Let me know if you've seen this before but I can't find anything similar and it seems simple enough to be considered a basic strategy. Look at Friday's (6/6/08) Mepham Group sudoku after the basic strategies fail. I have a 5 and 6 in squares A5 and A6 and a 3, 5 and 6 in squares H5 and H6. If the 3 in row H were in H4, H5 and H6 would only have a 5 and 6. Only 5s and 6s in A5, A6, H5 and H6 would lead to an ambiguous situation that would not produce a unique solution. Since the sudoku does have a unique solution H5 or H6 must be a 3. This means that H4 cannot be a 3. Now the sudoku can be solved to completion. I have several more examples with graphics on a Word file I would be glad to send anyone interested. My email address is jmeyer2@knology.net. If you have an archive you might also want to check out The Mepham Group's 1/11/08, 2/29/08, 3/28/08 and 4/25/08 before looking at my examples. 1/11/08 doesn't seem to lead to a solution but is still interesting.

John Meyer