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Ed
 Post subject: Assassin 412
PostPosted: Mon Feb 28, 2022 6:27 pm 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1043
Location: Sydney, Australia
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Assassin 412
A routine start but then quite resilient. JSudoku uses 5 advanced steps. I used two, both neat. SS gives it 1.60.

triple click code:
3x3::k:5120:5120:6145:6145:1794:1794:6915:6915:6915:4612:5120:4613:6145:1794:3590:5383:6915:6915:4612:4613:4613:6145:2824:3590:5383:2313:6915:4612:4612:4613:7434:2824:0000:5383:2313:4108:3341:7434:7434:7434:2574:0000:0000:0000:4108:3341:4111:3088:7434:2574:0000:5137:2578:4108:6931:4111:3088:7434:4116:0000:5137:2578:6421:6931:4111:4111:4875:4116:4875:5137:6421:6421:6931:6931:6931:4875:4875:4875:5137:5137:6421:
solution: 
+-------+-------+-------+
| 6 5 8 | 3 1 4 | 9 7 2 |
| 7 9 4 | 6 2 5 | 8 1 3 |
| 1 2 3 | 7 8 9 | 6 4 5 |
+-------+-------+-------+
| 4 6 9 | 8 3 2 | 7 5 1 |
| 8 1 2 | 5 6 7 | 4 3 9 |
| 5 3 7 | 9 4 1 | 2 8 6 |
+-------+-------+-------+
| 3 8 5 | 4 9 6 | 1 2 7 |
| 9 4 1 | 2 7 3 | 5 6 8 |
| 2 7 6 | 1 5 8 | 3 9 4 |
+-------+-------+-------+
Cheers
Ed
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Andrew
 Post subject: Re: Assassin 412
PostPosted: Fri Mar 04, 2022 2:09 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1894
Location: Lethbridge, Alberta, Canada
Thanks Ed. Maybe the hardest for some time. It resisted for a while but allowed some interesting steps, plus I only used two forcing steps.

Here's how I solved Assassin 412:
Prelims

a) R23C6 = {59/68}
b) R34C5 = {29/38/47/56}, no 1
c) R34C8 = {18/27/36/45}, no 9
d) R56C1 = {49/58/67}, no 1,2,3
e) R56C5 = {19/28/37/46}, no 5
f) R67C3 = {39/48/57}, no 1,2,6
g) R67C8 = {19/28/37/46}, no 5
h) R78C5 = {79}
i) 20(3) cage at R1C1 = {389/479/569/578}, no 1,2
j) 7(3) cage at R1C5 = {124}
k) 21(3) cage at R2C7 = {489/579/678}, no 1,2,3

1a. Naked pair {79} in R78C5, locked for C5 and N8, clean-up: no 2,4 in R34C5, no 1,3 in R56C5
1b. Naked triple {124} in 7(3) cage at R1C5, locked for N2
1c. 45 rule on C1234 2 innies R89C4 = 3 = {12}, locked for C4 and N8
1d. 45 rule on N8 2 innies R7C46 = 10 = {46}, locked for R7 and N8, clean-up: no 8 in R6C3, no 4,6 in R6C8
1e. Killer pair 6,8 in R34C5 and R56C5, locked for C5
1f. 8 in N8 only in R89C6, locked for C6, clean-up: no 6 in R23C6
1g. Naked pair {59} in R23C6, locked for C6 and N2, clean-up: no 6 in R4C5
1h. Naked pair {38} in R89C6, 3 locked for C6 and N8 -> R9C5 = 5, clean-up: no 6 in R3C5
1i. Naked pair {38} in R34C5, 8 locked for C5, clean-up: no 2 in R56C5
1j. Naked pair {46} in R56C5, locked for N5, 4 locked for C5
1k. Naked triple {127} in R456C6, 1,2 locked for C6, 7 locked for N5
1l. R1C6 = 4 -> R7C46 = [46]
1m. 24(4) cage at R1C3 = {3678} (only remaining combination)
1n. 45 rule on N2 1 outie R1C3 = 1 innie R3C5 = {38}
1o. 20(3) cage at R1C1 = {479/569/578} (cannot be {389} which clashes with R1C3), no 3

2a. 45 rule on N3 2 outies R4C78 = 12 = {48/57}/[93], no 6, no 1,2 in R4C8, clean-up: no 3,7,8 in R3C8
2b. 45 rule on N7 2 outies R6C23 = 10 = [19/37/73] (cannot be [64] which clashes with R6C5), R6C2 = {137}, no 4,5 in R6C3, clean-up: no 7,8 in R7C3
2c. 45 rule on N9 2 outies R6C78 = 10 = {19/28/37}, no 4,5,6 in R6C7

3a. 45 rule on C123 3(1+2) innies R1C3 + R5C23 = 11 = 3+8/8+3 = 3{26}/8{12} (cannot be 3{17} which clashes with R6C23, cannot be 3[35] which clashes with R456C4 ALS block) -> R5C23 = {12/26}, 2 locked for R5 and N4
3b. 45 rule on C789 2 innies R5C78 = 7 = {34} (cannot be {16} which clashes with R5C23), locked for R5 and N6 -> R56C5 = [64], R5C23 = {12}, 1 locked for R5 and N4 -> R5C6 = 7
3c. R1C3 + R5C23 = 11, R5C23 = 3 -> R1C3 = 8, R123C4 = {367}, 3 locked for C4 and N2 -> R34C5 = [83]
3d. R6C23 (step 2b) = {37}, 7 locked for R6 and N4
3e. Naked pair {37} in R6C23, CPE no 3,7 in R8C3
3f. R4C78 (step 2a) = {57}, locked for N6, 5 locked for R4
3g. R56C1 = {58} (only remaining combination), locked for C1, 8 locked for N4
3h. Naked triple {469} in R4C123, 6,9 locked for R4 -> R4C4 = 8
3i. R6C9 = 6 (hidden single in R6)
3j. 21(3) cage at R2C7 = {579/678} (cannot be {489} because R4C7 only contains 5,7), no 4, 7 locked for C7
3k. 20(3) cage at R1C1 = {479/569}, 9 locked for N1
Clean-ups: R3C8 = {24}, no 3 in R7C3, no 3,7 in R7C8

[Plenty of progress from that easy routine start, now things start to get harder.]
4a. R4C12 = {469} -> 18(4) cage at R2C1 = {17}{46}/{23}{49}/{12}{69}, no 4,6 in R23C1
4b. 16(4) cage at R6C2 = {1267/1348/1357/2347/2356} (cannot be {1249/1258/1456} because R6C2 only contains 3,7), no 9
4c. 16(4) cage = {1267/1348/1357/2347} (cannot be {2356} = 3{256} which clashes with R67C3 = [75])
4d. 6 of {1267} must be in R8C2 (R78C2 cannot be {12} which clashes with R5C2), no 6 in R8C3

5a. 25(4) cage at R7C9 = {1789/3589/3679/4579/4678} (cannot be {2689} which clashes with R5C9), no 2
5b. 6 of {3679} must be in R8C8, 1,3 of {1789/3589} must be in R789C9 because R789C9 = {589/789} clashes with R5C9), no 1,3 in R8C8
5c. 45 rule on N9 1 outie R6C7 = 1 innie R7C8 -> R45C9 + R67C8 form naked quad {1289}
5d. 25(4) cage at R7C9 = {3589/3679/4579/4678} (cannot be {1789} which clashes with R45C9 + R67C8), no 1
5e. 25(4) cage = {3589/3679/4579/4678} -> R7C78 + R8C7 + R9C78 = {12359/12368/12458/12467}
5f. Consider placements for R34C8 = [27/45]
R34C8 = [27] => R4C7 = 5 => R7C78 + R8C7 + R9C78 = {12368}
or R34C8 = [45] => R5C7 = 4 => R7C78 + R8C7 + R9C78 = {12359/12368}
-> R7C78 + R8C7 + R9C78 = {12359/12368}, no 4,7, 3 locked for N9
[With hindsight, this step would also have deleted R7C78 + R8C7 + R9C78 = {23456} and therefore 25(4) cage = {1789}; I enjoyed finding the way I eliminated it.]
5g. 7 in N9 only in 25(4) cage = {4579/4678}
5h. 6 of {4678} must be in R8C8 -> no 8 in R8C8
5i. 3 in C9 only in R123C9, locked for N3
5j. R5C7 = 4 (hidden single in C7) -> R5C8 = 3
5k. 45 rule on C89 1 outie R1C7 = 1 remaining innie R9C8 = {1269}

6a. 5 in N7 only in R78C23
6b. 45 rule on N7 4 innies R78C23 = 18 = {1359/1458/2358/2457/3456}
6c. 16(4) cage at R6C2 (step 4c) and R78C23 share 3 cells -> 16(4) cage = {1348/1357/2347} (cannot be {1267} which doesn’t contain any of 3,4,5), no 6, 3 locked for C8
6d. R78C23 = {1359/1458/2457} (cannot be {2358} which only contains two numbers in 16(4) cage)
6e. Variable hidden killer triple 1,2,3 in R23C1 and R789C1 for C1, R23C1 (step 4a) = {12}/{17}/{23} -> R789C1 must contain at least 1 or 3
6f. R78C23 = {1458/2457} (cannot be {1359} which clashes with R789C1), no 3,9, 4 locked for R8 and N7
6g. R7C3 = 5 -> R6C23 = [37]
6h. R9C9 = 4 (hidden single in R9}
6i. R4C1 = 4 (hidden single in C1)
6j. Naked pair {69} in R4C23, CPE no 6 in R3C2
6k. Hidden killer pair 1,2 in R23C1 and R789C1 for C1, 18(4) cage at R23C1 = {17}[49]/{23}[46] -> R789C1 must contain one of 1,2
6l. Killer pair 1,2 in R789C1 and R78C23, locked for N7
6m. 18(4) cage at R2C3 = {1467/2349/3456} (cannot be {1269/1359/2367/2457} which clash with R23C1), 4 locked for N1
6n. 20(3) cage at R1C1 = {569} (only remaining combination), 5,6 locked for N1
6o. 18(4) cage at R2C3 = {2349} (cannot be {1467} because R234C3 = {146} clashes with R58C3, ALS block)
6p. R4C23 = [69], R2C3 + R3C23 = {234}, 2,3 locked for N1, 3 locked for C3 -> R9C3 = 6
6q. Naked pair {59} in R12C2, 9 locked for R12C2 and N1 -> R1C1 = 6
6r. Naked pair {17} in R23C1, locked for C1
6s. 2 in C1 only in R789C1, locked for N7
6t. R9C2 = 7 (hidden single in R9)

7a. Combined half cage R67C8 and R9C8 = {19}2/{28}1/{28}9, 2 locked for C8
7b. R3C8 = 4 -> R4C78 = [75], R3C23 = [23], R5C23 = [12], R78C2 = [84], R8C3 = 1, R89C4 = [21]
7c. Naked pair {79} in R7C59, 9 locked for R7
7d. R7C8 = {12} -> R6C8 = {89}, R6C7 = {12} (step 2c)
7e. Killer pair 2,9 in R67C8 and R9C8, 9 locked for C8
7f. R9C8 = {29} -> R1C7 (step 5k) = {29}
7g. 25(4) cage at R7C9 = {4579/4678}
7h. Killer pair 8,9 in R5C9 and R78C9, locked for C9
7i. 9 in N3 only in R123C7, locked for C7

8a. Consider placements for R1C8 = {17}
R1C8 = 1 => R4C9 = 1 (hidden single in C9), R5C9 = 9 (cage sum) => R7C9 = 7
or R1C8 = 7
-> R8C8 = 6
8b. R8C8 = 6, R9C9 = 4 -> R78C9 = 15 = [78] -> R5C9 = 9, R4C9 = 1 (cage sum), R6C8 = 8 -> R7C8 = 2

and the rest is naked singles.
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Ed
 Post subject: Re: Assassin 412
PostPosted: Thu Mar 10, 2022 8:53 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1043
Location: Sydney, Australia
Haven't finished going through Andrew's WT yet. Floods here are playing havoc with my daily commute so no spare time for puzzles. But we used the same step which was my key (Andrew's step 6e and 6f, my 20a). I went a different way after that. I found this puzzle much easier than A410.
a412 WT:
Preliminaries by SudokuSolver
Cage 16(2) n8 - cells ={79}
Cage 14(2) n2 - cells only uses 5689
Cage 12(2) n47 - cells do not use 126
Cage 13(2) n4 - cells do not use 123
Cage 9(2) n36 - cells do not use 9
Cage 10(2) n69 - cells do not use 5
Cage 10(2) n5 - cells do not use 5
Cage 11(2) n25 - cells do not use 1
Cage 7(3) n2 - cells ={124}
Cage 21(3) n36 - cells do not use 123
Cage 20(3) n1 - cells do not use 12

This is a highly optimised solution. Any clean-up done is stated.
1. "45" on c1234: 2 innies r89c4 = 3 = {12}: both locked for c4 and n8

2. 16(2)n8 = {79}: both locked for c5 and n8

3. "45" on n8: 2 innies r7c46 = 10 = {46} only: both locked for r7 and n8

4. "45" on c5: 1 outie r1c6 + 1 = 1 innie r9c5
4a. = [23/45]

5. 8 in n8 only in c6: locked for c6

6. 14(2)n2 = {59} only: both locked for c6 and n2

7. r78c6 = {38}: 3 locked for c6 and n8
7a. -> r9c5 = 5
7b. -> r1c6 = 4 (iodc5=+1)
7c. -> r7c46 = [46]

8. 11(2)r3c5 = {38} only: both locked for c5

9. 10(2)r5c5 = {46} only: 6 locked for n5

10. 5 in c4 only in 29(6)r4c4: locked for that cage

11. "45" on n2: 2 outies r1c3 + r4c5 = 11 = {38} only

12. "45" on c123: 3 innies r1c3 + r5c23 = 11
12a. = [3]{17/26}/[8]{12}(no 3,8,9 in r5c23)
12b. ie, r5c23 = {17/26/12} = 1 or 6

13. "45" on c789: 2 innies r5c78 = 7
13a. but {16} blocked by r5c23
13b. and {25} blocked since combined with {17} in r5c23 clash with r5c6 = (127) (alternatively, killer single 2 in r5c236, locked for r5)
13c. = {34} only: both locked for r5 and n6
13d. -> r56c5 = [64]

14. naked triple 1,2,7 in r5c236: all locked for r5
14a. from step 12b, r5c23 = {17/12}: 1 locked for n4, r5

15. 13(2)n4 = {58} only combination: both locked for c1 and n4

16. "45" on n7: 2 outies r6c23 = 10 = {37} only combination: both locked for n4 and r6
16a. r8c3 sees both those so not there either (Common Peer Elimination CPE)
16b. r7c3 = (59)

17. naked pair {12} in r5c23: 2 locked for n4 and r5
17a. r5c23 = 3 -> r1c3 = 8 (3inn=11)
17b. -> r4c5 = 3 (2outtn2=11)
17c. r3c5 = 8
17d. r5c6 = 7

18. 20(3)n1 = {479/568}(no 3)
18a. 9 locked for n1

Now, the key find
19. 18(4)r2c1 must have two of {469} for r4c12
19a. = {1269/1467/2349} = 1 or 3 but not both
19b. can't have three of 4,6,9 -> no 4,6 in r23c1
19c. note: r789c1 must have one of 1,3 for c1 (hidden killer pair) (took me a long time to see that)

20. "45" on n7: 4 innies r78c23 = 18 and must have 5 for n7
20a. but {1359} blocked by r789c1 needing 1 or 3
20b. = {1458/2358/2457/3456}(no 9)
20c. r7c3 = 5
20d. r6c23 = [37]
20e. -> 16(4)r6c2 = [3]{148/247}(no 6)
20g. 4 locked for n7 and r8

21. r4c1 = 4 (hsingle c1)
21a. -> sp14(4)r2c1 = {17}[6]/{23}[9]

22. 2 & 3 in n1 are both in r23c1 or must have both in 18(4)r2c3 (or neither) (Andrew calls these 'variable hidden killer pair'. Great name!)
22a. -> 18(4)r2c3 = {1467/2349/2367}(no 5) (note: {2367} blocks r23c1 but I didn't use this)

23. 5 in n1 only in 20(3) = {569} only: 6 locked for n1.

24. another naked triple {569} in r124c2: 6,9 locked for c2
24a. -> r1c1 = r4c2
24b. -> r1c1 = r4c2 = r9c3 = (69)
24c. -> r4c2 + r9c3 = 12 or 18

25. "45" on c1: 1 innie r1c1 + 13 = 3 outies r4c2 + r9c23
25a. -> 3 outies = 19 or 22
25b. but no 4 in r9c2 -> 3 outies can't be 22 -> 3 outies = 19 = [676] only
25c. and r1c1 = 6
25d. r4c3 = 9
25e. -> r23c1 = 8 = {17}: 1 locked for n1 and c1

26. naked pair {59} in r2c26: both locked for r2

27. "45" on n3: 2 outies r4c78 = 12 = {57} only: both locked for n6, 5 for r4
27a. r3c8 = (24)

28. 16(3)n6 = [196/286] -> r6c9 = 6

29. naked triple {234} in r3c238: all locked for r3

final crack
30. 27(6)n3 must have 1 & 3 for n3, and at least one of 5 or 9 for r1 (hidden killer pair)
30a. but {123489} blocked by r3c8 = (24)
30b. and {134568} blocked by 4,6,8 only in r2c89
30c. = {123579} only (no 4,6,8): 2,5,7,9 locked for n3
30d. -> r3c78 = [64]

31. r8c8 = 6, r9c9 = 4 (hsingles n9,c9)
31a. -> r78c9 = 15 = {78} only: both locked for c9 and n9

Much easier now
Cheers
Ed
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wellbeback
 Post subject: Re: Assassin 412
PostPosted: Thu Mar 10, 2022 8:45 pm 
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 281
Location: California, out of London
Tough one to finish! Here's how I did it. Similar to Ed's I think.
Assassin 412 WT:
1. 16(2)n8 = {79}
Innies c1234 = r89c4 = +3(2) = {12}
-> Innies n8 = r7c46 = +10(2) = {46}
Innies c5 = r129c5 = +8(3)
-> 8 in n8 in r89c6
-> 14(2)n2 = {59}
-> (HS 5 in n8) r9c5 = 5 and r89c6 = {38}
-> r12c5 = {12}
-> r1c6 = 4
Also 10(2)c5 = {46}
-> 11(2)c5 = {38}
Also r456c6 = {127}
Also r7c6 = [46]
Also 24(4)r1c3 = {3678} with r1c3 from (38) and (67) in r123c4
Also r456c4 = {59(3|8)}

2. Outies n3 = r4c78 = +12(2)
Outies n9 = r6c78 = +10(2)
-> Remaining Innies n6 = r5c78 = +7(2)

3! Outies n7 = r6c23 = +10(2)
Innies c123 = r1c3,r5c23 = +11(3)
Whichever of (38) is in r1c3 is also in r456c4 so cannot be in r5c23
-> One of the following is correct:
(A) r1c3 = 8 puts r5c23 = {12} puts r5c78 = {34}
(B) r1c3 = 3 and r5c23 = {17} puts r5c6 = 2 and r5c78 = {34}
(C) r1c3 = 3 and r5c23 = {26} puts r5c78 = {34}
In all case r5c78 = {34}

3. Following on from that ...
10(2)c5 = [64]
Also -> (Outies n3) r4c78 = {57}
-> (HS 5 in n4) 13(2)n4 = {58}
Also 4 in n4 in r4 -> 6 in n4 also in r4 (Since r6c23 = +10(2))
-> Outies n12 = r4c1235 = +22(4) = [{469}3]
-> r3c5 = 8 and r1c3 = 8
-> r123c4 = {367}
-> r456c4 = [8{59}]
-> r5c23 = {12}
-> r6c23 = {37}
-> r456c6 = <172>
Also r6c9 = 6

4. 20(3)n1 from [{79}4] or {569} (with 5 in c2)
18(4)r2c1 only from [{17}{46}], [{23}{49}], [{12}{69}]
-> One each of (13) in c1 in n1 and n7

5 in n7 only in r78c23
Trying r6c23 = [73] puts r7c3 = 9 and 16(4)r6c2 = [7{135}] contradicting one of (13) in r789c1
-> r6c23 = [37] and r7c3 = 5

6. 16(4)r6c2 can only be [3{4(18|27)}
-> 4 in n7/r8 in r8c23
4 in n1 only in r23c23
-> (HS 4 in n4) r4c1 = 4
-> 18(4)r2c1 from [{17}46] or [{23}49]
Either way 20(3)n1 = {569}

7. For (27) in c1 - one goes in n1 and the other in n7
-> 16(4)r6c2 = [3{148}]
-> 18(4)r2c1 = [{17}46] and 18(4)r2c3 = [{234}9]
-> 20(3)n1 = [6{59}]
-> 27(5)n7 = [{239}76]

8. 4 in n9 in r9
(56) in n9 in r8
Since 25(4)n9 cannot contain both (56) -> One of (56) in r8c7 and the other in 25(4)n9
Since r4c7 from (57) -> 21(3)r2c7 cannot be {489} -> must have a 7 (locked for c7)
-> 7 in n9 in 25(4)
Since r2c26 = {59}, r4c78 = {57}, and r3c5 = 8
-> Either 21(3)r2c7 = [867] and 9(2)r3c8 = [45]
or 21(3)r2c7 = [795] and 9(2)r3c8 = [27]
In the former case this puts r5c7 = 4 and r9c9 = 4
In the latter case this puts 25(4)n9 = [{579}4]
Either way -> r9c9 = 4

9! 4 in n3 only in r23c8
6 in n3 only in r2c8 or r3c7
But 4 in r2c8 puts r3c7 = 6
and r34c8 = [45] puts 21(3)r2c7 = [867]
Either way 21(3)r2c7 = [867]
-> 9(2)r3c8 = [45]
-> r5c78 = [43]
-> (HS 6 in n9) r8c8 = 6
-> r789c9 = [{78}4]
-> 16(3)n6 = [196]
-> r123c9 = {235}
Also r6c78 = [28] -> r7c8 = 2
Also 13(2)n4 = [85]
-> r456c4 = [859]
Also r46c6 = [21]

10. Also (HS 1 in r3) r23c1 = [71]
-> (HS 7 in r3) 24(4)r1c3 = [8367]
Also 18(4)r2c3 = [4239]
Also (HS 9 in r3) 14(2)n2 = [59]
etc.

We could do with some of that rain. Less than 0.1 inches here this year so far! And it's supposed to be the rainy season!
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