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After a couple of monsters the last few Assassins, time for a....nother one!! I used a trick move to crack it. Turns out it is available almost from the start. However, the SSscoreV3.3.1 is 1.12 so it will have a more conventional solution, just couldn't find it. The ALT4 score (in next release of SS) is quite a bit higher so will be an interesting test.

Assassin 200
note:Killer X so 1-9 cannot repeat on the diagonals
It has a remote 26(4) in n5. Thanks to Børge for the pics. Much clearer than mine.
Cheers
Ed

Thank you for number 200.A very appropriate author.I dont know about a "trick" but I did do it very quickly and effectively used "insertion solving"(to use Andrews expression).I am happy to say how I solved I it but think I should wait a couple of days first.

This has been the first assassin I've done in ages, but I enjoyed it. I did do something different, but i'm not sure if it is the trick. I'll hide it to not spoil.



I don't really feel up to working out a whole walk-through yet, but figured I'd share something again. Not sure if this is what you were talking about, but I found it an interesting move. Don't really remember ever doing anything like this before, but I could just be suffering from memory issues as I don't exactly remember everything accurately anymore. I looked through some of my old walk-throughs recently and am still a bit shocked I thought of some of those moves back then. Half of that I wouldn't be able to come up with these days. Let's just say I'm a bit rusty. I might try a few more in the next weeks, will see.

greetings

Para

I suppose my way was pretty similar to paras but a lot less elegant.Anyway we both attacked the same corner
My brute force (as opposed to paras elegance) was as follows
Moderator edit: formatting

Hello, still here, still can't leave the killers alone. Thanks again to the setters; I don't mind if the puzzles are less frequent, it leaves time to do something else once in a while. I solved 200 in much the same way (But today's KSO weekly 247 gave me a lot more trouble than usual. Got it in the end, but all brute force and attrition, having to go deep to eliminate alternatives).

cheers

_________________
Joe

Thanks Ed for a challenging Assassin! Unlike the other solvers who have posted so far, I only got part way toward the trick (see comment at the end of my walkthrough) so I used a more conventional solving path.

Welcome back Para and thanks for a very clear explanation of how the trick works!


Here is my walkthrough for A200. Thanks Ed for your feedback comments. I've edited my walkthrough and also removed "at least" from my rating comment because step 32a has been simplified further.

Prelims

a) R12C3 = {19/28/37/46}, no 5
b) R12C7 = {59/68}
c) R23C5 = {18/27/36/45}, no 9
d) R78C5 = {59/68}
e) R89C3 = {13}
f) R89C7 = {19/28/37/46}, no 5
g) 9(3) cage in N3 = {126/135/234}, no 7,8,9
h) 22(3) cage in N6 = {589/679}
i) 10(3) cage at R5C2 = {127/136/145/235}, no 8,9
j) 9(3) cage at R6C9 = {126/135/234}, no 7,8,9
k) 22(3) cage in N9 = {589/679}
l) 26(4) disjoint cage in N5 = {2789/3689/4589/4679/5678}, no 1
m) 13(4) cage at R6C7 = {1237/1246/1345}, no 8,9

Steps resulting from Prelims
1a. Naked pair {13} in R89C3, locked for C3 and N7, clean-up: no 7,9 in R12C3
1b. 22(3) cage in N6 = {589/679}, 9 locked for N6
1c. 22(3) cage in N9 = {589/679}, 9 locked for N9, clean-up: no 1 in R89C7
1d. 9 in C7 only in R123C7, locked for N3

2. 45 rule on N2 2 innies R3C46 = 5 = {14/23}

3. 45 rule on N5 2 innies R5C46 = 7 = {16/25/34}, no 7,8,9

4. 45 rule on N8 2 innies R7C46 = 11 = [47/74/83/92] (cannot be {56} which clashes with R78C5), no 1,5,6, no 2,3 in R7C4
4a. 1 in R7 only in R7C789, locked for N9
4b. 13(4) cage at R6C7 = {1237/1246/1345}, 1 locked for C7

5. 45 rule on R12 3 innies R2C258 = 9 = {126/135/234}, no 7,8,9, clean-up: no 1,2 in R3C5
5a. 7 in N3 only in R3C789, locked for R3, clean-up: no 2 in R2C5

6. 45 rule on C789 3 outies R357C6 = 11 = {137/146/236/245}
6a. 5,6 of {146/236/245} only in R5C6 -> no 2,4 in R5C6, clean-up: no 3,5 in R5C4 (step 3)

7. 45 rule on C5 2 innies R19C5 = 10 = {19/28/37/46}, no 5

8. 9 in R5 only in R5C159
8a. 45 rule on R5 3 innies R5C159 = 19 = {289/379/469}, no 1,5

9. 45 rule on N3 3(2+1) outies R3C6 + R4C79 = 9
9a. Max R4C79 = 8, no 8 in R4C79
9b. Min R3C6 + R4C7 = 3 -> max R4C9 = 6

10. 45 rule on R789 4 outies R6C1379 = 12 = {1236/1245}, no 7,8,9, 1,2 locked for R6

11. 45 rule on N7 2 innies R7C3 + R8C2 = 1 outie R6C1 + 11
11a. Min R7C3 + R8C2 = 12, no 2 in R7C3 + R8C2

12. 45 rule on N7 3(2+1) outies R6C13 + R7C4 = 14
12a. R6C3 + R7C4 cannot total 8 -> no 6 in R6C1

13. 45 rule on N1 3(2+1) outies R3C4 + R4C13 = 15
13a. Max R3C4 = 4 -> min R4C13 = 11, no 1 in R4C1

14. 45 rule on N9 3(2+1) outies R6C79 + R7C6 = 9
14a. Min R6C79 = 3 -> no 7 in R7C6, clean-up: no 4 in R7C4 (step 4)

15. R357C6 (step 6) = {146/236/245}
15a. 5,6 only in R5C6 -> R5C6 = {56}, clean-up: no 4,6 in R5C4 (step 3)

16. 12(3) cage at R4C5 = {138/147/237/345} (cannot be {129} which clashes with R5C4, cannot be {156} which clashes with R5C6, cannot be {246} which clashes with R5C46), no 6,9
[Also cannot be {345} because cannot then place 7 for C5 but I’ll leave that for now and hope that I find a simpler step to eliminate this combination.]
16a. 1 of {138} must be in R4C5 -> no 8 in R4C5

17. 16(3) cage at R5C6 = {268/358/367/457} (cannot be {178} because R5C6 only contains 5,6), no 1
17a. R5C6 = {56} -> no 5,6 in R5C78
17b. Killer pair 7,8 in 16(3) cage at R5C6 and 22(3) cage in N6, locked for N6

18. 1 in R5 only in 10(3) cage at R5C2 = {127/145} (cannot be {136} = [361] which clashes with R5C46, CCC), no 3,6
[Ed pointed out this can also be obtained by
45 rule on N5 2 outies R5C23 = 1 innie R5C6 + 3, IOU no 3 in R5C2
10(3) cage at R5C2 = {127/145}, no 6]
18a. 16(3) cage at R5C6 (step 17) = {268/358/367} (cannot be {457} which clashes with 10(3) cage at R5C2), no 4

19. R5C159 (step 8a) = {289/379/469}
19a. 4 of {469} must be in R5C5 -> no 4 in R5C1

20. 1 on D/ only in R1C9 + R2C8, locked for N3
20a. 9(3) cage in N3 = {126/135/234}
20b. 1 of {126} only in R1C9 -> no 5,6 in R1C9

21. 7 in N3 only in R3C789
21a. 45 rule on N3 4 innies R2C8 + R3C789 = 22 = {1579/1678/2479/3478} (cannot be {2578/4567} which clash with R12C7)
21b. 1 of {1579/1678} must be in R2C8 -> no 5,6 in R2C8
21c. 9 of {1579/2479} must be in R3C7 -> no 2,5 in R3C7
21d. {2479} cannot be 49{27} (because min R3C89 = 10) -> no 2 in R3C89

22. 25(4) cage at R6C3 = {2689/4579/4678}
22a. Min R6C3 + R7C4 = [28] = 10 (cannot be [27] from combinations for 25(4) cage) -> max R6C1 = 4 (step 12)

23. 7 in N1 only in 17(3) cage = {179/278/467}, no 3,5
[I ought to have spotted this a lot earlier.]

24. 3 in N1 only in R2C2 + R3C12, CPE no 3 in R3C4, clean-up: no 2 in R3C6 (step 2)

25. 45 rule on C123 3 outies R357C4 = 12 = {129/147}, no 8, 1 locked for C4, clean-up: no 3 in R7C6 (step 4)
[Another step I ought to have spotted earlier.]

26. 3 in R7 only in R7C789, locked for N9, clean-up: no 7 in R89C7
26a. Killer pair 6,8 in R89C7 and 22(3) cage, locked for N9

27. R12C7 = {59} (cannot be {68} which clashes with R89C7), locked for C7 and N3

28. 9(3) cage in N3 = {126/234}, 2 locked for N3

29. 8 in N3 only in R3C789, locked for R3, clean-up: no 1 in R2C5
29a. Killer pair 3,4 in R23C5 and R3C46, locked for N2, clean-up: no 6,7 in R9C5 (step 7)

30. 12(3) cage at R4C5 (step 16) = {138/147/237} (cannot be {345} which clashes with R23C5), no 5
30a. Killer pair 1,2 in 12(3) cage and R5C4, locked for N5
30b. Killer pair 3,4 in R23C5 and 12(3) cage, locked for C5, clean-up: no 6,7 in R1C5 (step 7)
30c. Killer pair 8,9 in R19C5 and R78C5, locked for C5
30d. 7 in C5 only in 12(3) cage, locked for N5

31. 16(3) cage at R3C8 = {178/268/358/367/457}
31a. 5 of {457} must be in R4C9 -> no 4 in R4C9

32. Hidden killer pair 2,4 in R7C6 and 20(5) cage for N8, R7C6 = {24} -> 20(5) cage must contain one of 2,4
32a. Grouped hidden killer pair 2,4 in 16(3) cage in N7, 20(5) cage in N8, R89C7 and R8C28 for R89, 16(3) cage contains one of 2,4, 20(5) cage must contain one of 2,4, R89C3 contains one of 2,4 -> R8C28 must contain one of 2,4
32b. 45 rule on R89 3 innies R8C258 = 18 = {279/459/468} (cannot be {567} which doesn’t contain 2 or 4)
32c. 2 of {279} must be in R8C8 -> no 7 in R8C8
[I saw this a long time ago but it’s only useful now.
I’ve simplified steps 32 and 32a; when I first saw it R7C6 was still {234} with step 32 a hidden killer triple and step 32a a grouped hidden killer triple.
I’ve further simplified step 32a after Ed pointed out that I don’t need to use 1 in it.]

33. 13(4) cage at R6C7 = {1246/1345} (cannot be {1237} because 1,3,7 only in R67C7), no 7

34. 7 in N9 only in 22(3) cage = {679}, locked for N9, clean-up: no 4 in R89C7

35. Naked pair {28} in R89C7, locked for C7 and N9

36. 16(3) cage at R5C6 (step 18a) = {358/367} (cannot be {268} because 2,8 only in R5C8), no 2, 3 locked for R5 and N6, clean-up: no 7 in R5C159 (step 8a)

37. 13(4) cage at R6C7 (step 33) = {1246/1345}
37a. 4 of {1246} must be in R8C8, 4 of {1345} must be in R7C6 -> no 4 in R67C7

38. 4 in C7 only in R34C7, locked for 15(4) cage at R2C8, no 4 in R2C8 + R3C6, clean-up: no 1 in R3C4 (step 2)

39. R2C8 + R3C6 = {13} = 4 -> R34C7 = 11 = [74], 7 placed for D/, R5C7 = 3, R7C7 = 1, placed for D\, R6C7 = 6, clean-up: no 7 in 22(3) cage in N6
39a. Naked pair {13} in R2C8 + R3C6, CPE no 3 in R2C5, no 1 in R2C6, clean-up: no 6 in R3C5
[Thanks Ed for pointing out the CPE that I missed. A clean-up has been removed from step 41.]

40. Killer pair 1,3 in 9(3) cage and R2C8, locked for N3

41. R3C89 = {468} -> 16(3) cage at R3C8 (step 31) must contain three even numbers = {268} (only remaining combination) -> R4C9 = 2, R3C89 = {68}, 6 locked for R3 and N3

42. 9(3) cage in N3 (step 28) = {234} (only remaining combination) -> R1C8 = 2, R12C9 = {34}, locked for C9 and N3 -> R2C8 = 1, R3C6 = 3, R3C4 = 2 (step 2), R67C9 = [15], R8C8 = 4, placed for D\, R7C8 = 3, R5C5 = 2, placed for D/, R5C4 = 1, R5C6 = 6 (step 3), R5C8 = 7 (step 36), clean-up: no 8 in R2C3, no 6 in R2C5, no 4 in R6C5 (step 30), no 9 in R8C5, no 8 in R19C5 (step 7)

43. Naked pair {45} in R5C23, locked for N4 -> R6C3 = 2, R6C1 = 3, R46C5 = [37], clean-up: no 8 in R1C3

44. Naked pair {89} in R5C1 + R6C2, locked for N4, R4C2 = 1 (cage sum)

45. R6C3 = 2 -> 25(4) cage at R6C3 (step 22) = {2689} (only remaining combination) -> R7C4 = 9, R7C3 + R8C2 = {68}, locked for N7, R7C6 = 2 (step 4), R9C5 = 1, R1C5 = 9, R12C7 = [59], R89C3 = [13]

46. Naked pair {46} in R12C3, locked for C3 and N1 -> R5C3 = 5

and the rest is naked singles including some placements on the diagonals.

[I’ve had thoughts for some time that Ed’s trick might be
45 rule on N7 2 outies R6C3 + R7C4 = 2 innies R7C12, with 3 of the cells on R7
but haven’t worked out how to use this. I’d also noticed that 45 rule on N7 = 4 innies R7C123 + R8C2 = 25, sharing two cells with the 25(4) cage at R6C3. I’d realised that the 25(4) cage and the 25(4) innies probably had to have different combinations although I hadn’t fully worked out why. That's expressed very clearly in Para's message.

I don’t know why but I never thought of looking for similar relationships for N9. I can only guess it was because I studied N7 more than N9.]

This has turned into a really great puzzle and thread. A bit of a reunion! Very worthy of a milestone puzzle.

A very big thank you to Andrew for finding a really neat way through this puzzle with step 32a. That type of step is one of the new techniques in our arsenal the last 100. I know SS can't do that kind of step so still opportunity for someone to find another traditional way in.

Interesting that the rest of us focused on the same area to crack it. Love Para's way!! I really enjoy puzzles that have a unique one-off trick available. I hope we can keep Para around with continuing high quality puzzles, though irregular.

This is the trick I used (step 5). Works for the same reasons as Para's but a slightly more circuitous route to get there.
Cheers
Ed