With 45 cells resolved and 36 still to go the puzzle looked like this (I had only made the obvious moves up to this point, plus I had used some simple coloring to eliminate one "6" [at r8c5] and one "8" [at r5c8] from the matrix).
Code: Select all
6 8 2 9 5 7 4 9 1
1 4 3 8 26 9 5 26 7
5 7 9 124 1246 146 68 268 3
3 9 78 6 78 5 2 1 4
278 12 6 147 3 14 9 57 58
4 5 178 179 1789 2 78 3 6
278 26 5 2479 24679 3 1 478 89
27 3 4 1279 1279 8 67 567 59
9 16 178 5 1467 146 3 4678 2
r7c9 = 8 ==> r5c9 = 5 ==> r5c8 = 7 ==> no "7"s in rest of col 8
r7c9 = 9 ==> r8c9 = 5 ==> {6, 7} pair in row 8 ==> r7c8, r9c8 <> 7
So either way there can't be a "7" at r7c8 or r9c8 -- this leaves a naked quad and a "hidden" pair {5, 7} in column 8, and the "6" in r8c7 is revealed as unique in row 8 -- the rest of the puzzle just falls apart!
I'm sure there are other ways to solve this puzzle -- I don't think it's as hard as a typical "Nightmare." But I sort of like this way. It's cute. dcb