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Ed
 Post subject: Assassin 447
PostPosted: Fri Dec 01, 2023 6:45 pm 
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X-puzzle so 1-9 cannot repeat on either diagonal. There is a broken 16(3) cage in n12.

Assassin 447
Very satisfied to finally get this one out. One nice step and one difficult step amoung the many. Both SudokuSolver's (1.85) and JSudoku's logs are at their limits of (my) hardness.

triple click code:
3x3:d:k:2560:2560:3329:3329:3330:2563:2563:2820:2820:5381:5381:5381:4102:3330:2567:2567:2820:8968:5381:4102:6921:4102:3330:8458:3595:3595:8968:4620:6921:6921:6921:8458:8458:3595:8968:8968:4620:4620:6921:8458:8458:8458:8968:8968:8968:4620:10509:10509:10509:1806:1806:7439:2576:2576:10509:10509:10509:4113:7439:7439:7439:7439:4882:2323:10509:10509:4113:5396:7439:5396:4882:4882:2323:2325:2325:4113:4113:5396:3350:3350:3350:
solution: 
+-------+-------+-------+
| 7 3 8 | 5 2 1 | 9 6 4 |
| 2 4 6 | 9 8 7 | 3 1 5 |
| 9 1 5 | 6 3 4 | 2 7 8 |
+-------+-------+-------+
| 1 7 9 | 2 6 8 | 5 4 3 |
| 3 8 4 | 1 9 5 | 6 2 7 |
| 6 5 2 | 7 4 3 | 8 9 1 |
+-------+-------+-------+
| 8 9 3 | 4 7 6 | 1 5 2 |
| 4 6 1 | 3 5 2 | 7 8 9 |
| 5 2 7 | 8 1 9 | 4 3 6 |
+-------+-------+-------+
Cheers
Ed
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wellbeback
 Post subject: Re: Assassin 447
PostPosted: Sat Dec 09, 2023 12:02 am 
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Thanks Ed! Had to keep clearer thinking than usual on this one. Very easy to get confused...
Here's how I (eventually) did it.
Assassin 447 WT:
1. Innies r6789 -> r6c1 = 6
Outies n89 -> r6c7 = 8
No 4 in 41(8)r6c2
-> One of the 9(2)s in n7 = {45)
-> 5 in 41(8) in r6c234
-> 7(2)r6 = {34}
-> 10(2)r6 = {19}
-> r6c234 = {257}
-> The 9(2)s in n7 are {45} and {27}

2. Remaining Innies n36 = r12c7 = +12(2)
Since 8 already in c7 and 10(2) cannot contain a 5 -> r12c7 = {39}
-> r12c6 = {71}
Also no 9 in 35(7)r2c9
-> 35(7) = {2345678}
-> 8 in r23c9
-> 8 in r89c8
Also, remaining IOD n3 -> whatever value is in r4c7 goes in n3 in r23c9

3. Innies n12 = r3c26 = +9(2) (No 9)
1 already in n2 -> 9 not in 13(3)n2
-> 9 in n2 in r123c4

4. Since 7(2)r6c5 = {34} and r12c6 = {17} -> 13(3)n2 from {238} or {256}
I.e., 2 locked in r123c5

5. IOD c1234 -> r5c4 = r9c5
That value can only go in n2 in r12c6.
I.e., r5c4 = r9c5 and is from (17)

6. For the values (689) in r9 - at most one can be in 13(3)r9
-> Two of them are in r9c46
I.e., r9c4 from (68), r9c6 from (689)
-> 3 in r9 only in 13(3)n9 in r9c789

7. Given:
1 in r9 only in c5789
9 in r9 only in c6789
3 in r9 only in c789
8 in n9 in r8c8 or r9c8

Trying r9c8 = 8 puts 13(3)r9 = <283>, r9c456 = [619], and innies n9 = {157}
But this leaves no place for 8 in n8 since 21(3)r8c5 would be [{57}9]

-> r8c8 = 8
-> 8 in n5 in r4c56 or r5c6
-> 8 not in r3c6

8! r9c5 from (17)
But it cannot be 7 since that puts r789c4 = [{12}6] which leaves no place for all of (3489} in c4
-> r9c5 = r5c4 = 1

9. -> (Since 9 not in 13(3)r9) r9c6 = 9
Also 1 in n9 in r7c78
-> (Since r8c7 cannot be 3 or 9) Innies n9 = +13(3) = {157}
-> 19(3)n9 = <289>
-> 13(3)n9 = {346}
-> 9(2)r8c1 = [45], 9(2)r9c2 = {27}, r9c4 = 8
Also 21(3)r8c5 = [{57}9]
Also (since 8 in n2 only in c2) 13(3)n2 = {238}
-> 7(2)n5 = [43]
-> 3 in c4 in n8
-> r78c4 = [43]
-> (HS 4 in n2) r3c6 = 4
-> (Innies n12) r3c3 = 5
Also -> r123c4 = {569}
-> r46c4 = {27}
-> (HS 5 in r6) r6c2 = 5

10. 16(3)r2c4 cannot contain both (56)
-> 9 in r23c4
Since r6c34 = {27} this -> r69c3 = {27}
-> 6 not in r1c4
-> 13(2)r1c3 = [85] and 16(3)r2c4 = <619>
IOD r1 -> r1c5 = r2c8 + 1
Since r1c5 from (23) -> r2c8 from (12) -> r2c9 = 5
-> r4c7 = 5
-> r7c8 = 5
-> r7c7 = 1 and 21(3)r8c5 = [579]
-> 14(3)r3c7 = [275]
etc.
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Ed
 Post subject: Re: Assassin 447
PostPosted: Sun Dec 10, 2023 12:49 am 
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Ruudiculous view count on this thread. Any ideas why? It happens every couple of puzzles.

I used the same key steps as wellbeback but, as usual, for different reasons. [Thanks to Andrew for checking my WT. Couple of clarifications added]
a447 WT:
A447 WT
Preliminaries from SudokuSolver
Cage 7(2) n5 - cells do not use 789
Cage 13(2) n12 - cells do not use 123
Cage 9(2) n7 - cells do not use 9
Cage 9(2) n7 - cells do not use 9
Cage 10(2) n6 - cells do not use 5
Cage 10(2) n23 - cells do not use 5
Cage 10(2) n1 - cells do not use 5
Cage 10(2) n23 - cells do not use 5
Cage 21(3) n89 - cells do not use 123
Cage 11(3) n3 - cells do not use 9
Cage 19(3) n9 - cells do not use 1
Cage 41(8) n457 - cells ={12356789}

Note: no clean-up done unless stated.

1. "45" on n89: 1 outie r6c7 = 8

2. "45" on r6789: 1 innie r6c1 = 6
2a. no 4 in 10(2)n6

4. 4 in r6 only in 7(2)n5 = {34}: both locked for n5, 3 for r6
4a. -> 10(2)n6 = {19} only: both locked for r6 and n6

5. naked triple {257} in r6c234: all locked for 41(8)

6. 2,4,5,7 only in the two 9(2) cages n7

7. "45" on n36: 2 innies r12c6 = 12 = {39} only: both locked for c7 and n3
7a. -> r12c6 = 8 (cages sum) = {17} only (since r12c6 have 3 & 9 Locking Cages): both locked for n2 and c6
7b. no 6 in r1c3

Slow to see 8a in this step
8. "45" on c1234: 1 innie r5c4 = 1 outie r9c5
8a. they can't repeat in r3c6 since it's in the same cage as r5c4
8b. -> must only repeat in r12c6
8c. -> r5c4 = r9c5 = (17)

9. "45" on n1236: 2 remaining innies r3c36 = 9 (no 9)
9a. no 2,8 in r3c3

10. 33(6)r3c6: can't have both 3 & 4 since they are only in r3c6
10a. must have 1 or 7 for r5c4
10b. and 3 in r3c6 must also have 6 in the 33(6) since [63] is in h9(2)r3, and 6 in n5 only in r4c4 or 33(6)
10c. = {126789/145689/245679}(no 3)
10d. -> no 6 in r3c3 (h9(2)r3)
10e. 5 in 33(6) must also have 4 which is only in r3c6 -> no 5 in r3c6
10f. no 4 in r3c3 (h9(2)r3)

11. 13(3)r1c5: {346} blocked by r6c5 = (34)
11a. = {238/256}(no 4,9): 2 locked for n2 and c5
11b. no 7 r3c3 (h9(2)r3)

12. 35(7)r2c9 = {2345678}(no 1)
12a. 8 only in r23c9: locked for c9 and n3

13. "45" on r9: 4 innies r9c1456 = 23
13a. must have 1 or 7 for r9c5 and can't clash with the 9(2)r9c2 = {27/45}
13b. = {1589/1679/2678}(no 3,4)

14. 3 in r9 only in 13(3) = {139/238/346}(no 5,7)
14a. 3 locked for n9

key step
15. in 13(3)n9, {238} can only be [283]
15a. -> 4 in r6c6 and 1 in c7 only in r37c6
15b. but this blocks all permutations for the 9(2)r3c36 through D\ (ie, no 1,3 in r3c3, no 4 in r3c6)
15c. -> 13(3)n9 = {139/346}(no 2,8)

16. r8c8 = 8 (hsingle n9), 8 locked for d\

17. 8 in n5 only in 33(6) -> no 8 in r3c6
17a. no 1 in r3c3 (h9(2))

18. "45" on n236: 2 outies r1c3 + r3c2 - 5 = 1 remaining innie r3c6
18a. 1 innie sees one outie so must be unequal -> r1c3 <> 5 (Innie Outies Unequal IOU)
18b. no 8 in r1c4

19. "45" on n1: 3 innies r1c3 + r3c23 = 14
19a. must have 3 or 5 for r3c3
19b. but {356} blocked by none are in r1c3
19c. = {158/239/257/347}(no 6)
19d. 1 and 2 only in r3c2 -> r3c2 = (1247)

20. 8 in the broken 16(3)r2c4?
20a. {178} blocked by 1,7 only in r3c2
20b. {268} is blocked by 13(3)r1c5 needs one of 6 or 8 (step 11a)
20c. {358} blocked by none are in r3c2
20d. -> no 8 in r23c4

21. 8 in n2 only in 13(3) = {238} only: 3 & 8 locked for c5, 3 for n2
21a. r6c56 = [43], 3 locked for d\
21b. r3c3 = 5, locked for d\
21c. -> r3c6 = 4 (h9(2)r3)
21d. -> r1c3 + r3c2 = 9 (step 18) = [81/72]
21e. r1c4 = (56)

22. naked triple {569} in r123c4, all locked for c4

23. two remaining innies n5: r46c4 = 9 = {27}: both locked for c4 & n5
23a. r5c4 = 1 -> r9c5 = 1 (step 8c)
23b. r9c4 = 8

24. r9c8 = 3 (hsingle r9)
24a. -> r9c79 = 10 = {46}: both locked r9 and n9

25. naked pair {27} in r69c3: both locked for c3
25a. r1c34 = [85]

on from there
Cheers
Ed
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Andrew
 Post subject: Re: Assassin 447
PostPosted: Sat Dec 23, 2023 4:35 am 
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Joined: Wed Apr 23, 2008 6:04 pm
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Thanks Ed for this challenging puzzle, a real monster. There were some bits which were really stubborn to remove, it was so easy to overlook possibilities while doing forcing chains while some simpler looking forcing chains just fell short of achieving the desired effect.

Ed commented to me "You'll kick yourself about step 3n!! Much, much simpler way to make that elimination." so I went through my walkthrough again and quickly found it. Comment added below. So this wasn't as much of a monster as I thought when I originally solved it.

Here's how I solved Assassin 447:
Prelims

a) R1C12 = {19/28/37/46}, no 5
b) R1C34 = {49/58/67}, no 1,2,3
c) R1C67 = {19/28/37/46}, no 5
d) R2C67 = {19/28/37/46}, no 5
e) R6C56 = {16/25/34}, no 7,8,9
f) R6C89 = {19/28/37/46}, no 5
g) R89C1 = {18/27/36/45}, no 1
h) R9C23 = {18/27/36/45}, no 1
i) 11(3) cage at R1C8 = {128/137/146/236/245}, no 9
j) 19(3) cage at R7C9 = {289/379/469/478/568}, no 1
k) 21(3) cage at R8C5 = {489/579/678}, no 1,2,3
l) 41(8) cage at R6C2 = {12356789}, no 4

1a. 45 rule on N89 1 outie R6C7 = 8, clean-up: no 2 in R1C6, no 2 in R2C6, no 2 in R6C89
1b. 4 in N7 only in R89C1 = {45} or R9C23 = {45} (locking cages), 5 locked for N9
1c. 41(8) cage at R6C2 = {12356789}, 5 in R6C234, locked for R6, clean-up: no 2 in R6C56
1d. 41(8) cage at R6C2 = {12356789}, 8 in R7C123 + R8C23, locked for N7, clean-up: no 1 in R89C1, no 1 in R9C23
1e. 45 rule on N7 3 outies R6C234 = 14 contains 5 = {257/356}, no 1,9
1f. R6C89 = {19} (cannot be {37} which clashes with R6C234, cannot be {46} which clashes with R6C56), locked for R6 and N6, clean-up: no 6 in R6C56
1g. Naked pair {34} in R6C56, locked for R6 and N5, clean-up: no 6 in R6C234
1h. Naked triple {257} in R6C234, 2,7 locked for R6 and locked for 41(8) cage -> R6C1 = 6, clean-up: no 4 in R1C2
1i. 41(8) cage at R6C2 = {12356789}, 3,6 locked for N7
1j. 35(7) cage at R2C9 = {2345678} {cannot be {1235789/1245689/1345679} which clash with R6C9, no 1,9, 8 locked for C9 and N3
1k. 45 rule on C1234 1 innie R5C4 = 1 outie R9C5, no 3,4 in R9C5
1l. 45 rule on N36 2 innies R12C7 = 12 = {39}, locked for C7 and N3 -> R12C6 = {17}, locked for C6 and N2, clean-up: no 6 in R1C3
1m. 13(3) cage at R1C5 = {238/256} (cannot be {346} which clashes with R6C5), no 4,9, 2 locked for C5 and N2, clean-up: no 2 in R5C4
1n. 14(3) cage at R3C7 = {167/257}, no 4
1o. 45 rule on N6 2 outies R23C9 must contain 8 = 1 innie R4C7 + 8, no 4 in R4C7 -> no 4 in R23C9
1p. 4 in N3 only in 11(3) cage at R1C8 = {146/245}, no 7
1q. 45 rule on N9 3 innies R7C78 + R8C7 = 13 = {157/247/256/346} (cannot be {139} because R8C7 only contains 4,5,6,7), no 9
1r. 21(3) cage at R8C5 = {579/678}/{89}4, no 4 in R8C5 + R9C6
1s. 33(6) cage at R3C6 = {126789/135789/145689/235689} (cannot be {234789/345678} because 3,4 only in R3C6, cannot be {245679} because 1,8 in N5 cannot both be in R4C4)
1t. 3,4 of {135789/145689/235689} must be in R3C6 -> no 5 in R3C6
1u. 3 on D/ only in R7C3 + R8C2, locked for N7
1v. 9 in N2 only in R123C4 + R3C6, CPE no 9 in R5C4, clean-up: no 9 in R9C5
1w. 45 rule on R1 1 innie R1C5 = 1 outie R2C8 +1, no 8 in R1C5, no 6 in R2C8
1x. 45 rule on R1 3 innies R1C589 = 12 = {156/246/345}
1y. 2 in R1 only in R1C12 = {28} or R1C589 = {246} -> R1C12 = {19/28/37} (cannot be [46], locking-out cages), no 4,6

2a. R5C4 = R9C5 (step 1k)
2b. 45 rule on N5 2 innies R46C4 = 1 outie R3C6 + 5
2c. Min R3C6 = 3 -> min R46C4 = 8
2d. 16(4) cage at R7C4 = R5789C4 = {1249/1258/1267/1348/1357/1456/2356} (cannot be {2347} because R4C4 = 1 (hidden single in C4), R6C4 = 5 total less than 8)
2e. Consider combinations for 13(3) cage at R1C5 (step 1m) = {238/256}
13(3) cage = {238}, locked for N2 => min R3C6 = 4, min R46C4 = 9 => R5789C4 cannot be {2356} because R4C4 = 1 (hidden single in C4), R6C4 = 7 total less than 9)
or 13(3) cage = {256}, locked for C5 => 16(4) cage cannot be {2356} because R9C5 only contains 1,7,8)
-> 16(4) cage at R7C4 = R5789C4 = {1249/1258/1267/1348/1357/1456}, 1 locked for C4 and N8
2f. Hidden killer pair 4,9 in R123C4 and R3C6 for N2, R3C6 cannot contain both of 4,9 -> R123C4 must contain at least one of 4,9
2g. 16(4) cage at R7C4 = R5789C4 = {1258/1267/1348/1357/1456} (cannot be {1249} which clashes with R123C4), no 9
2h. 1 in N5 only in 33(6) cage at R3C6 (step 1s) = {126789/135789/145689}
2i. R12C6 = {17}, 13(3) cage at R1C5 = {238/256} -> R123C4 + R3C6 = {3489/4569}
2j. 6 of {4569} must be in R123C4 (cannot be {459}6 which clashes with R46C4 = 11 = [65/92]), no 6 in R3C6
2k. R3C6 = {3489} -> R46C4 = 8,9,13,14 = [62]/{27}/[85/95] (cannot be {67} which clashes with the 33(6) cage)
2l. 16(4) cage at R7C4 = R5789C4 = {1267/1348/1357/1456} (cannot be {1258} which clashes with R46C4

3a. R7C78 + R8C7 (step 1q) = {157/247/256/346}
3b. R6C7 = 8 -> 29(6) cage at R6C7 = {124589/125678/234578} (cannot be {123689} because R7C78 with 1 only contain 1,5,7, cannot be {134678} because 8{346}{17} clashes with 16(4) cage at R7C4)
3c. Consider placement for 9 in N8
9 in 29(6) cage = {124589} => R7C78 + R8C7 = {15}7
or 9 in 21(3) cage at R8C5 = {579}/{89}4
-> no 6 in R8C7
3d. Consider combinations for 29(6) cage
29(6) cage = {124589/125678} => R7C78 + R8C7 = {157}
or 29(6) cage = {234578}, no 6 => R7C78 + R8C7 = {247}
-> R7C78 + R8C7 = {157/247}, no 3,6, 7 locked for N9
3e. 19(3) cage at R7C9 = {289/469/568}, no 3
3f. 8 of {289/568} only in R8C8 -> no 2,5 in R8C8
3g. 3 in N9 only in 13(3) cage at R9C7, locked for R9
3h. 13(3) cage = {139/238/346}, no 5
3i. 1,2 of {139/238} must be in R9C7 -> no 1,2, in R9C89
3j. R46C4 (step 2k) = [62]/{27}/[85/95]
3k. Consider combinations for 13(3) cage
13(3) cage = {139}, 9 locked for N9 => 19(3) cage = {568}, 8 placed for D\ => R46C4 = {27}/[62/95]
or 13(3) cage = {238/346} => 1 in N9 in R7C78 + R8C7 = {157} => 29(6) cage = {124589/125678}, 2 locked for C6 and N8 => 2 in C4 only in R46C4
-> R46C4 = {27}/[62/95], no 8
3l. R46C4 = {27}/[62/95]= 8,9,14 -> R3C3 = {349} (step 2b)
3m. 45 rule on N1 3 innies R1C3 + R3C23 = 14
[Ignore the difficult step 3n. Much easier is R46C4 = 1 outie R3C6 + 5 (step 2b), 45 rule on N47 3(1+2) outies R3C3 + R46C4 = 14 which I eventually used at step 4f; I’d seen that 45 much earlier but not added a note under my worksheet as it hadn’t been useful then, I ought to have remembered it at this stage -> max R46C4 = 13 -> no 9 in R3C6
Even simpler is wellbeback’s 45 rule on N12 2 remaining innies R3C26 = 9, no 9 in R3C26]
3n. Consider combinations for R123C4 + R3C6 (step 2i) = {3489/4569}
R123C4 + R3C6 = {3489} cannot be 4{38}9 because then R1C3 = 9, R3C2 = 5 (cage sum) so R1C3 + R3C23 greater than 14
or R123C4 + R3C6 = {3489} cannot be 8{34}9 because then R1C1267 = {1379}, locked for R1, R6C9 = 1 (hidden single in C9), R6C8 = 9 and either 1 in R1C78 => 29(6) cage = {124589/125678}, 2 locked for C6 => R46C4 = {27}/[62], no 9 in R4C4 => no 9 in R3C6 or 1 in 13(3) cage = [139], 9 placed for D\, no 9 in R4C4 => no 9 in R3C6
or R123C4 + R3C6 = {4569} cannot be {456}9 which clashes with R46C4 = 14 = [95] -> 9 in R123C4, locked for C4 and N2
3o. R3C6 = {34} -> 33(6) cage (step 2h) = {135789/145689}, no 2, 5 locked for N5
3p. Naked pair {34} in R36C6, locked for C6
3q. 2 in C6 only in R78C6, locked for N8 and 29(6) cage, no 2 in R7C78
3r. R7C78 + R8C7 = {157} (only remaining combination), 1,5 locked for N9, 1 locked for R7 -> 19(3) cage = {289/469}, 9 locked for N9, 29(6) cage = {124589/125678}, no 3
[Cracked, it gets easier from here.]
3s. 3 in N8 only in R78C4, locked for C4
3s. 16(4) cage at R7C4 = R5789C4 (step 2l) = {1348/1357}, no 6
3t. R9C6 = 9 (hidden single in R9) -> 21(3) cage at R8C5 = {579}, R8C57 = {57} locked for R8, clean-up: no 2,4 in R9C1
3u. 29(6) cage = {125678}, no 4
3v. 8 in C6 only in R45C6, locked for N5, clean-up: no 8 in R9C5 (step 1k)
3w. R789C4 + R9C5 = {348}1 (hidden quad in N8), 4,8 locked for C4, R5C4 = 1 (hidden single in N5), clean-up: no 5,9 in R1C3
3x. Naked triple {569} in R123C4, 5,6 locked for C4 , 6,9 locked for N2
3y. Naked triple {238} in 13(3) cage at R1C5, 3 locked for C5 and N2 -> R3C6 = 4, R6C56 = [43], 3 placed for D\, clean-up: no 7 in R1C2
3z. R46C4 = {27} (hidden pair in C4), 7 locked for N5

4a. R9C8 = 3 (hidden single in R9) -> R9C79 = 10 = {46}, locked for N9, 4 locked for R9 -> R9C4 = 8, clean-up: no 5 in R9C23
4b. Naked pair {29} in R78C9, locked for C9, 9 locked for N9 -> R8C8 = 8, placed for D\, R6C89 = [91], clean-up: no 2 in R1C2
4c. R89C1 = [45] (hidden pair in N7), 5 placed for D/, R78C4 = [43]
4d. 5 in R6 only in R6C23, locked for N4
4e. R23C4 = {569} -> 16(3) disjoint cage at R2C4 = {169/259}, 9 locked for C4, R3C2 = {12}, clean-up: no 4 in R1C3
4f. 45 rule on N47 3(1+2) outies R3C3 + R46C4 = 14, R46C4 = {27} -> R3C3 = 5, placed for D\
4g. 9 in R1 only in R1C12 = {19} or R1C67 = [19] (locking cages), 1 locked for R1
4h. 4 in R1 only in R1C89, locked for N3
4i. R1C9 = 4 (hidden single on D/) -> R12C8 = 7 = [52/61], R9C9 = 6, placed for D\ -> R5C5 = 9, placed for both diagonals, R2C2 = 4 (hidden single on D\), clean-up: no 1 in R1C2
4j. R7C3 = 3 (hidden single on D/) -> R4C6 = 8 (hidden single on D/)
4k. 18(4) cage at R4C1 = {1368} (cannot be {1289} because 1,9 only in R4C1, cannot be {2378} which clashes with R6C3) -> R4C1 = 1, R5C12 = {38}, locked for N4, 3 locked for R5, clean-up: no 9 in R1C2
4l. R1C67 = [19] (hidden pair in R1) -> R2C67 = [73]
4m. R2C2 = 4, R2C3 = 6 (hidden single in N1) = 10, 9 in N1 only in R23C1 = 11 = {29}, 2 locked for N1, 9 locked for C1, R1C1 = 7, placed for D\, R1C3 = 8 -> R1C4 = 5, R1C89 = [64] -> R2C8 = 1 (cage sum), placed for D/
4n. R4C4 = 2 -> R6C4 = 7, placed for D/, R3C78 = [27] -> R4C7 = 5 (cage sum)

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