Btw, the transforms you list aren't the only maps preserving the puzzle structure. You should allow more equivalences IMO. :)
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As best I can tell the problem's full symmetry group is of order 3072, not a mere 192 :)
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Hm. *Thinks.* Still working in my head, I believe my proof generalizes, & there remains more than one solution taking into account the extra 4 symmetries and 16x isomorphisms.
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I can confirm there are at least 2 distinct solutions even with this (dunno if there are more). Did that with pencil and paper though!
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I guess it was a pretty successful nerdsnipe after all, even if not as originally intended...
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Take solution where n:th row is 1 2 3 4 cyclicly shifted by n. Observe every row and column is either increasing/decreasing by 1 (property A). Now swap first two columns. All columns still satisfy A. Rows no longer has satsify A. Rotation/reflection maintain the A property. 1/
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You can't "fix" the A property by permutation because if you fix row, column will break. Thus we have found at least two solutions. [I've yet to verify on paper] 2/
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lgtm
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Well? I'm curious. Is that correct?
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Think so
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I just went by "analogue of Sudoku". If you have 1234 going counterclockwise around a 4×4 grid and spiraling inward, you can reverse that spiral to be clockwise and still solve the puzzle.
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Beautiful. Rotation of a spiral remains a spiral. And reflection of a spiral remains a spiral. And symbol permutation of a spiral remains a spiral. Now if you can find a solution that's not a spiral you have proven there are at least two distinct solutions. :)
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Doesn't "treat permutations of the 4 symbols as equivalent" mean every solution is equivalent by definition?
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No. On a line "abab" would be equivalent to "baba", but not to "baab", so permutation does not means everything is equivalent to everything.
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Don't you need 4 *unique* symbols per row/column, i.e. "abcd" not "abab"?
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I was trying to illustrate permutation and not sudoku here. Better exampe abcd bcda is a permutation of bcda cdab (a always sent to b) But not of cdab dabc Since here a is sent to b and c on 1st and 2 nd line
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That's just not much of a nerdsnipe because I just quickly conclude I can't do that in my head.
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Yes. There is one solution where a symbol is never diagonally adjacent to itself and one where it always is (to 2 other copies). There are other solutions but harder to compute in your head.
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You sniped me well! Head: def. more than one. Paper: I know the number. I wont spoil. Pic gives solution. Thanks Eliezer!pic.twitter.com/wqftUOM4ha
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Well. Made a mistake. Heres the correct one.pic.twitter.com/BGSbN8VJG4
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