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A general result about density increments

2009-06-04T20:14:30Z

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==Terminology==

If A and B are two subsets of a finite set X, then we say that the ''density of'' A ''in'' B is <math>|A\cap B|/|B|.</math>

==Description of result==

Let us focus on the proof of DHJ(3), though what we say is much more general. That proof has two stages, which can be described as follows.

1. Prove that if <math>\mathcal{A}</math> is a subset of <math>[3]^n</math> of density <math>\delta</math> then there is a dense 12-subset <math>\mathcal{B}</math> of a subspace S of dimension tending to infinity, such that the density of <math>\mathcal{A}</math> in <math>\mathcal{B}</math> is at least <math>\delta+c(\delta),</math> where <math>c(\delta)>0.</math> (In fact, <math>c(\delta)</math> is proportional to <math>\delta^2.</math>)

2. Every 12-set can be almost entirely partitioned into m-dimensional subspaces, where m tends to infinity with n. Here, m depends only on the density of the part of the 12-set that is allowed not to be partitioned.

Once we have these two stages, we are basically done, for reasons that can be appreciated even if one does not know the definition of a 12-set. The reason is that if we partition all of a 12-set apart from a subset of measure at most <math>c(\delta)/2</math> into m-dimensional subspaces, then the density of <math>\mathcal{A}</math> in the partitioned part is at least <math>\delta+c(\delta)/2,</math> so by averaging <math>\mathcal{A}</math> has density at least <math>\delta+c(\delta)/2</math> in at least one of these subspaces. That gives us a density increment on a subspace, which is exactly what we need for a [[density-increment_strategies|density-increment strategy]].

Now let us generalize 2 very slightly. Given a finite set X, we define its ''characteristic measure'' <math>\xi</math> to be the function that takes the value <math>1/|X|</math> everywhere in X and 0 everywhere else. Given a set Y, we write <math>\xi(Y)</math> for <math>\sum_{y\in Y}\xi(y)=|X\cap Y|/|X|.</math> Let <math>\beta</math> be the characteristic measure of <math>\mathcal{B},</math> let <math>S_1,\dots,S_N</math> be a collection of subspaces, and for each i let <math>\sigma_i</math> be the characteristic measure of <math>S_i.</math> We are assuming that <math>\beta(\mathcal{A})\geq\delta+c(\delta).</math> If we can find a convex combination <math>\sum_{i=1}^N\lambda_i\sigma_i</math> such that <math>\|\sum_i\lambda_i\sigma_i-\beta\|\leq c(\delta)/2,</math> then <math>\sum_i\lambda_i\sigma_i(\mathcal{A})\geq\delta+c(\delta)/2.</math> It follows that there exists i such that <math>\sigma_i(\mathcal{A})\geq\delta+c(\delta)/2,</math> which is what we wanted.

The main result of this page is that a converse to this generalized step 2 is true as well. Loosely, this tells us that any proof that a density increase on a 12-set implies a density increase on a subspace must also show that a 12-set can be evenly covered with subspaces, up to a small error.

==The proof==

Suppose that we ''cannot'' find a convex combination <math>\sum_{i=1}^N\lambda_i\sigma_i</math> such that <math>\|\sum_i\lambda_i\sigma_i-\beta\|\leq c(\delta)/2.</math> Then the Hahn-Banach theorem provides us with a function F and non-negative reals <math>\lambda+\mu=1</math> such that <math>\mathbb{E}_{x\in\mathcal{B}}F(x)>1,</math> while <math>\|F\|_\infty\leq 2\lambda/c(\delta)</math> and <math>\sigma_i(F)=\mathbb{E}_{x\in S_i}F(x)\leq \mu</math> for every i. From this it follows that <math>\lambda>c(\delta)/2.</math>

Now let <math>G(x)=\delta(1+F(x)/\|F\|_\infty).</math> Then G takes values in <math>[0,2\delta],</math> and <math>\mathbb{E}_{x\in\mathcal{B}}G(x)>\delta(1+c(\delta)/2\lambda).</math> However, for each i we have <math>\sigma_i(G)\leq\delta(1+\mu/\|F\|_\infty)\leq\delta(1+c(\delta)^2\mu/2\lambda)</math><math></math><math></math><math></math><math></math><math></math><math></math><math></math>

Haven't quite finished this.

Obstructions to uniformity

2009-06-04T20:14:08Z

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Suppose that we have a definition of [[quasirandomness]] for subsets of some structure. As stated in the quasirandomness article, a definition of quasirandomness is not very useful unless one has some understanding of sets that are ''not'' quasirandom. Let S be a structure (such as a complete graph, <math>[3]^n,</math> a finite Abelian group, or <math>[n]^2</math>) and let A be a subset of S of density <math>\delta</math>. Let f be <math>1-\delta</math> on A and <math>-\delta</math> on the complement of A. Then the average of f is zero. Typically, one would like to show that if A, or equivalently f, is not quasirandom then there is some "structured subset" <math>S'\subset S</math> such that <math>\mathbb{E}_{x\in S'}f(x)\geq c</math> for some positive constant c that depends on <math>\delta</math> only. Better still, one would like a converse: that any function that has positive expectation on a substructure must fail to be quasirandom. If we can do this, then we say that we have a complete set of obstructions to uniformity. ("Uniformity" is sometimes used as a synonym for "quasirandomness".)

More generally, we can look not for structured sets but structured ''functions''. In this case we would like to find a set G of functions such that for every non-quasirandom function f there exists g in G such that <math>\mathbb{E}_{x\in S}f(x)g(x)\geq c,</math> and such that if there exists such a g then f is not quasirandom.

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==The relevance to the density Hales-Jewett theorem==

It is possible to think about obstructions to uniformity even in the absence of a precisely formulated definition of quasirandomness. For example, in the case of the density Hales-Jewett theorem, we know that we want quasirandom sets to contain roughly the expected number of combinatorial lines. Therefore, we can temporarily (and unsatisfactorily) ''define'' a quasirandom set to be one that contains roughly the expected number of combinatorial lines and try to classify "extreme examples" of sets that do not contain roughly the expected number. These extreme examples will typically be highly structured sets: for instance, the set of all sequences x such that <math>x_1=1</math> is easily shown to have more combinatorial lines than a random set of density 1/3, but it is also a highly structured subset of <math>[3]^n.</math> If these examples appear to belong to a limited number of types, we can then hypothesize that ''every'' set with the wrong number of combinatorial lines correlates with one of these extreme examples. It is these extreme examples that one calls obstructions to uniformity.

Once we have formulated a conjecture of this type, we would hope to prove it in two stages as follows.

*Every quasirandom set contains roughly as many combinatorial lines as a random set of the same density. Therefore, a set with the wrong number of combinatorial lines is not quasirandom.

*Every non-quasirandom set is noticeably correlated with an obstruction to uniformity.

==Possible candidates==

The obstructions to uniformity appear to be hard to describe if one uses the uniform measure on <math>[3]^n,</math> even if one localizes to a few slices. However, if one uses [[equal-slices measure]], then all known obstructions are sets of the following form. (The next couple of paragraphs are lifted from [http://michaelnielsen.org/polymath1/index.php?title=DHJ%283%29&section=4 the discussion of DHJ(1,3)].)

Let <math>\mathcal{U},\mathcal{V}</math> and <math>\mathcal{W}</math> be collections of subsets of <math>[n].</math> Define <math>\mathcal{A}(\mathcal{U},\mathcal{V},\mathcal{W})</math> to be the set of all triples <math>(U,V,W),</math> where <math>U,V,W</math> are disjoint, and <math>U\in\mathcal{U},V\in\mathcal{V},W\in\mathcal{W}.</math> These triples are in an obvious one-to-one correspondence with elements of <math>[3]^n.</math>

Call <math>\mathcal{A}</math> a ''set of complexity 1'' if it is of the form <math>\mathcal{A}(\mathcal{U},\mathcal{V},\mathcal{W})</math>. For instance, a [[slice]] <math>\Gamma_{a,b,c}</math> is of complexity 1, as is a union of slices of the form <math>\bigcup\{\Gamma_{a,b,c}:(a,b,c)\in A\times B\times C, a+b+c=n\}</math>

It is conceivable, though certainly not proved, that every set that does not contain roughly the expected number of combinatorial lines, given its density (where everything is with respect to equal-slices measure) correlates significantly with a set of complexity 1. In the opposite direction, sets of complexity 1 do give a large source of examples of sets with the wrong number of combinatorial lines.

A general result about density increments

2009-06-03T22:55:51Z

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==Terminology==

If A and B are two subsets of a finite set X, then we say that the ''density of'' A ''in'' B is <math>|A\cap B|/|B|.</math>

==Description of result==

Let us focus on the proof of DHJ(3), though what we say is much more general. That proof has two stages, which can be described as follows.

1. Prove that if <math>\mathcal{A}</math> is a subset of <math>[3]^n</math> of density <math>\delta</math> then there is a dense 12-subset <math>\mathcal{B}</math> of a subspace S of dimension tending to infinity, such that the density of <math>\mathcal{A}</math> in <math>\mathcal{B}</math> is at least <math>\delta+c(\delta),</math> where <math>c(\delta)>0.</math> (In fact, <math>c(\delta)</math> is proportional to <math>\delta^2.</math>)

2. Every 12-set can be almost entirely partitioned into m-dimensional subspaces, where m tends to infinity with n. Here, m depends only on the density of the part of the 12-set that is allowed not to be partitioned.

Once we have these two stages, we are basically done, for reasons that can be appreciated even if one does not know the definition of a 12-set. The reason is that if we partition all of a 12-set apart from a subset of measure at most <math>c(\delta)/2</math> into m-dimensional subspaces, then the density of <math>\mathcal{A}</math> in the partitioned part is at least <math>\delta+c(\delta)/2,</math> so by averaging <math>\mathcal{A}</math> has density at least <math>\delta+c(\delta)/2</math> in at least one of these subspaces. That gives us a density increment on a subspace, which is exactly what we need for a [[density-increment_strategies|density-increment strategy]].

Now let us generalize 2 very slightly. Given a finite set X, we define its ''characteristic measure'' <math>\xi</math> to be the function that takes the value <math>1/|X|</math> everywhere in X and 0 everywhere else. Given a set Y, we write <math>\xi(Y)</math> for <math>\sum_{y\in Y}\xi(y)=|X\cap Y|/|X|.</math> Let <math>\beta</math> be the characteristic measure of <math>\mathcal{B},</math> let <math>S_1,\dots,S_N</math> be a collection of subspaces, and for each i let <math>\sigma_i</math> be the characteristic measure of <math>S_i.</math> We are assuming that <math>\beta(\mathcal{A})\geq\delta+c(\delta).</math> If we can find a convex combination <math>\sum_{i=1}^N\lambda_i\sigma_i</math> such that <math>\|\sum_i\lambda_i\sigma_i-\beta\|\leq c(\delta)/2,</math> then <math>\sum_i\lambda_i\sigma_i(\mathcal{A})\geq\delta+c(\delta)/2.</math> It follows that there exists i such that <math>\sigma_i(\mathcal{A})\geq\delta+c(\delta)/2,</math> which is what we wanted.

The main result of this page is that a converse to this generalized step 2 is true as well. Loosely, this tells us that any proof that a density increase on a 12-set implies a density increase on a subspace must also show that a 12-set can be evenly covered with subspaces, up to a small error.

==The proof==

Suppose that we ''cannot'' find a convex combination <math>\sum_{i=1}^N\lambda_i\sigma_i</math> such that <math>\|\sum_i\lambda_i\sigma_i-\beta\|\leq c(\delta)/2.</math> Then the Hahn-Banach theorem provides us with a function F and non-negative reals <math>\lambda+\mu=1</math> such that <math>\mathbb{E}_{x\in\mathcal{B}}F(x)>1,</math> while <math>\|F\|_\infty\leq 2\lambda/c(\delta)</math> and <math>\sigma_i(F)=\mathbb{E}_{x\in S_i}F(x)\leq \mu</math> for every i. From this it follows that <math>\lambda>c(\delta)/2.</math>

Now let <math>G(x)=\delta(1+F(x)/\|F\|_\infty).</math> Then G takes values in <math>[0,2\delta],</math> and <math>\mathbb{E}_{x\in\mathcal{B}}G(x)>\delta(1+c(\delta)/2\lambda).</math> However, for each i we have <math>\sigma_i(G)\leq\delta(1+\mu/\|F\|_\infty)\leq\delta(1+c(\delta)^2\mu/2\lambda)</math><math></math><math></math><math></math><math></math><math></math><math></math><math></math>

Haven't quite finished this.

Obstructions to uniformity

2009-06-03T22:54:57Z

93.190.138.249: http://www.youtube.com/MikheilEstebe#1 levita, 1068256786,

Suppose that we have a definition of [[quasirandomness]] for subsets of some structure. As stated in the quasirandomness article, a definition of quasirandomness is not very useful unless one has some understanding of sets that are ''not'' quasirandom. Let S be a structure (such as a complete graph, <math>[3]^n,</math> a finite Abelian group, or <math>[n]^2</math>) and let A be a subset of S of density <math>\delta</math>. Let f be <math>1-\delta</math> on A and <math>-\delta</math> on the complement of A. Then the average of f is zero. Typically, one would like to show that if A, or equivalently f, is not quasirandom then there is some "structured subset" <math>S'\subset S</math> such that <math>\mathbb{E}_{x\in S'}f(x)\geq c</math> for some positive constant c that depends on <math>\delta</math> only. Better still, one would like a converse: that any function that has positive expectation on a substructure must fail to be quasirandom. If we can do this, then we say that we have a complete set of obstructions to uniformity. ("Uniformity" is sometimes used as a synonym for "quasirandomness".)

More generally, we can look not for structured sets but structured ''functions''. In this case we would like to find a set G of functions such that for every non-quasirandom function f there exists g in G such that <math>\mathbb{E}_{x\in S}f(x)g(x)\geq c,</math> and such that if there exists such a g then f is not quasirandom.

http://www.youtube.com/MikheilEstebe#1 levita, 1068256786,

==The relevance to the density Hales-Jewett theorem==

It is possible to think about obstructions to uniformity even in the absence of a precisely formulated definition of quasirandomness. For example, in the case of the density Hales-Jewett theorem, we know that we want quasirandom sets to contain roughly the expected number of combinatorial lines. Therefore, we can temporarily (and unsatisfactorily) ''define'' a quasirandom set to be one that contains roughly the expected number of combinatorial lines and try to classify "extreme examples" of sets that do not contain roughly the expected number. These extreme examples will typically be highly structured sets: for instance, the set of all sequences x such that <math>x_1=1</math> is easily shown to have more combinatorial lines than a random set of density 1/3, but it is also a highly structured subset of <math>[3]^n.</math> If these examples appear to belong to a limited number of types, we can then hypothesize that ''every'' set with the wrong number of combinatorial lines correlates with one of these extreme examples. It is these extreme examples that one calls obstructions to uniformity.

Once we have formulated a conjecture of this type, we would hope to prove it in two stages as follows.

*Every quasirandom set contains roughly as many combinatorial lines as a random set of the same density. Therefore, a set with the wrong number of combinatorial lines is not quasirandom.

*Every non-quasirandom set is noticeably correlated with an obstruction to uniformity.

==Possible candidates==

The obstructions to uniformity appear to be hard to describe if one uses the uniform measure on <math>[3]^n,</math> even if one localizes to a few slices. However, if one uses [[equal-slices measure]], then all known obstructions are sets of the following form. (The next couple of paragraphs are lifted from [http://michaelnielsen.org/polymath1/index.php?title=DHJ%283%29&section=4 the discussion of DHJ(1,3)].)

Let <math>\mathcal{U},\mathcal{V}</math> and <math>\mathcal{W}</math> be collections of subsets of <math>[n].</math> Define <math>\mathcal{A}(\mathcal{U},\mathcal{V},\mathcal{W})</math> to be the set of all triples <math>(U,V,W),</math> where <math>U,V,W</math> are disjoint, and <math>U\in\mathcal{U},V\in\mathcal{V},W\in\mathcal{W}.</math> These triples are in an obvious one-to-one correspondence with elements of <math>[3]^n.</math>

Call <math>\mathcal{A}</math> a ''set of complexity 1'' if it is of the form <math>\mathcal{A}(\mathcal{U},\mathcal{V},\mathcal{W})</math>. For instance, a [[slice]] <math>\Gamma_{a,b,c}</math> is of complexity 1, as is a union of slices of the form <math>\bigcup\{\Gamma_{a,b,c}:(a,b,c)\in A\times B\times C, a+b+c=n\}</math>

It is conceivable, though certainly not proved, that every set that does not contain roughly the expected number of combinatorial lines, given its density (where everything is with respect to equal-slices measure) correlates significantly with a set of complexity 1. In the opposite direction, sets of complexity 1 do give a large source of examples of sets with the wrong number of combinatorial lines.

A general result about density increments

2009-05-31T15:16:59Z

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==Terminology==

If A and B are two subsets of a finite set X, then we say that the ''density of'' A ''in'' B is <math>|A\cap B|/|B|.</math>

==Description of result==

Let us focus on the proof of DHJ(3), though what we say is much more general. That proof has two stages, which can be described as follows.

1. Prove that if <math>\mathcal{A}</math> is a subset of <math>[3]^n</math> of density <math>\delta</math> then there is a dense 12-subset <math>\mathcal{B}</math> of a subspace S of dimension tending to infinity, such that the density of <math>\mathcal{A}</math> in <math>\mathcal{B}</math> is at least <math>\delta+c(\delta),</math> where <math>c(\delta)>0.</math> (In fact, <math>c(\delta)</math> is proportional to <math>\delta^2.</math>)

2. Every 12-set can be almost entirely partitioned into m-dimensional subspaces, where m tends to infinity with n. Here, m depends only on the density of the part of the 12-set that is allowed not to be partitioned.

Once we have these two stages, we are basically done, for reasons that can be appreciated even if one does not know the definition of a 12-set. The reason is that if we partition all of a 12-set apart from a subset of measure at most <math>c(\delta)/2</math> into m-dimensional subspaces, then the density of <math>\mathcal{A}</math> in the partitioned part is at least <math>\delta+c(\delta)/2,</math> so by averaging <math>\mathcal{A}</math> has density at least <math>\delta+c(\delta)/2</math> in at least one of these subspaces. That gives us a density increment on a subspace, which is exactly what we need for a [[density-increment_strategies|density-increment strategy]].

Now let us generalize 2 very slightly. Given a finite set X, we define its ''characteristic measure'' <math>\xi</math> to be the function that takes the value <math>1/|X|</math> everywhere in X and 0 everywhere else. Given a set Y, we write <math>\xi(Y)</math> for <math>\sum_{y\in Y}\xi(y)=|X\cap Y|/|X|.</math> Let <math>\beta</math> be the characteristic measure of <math>\mathcal{B},</math> let <math>S_1,\dots,S_N</math> be a collection of subspaces, and for each i let <math>\sigma_i</math> be the characteristic measure of <math>S_i.</math> We are assuming that <math>\beta(\mathcal{A})\geq\delta+c(\delta).</math> If we can find a convex combination <math>\sum_{i=1}^N\lambda_i\sigma_i</math> such that <math>\|\sum_i\lambda_i\sigma_i-\beta\|\leq c(\delta)/2,</math> then <math>\sum_i\lambda_i\sigma_i(\mathcal{A})\geq\delta+c(\delta)/2.</math> It follows that there exists i such that <math>\sigma_i(\mathcal{A})\geq\delta+c(\delta)/2,</math> which is what we wanted.

The main result of this page is that a converse to this generalized step 2 is true as well. Loosely, this tells us that any proof that a density increase on a 12-set implies a density increase on a subspace must also show that a 12-set can be evenly covered with subspaces, up to a small error.

==The proof==

Suppose that we ''cannot'' find a convex combination <math>\sum_{i=1}^N\lambda_i\sigma_i</math> such that <math>\|\sum_i\lambda_i\sigma_i-\beta\|\leq c(\delta)/2.</math> Then the Hahn-Banach theorem provides us with a function F and non-negative reals <math>\lambda+\mu=1</math> such that <math>\mathbb{E}_{x\in\mathcal{B}}F(x)>1,</math> while <math>\|F\|_\infty\leq 2\lambda/c(\delta)</math> and <math>\sigma_i(F)=\mathbb{E}_{x\in S_i}F(x)\leq \mu</math> for every i. From this it follows that <math>\lambda>c(\delta)/2.</math>

Now let <math>G(x)=\delta(1+F(x)/\|F\|_\infty).</math> Then G takes values in <math>[0,2\delta],</math> and <math>\mathbb{E}_{x\in\mathcal{B}}G(x)>\delta(1+c(\delta)/2\lambda).</math> However, for each i we have <math>\sigma_i(G)\leq\delta(1+\mu/\|F\|_\infty)\leq\delta(1+c(\delta)^2\mu/2\lambda)</math><math></math><math></math><math></math><math></math><math></math><math></math><math></math>

Haven't quite finished this.

Obstructions to uniformity

2009-05-30T10:11:49Z

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Suppose that we have a definition of [[quasirandomness]] for subsets of some structure. As stated in the quasirandomness article, a definition of quasirandomness is not very useful unless one has some understanding of sets that are ''not'' quasirandom. Let S be a structure (such as a complete graph, <math>[3]^n,</math> a finite Abelian group, or <math>[n]^2</math>) and let A be a subset of S of density <math>\delta</math>. Let f be <math>1-\delta</math> on A and <math>-\delta</math> on the complement of A. Then the average of f is zero. Typically, one would like to show that if A, or equivalently f, is not quasirandom then there is some "structured subset" <math>S'\subset S</math> such that <math>\mathbb{E}_{x\in S'}f(x)\geq c</math> for some positive constant c that depends on <math>\delta</math> only. Better still, one would like a converse: that any function that has positive expectation on a substructure must fail to be quasirandom. If we can do this, then we say that we have a complete set of obstructions to uniformity. ("Uniformity" is sometimes used as a synonym for "quasirandomness".)

More generally, we can look not for structured sets but structured ''functions''. In this case we would like to find a set G of functions such that for every non-quasirandom function f there exists g in G such that <math>\mathbb{E}_{x\in S}f(x)g(x)\geq c,</math> and such that if there exists such a g then f is not quasirandom.

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==The relevance to the density Hales-Jewett theorem==

It is possible to think about obstructions to uniformity even in the absence of a precisely formulated definition of quasirandomness. For example, in the case of the density Hales-Jewett theorem, we know that we want quasirandom sets to contain roughly the expected number of combinatorial lines. Therefore, we can temporarily (and unsatisfactorily) ''define'' a quasirandom set to be one that contains roughly the expected number of combinatorial lines and try to classify "extreme examples" of sets that do not contain roughly the expected number. These extreme examples will typically be highly structured sets: for instance, the set of all sequences x such that <math>x_1=1</math> is easily shown to have more combinatorial lines than a random set of density 1/3, but it is also a highly structured subset of <math>[3]^n.</math> If these examples appear to belong to a limited number of types, we can then hypothesize that ''every'' set with the wrong number of combinatorial lines correlates with one of these extreme examples. It is these extreme examples that one calls obstructions to uniformity.

Once we have formulated a conjecture of this type, we would hope to prove it in two stages as follows.

*Every quasirandom set contains roughly as many combinatorial lines as a random set of the same density. Therefore, a set with the wrong number of combinatorial lines is not quasirandom.

*Every non-quasirandom set is noticeably correlated with an obstruction to uniformity.

==Possible candidates==

The obstructions to uniformity appear to be hard to describe if one uses the uniform measure on <math>[3]^n,</math> even if one localizes to a few slices. However, if one uses [[equal-slices measure]], then all known obstructions are sets of the following form. (The next couple of paragraphs are lifted from [http://michaelnielsen.org/polymath1/index.php?title=DHJ%283%29&section=4 the discussion of DHJ(1,3)].)

Let <math>\mathcal{U},\mathcal{V}</math> and <math>\mathcal{W}</math> be collections of subsets of <math>[n].</math> Define <math>\mathcal{A}(\mathcal{U},\mathcal{V},\mathcal{W})</math> to be the set of all triples <math>(U,V,W),</math> where <math>U,V,W</math> are disjoint, and <math>U\in\mathcal{U},V\in\mathcal{V},W\in\mathcal{W}.</math> These triples are in an obvious one-to-one correspondence with elements of <math>[3]^n.</math>

Call <math>\mathcal{A}</math> a ''set of complexity 1'' if it is of the form <math>\mathcal{A}(\mathcal{U},\mathcal{V},\mathcal{W})</math>. For instance, a [[slice]] <math>\Gamma_{a,b,c}</math> is of complexity 1, as is a union of slices of the form <math>\bigcup\{\Gamma_{a,b,c}:(a,b,c)\in A\times B\times C, a+b+c=n\}</math>

It is conceivable, though certainly not proved, that every set that does not contain roughly the expected number of combinatorial lines, given its density (where everything is with respect to equal-slices measure) correlates significantly with a set of complexity 1. In the opposite direction, sets of complexity 1 do give a large source of examples of sets with the wrong number of combinatorial lines.

DJH(1,3)

2009-05-28T10:09:20Z

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