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Oct 25, 2022 at 13:25 vote accept Keen-ameteur
Oct 14, 2022 at 16:34 history edited YCor CC BY-SA 4.0
fixed typo
Oct 14, 2022 at 16:13 answer added Ville Salo timeline score: 1
Oct 13, 2022 at 18:27 comment added Ville Salo I think that the periodic tiling problem where you require that all tiles are used, let's call this problem $P$, is easier to prove undecidable than the one where you don't require this. Namely, I think you don't even need the existence of an aperiodic tile set to show that $P$ is undecidable. The same is true for the usual tiling problem, where I think the requirement that all tiles are used in the configuration is roughly equivalent to requiring that a seed tile is used. (You do need some additional tricks to make sure your Turing machines use all their transitions.)
Oct 13, 2022 at 11:20 comment added Keen-ameteur @VilleSalo Okay, thanks for your answer. It is not the answer I wished for, but it settles this issue.
Oct 12, 2022 at 16:55 comment added Ville Salo Apologies, I did not check your link. Anyway it's undecidable whether or not we require all tiles are used.
Oct 12, 2022 at 16:48 comment added Keen-ameteur @VilleSalo It looked to me like that result there might be talking about having to use all non prohibited tiles.
Oct 12, 2022 at 15:53 comment added Ville Salo I didn't get the part about not having to use all tiles though
Oct 12, 2022 at 15:51 comment added Ville Salo The question of whether you can tile periodically with a given tile set is not the same as the domino problem (whether you can tile the plane). It is, however, also undecidable mathoverflow.net/questions/121483/…
Oct 12, 2022 at 15:15 history asked Keen-ameteur CC BY-SA 4.0