Re: [math-fun] sudoku, uniqueness, proofs
Michael Kleber's post led me to wonder about a related issue: Suppose a sudoku has *exactly* 2 solutions. Then what is the maximum number of boxes that can differ in the two solutions? I'd like this question a bit more if it were imo "purer": The whole sudoku idea could be played based instead on finishing merely a Latin square (each row and column of an NxN square must contain all N of a set of N distinct tokens). So here's a question: What is the set comp(N) of numbers C such that there exists a partially completed NxN Latin square having exactly C completions? I strongly suspect that if N >= 4, then 2 belongs to comp(N). In which case we can go back to the original question for each N and ask, Given an NxN partially completed Latin square having exactly 2 completions, what is the maximum number of squares that can differ between its completions? --Dan
Dan mused:
Suppose a sudoku has *exactly* 2 solutions. Then what is the maximum number of boxes that can differ in the two solutions?
Surely it's possible to attain 18: take a sudoku and erase all the 3s and 8s. As long as the permutation i -> the column containing the 8 in the row where the 3 was in column i is a 9-cycle, you can fill it back in as it was originally or with 3 and 8 swapped, and in no other way. I think many sudoku will have such a 9-cycle -- um, the sample that appears in the Wikipedia sudoku article does (for 3 and 8, in fact!), for example. (Of course, this will work for latin squares also.) Needless to say, I have no reason to think this is maximal. --Michael Kleber -- It is very dark and after 2000. If you continue you are likely to be eaten by a bleen.
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Michael Kleber