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Aug 26 at 3:49 comment added RavenclawPrefect 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.
Aug 22 at 5:06 comment added domotorp @Gerry I'm sure that it is the chromatic number of the graph, as you wrote.
Aug 22 at 2:53 comment added Gerry Myerson WHAT DOES $\chi(G)$ MEAN, PLEASE?
Aug 21 at 7:16 comment added Fedor Petrov Does not this bound follow from the straightforward pushforward of the coloring of $G$ to the coloring of tiles?
Aug 21 at 5:22 answer added domotorp timeline score: 2
Aug 20 at 22:55 comment added Gerry Myerson $\chi(G)$ being the chromatic number of the graph $G$?
Aug 20 at 22:50 history edited YCor CC BY-SA 4.0
removed capitals from title, added tag
S Aug 20 at 22:30 review First questions
Aug 21 at 4:45
S Aug 20 at 22:30 history asked Vincenco Fedor CC BY-SA 4.0