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Timeline for Prime lights out

Current License: CC BY-SA 4.0

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Aug 2, 2023 at 3:38 history bounty ended msh210
Aug 9, 2022 at 11:38 comment added Falco @WhatsUp you are right. I was thinking of a 2x2 grid, which can be turned to all 1 in a classical lights-out, but which will have all 3s when you try to get all equal. And this cannot be removed to all-1 even with negative moves, since you can only get multiples of 3 with a 2x2 grid and all equal.
Aug 5, 2022 at 16:00 comment added Ben Reiniger @WhatsUp, I think that argument requires negative moves? But that's fine, just crank up the binary representation grid with more moves.
Aug 5, 2022 at 14:56 comment added WhatsUp @Falco A $2\times 3$ grid can be filled with all-one by making moves in two opposite corners. With small modification, this method also works if one can produce an "all-equal" result, which is equivalent to saying that the "all-one" vector lies in the $\Bbb Q$-vector space generated by the basis vectors. I have checked every rectangular grid up to $20\times 20$ that this is always true. Perhaps there is a general argument to prove that.
Aug 5, 2022 at 8:59 comment added Dmitry Kamenetsky @Falco ok. Can we say that solutions exist for all grids NxN with N>=3? I can confirm that the 5x5 solution exists.
Aug 5, 2022 at 8:48 comment added Falco I especially like this answer, because it found the solution (16) without actually searching for any specific primes.
Aug 5, 2022 at 8:47 comment added Falco @DmitryKamenetsky I don't think that's true. This answer requires a set of moves which produce 1 on all cells. Such a move does not exist for any grid. E.g. a 2x3 grid cannot be filled with all 1s.
Aug 3, 2022 at 23:40 comment added Dmitry Kamenetsky This is a very nice generalisation. It means we can find solutions for a grid of any size.
Aug 3, 2022 at 17:11 history answered WhatsUp CC BY-SA 4.0