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Thanks Ed for the challenging V2 by omitting the 10(3) cage from HATMAN's V1.5. Even though a lot harder it was still a fun puzzle. With this extra cage omission it felt more like one of HATMAN's Human Solvables than an Assassin.


Here is my walkthrough for A192X V2. Thanks Ed for your comments on my draft partial walkthrough. Ed had found an alternative, and in my opinion better, key breakthrough which I've given at the end.

Prelims

a) 11(3) cage at R1C1 = {128/137/146/236/245}, no 9
b) 22(3) cage at R1C2 = {589/679}
c) 19(3) cage at R1C5 = {289/379/469/478/568}, no 1
d) 22(3) cage at R1C8 = {589/679}
e) 10(3) cage at R1C9 = {127/136/145/235}, no 8,9
f) 22(3) cage at R2C1 = {589/679}
g) 20(3) cage at R2C9 = {389/479/569/578}, no 1,2
h) 10(3) cage at R6C3 = {127/136/145/235}, no 8,9
i) 22(3) cage at R6C7 = {589/679}
j) 11(3) cage at R7C4 = {128/137/146/236/245}, no 9
k) 21(3) cage at R7C6 = {489/579/678}, no 1,2,3

Steps resulting from Prelims
1a. 22(3) cage at R1C2 = {589/679}, CPE no 9 in R3C12
1b. 22(3) cage at R1C8 = {589/679}, CPE no 9 in R3C89
1c. 22(3) cage at R2C1 = {589/679}, CPE no 9 in R12C3 + R456C1
1d. 22(3) cage at R6C7 = {589/679}, CPE no 9 in R789C7
1e. 9 in R3 only in R3C456, locked for N2

2. 45 rule on D/ 3 innies R4C6 + R5C5 + R6C4 = 22 = {589/679}, 9 locked for N5 and D/

3. 45 rule on D\ 3 innies R4C4 + R5C5 + R6C6 = 21 = {489/678} (cannot be {579} which clashes with R4C6 + R5C5 + R6C4), no 1,2,3,5, 8 locked for N5 and D\
3a. 8 of {678} must be in R5C5 (permutations with 6 or 7 in R5C5 clash with R4C6 + R5C5 + R6C4) -> no 6,7 in R5C5

4. 19(3) cage at R1C5 = {379/469/478/568} (cannot be {289} which clashes with R5C5), no 2
4a. 9 of {379} must be in R3C5 -> no 3 in R3C5
4b. Killer pair 8,9 in 19(3) cage and R5C5, locked for C5

5. 45 rule on R5 3 innies R5C456 = 13 = {139/148/238} (only combinations containing 8 or 9 for R5C5), no 5,6,7

6. Combined cage R4C46 + R5C5 + R6C46 = {46789/56789} (only ways to combine R4C4 + R5C5 + R6C6 and R4C6 + R5C5 + R6C4), 6,7 locked for N5
[Alternatively combined the overlapping 21(3) and 22(3) hidden cages = 43 – one of 8,9 = 34(5) or 35(5) = {46789/56789}. This is an alternative way to make R5C5 = {89} because maximum for 5-cell cage is 35.]

7. 13(3) cage at R7C7 = {139/157/256} (cannot be {247/346} which clash with R4C4 + R5C5 + R6C6), no 4

8. 13(3) cage at R7C3 = {148/238/247/346} (cannot be {157/256} which clash with R4C6 + R5C5 + R6C4), no 5

9. 45 rule on N1 2(1+1) outies R3C4 + R4C3 = 2 innies R1C3 + R3C1 + 10
9a. Max R3C4 + R4C3 = 18 -> max R1C3 + R3C1 = 8, no 8,
9b. One of the 22(3) cages in N1 must be {589} and the other {679} (cannot have the same combination because of ALS block), CPE no 5,6,7 in R3C3
9c. 8 in N1 only in one of the 22(3) cages -> no 8 in R3C4 + R4C3
9d. 9 in N1 only in one of the 22(3) cages -> the other 9 for the 22(3) cages must be in R3C4 or R4C3
9e. Max R3C4 + R4C3 = {79} = 16 -> max R1C3 + R3C1 = 6, no 6,7
9f. For R3C4 + R4C3 = {79} then one of the 22(3) cage in N1 must contain 5 within N1 -> no 5 in R1C3 + R3C1
[Note that this doesn’t eliminate 5 from 11(3) cage at R1C1 because R3C4 + R4C3 can still be {59}.]

10. 45 rule on N9 2(1+1) outies R6C7 + R7C6 = 2 innies R7C9 + R9C7 + 11
10a. Min R7C9 + R9C7 = 4 (cannot be {12} which clashes with 13(3) cage at R7C7) -> min R6C7 + R7C6 = 15, no 5 in R6C7, no 4,5 in R7C6
10b. Max R6C7 + R7C6 = 18 -> max R7C9 + R9C7 = 7, no 7,8,9
10c. Max R6C7 + R7C6 = 18 = [99] => 13(3) cage at R7C7 = {139} (only other place for 9 in N9) -> max R7C9 + R9C7 = 7 cannot be {16} -> no 6 in R7C9 + R9C7

11. 45 rule on N7 2 innies R7C1 + R9C3 = 2(1+1) outies R6C3 + R7C4 + 11
11a. Min R6C3 + R7C4 = 2 -> min R7C1 + R9C3 = 13, no 1,2,3
11b. Max R7C1 + R9C3 = 17 -> max R6C3 + R7C4 = 6, no 6,7,8

12. 9 in N7 only in R7C1 + R9C3
12a. R7C1 + R9C3 cannot be {49}, here’s how
R7C1 + R9C3 = {49} = 13 => R6C3 + R7C4 (step 11) = 2 = [11] => 13(3) cage in N7 (only other place for 1 in N7) = {148} clashes with R7C1 + R9C3
-> no 4 in R7C1 + R9C3

13. R4C6 + R5C5 + R6C4 (step 2) = {589/679}
13a. Either R4C6 + R5C5 + R6C4 = {589} with 8 in R5C5 or 5 is in R46C5 (this might be described as a hidden killer single 5 for N5)
13b. 19(3) cage at R1C5 (step 4) = {379/469/478} (cannot be {568} which clashes with R456C5), no 5
13c. 9 of {469} must be in R3C5 -> no 6 in R3C5

14. 13(3) cage at R7C7 (step 7) = {139/157/256}
14a. 21(3) cage at R7C6 = {489/579/678} cannot be {579}, here’s how
For 13(3) cage at R7C7 = {139} => 9 locked for N9 => 22(3) cage at R6C7 = 9{58/67} => 21(3) cage at R7C6 = 9{57} clashes with 22(3) cage
For 13(3) cage at R7C7 = {157/256} => 5 locked for N9
14b. -> 21(3) cage at R7C6 = {489/678}, no 5, CPE no 8 in R7C8

15. 13(3) cage at R7C7 (step 7) = {139/157/256}
15a. 22(3) cage at R6C7 = {589/679}
15b. {589} can only be [958] (cannot be 8{59} which clashes with the 13(3) cage)
-> no 8 in R6C7, no 5 in R8C9
[Thanks Ed for pointing out this way, which is simpler than my original steps 15a and 15b.]

16. Caged X-Wing for 9 in 22(3) cage at R2C1 and hidden cage R7C1 + R9C3, no other 9 in C3

17. R5C456 (step 5) = {139/148/238}
17a. 16(3) cage at R5C1 = {169/259/268/367/457} (cannot be {178/349/358} which clash with R5C456)
17b. 9 of {169} must be in R5C2 -> no 1 in R5C2

18. 9 in R3 only in R3C456
18a. For R3C4 = 9 => R2C1 = 9 => R1C8 = 9
For R3C5 = 9 => three 9s in 22(3) cage at R1C2, 22(3) cage at R1C8 and R3C5, locked for R123
For R3C6 = 9 => R1C2 = 9
18b. -> 9 locked in R1C28, locked for R1

19. 9 in R3 only in R3C456
19a. For 9 in R3C46 => R5C5 = 9 (hidden single in C5) => no 9 in R4C6
For R3C5 = 9 => R1C2 = 9 => R4C3 = 9 => no 9 in R4C6
19b. -> no 9 in R4C6
19c. R4C6 + R5C5 + R6C4 (step 2) = {589/679}
19d. 5 of {589} must be in R4C6 -> no 5 in R6C4

[Ed found another forcing chain from a position similar to this, see note after my walkthrough.]

20. R3C5 cannot be 9, here’s how
20a. R3C5 = 9 => R5C5 = 8, R1C2 = 9 => R2C7 = 9 => 9 in 22(3) cage at R6C7 must be in R7C8 + R8C9, locked for N9 => no 9 on D\
20b. -> 19(3) cage at R1C5 (step 13b) = {478} (only remaining combination), locked for C5 and N2
[I ought to have spotted this step earlier but I’d been focussing on trying to find forcing chains based on the position of 9 in R3C456, rather than looking for a contradiction move.]

21. R5C5 = 9, placed for D\
21a. R4C6 + R5C5 + R6C4 (step 2) = {679} (only remaining combination), 6,7 locked for N5 and D/
21b. Naked pair {48} in R4C4 + R6C6, locked for N5 and D\

22. R5C456 (step 5) = {139} (only remaining combination), locked for R5 and N5
22a. Naked pair {25} in R46C5, locked for C5
22b. Naked triple {136} in R789C5, locked for N8

This puzzle has now been reduced to part way through the V1.5, which had a 10(3) cage at R7C5.

23. 13(3) cage at R7C7 (step 7) = {157/256}, no 3, 5 locked for N9 and D\
23a. 3 on D\ only in 11(3) cage at R1C1, locked for N1

24. 22(3) cage at R6C7 = {679} (only remaining combination), no 8, CPE no 6,7 in R78C7
24a. 9 in 22(3) cage must be in R7C8 + R8C9 (R7C8 + R8C9 cannot be {67} which clashes with 13(3) cage at R7C7), 9 locked for N9 and 22(3) cage, no 9 in R6C7
24b. Killer pair 6,7 in 13(3) cage and 22(3) cage, locked for N9

25. Naked pair {48} in R8C7 and R9C8, locked for N9, R7C6 = 9 (step 14b)
25a. R8C9 = 9 (hidden single in N9)
25b. Naked pair {67} in R6C7 + R7C8, CPE no 6,7 in R6C8

26. R3C4 = 9 (hidden single in N2)
26a. R2C1 = 9 (hidden single in N1)
26b. R1C8 = 9 (hidden single in N3)
26c. R4C7 = 9 (hidden single in C7)
26d. R6C2 = 9 (hidden single in N4)
26e. R9C3 = 9 (hidden single in N7)

27. 22(3) cage at R2C1 = {589/679}
27a. 8 of {589} must be in R3C2 -> no 5 in R3C2
27b. 5 in N1 only in 22(3) cage at R1C2 = {589}, 8 locked for N1
27c. 22(3) cage at R2C1 = {679} (only remaining combination), no 5, CPE no 6,7 in R45C2
27d. Naked pair {67} in R4C36, locked for R4

28. 5 on D/ only in 10(3) cage at R1C9, locked for N3
28a. 10(3) cage = {145/235}

29. 22(3) cage at R1C8 = {589/679}
29a. 7,8 only in R2C7 -> R2C7 = {78}

30. 20(3) cage at R2C9 = {389/479}, no 6
30a. Killer pair 7,8 in R2C7 and 20(3) cage, locked for N3
30b. Killer pair 3,4 in 10(3) cage and 20(3) cage, locked for N3

[I remembered that Mike had used 45s on N3 to getter a quicker finish for A192X V1.5 than I did, so ...]
31. 45 rule on N3 2 innies R1C7 + R3C9 = 1 remaining outie R3C6 + 2, CPE no 2 in R1C7
31a. 45 rule on N3 3 innies R12C7 + R3C9 = 15 = [186/681] (cannot be [672] which clashes with R6C7) -> R2C7 = 8, R3C6 = 5 (step 29), R1C7 + R3C9 = {16}, locked for N3, R2C3 = 5, R1C2 = 8, R8C7 = 4, R9C8 = 8
31b. R3C5 = 8 (hidden single in N2)

32. 10(3) cage (step 28a) = {235} (only remaining combination) -> R1C9 = 5, R2C8 + R3C7 = {23}, locked for N3 and D/ -> R8C2 = 1, R9C1 = 4, R7C3 = 8

33. Naked triple {123} in R3C137, locked for R3 -> R3C9 = 6, R1C7 = 1, R3C2 = 7, R4C3 = 6, R3C8 = 4, R2C9 = 7, R12C5 = [74], R4C6 = 7, R6C4 = 6, R6C7 = 7, R7C8 = 6, R9C6 = 2, R9C9 = 1, placed for D\

34. 7 in R5 only in 16(3) cage at R5C1 = {457} (only remaining combination), locked for R5 and N4 -> R5C8 = 2

and the rest is naked singles without using the diagonals.


From the position after step 19.

I looked at it and my way of expressing it was
9 in N5 only in R5C5 + R6C4
R5C5 = 9, placed for D\ => no 9 in R8C8 + R9C9
R6C4 = 9 => no 9 in R6C7 => 9 in 22(3) cage at R6C7 must be in N9, locked for N9 => no 9 in R8C8 + R9C9
R5C5 = 9 (hidden single on D\)

I prefer this to my contradiction move. I’d been too focussed on forcing chains based on the 9s in R3C456 that I never thought of looking for a forcing chain based on the 9s in N5.