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Oct 24, 2018 at 23:04 comment added Gerhard Paseman Not much more. There are some universal patterns (33 squares, I think) which guarantee a Sudoku solution if it exists. I think it is easy to find these patterns, and possibly work ones way down. Gerhard "Unpublished And Half Remembered Work" Paseman, 2018.10.24.
Oct 24, 2018 at 22:16 comment added David G. Stork @GerhardPaseman: What is becoming clear is that the assumption that a fixed $17$ cells together with the minimum-information assignment of digits to those cells cannot lead to a tight bound on the number of bits needed to describe a latin square. ZachTeitler's improvement of the question is: How few cells need be specified ahead of time ($k = 23?$) such that an assignment of digits leads to a unique solution such that the total information (in bits) of the cell sub-selection and digit assignments can describe all Latin squares. Presumably the total information will be $\log_2 N$.
Oct 24, 2018 at 21:54 history edited David G. Stork CC BY-SA 4.0
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Oct 24, 2018 at 21:46 comment added David G. Stork @ZachTeitler: Good point: Some of the assignments of digits to $17$ cells will lead to unsolvable puzzles. Indeed, perhaps the better question is how few ($23$?) cells suffice such that some subset can be chosen (with digit assignments) such that the expected information content (in bits) is minimal. (I'll revise the question.) Thanks.
Oct 24, 2018 at 19:19 comment added Zach Teitler No single selection of $17$ locations will come close to determining all puzzles. Your bit count shows this. On top of that, a lot of bits are "wasted" on unsolvable clues (e.g., two $1$s in the same row). Here's a question. Is there some set of, say, $23$ squares that does uniquely determine all puzzles? $23$ is the least integer greater than $\log_9(N)$, so there are $>N$ ways to fill clues into a set of $23$ squares. My guess is no, you would need more than $23$.
Oct 24, 2018 at 19:07 comment added David G. Stork @ZachTeitler: My guess/ansatz was that a unique selection of 17 cell locations might uniquely determine all sudoku puzzles, but that set would be arranged like the set in my example, certainly not a row and a column. (Frankly, that would be one of the least constraining arrangements.
Oct 24, 2018 at 18:54 comment added Zach Teitler ... counting these seems like a challenge, to put it mildly. It will certainly depend on which $17$ squares are chosen for the clue locations. E.g., if the $17$ squares are the first row and first column, then very, very few clue fillings will uniquely determine puzzle solutions!
Oct 24, 2018 at 18:53 comment added Zach Teitler 1. Can you please explain where $\binom{17}{9}/(2!)^8$ comes from? The number of ways to put $1,1,2,2,\dotsc,8,8,9$ into $17$ squares is $17!/(2!)^8$. If you allow a choice of which digit is the "odd" one ($9$) then you get $9 \cdot 17!/(2!)^8$. Have I misunderstood what you are counting? 2. Presumably the large majority of placements into those $17$ squares fail to uniquely determine a puzzle solution. Either they admit more than one solution (failure of uniqueness) or they admit no solutions — e.g., if the $17$ clue squares have a duplicated digit in a row, column, or region. ...
Oct 24, 2018 at 17:59 comment added Gerhard Paseman Suppose I give you k bits of information to represent a Sudoku solution. If k is less than log_2 N, how would I use k bits or fewer to distinguish each of the other N-1 solutions? Are you asking about equivalence classes of solutions instead? Gerhard "Finds The Question Itself Puzzling" Paseman, 2018.10.24.
Oct 24, 2018 at 17:27 history edited David G. Stork CC BY-SA 4.0
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Oct 24, 2018 at 0:22 history edited David G. Stork CC BY-SA 4.0
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Oct 23, 2018 at 22:23 comment added j.c. You might enjoy reading this old question and its answers: mathoverflow.net/questions/129143
Oct 23, 2018 at 21:43 history asked David G. Stork CC BY-SA 4.0