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Let $T$ be a finite set of tiles in $\mathbb{R}^d$. A tiling of $\mathbb{R}^d$ by $T$ is a collection of disjoint translates of tiles in $T$ whose union is $\mathbb{R}^d$. A tiling is $k$-chromatic if its tiles can be colored with $k$ colors such that no two tiles sharing a $(d-1)$-dimensional face have the same color. The chromatic tiling complexity $\chi_T(d)$ is the smallest $k$ such that every tiling by $T$ is $k$-chromatic.

Let $G$ be the graph with vertices corresponding to tiles in $T$, with edges between tiles that can share a $(d-1)$-dimensional face in a tiling.

Is it true that $\chi_T(d) \leq \chi(G)$.

I suspect the conjecture holds, but strict inequality might arise from intricate tile shapes or specific tiling patterns that force long-range color dependencies not captured by the adjacency graph.

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  • $\begingroup$ $\chi(G)$ being the chromatic number of the graph $G$? $\endgroup$ Commented Aug 20 at 22:55
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    $\begingroup$ Does not this bound follow from the straightforward pushforward of the coloring of $G$ to the coloring of tiles? $\endgroup$ Commented Aug 21 at 7:16
  • $\begingroup$ WHAT DOES $\chi(G)$ MEAN, PLEASE? $\endgroup$ Commented Aug 22 at 2:53
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    $\begingroup$ @Gerry I'm sure that it is the chromatic number of the graph, as you wrote. $\endgroup$
    – domotorp
    Commented Aug 22 at 5:06
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    $\begingroup$ How does this work if there is a tile that can share a face with itself? E.g., what if $T$ is just a single unit square? If $G$ has a loop, $\chi(G)$ is infinite, but if it doesn't then $\chi(G)=1$ while $\chi_T(d)$ is at least $3$ so the inequality trivially breaks. $\endgroup$ Commented Aug 26 at 3:49

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It can be strict already for $d=2$. Let $T$ consist of five tiles: a square, and four very silly shapes that nevertheless can be put together such that they pairwise share a $1$-dimensional face (thereby forming a $K_4$). Make the other sides of these four shapes so that nothing else can fit to them. This way the only tilings by $T$ can only use squares, so $\chi_T(2)=3$.

ps. And I recommend using the notation $\chi(T)$ instead of $\chi_T(d)$.

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