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Ed
 Post subject: Assassin 436
PostPosted: Thu Jun 15, 2023 6:35 pm 
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Assassin 436
Another one out of the (mostly) Cross+A puzzle generator. This started out as a 1.60 from SudokuSolver. Was a bit easy. This version gets 1.70. JSudoku has some trouble, but it had even more with the easy one! Has some interesting moves possible. Perfect.
triple click code:
3x3::k:4352:4352:4352:769:769:4866:4866:3843:3843:4352:5380:5380:2309:2309:4866:3590:3590:3590:1799:1799:5380:6408:6408:6408:3593:3593:3850:3595:3595:3595:8716:6408:1549:5134:3593:3850:3599:3344:3344:8716:8716:1549:5134:5134:3850:3599:3089:8716:8716:8716:5138:5138:5138:3850:3599:3089:3859:3859:8716:5138:5138:5913:3349:4630:3607:3607:3607:4632:3092:3092:5913:3349:4630:4630:4630:4632:4632:5913:5913:5913:5913:
solution: 
+-------+-------+-------+
| 7 5 3 | 1 2 8 | 4 6 9 |
| 2 4 9 | 3 6 7 | 5 8 1 |
| 6 1 8 | 4 5 9 | 2 7 3 |
+-------+-------+-------+
| 8 2 4 | 9 7 1 | 3 5 6 |
| 1 7 6 | 2 3 5 | 8 9 4 |
| 9 3 5 | 6 8 4 | 7 1 2 |
+-------+-------+-------+
| 4 9 7 | 8 1 2 | 6 3 5 |
| 5 6 1 | 7 4 3 | 9 2 8 |
| 3 8 2 | 5 9 6 | 1 4 7 |
+-------+-------+-------+
Cheers
Ed
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Andrew
 Post subject: Re: Assassin 436
PostPosted: Thu Jun 22, 2023 1:54 am 
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Thanks Ed for your latest Assassin! A fairly tough one; none of my steps seemed particularly interesting so I hope that others found some better ones.

Thanks Ed for your detailed corrections.
Here's how I solved Assassin 436:
Prelims

a) R1C45 = {12}
b) R1C89 = {69/78}
c) R2C45 = {18/27/36/45}, no 9
d) R3C12 = {16/25/34}, no 7,8,9
e) R45C6 = {15/24}
f) R5C23 = {49/58/67}, no 1,2,3
g) R67C2 = {39/48/57}, no 1,2,6
h) R7C34 = {69/78}
i) R78C9 = {49/58/67}, no 1,2,3
j) R8C67 = {39/48/57}, no 1,2,6
k) 19(3) cage at R1C6 = {289/379/469/478/568}, no 1
l) 21(3) cage at R2C2 = {489/579/678}, no 1,2,3
m) 20(3) cage at R4C7 = {389/479/569/578}, no 1,2
n) 23(6) cage at R7C8 = {123458/123467}, no 9

1a. Naked pair {12} in R1C45, locked for R1 and N2, clean-up: no 7,8 in R2C45
1b. 17(4) cage at R1C1 must contain at least one of 1,2 -> R2C1 = {12}
1c. 45 rule on R1 2 outies R2C16 = 9 = [18/27]
1d. 45 rule on R12 1 outie R3C3 = 8 -> R2C23 = 13 = {49/67}, no 5, clean-up: no 5 in R5C2, no 7 in R7C4
1e. Killer pair 4,6 in R2C23 and R2C45, locked for R2
1f. Hidden killer pair 1,2 in R2C1 and R3C12 for N1, R2C1 = {12} -> R3C12 must contain one of 1,2 = {16/25}, no 3,4
1g. 8 in N2 only in R12C6, locked for C6 and 19(3) cage at R1C6, no 8 in R1C7 clean-up: no 4 in R8C7
1h. 19(3) cage contains 8 = {478/568}, no 3,9
1i. 45 rule on C6789 1 innie R3C6 = 9, clean-up: no 3 in R8C7
1j. 45 rule on N2 2(1+1) outies R1C7 + R4C5 = 11 = {47/56}
1k. 45 rule on R89 2 outies R7C89 = 8 = [17/26/35], clean-up: no 4,5,9 in R8C9
1l. 9 in N9 only in R78C7, locked for C7
1m. 45 rule on N9 2 innies R78C7 = 1 outie R9C6 + 9
1n. R78C7 contains 9, no 8 in R9C6 -> no 8 in R78C7, clean-up: no 4 in R8C6
1o. 20(3) cage at R4C7 = {389/479/569/578}
1p. 9 of {389/479/569} must be in R5C8 -> no 3,4,6 in R5C8
1q. 3 in R1 only in 17(4) cage at R1C1 = {1349/2357} (cannot be {1367} = {367}1 which clashes with R1C89), no 6
1r. Killer pair 7,9 in 17(4) cage and R1C89, locked for R1, clean-up: no 4 in R4C5
1s. 19(3) cage = {478/568} = [847]/{56}8, no 4 in R1C6
1t. Consider placement for 4 in N3
R1C7 = 4 => R12C6 = [87]
or 4 in R3C789, locked for R3 => 4 in N2 only in R2C45 = {45}, 5 locked for N2
-> no 5 in R1C6
1u. 19(3) cage = [658/847], no 6 in R1C7, clean-up: no 5 in R4C5
1v. R8C67 = [39/57] (cannot be [75] which clashes with 19(3) cage)

2a. 15(4) cage at R3C9 = {1239/1248/1347/2346} (cannot be {1257/1356} which clash with R78C9), no 5
2b. 45 rule on C9 3 innies R129C9 = 17 = {179/269/359/458/467} (cannot be {278/368} which clash with R78C9)
2c. 4 of {458} only in R9C9 -> no 8 in R9C9
2d. 45 rule on R123 2 outies R4C58 = 1 innie R3C9 + 9
2e. Min R4C58 = 10, max R4C5 = 7 -> min R4C8 = 3
2f. 8,9 in N5 only in R4C4 + R6C45, locked for 34(7) cage at R4C4, no 8,9 in R6C3 + R7C5
2g. Hidden killer pair 8,9 in R1C89 and 14(3) cage at R2C7 for N3, R1C89 contains one of 8,9 -> 14(3) cage must contain one of 8,9 = {158/239}, no 7
2h. 45 rule on N47 2 outies R78C4 = 1 innie R6C3 + 10
2i. Min R78C4 = 11 -> no 1 in R8C4
2j. 45 rule on N36 3(1+2) innies R1C7 + R6C78 = 12, min R1C7 = 4 -> max R6C78 = 8, no 8,9 in R6C78
2k. 45 rule on N3 4 innies R1C7 + R3C789 = 16 must contain 4 for N3 = {1456/2347} = 4{237}/5{146} (cannot be 4{156} which clashes with R3C12), no 5 in R3C78
2l. 14(3) cage at R3C7 = {149/167/239/257/347} (cannot be {158} because 5,8 only in R4C8, cannot be {248} because R3C78 cannot contain both of 2,4, cannot be {356} because R3C78 cannot contain both of 3,6), no 8
2m. 6 of {167} must be in R3C78 (cannot be {17}6) -> no 6 in R4C8
2n. 4 of {347} must be in R4C8 (cannot be {47}3), 9 of {239} must be in R4C8 -> no 3 in R4C8
2o. R3C789 = {14}6/{16}4/{23}7/{27}3/{37}2 (14(3) cage cannot be {46}4), no 1 in R3C9
2p. 1 in R3 in R3C12 = {16} or R3C789 = {146} (locking cages), 6 locked for R3

3a. R129C9 (step 2b) = {179/269/359/458/467}, 14(3) cage at R2C7 (step 2g) = {158/239}, 15(4) cage at R3C9 (step 2a) = {1239/1248/1347/2346}
3b. Consider combinations for R1C89 = {69/78}
R1C89 = {69}, locked for N3 => 14(3) cage = {158} => R129C9 = {179/359} (cannot be {269/467} which don’t contain 1,5,8 for R2C9, cannot be {458} which doesn’t contain 6,9 for R1C9)
or R1C89 = {78}, locked for N3 => 14(3) cage = {239}, killer pair 7,8 in R1C9 and R78C9, locked for C9, 15(4) cage = {2346} (cannot be {1239} which clashes with R2C9), locked for C9, R78C9 = [58], R129C9 = {179}
-> R129C9 = {179/359}, no 2,4,6,8, 9 locked for C9 and N3, clean-up: no 7 in R1C8
3c. Killer pair 5,7 in R129C9 and R78C9, locked for C9
3d. R1C7 + R6C78 = 12 (step 2j)
3e. Consider placements for 4 in C9
R3C9 = 4 => R1C7 = 5, R3C78 = 7 = {16}, R4C8 = 7 (cage sum), 20(3) cage at R4C8 = {389} (cannot be {569} = {56}9 which clashes with R1C7)
or 4 in R456C9, locked for N6
-> R4C8 = {579}, 20(3) cage = {389/569/578}, no 4
3f. Consider placements for R1C7 = {45}
R1C7 = 4 => R3C789 (step 2k) = {237}, R3C78 = {27} (cannot be {37} because no 4 in R4C8), R4C8 = 5 (cage sum) => 20(3) cage = {389}
or R1C7 = 5 => 20(3) cage = {389/578} (cannot be {569} = {56}9)
-> 20(3) cage = {389/578}, no 6, 8 locked for N6
3g. R8C9 = 8 (hidden single in C9) -> R7C9 = 5, R7C8 = 3 (step 1k), clean-up: no 7,9 in R6C2
3h. R129C9 = {179} (only remaining combination), no 3, 1 locked for C9
3i. 23(6) cage at R7C8 = {123467} (only remaining combination), no 5
3j. Deleted

4a. 1 in N6 only in R6C78, locked for R6 and 20(5) cage at R6C6, no 1 in R7C67, clean-up: no 1 in R9C6 (step 1m)
4b. R1C7 + R6C78 = 12 (step 2j) with 1 in R6C78 = 4{17}/5{16}
4c. 20(5) cage = {12467} (only remaining combination, cannot be {13456} because 3,5 only in R6C6), no 3,5,9
4d. R8C7 = 9 (hidden single in N9) -> R8C6 = 3
4e. R45C6 = {15} (hidden pair in C6), locked for N5
4f. 34(7) cage at R4C4 contains 8,9 (step 2f) = {1234789/1235689} -> R7C5 = 1, R1C45 = [12]
4g. Hidden killer pair 8,9 in R7C4 and 18(3) cage at R8C5 for N8, 18(3) cage without 1 cannot contain both of 8,9 -> R7C4 = {89}, 18(3) cage must contain one of 8,9, clean-up: no 9 in R7C3
4h. 14(3) cage at R8C2 = {167/257}, no 4, 7 locked for R8
4i. 23(6) cage at R7C8 (step 3i) = {123467}, 7 locked for R9
4j. 18(3) cage = {459/468}, no 2, 4 locked for N8

5a. R6C6 = 4 (hidden single in C6), clean-up: no 8 in R7C2
[Cracked. The rest is straightforward.]
5b. 2 in C6 only in R79C6, locked for N8
5c. 34(7) cage at R4C4 (step 4f) = {1235689} (only remaining combination), no 7, R6C3 = 5, clean-up: no 8 in R5C2, no 7 in R7C2
5d. R4C5 = 7 (hidden single in N5), R3C6 = 9 -> R3C45 = 9 = {45}, locked for R3 and N2, naked pair {36} in R2C45, locked for R2, 6 locked for N2, R12C6 = [87], R1C7 = 4 (cage sum), R1C8 = 6 -> R1C9 = 9, R29C9 = [17], R2C1 = 2
5e. R1C7 = 4 -> R6C78 (step 4b) = 8 = {17}, 7 locked for R6, N6 and 20(5) cage at R6C6, no 7 in R7C67
5f. Naked pair {26} in R7C67, locked for R7, R7C3 = 7 -> R7C4 = 8, clean-up: no 6 in R5C2
5g. 20(3) cage at R4C7 (step 3f) = {389} (only remaining combination) -> R5C8 = 9, 3 locked for C4 and N6, R4C8 = 5, R45C6 = [15], clean-up: no 4 in R5C23
5h. R5C23 = [76], R1C123 = [753]
5i. Naked pair {38} in R5C57, locked for R5 -> R5C49 = [24]
5j. R5C1 = 1 -> R67C1 = 13 = [94], R7C2 = 9 -> R6C2 = 3
5k. R89C1 = [53] (hidden pair in C1) = 8 -> R9C23 = 10 = [82]

and the rest is naked singles.
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Ed
 Post subject: Re: Assassin 436
PostPosted: Sun Jun 25, 2023 7:27 am 
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Joined: Wed Apr 16, 2008 1:16 am
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I followed a similar path to Andrew but for different reasons. Not as pure. But you did find some interesting steps in 4, 5 and 14; I've a feeling, without rechecking my solving path, that I may have missed that last one. Andrew [Thanks Andrew! Also, thanks for picking up the mistake in step 17. Fixed now]
my middle to a436: 
.-------------------------------.-------------------------------.-------------------------------.
| 34579     34579     34579     | 12        12        568       | 456       6789      6789      |
| 12        4679      4679      | 3456      3456      78        | 123578    1235789   1235789   |
| 1256      1256      8         | 34567     34567     9         | 1234567   1234567   1234567   |
:-------------------------------+-------------------------------+-------------------------------:
| 123456789 123456789 12345679  | 123456789 567       1245      | 345678    123456789 123456789 |
| 123456789 46789     45679     | 123456789 123456789 1245      | 345678    5789      123456789 |
| 123456789 345789    12345679  | 123456789 123456789 1234567   | 12345678  123456789 123456789 |
:-------------------------------+-------------------------------+-------------------------------:
| 123456789 345789    679       | 689       123456789 1234567   | 12345679  123       567       |
| 123456789 123456789 12345679  | 123456789 123456789 357       | 579       12345678  678       |
| 123456789 123456789 12345679  | 123456789 123456789 1234567   | 12345678  12345678  12345678  |
'-------------------------------.-------------------------------.-------------------------------'
Paste into A436 in SudokuSolver

Above to Andrew's step 1s. Note: no clean-up done unless stated

2. 14(3)r2c7 must have one of 8 or 9 for n3 since the 15(2)n3 can only have one off (hidden killer pair)
2a. = {158/239}(no 7)

3. "45" on n3: 4 innies r1c7 + r3c789 = 16 and must have 4 for n3 and only one of 1,2 for n3 since the 14(3)n3 has one of 1,2 (hidden killer pair)
3a. = {1456/2347}
3b. but [6]{145} is blocked by 7(2)n1 needs 1 or 5
3c. -> no 6 in r1c7
3d. -> no 5 in r4c5 (outiesn2=11)
3e. also [4]{156} blocked by 7(2)n1 = 1 or 5
3e. -> h16(4)n3 = [4]{237}/[5]{146}(no 5 in r3c789)
3f. note: if it has 5 in r1c7 then no 2 in r3c9

Still get a buzz finding these and the CCCs in the next step
4. "45" on c9: 1 outie r1c8 + 2 = 2 innies r29c9
4a. since 1 innie and 1 outie are in n3 -> r9c9 cannot be 2 (Innie & Outie Unequal IOU)

5. "45" on c9: 3 innies r129c9 = 17
5a. but {278/368} are blocked by 13(2)n9 needing 7 or 8, or 6 or 8
5b. {269} as [926] only, blocked by [39] in r2c78 (step 2a) (Combo Crossover Clash CCC)
5c. {458} as [854] only, blocked by {18} in r2c78 (step 2a) (CCC)
5d. {467} blocked by none are in r2c9
5e. = {179/359}(no 2468) = 5 or 7
5f. 9 locked for n3 and c9

6. killer pair 5,7 between h17(3)c9 and 13(2)r7c9: both locked for c9

7. from step 2a. 14(3)r2c7 = {158/239}
7a. 9 in {239} must be in r2c9 -> no 3 in r2c9

8. from step 5e. h17(4)c9 = {179/359}
8a. = [791/917/953]
8b. r9c9 = (137)

9. 2 & 4 in c9 only in 15(4)r3c9 = {1248/2346}
9a. note: {179} in h17(4) -> 15(4) = {2346}
9b. -> c9 has 5 in r2c9 (step 8a) or at least one of (34) in r456c9 (no eliminations yet)
9c. note: 15(4) has 8 in n6 or has both {26}

Key step. Difficult. Not as pure as the way Andrew did this but I really wanted to avoid the 14(3)r3c7
10. "45" on n36: 3 innies r1c7 + r6c78 = 12
10a. but [4]{26} (with 8 in r456c9, step 9c) blocks 20(3)n6 which needs 4 in c7 or 6/8 in n6
10b. and [4]{35} blocked by 20(3)n6 needing one of 3,4 (both in c7), or 5
10c. [5][25] blocked by no 2 left for 15(4) (step 3f)
10d. [5]{34} blocked by step 9b
10e. -> 3 innies n36 = [4]{17}/[5]{16}
10f. -> r6c78 = {16/17}, must have 1, locked for r6, n6 and 20(5) cage

11. 1 in n9 only in 23(6) -> no 1 in r9c6

12. 1 in c6 only in 6(2) = {15} only: both locked for n5, 5 for c6

13. 34(7)r4c4 must have 1
13a. -> r7c5 = 1
13b. r1c45 = [12]

14. "45" on n5: 2 innies r4c5 + r6c6 - 6 = r6c3
14a. -> no 6 in r4c5 (IOU)
14b. r4c5 = 7
14c. -> r1c7 = 4 (outiesn2=13)

15. also r3c789 = {237} (step 3e), all locked for n3
15a. 14(3)r3c7 must have 7 = {257/347}
15b. -> r4c8 = (45)

16. also r6c78 = {17} (inniesn36=12): 7 locked for n6, r6 and 20(5) cage

17. 34(7)r4c4 can't have 7 = {235689}[1] only (no 4)
17a. must have 5 -> r6c3 = 5
17b. r6c6 = 4 (hsingle n5)

18. r6c678 = 12 -> r7c67 = 8 (no 9)
18a. -> r8c7 = 9 (hsingle n9), r8c6 = 3
18b. -> r7c67 = 8 = {26} only: both locked for r7
18c. -> r7c89 = [35] (h8(2))
18d. r7c34 = [78]

19. "45" on n47: 1 remaining outie r8c4 = 7

on from there
Cheers
Ed
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wellbeback
 Post subject: Re: Assassin 436
PostPosted: Thu Jun 29, 2023 7:41 pm 
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My path had a lot in common with Ed's. Took me a while to get here since I spent too much time at first looking into IOD n4 and IOD n47.
I especially liked the inferences around n5! Thanks Ed!
Assassin 436 WT:
1. Outies r12 -> r3c3 = 8
Innies c6789 -> r3c6 = 9
3(2)n2 = {12}
Given the above it is straightforward to show n123 consists of one of the following two alternatives:
(A)
n1: 17(4) = [{349}1], 21(3) = [{67}8], 7(2) = {25}
n2: 3(2) = {12}, 9(2) = {45}, 19(3) = [658], 25(4) = [{37}96]
n3: 15(2) = {78}, 14(3) = {239}, r3c789 = {146}
(B)
n1: 17(4) = [{357}2], 21(3) = [{49}8], 7(2) = {16}
n2: 3(2) = {12}, 9(2) = {36}, 19(3) = [847], 25(4) = [{45}97]
n3: 15(2) = {69}, 14(3) = {158}, r3c789 = {237}

2. (89) in n5 in 34(7)
-> 34(7)n5 = {12389(47|56)}
-> One of the pairs (56) and (47) in n5 not in 34(7)

3. 23(6)n9 = {1234(58|67)} (No 9)
Outies r89 = r7c89 = +8(2) (No 489)
-> 13(2)n9 not {49}
-> 9 in n9 in r78c7
-> (Remaining IOD n9) whatever is in r9c6 is aslo in r78c7

4! 13(2)n9 from {67} or [58]
If the former this puts Innies c9 = r129c9 = +17(3) = [953]
(This is the only solution given Step 1 (A) or (B). In this case Step 1 (B) would be true.)
But this puts 14(3)r3c7 = [{37}4] which leaves no place for 4 in c9
-> 13(2)n9 = [58]
-> r7c8 = 3

5! Innies c9 = r129c9 = +17(3) only from [791] (Step 1(A)) or [917] (Step 1(B))
-> 15(4)c9 = {2346}
Max r3c78 = +10 -> Min r4c8 = 4
-> 1 in n6 in r6c78
-> 1 in c6 in r45c6 -> 6(2)n5 = {15}
-> (HS 1 in 34(7)) r7c5 = 1

6. 3 not in r6c6 since that leaves no place for 3 in 34(7)
-> (HS 3 in c6) 12(2)r8 = [39]
-> r7c7 = r9c6

7! Whatever is in r6c6 is not in 34(7)
-> Either (56) not in 34(7) which puts r6c6 = 6 and r4c5 = 7
Or (47) not in 34(7) which puts r6c6 = 4 and (again) r4c5 = 7
Either way r4c5 = 7
-> Step 1(B) is correct.

Easier from here.
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