Solving “Arrow Head”
Step 1
45 on N3 -> R13C6 = 13
45 on N6 -> R6C789 = 16
45 on N7 -> R6C13 = 5
45 on N8 -> R789C6 = 12
Step 2
Overlap and Innies of R6+C6. Here we have both an overlaping cell (R6C6) and two innies (R2C6+R6C2) using the split cages we computed at step 1.
-> R2C6+R6C2+R6C6 = 2*45 - (13-16-5-12-19) = 25 = [889|997]
-> R2C6=R6C2={89}, R6C6={79}
must have 9 in these cells -> nowhere else in R6, C6
Step 3
45 on R6789 -> R45C6 = R6C2 - 5
Since R6C2 = {89} -> R45C6 = (3|4)/2 = {12|13} = {1(2|3)}
must have 1 in R45C6 -> nowhere else in C6, N5
Step 4
R13C6 = {58|67} = {(7|8)…}
Since R26C6 = {79|89} = {(7|8)…}
This forms a killer naked pair on {78} -> nowhere else in C6
-> R789C6 = {246|345} = {4(26|35)}
must have 4 in R789C6 -> nowhere else in N8
must have 8 in R123C6 -> nowhere else in N2
Step 5
45 on C6789 -> R6C45 = R2C6 - 1
Since R2C6 = {89} -> R6C45 = (7|8)/2 = {25|34|26|35} = {(2|3)…}
This forms a killer naked pair with R45C6 on {23}
-> nowhere else in N5
Step 6
Since R6C45 = {(2|3)…}
-> R6C13 = {14}
-> R6C45 = {25|26|35}
-> Cage 19 in N5 = {12367}
-> R6C6 = 7, R2C6=R6C2=9
-> R45C6 = {13}, R6C45 = {26}
Step 7
R13C6 = 13/2 = {58}
-> R789C6 = {246}
R6C789 = {358}
Step 8
Cage 21/4 in N4 = {2379}
R4C23+R5C3 = {568}
Step 9
45 on N2 -> R2C4+R3C45 = 8/3 = {134}
-> Cage 24/4 in N2 = {2679}
Step 10
45 on N1 -> R2C4+R4C2 = 12
-> R2C4 = 4, R4C2 = 8
-> R3C45 = {13}, R45C3 = {56}
Step 11
Cage 15/3 at R3C4 with R3C4 = {13} & R4C3 = {56}
-> Cage 15/3 = {159}
-> R3C45 = [13], R4C4 = 9, R45C3 = [56]
-> R4C5 = 4, R5C45 = {58}
Step 12
Cage 16/3 in N6 = {169}
-> R5C9 = 9, R4C89 = {16}
-> R45C6 = [31]
Cage 13/3 in N6 = {247}
Step 12
R16C4 forms a hidden pair on {26} within C4 -> R16C4 = {26}
-> 7 of N2 in R12C5 -> nowhere else in C5
Cage 18/3 in N8 = {189|378}
Since R34C4 = [19] -> ¬{189}
-> R9C5 = 8, R89C4 = {37}
-> R7C4 = 5, R78C5 = {19}
-> R5C45 = [85]
Step 13
Cage 17/3 at R3C678 = {278|458} = {8…}
Cage 14/4 in N3 can’t have 9
-> 9 of N3 in R1C78
-> Cage 27/4 at R1C6 = {3789|4689|5679}
-> 1 of N3 in Cage 14/4
-> 1 of N9 in R789C7
Step 14
In N1, R2C3+R3C23 = 27-4-8 = 15
-> R3C2 = {56}
45 on C12 -> R19C3 = R3C2 + 7
-> R19C3 = (12|13)/2 = {39|48|49}
-> R9C3 = {49}, R1C3 = {348}
Step 15
Cage 17/3 at R6C8 with R6C89 = {358}
Since R5C9 = 9 -> ¬{359}
-> {368|458} = {8(36|45)} -> R7C9 = {46}
-> R6C7 = {35}
45 on N8 -> R9C7 = R7C6 - 1 -> R9C7 = {135}
45 on N9 -> R7C79+R9C7 = 13/3
Min R7C9+R9C7 = 1+4 = 5 -> Max R7C7 = 13-5 = 8 -> can’t have 9
Step 16
45 on C9 -> R2469C8 = 16/4
The only combination of 16/4 with a 9 is {1249}
Since R246C8 can’t have 4 -> R9C8 can’t have 9
-> 9 of N9 must go in Cage 18/3
-> 9 of R9 in Cage 19/4 of N7 -> nowhere else in N7
-> Cage 12/3 at R6C3 = {138|147}
-> 2 of C3 in R23C3 -> nowhere else in N1
-> R2C3+R3C23 = {267} -> R3C2 = 6, R23C3 = {27}
-> R9C3 = 9 (hidden single in C3)
-> R3C1 = 9 (hidden single in N1)
-> Cage 12/3 at R6C3 = {138} -> R6C13 = [41], R78C3 = {38}, R6C1 = 4
Cage 19/4 at R6C1 = {2467}
Cage 19/4 at R8C2 = {1459}
Step 17
45 on R12 -> R3C9 = R2C3 -> R3C9 = {27}
-> Cage 17/3 in R3 = {458}
Cage 14/4 in N3 = {1238|1256} -> R3C9 = 2, R23C3 = [27]
4 of C9 in R789C9 -> nowhere else in N9
…
Here is the unique solution
