SudokuSolver Forum

7 6 5 0 1 4 3 0 2
4 3 2 6 7 0 0 5 1
0 1 0 3 5 2 7 6 4
3 7 4 1 2 6 0 0 5
5 0 1 7 4 0 2 3 6
6 2 0 0 3 5 1 4 7
2 5 3 4 0 7 6 1 0
0 0 7 5 6 1 4 2 3
1 4 6 2 0 3 5 7 0

This is a Zero Killer-X on the 007 theme. The candidates are two 0s and 1-7; no repeats on the diagonals. It’s a “twin” puzzle; both grids have to be used.

1. 1(3) cage at R3C1 must be {001}, locked for R3 and N1

2. 5(3) cage at R3C3 and 5(3) cage at R4C3 overlap -> R3C3 = R6C3 = {00} (only way to repeat a number), locked for C3, CPE no 0s in R6C6 using D\
[Note. R6C3 = 0 eliminates one possible 0 from R7C123.]
2a. R45C3 = 5 = {14/23}, no 5,6,7

3. 16(5) cage at R6C1 and 10(3) cage at R7C1 overlap -> R6C13 = 6(2) split cage, R6C3 = 0 -> R6C1 = 6, locked for 16(5) cage, no 6 in R7C123

4. 9(4) cage at R3C1 = {01)8 = {01}{35} (only remaining combination), 3,5 locked for C1 and N4
4a. R45C3 = {14} (only remaining combination), locked for C3 and N4
[No need to use the 28 rule on N4.
There is one other place where I spotted the possible use of the 28 rule, but I didn't need to use it.]

5. Naked triple {027} in R456C2, locked for C2 -> R3C2 = 1, R3C1 = 0

6. 10(3) cage at R7C1 = {037/145/235} (cannot be {127} because no 1,2,7 in R7C2)
6a. 1,2 of {145/235} must be in R7C1 -> no 4 in R7C1, no 2 in R7C3

7. 22(4) cage at R3C7 = {4567} (only possible combination), CPE no 4,5,6,7 in R12C9
7a. 2,3 in R3 only in R3C456, locked for N2
7b. 2,3 in R3C456, locked for 28(9) cage at R3C4, no 2,3 in R4C46 + R56C4 + R7C45

8. 7(2) cage at R5C4 = [07/61/70]
8a. 2,3 in R3 only in R3C456 (step 7a) -> 9(3) cage at R3C4 = {27}0/{26}1/{36}0/{35}1/{23}4, no 4 in R3C45, R4C4 = {014}
[Note that {36}0 is not blocked by the 7(2) cage, because the 28(9) cage contains two 0s.]

9. 24(7) cage at R3C6 is missing two candidates totalling 4 (because all nine candidates total 28), missing candidates must be {04/13} -> must contain 2, locked for C6

10. 9(3) cage at R3C4 (step 8a) = {27}0/{26}1/{36}0/{35}1/{23}4 and 7(2) cage at R5C4 (step 8) = [07/61/70] are both are part of 28(9) cage at R3C4 -> combined cage = {00367/01267/01357/02347/12346}
10a. Rest of 28(9) cage R34C6 + R7C45 = {0057/0156/0246} (cannot be {1245} because 24(7) cage at R3C6 is missing one of 1,4, step 9, cannot be {0345} because 24(7) cage at R3C6 is missing one of 3,4, step 9), no 3
10b. 2 of {0246} must be in R3C6 -> no 4 in R3C6
10c. 3 in R3 only in 9(3) cage = {35}1/{23}4 (cannot be {36}0 because R34C6 + R7C45 = {1245} was eliminated in step 10a)
10d. Naked pair {14} in R4C34, locked for R4
10e. 4 in R3 only in R3C789, locked for N3

11. 24(7) cage at R3C6 is missing two candidates totalling 4 (because all nine candidates total 28), missing candidates must be {04/13} (step 9)
11a. R34C6 + R7C45 (step 10a) = {0057/0156/0246} -> R567C6 = {246/237/057}, no 1
11b. R34C6 + R7C45 cannot be {0057} because R456C4 = [461] (using 28(9) cage at R3C6) clashes with R567C6 = {246}
11c. R34C6 + R7C45 = {0156/0246}, no 7, R567C6 = {237/057}, no 4,6, 7 locked for C6 and 24(6) cage at R3C6, no 7 in R7C45
11d. 7 in 28(9) cage at R3C4 only in 7(2) cage at R5C4 = {07}, locked for R5, N5 and 28(9) cage at R3C4
11e. R567C6 = {237/057} -> R7C6 = 7
11f. 5 of {057} must be in R6C6 -> no 5 in R5C6
11g. 7 in R3 only in R3C789, locked for N3 and 22(4) cage at R3C7, no 7 in R4C9
11h. R34C6 + R7C45 = {0156/0246}, CPE no 6 in R89C6 (also eliminates one of the 0s from R89C6)

12. 10(3) cage at R7C1 (step 6) = {235} (only remaining combination, cannot be {145} which clashes with R34C6 + R7C45, which must have one of 1,4 in R7C45) -> R7C1 = 2, R7C23 = {35}, locked for R7 and N7
12a. Naked pair {47} in R12C1, locked for C1 and N1
12b. Naked pair {67} in R89C3, locked for C3 and N7

13. 13(3) cage at R5C9 = {067/157/247/346} (cannot be {256} which clashes with R4C9)
13a. 1 of {157} must be in R7C9 -> no 1 in R56C9
13b. Combined cage 13(3) cage + R34C9 = {067}[45]/{157}[46]/{247}{56}/{346}[75], 4,5,6,7 locked for C9

14. Consider combinations for R567C6 (step 11c) = {237/057}
R567C6 = {237} => 9(3) cage at R3C4 = {23}4 => 1 in N5 only in R56C5 => R7C4 = 1 (only remaining place for 1 in 28(9) cage at R3C4)
or R567C6 = {057} => 9(3) cage at R3C4 = {35}1 => R456C5 = {234} => R7C4 = 4 (only remaining place for 1 in 28(9) cage at R3C4)
-> R7C4 = {14}
14a. Naked pair {14} in R47C4, locked for C4 and 28(9) cage at R3C4, no 1,4 in R7C5
14b. Naked pair {14} in R47C4, CPE no 1,4 in R7C7 using D\

15. Naked pair {06} in R7C57, locked for R7
15a. Naked pair {06} in R7C57, CPE no 0,6 in R5C5 using D\
15b. Naked pair {06} in R7C57 -> the other 0 in R7 must be in R7C89
15c. R3C3 = 0 -> the other 0 on D\ must be in R7C7 + R8C8 + R9C9
15d. Grouped X-Wing for 0 in R7C89 and R7C7 + R8C8 + R9C9, no other 0 in N9

16. 13(3) cage at R5C9 = {067/157/247/346}
16a. R7C9 = {014} -> no 0,4 in R56C9
16b. 6 of {346} must be in R5C9 -> no 3 in R5C9

17. Consider combinations for R567C6 (step 11c) = {237/057}
R567C6 = {237}, locked for N5
or R567C6 = {057} => 9(3) cage at R3C4 = {35}1 => R456C5 = {234}, R34C6 = [26], R4C9 = 5, R4C1 = 3 => R4C5 = 2, R56C5 = {34}
-> no 3 in R4C5, no 2 in R56C5
17a. 2 on D/ only in R1C9 + R2C8, locked for N3

18. The two 0s in 28(9) cage at R3C4 are in R4C6 + R56C4 + R7C5, CPE no 0 in R4C5

19. 6 in N5 only in R4C56, locked for R4 -> R4C9 = 5, R45C1 = [35]
19a. Naked triple {467} in R3C789, locked for R3 and N3
19b. Naked triple {235} in R3C456, locked for N2
19c. Killer pair 2,5 in R3C6 and R56C6, locked for C6

20. Naked pair {06} in R12C4, 6 locked for C4 and N2
20a. Killer pair 0,0 in R12C4 and R56C4, locked for C4
[Step 20a is in case one hasn’t already eliminated both of these 0s using steps 11d and 20.]

21. 13(3) cage at R5C9 (step 16) = {067/247/346}, no 1

[I was a bit slow in spotting this, after which the puzzle is cracked.]
22. R34C6 + R7C45 (step 11c) = {0156/0246}, CPE no 6 in R4C5 -> R4C5 = 2
22a. R567C6 (step 11c) = {057} (only remaining combination) -> R5C6 = 0, R6C6 = 5, placed for D\, R3C6 = 2, R3C45 = {35} -> R4C4 = 1 (cage sum), placed for D\, R4C6 = 6 (hidden single in N5), placed for D, R7C45 = [40], R7C9 = 0, R7C8 = 1, R7C7 = 6, placed for D\, R2C2 = 3, placed for D\, R5C5 = 4, placed for both diagonals, R1C1 = 7, placed for D\, R3C7 = 7, placed for D/, R6C4 = 0, R8C2 = 0, no other 0s on D/

23. Naked pair {00} in R6C34, locked for R6

24. R9C1 = 1, placed for D/
24a. Naked pair {00} in R8C23, locked for R8 and N9 -> R8C8 = 2

and the rest is naked singles, without using the diagonals.

I'll rate my walkthrough at 1.5. I used a couple of forcing chains; well actually the same one twice with some other steps intervening.