by Jean-Christophe » Sun Aug 13, 2006 7:39 pm
Here is my walkthrough in tiny text you may copy-paste to a text editor like note pad
Step 1
R7C8 = 4 -> R8C8 <> {35}, R8C7 = 2 -> R8C8 <> {13}
7 of N9 locked in R89C7 -> R89C7 <> {68}
-> R8C8 = 7
Step 2
R3C8 = 9 -> R3C79 <> 8, R2C7 = 7 -> R1C7 <> 8
-> 8 of N3 locked in R1C89, 8 of N2 locked in R3C456
-> R3C5 <> 7
R5C5 = 6 -> R46C5 <> 7
-> R7C5 = 7 (hidden single in C5)
Step 3
R7C5 = 7 -> R8C5 <> 8, R9C5 = 4 -> R8C5 <> 3
-> R8C5 = 1
Step 4
R2C3 = 8 -> R2C2 <> 9, R3C2 = 5 -> R24C2, R3C13 <> {46}
-> R2C2 = {23}, R4C2 = {23789}
7 of N7 locked in R9C123-> R9C2 <> {68} = {27}
R8C3 = 9 -> R8C2 <> 8
-> 8 of N7 locked in R789C1, 8 of N4 locked in R456C3
-> R5C2 <> 9 = {238}
Step 5
Either R5C2 = 8 -> R4C2 <> {789} = {23}
Or else R5C2 = {23}
-> R45C2 must hold one of {23}
This forms a complex naked pair on {23} with R2C2
-> not elsewhere in C2
-> R9C2 = 7, R8C2 = 4
Step 6
R2C3 = 8 -> R13C3 <> 7
7 of N1 locked in R123C1 -> R2C1 <> 6
R2C7 = 7 -> R2C68, R13C7 <> 6
5 of N3 locked in R2C89 -> R2C89 <> {46}
-> R2C4 = 6 (hidden single in R2)
Step 7
R3C9 = 6 (hidden single in R3)
-> R2C8 = 5
6 of N9 locked in R79C7, 6 of N6 locked in R46C8
7 of N2 locked in R13C6, 7 of N5 locked in R46C4 -> R5C4 <> 8
Step 8
R7C2 = 1 -> R7C13 <> 2
-> 2 of N7 locked in R9C13, 2 of N8 locked in R7C46
R9C5 = 4 -> R9C46 <> {35}, R9C4 = {89}, R9C6 = {689}
6 of N8 locked in R789C6 -> R8C6 <> 5
R9C46, R8C6 = naked triplet on {689}
Step 9
R7C4 = {235}, R8C4 = {35}
Either R8C4 = 5 or else R8C4 = 3 -> R7C4 <> {23} = 5
-> 5 of N8 locked in R78C4, 5 of N5 locked in R46C6 -> R5C6 <> 4
5 of R7 locked in R7C34 -> R7C3 <> 6 = {35}
R7C346 = naked triplet on {235}
R78C1 = {68} = nacked pair
Step 10
6 of R9 locked in R9C67 -> R9C7 <> 5 = {68}
R9C467 = naked triplet on {689} -> R9C9 = {135}
5 of N9 locked in R89C9, 5 of N6 locked in R46C6 -> R5C6 <> 4
5 of R6 locked in R6C67 -> R6C67 <> 4
5 of R4 locked in R4C67 -> R4C67 <> 4
Step 11
4 of N6 locked in R46C9, 4 of N3 locked in R13C7
R2C9 = {123} -> R1C9 <> 2
2 of N3 locked in R1C8, R2C9 -> R1C9 <> {13}
-> R1C9 = 8, R6C8 = 8, R7C9 = 9, R4C8 = 6, R6C7 = 5
R79C7 = [86], R78C1 = [68], R8C6 = 6
R4C6 = 5, R3C7 = 4, R2C6 = 4, R1C1 = 4
R1C6 = 7, R3C1 = 7
R1C7 = {13} -> R1C8 <> 2 -> R2C9 = 2
...
Unique solution :
496257318
138694752
752831496
927485163
584163927
361729584
613572849
849316275
275948631