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Ed
 Post subject: Assassin 437
PostPosted: Sat Jul 01, 2023 8:48 pm 
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a437.png
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note the broken 30(7)r1c5

Assassin 437

I found this one really resistant. Found a couple of different ways, none easy. Loved the challenge! It gets 1.55.
triple click code:
3x3::k:6400:2305:2305:5890:7683:7683:4356:4356:7683:6400:6400:5890:5890:5890:7683:7683:4356:7683:6400:6400:4869:5890:6406:7683:4359:4359:4616:2057:2057:4869:6406:6406:8458:4359:4359:4616:2827:5644:4869:6406:8458:8458:8458:8458:4616:2827:5644:5644:5644:8458:8458:3341:3341:3341:3598:3598:5903:5903:5903:5903:5392:2833:2833:2066:3598:3603:2580:2580:5392:5392:2833:3605:2066:3603:3603:2580:5142:5142:5142:2833:3605:
solution: 
+-------+-------+-------+
| 5 7 2 | 4 9 6 | 3 8 1 |
| 9 1 3 | 8 7 2 | 5 6 4 |
| 4 6 8 | 1 5 3 | 2 9 7 |
+-------+-------+-------+
| 6 2 7 | 3 8 4 | 1 5 9 |
| 3 5 4 | 9 6 1 | 8 7 2 |
| 8 9 1 | 7 2 5 | 6 4 3 |
+-------+-------+-------+
| 2 8 9 | 6 1 7 | 4 3 5 |
| 7 4 5 | 2 3 8 | 9 1 6 |
| 1 3 6 | 5 4 9 | 7 2 8 |
+-------+-------+-------+
Cheers
Ed
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wellbeback
 Post subject: Re: Assassin 437
PostPosted: Sun Jul 09, 2023 12:56 am 
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Another fun one! Thanks Ed!
Clarifications thanks to Ed.
Assassin 437 WT:
1. Innies n7 -> r7c3 = 9
-> 19(3)c3 = {8(47|56)}
Innies n1 = r23c3 = +11(2)
-> 8 not in r45c3 since that would put r2c3 = other value in r45c3
-> r3c3 = 8
-> r2c3 = 3
-> (Outies n4) r6c4 = 7
(89) in n4 in r45c12
22(4) cannot contain all of (789) -> 11(2)n4 from {29} or {38}

2. 11(4)n9 = {1235}
-> 14(2)n9 = {68}
-> r789c7 = {479}
-> Either 21(3)n9 = [489] and 20(3)r9 = [{49}7]
or 21(3)n9 = [759] and 20(3)r9 = [{79}4]

3. Innies c9 = r1267c9 = +13(4) (No 9)
-> 18(3)c9 = {9(27|45)}

4. Remaining Outies n5 -> r3c5 + r5c7 + r5c8 = +20(3)
That cannot be {479} since r789c7 = {479}
If r3c5 not 7 the value in r3c5 goes in n5 in c6
-> Since 9 in r9c56 -> r3c5 not 9
-> r3c5 is max 7

5. Either r5c8 is 7 or it is not 7.
If it is not 7 the values in r5c78 must go in n5 in r4c45 and since those two + r3c5 = 20 this puts r5c4 = 5
In that case outies n5 can only be = [3{89}]
Or r5c8 = 7 in which case outies n5 from [587] or [767]

6. 30(7)n23 is missing +15(2) = (69) or (78)
r3c5 from (357)
If r3c5 from (35) that puts the missing values from 30(7) into the 23(5)n2 in n2
If r3c5 = 7 (and r5c78 = [67]) that puts 7 in n3 in r12c9 which puts (69) into the 23(5)n2 in n2
Either way 23(5)n2 contains a +15(2) in n2
-> the other two values in 23(5) in n2 are (14)
-> 4 in 30(7)n23 in r12c9
-> 18(3)c9 = {279}

7. -> Outies n5 not [767] since that leaves no solution for 18(3)c9
-> Either Outies n5 = [3{89}] -> 18(3)c9 = [9{27}]
Or Outies n5 = [587] -> 18(3)c9 = [7{29}]
I.e., r3c59 = [57] or [39] and r5c78 + r45c9 = {2789}
-> 13(3)n6 = {346} and r4c78 = {15}
-> r3c78 = [29]
-> 18(3)c9 = [7{29}]
-> r5c78 = [78] and (Outies n5) r3c5 = 5
-> 8 in r4c45
-> 25(4)r3c5 = [5{389}]

8. Continuing...
13(3)n6 = [643]
-> 17(3)n3 = [3{68}]
Also 5 in n5 in c6
-> 21(3)r7c7 = [489] and 20(3)r9 = [{49}7]
-> 14(2)n9 = [68]
Also (since r4c78 = {15}) 8(2)n4 = {26}
-> r45c3 = [74]
Also 11(2)n4 = [38]
-> 14(3)r7c1 = [284]
Also 8(2)n7 = [71]
Also (Remaining innies r7) 11(4)n9 = [3512]
etc.


Last edited by wellbeback on Sat Aug 05, 2023 4:56 pm, edited 3 times in total.

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Andrew
 Post subject: Re: Assassin 437
PostPosted: Tue Jul 11, 2023 4:48 am 
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Thanks Ed for another challenging Assassin. At a glance, it looks like wellbeback found a more direct solution than I did.

Here's how I solved Assassin 437:
Prelims

a) R1C23 = {18/27/36/45}, no 9
b) R4C12 = {17/26/35}, no 4,8,9
c) R56C1 = {29/38/47/56}, no 1
d) R89C1 = {17/26/35}, no 4,8,9
e) R89C9 = {59/68}
f) 19(3) cage at R3C3 = {289/379/469/478/568}, no 1
g) 21(3) cage at R7C7 = {489/579/678}, no 1,2,3
h) 10(3) cage at R8C4 = {127/136/145/235}, no 8,9
i) 20(3) cage at R9C5 = {389/479/569/578}, no 1,2
j) 11(4) cage at R7C8 = {1235}

1a. Naked quad {1235} in 11(4) cage at R7C8, locked for N9, clean-up: no 9 in R89C9
1b. Naked pair {68} in R89C9, locked for C9 and N9
1c. Naked triple {479} in R789C7, locked for C7
1d. 18(3) cage at R3C9 = {279/459}, no 1,3, 9 locked for C9
1e. 21(3) cage at R7C7 = {489/579} (cannot be {678} because 6,8 only in R8C6) -> R8C6 = {58}
1f. 45 rule on N9 3 outies R8C6 + R9C56 = 21 = {489/579/678}, no 3
1g. R8C6 = {58} -> no 5,8 in R9C56
1h. 20(3) cage at R9C5 = {479} (only remaining combination), locked for R9, clean-up: no 1 in R8C1
1i. R8C6 + R9C56 = {489/579}, 9 locked for R9 and N8
1j. 10(3) cage at R8C4 = {127/136/235} (cannot be {145} which clashes with R8C6 + R9C56), no 4
1k. 45 rule on N7 1 innie R7C3 = 9
1l. 19(3) cage at R3C3 = {478/568}, no 2,3, 8 locked for C3, clean-up: no 1 in R1C2
1m. R8C7 = 9 (hidden single in R8)
1n. 45 rule on N1 2 innies R23C3 = 11 = [38] (cannot be {47/56} which clash with 19(3) cage, combo crossover clash), clean-up: no 1,6 in R1C23
1o. 45 rule on N4 1 remaining outie R6C4 = 7, clean-up: no 4 in R5C1

2a. 45 rule on N5 3(1+2) outies R3C5 + R5C78 = 20
2b. Max R5C78 = 17 -> min R3C5 = 3
2c. Min R5C78 = 11, no 1 in R5C7, no 1,2 in R5C8
2d. 33(7) cage at R4C6 must contain 1, locked for N5
2e. 33(7) cage must contain 2,6, CPE no 2,6 in R5C4

3a. Hidden killer pair 8,9 in R56C1 and 22(4) cage at R5C2 for N4, R56C1 and 22(4) cage with R6C4 =7 cannot contain both of 8,9 -> each must contain one of 8,9
3b. R56C1 = {29/38}, no 4,5,6,7
3c. 22(4) cage with 7 and one of 8,9 = {1579/1678/2479/3478} (cannot be {2578} which clashes with R56C1)

4a. Hidden killer pair 6,8 in R7C12 and 23(4) cage at R7C3 for R7, 14(3) cage at R7C1 and 23(4) cage with R7C3 = 9 cannot contain both of 6,8 -> R7C12 and 23(4) cage must each contain one of 6,8
4b. 14(3) cage at R7C1 = {158/167/248/356} (cannot be {257/347} which don’t contain one of 6,8)
4c. 6,8 of 14(3) cage must be in R7C12 -> no 6,8 in R8C2
4d. 23(4) cage at R7C3 = {1679/2489/3569} (cannot be {1589} which clashes with R8C6, cannot be {2579/3479} which don’t contain one of 6,8)
4e. Consider permutations for R7C7 + R8C6 = [48/75]
R7C7 + R8C6 = [48] => 8 in R7 only in 14(3) cage = {158/248}, no 7 => 7 in R7 only in 23(4) cage = {1679}
or R7C7 + R8C6 = [75] => 23(4) cage = {2489}
-> 23(4) cage = {1679/2489}, no 3,5
4f. Killer pair 4,7 in 23(4) cage and R7C7, locked for R7
4g. 10(3) cage at R8C4 (step 1j) = {136/235} (cannot be {127} which clashes with 23(4) cage), no 7
4h. 14(3) cage at R8C3 = {158/167/248/356} (cannot be {257} which clashes with R89C1, cannot be {347} because 4,7 only in R8C3)
4i. 7 in N7 only in 14(3) cage at R7C1 = {167} or 14(3) cage at R8C3 = {167} or R89C1 = [71] -> R89C1 = {35}/[71] (cannot be {26} which clashes with {167}, blocking cages), no 2,6
4j. 14(3) cage at R7C1 = {167/248/356} (cannot be {158} which clashes with R89C1)
4k. 4,7 of {167/248} must be in R8C2 -> no 1,2 in R8C2
4l. 14(3) cage at R8C3 = {167/248/356} (cannot be {158} which clashes with R89C1)
4m. 4,7 of {167/248} must be in R8C3 -> no 1,2 in R8C3
4n. 3,8 of {248/356} must be in R9C2 -> no 2,5 in R9C2
4o. 1 of {167} must be in R9C3 (R89C3 cannot be [76] which clashes with R45C3), no 1 in R9C2
4p. Hidden killer pair 1,2 in R8C45 and R8C8 for R8, 10(3) cage at R8C4 and R8C8 cannot contain both of 1,2 -> R8C45 and R8C8 must each contain one of 1,2
4q. R8C8 = {12}
4r. 1,2 of 10(3) cage must be in R8C45 -> no 1,2 in R9C4

5a. 4 in C1 only in R123C1, locked for N1, clean-up: no 5 in R1C23
5b. Naked pair {27} in R1C23, locked for R1 and N1
5c. 30(7) disjoint cage at R1C5 must contain 2, CPE no 2 in R2C45
5d. 45 rule on R123 4 innies R3C5789 = 23 = {1679/2579/3479/3569}, 9 locked for R3

6a. 22(4) cage at R5C2 (step 3c) = {1579/1678/2479/3478}, 14(3) cage at R8C3 (step 4l) = {167/248/356}
6b. Consider combinations for R45C3 = {47/56}
R45C3 = {47}, locked for C3 => 14(3) cage = {356}
or R45C3 = {56}, locked for N4 and 22(4) cage = {2479}, caged X-Wing for 2 in R1C23 and 22(4) cage, no other 2 in C23 => 14(3) cage = {167/356}
or R45C3 = {56}, locked for N4 and 22(4) cage = {3478} => R6C3 = 4 => 14(3) cage = {167/356}
-> 14(3) cage = {167/356}, no 2,4,8
6c. R7C12 = {28}, R8C2 = 4 (hidden triple in N7), 2,8 locked for R7
6d. Killer pair 2,8 in R56C1 and R7C1, locked for C1, clean-up: no 6 in R4C2
6e. 23(4) cage at R7C3 (step 4e) = {1679} (only remaining combination), locked for N8, 1,7 locked for R7, R7C7 = 4, R8C7 = 9 -> R8C6 = 8 (cage sum), R89C9 = [68]
6f. R9C8 = 2 (hidden single in R9) -> R8C8 = 1
6g. Killer pair 5,7 in R45C3 and R8C3, locked for C3 -> R1C23 = [72], clean-up: no 1 in R4C1
6h. 2 in R2 only in R2C679, locked for 30(7) cage at R1C5, no 2 in R3C6
6i. 4,9 in C4 only in R12345C4, CPE no 4,9 in R3C5
6j. 9 in R3 only in R3C89, locked for N3
6k. R3C5 + R5C78 = 20 (step 2a)
6l. Max R3C5 = 7 -> min R5C78 = 13, no 2,3,4
6m. 33(7) cage at R4C6 contains 2, locked for N5

7a. 17(3) cage at R1C7 = {368/458/467}, no 1
7b. 1 in N3 only in R1C9 + R2C79 + R3C7, CPE no 1 in R3C6
7c. R3C5789 (step 5d) = {1679/2579/3569} (cannot be {3479} = [73]{49} which clashes with 17(3) cage), no 4
7d. 3 of {3569} must be in R3C78 (cannot be 3{569} which clashes with 17(3) cage), no 3 in R3C5
7e. 3 in R3 only in R3C678, CPE no 3 in R1C9
7f. 25(4) cage at R3C5 = {3589/3679/4579/4678}
7g. 7 of {3679/4678} must be in R3C5 -> no 6 in R3C5
7h. Consider combinations for 17(3) cage = {368/458/467}
17(3) cage = {368/467}, 6 locked for N3 => R3C5789 = {2579}
or 17(3) cage = {458}, 5 locked for N3, 30(7) cage at R1C5 must contain 5, locked for N2 => R3C5 = 7 => R3C5789 = {1679/2579}
-> R3C5789 = {1679/2579}, no 3
7i. R3C6 = 3 (hidden single in R3)
[Cracked, fairly straightforward from here; no more routine clean-ups.]
7j. 3 in N3 only in 17(3) cage = {368}, 6,8 locked for N3
7k. R3C5789 = {2579} (only remaining combination), 2,5 locked for R3, 2 locked for N3
7l. 4 in N3 only in R12C9, locked for C9
7m. 18(3) cage at R3C9 = {279} (only remaining combination), 2,7 locked for C9
7n. Naked triple {145} in R1C9 + R2C79, locked for 30(7) cage at R1C5, 5 locked for N3 -> R3C7 = 2
7o. R3C5 = 5 (hidden single in R3)

8a. R2C6 = 2 (hidden single in N2)
8b. R23C6 = [23] = 5, R1C9 + R2C79 = {145} = 10 -> R1C56 = 15 = {69}, locked for R1 and N2
8c. Naked triple {148} in R123C4, locked for C4 and N2 -> R2C5 = 7, R7C456 = [617]
8d. Naked pair {38} in R1C78, 8 locked for R1 and N3
8e. Naked pair {39} in R45C4, locked for N5, 3 locked for C4
8f. R3C5 = 5, R45C4 = {39} = 12 -> R4C5 = 8 (cage sum)
8g. 33(7) cage at R4C6 = {1245678}, no 9
8h. R3C5 = 5 -> R5C78 = 15 (step 2a) = [87], R1C7 = 3
8i. Naked pair {29} in R45C9, locked for N6
8j. R3C78 = [29] = 11 -> R4C78 = 6 = [15]
8k. R4C12 = [62] (only remaining permutation) -> R56C1 = [38] (only remaining permutation), R8C1 = 7 (hidden single in C1)

and the rest is naked singles.
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Ed
 Post subject: Re: Assassin 437
PostPosted: Fri Jul 14, 2023 8:26 pm 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Huge shortcut by wellbeback! I particularly enjoyed the logic in his step 6. I used part of his step 5 as my key step. [Thanks Andrew for some comments and corrections.]
a437 WT:
Enjoyed this way. But very long compared to wellbeback's. It's not fully optimized but don't have time to go through it again.

Preliminaries
Cage 14(2) n9 - cells only uses 5689
Cage 8(2) n4 - cells do not use 489
Cage 8(2) n7 - cells do not use 489
Cage 9(2) n1 - cells do not use 9
Cage 11(2) n4 - cells do not use 1
Cage 21(3) n89 - cells do not use 123
Cage 10(3) n8 - cells do not use 89
Cage 20(3) n89 - cells do not use 12
Cage 19(3) n14 - cells do not use 1
Cage 11(4) n9 - cells ={1235}

1. "45" on n7: 1 innie r7c3 = 9

2. "45" on n1: 1 innie r2c3 + 8 = 2 outies r45c3
2a. 1 innie and each of outies see each other -> can't be the same -> no 8 in r45c3 (Innie Outies Unequal IOU)

3. 19(3)r3c3 = {478/568}
3a. -> r3c3 = 8
3b. r45c3 = {47/56}

4. "45" on n1: 1 innie r2c3 = 3
4a. "45" on c123: 1 remaining outie r6c4 = 7
4b. 9(2)n1 = {27/45}(no 1,6)

5. 11(4)n9 = {1235}: all locked for n9

6. 14(2)n9 = {68}: both locked for n9 and c9

7. 21(3)r7c7 must have two of {479} for r78c7
7a. = [489/759]
7b. ie r8c7 = 9

8. naked pair {47} in r79c7: locked for c7

9. 20(3)r9c5 must have 4 or 7 for r9c7, and 9 for n8
9a. = {79}[4]/{49}[7]
9b. 4 and 7 locked for r9

10. "45" on c12: 2 remaining outies r16c3 = 1 innie r9c2
10a. -> min r9c2 = 3

11. r5c2 + r6c23 = 15 (cage sum)
11a. but {456} blocked by r45c3 needs one of 4,5 (step 3b)
11b. = {159/168/249/258/348}
11c. note: if it has 1 must have 5 or 6 -> must have {47} in r45c3

12. sp15(3) can only have one of 8 or 9 (step 11b)
12a. -> 11(2)n4 must have one of 8 or 9 for n4 (hidden killer pair)
12b. = {29/38}(no 4,5,6,7)

13. 14(3)r8c3
13a. {347} blocked by 4,7 only in r8c3
13b. {257} blocked by 8(2)n7 needs one of those
13c. = {158/167/248/356}
13d. note: 4 in r8c3 must have 2 in r9c3

key step for the start
14. "45" on c12: 4 remaining outies r1689c3 = 14
14a. = {1247/1256}
14b. but [7142] blocked by {47} in r45c3 (step 11c)
14c. -> no 4 in r8c3

15. 4 in n7 only in 14(3)r7c1 = {248/347}(no 1,5,6)

16. 6 in r7 only in sp14(3)r7c456 = {167/356}(no 2,4,8) = 7 or 5
16a. -> r7c7 + r8c6 can't be [75]
16b. -> r7c7 = 4, r8c6 = 8
16c. r9c7 = 7
16d. r89c9 = [68]
16e. r8c2 = 4 (hsingle n7)
16f. -> r7c12 = 10 and must have 8 for r7 = {28} only: 2 locked for r7, n9
16g. no 5 in r1c3
16h. no 1 in r8c1
16i. 14(3)r8c3 = [761]/[536](no 1 in r8c3, no 5 in r9c2)

17. sp14(3)r7c4: (step 16.): must have both 6 & 7 for r7
17a. = {167} only: 1 locked for r7 & n8, 7 for n8
17b. -> r89c8 = [12](hsingle 1 in r8, hsingle 2 n9)

18. "45" on c12: 4 innies r1569c2 = 24 and must have (257) for r1c2 and (36) for r9c2
18a. = {2679/3579/3678}(no 1)
18b. must have 7 -> r1c2 = 7, r1c3 = 2

19. "45" on c12: 1 remaining outie r6c3 + 2 = 1 remaining innie r9c2 = [13/46]

20. 18(3)r3c9 = {279/459}(no 1,3)
20a. 9 locked for c9

21. 9 in c4 only in r12345c4 -> no 9 in r3c5 since it sees all those (Common Peer Elimination CPE)

22. "45" on n5: 3 outies r3c5 + r5c78 = 20
22a. max. r3c5 = 7 -> min. r5c78 = 13 (no 1,2,3,4 (since no 9 in r5c7))
22b. also min. r3c5 = 3

23. "45" on r123: 4 remaining innies r3c5789 = 23 = {1679/2579/3479/3569}
23a. 9 locked for n3, r3

24. 17(3)n3 = {368/458/467}(no 1) = 3 or 4, 5 or 6

25. from step 23, h23r3 = {1679/2579/3479/3569}
25a. 3 in {3479} must be in r3c7
25b. 3 in {3569} can't be in r3c5 since {56} in n3 is blocked by 17(3) (step 23)
25e. -> no 3 in r3c5

Really enjoyed finding this step
26. can both of r5c78 fit into r4c45?
26a. 33(7)r4c6 = {1234689/1235679/1245678}
26b. ie, if it has 7 which must be in r5c8, cannot be in r4c45, or it is {1234689} with {689} only in r5c78 and no 5 in that cage
26c. -> 5 in n5 would have to be in r5c4 if both r5c78 is in r4c45
(alternatively: "45" on n5: 2 outies r5c78 + 5 = 3 innies r4c45 + r5c4 -> if the two outies = two of the innies the remaining cell must be the IOD of 5)
[edit: alternative 2: thinking about wellbeback's step 5....the overlap of the 20(3) outies n5 and the 25(4) cage means the extra cell r5c4 = 25 - 20 = 5. Nice!]
26d. -> 25(4)r3c5 can't be {389}[5] since no 3,8,9 in r3c5
26e. -> = [7]{49}[5] only
26f. but no 4 in r5c78
26g. -> both of r5c78 can't fit into r4c45
26h. -> one of r5c78 must go in r6c4
26i. -> r5c8 = 7
26j. -> remaining two outies n5 = 13 = [76/58]
[edit: by combining the outies n5 with this step, Andrew noticed that "r5c78 cannot be {68} since r3c5 = 6, no 6 in n5 so r5c8 must be {79} which probably simplifies the rest of step 26." It does! Thanks]

27. "45" on c9: 4 innies r1267c9 = 13 and must have 1 & 3 for c9 = {1237/1345}
27a. = [1723]/{1345}(no eliminations yet)
27b. note: if it has 7 in r2c9 must have 1 in r1c9

28. 7 in n3 only in r23c9
28a. from step 25, h23r3 = {1679/2579/3479/3569}
28b. but [7]{169} leaves no 7 for n3 (by step 27b)(locking out cages)
28b. -> h23r3 = {2579/3479/3569} (no 1)
28c. also [7]{349} blocked by 17(3)n3 needs one of 3 or 4 (step 24)
28d. -> h23r3 = {2579/3569} (no 4)
28e. 5 locked for r3

29. 1 in n3 only in r12c9 + r2c7 in 30(7): 1 locked for 30(7)

30. 30(7)r1c5 = {12345}{69/78}
30a. caged x-wing on 5 with h23(4)r3: 5 locked for n23

31. 17(3)n3 = {368} only
31a. 6 & 8 locked for n3, 3 for r1

32. h23(4)r3 = {2579} only: 2 & 7 locked for r3, 2 locked for n3

33. 4 in n3 only in h13(3)c9 = {1345}: 4 and 5 locked for c9
33a. 4 locked for 30(7)r1c5 and c9

34. r3c6 = 3 (hsingle r3)

35. naked triple {145} in r1c9 + r2c79: 5 locked for n3 and 30(7) cage
35a. -> r3c5789 = [5297]
35b. -> r5c7 = 8 (outies n5=20)
35c. -> 33(7)r4c6 = {12456}[87](no 3,9)
35d. -> 25(4)r3c5 must have both 3,9 = [5]{389}

36. r4c78 = 6 (cage sum) = [15] only

37. 8(2)n4 = {26} only: both locked for r4 and n4

on from there
Cheers
Ed
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