In single-solution regular Sudoku puzzles, sometimes you may encounter a combination of possible cell solutions such as:

  • {2,7} in A1 and A7 (same row)
  • {2,7} in A7 and B8 (same box)
  • {2,7} in B8 and E8 (same column)
  • {2,3,5,7} in E1 (same column as A1 and same row as E8)

I call patterns like the one above conjugate-pair pseudocycles because they're made of conjugate pairs that link to each other by common cells to form a chain, but one node has more solutions possible than just that pair. How do the cells with just that pair influence which digits are removed from the possibilities for that cell?

I have answered below:


1 Answer 1


As it turns out, if the Sudoku is a regular puzzle with one solution, if such a pattern is present, the two digits involved in the conjugate pairs should always be removed from the notes of the cell in question. This is true whether there is an odd or even number of such cells—if it is one parity, i.e. odd, the combination may result in a multiple-solution impasse; if it is the other, i.e. even, the digits will be eliminated anyway as solutions are assigned to the other cells in the pseudocycle. In this case, {2,7} would be removed from E1, leaving only {3,5} as the remaining possibilities.

Let's examine the case above more closely: if A1 is 2, then A7 is 7, B8 is 2, and E8 is 7, which eliminates {2,7} from E1 because both digits are already assigned. If you pick 7 for A1, E8 must likewise be 2, also eliminating both digits. Therefore E1 must be reduced to {3,5}. As another example,

  • {2,7} in A1 and A7
  • {2,7} in A7 and E7
  • {2,3,5,7} in E1

If A1 is 2, then A7 is 7 and E7 is also 2, so E1 could be 7. But if A1 is 7, E7 is likewise 7, so E1 could be 2. Now, reducing E1 to {2,7} will not work because the puzzle has only one solution, and doing so would make 2 solutions possible! So we must reduce E1 to {3,5} in this case as well.

  • 1
    $\begingroup$ Could you expand upon your explanation? Right now you're just claiming that this is true, but not proving it. Especially the remark about the parity could use some explanation. Should E1 have been reduced to {3,5} if {2,7} had been in A1, A7, and E7? $\endgroup$
    – SQB
    Commented May 21, 2014 at 19:43
  • $\begingroup$ @SQB I have taken your suggestion and updated my answer. $\endgroup$ Commented May 21, 2014 at 22:44
  • $\begingroup$ @SQB by parity I mean the mathematical sense of the word: whether the number of cells having only that pair is even or odd. $\endgroup$ Commented May 21, 2014 at 22:58

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