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Merry Christmas! Found a really nice way to crack this one which might have to go into the techniques forum. Love the way hard killers keep surprising with the variety of ways to attack them.

Assassin 204
Note: 1-9 cannot repeat on the diagonals

Cheers
Ed

Nice one, Ed. Found a nice logical way into it. Don't know if it's the same as yours, but very satisfying anyway.
Have a Happy Christmas.

_________________
Joe

Thanks Ed for a fun Assassin!

I don't know whether I found the same way as Ed did to crack this puzzle; my key breakthrough was in step 12 (with hindsight it was possible earlier than that) and I used another interesting move in step 19 to get a quick finish.


Here is my walkthrough for A204. I forgot until the final stages that this is a Killer-X so if you work through my steps using a software solver in editor mode please switch off the Killer-X option.

Prelims

a) R1C67 = {12}
b) R1C89 = {18/27/36/45}, no 9
c) R23C7 = {18/27/36/45}, no 9
d) R5C12 = {18/27/36/45}, no 9
e) R56C5 = {29/38/47/56}, no 1
f) R56C8 = {12}
g) R67C1 = {18/27/36/45}, no 9
h) R9C56 = {69/78}
i) 10(3) cage in N1 = {127/136/145/235}, no 8,9
j) 21(3) cage at R3C9 = {489/579/678}, no 1,2,3
k) 9(3) cage in N7 = {126/135/234}, no 7,8,9
l) 19(3) cage in N9 = {289/379/469/478/568}, no 1
m) 27(4) cage at R6C2 = {3789/4689/5679}, no 1,2
n) 34(5) cage at R1C4 = {46789}, no 1,2,3,5

Steps resulting from Prelims
1a. Naked pair {12} in R1C67, locked for R1, clean-up: no 7,8 in R1C89
1b. Naked pair {12} in R56C8, locked for C8 and N6
1c. 34(5) cage at R1C4 = {46789}, CPE no 4,6,7,8,9 in R3C56

2. 45 rule on N3 2 innies R1C7 + R3C9 = 11 = [29], R1C6 = 1, clean-up: no 7 in R23C7
2a. 21(3) cage at R3C9 = {489/579}, no 6
2b. 10(3) cage in N1 = {127/136/145/235}
2c. 7 of {127} must be in R1C3 -> no 7 in R2C23

3. 20(5) cage at R2C5 = {23456} (only remaining combination), no 7,8,9

4. 45 rule on R9 4 innies R9C1234 = 11 = {1235}, locked for R9
4a. 4 in R9 only in 19(3) cage, locked for N9
4b. 19(3) cage = {469/478}
4c. 9(3) cage in N7 = {126/135/234}
4d. 4,6 of {126/234} must be in R8C1 -> no 2 in R8C1

5. 7,8 in N2 only in R1C45 + R23C4, locked for 34(5) cage at R1C4, no 7,8 in R3C3
5a. 45 rule on N2 2(1+1) outies R3C3 + R4C6 = 10 = [46/64], CPE no 4,6 in R4C3

6. 45 rule on N1 3 innies R3C123 = 17 = {368/458/467} (cannot be {278} because R3C3 only contains 4,6), no 1,2
6a. 45 rule on N1 2 outies R4C12 = 1 innie R3C3 + 7
6b. Min R3C3 = 4 -> min R4C12 = 11, no 1 in R4C12

7. R3C7 = 1 (hidden single in R3), R2C7 = 8
7a. 2 in R3 only in R3C56, locked for N2

8. R78C9 = {12} (hidden pair in C9)
8a. R6C9 + R7C8 = 12 = [39/48/57/75], no 6,8 in R6C9, no 3,6 in R7C8

9. 45 rule on N9 2(1+1) outies R6C9 + R8C6 = 12 = [39/48/57/75], no 2,3,4,6 in R8C6

10. 45 rule on C12 1 outie R8C3 = 1 innie R2C2 + 6 -> R2C2 = {123}, R8C3 = {789}

11. 45 rule on N6 1 remaining outie R5C6 = 1 innie R6C9, no 2,6,8,9 in R5C6
11a. 9 in N6 only in 30(5) cage at R4C7 = {34689/35679}
11b. 8 of {34689} must be in R4C8 -> no 4 in R4C8

12. 30(7) cage at R6C6 = {1234569/1234578}, R9C56 = {69/78}, CPE no 7,8,9 in R8C6 (R8C6 “sees” all 6,7,8,9 in the 30(7) cage and R9C56, which cannot both contain 6,9 and cannot both contain 7,8)
12a. R8C6 = 5
12b. R9C3 = 5 (only remaining place for 5 in 30(7) cage), clean-up: no 4 in R6C1
12c. R7C8 = 5 (hidden single in N9), R6C9 = 7 (step 8a), R5C6 = 7 (step 11), clean-up: no 4 in R1C9, no 5 in R45C9, no 2 in R5C12, no 4 in R56C5, no 2 in R7C1, no 8 in R9C5

13. Naked pair {48} in R45C9, locked for C9 and N6 -> R9C9 = 6, clean-up: no 3 in R1C8, no 9 in R9C56

14. R9C56 = [78]
14a. R9C7 = 4 (hidden single in C7), R9C8 = 9
14b. Naked pair {37} in R78C7, locked for C7 and N9 -> R8C8 = 8, clean-up: no 2 in R2C2 (step 10)
14c. R4C8 = 3 (hidden single in N6)

15. 9 in C6 only in R67C6, locked for 30(7) cage at R6C6, no 9 in R7C5 + R8C45
15a. 1 in 30(7) cage at R6C6 only in R7C5 + R8C45 + R9C4, locked for N8

16. 5 in N2 only in R23C5, locked for C5, clean-up: no 6 in R56C5

17. 10(3) cage in N1 = {127/136}, no 4, 1 locked for N1
17a. R3C123 (step 6) = {368/458} (cannot be {467} which clashes with 10(3) cage), no 7, 8 locked for R3, N1 and 24(4) cage at R3C1, no 8 in R4C12

18. Naked pair {467} in R3C348, locked for R3

19. 10(3) cage in N1 (step 17) = {136} (only remaining combination, cannot be {127} = [712] which clashes with R2C2 + R8C3 = [17], step 10), locked for N1, 6 locked for C3 -> R3C3 = 4

20. R2C1 = 2 (hidden single in N1), clean-up: no 7 in R7C1
20a. Naked pair {58} in R3C12, locked for R3, N1 and 24(4) cage at R3C1, no 5 in R4C12

21. Naked pair {79} in R1C12, locked for R1
21a. Naked pair {68} in R1C45, locked for R1 and N2 -> R1C3 = 3, R2C23 = [16], R8C3 = 7 (step 10), R1C89 = [45], R2C89 = [73], R3C8 = 6, R2C56 = [54], R4C6 = 6, R23C4 = [97], R78C7 = [73], clean-up: no 8 in R5C1
21b. Killer pair 2,3 in R3C5 and R56C5, locked for C5

22. R8C2 = 9 (hidden single in R8), R1C12 = [97]

23. R3C12 = {58} = 13 -> R4C12 = 11 = [74], R45C9 = [84], clean-up: no 5 in R5C12

24. 1,2 in R4 only in 16(4) cage = {1258} (only remaining combination) -> R5C4 = 8, R4C4 = 5, R5C7 = 9, R5C5 = 1, R5C3 = 2, R1C45 = [68], clean-up: no 1 in R5C1, no 3 in R56C5
24a. Naked pair {36} in R5C12, locked for R5 and N4 -> R56C7 = [56], clean-up: no 3,6 in R7C1
24b. Naked pair {29} in R56C5, locked for C5 and N5 -> R3C56 = [32], R6C6 = 3, R7C6 = 9

25. R6C4 = 4, R567C3 = {189} = 18 -> R7C4 = 3 (cage sum)

26. 27(4) cage at R6C2 = {5679} (only remaining combination) -> R6C2 = 5, R7C2 = 6

It was only at this stage, when I started checking whether I’d reached naked singles, that I saw at the top of my worksheet the note that this puzzle is a Killer-X. Up to this stage I haven’t made any eliminations resulting from placements on the diagonals.

So close to being a unique puzzle without using the diagonals!

27. R5C5 = 2 (hidden single on D\), R7C3 = 8 (hidden single on D/)

and the rest is naked singles.
Happy Christmas to all forum members!

Joe

Very neat but let me check out that jump to 555
Outies and innies on N9 r8c6 = r7c8 = 5/7/8/9 and r9c3 already = 1/2/3/5
R9c6 sees all of 30(7) so r9c56 contains the missing 15 hence they form a 45 cage
LOL with N8 -> r6c6 = r7c4 & r9c3=r8c6 hence 555

Thanks. I gather that my post was a bit too elliptical.

I still don't do walk-throughs. But sometimes I want to substantiate my claim to have solved the puzzle; or to let other solvers compare their route; or to offer a hint and encouragement to anyone who hasn't solved it. That, and my natural indolence, is why I'm a bit elliptical.

cheers

_________________
Joe

Joe

I have similar views on walkthroughs, perhaps the phase skip-through instead?

Maurice

Very interesting! Andrew got my way with his step 12 but his is better because it locks four candidates whereas I just saw the two highest ones. It looks like the Blocking Cages technique (see several posts starting at the bottom of this post). Great walkthrough Andrew and love the rating!

I still can't work out how Joe Casey and HATMAN get their way to work but know it is correct. [edit: got it now - thanks to Andrew and Joe for helpful PMs] Checked a number of puzzles with this same pattern (generated by JSudoku) and they all have the same feature with that sum of 45. Well done! Another way to get the same result: Cheers
Ed