Dhj-lown.tex
\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. See above that for $B_{0,6}$, the excluded sets are all permutations of (0,1,5). So the included sets are all 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{align*}
(2,3,4),(1,3,5),(0,4,5) and permutations \\
(3,4,5),(2,4,6),(1,5,6),(0,2,10),(0,5,7) and permutations \\
(4,5,6),(3,5,7),(2,6,7),(1,3,11),(1,6,8),(0,4,11),(0,7,8) and permutations
\end{align*}
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), 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 (abc) triangle: $$M+(-8-2x,-6-2x,14+4x),M+(-8-2x,-3-2x,11+4x),M+(-8-2x,x,8+x),M+(-8-2x,3+x,5+x) and permutations (0\le 2x \le M-8)\\ M+(-9-2x,-5-2x,14+4x),M+(-9-2x,-2-2x,11+4x),M+(-9-2x,1+x,8+x),M+(-9-2x,4+x,5+x) and permutations (0\le 2x \le M-9$$
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 $3^n/2$ points.
The following solution is believed to give more points for n>1000, but not for moderate n:
\begin{align*} 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 $1,2,7,8,19,20,25,26,55, \ldots$\\ List all the (abc) triples for which the larger two differ by a number from the sequence;\\ Exclude the case when the smaller two differ by 1;\\ Include the case when (a,b,c) is a permutation of $n/3+(-1,0,1)$. \end{align*}
Now we prove Theorem \ref{dhj-lower}.
%\begin{proof}[Proof of Theorem \ref {dhj-lower}] %Let $S$ be a subset of the interval $[-\sqrt {n}/2, \sqrt {n}/2)$ that contains no arithmetic %progressions of length 3, and let $B\subset\Delta_{n, 3}$ be the set % \[ B := \{(n-s-t,s-2t,-2s+t) : s,t \in S\}.\] %The map $(a,b,c)\mapsto (b+2c,2b+c)$ takes simplices to nonconstant arithmetic progressions, %and takes $B$ to $\{-3(s,t)\colon s,t \in S\}$, 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 $a,b,c$ are within $C_1\sqrt{n}$ of $n/3$, then $ | \Gamma_ {a, b, c} | \geq C 3^n/n$ %(where $C$ depends on $C_1$) by the central limit theorem. By %our choice of $S$ and applying~\eqref{cn3}, we obtain % $$ c_ {n, 3}\geq C |S| ^2 3^n/n. $$ %One can take $S$ to have cardinality $r_ 3 (\sqrt {n}) $, which from the results of Elkin~\cite {elkin} %(see also~\cite{greenwolf,obryant}) satisfies (for all sufficiently %large $n$) % $$ r_ 3 (\sqrt {n})\geq\sqrt {n} (\log n)^{1/4}\exp_ 2 (-2\sqrt {\log_ 2 n}),$$ %which completes the proof. %\end {proof}
\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}
