For integers $n$ and $k$, the density Hales-Jewett number $c_{n,k}$ isdefined as the maximal size of a subset of $[k]^n$ that contains nocombinatorial line. We prove a lower bound on $c_{n,k}$, similar to the lowerbound in \cite{polymath2010moser}, but with better dependency on $k$. The boundin \cite{polymath2010moser} is roughly $c_{n,k}/k^n \ge exp(-O(\logn)^{1/\lceil \log_2 k\rceil})$ and we show $c_{n,k}/k^n \ge exp(-O(\logn)^{1/(k-1)})$. The proof of the bound uses the well-known construction ofBehrend \cite{behrend1946sets} and Rankin \cite{rankin1965sets}, as the one in\cite{polymath2010moser}, but does not require the recent refinements\cite{elkin2010improved, green2010note, o2011sets}. Instead our proof relies onan argument from communication complexity. In addition, we show that for $k \ge 3$ the density Hales-Jewett number$c_{n,k}$ is equal to the maximal size of a cylinder intersection in theproblem $Part_{n,k}$ of testing whether $k$ subsets of $[n]$ form a partition.It follows that the communication complexity, in the Number On the Forehead(NOF) model, of $Part_{n,k}$, is equal to the minimal size of a partition of$[k]^n$ into subsets that do not contain a combinatorial line. Thus, Tesson'sbound on the problem using the Hales-Jewett theorem \cite{tessonapplication} isin fact tight, and the density Hales-Jewett number can be thought of as aquantity in communication complexity.
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机译:对于整数$ n $和$ k $,密度Hales-Jewett数$ c_ {n,k} $定义为包含无组合线的$ [k] ^ n $子集的最大大小。我们证明了$ c_ {n,k} $的下界,类似于\ cite {polymath2010moser}中的下界,但对$ k $的依赖性更好。边界\ cite {polymath2010moser}大约为$ c_ {n,k} / k ^ n \ ge exp(-O(\ logn)^ {1 / \ lceil \ log_2 k \ rceil})$,我们显示$ c_ { n,k} / k ^ n \ ge exp(-O(\ logn)^ {1 /(k-1)})$。边界的证明使用了著名的Behrend \ cite {behrend1946sets}和Rankin \ cite {rankin1965sets}的构造作为in \ cite {polymath2010moser}的构造,但不需要最近的改进\ cite {elkin2010improved,green2010note,o2011sets }。相反,我们的证明依赖于通信复杂性的论证。此外,我们证明了对于$ k \ ge 3 $,密度Hales-Jewett数$ c_ {n,k} $等于问题$ Part_ {n,k} $中的圆柱交点的最大尺寸$ [n] $的$ k $个子集形成一个分区。因此,在$ Part_ {n,k} $的额头数字(NOF)模型中,通信复杂度等于a的最小大小。将$ [k] ^ n $划分为不包含组合线的子集。因此,Tesson使用Hales-Jewett定理\ cite {tessonapplication}解决这个问题实际上是紧密的,并且可以将密度Hales-Jewett数视为通信复杂性的数量。
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