“This is a simple killer however the solution is semi-symmetrical. There are four unknown number pairs and an odd number (which of course goes in r5c5).”
Prelims
a) R12C5 = {19/28/37/46}, no 5
b) R3C34 = {18/27/36/45}, no 9
c) R3C67 = {16/25/34}, no 7,8,9
d) R4C67 = {19/28/37/46}, no 5
e) R5C34 = {15/24}
f) R5C67 = {14/23}
g) R67C5 = {16/25/34}, no 7,8,9
h) R7C34 = {15/24}
i) R7C67 = {19/28/37/46}, no 5
j) R89C5 = {59/68}
k) 19(3) cage at R4C2 = {289/379/469/478/568}, no 1
1a. R5C34 = {15} (cannot be {24} which clashes with R5C67), locked for R5, clean-up: no 4 in R5C67
1b. R5C67 = {23}, locked for R5
1c. R5C34 and R5C67 are corresponding pairs of cells -> 1,5 and 2,3 must be paired
1d. Hidden killer triple 4,5,6 in R12C9, R67C5 and R89C5 for C5, R67C5 and R89C5 each contain one of 4,5,6 -> R12C9 cannot contain more than one of 4,5,6 -> R12C9 = {19/28/37} (cannot be {46}), no 4,6
2. R12C5 corresponds with R89C5
2a. R12C5 (step 1d) = {28/37} (cannot be {19} because 1 is paired with one of 2,3 so cannot correspond with {59/68})
2b. R12C5 = {28} can correspond with R89C5 = {59} if 8 is paired with 9 (cannot correspond with {68} because 3 must be paired with one of 1,5),
R12C5 = {37} can correspond with R89C5 = {59} if 7 is paired with 9 (cannot correspond with {68} because 3 must be paired with one of 1,5)
-> R89C5 = {59}, locked for C5 and N8, clean-up: no 2 in R67C5, no 1 in R7C3, no 1 in R7C7
2c. 9 must be paired with one of 7,8
3. R7C34 correspond with R3C67
3a. R7C34 = {24} can only correspond with R3C67 = {16},
R7C34 = [51] can only correspond with R3C67 = [25]
-> R3C67 = {16}/[25], no 3,4 , no 5 in R3C6, no 2 in R3C7
3b. 1 is paired with 2 -> 3 is paired with 5
3c. R12C5 corresponds with R89C5, 3 paired with 5 -> R12C5 = {37}, locked for C5 and N2, clean-up: no 2,6 in R3C3, no 4 in R67C5
3d. 7 is paired with 9
3e. Naked pair {16} in R67C5, locked for C5
3f. From step 3a and using R5C34 = {15} and R5C67 = {23}, no 1 in R3C34, no 2 in R7C67, clean-up: no 8 in R3C34, no 8 in R7C67
4. R3C34 correspond with R7C67
4a. R3C34 = [72] (cannot be [36]/{45} because no 5 in R7C67 to pair with 3 and {37} would need a pair with 9), clean-up: no 5 in R3C7
4b. Naked pair {16} in R3C67, locked for R3 -> R7C34 = [24] (step 3a)
4c. R3C4 = 2 -> R7C6 = 1, R7C7 = 9, R3C67 = [61], R67C5 = [16], R5C34 = [15]
4d. R3C6 corresponds with R7C4 -> 4 paired with 6
4e. R7C5 corresponds with R3C5 -> R3C5 = 4, R45C5 = [28], R5C67 = [32]
4f. R4C67 = [46/73]
[Note that Unique Rectangles, which I refuse to use, doesn’t apply here because R5C34 and R7C34 are each split between different nonets. Therefore I had to work quite hard to determine R7C34.]
5. 45 rule on N7 3 innies R7C2 + R8C23 = 22 = {589/679}, 9 locked for R8 and N7 -> R89C5 = [59] -> R7C2 = {57}, R8C23 = {689}
5a. R89C5 corresponds with R12C5, 5 paired with 3, 9 paired with 7 -> R12C5 = [73]
6. 45 rule on N1 3 innies R2C23 + R3C2 = 9 = {135/234} (cannot be {126} because 1,2 only in R2C2) -> R2C2 = {12}, R3C2 = 3, R2C3 = {45}
6a. R2C2 corresponds with R8C8, 1 paired with 2 -> R8C8 = {12}
6b. R2C3 corresponds with R8C7, 4 paired with 6, 5 paired with 3 4 also in R8C7 -> R8C7 = {346}
6c. R3C2 corresponds with R7C8, 3 paired with 5, 3 also in R7C8 -> R7C8 = {35}
6d. 8 can only be in R5C5 and in pairs of corresponding cells, no 8 in R7C2 -> no 8 in R3C8
6e. 8 in R3 only in R3C19 which correspond with R7C19 -> 8 in R37C1, locked for C1, 8 in R37C9, locked for C9
[I realised latter that this is X-Wing for 8 in R3C19 and R7C19, without using semi-symmetry.]
7. R8C23 correspond with R2C78, 6 is paired with 4, 9 is paired with 7, 8 can be in either pair -> no 2,5 in R2C78
7a. 45 rule on N3 3 innies R2C78 + R3C8 = 20 = {479/569/578}
7b. R2C78 + R3C8 corresponds with R7C2 + R8C23 (step 5) = 5{89}/7{69}
7c. R2C78 + R3C8 = {479/578} (cannot be {569} because {69}5 cannot correspond with 5{89}), no 6, 7 locked for N3
7d. R2C78 + R3C8 = {47}9/{78}5
8. 45 rule on N9 3 innies R7C8 + R8C78 = 10 = {136/145/235}
8a. Consider combinations for R2C78 + R3C8 (step 7d) = {47}9/{78}5 and using corresponding pairs
R2C78 + R3C8 = {47}9, 4 locked for R2 => R2C3 = 5 => R8C7 = 3, R7C8 = 5, R7C2 = 7 => R8C23 = {69}
or R2C78 + R3C8 = {78}5 => R7C2 = 5, R7C8 = 3, R8C78 = [61] => R2C3 = 4
-> R7C8 + R8C78 = {136/235}, no 4, 3 locked for N9
4 in R2C3 + R2C78, locked for R2
5 in R37C8, locked for C8
5 in R7C28, locked for R7
6 in R8C23 + R8C7, locked for R8
8b. R3C1 corresponds with R7C9, no 3,5 in R7C9 -> no 5 in R3C1
8c. Consider combinations for R2C78 + R3C8 = {47}9/{78}5
R2C78 + R3C8 = {47}9 => R3C19 = [85]
or R2C78 + R3C8 = {78}5 => R3C9 = 9
-> R3C89 = {59}, locked for R3 and N3
8d. R3C1 = 8, R3C1 corresponds with R7C9 -> R7C9 = 8
8e. 7 in R7 only in R7C12, locked for N7
9. 16(3) cage at R4C8 = {169/178} (cannot be {349} which clashes with R37C8, ALS block, cannot be {367} which clashes with R4C7) -> R4C8 = 1, R56C8 = {69}/[78]
9a. R8C8 = 2 -> R7C8 + R8C78 (step 8a) = {235}, R7C8 = 5, R8C7 = 3, R3C89 = [95], R7C12 = [37], R4C7 = 6 -> R4C6 = 4
9b. 16(3) cage corresponds with 19(3) cage at R4C2, but the other way up, R456C8 = [178] -> R456C2 = [892]
and the rest is naked singles, without using corresponding pairs.
