by **RodHagglund** » Wed Dec 28, 2005 5:04 am

Actually there's a trick to solve this one in fewer steps. After you use the naked pair in r3c5 and r3c6 to remove a 5 from r1c3, you can use the uniqueness condition to exclude the possibility of a 1 in r5c3.

A 1 placed there would remove the 1 from r4c1, and also place a 3 in r1c3, which in turn would place a 5 in r6c3, removing the 5 from r6c5. This would leave the numbers 2 and 3 as possible candidates for r4c1, r6c1, r4c5 and r6c5. Since the 2's and 3's can be placed in those cells in two different ways without affecting the reasoning for any other cell, there would be at least two solutions to the sudoku; the uniqueness assertion is that if we trust this is a valid sudoku and has only one solution, we can conclude that there can't be a 1 in r4c1.

That leaves only r1c3 as a place for a 1 in column 3, and from there the puzzle solves.

Rod Hagglund