by **Paranoia** » Thu Sep 08, 2005 7:33 pm

Ok, IIRC, for bottom cell of R(16), I had 3 candidates: 4,7,8.

For the bottom right square, I had a total value (aiming for 45), of 34 (22+12) + the bottom cell of R(16) + the right two cells of $(16).

If BCR(16) (bottom cell), was 8, then the RC$(16) (right cells) must equal 3, as the total square must equal 45. Because of the 2 in E(10), this was not possible.

If BCR(16) was 7, then RC$(16) must equal 4, and due to the 3 in O(11), that is not possible.

BCR(16) must therefore equal 4, and RC$(16) must total 7. With 2,3,4,5 all elimated from cells higher in the collumn, this makes a pair of 1-6 6-1, eliminating those candidates from the rest of the bottom right square.

In F(12), I have a pair of candiates: 3-9, 9-3 (please tell me you had these too :p), so neither 3 or 9 can go in the right side of the bottom right square. so 2-8, 8-2 must go in the right two cells of %(22).

The left two cells of %(22) must now total 12. 4-8, 8-4 is not possible because of the BCR(16) being a 4, and 5-7, 7-5 is not possible because of the 7 in H(10). This leaves only 3-9, 9-3. A pair, and thus these candidates can be elimated from other cells in the bottom right square, leaving only 7-5 for X(12).

I *think* that's how I did it. I can't remember totally, so I tried to remember and make it up as I go along. There might be some breakdown in my logic somewhere, but I hope thats correct and you can follow it.