\]
With k=4
\[
\begin{array}{llll}
c'_{10,3}&\geq &\binom{10}{0}A(10,5)+\binom{10}{1}A(9,4)+\binom{10}{2}A(8,3)
+ \binom{10}{3}A(7,2)+\binom{10}{4}A(6,1)&=22232.\\
c'_{11,3}&\geq &\binom{11}{0}A(11,5)+\binom{11}{1}A(10,4)+\binom{11}{2}A(9,3)
+ \binom{11}{3}A(8,2)+\binom{11}{4}A(7,1)&=66024.\\
c'_{12,3}&\geq &\binom{12}{0}A(12,5)+\binom{12}{1}A(11,4)+\binom{12}{2}A(10,3)
+ \binom{12}{3}A(9,2)+\binom{12}{4}A(8,1)&=188688.\\
\end{array}\]
With $k=5$
\[ c'_{13,3}\geq 539168.\]

It should be pointed out that these bounds are even numbers, so that
$c'_{4,3}=43$ shows that one cannot generally expect this lower bound
gives the optimum.

The maximum value appears to occur for $k=\lfloor\frac{n+2}{3}\rfloor$,
so that using
Stirling's formula and explicit bounds on $A(n,d)$ the
best possible value known to date
of the constant $C$ in
equation \eqref{cpn3} can be worked out, but we refrain from doing this here.
Using the Singleton bound $A(n,d)\leq 2^{n-d+1}$ Chv\'{a}tal
\cite{chvatal1} proved that the expression on the right hand side of
\eqref{cnchvatal} is also
$O\left( \frac{3^n}{\sqrt{n}}\right)$, so that the refinement described
above gains a constant factor over the initial construction only.

For $n=4$ the above does not yet give the exact value.
The value $c'_{4,3}=43$ was first proven by Chandra \cite{chandra}.
A uniform way of describing examples for the optimum values of
$c'_{4,3}=43$ and $c'_{5,3}=124$ is the following:

Let us consider the sets $$ A := S_{i-1,n}
\cup S_{i,n}^e \cup A'$$ where $A' \subset S_{i+1,n}$ has the property
that any two elements in $A'$ are separated by a Hamming distance of at
least three, or have a Hamming distance of exactly one but their
midpoint lies in $S_{i,n}^o$. By the previous discussion we see that
this is a Moser set, and we have the lower bound
\begin{equation}\label{cnn}
c'_{n,3} \geq \binom{n+1}{i} 2^{i-1} + |A'|.
\end{equation} This gives some improved lower bounds for $c'_{n,3}$:

\begin{itemize} \item By taking $n=4$, $i=3$, and $A' = \{ 1111, 3331,
3333\}$, we obtain $c'_{4,3} \geq 43$; \item By taking $n=5$, $i=4$, and
$A' = \{ 11111, 11333, 33311, 33331 \}$, we obtain $c'_{5,3} \geq 124$.
\item By taking $n=6$, $i=5$, and $A' = \{ 111111, 111113, 111331,
111333, 331111, 331113\}$, we obtain $c'_{6,3} \geq 342$. \end{itemize}

This gives the lower bounds in Theorem \ref{moser} up to $n=5$, but
the bound for $n=6$ is inferior to the lower bound $c'_{6,3}\geq 344$ given above.

\subsection{Higher $k$ values}
One can consider subsets of $[k]{}^n$ that contain no geometric lines. Section \ref{moser-lower-sec} has considered the case $k=3$. Let $c'_{n,k}$ be the greatest number of points in $[k]{}^n$ with no geometric line. For example, $c'_{n,3} = c'_n$. We have the following lower bounds:

$c'_{n,4} \ge \binom{n}{n/2}2^n$.
The set of points with $a$ $1$s,$b$ $2$s,$c$ $3$s and $d$ $4$s, where $a+d$ has the constant value $n/2$, does not form geometric lines because points at the ends of a geometric line have more $a$ or $d$ values than points in the middle of the line.

One can show a lower bound that, asymptotically, is twice as large. Take all points with $a$ $1$s, $b$ $2$s, $c$ $3$s and $d$ $4$s, for which:

\begin{itemize}
\item Either $a+d = q$ or $q-1$, $a$ and $b$ have the same parity; or
\item $a+d = q-2$ or $q-3$, $a$ and $b$ have opposite parity.
\end{itemize}

This includes half the points of four adjacent layers, and therefore may include $(1+o(1))\binom{n}{n/2}2^{n+1}$ points.

We also have a DHJ(3)-like lower bound for $c'_{n,5}$, namely $c'_{n,5} = 5^{n-O(\sqrt{\log n})}$.
Consider points with $a$ $1$s, $b$ $2$s, $c$ $3$s, $d$ $4$s and $e$ $5$s. For each point, take the value $a+e+2(b+d)+3c$. The first three points in any geometric line give values that form an arithmetic progression of length three.

Select a set of integers with no arithmetic progression of length 3. Select all points whose value belongs to that sequence; there will be no geometric line among those points. By Behrend theory, it is possible to choose these points with density $\exp{-O(\sqrt{\log n})}$.

The $k=6$ version of Moser implies DHJ(3). Indeed, any $k=3$ combinatorial line-free set can be "doubled up" into a $k=6$ geometric line-free set of the same density by pulling back the set from the map that maps 1, 2, 3, 4, 5, 6 to 1, 2, 3, 3, 2, 1 respectively; note that this map sends $k=6$ geometric lines to $k=3$ combinatorial lines. So $c'_{k,6} \geq 2^k c_k$, and more generally, $c'_{k,2n} \geq 2^k c_{k,n}$.

Outline of second paper

2009-07-13T08:28:52Z

Mpeake: /* Abstract */

Here is a proposed outline of the second paper, which will focus on the new bounds on DHJ(3) and Moser numbers, and related quantities.

== Metadata ==

* '''Author''': D.H.J. Polymath
* '''Address''': http://michaelnielsen.org/polymath1/index.php (Do we need a more stable address?)
* '''Email''': ???
* '''Title''': Density Hales-Jewett and Moser numbers
* '''AMS Subject classification''': ???

== Abstract ==

(A draft proposal - please edit)

For any <math>n \geq 0</math> and <math>k \geq 1</math>, the density Hales-Jewett number <math>c_{n,k}</math> is defined as the size of the largest subset of the cube <math>[k]^n := \{1,\ldots,k\}^n</math> which contains no combinatorial line; similarly, the Moser number <math>c'_{n,k}</math> is the largest subset of the cube <math>[k]^n</math> which contains no geometric line. A deep theorem of Furstenberg and Katznelson [cite] shows that <math>c_{n,k} = o(k^n)</math> as <math>n \to \infty</math> (which implies a similar claim for <math>c'_{n,k}</math>; this is already non-trivial for <math>k=3</math>. Several new proofs of this result have also been recently established [cite Polymath], [cite Austin].

Using both human and computer-assisted arguments, we compute several values of <math>c_{n,k}</math> and <math>c'_{n,k}</math> for small <math>n,k</math>. For instance the sequence <math>c_{n,3}</math> for <math>n=0,\ldots,6</math> is <math>1,2,6,18,52,150,450</math>, while the sequence <math>c'_{n,3}</math> for <math>n=0,\ldots,6</math> is <math>1,2,6,16,43,124,353</math>. We also establish some results for higher <math>k</math>, showing for instance that an analogue of the LYM inequality (which relates to the <math>k=2</math> case) does not hold for higher <math>k</math>.

== Sections ==

=== Introduction ===

Basic definitions. Definitions and notational conventions include

* [k] = {1, 2, ..., k}
* Subsets of [k]^n are called A
* definition of combinatorial line, geometric line
* Hales-Jewett numbers, Moser numbers

History of and motivation for the problem:

* Sperner's theorem
* Density Hales-Jewett theorem, including new proofs
* Review literature on Moser problem

New results

* Computation of several values of <math>c_{n,3}</math>
* Computation of several values of <math>c'_{n,3}</math>
* Asymptotic lower bounds for <math>c_{n,k}</math>
* Genetic algorithm lower bounds
* Some bounds for <math>c_{n,k}</math> for low n and large k
* Connection between Moser(2k) and DHJ(k)
* Hyper-optimistic conjecture, and its failure
* New bounds for colouring Hales-Jewett numbers

=== Lower bounds for density Hales-Jewett ===

Fujimura implies DHJ lower bounds; some selected numerics (e.g. lower bounds up to 10 dimensions, plus a few dimensions afterwards).

The precise asymptotic bound of <math>c_{n,k} > C k^{n - \alpha(k)\sqrt[\ell]{\log n}+\beta(k) \log \log n}</math>

Discussion of genetic algorithm

=== Low-dimensional density Hales-Jewett numbers ===

==== Very small n ====

<math>n=0,1,2</math> are trivial. But the six-point examples will get mentioned a lot.

For <math>n=3</math>, one needs to classify the 17-point and 18-point examples.

==== n=4 ====

One needs to classify the 50-point, 51-point, and 52-point examples.

==== n=5 ====

This is the big section, showing there are no 151-point examples.

==== n=6 ====

Easy corollary of n=5 theory

=== Higher k DHJ numbers ===

Exact computations of <math>c_{2,k}, c_{3,k}</math>

Connection between Moser<math>(n,2k)</math> and DHJ<math>(n,k)</math>

Numerics

Failure of hyper-optimistic conjecture

=== Lower bounds for Moser ===

Using Gamma sets to get lower bounds

Adding extra points from degenerate triangles

Higher k; Implications between Moser and DHJ

=== Moser in low dimensions ===

There is some general slicing lemma that needs to be proved here that allows inequalities for low-dim Moser to imply inequalities for higher dim.

For n=0,1,2 the theory is trivial.

For n=3 we need the classification of Pareto optimal configurations etc. So far this is only done by computer brute force search; we may have to find a human version.

n=4 theory: include both computer results and human results

n=5: we have a proof using the n=4 computer data; we should keep looking for a purely human proof.

n=6: we can give the partial results we have.

=== Fujimura's problem ===

=== Coloring DHJ ===

== Files ==

* [[polymath.tex]]
* [[introduction.tex]]
* [[dhj-lown.tex]]
* [[dhj-lown-lower.tex]]
* [[moser.tex]]
* [[moser-lower.tex]]
* [[fujimura.tex]]
* [[higherk.tex]]
* [[genetic.tex]]
* [[integer.tex]]
* [[coloring.tex]]
* [http://terrytao.wordpress.com/files/2009/07/alllinesin3n.pdf A figure depicting combinatorial lines]
* [http://terrytao.wordpress.com/files/2009/07/allgeomlinesin3n.pdf A figure depicting geometric lines]
* [http://thomas1111.files.wordpress.com/2009/06/moser353new.png A figure depicting a 353-point 6D Moser set]

The above are the master copies of the LaTeX files. Below are various compiled versions of the source:

* [http://terrytao.wordpress.com/files/2009/05/polymath.pdf May 24 version]
* [http://terrytao.wordpress.com/files/2009/05/polymath1.pdf May 25 version]
* [http://terrytao.files.wordpress.com/2009/05/polymath2.pdf May 27 version]
* [http://terrytao.wordpress.com/files/2009/05/polymath3.pdf Jun 1 version]
* [http://terrytao.wordpress.com/files/2009/06/polymath.pdf Jun 3 version]
* [http://terrytao.files.wordpress.com/2009/06/polymath2.pdf Jun 18 version]
* [http://terrytao.wordpress.com/files/2009/07/polymath.pdf Jul 9 version]

Fujimura.tex

2009-07-13T04:20:15Z

Mpeake:

\section{Fujimura's problem}\label{fujimura-sec}
Let $\overline{c}^\mu_n$ be the size of the largest subset of the trianglular grid
$$\Delta_n := \{(a,b,c)\in {\mathbb Z}^3_+ : a+b+c = n\}$$
which contains no equilateral triangles $(a+r,b,c), (a,b+r,c), (a,b,c+r)$ with $r>0$. These are upward-pointing equilateral triangles. We shall refer to such sets as 'triangle-free'.
(Kobon Fujimura is a prolific inventor of puzzles, and in [http://www.puzzles.com/PuzzlePlayground/CoinsAndTriangles/CoinsAndTriangles.htm this] puzzle asked the related question of eliminating all equilateral triangles.)

The following table was formed mostly by computer searches for optimal solutions. We also found human proofs for most of them (see [http://michaelnielsen.org/polymath1/index.php?title=Fujimura's_problem here]).

\begin{figure}\centerline{
\begin{tabular}[l|lllllllllllll]
$n$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13\\
\hline $\overline{c}^\mu_n$ & 1 & 2 & 4 & 6 & 9 & 12 & 15 & 18 & 22 & 26 & 31 & 35 & 40 & 46
\end{tabular}\label{lowFujimura}\caption{Fujimura numbers}}

For any equilateral triangle $(a+r,b,c)$,$(a,b+r,c)$ and $(a,b,c+r)$, the value $y+2z$ forms an arithmetic progression of length 3. A Behrend set is a finite set of integers with no arithmetic progression of length 3 (see [http://arxiv.org/PS_cache/arxiv/pdf/0811/0811.3057v2.pdf this paper]). By looking at those triples (a,b,c) with a+2b inside a Behrend set, one can obtain the lower bound of $\overline{c}^\mu_n \geq n^2 exp(-O(\sqrt{\log n}))$.

It can be shown that $\overline{c}^\mu_n = o(n^2)$ as $n \rightarrow \infty$, but the details are outside the scope of this paper.

Moser-lower.tex

2009-07-12T14:18:31Z

Mpeake:

\section{Lower bounds for the Moser problem}\label{moser-lower-sec}

Just as for the DHJ problem, we found that Gamma sets $\Gamma_{a,b,c}$ were useful in providing large lower bounds for the Moser problem. This is despite the fact that the symmetries of the cube do not respect Gamma sets.

Observe that if $B \subset \Delta_n$, then the set $A_B := \bigcup_{\vec a \in B} \Gamma_{a,b,c}$ is a Moser set as long as $B$ does not contain any ``isosceles triangles'' $(a+r,b,c+s), (a+s,b,c+r), (a,b+r+s,c)$ for any $r,s \geq 0$ not both zero; in particular, $B$ cannot contain any ``vertical line segments'' $(a+r,b,c+r), (a,b+2r,c)$. An example of such a set is provided by selecting $0 \leq i \leq n-3$ and letting $B$ consist of the triples $(a, n-i, i-a)$ when $a \neq 3 \mod 3$, $(a,n-i-1,i+1-a)$ when $a \neq 1 \mod 3$, $(a,n-i-2,i+2-a)$ when $a=0 \mod 3$, and $(a,n-i-3,i+3-a)$ when $a=2 \mod 3$. Asymptotically, this set includes about two thirds of the spheres $S_{n,i}$, $S_{n,i+1}$ and one third of the spheres $S_{n,i+2}, S_{n,i+3}$ and (setting $i$ close to $n/3$) gives a lower bound \eqref{cpn3} with $C = 2 \times \sqrt{\frac{9}{4\pi}}$.

An integer program was run to obtain the optimal lower bounds achievable by the $A_B$ construction (using \eqref{cn3}, of course). The results for
$1 \leq n \leq 20$ are displayed in Figure \ref{nlow-moser}:

\begin{figure}[tb]
\centerline{
\begin{tabular}{|ll|ll|}
\hline
$n$ & lower bound & $n$ & lower bound \\
\hline
1 & 2 &11& 71766\\
2 & 6 & 12& 212423\\
3 & 16 & 13& 614875\\
4 & 43 & 14& 1794212\\
5 & 122& 15& 5321796\\
6 & 353& 16& 15455256\\
7 & 1017& 17& 45345052\\
8 & 2902&18& 134438520\\
9 & 8622&19& 391796798\\
10& 24786& 20& 1153402148\\
\hline
\end{tabular}}
\caption{Lower bounds for $c'_n$ obtained by the $A_B$ construction.}
\label{nlow-moser}
\end{figure}

More complete data, including the list of optimisers, can be found at
\centerline{{\tt http://abel.math.umu.se/~klasm/Data/HJ/}.}

Note that the lower bound $c'_{6,3} \geq 353$ was first located by a genetic algorithm: see Appendix \ref{genetic-alg}.

\begin{figure}[tb]
\centerline{\includegraphics{moser353new.png}}
\caption{One of the examples of $353$-point sets in $[3]^6$ (elements of the set being indicated by white squares).}
\label{moser353-fig}
\end{figure}

However, we have been unable to locate a lower bound which is asymptotically better than \eqref{cpn3}. Indeed, any method based purely on the $A_B$ construction cannot do asymptotically better than the previous constructions:

\begin{proposition} Let $B \subset \Delta_n$ be such that $A_B$ is a Moser set. Then $|A_B| \leq (2 \sqrt{\frac{9}{4\pi}} + o(1)) \frac{3^n}{\sqrt{n}}$.
\end{proposition}

\begin{proof} By the previous discussion, $B$ cannot contain any pair of the form $(a,b+2r,c), (a+r,b,c+r)$ with $r>0$. In other words, for any $-n \leq h \leq n$, $B$ can contain at most one triple $(a,b,c)$ with $c-a=h$. From this and \eqref{cn3}, we see that
$$ |A_B| \leq \sum_{h=-n}^n \max_{(a,b,c) \in \Delta_n: c-a=h} \frac{n!}{a! b! c!}.$$
From the Chernoff inequality (or the Stirling formula computation below) we see that $\frac{n!}{a! b! c!} \leq \frac{1}{n^{10}} 3^n$ unless $a,b,c = n/3 + O( n^{1/2} \log^{1/2} n )$, so we may restrict to this regime, which also forces $h = O( n^{1/2}\log^{1/2} n)$. If we write $a = n/3 + \alpha$, $b = n/3 + \beta$, $c = n/3+\gamma$ and apply Stirling's formula $n! = (1+o(1)) \sqrt{2\pi n} n^n e^{-n}$, we obtain
$$ \frac{n!}{a! b! c!} = (1+o(1)) \frac{3^{3/2}}{2\pi n} 3^n \exp( - (\frac{n}{3}+\alpha) \log (1 + \frac{3\alpha}{n} ) - (\frac{n}{3}+\beta) \log (1 + \frac{3\beta}{n} ) - (\frac{n}{3}+\gamma) \log (1 + \frac{3\gamma}{n} ) ).$$
From Taylor expansion one has
$$ -(\frac{n}{3}+\alpha) \log (1 + \frac{3\alpha}{n} ) = -\alpha - \frac{3}{2} \frac{\alpha^2}{n} + o(1)$$
and similarly for $\beta,\gamma$; since $\alpha+\beta+\gamma=0$, we conclude that
$$ \frac{n!}{a! b! c!} = (1+o(1)) \frac{3^{3/2}}{2\pi n} 3^n \exp( - \frac{3}{2n} (\alpha^2+\beta^2+\gamma^2) ).$$
If $c-a=h$, then $\alpha^2+\beta^2+\gamma^2 = \frac{3\beta^2}{2} + \frac{h^2}{2}$. Thus we see that
$$ \max_{(a,b,c) \in \Delta_n: c-a=h} \frac{n!}{a! b! c!} \leq (1+o(1)) \frac{3^{3/2}}{2\pi n} 3^n \exp( - \frac{3}{4n} h^2 ).$$
Using the integral test, we thus have
$$ |A_B| \leq (1+o(1)) \frac{3^{3/2}}{2\pi n} 3^n \int_\R \exp( - \frac{3}{4n} x^2 )\ dx.$$
Since $\int_\R \exp( - \frac{3}{4n} x^2 )\ dx = \sqrt{\frac{4\pi n}{3}}$, we obtain the claim.
\end{proof}

Actually it is possible to improve upon these bounds by a slight amount. Observe that if $B$ is a maximiser for the right-hand side of \eqref{cn3} (subject to $B$ not containing isosceles triangles), then any triple $(a,b,c)$ not in $B$ must be the vertex of a (possibly degenerate) isosceles triangle with the other vertices in $B$. If this triangle is non-degenerate, or if $(a,b,c)$ is the upper vertex of a degenerate isosceles triangle, then no point from $\Gamma_{a,b,c}$ can be added to $A_B$ without creating a geometric line. However, if $(a,b,c) = (a'+r,b',c'+r)$ is only the lower vertex of a degenerate isosceles triangle $(a'+r,b',c'+r), (a',b'+2r,c')$, then one can add any subset of $\Gamma_{a,b,c}$ to $A_B$ and still have a Moser set as long as no pair of elements in that subset is separated by Hamming distance $2r$. For instance, in the $n=5$ case, we can start with the 122-point set built from
$$B = \{ (0 0 5),(0 2 3),(1 1 3 ),(1 2 2 ),(2 2 1 ),(3 1 1 ),(3 2 0 ),(5 0 0))\}$$, and add a point each from (1 0 4) and (4 0 1).
This gives an example of the maximal, 124-point solution. Again, in the $n=10$ case, the set
\begin{align*}
B &= \{(0 0 10),(0 2 8 ),(0 3 7 ),(0 4 6 ),(1 4 5 ),(2 1 7 ),(2 3 5 ),
(3 2 5 ),(3 3 4 ),(3 4 3 ),(4 4 2 ),(5 1 4 ), \\
&\quad (5 3 2 ),(6 2 2 ),
(6 3 1 ),(6 4 0 ),(8 1 1 ),(9 0 1 ),(9 1 0 ) \}
\end{align*}

generates the lower bound $c'_{10,3} \geq 24786$ given above (and, up to reflection $a \leftrightarrow c$, is the only such set that does so); but by adding the following twelve elements from $\Gamma_{5,0,5}$ one can increase the lower bound slightly to $24798$: $1111133333$, $1111313333$,
$1113113333$, $1133331113$, $1133331131$, $1133331311$, $3311333111$,
$3313133111$, $3313313111$, $3331111133$, $3331111313$, $3331111331$

A more general form goes with the $B$ set described at the start of this section. Include points from $\Gamma_{(a,n-i-4,i+4-a)}$ when $a=1 (\operatorname{mod}\ 3)$, subject to no two points being included if they differ by the interchange of a $1$ and a $3$. Each of these Gamma sets is the feet of a degenerate isosceles triangle with vertex $\Gamma_{(a-1,n-i-2,a+3-a)}$.

\begin{lemma} If $A$ is a subset of $\Gamma_{(a,b,c)}$ such that no two points of $A$ differ by the interchange of a $1$ and a $3$, then $|A| \leq |\Gamma_{a,b,c}|/(1+\max(a,c))$.
\end{lemma}
\begin{proof}
Say that two points of $\Gamma_{a,b,c}$ are neighbours if they differ by the exchange of a $1$ and a $3$. Each point of $A$ has $ac$ neighbours, none of which are in $A$. Each point of $\Gamma_{(a,b,c)}\backslash A$ has $ac$ neighbours, but only $min(a,c)$ of them may be in $A$. So for each point of $A$ there are on average $ac/\min(a,c) = \max(a,c)$ points not in $A$. So the proportion of points of $\Gamma_{(a,b,c)}$ that are in $A$ is at most one in $1+\max(a,c)$.
\end{proof}

The proportion of extra points for each of the cells $\Gamma_{(a,n-i-4,i+4-a)}$ is no more than $2/(i+6)$. Only one cell in three is included from the $b=n-i-4$ layer, so we expect no more than $\binom{n,i+4}2^{i+5}/(3i+18)$ new points, all from $S_{n,i+4}$. One can also find extra points from $S_{n,i+5}$ and higher spheres.

Earlier solutions may also give insight into the problem. In this section we discuss lower bounds for $c'_{n,3}$. Clearly we have
$c'_{0,3}=1$ and $c'_{1,3}=2$, so we focus on the case $n \ge 2$.
The first lower bounds may be due to Koml\'{o}s \cite{komlos}, who observed
that the sphere $S_{i,n}$ of elements with exactly $n-i$ 2 entries
(see Section \ref{notation-sec} for definition),
is a Moser set, so that
\begin{equation}\label{cin}
c'_{n,3}\geq \vert S_{i,n}\vert
\end{equation}
holds for all $i$. Choosing $i=\lfloor \frac{2n}{3}\rfloor$ and
applying Stirling's formula, we see that this lower bound takes the form
\begin{equation}\label{cpn3}
c'_{n,3} \geq (C-o(1)) 3^n / \sqrt{n}
\end{equation}
for some absolute constant $C>0$; in fact \eqref{cin} gives \eqref{cpn3} with $C := \sqrt{\frac{9}{4\pi}}$.
In particular $c'_{3,3} \geq 12, c'_{4,3}\geq 24, c'_{5,3}\geq 80,
c'_{6,3}\geq 240$.

These values can be improved by studying combinations of several spheres or
semispheres or applying elementary results from coding theory.

Observe that if $\{w(1),w(2),w(3)\}$ is a geometric line in $[3]^n$,
then $w(1), w(3)$ both lie in the same sphere $S_{i,n}$,
and that $w(2)$ lies in a lower
sphere $S_{i-r,n}$ for some $1 \leq r \leq i \leq n$. Furthermore,
$w(1)$ and $w(3)$ are separated by Hamming distance $r$.

As a consequence, we see that $S_{i-1,n} \cup S_{i,n}^e$ (or $S_{i-1,n}
\cup S_{i,n}^o$) is a Moser set for any $1 \leq i \leq n$, since any two
distinct elements $S_{i,n}^e$ are separated by a Hamming distance of at
least two.
(Recall Section \ref{notation-sec} for definitions),
This leads to the lower bound
\begin{equation}\label{cn3-low}
c'_{n,3} \geq \binom{n}{i-1}
2^{i-1} + \binom{n}{i} 2^{i-1} = \binom{n+1}{i} 2^{i-1}.
\end{equation}
It is not
hard to see that $\binom{n+1}{i+1} 2^{i} > \binom{n+1}{i} 2^{i-1}$ if
and only if $3i < 2n+1$, and so this lower bound is maximised when $i =
\lfloor \frac{2n+1}{3} \rfloor$ for $n \geq 2$, giving the formula
\eqref{binom}. This leads to the lower bounds $$ c'_{2,3} \geq 6;
c'_{3,3} \geq 16; c'_{4,3} \geq 40; c'_{5,3} \geq 120; c'_{6,3} \geq
336$$ which gives the right lower bounds for $n=2,3$, but is slightly
off for $n=4,5$. Asymptotically, Stirling's formula and \eqref{cn3-low} then
give the lower bound \eqref{cpn3} with $C = \frac{3}{2} \times \sqrt{\frac{9}{4\pi}}$, which is asymptotically $50\%$ better than the bound \eqref{cin}.

The work of Chv\'{a}tal \cite{chvatal1}
already contained a refinement of this idea
which we here translate into the usual notation of coding theory:
Let $A(n,d)$ denote the size of the largest binary code of length $n$
and minimal distance $d$.

Then
\begin{equation}\label{cnchvatal}
c'_{n,3}\geq \max_k \left( \sum_{j=0}^k \binom{n}{j} A(n-j, k-j+1)\right).
\end{equation}

With the following values for $A(n,d)$:
{\tiny{
\[
\begin{array}{llllllll}
A(1,1)=2&&&&&&&\\
A(2,1)=4& A(2,2)=2&&&&&&\\
A(3,1)=8&A(3,2)=4&A(3,3)=2&&&&&\\
A(4,1)=16&A(4,2)=8& A(4,3)=2& A(4,4)=2&&&&\\
A(5,1)=32&A(5,2)=16& A(5,3)=4& A(5,4)=2&A(5,5)=2&&&\\
A(6,1)=64&A(6,2)=32& A(6,3)=8& A(6,4)=4&A(6,5)=2&A(6,6)=2&&\\
A(7,1)=128&A(7,2)=64& A(7,3)=16& A(7,4)=8&A(7,5)=2&A(7,6)=2&A(7,7)=2&\\
A(8,1)=256&A(8,2)=128& A(8,3)=20& A(8,4)=16&A(8,5)=4&A(8,6)=2
&A(8,7)=2&A(8,8)=2\\
A(9,1)=512&A(9,2)=256& A(9,3)=40& A(9,4)=20&A(9,5)=6&A(9,6)=4
&A(9,7)=2&A(9,8)=2\\
A(10,1)=1024&A(10,2)=512& A(10,3)=72& A(10,4)=40&A(10,5)=12&A(10,6)=6
&A(10,7)=2&A(10,8)=2\\
A(11,1)=2048&A(11,2)=1024& A(11,3)=144& A(11,4)=72&A(11,5)=24&A(11,6)=12
&A(11,7)=2&A(11,8)=2\\
A(12,1)=4096&A(12,2)=2048& A(12,3)=256& A(12,4)=144&A(12,5)=32&A(12,6)=24
&A(12,7)=4&A(12,8)=2\\
A(13,1)=8192&A(13,2)=4096& A(13,3)=512& A(13,4)=256&A(13,5)=64&A(12,6)=32
&A(13,7)=8&A(13,8)=4\\
\end{array}
\]
}}

Generally, $A(n,1)=2^n, A(n,2)=2^{n-1}, A(n-1,2e-1)=A(n,2e), A(n,d)=2$,
if $d>\frac{2n}{3}$.
The values were taken or derived from Andries Brower's table at
\centerline{\tt http://www.win.tue.nl/$\sim$aeb/codes/binary-1.html}
\textbf{include to references? or other book with explicit values of $A(n,d)$ }

For $c'_{n,3}$ we obtain the following lower bounds:
with $k=2$
\[
\begin{array}{llll}
c'_{4,3}&\geq &\binom{4}{0}A(4,3)+\binom{4}{1}A(3,2)+\binom{4}{2}A(2,1)
=1\cdot 2+4 \cdot 4+6\cdot 4&=42.\\
c'_{5,3}&\geq &\binom{5}{0}A(5,3)+\binom{5}{1}A(4,2)+\binom{5}{2}A(3,1)
=1\cdot 4+5 \cdot 8+10\cdot 8&=124.\\
c'_{6,3}&\geq &\binom{6}{0}A(6,3)+\binom{6}{1}A(5,2)+\binom{6}{2}A(4,1)
=1\cdot 8+6 \cdot 16+15\cdot 16&=344.
\end{array}
\]
With k=3
\[
\begin{array}{llll}
c'_{7,3}&\geq& \binom{7}{0}A(7,4)+\binom{7}{1}A(6,3)+\binom{7}{2}A(5,2)
+ \binom{7}{3}A(4,1)&=960.\\
c'_{8,3}&\geq &\binom{8}{0}A(8,4)+\binom{8}{1}A(7,3)+\binom{8}{2}A(6,2)
+ \binom{8}{3}A(5,1)&=2832.\\
c'_{9,3}&\geq & \binom{9}{0}A(9,4)+\binom{9}{1}A(8,3)+\binom{9}{2}A(7,2)
+ \binom{9}{3}A(6,1)&=7880.
\end{array}\]
With k=4
\[
\begin{array}{llll}
c'_{10,3}&\geq &\binom{10}{0}A(10,5)+\binom{10}{1}A(9,4)+\binom{10}{2}A(8,3)
+ \binom{10}{3}A(7,2)+\binom{10}{4}A(6,1)&=22232.\\
c'_{11,3}&\geq &\binom{11}{0}A(11,5)+\binom{11}{1}A(10,4)+\binom{11}{2}A(9,3)
+ \binom{11}{3}A(8,2)+\binom{11}{4}A(7,1)&=66024.\\
c'_{12,3}&\geq &\binom{12}{0}A(12,5)+\binom{12}{1}A(11,4)+\binom{12}{2}A(10,3)
+ \binom{12}{3}A(9,2)+\binom{12}{4}A(8,1)&=188688.\\
\end{array}\]
With $k=5$
\[ c'_{13,3}\geq 539168.\]

It should be pointed out that these bounds are even numbers, so that
$c'_{4,3}=43$ shows that one cannot generally expect this lower bound
gives the optimum.

The maximum value appears to occur for $k=\lfloor\frac{n+2}{3}\rfloor$,
so that using
Stirling's formula and explicit bounds on $A(n,d)$ the
best possible value known to date
of the constant $C$ in
equation \eqref{cpn3} can be worked out, but we refrain from doing this here.
Using the Singleton bound $A(n,d)\leq 2^{n-d+1}$ Chv\'{a}tal
\cite{chvatal1} proved that the expression on the right hand side of
\eqref{cnchvatal} is also
$O\left( \frac{3^n}{\sqrt{n}}\right)$, so that the refinement described
above gains a constant factor over the initial construction only.

For $n=4$ the above does not yet give the exact value.
The value $c'_{4,3}=43$ was first proven by Chandra \cite{chandra}.
A uniform way of describing examples for the optimum values of
$c'_{4,3}=43$ and $c'_{5,3}=124$ is the following:

Let us consider the sets $$ A := S_{i-1,n}
\cup S_{i,n}^e \cup A'$$ where $A' \subset S_{i+1,n}$ has the property
that any two elements in $A'$ are separated by a Hamming distance of at
least three, or have a Hamming distance of exactly one but their
midpoint lies in $S_{i,n}^o$. By the previous discussion we see that
this is a Moser set, and we have the lower bound
\begin{equation}\label{cnn}
c'_{n,3} \geq \binom{n+1}{i} 2^{i-1} + |A'|.
\end{equation} This gives some improved lower bounds for $c'_{n,3}$:

\begin{itemize} \item By taking $n=4$, $i=3$, and $A' = \{ 1111, 3331,
3333\}$, we obtain $c'_{4,3} \geq 43$; \item By taking $n=5$, $i=4$, and
$A' = \{ 11111, 11333, 33311, 33331 \}$, we obtain $c'_{5,3} \geq 124$.
\item By taking $n=6$, $i=5$, and $A' = \{ 111111, 111113, 111331,
111333, 331111, 331113\}$, we obtain $c'_{6,3} \geq 342$. \end{itemize}

This gives the lower bounds in Theorem \ref{moser} up to $n=5$, but
the bound for $n=6$ is inferior to the lower bound $c'_{6,3}\geq 344$ given above.

\subsection{Higher $k$ values}
One can consider subsets of $[k]{}^n$ that contain no geometric lines. Section \ref{moser-lower-sec} has considered the case $k=3$. Let $c'_{n,k}$ be the greatest number of points in $[k]{}^n$ with no geometric line. For example, $c'_{n,3} = c'_n$. We have the following lower bounds:

$c'_{n,4} \ge \binom{n}{n/2}2^n$.
The set of points with $a$ $1$s,$b$ $2$s,$c$ $3$s and $d$ $4$s, where $a+d$ has the constant value $n/2$, does not form geometric lines because points at the ends of a geometric line have more $a$ or $d$ values than points in the middle of the line.

One can show a lower bound that, asymptotically, is twice as large. Take all points with $a$ $1$s, $b$ $2$s, $c$ $3$s and $d$ $4$s, for which:

\begin{itemize}
\item Either $a+d = q$ or $q-1$, $a$ and $b$ have the same parity; or
\item $a+d = q-2$ or $q-3$, $a$ and $b$ have opposite parity.
\end{itemize}

This includes half the points of four adjacent layers, and therefore may include $(1+o(1))\binom{n}{n/2}2^{n+1}$ points.

We also have a DHJ(3)-like lower bound for $c'_{n,5}$, namely $c'_{n,5} = 5^{n-O(\sqrt{\log n})}$.
Consider points with $a$ $1$s, $b$ $2$s, $c$ $3$s, $d$ $4$s and $e$ $5$s. For each point, take the value $a+e+2(b+d)+3c$. The first three points in any geometric line give values that form an arithmetic progression of length three.

Select a set of integers with no arithmetic progression of length 3. Select all points whose value belongs to that sequence; there will be no geometric line among those points. By Behrend theory, it is possible to choose these points with density $\exp{-O(\sqrt{\log n})}$.

The $k=6$ version of Moser implies DHJ(3). Indeed, any $k=3$ combinatorial line-free set can be "doubled up" into a $k=6$ geometric line-free set of the same density by pulling back the set from the map that maps 1, 2, 3, 4, 5, 6 to 1, 2, 3, 3, 2, 1 respectively; note that this map sends $k=6$ geometric lines to $k=3$ combinatorial lines. So $c'_{k,6} \geq 2^k c_k$, and more generally, $c'_{k,2n} \geq 2^k c_{k,n}$.

Moser-lower.tex

2009-07-12T14:14:26Z

Mpeake:

\section{Lower bounds for the Moser problem}\label{moser-lower-sec}

Just as for the DHJ problem, we found that Gamma sets $\Gamma_{a,b,c}$ were useful in providing large lower bounds for the Moser problem. This is despite the fact that the symmetries of the cube do not respect Gamma sets.

Observe that if $B \subset \Delta_n$, then the set $A_B := \bigcup_{\vec a \in B} \Gamma_{a,b,c}$ is a Moser set as long as $B$ does not contain any ``isosceles triangles'' $(a+r,b,c+s), (a+s,b,c+r), (a,b+r+s,c)$ for any $r,s \geq 0$ not both zero; in particular, $B$ cannot contain any ``vertical line segments'' $(a+r,b,c+r), (a,b+2r,c)$. An example of such a set is provided by selecting $0 \leq i \leq n-3$ and letting $B$ consist of the triples $(a, n-i, i-a)$ when $a \neq 3 \mod 3$, $(a,n-i-1,i+1-a)$ when $a \neq 1 \mod 3$, $(a,n-i-2,i+2-a)$ when $a=0 \mod 3$, and $(a,n-i-3,i+3-a)$ when $a=2 \mod 3$. Asymptotically, this set includes about two thirds of the spheres $S_{n,i}$, $S_{n,i+1}$ and one third of the spheres $S_{n,i+2}, S_{n,i+3}$ and (setting $i$ close to $n/3$) gives a lower bound \eqref{cpn3} with $C = 2 \times \sqrt{\frac{9}{4\pi}}$.

An integer program was run to obtain the optimal lower bounds achievable by the $A_B$ construction (using \eqref{cn3}, of course). The results for
$1 \leq n \leq 20$ are displayed in Figure \ref{nlow-moser}:

\begin{figure}[tb]
\centerline{
\begin{tabular}{|ll|ll|}
\hline
$n$ & lower bound & $n$ & lower bound \\
\hline
1 & 2 &11& 71766\\
2 & 6 & 12& 212423\\
3 & 16 & 13& 614875\\
4 & 43 & 14& 1794212\\
5 & 122& 15& 5321796\\
6 & 353& 16& 15455256\\
7 & 1017& 17& 45345052\\
8 & 2902&18& 134438520\\
9 & 8622&19& 391796798\\
10& 24786& 20& 1153402148\\
\hline
\end{tabular}}
\caption{Lower bounds for $c'_n$ obtained by the $A_B$ construction.}
\label{nlow-moser}
\end{figure}

More complete data, including the list of optimisers, can be found at
\centerline{{\tt http://abel.math.umu.se/~klasm/Data/HJ/}.}

Note that the lower bound $c'_{6,3} \geq 353$ was first located by a genetic algorithm: see Appendix \ref{genetic-alg}.

\begin{figure}[tb]
\centerline{\includegraphics{moser353new.png}}
\caption{One of the examples of $353$-point sets in $[3]^6$ (elements of the set being indicated by white squares).}
\label{moser353-fig}
\end{figure}

However, we have been unable to locate a lower bound which is asymptotically better than \eqref{cpn3}. Indeed, any method based purely on the $A_B$ construction cannot do asymptotically better than the previous constructions:

\begin{proposition} Let $B \subset \Delta_n$ be such that $A_B$ is a Moser set. Then $|A_B| \leq (2 \sqrt{\frac{9}{4\pi}} + o(1)) \frac{3^n}{\sqrt{n}}$.
\end{proposition}

\begin{proof} By the previous discussion, $B$ cannot contain any pair of the form $(a,b+2r,c), (a+r,b,c+r)$ with $r>0$. In other words, for any $-n \leq h \leq n$, $B$ can contain at most one triple $(a,b,c)$ with $c-a=h$. From this and \eqref{cn3}, we see that
$$ |A_B| \leq \sum_{h=-n}^n \max_{(a,b,c) \in \Delta_n: c-a=h} \frac{n!}{a! b! c!}.$$
From the Chernoff inequality (or the Stirling formula computation below) we see that $\frac{n!}{a! b! c!} \leq \frac{1}{n^{10}} 3^n$ unless $a,b,c = n/3 + O( n^{1/2} \log^{1/2} n )$, so we may restrict to this regime, which also forces $h = O( n^{1/2}\log^{1/2} n)$. If we write $a = n/3 + \alpha$, $b = n/3 + \beta$, $c = n/3+\gamma$ and apply Stirling's formula $n! = (1+o(1)) \sqrt{2\pi n} n^n e^{-n}$, we obtain
$$ \frac{n!}{a! b! c!} = (1+o(1)) \frac{3^{3/2}}{2\pi n} 3^n \exp( - (\frac{n}{3}+\alpha) \log (1 + \frac{3\alpha}{n} ) - (\frac{n}{3}+\beta) \log (1 + \frac{3\beta}{n} ) - (\frac{n}{3}+\gamma) \log (1 + \frac{3\gamma}{n} ) ).$$
From Taylor expansion one has
$$ -(\frac{n}{3}+\alpha) \log (1 + \frac{3\alpha}{n} ) = -\alpha - \frac{3}{2} \frac{\alpha^2}{n} + o(1)$$
and similarly for $\beta,\gamma$; since $\alpha+\beta+\gamma=0$, we conclude that
$$ \frac{n!}{a! b! c!} = (1+o(1)) \frac{3^{3/2}}{2\pi n} 3^n \exp( - \frac{3}{2n} (\alpha^2+\beta^2+\gamma^2) ).$$
If $c-a=h$, then $\alpha^2+\beta^2+\gamma^2 = \frac{3\beta^2}{2} + \frac{h^2}{2}$. Thus we see that
$$ \max_{(a,b,c) \in \Delta_n: c-a=h} \frac{n!}{a! b! c!} \leq (1+o(1)) \frac{3^{3/2}}{2\pi n} 3^n \exp( - \frac{3}{4n} h^2 ).$$
Using the integral test, we thus have
$$ |A_B| \leq (1+o(1)) \frac{3^{3/2}}{2\pi n} 3^n \int_\R \exp( - \frac{3}{4n} x^2 )\ dx.$$
Since $\int_\R \exp( - \frac{3}{4n} x^2 )\ dx = \sqrt{\frac{4\pi n}{3}}$, we obtain the claim.
\end{proof}

Actually it is possible to improve upon these bounds by a slight amount. Observe that if $B$ is a maximiser for the right-hand side of \eqref{cn3} (subject to $B$ not containing isosceles triangles), then any triple $(a,b,c)$ not in $B$ must be the vertex of a (possibly degenerate) isosceles triangle with the other vertices in $B$. If this triangle is non-degenerate, or if $(a,b,c)$ is the upper vertex of a degenerate isosceles triangle, then no point from $\Gamma_{a,b,c}$ can be added to $A_B$ without creating a geometric line. However, if $(a,b,c) = (a'+r,b',c'+r)$ is only the lower vertex of a degenerate isosceles triangle $(a'+r,b',c'+r), (a',b'+2r,c')$, then one can add any subset of $\Gamma_{a,b,c}$ to $A_B$ and still have a Moser set as long as no pair of elements in that subset is separated by Hamming distance $2r$. For instance, in the $n=5$ case, we can start with the 122-point set built from
$$B = \{ (0 0 5),(0 2 3),(1 1 3 ),(1 2 2 ),(2 2 1 ),(3 1 1 ),(3 2 0 ),(5 0 0))\}$$, and add a point each from (1 0 4) and (4 0 1).
This gives an example of the maximal, 124-point solution. Again, in the $n=10$ case, the set
\begin{align*}
B &= \{(0 0 10),(0 2 8 ),(0 3 7 ),(0 4 6 ),(1 4 5 ),(2 1 7 ),(2 3 5 ),
(3 2 5 ),(3 3 4 ),(3 4 3 ),(4 4 2 ),(5 1 4 ), \\
&\quad (5 3 2 ),(6 2 2 ),
(6 3 1 ),(6 4 0 ),(8 1 1 ),(9 0 1 ),(9 1 0 ) \}
\end{align*}

generates the lower bound $c'_{10,3} \geq 24786$ given above (and, up to reflection $a \leftrightarrow c$, is the only such set that does so); but by adding the following twelve elements from $\Gamma_{5,0,5}$ one can increase the lower bound slightly to $24798$: $1111133333$, $1111313333$,
$1113113333$, $1133331113$, $1133331131$, $1133331311$, $3311333111$,
$3313133111$, $3313313111$, $3331111133$, $3331111313$, $3331111331$

A more general form goes with the $B$ set described at the start of this section. Include points from $\Gamma_{(a,n-i-4,i+4-a)}$ when $a=1 (\operatorname{mod}\ 3)$, subject to no two points being included if they differ by the interchange of a $1$ and a $3$. Each of these Gamma sets is the feet of a degenerate isosceles triangle with vertex $\Gamma_{(a-1,n-i-2,a+3-a)}$.

\begin{lemma} If $A$ is a subset of $\Gamma_{(a,b,c)}$ such that no two points of $A$ differ by the interchange of a $1$ and a $3$, then $|A| \leq |\Gamma_{a,b,c}|/(1+\max(a,c))$.
\end{lemma}
\begin{proof}
Say that two points of $\Gamma_{a,b,c}$ are neighbours if they differ by the exchange of a $1$ and a $3$. Each point of $A$ has $ac$ neighbours, none of which are in $A$. Each point of $\Gamma_{(a,b,c)}\backslash A$ has $ac$ neighbours, but only $min(a,c)$ of them may be in $A$. So for each point of $A$ there are on average $ac/\min(a,c) = \max(a,c)$ points not in $A$. So the proportion of points of $\Gamma_{(a,b,c)}$ that are in $A$ is at most one in $1+\max(a,c)$.
\end{proof}

The proportion of extra points for each of the cells $\Gamma_{(a,n-i-4,i+4-a)}$ is no more than $2/(i+6)$. Only one cell in three is included from the $b=n-i-4$ layer, so we expect no more than $\binom{n,i+4}2^{i+5}/(3i+18)$ new points, all from $S_{n,i+4}$. One can also find extra points from $S_{n,i+5}$ and higher spheres.

Earlier solutions may also give insight into the problem. In this section we discuss lower bounds for $c'_{n,3}$. Clearly we have
$c'_{0,3}=1$ and $c'_{1,3}=2$, so we focus on the case $n \ge 2$.
The first lower bounds may be due to Koml\'{o}s \cite{komlos}, who observed
that the sphere $S_{i,n}$ of elements with exactly $n-i$ 2 entries
(see Section \ref{notation-sec} for definition),
is a Moser set, so that
\begin{equation}\label{cin}
c'_{n,3}\geq \vert S_{i,n}\vert
\end{equation}
holds for all $i$. Choosing $i=\lfloor \frac{2n}{3}\rfloor$ and
applying Stirling's formula, we see that this lower bound takes the form
\begin{equation}\label{cpn3}
c'_{n,3} \geq (C-o(1)) 3^n / \sqrt{n}
\end{equation}
for some absolute constant $C>0$; in fact \eqref{cin} gives \eqref{cpn3} with $C := \sqrt{\frac{9}{4\pi}}$.
In particular $c'_{3,3} \geq 12, c'_{4,3}\geq 24, c'_{5,3}\geq 80,
c'_{6,3}\geq 240$.

These values can be improved by studying combinations of several spheres or
semispheres or applying elementary results from coding theory.

Observe that if $\{w(1),w(2),w(3)\}$ is a geometric line in $[3]^n$,
then $w(1), w(3)$ both lie in the same sphere $S_{i,n}$,
and that $w(2)$ lies in a lower
sphere $S_{i-r,n}$ for some $1 \leq r \leq i \leq n$. Furthermore,
$w(1)$ and $w(3)$ are separated by Hamming distance $r$.

As a consequence, we see that $S_{i-1,n} \cup S_{i,n}^e$ (or $S_{i-1,n}
\cup S_{i,n}^o$) is a Moser set for any $1 \leq i \leq n$, since any two
distinct elements $S_{i,n}^e$ are separated by a Hamming distance of at
least two.
(Recall Section \ref{notation-sec} for definitions),
This leads to the lower bound
\begin{equation}\label{cn3-low}
c'_{n,3} \geq \binom{n}{i-1}
2^{i-1} + \binom{n}{i} 2^{i-1} = \binom{n+1}{i} 2^{i-1}.
\end{equation}
It is not
hard to see that $\binom{n+1}{i+1} 2^{i} > \binom{n+1}{i} 2^{i-1}$ if
and only if $3i < 2n+1$, and so this lower bound is maximised when $i =
\lfloor \frac{2n+1}{3} \rfloor$ for $n \geq 2$, giving the formula
\eqref{binom}. This leads to the lower bounds $$ c'_{2,3} \geq 6;
c'_{3,3} \geq 16; c'_{4,3} \geq 40; c'_{5,3} \geq 120; c'_{6,3} \geq
336$$ which gives the right lower bounds for $n=2,3$, but is slightly
off for $n=4,5$. Asymptotically, Stirling's formula and \eqref{cn3-low} then
give the lower bound \eqref{cpn3} with $C = \frac{3}{2} \times \sqrt{\frac{9}{4\pi}}$, which is asymptotically $50\%$ better than the bound \eqref{cin}.

The work of Chv\'{a}tal \cite{chvatal1}
already contained a refinement of this idea
which we here translate into the usual notation of coding theory:
Let $A(n,d)$ denote the size of the largest binary code of length $n$
and minimal distance $d$.

Then
\begin{equation}\label{cnchvatal}
c'_{n,3}\geq \max_k \left( \sum_{j=0}^k \binom{n}{j} A(n-j, k-j+1)\right).
\end{equation}

With the following values for $A(n,d)$:
{\tiny{
\[
\begin{array}{llllllll}
A(1,1)=2&&&&&&&\\
A(2,1)=4& A(2,2)=2&&&&&&\\
A(3,1)=8&A(3,2)=4&A(3,3)=2&&&&&\\
A(4,1)=16&A(4,2)=8& A(4,3)=2& A(4,4)=2&&&&\\
A(5,1)=32&A(5,2)=16& A(5,3)=4& A(5,4)=2&A(5,5)=2&&&\\
A(6,1)=64&A(6,2)=32& A(6,3)=8& A(6,4)=4&A(6,5)=2&A(6,6)=2&&\\
A(7,1)=128&A(7,2)=64& A(7,3)=16& A(7,4)=8&A(7,5)=2&A(7,6)=2&A(7,7)=2&\\
A(8,1)=256&A(8,2)=128& A(8,3)=20& A(8,4)=16&A(8,5)=4&A(8,6)=2
&A(8,7)=2&A(8,8)=2\\
A(9,1)=512&A(9,2)=256& A(9,3)=40& A(9,4)=20&A(9,5)=6&A(9,6)=4
&A(9,7)=2&A(9,8)=2\\
A(10,1)=1024&A(10,2)=512& A(10,3)=72& A(10,4)=40&A(10,5)=12&A(10,6)=6
&A(10,7)=2&A(10,8)=2\\
A(11,1)=2048&A(11,2)=1024& A(11,3)=144& A(11,4)=72&A(11,5)=24&A(11,6)=12
&A(11,7)=2&A(11,8)=2\\
A(12,1)=4096&A(12,2)=2048& A(12,3)=256& A(12,4)=144&A(12,5)=32&A(12,6)=24
&A(12,7)=4&A(12,8)=2\\
A(13,1)=8192&A(13,2)=4096& A(13,3)=512& A(13,4)=256&A(13,5)=64&A(12,6)=32
&A(13,7)=8&A(13,8)=4\\
\end{array}
\]
}}

Generally, $A(n,1)=2^n, A(n,2)=2^{n-1}, A(n-1,2e-1)=A(n,2e), A(n,d)=2$,
if $d>\frac{2n}{3}$.
The values were taken or derived from Andries Brower's table at
\centerline{\tt http://www.win.tue.nl/$\sim$aeb/codes/binary-1.html}
\textbf{include to references? or other book with explicit values of $A(n,d)$ }

For $c'_{n,3}$ we obtain the following lower bounds:
with $k=2$
\[
\begin{array}{llll}
c'_{4,3}&\geq &\binom{4}{0}A(4,3)+\binom{4}{1}A(3,2)+\binom{4}{2}A(2,1)
=1\cdot 2+4 \cdot 4+6\cdot 4&=42.\\
c'_{5,3}&\geq &\binom{5}{0}A(5,3)+\binom{5}{1}A(4,2)+\binom{5}{2}A(3,1)
=1\cdot 4+5 \cdot 8+10\cdot 8&=124.\\
c'_{6,3}&\geq &\binom{6}{0}A(6,3)+\binom{6}{1}A(5,2)+\binom{6}{2}A(4,1)
=1\cdot 8+6 \cdot 16+15\cdot 16&=344.
\end{array}
\]
With k=3
\[
\begin{array}{llll}
c'_{7,3}&\geq& \binom{7}{0}A(7,4)+\binom{7}{1}A(6,3)+\binom{7}{2}A(5,2)
+ \binom{7}{3}A(4,1)&=960.\\
c'_{8,3}&\geq &\binom{8}{0}A(8,4)+\binom{8}{1}A(7,3)+\binom{8}{2}A(6,2)
+ \binom{8}{3}A(5,1)&=2832.\\
c'_{9,3}&\geq & \binom{9}{0}A(9,4)+\binom{9}{1}A(8,3)+\binom{9}{2}A(7,2)
+ \binom{9}{3}A(6,1)&=7880.
\end{array}\]
With k=4
\[
\begin{array}{llll}
c'_{10,3}&\geq &\binom{10}{0}A(10,5)+\binom{10}{1}A(9,4)+\binom{10}{2}A(8,3)
+ \binom{10}{3}A(7,2)+\binom{10}{4}A(6,1)&=22232.\\
c'_{11,3}&\geq &\binom{11}{0}A(11,5)+\binom{11}{1}A(10,4)+\binom{11}{2}A(9,3)
+ \binom{11}{3}A(8,2)+\binom{11}{4}A(7,1)&=66024.\\
c'_{12,3}&\geq &\binom{12}{0}A(12,5)+\binom{12}{1}A(11,4)+\binom{12}{2}A(10,3)
+ \binom{12}{3}A(9,2)+\binom{12}{4}A(8,1)&=188688.\\
\end{array}\]
With $k=5$
\[ c'_{13,3}\geq 539168.\]

It should be pointed out that these bounds are even numbers, so that
$c'_{4,3}=43$ shows that one cannot generally expect this lower bound
gives the optimum.

The maximum value appears to occur for $k=\lfloor\frac{n+2}{3}\rfloor$,
so that using
Stirling's formula and explicit bounds on $A(n,d)$ the
best possible value known to date
of the constant $C$ in
equation \eqref{cpn3} can be worked out, but we refrain from doing this here.
Using the Singleton bound $A(n,d)\leq 2^{n-d+1}$ Chv\'{a}tal
\cite{chvatal1} proved that the expression on the right hand side of
\eqref{cnchvatal} is also
$O\left( \frac{3^n}{\sqrt{n}}\right)$, so that the refinement described
above gains a constant factor over the initial construction only.

For $n=4$ the above does not yet give the exact value.
The value $c'_{4,3}=43$ was first proven by Chandra \cite{chandra}.
A uniform way of describing examples for the optimum values of
$c'_{4,3}=43$ and $c'_{5,3}=124$ is the following:

Let us consider the sets $$ A := S_{i-1,n}
\cup S_{i,n}^e \cup A'$$ where $A' \subset S_{i+1,n}$ has the property
that any two elements in $A'$ are separated by a Hamming distance of at
least three, or have a Hamming distance of exactly one but their
midpoint lies in $S_{i,n}^o$. By the previous discussion we see that
this is a Moser set, and we have the lower bound
\begin{equation}\label{cnn}
c'_{n,3} \geq \binom{n+1}{i} 2^{i-1} + |A'|.
\end{equation} This gives some improved lower bounds for $c'_{n,3}$:

\begin{itemize} \item By taking $n=4$, $i=3$, and $A' = \{ 1111, 3331,
3333\}$, we obtain $c'_{4,3} \geq 43$; \item By taking $n=5$, $i=4$, and
$A' = \{ 11111, 11333, 33311, 33331 \}$, we obtain $c'_{5,3} \geq 124$.
\item By taking $n=6$, $i=5$, and $A' = \{ 111111, 111113, 111331,
111333, 331111, 331113\}$, we obtain $c'_{6,3} \geq 342$. \end{itemize}

This gives the lower bounds in Theorem \ref{moser} up to $n=5$, but
the bound for $n=6$ is inferior to the lower bound $c'_{6,3}\geq 344$ given above.

\subsection{Higher $k$ values}
One can consider subsets of $[k]{}^n$ that contain no geometric lines. Section \ref{moser-lower-sec} has considered the case $k=3$. Let $c'_{n,k}$ be the greatest number of points in $[k]{}^n$ with no geometric line. For example, $c'_{n,3} = c'_n$. We have the following lower bounds:

$c'_{n,4} \ge \binom{n}{n/2}2^n$.
The set of points with $a$ $1$s,$b$ $2$s,$c$ $3$s and $d$ $4$s, where $a+d$ has the constant value $n/2$, does not form geometric lines because points at the ends of a geometric line have more $a$ or $d$ values than point in the middle of the line.

One can show a lower bound that, asymptotically, is twice as large. Take all points with $a$ $1$s, $b$ $2$s, $c$ $3$s and $d$ $4$s, for which:

\begin{itemize}
\item Either $a+d = q$ or $q-1$, $a$ and $b$ have the same parity; or
\item $a+d = q-2$ or $q-3$, $a$ and $b$ have opposite parity.
\end{itemize}

This includes half the points of four adjacent layers, and therefore may include $(1+o(1))\binom{n}{n/2}2^{n+1}$ points.

We also have a DHJ(3)-like lower bound for $c'_{n,5}$, namely $c'_{n,5} = 5^{n-O(\sqrt{\log n})}$.
Consider points with $a$ $1$s, $b$ $2$s, $c$ $3$s, $d$ $4$s and $e$ $5$s. For each point, take the value $a+e+2(b+d)+3c$. The first three points in any geometric line give values that form an arithmetic progression of length three.

Select a set of integers with no arithmetic progression of length 3. Select all points whose value belongs to that sequence; there will be no geometric line among those points. By Behrend theory, it is possible to choose these points with density $\exp{-O(\sqrt{\log n})}$.

The $k=6$ version of Moser implies DHJ(3). Indeed, any $k=3$ combinatorial line-free set can be "doubled up" into a $k=6$ geometric line-free set of the same density by pulling back the set from the map that maps 1, 2, 3, 4, 5, 6 to 1, 2, 3, 3, 2, 1 respectively; note that this map sends $k=6$ geometric lines to $k=3$ combinatorial lines. So $c'_{k,6} \geq 2^k c_k$, and more generally, $c'_{k,2n} \geq 2^k c_{k,n}$.

Dhj-lown-lower.tex

2009-07-12T14:11:16Z

Mpeake:

\section{Lower bounds for the density Hales-Jewett problem}\label{dhj-lower-sec}

The purpose of this section is to establish various lower bounds for $c_{n,3}$, in particular establishing Theorem \ref{dhj-lower} and the lower bound component of Theorem \ref{dhj-upper}.

As observed in the introduction, if $B \subset \Delta_{3,n}$ is a Fujimura set (i.e. a subset of $\Delta_{3,n} = \{ (a,b,c) \in \N^3: a+b+c=n\}$ which contains no upward equilateral triangles $(a+r,b,c), (a,b+r,c), (a,b,c+r)$), then the set $A_B := \bigcup_{\vec a \in B} \Gamma_{a,b,c}$ is a line-free subset of $[3]^n$, which gives the lower bound
\begin{equation}\label{cn3}
c_{n,3} \geq |A_B| = \sum_{(a,b,c) \in B} \frac{n!}{a! b! c!}.
\end{equation}
All of the lower bounds for $c_{n,3}$ in this paper will be constructed via this device.

In order to use \eqref{cn3}, one of course needs to build Fujimura sets $B$ which are ``large'' in the sense that the right-hand side of \eqref{cn3} is large. A fruitful starting point for this goal is the sets
$$B_{j,n} := \{ (a,b,c) \in \Delta_{3,n}: a + 2b \neq j \hbox{ mod } 3 \}$$
for $j=0,1,2$. Observe that in order for a triangle $(a+r,b,c), (a,b+r,c), (a,b,c+r)$ to lie in $B_{j,n}$, the length $r$ of the triangle must be a multiple of $3$. This already makes $B_{j,n}$ a Fujimura set for $n < 3$ (and $B_{0,n}$ a Fujimura set for $n = 3$).

When $n$ is not a multiple of $3$, the $B_{j,n}$ are all rotations of each other and give equivalent sets (of size $2 \times 3^{n-1}$). When $n$ is a multiple of $3$, the sets $B_{1,n}$ and $B_{2,n}$ are reflections of each other, but $B_{0,n}$ is not equivalent to the other two sets (in particular, it omits all three corners of $\Delta_{3,n}$); the associated set $A_{B_{0,n}}$ is slightly larger than $A_{B_{1,n}}$ and $A_{B_{2,n}}$ and thus is slightly better for constructing line-free sets.

As mentioned already, $B_{0,n}$ is a Fujimura set for $n \leq 3$, and hence $A_{B_{0,n}}$ is line-free for $n \leq 3$. Applying \eqref{cn3} one obtains the lower bounds
$$ c_{0,3} \geq 1; c_{1,3} \geq 2; c_{2,3} \geq 6; c_{3,3} \geq 18.$$

For $n>3$, $B_{0,n}$ contains some triangles $(a+r,b,c), (a,b+r,c), (a,b,c+r)$ and so is not a Fujimura set, but one can remove points from this set to recover the Fujimura property. For instance, for $n \leq 6$, the only triangles in $B_{0,n}$ have side length $r=3$. One can ``delete'' these triangles by removing one vertex from each; in order to optimise the bound \eqref{cn3} it is preferable to delete vertices near the corners of $\Delta_{3,n}$ rather than near the centre. These considerations lead to the Fujimura sets
\begin{align*}
B_{0,4} &\backslash \{ (0,0,4), (0,4,0), (4,0,0) \}\\
B_{0,5} &\backslash \{ (0,4,1), (0,5,0), (4,0,1), (5,0,0) \}\\
B_{0,6} &\backslash \{ (0,1,5), (0,5,1), (1,0,5), (0,1,5), (1,5,0), (5,1,0) \}
\end{align*}
which by \eqref{cn3} gives the lower bounds
$$ c_{4,3} \geq 52; c_{5,3} \geq 150; c_{6,3} \geq 450.$$
Thus we have established all the lower bounds needed for Theorem \ref{dhj-upper}.

One can of course continue this process by hand, for instance the set
$$ B_{0,7} \backslash \{(0,1,6),(1,0,6),(0,5,2),(5,0,2),(1,5,1),(5,1,1),(1,6,0),(6,1,0) \}$$
gives the lower bound $c_{7,3} \geq 1302$, which we tentatively conjecture to be the correct bound.

A simplification was found when $n$ is a multiple of $3$. Observe that for $n=6$, the sets excluded from $B_{0,6}$ are all permutations of $(0,1,5)$. So the remaining sets are all the permutations of $(1,2,3)$ and $(0,2,4)$. In the same way, sets for $n=9$, $12$ and $15$ can be described as:
\begin{itemize}
\item $n=9$: $(2,3,4),(1,3,5),(0,4,5)$ and permutations;
\item $n=12$: $(3,4,5),(2,4,6),(1,5,6),(0,2,10),(0,5,7)$ and permutations;
\item $n=15$: $(4,5,6),(3,5,7),(2,6,7),(1,3,11),(1,6,8),(0,4,11),(0,7,8)$ and permutations.
\end{itemize}

When $n$ is not a multiple of $3$, say $n=3m-1$ or $n=3m-2$, one first finds a solution for $n=3m$. Then for $n=3m-1$, one restricts the first digit of the $3m$ sequence to equal $1$. This leaves exactly one-third as many points for $3m-1$ as for $3m$. For $n=3m-1$, one restricts the first two digits of the $3m$ sequence to be $12$. This leaves roughly one-ninth as many points for $3m-2$ as for $3m$.

The following is an effective method to find good, though not optimal, solutions for any $n=3m$. (For $n<21$, ignore any triple with a negative entry.)

Start with thirteen groups of points in the centre, formed from adding one of the following points, or its permutation, to $M := (m,m,m)$, when $n=3m$:
$$(-7,-3,+10), (-7, 0,+7),(-7,+3,+4),(-6,-4,+10),(-6,-1,+7),(-6,+2,+4)$$
$$(-5,-1,+6),(-5,+2,+3),(-4,-2,+6),(-4,+1,+3),(-3,+1,+2),(-2,0,+2),(-1,0,+1)$$
Then include eights string of points, stretching to the edges of the triangle $\Delta_n$;
\begin{itemize}
\item $M+(-8-2x,-6-2x,14+4x)$, $M+(-8-2x,-3-2x,11+4x)$, $M+(-8-2x,x,8+x)$, $M+(-8-2x,3+x,5+x)$ and permutations ($0\le 2x \le M-8$);
\item $M+(-9-2x,-5-2x,14+4x)$, $M+(-9-2x,-2-2x,11+4x)$, $M+(-9-2x,1+x,8+x)$, $M+(-9-2x,4+x,5+x)$ and permutations ($0\le 2x \le M-9$).
\end{itemize}
This solution gives $O(2.7 \sqrt(\log (n)/n)3^n$ points for values of $n$ up to around $1000$. This is asymptotically smaller than the known optimum of $3^{n-O(\sqrt(\log(n)))}$. When $n=99$, it gives more than $3^n/3$ points.

The following solution gives more points for $n>1000$, but not for moderate $n$:

\begin{itemize}
\item Define a sequence, of all positive numbers which, in base 3, do not contain a 1. Add 1 to all multiples of 3 in this sequence. This sequence does not contain a length-3 arithmetic progression. It starts $1,2,7,8,19,20,25,26,55, \ldots$;
\item List all the $(abc)$ triples that sum to $n$, for which the larger two differ by a number from the sequence;
\item Exclude the case when the smaller two differ by 1;
\item Include the case when $(a,b,c)$ is a permutation of $n/3+(-1,0,1)$.
\end{itemize}

An integer program was run obtain the maximum lower bound one could establish from \eqref{cn3} (see Appendix \ref{integer-sec}). The results for $1 \leq n \leq 20$ are displayed in Figure \ref{nlow}:

\begin{figure}[tb]
\centerline{
\begin{tabular}{|ll|ll|}
\hline
$n$ & lower bound & $n$ & lower bound \\
\hline
1 & 2 &11& 96338\\
2 & 6 & 12& 287892\\
3 & 18 & 13& 854139\\
4 & 52 & 14& 2537821\\
5 & 150& 15& 7528835\\
6 & 450& 16& 22517082\\
7 & 1302& 17& 66944301\\
8 & 3780&18& 198629224\\
9 & 11340&19& 593911730\\
10& 32864& 20& 1766894722\\
\hline
\end{tabular}}
\caption{Lower bounds for $c_n$ obtained by the $A_B$ construction.}
\label{nlow}
\end{figure}

More complete data, including the list of optimisers, can be found at
\centerline{{\tt http://abel.math.umu.se/~klasm/Data/HJ/}.}

If $B$ maximises the right-hand side of \eqref{cn3}, it is easy to see that $A_B$ is a line-free set which is maximal in the sense that the addition of any further point to $A_B$ will create a combinatorial line. Thus one might conjecture that the maximal value of the right-hand side \eqref{cn3} is in fact equal to $c_{n,3}$ for all $n$; Theorem \ref{dhj-upper} asserts that this conjecture is true for $n \leq 6$.

Now we prove Theorem \ref{dhj-lower}.

\begin{proof}[Proof of Theorem \ref {dhj-lower}]
Let $M$ be the circulant matrix with first row $(1,2,\ldots,k-1)$, second row $(k,1,2,\dots,k-1)$, and so on. Note that $M$ has nonzero determinant by well-known properties of circulant matrices.

Let $S$ be a subset of the interval $[-\sqrt {n}/2, \sqrt {n}/2)$ that contains no nonconstant arithmetic progressions of length k, and let $B\subset\Delta_{n, k}$ be the set
\[ B := \{(n-\sum_{i=1}^{k-1} a_i ,a_1,a_2,\dots, a_{k-1}) :
(a_1,\dots,a_{k-1})= \det(M) M^{-1}\vec{s} , \vec{s}\in S^{k-1}\}.\]
The map $(m,a_1,\dots,a_{k-1}) \mapsto M (a_1,\dots,a_{k-1})$ takes simplices to nonconstant arithmetic progressions in ${\mathbb Z}^{k-1}$, and takes $B$ to $\{det(M) \, \vec{s} \colon \vec{s} \in S^{k-1}\}$, which is a set containing no nonconstant arithmetic progressions. Thus, $B$ is a Fujimora set and so does not contain any combinatorial lines.

If all of $a_1,\ldots,a_k$ are within $C_1\sqrt{n}$ of $n/k$, then $|\Gamma_{\vec{a}}| \geq C k^n/n^{(k-1)/2}$ (where $C$ depends on $C_1$) by the central limit theorem. By our choice of $S$ and applying~\eqref{cn3}, we obtain
$$ c_ {n, k}\geq C k^n/n^{(k-1)/2} |S|^{k-1} = C k^n \left( \frac{|S|}{\sqrt{n}} \right)^{k-1}. $$
One can take $S$ to have cardinality $r_ k (\sqrt {n}) $, which from the results of O'Bryant~\cite {obryant}) satisfies (for all sufficiently large $n$, some $C>0$, and $\ell$ the largest integer satisfying $k> 2^{\ell-1}$)
$$ \frac{r_k (\sqrt{n})}{\sqrt{n}} \geq C (\log n)^{1/(2\ell)}\exp_2 (-\ell 2^{(\ell-1)/2-1/\ell} \sqrt[\ell]{\log_2 n}),$$
which completes the proof.
\end {proof}

Coloring.tex

2009-07-12T14:07:17Z

Mpeake:

\section{The coloring Hales-Jewett problem}\label{coloring-sec}
The Hales-Jewett theorem states that for every $k$ and every $r$ there exists an $n = HJ(k,r)$ such that if you colour the elements of $[k]{}^n$ with $r$ colours, then there must be a combinatorial line with all its points of the same colour.

This is a consequence of the Density Hales-Jewett theorem, since there must be a set of density at least $r^{-1}$ of elements of $[k]{}^n$ all of whose elements have the same colour. It also follows from the Graham-Rothschild theorem [\cite{}].
By iterating the Hales-Jewett theorem, one can also show that one of the color classes contains an $m$-dimensional combinatorial subspace, if $n$ is sufficiently large depending on $k$, $r$ and $m$.

A related idea is that of the Van der Waerden number $W(k,r)$ [\cite{waerden}], which is the length of the longest sequence $1,2,\ldots,n$ which can be $r$-coloured without a monochromatic arithmetic progression of length $k$. These numbers were used in this project to construct $r$-colourings of $[k]{}^m$ without combinatorial lines, and so provide new lower bounds for $HJ(k,r)$.

\begin{lemma} $HJ(k,r) \ge \lfloor (W(k,r)-1)/(k-1)\rfloor$ \end{lemma}
\begin{proof}
Let $m = \lfloor (W(k,r)-1)/(k-1) \rfloor$. Suppose we have a Van der Warden colouring of $1, 2, \ldots W(k,r)$. Give each point $(a_1,a_2,\ldots,a_m) \in [k]{}^m$ the colour that corresponds to $1+\sum (a_j-1)$. The points of a combinatorial line give sums that form an arithmetic progression, and will therefore not be monochromatic.
\end{proof}

A recent table of lower bounds for the Van der Waerden numbers is at this website: [\cite{heule}] [http://www.st.ewi.tudelft.nl/sat/slides/waerden.pdf].
These numbers were used to produce the following table of lower bounds for colouring Hales-Jewett numbers.

\begin{figure}[tb] \centerline{\begin{tabular}[l|lllll]
k\r & 2 & 3 & 4 & 5 & 6 \\
\hline
3 & & 13 & 37 & 84 & 103 \\
4 & 11 & 97 & 349 & 751 & 3259 \\
5 & 59 & 302 & 2609 & 6011 & 14173 \\
6 & 226 & 1777 & 18061 & 49391 & 120097 \\
7 & 617 & 7309 & 64661 & & \\
8 & 1069 & 34057 & & & \\
9 & 3389 & & & &
\end{tabular}}\caption{Lower bounds for colouring Hales-Jewett numbers
\label{HJcolour}\end{figure}

Polymath.tex

2009-07-12T14:06:12Z

Mpeake:

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\usepackage{amssymb}
\usepackage{amscd}
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\newtheorem{principle}[theorem]{Principle}
\newtheorem{claim}[theorem]{Claim}

\theoremstyle{definition}

%\newtheorem{roughdef}[subsection]{Rough Definition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{remarks}[theorem]{Remarks}
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%\newtheorem{problem}[subsection]{Problem}
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\parindent 0mm
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\begin{document}

\title{Density Hales-Jewett and Moser numbers}

\author{D.H.J. Polymath}
\address{http://michaelnielsen.org/polymath1/index.php}
\email{???}

\subjclass{???}

\begin{abstract}
For any $n \geq 0$ and $k \geq 1$, the density Hales-Jewett number $c_{n,k}$ is defined as the size of the largest subset of the cube $[k]^n$ := $\{1,\ldots,k\}^n$ which contains no combinatorial line; similarly, the Moser number $c'_{n,k}$ is the largest subset of the cube $[k]^n$ which contains no geometric line. A deep theorem of Furstenberg and Katznelson \cite{fk1}, \cite{fk2}, \cite{mcc} shows that $c_{n,k}$ = $o(k^n)$ as $n \to \infty$ (which implies a similar claim for $c'_{n,k}$; this is already non-trivial for $k = 3$. Several new proofs of this result have also been recently established \cite{poly}, \cite{austin}.

Using both human and computer-assisted arguments, we compute several values of $c_{n,k}$ and $c'_{n,k}$ for small $n,k$. For instance the sequence $c_{n,3}$ for $n=0,\ldots,6$ is $1,2,6,18,52,150,450$, while the sequence $c'_{n,3}$ for $n=0,\ldots,6$ is $1,2,6,16,43,124,353$. We also establish some results for higher $k$, showing for instance that an analogue of the LYM inequality (which relates to the $k = 2$ case) does not hold for higher $k$.
\end{abstract}

\maketitle
%\today

\setcounter{tocdepth}{1}
\tableofcontents

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\include{introduction}
\include{dhj-lown-lower}
\include{dhj-lown}
\include{moser-lower}
\include{moser}
\include{fujimura}
\include{higherk}
\include{coloring}

\appendix

\include{genetic}
\include{integer}

\begin{thebibliography}{10}

\bibitem{austin} T. Austin, \emph{Deducing the density Hales-Jewett theorem from an infinitary removal lemma}, preprint.

\bibitem{behrend}
F. Behrend, \emph{On the sets of integers which contain no three in arithmetic progression}, Proceedings of the National Academy of Sciences \textbf{23} (1946), 331–-332.

\bibitem{chandra}
A. Chandra, \emph{On the solution of Moser's problem in four dimensions}, Canad. Math. Bull. \textbf{16} (1973), 507--511.

\bibitem{chvatal1} V. Chv\'{a}tal, \emph{Remarks on a problem of Moser}, Canadian Math Bulletin, Vol 15, 1972, 19--21.

\bibitem{chvatal2} V. Chv\'{a}tal, \emph{Edmonds polytopes and a hierarchy of combinatorial problems}, Discrete Math. 4 (1973) 305-337.

\bibitem{elkin}
M. Elkin, \emph{An Improved Construction of Progression-Free Sets}, preprint.

\bibitem{fk1} H. Furstenberg, Y. Katznelson, \emph{A density version of the Hales-Jewett theorem for $k = 3$}, Graph Theory and Combinatorics (Cambridge, 1988). Discrete Math. 75 (1989), no. 1-3, 227–-241.

\bibitem{fk2} H. Furstenberg, Y. Katznelson, \emph{A density version of the Hales-Jewett theorem}, J. Anal. Math. 57 (1991), 64–-119. MR1191743

\bibitem{greenwolf}
B. Green, J. Wolf, \emph{A note on Elkin's improvement of Behrend's construction}, preprint.

\bibitem{heule} Marijn Heule, presentation at {\tt http://www.st.ewi.tudelft.nl/sat/slides/waerden.pdf}

\bibitem{komlos}
J. Koml\'{o}s, solution to problem P.170 by Leo Moser, Canad. Math.. Bull. vol {\bf (??check)} (1972), 312--313, 1970.

\bibitem{Krisha} K. Krishna, M. Narasimha Murty, "Genetic K-means algorithm," Systems, Man, and Cybernetics, Part B: Cybernetics, IEEE Transactions on , vol.29, no.3, pp.433-439, Jun 1999

\bibitem{moser} L. Moser, Problem P.170 in Canad. Math. Bull. 13 (1970), 268.

\bibitem{mcc} R. McCutcheon, \emph{The conclusion of the proof of the density Hales-Jewett theorem for k=3}, unpublished.

\bibitem{obryant}
K. O'Bryant, \emph{Sets of integers that do not contain long arithmetic progressions}, preprint.

\bibitem{oeis}
N. J. A. Sloane, Ed. (2008), The On-Line Encyclopedia of Integer Sequences, {\tt www.research.att.com/~njas/sequences/}

\bibitem{potenchin}
A. Potechin, \emph{Maximal caps in $AG(6, 3)$}, Journal Designs, Codes and Cryptography, Volume 46, Number 3 / March, 2008.

\bibitem{poly} D.H.J. Polymath, ???, preprint. {\bf need title}

\bibitem{rankin}
R. A. Rankin, Sets of integers containing not more than a given number of terms in arithmetical progression, Proc. Roy. Soc. Edinburgh Sect. A 65 (1960/1961), 332–344 (1960/61). MR 0142526 (26 \#95)

\bibitem{roth}
K. Roth, \emph{On certain sets of integers, I}, Journal of the London Mathematical Society \textbf{28} (1953), 104-–109.

\bibitem{Rothlauf} F. Rothlauf, D. E. Goldberg, Representations for Genetic and Evolutionary Algorithms. Physica-Verlag, 2002.

\bibitem{sperner}
E. Sperner, \emph{Ein Satz \"uber Untermengen einer endlichen Menge}, Mathematische Zeitschrift \textbf{27} (1928), 544-–548.

\bibitem{szem}
E. Szemer\'edi, \emph{On sets of integers containing no $k$ elements in arithmetic progression}, Acta Arithmetica \textbf{27} (1975), 199-–245.

\bibitem{waerden} {\tt http://en.wikipedia.org/wiki/Van_der_Waerden_number}

\end{thebibliography}

\end{document}

Coloring.tex

2009-07-12T14:02:18Z

Mpeake:

\section{The coloring Hales-Jewett problem}\label{coloring-sec}
The Hales-Jewett theorem states that for every $k$ and every $r$ there exists an $n = HJ(k,r)$ such that if you colour the elements of $[k]{}^n$ with $r$ colours, then there must be a combinatorial line with all its points of the same colour.

This is a consequence of the Density Hales-Jewett theorem, since there must be a set of density at least $r^{-1}$ of elements of $[k]{}^n$ all of whose elements have the same colour. It also follows from the Graham-Rothschild theorem [\cite{}].
By iterating the Hales-Jewett theorem, one can also show that one of the color classes contains an $m$-dimensional combinatorial subspace, if $n$ is sufficiently large depending on $k$, $r$ and $m$.

A related idea is that of the Van der Waerden number $W(k,r)$ [\cite{waerden}], which is the length of the longest sequence $1,2,\ldots,n$ which can be $r$-coloured without a monochromatic arithmetic progression of length $k$. These numbers were used in this project to construct $r$-colourings of $[k]{}^m$ without combinatorial lines, and so provide new lower bounds for $HJ(k,r)$.

\begin{lemma} $HJ(k,r) \ge \lfloor (W(k,r)-1)/(k-1)\rfloor$ \end{lemma}
\begin{proof}
Let $m = \lfloor (W(k,r)-1)/(k-1) \rfloor$. Suppose we have a Van der Warden colouring of $1, 2, \ldots W(k,r)$. Give each point $(a_1,a_2,\ldots,a_m) \in [k]{}^m$ the colour that corresponds to $1+\sum (a_j-1)$. The points of a combinatorial line give sums that form an arithmetic progression, and will therefore not be monochromatic.
\end{proof}

A recent table of lower bounds for the Van der Waerden numbers is at this website: [http://www.st.ewi.tudelft.nl/sat/slides/waerden.pdf].
These numbers were used to produce the following table of lower bounds for colouring Hales-Jewett numbers.

\begin{figure}[tb] \centerline{\begin{tabular}[l|lllll]
k\r & 2 & 3 & 4 & 5 & 6 \\
\hline
3 & & 13 & 37 & 84 & 103 \\
4 & 11 & 97 & 349 & 751 & 3259 \\
5 & 59 & 302 & 2609 & 6011 & 14173 \\
6 & 226 & 1777 & 18061 & 49391 & 120097 \\
7 & 617 & 7309 & 64661 & & \\
8 & 1069 & 34057 & & & \\
9 & 3389 & & & &
\end{tabular}}\caption{Lower bounds for colouring Hales-Jewett numbers
\label{HJcolour}\end{figure}

Polymath.tex

2009-07-12T14:01:19Z

Mpeake:

\documentclass[12pt,a4paper,reqno]{amsart}
\usepackage{amssymb}
\usepackage{amscd}
%\usepackage{psfig}
\usepackage{graphicx}
%\usepackage{showkeys}
% uncomment this when editing cross-references
\numberwithin{equation}{section}

\addtolength{\textwidth}{3 truecm}
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\setlength{\hoffset}{-1.3 truecm}

\theoremstyle{plain}

\newtheorem{theorem}{Theorem}[section]
%\newtheorem{theorem}[theorem]{Theorem}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{principle}[theorem]{Principle}
\newtheorem{claim}[theorem]{Claim}

\theoremstyle{definition}

%\newtheorem{roughdef}[subsection]{Rough Definition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{remarks}[theorem]{Remarks}
\newtheorem{example}[theorem]{Example}
\newtheorem{examples}[theorem]{Examples}
%\newtheorem{problem}[subsection]{Problem}
%\newtheorem{question}[subsection]{Question}

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\newcommand\R{\mathbb{R}}
\newcommand\Z{\mathbb{Z}}
\newcommand\N{\mathbb{N}}
\renewcommand\P{\mathbb{P}}

\parindent 0mm
\parskip 5mm

\begin{document}

\title{Density Hales-Jewett and Moser numbers}

\author{D.H.J. Polymath}
\address{http://michaelnielsen.org/polymath1/index.php}
\email{???}

\subjclass{???}

\begin{abstract}
For any $n \geq 0$ and $k \geq 1$, the density Hales-Jewett number $c_{n,k}$ is defined as the size of the largest subset of the cube $[k]^n$ := $\{1,\ldots,k\}^n$ which contains no combinatorial line; similarly, the Moser number $c'_{n,k}$ is the largest subset of the cube $[k]^n$ which contains no geometric line. A deep theorem of Furstenberg and Katznelson \cite{fk1}, \cite{fk2}, \cite{mcc} shows that $c_{n,k}$ = $o(k^n)$ as $n \to \infty$ (which implies a similar claim for $c'_{n,k}$; this is already non-trivial for $k = 3$. Several new proofs of this result have also been recently established \cite{poly}, \cite{austin}.

Using both human and computer-assisted arguments, we compute several values of $c_{n,k}$ and $c'_{n,k}$ for small $n,k$. For instance the sequence $c_{n,3}$ for $n=0,\ldots,6$ is $1,2,6,18,52,150,450$, while the sequence $c'_{n,3}$ for $n=0,\ldots,6$ is $1,2,6,16,43,124,353$. We also establish some results for higher $k$, showing for instance that an analogue of the LYM inequality (which relates to the $k = 2$ case) does not hold for higher $k$.
\end{abstract}

\maketitle
%\today

\setcounter{tocdepth}{1}
\tableofcontents

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\include{introduction}
\include{dhj-lown-lower}
\include{dhj-lown}
\include{moser-lower}
\include{moser}
\include{fujimura}
\include{higherk}
\include{coloring}

\appendix

\include{genetic}
\include{integer}

\begin{thebibliography}{10}

\bibitem{austin} T. Austin, \emph{Deducing the density Hales-Jewett theorem from an infinitary removal lemma}, preprint.

\bibitem{behrend}
F. Behrend, \emph{On the sets of integers which contain no three in arithmetic progression}, Proceedings of the National Academy of Sciences \textbf{23} (1946), 331–-332.

\bibitem{chandra}
A. Chandra, \emph{On the solution of Moser's problem in four dimensions}, Canad. Math. Bull. \textbf{16} (1973), 507--511.

\bibitem{chvatal1} V. Chv\'{a}tal, \emph{Remarks on a problem of Moser}, Canadian Math Bulletin, Vol 15, 1972, 19--21.

\bibitem{chvatal2} V. Chv\'{a}tal, \emph{Edmonds polytopes and a hierarchy of combinatorial problems}, Discrete Math. 4 (1973) 305-337.

\bibitem{elkin}
M. Elkin, \emph{An Improved Construction of Progression-Free Sets}, preprint.

\bibitem{fk1} H. Furstenberg, Y. Katznelson, \emph{A density version of the Hales-Jewett theorem for $k = 3$}, Graph Theory and Combinatorics (Cambridge, 1988). Discrete Math. 75 (1989), no. 1-3, 227–-241.

\bibitem{fk2} H. Furstenberg, Y. Katznelson, \emph{A density version of the Hales-Jewett theorem}, J. Anal. Math. 57 (1991), 64–-119. MR1191743

\bibitem{greenwolf}
B. Green, J. Wolf, \emph{A note on Elkin's improvement of Behrend's construction}, preprint.

\bibitem{komlos}
J. Koml\'{o}s, solution to problem P.170 by Leo Moser, Canad. Math.. Bull. vol {\bf (??check)} (1972), 312--313, 1970.

\bibitem{Krisha} K. Krishna, M. Narasimha Murty, "Genetic K-means algorithm," Systems, Man, and Cybernetics, Part B: Cybernetics, IEEE Transactions on , vol.29, no.3, pp.433-439, Jun 1999

\bibitem{moser} L. Moser, Problem P.170 in Canad. Math. Bull. 13 (1970), 268.

\bibitem{mcc} R. McCutcheon, \emph{The conclusion of the proof of the density Hales-Jewett theorem for k=3}, unpublished.

\bibitem{obryant}
K. O'Bryant, \emph{Sets of integers that do not contain long arithmetic progressions}, preprint.

\bibitem{oeis}
N. J. A. Sloane, Ed. (2008), The On-Line Encyclopedia of Integer Sequences, {\tt www.research.att.com/~njas/sequences/}

\bibitem{potenchin}
A. Potechin, \emph{Maximal caps in $AG(6, 3)$}, Journal Designs, Codes and Cryptography, Volume 46, Number 3 / March, 2008.

\bibitem{poly} D.H.J. Polymath, ???, preprint. {\bf need title}

\bibitem{rankin}
R. A. Rankin, Sets of integers containing not more than a given number of terms in arithmetical progression, Proc. Roy. Soc. Edinburgh Sect. A 65 (1960/1961), 332–344 (1960/61). MR 0142526 (26 \#95)

\bibitem{roth}
K. Roth, \emph{On certain sets of integers, I}, Journal of the London Mathematical Society \textbf{28} (1953), 104-–109.

\bibitem{Rothlauf} F. Rothlauf, D. E. Goldberg, Representations for Genetic and Evolutionary Algorithms. Physica-Verlag, 2002.

\bibitem{sperner}
E. Sperner, \emph{Ein Satz \"uber Untermengen einer endlichen Menge}, Mathematische Zeitschrift \textbf{27} (1928), 544-–548.

\bibitem{szem}
E. Szemer\'edi, \emph{On sets of integers containing no $k$ elements in arithmetic progression}, Acta Arithmetica \textbf{27} (1975), 199-–245.

\bibitem{waerden} {\tt http://en.wikipedia.org/wiki/Van_der_Waerden_number}

\end{thebibliography}

\end{document}

Outline of second paper

2009-07-12T07:53:39Z

Mpeake: /* Abstract */

Here is a proposed outline of the second paper, which will focus on the new bounds on DHJ(3) and Moser numbers, and related quantities.

== Metadata ==

* '''Author''': D.H.J. Polymath
* '''Address''': http://michaelnielsen.org/polymath1/index.php (Do we need a more stable address?)
* '''Email''': ???
* '''Title''': Density Hales-Jewett and Moser numbers
* '''AMS Subject classification''': ???

== Abstract ==

(A draft proposal - please edit)

For any <math>n \geq 0</math> and <math>k \geq 1</math>, the density Hales-Jewett number <math>c_{n,k}</math> is defined as the size of the largest subset of the cube <math>[k]^n := \{1,\ldots,k\}^n</math> which contains no combinatorial line; similarly, the Moser number <math>c'_{n,k}</math> is the largest subset of the cube <math>[k]^n</math> which contains no geometric line. A deep theorem of Furstenberg and Katznelson [cite] shows that <math>c_{n,k} = o(k^n)</math> as <math>n \to \infty</math> (which implies a similar claim for <math>c'_{n,k}</math>; this is already non-trivial for <math>k=3</math>. Several new proofs of this result have also been recently established [cite Polymath], [cite Austin].

Using both human and computer-assisted arguments, we compute several values of <math>c_{n,k}</math> and <math>c'_{n,k}</math> for small <math>n,k</math>. For instance the sequence <math>c_{n,3}</math> for <math>n=0,\ldots,6</math> is <math>1,2,6,18,52,150,450</math>, while the sequence <math>c'_{n,3}</math> for <math>n=0,\ldots,5</math> is <math>1,2,6,16,43,124,353</math>. We also establish some results for higher <math>k</math>, showing for instance that an analogue of the LYM inequality (which relates to the <math>k=2</math> case) does not hold for higher <math>k</math>.

== Sections ==

=== Introduction ===

Basic definitions. Definitions and notational conventions include

* [k] = {1, 2, ..., k}
* Subsets of [k]^n are called A
* definition of combinatorial line, geometric line
* Hales-Jewett numbers, Moser numbers

History of and motivation for the problem:

* Sperner's theorem
* Density Hales-Jewett theorem, including new proofs
* Review literature on Moser problem

New results

* Computation of several values of <math>c_{n,3}</math>
* Computation of several values of <math>c'_{n,3}</math>
* Asymptotic lower bounds for <math>c_{n,k}</math>
* Genetic algorithm lower bounds
* Some bounds for <math>c_{n,k}</math> for low n and large k
* Connection between Moser(2k) and DHJ(k)
* Hyper-optimistic conjecture, and its failure
* New bounds for colouring Hales-Jewett numbers

=== Lower bounds for density Hales-Jewett ===

Fujimura implies DHJ lower bounds; some selected numerics (e.g. lower bounds up to 10 dimensions, plus a few dimensions afterwards).

The precise asymptotic bound of <math>c_{n,k} > C k^{n - \alpha(k)\sqrt[\ell]{\log n}+\beta(k) \log \log n}</math>

Discussion of genetic algorithm

=== Low-dimensional density Hales-Jewett numbers ===

==== Very small n ====

<math>n=0,1,2</math> are trivial. But the six-point examples will get mentioned a lot.

For <math>n=3</math>, one needs to classify the 17-point and 18-point examples.

==== n=4 ====

One needs to classify the 50-point, 51-point, and 52-point examples.

==== n=5 ====

This is the big section, showing there are no 151-point examples.

==== n=6 ====

Easy corollary of n=5 theory

=== Higher k DHJ numbers ===

Exact computations of <math>c_{2,k}, c_{3,k}</math>

Connection between Moser<math>(n,2k)</math> and DHJ<math>(n,k)</math>

Numerics

Failure of hyper-optimistic conjecture

=== Lower bounds for Moser ===

Using Gamma sets to get lower bounds

Adding extra points from degenerate triangles

Higher k; Implications between Moser and DHJ

=== Moser in low dimensions ===

There is some general slicing lemma that needs to be proved here that allows inequalities for low-dim Moser to imply inequalities for higher dim.

For n=0,1,2 the theory is trivial.

For n=3 we need the classification of Pareto optimal configurations etc. So far this is only done by computer brute force search; we may have to find a human version.

n=4 theory: include both computer results and human results

n=5: we have a proof using the n=4 computer data; we should keep looking for a purely human proof.

n=6: we can give the partial results we have.

=== Fujimura's problem ===

=== Coloring DHJ ===

== Files ==

* [[polymath.tex]]
* [[introduction.tex]]
* [[dhj-lown.tex]]
* [[dhj-lown-lower.tex]]
* [[moser.tex]]
* [[moser-lower.tex]]
* [[fujimura.tex]]
* [[higherk.tex]]
* [[genetic.tex]]
* [[integer.tex]]
* [[coloring.tex]]
* [http://terrytao.wordpress.com/files/2009/07/alllinesin3n.pdf A figure depicting combinatorial lines]
* [http://terrytao.wordpress.com/files/2009/07/allgeomlinesin3n.pdf A figure depicting geometric lines]
* [http://thomas1111.files.wordpress.com/2009/06/moser353new.png A figure depicting a 353-point 6D Moser set]

The above are the master copies of the LaTeX files. Below are various compiled versions of the source:

* [http://terrytao.wordpress.com/files/2009/05/polymath.pdf May 24 version]
* [http://terrytao.wordpress.com/files/2009/05/polymath1.pdf May 25 version]
* [http://terrytao.files.wordpress.com/2009/05/polymath2.pdf May 27 version]
* [http://terrytao.wordpress.com/files/2009/05/polymath3.pdf Jun 1 version]
* [http://terrytao.wordpress.com/files/2009/06/polymath.pdf Jun 3 version]
* [http://terrytao.files.wordpress.com/2009/06/polymath2.pdf Jun 18 version]
* [http://terrytao.wordpress.com/files/2009/07/polymath.pdf Jul 9 version]

Higherk.tex

2009-07-11T04:30:26Z

Mpeake:

\section{Higher-k DHJ numbers}\label{higherk-sec}

For any $n$, $k$ let $c_{n,k}$ denote the cardinality of the largest subset of $[k]{}^n$ that does not contain a combinatorial line. When $k=3$, the quantity $c_{n,k} = c_n$ is studied in Sections \ref{dhj-lower-sec}, \ref{dhj-upper-sec}. The density Hales-Jewett theorem asserts that for any fixed $k$, $\lim_{n\rightarrow\infty} c_n/k^n = 0$ .
We trivially have $c_{n,1} = 0$ for $n > 0$, and Sperner's theorem tells us that
$$c_{n,2} = \binom{n}{\lfloor n/2 \rfloor}$$.
Now we look at the opposite regime, in which $n$ is small and $k$ is large. We easily have $c_{1,k} = k-1$ together with the trivial bound $c_{n+1,k} \leq kc_{n,k}$. This implies that
$$c_{n,k}\leq (k-1)k^{n-1}$$
for any $n\geq 1$. Let us call a pair $(n,k)$ with $n > 0$ \emph{saturated} if $c_{n,k}=(k-1)k^{n-1}$, thus there exists a line-free set with exactly one point omitted from every row and column.
The question naturally arises, Which pairs $(n,k)$ are saturated? From the above discussion we see that $(1,k)$ is saturated for all $k \geq 1$, and $(n,1)$ is (rather trivially) saturated for all $n$. Sperner's theorem tells us that $(n,2)$ is saturated only for $n= 1, 2$. Note that if $(n,k)$ is unsaturated then $(n',k)$ will be unsaturated for all $n' > n$.
A computer search has found the following $c_{n,k}$ values for different values of dimension $n$ and edgelength $k$. Several of these values reach the upper bound of $(k - 1)k^{n-1}$.

\begin{tabular}{|l|cccccc}
$n\backslash k$ & 2& 3 & 4 & 5 & 6 & 7 \\
\hline
2 & 2 & 6 & 12 & 20 & 30 & 42\\
3 & 3 & 18 & 48 & 100 & 180 & 294\\
4 & 6 & 52 & 183 & 500 & 1051-1079 & 2058\\
5 & 10 & 150 & 712-732 & 2500 & 6325-6480 & 14406
\end{tabular}

$(2,k)$ is saturated when $k$ is at least $1$. In dimension two the maximal set size is $k(k-1)$. This can be done by removing the diagonal values $11, 22, 33, \ldots, kk$. Since they are in disjoint lines this removal is minimal.
The $k$ missing points are one per line and one per column. So their $y$-coordinates are a shuffle of their $x$-coordinates. There are $k!$ rearrangements of the numbers $1$ to $k$. The $k$ points include a point on the diagonal, so this shuffle is not a derangement. There are $k!/e$ derangements of the numbers $1$ to $k$, so $k!(1-1/e)$ optimal solutions. This number of optimal solutions is sequence A002467 from the Online Encyclopedia of Integer Sequences.

$(3,k)$ is saturated when $k>2$. Let $S$ be a latin square of side $k$ on the symbols $1…k$, with colour $i$ in position $(i,i)$ (This is not possible for $k=2$)
Let axis one in $S$ correspond to coordinate $1$ in $[k]{}^3$, axis two to coordinate $2$ and interpret the colour in position $(i,j)$ as the third coordinate. Delete the points so defined.
The line with three wild cards has now been removed. A line with two wildcards will be missing the point corresponding to the diagonal in $S$. A line with a single wildcard will be missing a point corresponding to an off diagonal point in $S$.

\subsection{$(n,k)$ is saturated when all prime divisors of $k$ are at least $n$}
First consider the case when $k$ is prime and at least $n$: Delete those points whose coordinates add up to a multiple of $k$. Every combinatorial line has one point deleted, except for the major diagonal of $n=k$, which has all points deleted.
Now consider for instance the case $(n,k) = (4,35)$. Select one value modulo $35$ and eliminate it. Combinatorial lines with one, two, three or four moving coordinates will realize all values modulo $35$ as one, two, three, or four are units modulo $35$, thus $(4,35)$ is saturated.
The same argument tells us that $(n,k)$ is saturated when all prime divisors of $k$ are at least $n$.
On the other hand, computer data shows that $(4,4)$ and $(4,6)$ are not saturated.

\subsection{Failure of Hyper-optimistic conjecture}
Let $\overline{c}^\mu_{n,4}$ be the largest subset of the tetrahedral grid:
$$\{(a,b,c,d)\in\mathbb{Z}^4_+ : a+b+c+d = n\}$$
which contains no tetrahedrons $(a + r,b,c,d)$,$(a,b + r,c,d)$,$(a,b,c + r,d)$,$(a,b,c,d + r)$ with $r > 0$; call such sets tetrahedron-free.
The first few values, for $n$ from 0 to 7, were found by an integer programming routine, and are given in Figure \ref{Fujimura4}.

\begin{figure}[tb]\centerline{
\begin{tabular}[lllllllll]
n & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
$\overline{c}^\mu_{n,4}$ & 1 & 3 & 7 & 14 & 24 & 37 & 55 & 78
\end{tabular}
\label{Fujimura4}\caption{Fujimura numbers $\overline{c}^\mu_{n,4}$}}
\end{figure}

Recall the 'hyper-optimistic' Conjecture \ref{hoc} that the weighted sum of the points in any combinatorial line-free set is at most $\overline{c}^\mu_{n,4}$. The following example with $n=2$ and $k=4$ has a weighted sum of 7.5, and therefore disproves the conjecture for these values of $n$ and $k$. (Note that the case $k=3$, which is at the centre of most of Polymath's interest, is still open.)

$$\{(1, 1), (1, 2), (1, 3), (2, 1), (2, 3), (2, 4), (3, 2), (3, 3), (3,
4), (4, 1), (4, 2), (4, 4)\}$$

Coloring.tex

2009-07-11T04:00:24Z

Mpeake:

\section{The coloring Hales-Jewett problem}\label{coloring-sec}
The Hales-Jewett theorem states that for every $k$ and every $r$ there exists an $n = HJ(k,r)$ such that if you colour the elements of $[k]{}^n$ with $r$ colours, then there must be a combinatorial line with all its points of the same colour.

This is a consequence of the Density Hales-Jewett theorem, since there must be a set of density at least $r^{-1}$ of elements of $[k]{}^n$ all of whose elements have the same colour. It also follows from the Graham-Rothschild theorem [\cite{}].
By iterating the Hales-Jewett theorem, one can also show that one of the color classes contains an $m$-dimensional combinatorial subspace, if $n$ is sufficiently large depending on $k$, $r$ and $m$.

A related idea is that of the Van der Waerden number $W(k,r)$ [\cite{}], which is the length of the longest sequence $1,2,\ldots,n$ which can be $r$-coloured without a monochromatic arithmetic progression of length $k$. These numbers were used in this project to construct $r$-colourings of $[k]{}^m$ without combinatorial lines, and so provide new lower bounds for $HJ(k,r)$.

\begin{lemma} $HJ(k,r) \ge \lfloor (W(k,r)-1)/(k-1)\rfloor$ \end{lemma}
\begin{proof}
Let $m = \lfloor (W(k,r)-1)/(k-1) \rfloor$. Suppose we have a Van der Warden colouring of $1, 2, \ldots W(k,r)$. Give each point $(a_1,a_2,\ldots,a_m) \in [k]{}^m$ the colour that corresponds to $1+\sum (a_j-1)$. The points of a combinatorial line give sums that form an arithmetic progression, and will therefore not be monochromatic.
\end{proof}

A recent table of lower bounds for the Van der Waerden numbers is at this website: [http://www.st.ewi.tudelft.nl/sat/slides/waerden.pdf].
These numbers were used to produce the following table of lower bounds for colouring Hales-Jewett numbers.

\begin{figure}[tb] \centerline{\begin{tabular}[l|lllll]
k\r & 2 & 3 & 4 & 5 & 6 \\
\hline
3 & & 13 & 37 & 84 & 103 \\
4 & 11 & 97 & 349 & 751 & 3259 \\
5 & 59 & 302 & 2609 & 6011 & 14173 \\
6 & 226 & 1777 & 18061 & 49391 & 120097 \\
7 & 617 & 7309 & 64661 & & \\
8 & 1069 & 34057 & & & \\
9 & 3389 & & & &
\end{tabular}}\caption{Lower bounds for colouring Hales-Jewett numbers
\label{HJcolour}\end{figure}

Moser-lower.tex

2009-07-10T17:57:41Z

Mpeake:

\section{Lower bounds for the Moser problem}\label{moser-lower-sec}

Just as for the DHJ problem, we found that Gamma sets $\Gamma_{a,b,c}$ were useful in providing large lower bounds for the Moser problem. This is despite the fact that the symmetries of the cube do not respect Gamma sets.

Observe that if $B \subset \Delta_n$, then the set $A_B := \bigcup_{\vec a \in B} \Gamma_{a,b,c}$ is a Moser set as long as $B$ does not contain any ``isosceles triangles'' $(a+r,b,c+s), (a+s,b,c+r), (a,b+r+s,c)$ for any $r,s \geq 0$ not both zero; in particular, $B$ cannot contain any ``vertical line segments'' $(a+r,b,c+r), (a,b+2r,c)$. An example of such a set is provided by selecting $0 \leq i \leq n-3$ and letting $B$ consist of the triples $(a, n-i, i-a)$ when $a \neq 3 \mod 3$, $(a,n-i-1,i+1-a)$ when $a \neq 1 \mod 3$, $(a,n-i-2,i+2-a)$ when $a=0 \mod 3$, and $(a,n-i-3,i+3-a)$ when $a=2 \mod 3$. Asymptotically, this set includes about two thirds of the spheres $S_{n,i}$, $S_{n,i+1}$ and one third of the spheres $S_{n,i+2}, S_{n,i+3}$ and (setting $i$ close to $n/3$) gives a lower bound \eqref{cpn3} with $C = 2 \times \sqrt{\frac{9}{4\pi}}$.

An integer program was run to obtain the optimal lower bounds achievable by the $A_B$ construction (using \eqref{cn3}, of course). The results for
$1 \leq n \leq 20$ are displayed in Figure \ref{nlow-moser}:

\begin{figure}[tb]
\centerline{
\begin{tabular}{|ll|ll|}
\hline
$n$ & lower bound & $n$ & lower bound \\
\hline
1 & 2 &11& 71766\\
2 & 6 & 12& 212423\\
3 & 16 & 13& 614875\\
4 & 43 & 14& 1794212\\
5 & 122& 15& 5321796\\
6 & 353& 16& 15455256\\
7 & 1017& 17& 45345052\\
8 & 2902&18& 134438520\\
9 & 8622&19& 391796798\\
10& 24786& 20& 1153402148\\
\hline
\end{tabular}}
\caption{Lower bounds for $c'_n$ obtained by the $A_B$ construction.}
\label{nlow-moser}
\end{figure}

More complete data, including the list of optimisers, can be found at
\centerline{{\tt http://abel.math.umu.se/~klasm/Data/HJ/}.}

Note that the lower bound $c'_{6,3} \geq 353$ was first located by a genetic algorithm: see Appendix \ref{genetic-alg}.

\begin{figure}[tb]
\centerline{\includegraphics{moser353new.png}}
\caption{One of the examples of $353$-point sets in $[3]^6$ (elements of the set being indicated by white squares).}
\label{moser353-fig}
\end{figure}

However, we have been unable to locate a lower bound which is asymptotically better than \eqref{cpn3}. Indeed, any method based purely on the $A_B$ construction cannot do asymptotically better than the previous constructions:

\begin{proposition} Let $B \subset \Delta_n$ be such that $A_B$ is a Moser set. Then $|A_B| \leq (2 \sqrt{\frac{9}{4\pi}} + o(1)) \frac{3^n}{\sqrt{n}}$.
\end{proposition}

\begin{proof} By the previous discussion, $B$ cannot contain any pair of the form $(a,b+2r,c), (a+r,b,c+r)$ with $r>0$. In other words, for any $-n \leq h \leq n$, $B$ can contain at most one triple $(a,b,c)$ with $c-a=h$. From this and \eqref{cn3}, we see that
$$ |A_B| \leq \sum_{h=-n}^n \max_{(a,b,c) \in \Delta_n: c-a=h} \frac{n!}{a! b! c!}.$$
From the Chernoff inequality (or the Stirling formula computation below) we see that $\frac{n!}{a! b! c!} \leq \frac{1}{n^{10}} 3^n$ unless $a,b,c = n/3 + O( n^{1/2} \log^{1/2} n )$, so we may restrict to this regime, which also forces $h = O( n^{1/2}\log^{1/2} n)$. If we write $a = n/3 + \alpha$, $b = n/3 + \beta$, $c = n/3+\gamma$ and apply Stirling's formula $n! = (1+o(1)) \sqrt{2\pi n} n^n e^{-n}$, we obtain
$$ \frac{n!}{a! b! c!} = (1+o(1)) \frac{3^{3/2}}{2\pi n} 3^n \exp( - (\frac{n}{3}+\alpha) \log (1 + \frac{3\alpha}{n} ) - (\frac{n}{3}+\beta) \log (1 + \frac{3\beta}{n} ) - (\frac{n}{3}+\gamma) \log (1 + \frac{3\gamma}{n} ) ).$$
From Taylor expansion one has
$$ -(\frac{n}{3}+\alpha) \log (1 + \frac{3\alpha}{n} ) = -\alpha - \frac{3}{2} \frac{\alpha^2}{n} + o(1)$$
and similarly for $\beta,\gamma$; since $\alpha+\beta+\gamma=0$, we conclude that
$$ \frac{n!}{a! b! c!} = (1+o(1)) \frac{3^{3/2}}{2\pi n} 3^n \exp( - \frac{3}{2n} (\alpha^2+\beta^2+\gamma^2) ).$$
If $c-a=h$, then $\alpha^2+\beta^2+\gamma^2 = \frac{3\beta^2}{2} + \frac{h^2}{2}$. Thus we see that
$$ \max_{(a,b,c) \in \Delta_n: c-a=h} \frac{n!}{a! b! c!} \leq (1+o(1)) \frac{3^{3/2}}{2\pi n} 3^n \exp( - \frac{3}{4n} h^2 ).$$
Using the integral test, we thus have
$$ |A_B| \leq (1+o(1)) \frac{3^{3/2}}{2\pi n} 3^n \int_\R \exp( - \frac{3}{4n} x^2 )\ dx.$$
Since $\int_\R \exp( - \frac{3}{4n} x^2 )\ dx = \sqrt{\frac{4\pi n}{3}}$, we obtain the claim.
\end{proof}

Actually it is possible to improve upon these bounds by a slight amount. Observe that if $B$ is a maximiser for the right-hand side of \eqref{cn3} (subject to $B$ not containing isosceles triangles), then any triple $(a,b,c)$ not in $B$ must be the vertex of a (possibly degenerate) isosceles triangle with the other vertices in $B$. If this triangle is non-degenerate, or if $(a,b,c)$ is the upper vertex of a degenerate isosceles triangle, then no point from $\Gamma_{a,b,c}$ can be added to $A_B$ without creating a geometric line. However, if $(a,b,c) = (a'+r,b',c'+r)$ is only the lower vertex of a degenerate isosceles triangle $(a'+r,b',c'+r), (a',b'+2r,c')$, then one can add any subset of $\Gamma_{a,b,c}$ to $A_B$ and still have a Moser set as long as no pair of elements in that subset is separated by Hamming distance $2r$. For instance, in the $n=5$ case, we can start with the 122-point set built from
$$B = \{ (0 0 5),(0 2 3),(1 1 3 ),(1 2 2 ),(2 2 1 ),(3 1 1 ),(3 2 0 ),(5 0 0))\}$$, and add a point each from (1 0 4) and (4 0 1).
This gives an example of the maximal, 124-point solution. Again, in the $n=10$ case, the set
\begin{align*}
B &= \{(0 0 10),(0 2 8 ),(0 3 7 ),(0 4 6 ),(1 4 5 ),(2 1 7 ),(2 3 5 ),
(3 2 5 ),(3 3 4 ),(3 4 3 ),(4 4 2 ),(5 1 4 ), \\
&\quad (5 3 2 ),(6 2 2 ),
(6 3 1 ),(6 4 0 ),(8 1 1 ),(9 0 1 ),(9 1 0 ) \}
\end{align*}

generates the lower bound $c'_{10,3} \geq 24786$ given above (and, up to reflection $a \leftrightarrow c$, is the only such set that does so); but by adding the following twelve elements from $\Gamma_{5,0,5}$ one can increase the lower bound slightly to $24798$: $1111133333$, $1111313333$,
$1113113333$, $1133331113$, $1133331131$, $1133331311$, $3311333111$,
$3313133111$, $3313313111$, $3331111133$, $3331111313$, $3331111331$

A more general form goes with the $B$ set described at the start of this section. Include points from $\Gamma_{(a,n-i-4,i+4-a)}$ when $a=1 (\operatorname{mod}\ 3)$, subject to no two points being included if they differ by the interchange of a $1$ and a $3$. Each of these Gamma sets is the feet of a degenerate isosceles triangle with vertex $\Gamma_{(a-1,n-i-2,a+3-a)}$.

\begin{lemma} If $A$ is a subset of $\Gamma_{(a,b,c)}$ such that no two points of $A$ differ by the interchange of a $1$ and a $3$, then $|A| \leq |\Gamma_{a,b,c}|/(1+\max(a,c))$.
\end{lemma}
\begin{proof}
Say that two points of $\Gamma_{a,b,c}$ are neighbours if they differ by the exchange of a $1$ and a $3$. Each point of $A$ has $ac$ neighbours, none of which are in $A$. Each point of $\Gamma_{(a,b,c)}\backslash A$ has $ac$ neighbours, but only $min(a,c)$ of them may be in $A$. So for each point of $A$ there are on average $ac/\min(a,c) = \max(a,c)$ points not in $A$. So the proportion of points of $\Gamma_{(a,b,c)}$ that are in $A$ is at most one in $1+\max(a,c)$.
\end{proof}

The proportion of extra points for each of the cells $\Gamma_{(a,n-i-4,i+4-a)}$ is no more than $2/(i+6)$. Only one cell in three is included from the $b=n-i-4$ layer, so we expect no more than $\frac{2}{3(i+6)}\binom(n,i+4)2^{i+4}$ new points, all from $S_{n,i+4}$. One can also find extra points from $S_{n,i+5}$ and higher spheres.

Earlier solutions may also give insight into the problem. In this section we discuss lower bounds for $c'_{n,3}$. Clearly we have
$c'_{0,3}=1$ and $c'_{1,3}=2$, so we focus on the case $n \ge 2$.
The first lower bounds may be due to Koml\'{o}s \cite{komlos}, who observed
that the sphere $S_{i,n}$ of elements with exactly $n-i$ 2 entries
(see Section \ref{notation-sec} for definition),
is a Moser set, so that
\begin{equation}\label{cin}
c'_{n,3}\geq \vert S_{i,n}\vert
\end{equation}
holds for all $i$. Choosing $i=\lfloor \frac{2n}{3}\rfloor$ and
applying Stirling's formula, we see that this lower bound takes the form
\begin{equation}\label{cpn3}
c'_{n,3} \geq (C-o(1)) 3^n / \sqrt{n}
\end{equation}
for some absolute constant $C>0$; in fact \eqref{cin} gives \eqref{cpn3} with $C := \sqrt{\frac{9}{4\pi}}$.
In particular $c'_{3,3} \geq 12, c'_{4,3}\geq 24, c'_{5,3}\geq 80,
c'_{6,3}\geq 240$.

These values can be improved by studying combinations of several spheres or
semispheres or applying elementary results from coding theory.

Observe that if $\{w(1),w(2),w(3)\}$ is a geometric line in $[3]^n$,
then $w(1), w(3)$ both lie in the same sphere $S_{i,n}$,
and that $w(2)$ lies in a lower
sphere $S_{i-r,n}$ for some $1 \leq r \leq i \leq n$. Furthermore,
$w(1)$ and $w(3)$ are separated by Hamming distance $r$.

As a consequence, we see that $S_{i-1,n} \cup S_{i,n}^e$ (or $S_{i-1,n}
\cup S_{i,n}^o$) is a Moser set for any $1 \leq i \leq n$, since any two
distinct elements $S_{i,n}^e$ are separated by a Hamming distance of at
least two.
(Recall Section \ref{notation-sec} for definitions),
This leads to the lower bound
\begin{equation}\label{cn3-low}
c'_{n,3} \geq \binom{n}{i-1}
2^{i-1} + \binom{n}{i} 2^{i-1} = \binom{n+1}{i} 2^{i-1}.
\end{equation}
It is not
hard to see that $\binom{n+1}{i+1} 2^{i} > \binom{n+1}{i} 2^{i-1}$ if
and only if $3i < 2n+1$, and so this lower bound is maximised when $i =
\lfloor \frac{2n+1}{3} \rfloor$ for $n \geq 2$, giving the formula
\eqref{binom}. This leads to the lower bounds $$ c'_{2,3} \geq 6;
c'_{3,3} \geq 16; c'_{4,3} \geq 40; c'_{5,3} \geq 120; c'_{6,3} \geq
336$$ which gives the right lower bounds for $n=2,3$, but is slightly
off for $n=4,5$. Asymptotically, Stirling's formula and \eqref{cn3-low} then
give the lower bound \eqref{cpn3} with $C = \frac{3}{2} \times \sqrt{\frac{9}{4\pi}}$, which is asymptotically $50\%$ better than the bound \eqref{cin}.

The work of Chv\'{a}tal \cite{chvatal1}
already contained a refinement of this idea
which we here translate into the usual notation of coding theory:
Let $A(n,d)$ denote the size of the largest binary code of length $n$
and minimal distance $d$.

Then
\begin{equation}\label{cnchvatal}
c'_{n,3}\geq \max_k \left( \sum_{j=0}^k \binom{n}{j} A(n-j, k-j+1)\right).
\end{equation}

With the following values for $A(n,d)$:
{\tiny{
\[
\begin{array}{llllllll}
A(1,1)=2&&&&&&&\\
A(2,1)=4& A(2,2)=2&&&&&&\\
A(3,1)=8&A(3,2)=4&A(3,3)=2&&&&&\\
A(4,1)=16&A(4,2)=8& A(4,3)=2& A(4,4)=2&&&&\\
A(5,1)=32&A(5,2)=16& A(5,3)=4& A(5,4)=2&A(5,5)=2&&&\\
A(6,1)=64&A(6,2)=32& A(6,3)=8& A(6,4)=4&A(6,5)=2&A(6,6)=2&&\\
A(7,1)=128&A(7,2)=64& A(7,3)=16& A(7,4)=8&A(7,5)=2&A(7,6)=2&A(7,7)=2&\\
A(8,1)=256&A(8,2)=128& A(8,3)=20& A(8,4)=16&A(8,5)=4&A(8,6)=2
&A(8,7)=2&A(8,8)=2\\
A(9,1)=512&A(9,2)=256& A(9,3)=40& A(9,4)=20&A(9,5)=6&A(9,6)=4
&A(9,7)=2&A(9,8)=2\\
A(10,1)=1024&A(10,2)=512& A(10,3)=72& A(10,4)=40&A(10,5)=12&A(10,6)=6
&A(10,7)=2&A(10,8)=2\\
A(11,1)=2048&A(11,2)=1024& A(11,3)=144& A(11,4)=72&A(11,5)=24&A(11,6)=12
&A(11,7)=2&A(11,8)=2\\
A(12,1)=4096&A(12,2)=2048& A(12,3)=256& A(12,4)=144&A(12,5)=32&A(12,6)=24
&A(12,7)=4&A(12,8)=2\\
A(13,1)=8192&A(13,2)=4096& A(13,3)=512& A(13,4)=256&A(13,5)=64&A(12,6)=32
&A(13,7)=8&A(13,8)=4\\
\end{array}
\]
}}

Generally, $A(n,1)=2^n, A(n,2)=2^{n-1}, A(n-1,2e-1)=A(n,2e), A(n,d)=2$,
if $d>\frac{2n}{3}$.
The values were taken or derived from Andries Brower's table at
\centerline{\tt http://www.win.tue.nl/$\sim$aeb/codes/binary-1.html}
\textbf{include to references? or other book with explicit values of $A(n,d)$ }

For $c'_{n,3}$ we obtain the following lower bounds:
with $k=2$
\[
\begin{array}{llll}
c'_{4,3}&\geq &\binom{4}{0}A(4,3)+\binom{4}{1}A(3,2)+\binom{4}{2}A(2,1)
=1\cdot 2+4 \cdot 4+6\cdot 4&=42.\\
c'_{5,3}&\geq &\binom{5}{0}A(5,3)+\binom{5}{1}A(4,2)+\binom{5}{2}A(3,1)
=1\cdot 4+5 \cdot 8+10\cdot 8&=124.\\
c'_{6,3}&\geq &\binom{6}{0}A(6,3)+\binom{6}{1}A(5,2)+\binom{6}{2}A(4,1)
=1\cdot 8+6 \cdot 16+15\cdot 16&=344.
\end{array}
\]
With k=3
\[
\begin{array}{llll}
c'_{7,3}&\geq& \binom{7}{0}A(7,4)+\binom{7}{1}A(6,3)+\binom{7}{2}A(5,2)
+ \binom{7}{3}A(4,1)&=960.\\
c'_{8,3}&\geq &\binom{8}{0}A(8,4)+\binom{8}{1}A(7,3)+\binom{8}{2}A(6,2)
+ \binom{8}{3}A(5,1)&=2832.\\
c'_{9,3}&\geq & \binom{9}{0}A(9,4)+\binom{9}{1}A(8,3)+\binom{9}{2}A(7,2)
+ \binom{9}{3}A(6,1)&=7880.
\end{array}\]
With k=4
\[
\begin{array}{llll}
c'_{10,3}&\geq &\binom{10}{0}A(10,5)+\binom{10}{1}A(9,4)+\binom{10}{2}A(8,3)
+ \binom{10}{3}A(7,2)+\binom{10}{4}A(6,1)&=22232.\\
c'_{11,3}&\geq &\binom{11}{0}A(11,5)+\binom{11}{1}A(10,4)+\binom{11}{2}A(9,3)
+ \binom{11}{3}A(8,2)+\binom{11}{4}A(7,1)&=66024.\\
c'_{12,3}&\geq &\binom{12}{0}A(12,5)+\binom{12}{1}A(11,4)+\binom{12}{2}A(10,3)
+ \binom{12}{3}A(9,2)+\binom{12}{4}A(8,1)&=188688.\\
\end{array}\]
With $k=5$
\[ c'_{13,3}\geq 539168.\]

It should be pointed out that these bounds are even numbers, so that
$c'_{4,3}=43$ shows that one cannot generally expect this lower bound
gives the optimum.

The maximum value appears to occur for $k=\lfloor\frac{n+2}{3}\rfloor$,
so that using
Stirling's formula and explicit bounds on $A(n,d)$ the
best possible value known to date
of the constant $C$ in
equation \eqref{cpn3} can be worked out, but we refrain from doing this here.
Using the Singleton bound $A(n,d)\leq 2^{n-d+1}$ Chv\'{a}tal
\cite{chvatal1} proved that the expression on the right hand side of
\eqref{cnchvatal} is also
$O\left( \frac{3^n}{\sqrt{n}}\right)$, so that the refinement described
above gains a constant factor over the initial construction only.

For $n=4$ the above does not yet give the exact value.
The value $c'_{4,3}=43$ was first proven by Chandra \cite{chandra}.
A uniform way of describing examples for the optimum values of
$c'_{4,3}=43$ and $c'_{5,3}=124$ is the following:

Let us consider the sets $$ A := S_{i-1,n}
\cup S_{i,n}^e \cup A'$$ where $A' \subset S_{i+1,n}$ has the property
that any two elements in $A'$ are separated by a Hamming distance of at
least three, or have a Hamming distance of exactly one but their
midpoint lies in $S_{i,n}^o$. By the previous discussion we see that
this is a Moser set, and we have the lower bound
\begin{equation}\label{cnn}
c'_{n,3} \geq \binom{n+1}{i} 2^{i-1} + |A'|.
\end{equation} This gives some improved lower bounds for $c'_{n,3}$:

\begin{itemize} \item By taking $n=4$, $i=3$, and $A' = \{ 1111, 3331,
3333\}$, we obtain $c'_{4,3} \geq 43$; \item By taking $n=5$, $i=4$, and
$A' = \{ 11111, 11333, 33311, 33331 \}$, we obtain $c'_{5,3} \geq 124$.
\item By taking $n=6$, $i=5$, and $A' = \{ 111111, 111113, 111331,
111333, 331111, 331113\}$, we obtain $c'_{6,3} \geq 342$. \end{itemize}

This gives the lower bounds in Theorem \ref{moser} up to $n=5$, but
the bound for $n=6$ is inferior to the lower bound $c'_{6,3}\geq 344$ given above.

\subsection{Higher $k$ values}
One can consider subsets of $[k]{}^n$ that contain no geometric lines. Section \ref{moser-lower-sec} has considered the case $k=3$. Let $c'_{n,k}$ be the greatest number of points in $[k]{}^n$ with no geometric line. For example, $c'_{n,3} = c'_n$. We have the following lower bounds:

$c'_{n,4} \ge \binom{n}{n/2}2^n$.
The set of points with $a$ $1$s,$b$ $2$s,$c$ $3$s and $d$ $4$s, where $a+d$ has the constant value $n/2$, does not form geometric lines because points at the ends of a geometric line have more $a$ or $d$ values than point in the middle of the line.

One can show a lower bound that, asymptotically, is twice as large. Take all points with $a$ $1$s, $b$ $2$s, $c$ $3$s and $d$ $4$s, for which:

\begin{itemize}
\item Either $a+d = q$ or $q-1$, $a$ and $b$ have the same parity; or
\item $a+d = q-2$ or $q-3$, $a$ and $b$ have opposite parity.
\end{itemize}

This includes half the points of four adjacent layers, and therefore may include $(1+o(1))\binom{n}{n/2}2^{n+1}$ points.

We also have a DHJ(3)-like lower bound for $c'_{n,5}$, namely $c'_{n,5} = 5^{n-O(\sqrt{\log n})}$.
Consider points with $a$ $1$s, $b$ $2$s, $c$ $3$s, $d$ $4$s and $e$ $5$s. For each point, take the value $a+e+2(b+d)+3c$. The first three points in any geometric line give values that form an arithmetic progression of length three.

Select a set of integers with no arithmetic progression of length 3. Select all points whose value belongs to that sequence; there will be no geometric line among those points. By Behrend theory, it is possible to choose these points with density $\exp{-O(\sqrt{\log n})}$.

The $k=6$ version of Moser implies DHJ(3). Indeed, any $k=3$ combinatorial line-free set can be "doubled up" into a $k=6$ geometric line-free set of the same density by pulling back the set from the map that maps 1, 2, 3, 4, 5, 6 to 1, 2, 3, 3, 2, 1 respectively; note that this map sends $k=6$ geometric lines to $k=3$ combinatorial lines. So $c'_{k,6} \geq 2^k c_k$, and more generally, $c'_{k,2n} \geq 2^k c_{k,n}$.

Moser-lower.tex

2009-07-10T17:55:30Z

Mpeake:

\section{Lower bounds for the Moser problem}\label{moser-lower-sec}

Just as for the DHJ problem, we found that Gamma sets $\Gamma_{a,b,c}$ were useful in providing large lower bounds for the Moser problem. This is despite the fact that the symmetries of the cube do not respect Gamma sets.

Observe that if $B \subset \Delta_n$, then the set $A_B := \bigcup_{\vec a \in B} \Gamma_{a,b,c}$ is a Moser set as long as $B$ does not contain any ``isosceles triangles'' $(a+r,b,c+s), (a+s,b,c+r), (a,b+r+s,c)$ for any $r,s \geq 0$ not both zero; in particular, $B$ cannot contain any ``vertical line segments'' $(a+r,b,c+r), (a,b+2r,c)$. An example of such a set is provided by selecting $0 \leq i \leq n-3$ and letting $B$ consist of the triples $(a, n-i, i-a)$ when $a \neq 3 \mod 3$, $(a,n-i-1,i+1-a)$ when $a \neq 1 \mod 3$, $(a,n-i-2,i+2-a)$ when $a=0 \mod 3$, and $(a,n-i-3,i+3-a)$ when $a=2 \mod 3$. Asymptotically, this set includes about two thirds of the spheres $S_{n,i}$, $S_{n,i+1}$ and one third of the spheres $S_{n,i+2}, S_{n,i+3}$ and (setting $i$ close to $n/3$) gives a lower bound \eqref{cpn3} with $C = 2 \times \sqrt{\frac{9}{4\pi}}$.

An integer program was run to obtain the optimal lower bounds achievable by the $A_B$ construction (using \eqref{cn3}, of course). The results for
$1 \leq n \leq 20$ are displayed in Figure \ref{nlow-moser}:

\begin{figure}[tb]
\centerline{
\begin{tabular}{|ll|ll|}
\hline
$n$ & lower bound & $n$ & lower bound \\
\hline
1 & 2 &11& 71766\\
2 & 6 & 12& 212423\\
3 & 16 & 13& 614875\\
4 & 43 & 14& 1794212\\
5 & 122& 15& 5321796\\
6 & 353& 16& 15455256\\
7 & 1017& 17& 45345052\\
8 & 2902&18& 134438520\\
9 & 8622&19& 391796798\\
10& 24786& 20& 1153402148\\
\hline
\end{tabular}}
\caption{Lower bounds for $c'_n$ obtained by the $A_B$ construction.}
\label{nlow-moser}
\end{figure}

More complete data, including the list of optimisers, can be found at
\centerline{{\tt http://abel.math.umu.se/~klasm/Data/HJ/}.}

Note that the lower bound $c'_{6,3} \geq 353$ was first located by a genetic algorithm: see Appendix \ref{genetic-alg}.

\begin{figure}[tb]
\centerline{\includegraphics{moser353new.png}}
\caption{One of the examples of $353$-point sets in $[3]^6$ (elements of the set being indicated by white squares).}
\label{moser353-fig}
\end{figure}

However, we have been unable to locate a lower bound which is asymptotically better than \eqref{cpn3}. Indeed, any method based purely on the $A_B$ construction cannot do asymptotically better than the previous constructions:

\begin{proposition} Let $B \subset \Delta_n$ be such that $A_B$ is a Moser set. Then $|A_B| \leq (2 \sqrt{\frac{9}{4\pi}} + o(1)) \frac{3^n}{\sqrt{n}}$.
\end{proposition}

\begin{proof} By the previous discussion, $B$ cannot contain any pair of the form $(a,b+2r,c), (a+r,b,c+r)$ with $r>0$. In other words, for any $-n \leq h \leq n$, $B$ can contain at most one triple $(a,b,c)$ with $c-a=h$. From this and \eqref{cn3}, we see that
$$ |A_B| \leq \sum_{h=-n}^n \max_{(a,b,c) \in \Delta_n: c-a=h} \frac{n!}{a! b! c!}.$$
From the Chernoff inequality (or the Stirling formula computation below) we see that $\frac{n!}{a! b! c!} \leq \frac{1}{n^{10}} 3^n$ unless $a,b,c = n/3 + O( n^{1/2} \log^{1/2} n )$, so we may restrict to this regime, which also forces $h = O( n^{1/2}\log^{1/2} n)$. If we write $a = n/3 + \alpha$, $b = n/3 + \beta$, $c = n/3+\gamma$ and apply Stirling's formula $n! = (1+o(1)) \sqrt{2\pi n} n^n e^{-n}$, we obtain
$$ \frac{n!}{a! b! c!} = (1+o(1)) \frac{3^{3/2}}{2\pi n} 3^n \exp( - (\frac{n}{3}+\alpha) \log (1 + \frac{3\alpha}{n} ) - (\frac{n}{3}+\beta) \log (1 + \frac{3\beta}{n} ) - (\frac{n}{3}+\gamma) \log (1 + \frac{3\gamma}{n} ) ).$$
From Taylor expansion one has
$$ -(\frac{n}{3}+\alpha) \log (1 + \frac{3\alpha}{n} ) = -\alpha - \frac{3}{2} \frac{\alpha^2}{n} + o(1)$$
and similarly for $\beta,\gamma$; since $\alpha+\beta+\gamma=0$, we conclude that
$$ \frac{n!}{a! b! c!} = (1+o(1)) \frac{3^{3/2}}{2\pi n} 3^n \exp( - \frac{3}{2n} (\alpha^2+\beta^2+\gamma^2) ).$$
If $c-a=h$, then $\alpha^2+\beta^2+\gamma^2 = \frac{3\beta^2}{2} + \frac{h^2}{2}$. Thus we see that
$$ \max_{(a,b,c) \in \Delta_n: c-a=h} \frac{n!}{a! b! c!} \leq (1+o(1)) \frac{3^{3/2}}{2\pi n} 3^n \exp( - \frac{3}{4n} h^2 ).$$
Using the integral test, we thus have
$$ |A_B| \leq (1+o(1)) \frac{3^{3/2}}{2\pi n} 3^n \int_\R \exp( - \frac{3}{4n} x^2 )\ dx.$$
Since $\int_\R \exp( - \frac{3}{4n} x^2 )\ dx = \sqrt{\frac{4\pi n}{3}}$, we obtain the claim.
\end{proof}

Actually it is possible to improve upon these bounds by a slight amount. Observe that if $B$ is a maximiser for the right-hand side of \eqref{cn3} (subject to $B$ not containing isosceles triangles), then any triple $(a,b,c)$ not in $B$ must be the vertex of a (possibly degenerate) isosceles triangle with the other vertices in $B$. If this triangle is non-degenerate, or if $(a,b,c)$ is the upper vertex of a degenerate isosceles triangle, then no point from $\Gamma_{a,b,c}$ can be added to $A_B$ without creating a geometric line. However, if $(a,b,c) = (a'+r,b',c'+r)$ is only the lower vertex of a degenerate isosceles triangle $(a'+r,b',c'+r), (a',b'+2r,c')$, then one can add any subset of $\Gamma_{a,b,c}$ to $A_B$ and still have a Moser set as long as no pair of elements in that subset is separated by Hamming distance $2r$. For instance, in the $n=5$ case, we can start with the 122-point set built from
$$B = \{ (0 0 5),(0 2 3),(1 1 3 ),(1 2 2 ),(2 2 1 ),(3 1 1 ),(3 2 0 ),(5 0 0))\}$$, and add a point each from (1 0 4) and (4 0 1).
This gives an example of the maximal, 124-point solution. Again, in the $n=10$ case, the set
\begin{align*}
B &= \{(0 0 10),(0 2 8 ),(0 3 7 ),(0 4 6 ),(1 4 5 ),(2 1 7 ),(2 3 5 ),
(3 2 5 ),(3 3 4 ),(3 4 3 ),(4 4 2 ),(5 1 4 ), \\
&\quad (5 3 2 ),(6 2 2 ),
(6 3 1 ),(6 4 0 ),(8 1 1 ),(9 0 1 ),(9 1 0 ) \}
\end{align*}

generates the lower bound $c'_{10,3} \geq 24786$ given above (and, up to reflection $a \leftrightarrow c$, is the only such set that does so); but by adding the following twelve elements from $\Gamma_{5,0,5}$ one can increase the lower bound slightly to $24798$: $1111133333$, $1111313333$,
$1113113333$, $1133331113$, $1133331131$, $1133331311$, $3311333111$,
$3313133111$, $3313313111$, $3331111133$, $3331111313$, $3331111331$

A more general form goes with the $B$ set described at the start of this section. Include points from $\Gamma_{(a,n-i-4,i+4-a)}$ when $a=1 (\operatorname{mod}\ 3)$, subject to no two points being included if they differ by the interchange of a $1$ and a $3$. Each of these Gamma sets is the feet of a degenerate isosceles triangle with vertex $\Gamma_{(a-1,n-i-2,a+3-a)}$.

\begin{lemma} If $A$ is a subset of $\Gamma_{(a,b,c)}$ such that no two points of $A$ differ by the interchange of a $1$ and a $3$, then $|A| \leq |\Gamma_{a,b,c}|/(1+\max(a,c))$.
\end{lemma}
\begin{proof}
Say that two points of $\Gamma_{a,b,c}$ are neighbours if they differ by the exchange of a $1$ and a $3$. Each point of $A$ has $ac$ neighbours, none of which are in $A$. Each point of $\Gamma_{(a,b,c)}\backslash A$ has $ac$ neighbours, but only $min(a,c)$ of them may be in $A$. So for each point of $A$ there are on average $ac/\min(a,c) = \max(a,c)$ points not in $A$. So the proportion of points of $\Gamma_{(a,b,c)}$ that are in $A$ is at most one in $1+\max(a,c)$.
\end{proof}

The proportion of extra points for each of the cells $\Gamma_{(a,n-i-4,i+4-a)}$ is no more than $2/(i+6)$. Only one cell in three is included from the $b=n-i-4$ layer, so we expect no more than $\frac{2}{3(i+6)}\binom(n,i+4)2^{i+4}$ new points, all from $S_{n,i+4}$. One can also find extra points from $S_{n,i+5}$ and higher spheres.

Earlier solutions may also give insight into the problem. In this section we discuss lower bounds for $c'_{n,3}$. Clearly we have
$c'_{0,3}=1$ and $c'_{1,3}=2$, so we focus on the case $n \ge 2$.
The first lower bounds may be due to Koml\'{o}s \cite{komlos}, who observed
that the sphere $S_{i,n}$ of elements with exactly $n-i$ 2 entries
(see Section \ref{notation-sec} for definition),
is a Moser set, so that
\begin{equation}\label{cin}
c'_{n,3}\geq \vert S_{i,n}\vert
\end{equation}
holds for all $i$. Choosing $i=\lfloor \frac{2n}{3}\rfloor$ and
applying Stirling's formula, we see that this lower bound takes the form
\begin{equation}\label{cpn3}
c'_{n,3} \geq (C-o(1)) 3^n / \sqrt{n}
\end{equation}
for some absolute constant $C>0$; in fact \eqref{cin} gives \eqref{cpn3} with $C := \sqrt{\frac{9}{4\pi}}$.
In particular $c'_{3,3} \geq 12, c'_{4,3}\geq 24, c'_{5,3}\geq 80,
c'_{6,3}\geq 240$.

These values can be improved by studying combinations of several spheres or
semispheres or applying elementary results from coding theory.

Observe that if $\{w(1),w(2),w(3)\}$ is a geometric line in $[3]^n$,
then $w(1), w(3)$ both lie in the same sphere $S_{i,n}$,
and that $w(2)$ lies in a lower
sphere $S_{i-r,n}$ for some $1 \leq r \leq i \leq n$. Furthermore,
$w(1)$ and $w(3)$ are separated by Hamming distance $r$.

As a consequence, we see that $S_{i-1,n} \cup S_{i,n}^e$ (or $S_{i-1,n}
\cup S_{i,n}^o$) is a Moser set for any $1 \leq i \leq n$, since any two
distinct elements $S_{i,n}^e$ are separated by a Hamming distance of at
least two.
(Recall Section \ref{notation-sec} for definitions),
This leads to the lower bound
\begin{equation}\label{cn3-low}
c'_{n,3} \geq \binom{n}{i-1}
2^{i-1} + \binom{n}{i} 2^{i-1} = \binom{n+1}{i} 2^{i-1}.
\end{equation}
It is not
hard to see that $\binom{n+1}{i+1} 2^{i} > \binom{n+1}{i} 2^{i-1}$ if
and only if $3i < 2n+1$, and so this lower bound is maximised when $i =
\lfloor \frac{2n+1}{3} \rfloor$ for $n \geq 2$, giving the formula
\eqref{binom}. This leads to the lower bounds $$ c'_{2,3} \geq 6;
c'_{3,3} \geq 16; c'_{4,3} \geq 40; c'_{5,3} \geq 120; c'_{6,3} \geq
336$$ which gives the right lower bounds for $n=2,3$, but is slightly
off for $n=4,5$. Asymptotically, Stirling's formula and \eqref{cn3-low} then
give the lower bound \eqref{cpn3} with $C = \frac{3}{2} \times \sqrt{\frac{9}{4\pi}}$, which is asymptotically $50\%$ better than the bound \eqref{cin}.

The work of Chv\'{a}tal \cite{chvatal1}
already contained a refinement of this idea
which we here translate into the usual notation of coding theory:
Let $A(n,d)$ denote the size of the largest binary code of length $n$
and minimal distance $d$.

Then
\begin{equation}\label{cnchvatal}
c'_{n,3}\geq \max_k \left( \sum_{j=0}^k \binom{n}{j} A(n-j, k-j+1)\right).
\end{equation}

With the following values for $A(n,d)$:
{\tiny{
\[
\begin{array}{llllllll}
A(1,1)=2&&&&&&&\\
A(2,1)=4& A(2,2)=2&&&&&&\\
A(3,1)=8&A(3,2)=4&A(3,3)=2&&&&&\\
A(4,1)=16&A(4,2)=8& A(4,3)=2& A(4,4)=2&&&&\\
A(5,1)=32&A(5,2)=16& A(5,3)=4& A(5,4)=2&A(5,5)=2&&&\\
A(6,1)=64&A(6,2)=32& A(6,3)=8& A(6,4)=4&A(6,5)=2&A(6,6)=2&&\\
A(7,1)=128&A(7,2)=64& A(7,3)=16& A(7,4)=8&A(7,5)=2&A(7,6)=2&A(7,7)=2&\\
A(8,1)=256&A(8,2)=128& A(8,3)=20& A(8,4)=16&A(8,5)=4&A(8,6)=2
&A(8,7)=2&A(8,8)=2\\
A(9,1)=512&A(9,2)=256& A(9,3)=40& A(9,4)=20&A(9,5)=6&A(9,6)=4
&A(9,7)=2&A(9,8)=2\\
A(10,1)=1024&A(10,2)=512& A(10,3)=72& A(10,4)=40&A(10,5)=12&A(10,6)=6
&A(10,7)=2&A(10,8)=2\\
A(11,1)=2048&A(11,2)=1024& A(11,3)=144& A(11,4)=72&A(11,5)=24&A(11,6)=12
&A(11,7)=2&A(11,8)=2\\
A(12,1)=4096&A(12,2)=2048& A(12,3)=256& A(12,4)=144&A(12,5)=32&A(12,6)=24
&A(12,7)=4&A(12,8)=2\\
A(13,1)=8192&A(13,2)=4096& A(13,3)=512& A(13,4)=256&A(13,5)=64&A(12,6)=32
&A(13,7)=8&A(13,8)=4\\
\end{array}
\]
}}

Generally, $A(n,1)=2^n, A(n,2)=2^{n-1}, A(n-1,2e-1)=A(n,2e), A(n,d)=2$,
if $d>\frac{2n}{3}$.
The values were taken or derived from Andries Brower's table at
\centerline{\tt http://www.win.tue.nl/$\sim$aeb/codes/binary-1.html}
\textbf{include to references? or other book with explicit values of $A(n,d)$ }

For $c'_{n,3}$ we obtain the following lower bounds:
with $k=2$
\[
\begin{array}{llll}
c'_{4,3}&\geq &\binom{4}{0}A(4,3)+\binom{4}{1}A(3,2)+\binom{4}{2}A(2,1)
=1\cdot 2+4 \cdot 4+6\cdot 4&=42.\\
c'_{5,3}&\geq &\binom{5}{0}A(5,3)+\binom{5}{1}A(4,2)+\binom{5}{2}A(3,1)
=1\cdot 4+5 \cdot 8+10\cdot 8&=124.\\
c'_{6,3}&\geq &\binom{6}{0}A(6,3)+\binom{6}{1}A(5,2)+\binom{6}{2}A(4,1)
=1\cdot 8+6 \cdot 16+15\cdot 16&=344.
\end{array}
\]
With k=3
\[
\begin{array}{llll}
c'_{7,3}&\geq& \binom{7}{0}A(7,4)+\binom{7}{1}A(6,3)+\binom{7}{2}A(5,2)
+ \binom{7}{3}A(4,1)&=960.\\
c'_{8,3}&\geq &\binom{8}{0}A(8,4)+\binom{8}{1}A(7,3)+\binom{8}{2}A(6,2)
+ \binom{8}{3}A(5,1)&=2832.\\
c'_{9,3}&\geq & \binom{9}{0}A(9,4)+\binom{9}{1}A(8,3)+\binom{9}{2}A(7,2)
+ \binom{9}{3}A(6,1)&=7880.
\end{array}\]
With k=4
\[
\begin{array}{llll}
c'_{10,3}&\geq &\binom{10}{0}A(10,5)+\binom{10}{1}A(9,4)+\binom{10}{2}A(8,3)
+ \binom{10}{3}A(7,2)+\binom{10}{4}A(6,1)&=22232.\\
c'_{11,3}&\geq &\binom{11}{0}A(11,5)+\binom{11}{1}A(10,4)+\binom{11}{2}A(9,3)
+ \binom{11}{3}A(8,2)+\binom{11}{4}A(7,1)&=66024.\\
c'_{12,3}&\geq &\binom{12}{0}A(12,5)+\binom{12}{1}A(11,4)+\binom{12}{2}A(10,3)
+ \binom{12}{3}A(9,2)+\binom{12}{4}A(8,1)&=188688.\\
\end{array}\]
With $k=5$
\[ c'_{13,3}\geq 539168.\]

It should be pointed out that these bounds are even numbers, so that
$c'_{4,3}=43$ shows that one cannot generally expect this lower bound
gives the optimum.

The maximum value appears to occur for $k=\lfloor\frac{n+2}{3}\rfloor$,
so that using
Stirling's formula and explicit bounds on $A(n,d)$ the
best possible value known to date
of the constant $C$ in
equation \eqref{cpn3} can be worked out, but we refrain from doing this here.
Using the Singleton bound $A(n,d)\leq 2^{n-d+1}$ Chv\'{a}tal
\cite{chvatal1} proved that the expression on the right hand side of
\eqref{cnchvatal} is also
$O\left( \frac{3^n}{\sqrt{n}}\right)$, so that the refinement described
above gains a constant factor over the initial construction only.

For $n=4$ the above does not yet give the exact value.
The value $c'_{4,3}=43$ was first proven by Chandra \cite{chandra}.
A uniform way of describing examples for the optimum values of
$c'_{4,3}=43$ and $c'_{5,3}=124$ is the following:

Let us consider the sets $$ A := S_{i-1,n}
\cup S_{i,n}^e \cup A'$$ where $A' \subset S_{i+1,n}$ has the property
that any two elements in $A'$ are separated by a Hamming distance of at
least three, or have a Hamming distance of exactly one but their
midpoint lies in $S_{i,n}^o$. By the previous discussion we see that
this is a Moser set, and we have the lower bound
\begin{equation}\label{cnn}
c'_{n,3} \geq \binom{n+1}{i} 2^{i-1} + |A'|.
\end{equation} This gives some improved lower bounds for $c'_{n,3}$:

\begin{itemize} \item By taking $n=4$, $i=3$, and $A' = \{ 1111, 3331,
3333\}$, we obtain $c'_{4,3} \geq 43$; \item By taking $n=5$, $i=4$, and
$A' = \{ 11111, 11333, 33311, 33331 \}$, we obtain $c'_{5,3} \geq 124$.
\item By taking $n=6$, $i=5$, and $A' = \{ 111111, 111113, 111331,
111333, 331111, 331113\}$, we obtain $c'_{6,3} \geq 342$. \end{itemize}

This gives the lower bounds in Theorem \ref{moser} up to $n=5$, but
the bound for $n=6$ is inferior to the lower bound $c'_{6,3}\geq 344$ given above.

\subsection{Higher $k$ values}
One can consider subsets of $[k]{}^n$ that contain no geometric lines. Section \ref{moser-lower-sec} has considered the case $k=3$. Let $c'_{n,k}$ be the greatest number of points in $[k]{}^n$ with no geometric line. For example, $c'_{n,3} = c'_n$. We have the following lower bounds:

$c'_{n,4} \ge \binom{n}{n/2}2^n$.
The set of points with $a$ $1$s,$b$ $2$s,$c$ $3$s and $d$ $4$s, where $a+d$ has the constant value $n/2$, does not form geometric lines because points at the ends of a geometric line have more $a$ or $d$ values than point in the middle of the line.

One can show a lower bound that, asymptotically, is twice as large. Take all points with $a$ $1$s, $b$ $2$s, $c$ $3$s and $d$ $4$s, for which:

\begin{itemize}
\item Either $a+d = q$ or $q-1$, $a$ and $b$ have the same parity; or
\item $a+d = q-2$ or $q-3$, $a$ and $b$ have opposite parity.
\end{itemize}

This includes half the points of four adjacent layers, and therefore may include $(1+o(1))\binom{n}{n/2}2^{n+1}$ points.

We also have a DHJ(3)-like lower bound for $c'_{n,5}$, namely $c'_{n,5} = 5^{n-O(\sqrt{\log n})}$.
Consider points with $a$ $1$s, $b$ $2$s, $c$ $3$s, $d$ $4$s and $e$ $5$s. For each point, take the value $a+e+2(b+d)+3c$. The first three points in any geometric line give values that form an arithmetic progression of length three.

Select a set of integers with no arithmetic progression of length 3. Select all points whose value belongs to that sequence; there will be no geometric line among those points. By Behrend theory, it is possible to choose these points with density $\exp{-O(\sqrt{\log n})}$.

The $k=6$ version of Moser implies DHJ(3). Indeed, any $k=3$ combinatorial line-free set can be "doubled up" into a $k=6$ geometric line-free set of the same density by pulling back the set from the map that maps 1, 2, 3, 4, 5, 6 to 1, 2, 3, 3, 2, 1 respectively; note that this map sends $k=6$ geometric lines to $k=3$ combinatorial lines. So $c'_{k,6} \geq 2^k c_k$.

Moser-lower.tex

2009-07-10T17:52:43Z

Mpeake:

\section{Lower bounds for the Moser problem}\label{moser-lower-sec}

Just as for the DHJ problem, we found that Gamma sets $\Gamma_{a,b,c}$ were useful in providing large lower bounds for the Moser problem. This is despite the fact that the symmetries of the cube do not respect Gamma sets.

Observe that if $B \subset \Delta_n$, then the set $A_B := \bigcup_{\vec a \in B} \Gamma_{a,b,c}$ is a Moser set as long as $B$ does not contain any ``isosceles triangles'' $(a+r,b,c+s), (a+s,b,c+r), (a,b+r+s,c)$ for any $r,s \geq 0$ not both zero; in particular, $B$ cannot contain any ``vertical line segments'' $(a+r,b,c+r), (a,b+2r,c)$. An example of such a set is provided by selecting $0 \leq i \leq n-3$ and letting $B$ consist of the triples $(a, n-i, i-a)$ when $a \neq 3 \mod 3$, $(a,n-i-1,i+1-a)$ when $a \neq 1 \mod 3$, $(a,n-i-2,i+2-a)$ when $a=0 \mod 3$, and $(a,n-i-3,i+3-a)$ when $a=2 \mod 3$. Asymptotically, this set includes about two thirds of the spheres $S_{n,i}$, $S_{n,i+1}$ and one third of the spheres $S_{n,i+2}, S_{n,i+3}$ and (setting $i$ close to $n/3$) gives a lower bound \eqref{cpn3} with $C = 2 \times \sqrt{\frac{9}{4\pi}}$.

An integer program was run to obtain the optimal lower bounds achievable by the $A_B$ construction (using \eqref{cn3}, of course). The results for
$1 \leq n \leq 20$ are displayed in Figure \ref{nlow-moser}:

\begin{figure}[tb]
\centerline{
\begin{tabular}{|ll|ll|}
\hline
$n$ & lower bound & $n$ & lower bound \\
\hline
1 & 2 &11& 71766\\
2 & 6 & 12& 212423\\
3 & 16 & 13& 614875\\
4 & 43 & 14& 1794212\\
5 & 122& 15& 5321796\\
6 & 353& 16& 15455256\\
7 & 1017& 17& 45345052\\
8 & 2902&18& 134438520\\
9 & 8622&19& 391796798\\
10& 24786& 20& 1153402148\\
\hline
\end{tabular}}
\caption{Lower bounds for $c'_n$ obtained by the $A_B$ construction.}
\label{nlow-moser}
\end{figure}

More complete data, including the list of optimisers, can be found at
\centerline{{\tt http://abel.math.umu.se/~klasm/Data/HJ/}.}

Note that the lower bound $c'_{6,3} \geq 353$ was first located by a genetic algorithm: see Appendix \ref{genetic-alg}.

\begin{figure}[tb]
\centerline{\includegraphics{moser353new.png}}
\caption{One of the examples of $353$-point sets in $[3]^6$ (elements of the set being indicated by white squares).}
\label{moser353-fig}
\end{figure}

However, we have been unable to locate a lower bound which is asymptotically better than \eqref{cpn3}. Indeed, any method based purely on the $A_B$ construction cannot do asymptotically better than the previous constructions:

\begin{proposition} Let $B \subset \Delta_n$ be such that $A_B$ is a Moser set. Then $|A_B| \leq (2 \sqrt{\frac{9}{4\pi}} + o(1)) \frac{3^n}{\sqrt{n}}$.
\end{proposition}

\begin{proof} By the previous discussion, $B$ cannot contain any pair of the form $(a,b+2r,c), (a+r,b,c+r)$ with $r>0$. In other words, for any $-n \leq h \leq n$, $B$ can contain at most one triple $(a,b,c)$ with $c-a=h$. From this and \eqref{cn3}, we see that
$$ |A_B| \leq \sum_{h=-n}^n \max_{(a,b,c) \in \Delta_n: c-a=h} \frac{n!}{a! b! c!}.$$
From the Chernoff inequality (or the Stirling formula computation below) we see that $\frac{n!}{a! b! c!} \leq \frac{1}{n^{10}} 3^n$ unless $a,b,c = n/3 + O( n^{1/2} \log^{1/2} n )$, so we may restrict to this regime, which also forces $h = O( n^{1/2}\log^{1/2} n)$. If we write $a = n/3 + \alpha$, $b = n/3 + \beta$, $c = n/3+\gamma$ and apply Stirling's formula $n! = (1+o(1)) \sqrt{2\pi n} n^n e^{-n}$, we obtain
$$ \frac{n!}{a! b! c!} = (1+o(1)) \frac{3^{3/2}}{2\pi n} 3^n \exp( - (\frac{n}{3}+\alpha) \log (1 + \frac{3\alpha}{n} ) - (\frac{n}{3}+\beta) \log (1 + \frac{3\beta}{n} ) - (\frac{n}{3}+\gamma) \log (1 + \frac{3\gamma}{n} ) ).$$
From Taylor expansion one has
$$ -(\frac{n}{3}+\alpha) \log (1 + \frac{3\alpha}{n} ) = -\alpha - \frac{3}{2} \frac{\alpha^2}{n} + o(1)$$
and similarly for $\beta,\gamma$; since $\alpha+\beta+\gamma=0$, we conclude that
$$ \frac{n!}{a! b! c!} = (1+o(1)) \frac{3^{3/2}}{2\pi n} 3^n \exp( - \frac{3}{2n} (\alpha^2+\beta^2+\gamma^2) ).$$
If $c-a=h$, then $\alpha^2+\beta^2+\gamma^2 = \frac{3\beta^2}{2} + \frac{h^2}{2}$. Thus we see that
$$ \max_{(a,b,c) \in \Delta_n: c-a=h} \frac{n!}{a! b! c!} \leq (1+o(1)) \frac{3^{3/2}}{2\pi n} 3^n \exp( - \frac{3}{4n} h^2 ).$$
Using the integral test, we thus have
$$ |A_B| \leq (1+o(1)) \frac{3^{3/2}}{2\pi n} 3^n \int_\R \exp( - \frac{3}{4n} x^2 )\ dx.$$
Since $\int_\R \exp( - \frac{3}{4n} x^2 )\ dx = \sqrt{\frac{4\pi n}{3}}$, we obtain the claim.
\end{proof}

Actually it is possible to improve upon these bounds by a slight amount. Observe that if $B$ is a maximiser for the right-hand side of \eqref{cn3} (subject to $B$ not containing isosceles triangles), then any triple $(a,b,c)$ not in $B$ must be the vertex of a (possibly degenerate) isosceles triangle with the other vertices in $B$. If this triangle is non-degenerate, or if $(a,b,c)$ is the upper vertex of a degenerate isosceles triangle, then no point from $\Gamma_{a,b,c}$ can be added to $A_B$ without creating a geometric line. However, if $(a,b,c) = (a'+r,b',c'+r)$ is only the lower vertex of a degenerate isosceles triangle $(a'+r,b',c'+r), (a',b'+2r,c')$, then one can add any subset of $\Gamma_{a,b,c}$ to $A_B$ and still have a Moser set as long as no pair of elements in that subset is separated by Hamming distance $2r$. For instance, in the $n=5$ case, we can start with the 122-point set built from
$$B = \{ (0 0 5),(0 2 3),(1 1 3 ),(1 2 2 ),(2 2 1 ),(3 1 1 ),(3 2 0 ),(5 0 0))\}$$, and add a point each from (1 0 4) and (4 0 1).
This gives an example of the maximal, 124-point solution. Again, in the $n=10$ case, the set
\begin{align*}
B &= \{(0 0 10),(0 2 8 ),(0 3 7 ),(0 4 6 ),(1 4 5 ),(2 1 7 ),(2 3 5 ),
(3 2 5 ),(3 3 4 ),(3 4 3 ),(4 4 2 ),(5 1 4 ), \\
&\quad (5 3 2 ),(6 2 2 ),
(6 3 1 ),(6 4 0 ),(8 1 1 ),(9 0 1 ),(9 1 0 ) \}
\end{align*}

generates the lower bound $c'_{10,3} \geq 24786$ given above (and, up to reflection $a \leftrightarrow c$, is the only such set that does so); but by adding the following twelve elements from $\Gamma_{5,0,5}$ one can increase the lower bound slightly to $24798$: $1111133333$, $1111313333$,
$1113113333$, $1133331113$, $1133331131$, $1133331311$, $3311333111$,
$3313133111$, $3313313111$, $3331111133$, $3331111313$, $3331111331$

A more general form goes with the $B$ set described at the start of this section. Include points from $\Gamma_{(a,n-i-4,i+4-a)}$ when $a=1 (\operatorname{mod}\ 3)$, subject to no two points being included if they differ by the interchange of a $1$ and a $3$. Each of these Gamma sets is the feet of a degenerate isosceles triangle with vertex $\Gamma_{(a-1,n-i-2,a+3-a)}$.

\begin{lemma} If $A$ is a subset of $\Gamma_{(a,b,c)}$ such that no two points of $A$ differ by the interchange of a $1$ and a $3$, then $|A| \leq |\Gamma_{a,b,c}|/(1+\max(a,c))$.
\end{lemma}
\begin{proof}
Say that two points of $\Gamma_{a,b,c}$ are neighbours if they differ by the exchange of a $1$ and a $3$. Each point of $A$ has $ac$ neighbours, none of which are in $A$. Each point of $\Gamma_{(a,b,c)}\backslash A$ has $ac$ neighbours, but only $min(a,c)$ of them may be in $A$. So for each point of $A$ there are on average $ac/\min(a,c) = \max(a,c)$ points not in $A$. So the proportion of points of $\Gamma_{(a,b,c)}$ that are in $A$ is at most one in $1+\max(a,c)$.
\end{proof}

The proportion of extra points for each of the cells $\Gamma_{(a,n-i-4,i+4-a)}$ is no more than $2/(i+6)$. Only one cell in three is included from the $b=n-i-4$ layer, so we expect no more than $\frac{2}{3(i+6)}\binom(n,i+4)2^{i+4}$ new points, all from $S_{n,i+4}$. One can also find extra points from $S_{n,i+5}$ and higher spheres.

Earlier solutions may also give insight into the problem. In this section we discuss lower bounds for $c'_{n,3}$. Clearly we have
$c'_{0,3}=1$ and $c'_{1,3}=2$, so we focus on the case $n \ge 2$.
The first lower bounds may be due to Koml\'{o}s \cite{komlos}, who observed
that the sphere $S_{i,n}$ of elements with exactly $n-i$ 2 entries
(see Section \ref{notation-sec} for definition),
is a Moser set, so that
\begin{equation}\label{cin}
c'_{n,3}\geq \vert S_{i,n}\vert
\end{equation}
holds for all $i$. Choosing $i=\lfloor \frac{2n}{3}\rfloor$ and
applying Stirling's formula, we see that this lower bound takes the form
\begin{equation}\label{cpn3}
c'_{n,3} \geq (C-o(1)) 3^n / \sqrt{n}
\end{equation}
for some absolute constant $C>0$; in fact \eqref{cin} gives \eqref{cpn3} with $C := \sqrt{\frac{9}{4\pi}}$.
In particular $c'_{3,3} \geq 12, c'_{4,3}\geq 24, c'_{5,3}\geq 80,
c'_{6,3}\geq 240$.

These values can be improved by studying combinations of several spheres or
semispheres or applying elementary results from coding theory.

Observe that if $\{w(1),w(2),w(3)\}$ is a geometric line in $[3]^n$,
then $w(1), w(3)$ both lie in the same sphere $S_{i,n}$,
and that $w(2)$ lies in a lower
sphere $S_{i-r,n}$ for some $1 \leq r \leq i \leq n$. Furthermore,
$w(1)$ and $w(3)$ are separated by Hamming distance $r$.

As a consequence, we see that $S_{i-1,n} \cup S_{i,n}^e$ (or $S_{i-1,n}
\cup S_{i,n}^o$) is a Moser set for any $1 \leq i \leq n$, since any two
distinct elements $S_{i,n}^e$ are separated by a Hamming distance of at
least two.
(Recall Section \ref{notation-sec} for definitions),
This leads to the lower bound
\begin{equation}\label{cn3-low}
c'_{n,3} \geq \binom{n}{i-1}
2^{i-1} + \binom{n}{i} 2^{i-1} = \binom{n+1}{i} 2^{i-1}.
\end{equation}
It is not
hard to see that $\binom{n+1}{i+1} 2^{i} > \binom{n+1}{i} 2^{i-1}$ if
and only if $3i < 2n+1$, and so this lower bound is maximised when $i =
\lfloor \frac{2n+1}{3} \rfloor$ for $n \geq 2$, giving the formula
\eqref{binom}. This leads to the lower bounds $$ c'_{2,3} \geq 6;
c'_{3,3} \geq 16; c'_{4,3} \geq 40; c'_{5,3} \geq 120; c'_{6,3} \geq
336$$ which gives the right lower bounds for $n=2,3$, but is slightly
off for $n=4,5$. Asymptotically, Stirling's formula and \eqref{cn3-low} then
give the lower bound \eqref{cpn3} with $C = \frac{3}{2} \times \sqrt{\frac{9}{4\pi}}$, which is asymptotically $50\%$ better than the bound \eqref{cin}.

The work of Chv\'{a}tal \cite{chvatal1}
already contained a refinement of this idea
which we here translate into the usual notation of coding theory:
Let $A(n,d)$ denote the size of the largest binary code of length $n$
and minimal distance $d$.

Then
\begin{equation}\label{cnchvatal}
c'_{n,3}\geq \max_k \left( \sum_{j=0}^k \binom{n}{j} A(n-j, k-j+1)\right).
\end{equation}

With the following values for $A(n,d)$:
{\tiny{
\[
\begin{array}{llllllll}
A(1,1)=2&&&&&&&\\
A(2,1)=4& A(2,2)=2&&&&&&\\
A(3,1)=8&A(3,2)=4&A(3,3)=2&&&&&\\
A(4,1)=16&A(4,2)=8& A(4,3)=2& A(4,4)=2&&&&\\
A(5,1)=32&A(5,2)=16& A(5,3)=4& A(5,4)=2&A(5,5)=2&&&\\
A(6,1)=64&A(6,2)=32& A(6,3)=8& A(6,4)=4&A(6,5)=2&A(6,6)=2&&\\
A(7,1)=128&A(7,2)=64& A(7,3)=16& A(7,4)=8&A(7,5)=2&A(7,6)=2&A(7,7)=2&\\
A(8,1)=256&A(8,2)=128& A(8,3)=20& A(8,4)=16&A(8,5)=4&A(8,6)=2
&A(8,7)=2&A(8,8)=2\\
A(9,1)=512&A(9,2)=256& A(9,3)=40& A(9,4)=20&A(9,5)=6&A(9,6)=4
&A(9,7)=2&A(9,8)=2\\
A(10,1)=1024&A(10,2)=512& A(10,3)=72& A(10,4)=40&A(10,5)=12&A(10,6)=6
&A(10,7)=2&A(10,8)=2\\
A(11,1)=2048&A(11,2)=1024& A(11,3)=144& A(11,4)=72&A(11,5)=24&A(11,6)=12
&A(11,7)=2&A(11,8)=2\\
A(12,1)=4096&A(12,2)=2048& A(12,3)=256& A(12,4)=144&A(12,5)=32&A(12,6)=24
&A(12,7)=4&A(12,8)=2\\
A(13,1)=8192&A(13,2)=4096& A(13,3)=512& A(13,4)=256&A(13,5)=64&A(12,6)=32
&A(13,7)=8&A(13,8)=4\\
\end{array}
\]
}}

Generally, $A(n,1)=2^n, A(n,2)=2^{n-1}, A(n-1,2e-1)=A(n,2e), A(n,d)=2$,
if $d>\frac{2n}{3}$.
The values were taken or derived from Andries Brower's table at
\centerline{\tt http://www.win.tue.nl/$\sim$aeb/codes/binary-1.html}
\textbf{include to references? or other book with explicit values of $A(n,d)$ }

For $c'_{n,3}$ we obtain the following lower bounds:
with $k=2$
\[
\begin{array}{llll}
c'_{4,3}&\geq &\binom{4}{0}A(4,3)+\binom{4}{1}A(3,2)+\binom{4}{2}A(2,1)
=1\cdot 2+4 \cdot 4+6\cdot 4&=42.\\
c'_{5,3}&\geq &\binom{5}{0}A(5,3)+\binom{5}{1}A(4,2)+\binom{5}{2}A(3,1)
=1\cdot 4+5 \cdot 8+10\cdot 8&=124.\\
c'_{6,3}&\geq &\binom{6}{0}A(6,3)+\binom{6}{1}A(5,2)+\binom{6}{2}A(4,1)
=1\cdot 8+6 \cdot 16+15\cdot 16&=344.
\end{array}
\]
With k=3
\[
\begin{array}{llll}
c'_{7,3}&\geq& \binom{7}{0}A(7,4)+\binom{7}{1}A(6,3)+\binom{7}{2}A(5,2)
+ \binom{7}{3}A(4,1)&=960.\\
c'_{8,3}&\geq &\binom{8}{0}A(8,4)+\binom{8}{1}A(7,3)+\binom{8}{2}A(6,2)
+ \binom{8}{3}A(5,1)&=2832.\\
c'_{9,3}&\geq & \binom{9}{0}A(9,4)+\binom{9}{1}A(8,3)+\binom{9}{2}A(7,2)
+ \binom{9}{3}A(6,1)&=7880.
\end{array}\]
With k=4
\[
\begin{array}{llll}
c'_{10,3}&\geq &\binom{10}{0}A(10,5)+\binom{10}{1}A(9,4)+\binom{10}{2}A(8,3)
+ \binom{10}{3}A(7,2)+\binom{10}{4}A(6,1)&=22232.\\
c'_{11,3}&\geq &\binom{11}{0}A(11,5)+\binom{11}{1}A(10,4)+\binom{11}{2}A(9,3)
+ \binom{11}{3}A(8,2)+\binom{11}{4}A(7,1)&=66024.\\
c'_{12,3}&\geq &\binom{12}{0}A(12,5)+\binom{12}{1}A(11,4)+\binom{12}{2}A(10,3)
+ \binom{12}{3}A(9,2)+\binom{12}{4}A(8,1)&=188688.\\
\end{array}\]
With $k=5$
\[ c'_{13,3}\geq 539168.\]

It should be pointed out that these bounds are even numbers, so that
$c'_{4,3}=43$ shows that one cannot generally expect this lower bound
gives the optimum.

The maximum value appears to occur for $k=\lfloor\frac{n+2}{3}\rfloor$,
so that using
Stirling's formula and explicit bounds on $A(n,d)$ the
best possible value known to date
of the constant $C$ in
equation \eqref{cpn3} can be worked out, but we refrain from doing this here.
Using the Singleton bound $A(n,d)\leq 2^{n-d+1}$ Chv\'{a}tal
\cite{chvatal1} proved that the expression on the right hand side of
\eqref{cnchvatal} is also
$O\left( \frac{3^n}{\sqrt{n}}\right)$, so that the refinement described
above gains a constant factor over the initial construction only.

For $n=4$ the above does not yet give the exact value.
The value $c'_{4,3}=43$ was first proven by Chandra \cite{chandra}.
A uniform way of describing examples for the optimum values of
$c'_{4,3}=43$ and $c'_{5,3}=124$ is the following:

Let us consider the sets $$ A := S_{i-1,n}
\cup S_{i,n}^e \cup A'$$ where $A' \subset S_{i+1,n}$ has the property
that any two elements in $A'$ are separated by a Hamming distance of at
least three, or have a Hamming distance of exactly one but their
midpoint lies in $S_{i,n}^o$. By the previous discussion we see that
this is a Moser set, and we have the lower bound
\begin{equation}\label{cnn}
c'_{n,3} \geq \binom{n+1}{i} 2^{i-1} + |A'|.
\end{equation} This gives some improved lower bounds for $c'_{n,3}$:

\begin{itemize} \item By taking $n=4$, $i=3$, and $A' = \{ 1111, 3331,
3333\}$, we obtain $c'_{4,3} \geq 43$; \item By taking $n=5$, $i=4$, and
$A' = \{ 11111, 11333, 33311, 33331 \}$, we obtain $c'_{5,3} \geq 124$.
\item By taking $n=6$, $i=5$, and $A' = \{ 111111, 111113, 111331,
111333, 331111, 331113\}$, we obtain $c'_{6,3} \geq 342$. \end{itemize}

This gives the lower bounds in Theorem \ref{moser} up to $n=5$, but
the bound for $n=6$ is inferior to the lower bound $c'_{6,3}\geq 344$ given above.

\subsection{Higher $k$ values}
One can consider subsets of $[k]{}^n$ that contain no geometric lines. Section \ref{moser-lower-sec} has considered the case $k=3$. Let $c'_{n,k}$ be the greatest number of points in $[k]{}^n$ with no geometric line. For example, $c'_{n,3} = c'_n$. We have the following lower bounds:

$c'_{n,4} \ge \binom{n}{n/2}2^n$.
The set of points with $a$ $1$s,$b$ $2$s,$c$ $3$s and $d$ $4$s, where $a+d$ has the constant value $n/2$, does not form geometric lines because points at the ends of a geometric line have more $a$ or $d$ values than point in the middle of the line.

One can show a lower bound that, asymptotically, is twice as large. Take all points with $a$ $1$s, $b$ $2$s, $c$ $3$s and $d$ $4$s, for which:

\begin{itemize}
\item Either $a+d = q$ or $q-1$, $a$ and $b$ have the same parity; or
\item $a+d = q-2$ or $q-3$, $a$ and $b$ have opposite parity.
\end{itemize}

This includes half the points of four adjacent layers, and therefore may include $(1+o(1))\binom{n}{n/2}2^{n+1}$ points.

We also have a DHJ(3)-like lower bound for $c'_{n,5}$, namely $c'_{n,5} = 5^{n-O(\sqrt{\log n})}$.
Consider points with $a$ $1$s, $b$ $2$s, $c$ $3$s, $d$ $4$s and $e$ $5$s. For each point, take the value $a+e+2(b+d)+3c$. The first three points in any geometric line give values that form an arithmetic progression of length three.

Select a set of integers with no arithmetic progression of length 3. Select all points whose value belongs to that sequence; there will be no geometric line among those points. By Behrend theory, it is possible to choose these points with density $\exp{-O(\sqrt{\log n})}$.

The $k=6$ version of Moser implies DHJ(3). Indeed, any $k=3$ combinatorial line-free set can be "doubled up" into a $k=6$ geometric line-free set of the same density by pulling back the set from the map that maps 1, 2, 3, 4, 5, 6 to 1, 2, 3, 3, 2, 1 respectively; note that this map sends $k=6$ geometric lines to $k=3$ combinatorial lines.

Fujimura.tex

2009-07-10T17:46:43Z

Mpeake:

\section{Fujimura's problem}\label{fujimura-sec}
Let $\overline{c}^\mu_n$ be the size of the largest subset of the trianglular grid
$$\Delta_n := \{(a,b,c)\in \mathbb{Z}^3_+ : a+b+c = n\}$$
which contains no equilateral triangles $(a+r,b,c), (a,b+r,c), (a,b,c+r)$ with $r>0$. These are upward-pointing equilateral triangles. We shall refer to such sets as 'triangle-free'.
(Kobon Fujimura is a prolific inventor of puzzles, and in [http://www.puzzles.com/PuzzlePlayground/CoinsAndTriangles/CoinsAndTriangles.htm this] puzzle asked the related question of eliminating all equilateral triangles.)

The following table was formed mostly by computer searches for optimal solutions. We also found human proofs for most of them (see [http://michaelnielsen.org/polymath1/index.php?title=Fujimura's_problem here]).

\begin{figure}\centerline{
\begin{tabular}[l|lllllllllllll]
$n$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13\\
\hline $\overline{c}^\mu_n$ & 1 & 2 & 4 & 6 & 9 & 12 & 15 & 18 & 22 & 26 & 31 & 35 & 40 & 46
\end{tabular}\label{lowFujimura}\caption{Fujimura numbers}}

For any equilateral triangle $(a+r,b,c)$,$(a,b+r,c)$ and $(a,b,c+r)$, the value $y+2z$ forms an arithmetic progression of length 3. A Behrend set is a finite set of integers with no arithmetic progression of length 3 (see [http://arxiv.org/PS_cache/arxiv/pdf/0811/0811.3057v2.pdf this paper]). By looking at those triples (a,b,c) with a+2b inside a Behrend set, one can obtain the lower bound of $\overline{c}^\mu_n \geq n^2 exp(-O(\sqrt{\log n}))$.

It can be shown that $\overline{c}^\mu_n = o(n^2)$ as $n \rightarrow \infty$, but the details are outside the scope of this paper.

Coloring.tex

2009-07-10T16:39:32Z

Mpeake:

\section{The coloring Hales-Jewett problem}\label{coloring-sec}
The Hales-Jewett theorem states that for every $k$ and every $r$ there exists an $n = HJ(k,r)$ such that if you colour the elements of $[k]{}^n$ with $r$ colours, then there must be a combinatorial line with all its points of the same colour.

This is a consequence of the Density Hales-Jewett theorem, since there must be a set of density at least $r^{-1}$ of elements of $[k]{}^n$ all of whose elements have the same colour. It also follows from the Graham-Rothschild theorem [\cite{}].
By iterating the Hales-Jewett theorem, one can also show that one of the color classes contains an $m$-dimensional combinatorial subspace, if $n$ is sufficiently large depending on $k$, $r$ and $m$.

A related idea is that of the Van der Waerden number $W(k,r)$ [\cite{}], which is the length of the longest sequence $1,2,\ldots,n$ which can be $r$-coloured without a monochromatic arithmetic progression of length $k$. These numbers were used in this project to construct $r$-colourings of $[k]{}^m$ without combinatorial lines, and so provide new lower bounds for $HJ(k,r)$.

\begin{lemma} $HJ(k,r) \ge \lfloor (W(k,r)-1)/(k-1)\rfloor$ \end{lemma}
\begin{proof}
Let $m = \lfloor (W(k,r)-1)/(k-1) \rfloor$. Suppose we have a Van der Warden colouring of $1, 2, \ldots W(k,r)$. Give each point $(a_1,a_2,\ldots,a_m) \in [k]{}^m$ the colour that corresponds to $1+\sum a_j$. The points of a combinatorial line give sums that form an arithmetic progression, and will therefore not be monochromatic.
\end{proof}

A recent table of lower bounds for the Van der Waerden numbers is at this website: [http://www.st.ewi.tudelft.nl/sat/slides/waerden.pdf].
These numbers were used to produce the following table of lower bounds for colouring Hales-Jewett numbers.

\begin{figure}[tb] \centerline{\begin{tabular}[l|lllll]
k\r & 2 & 3 & 4 & 5 & 6 \\
\hline
3 & & 13 & 37 & 84 & 103 \\
4 & 11 & 97 & 349 & 751 & 3259 \\
5 & 59 & 302 & 2609 & 6011 & 14173 \\
6 & 226 & 1777 & 18061 & 49391 & 120097 \\
7 & 617 & 7309 & 64661 & & \\
8 & 1069 & 34057 & & & \\
9 & 3389 & & & &
\end{tabular}}\caption{Lower bounds for colouring Hales-Jewett numbers
\label{HJcolour}\end{figure}