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 Post subject: Assassin 422
PostPosted: Tue Aug 02, 2022 2:16 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 947
Location: Sydney, Australia
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Assassin 422

Really enjoyed this one. Interesting start then a bit of a bump at the end. What I always aim for!! SudokuSolver gives it 1.70
code: triple click:
3x3::k:4864:4864:4864:2561:2561:2561:7938:4611:4611:4612:4612:4612:3077:7938:7938:7938:4611:5126:2567:1800:4612:3077:7938:8969:4611:4611:5126:2567:1800:7946:8969:8969:8969:8969:5126:5126:4363:7946:7946:7946:7692:8969:3341:3341:5390:4363:4363:7946:7692:7692:7692:7692:3341:5390:5397:5397:7946:4112:7692:4369:4369:2834:5390:5397:5396:4112:4112:4371:4371:4369:2834:2575:5397:5396:5396:5396:4371:4371:4369:2834:2575:
solution:
+-------+-------+-------+
| 6 9 4 | 1 2 7 | 8 5 3 |
| 1 7 8 | 4 5 3 | 9 2 6 |
| 3 5 2 | 8 6 9 | 1 7 4 |
+-------+-------+-------+
| 7 2 6 | 5 3 8 | 4 9 1 |
| 8 1 9 | 7 4 6 | 2 3 5 |
| 5 4 3 | 2 9 1 | 6 8 7 |
+-------+-------+-------+
| 4 6 5 | 3 8 2 | 7 1 9 |
| 9 8 7 | 6 1 5 | 3 4 2 |
| 2 3 1 | 9 7 4 | 5 6 8 |
+-------+-------+-------+
Cheers
Ed


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 Post subject: Re: Assassin 422
PostPosted: Sun Aug 07, 2022 5:31 pm 
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 251
Location: California, out of London
I my not have found the most efficient path here, but I found it very interesting! Thanks Ed!
Assassin 422 WT:
1. Innies n3 = +27(4)
Innies c9 = r1234c9 = +14(4)
-> r23c9 = Max +11(2)
-> r12c7 is Min +16(2)

IOD n2 -> r12c7 = r3c6 + 8

-> One of:
(A) r12c7 = {79}, r3c6 = 8, r23c9 = +11(2), r14c9 = {12}
(B) r12c7 = {89}, r3c6 = 9, r23c9 = +10(2), r14c9 = {13}

But in case (A) there is no place for both 7 and 9 in n2
-> r12c7 = {89}, r3c6 = 9, r23c9 = +10(2), r14c9 = {13}

2. 9 in c8 not in 18(5)n2 nor in 11(3)n9
-> 9 in c8 in n6
Since r4c9 from (13) -> 9 not in 13(3)n6
-> (HS 9 in c8) r4c8 = 9
-> r14c9 = [31]
-> (HS 9 in c9) r7c9 = 9
-> (IOD n9) r7c6 = 2
Also (HS 2 in c9) 10(2)n9 = {28}
-> 21(3)c9 = [{57}9]
-> r23c9 = {46}
Also (Outies c89) r35c7 = [12]
-> r123c8 = {257}
-> 13(3)n6 = [2{38}]
-> 11(3)n9 = {146}
-> r789c7 = {357}
Also r46c7 = {46}

3. Innies r1 = r1c789 = +16(3)
This can only be [853]
-> r2c7 = 9
-> 9 in r1 in r1c123
-> 19(3)n1 = {469}
-> 10(3)r1 = {127}
-> 12(2)n2 = {48}
Also Innies n1 = r3c12 = +(8)2 can only be [35]
-> r4c12 = [72]
Also 31(5)n23 = [8{35}96]
-> r23c9 = [64]
-> 12(2)n2 = [48]
-> 18(4)n1 = {1278} with (18) in r2.

4. Remaining Innies n78 = r7c35 = +13(2) = {58} or {67}
Outies n14 = r5c4 + r7c3 = +12(2)
This cannot be {39} since r7c3 from (5678)
Also cannot be [66] since they are in the same cage.
Also cannot be {48} since (48) already in c4
-> [r5c4,r7c3] = {57}

5! r46c7 = {46}
Since r5c4 from (57) -> Whichever of (46) is in r4c7 must be in n5 in the 30(6) cage.
Similarly whichever of (46) is in r6c7 is in n5 in the 35(6) cage.
-> Both (46) are in 30(6) cage in n56 and also both in 35(6) cage in n56.
-> r7c5 cannot be 6
-> r7c35 = [58]
-> r5c4 = 7
Also 8 in n5 in r45c6
Also r56c9 = [57]

6. Innies c1234 = r146c4 = +8(3)
Since 4 already in c4 and (12) already in r4 -> r146c4 = <152>
-> 35(6) = [95{3468}] with r4c7 from (46)
Also 30(6)n5 = [{12469}8] with r6c7 from (64)
Also (HS) r6c1 = 5

7. Remaining cells c4 = r789c4 = {369}
-> 17(4)n8 = {1457}
Also r8c3 from (147)

8. 17(4)n4 cannot be [953] since that would put all of (369) in 21(4)r8c2
-> 17(3)n4 = <458>

9! (Complicated step!)
Consider where (369) go in n7
None go in r78c3.
Neither 21(4)r7c1 nor 21(4)r8c2 can contain all of (369)
Since one of (369) already in r9c4 -> two of (369) are in each of the 21(4)s.

Since 3, 6, 9, and 21 are all divisble by 3 -> the sum of the other two cells in each of the 21(4)s must also be divisble by 3. (E.g., (24), (18) etc.)

Conclusions from that:
(A) 21(4)r7c1 cannot contain both (14)
-> in r7 one of (14) in r7c12 and the other in r7c8
-> 6 in r89c8

(B) If 3 in 21(4)r7c1 it must be [13{89}] (Cannot be [43{68}] since one of (48) already in c1).
If 3 not in 21(4)r7c1 -> that cage must have (69) so can only be [{46}{29}]
-> 9 in r89c1
Also (HS 7 in r7) r7c7 = 7

(C) 21(4)r8c2 cannot contain (36) since the other two cells in that cage in n7 cannnot be either {57} or {48}
-> 21(4)4r8c2 must contain a 9 in r9c4
-> r78c4 = {36} and r8c3 = 7

10! -> r8c1 = 9
-> 2 in n7 only in r9c13
-> 10(2)n9 = [28]
-> (HS 8 in r8) r8c2 = 8
-> 21(4)r8c2 = [8{13}9]
Also -> 17(3)n4 = [854]
-> 21(4)r7c1 = [4692]
-> 19(3)n1 = [694]
Also 17(4)n1 = [17{28}]
Also r56c8 = [38]
-> r4c6 = 8
Also r46c7 = [46]
-> 35(6) = [953846]
-> r6c3 = 3
-> r9c23 = [31]
-> 31(6)r4c3 = [619735]
etc.

First time for ...:
I have used even/odd in a number of WTs - but don't remember ever using divisible by 3 before!


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 Post subject: Re: Assassin 422
PostPosted: Sun Aug 07, 2022 6:46 pm 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1820
Location: Lethbridge, Alberta, Canada
Ed wrote:
Interesting start then a bit of a bump at the end. What I always aim for!!
Hope you'll show us your interesting start; wellbeback has a very powerful start, mine was fairly routine.

I've slightly simplified how I dealt with the bump at the end.
Here's how I solved Assassin 422:
Prelims

a) R23C4 = {39/48/57}, no 1,2,6
b) R34C1 = {19/28/37/46}, no 5
c) R34C2 = {16/25/34}, no 7,8,9
d) R89C9 = {19/28/37/46}, no 5
e) 19(3) cage at R1C1 = {289/379/469/478/568}, no 1
f) 10(3) cage at R1C4 = {127/136/145/235}, no 8,9
g) 21(3) cage at R5C9 = {489/579/678}, no 1,2,3
h) 11(3) cage at R7C8 = {128/137/146/236/245}, no 9
i) 18(5) cage at R1C8 = {12348/12357/12456}, no 9

1a. 18(5) cage at R1C8 = {12348/12357/12456}, 1,2 locked for N3
1b. 45 rule on N1 2 innies R3C12 = 8 = [26/35/62/71] -> R4C1 = {3478}, R4C2 = {1256}
1c. 45 rule on R789 3 innies R7C359 = 22 = {589/679}, 9 locked for R7
1d. 45 rule on N14 2(1+1) outies R5C4 + R7C3 = 12 = [39/48/57/75]
1e. 45 rule on C9 4 innies R1234C9 = 14 = {1238/1256/1346/2345} (cannot be {1247} which clashes with R89C9), no 7,9
1f. 45 rule on C9 1 outie R4C8 = 1 innie R1C9 + 6 -> R1C9 = {123}, R4C8 = {789}
1g. 45 rule on R1 3 innies R1C789 = 16 = {169/178/259/268/349/358/367} (cannot be {457} because R19 only contains 1,2,3)
1h. R1C9 = {123} -> no 1,2,3 in R1C78
1i. 9 of {349} must be in R1C7 -> no 4 in R1C7
1j. Min R23C9 + R4C8 = 14 -> max R4C9 = 6
1k. 45 rule on C89 2 outies R35C7 = 3 = {12}, locked for C7
1l. 13(3) cage at R5C7 = {139/148/157/238/247/256} (cannot be {346} because R5C7 only contains 1,2)
1m. R5C7 = {12} -> no 1,2 in R56C8
1n. R4C9 + R5C7 = {12} (hidden pair in N6)
1o. 45 rule on N9 1 innie R7C9 = 1 outie R7C6 + 7 -> R7C6 = {12}, R7C9 = {89}

2a. 9 in N3 only in R12C7, locked for C7 and 31(5) cage at R1C7
2b. 9 in N9 only in R789C9, locked for C9
2c. 45 rule on N23 2 outies R4C89 = 1 innie R3C6 + 1
2d. Min R4C89 = 8 -> min R3C6 = 7
2e. 20(4) cage at R1C9 = {1469/1568/2378/2459/2567} (cannot be {1289} because 1,2 only in R4C9, cannot be {1379} because 7,9 only in R4C8, cannot be {1478} which clashes with R1C9 + R4C8 = [17] , cannot be {2369} which clashes with R1C9 + R4C8 = [39], cannot be {2468} which clashes with R1C9 + R4C8 = [28], cannot be {3458/3467} because R4C9 only contains 1,2)
2f. 20(4) cage = {1469/2459/2567} (cannot be {1568/2378} which clash with 21(3) cage at R5C9), no 3,8, clean-up: no 2 in R1C9
2g. Killer triple 4,5,6 in 20(4) cage and 21(3) cage, locked for C9
2h. 18(5) cage at R1C8 = {12348/12357} (cannot be {12456} which clashes with 20(4) cage), no 6, 3 locked for N3
2i. 20(4) cage = {1469/2567} (cannot be {2459} which clashes with 18(5) cage), 6 locked for C9 and N3
2j. Killer pair 4,5 in 18(5) cage and 20(4) cage, locked for N3
2k. 21(3) cage at R5C9 = {489/579} -> R7C9 = 9, R7C6 = 2, clean-up: no 3 in R5C4, no 1 in R89C9
2l. 10(3) cage at R1C4 = {127/145/235} (cannot be {136} which clashes with R1C9), no 6
2m. 6 in R1 only in R1C123, locked for N1, clean-up: no 2 in R3C12 (step 1b), no 4,8 in R4C1, no 1,5 in R4C2
2n. 19(3) cage at R1C1 contains 6 = {469/568}, no 2,3,7
2o. Naked pair {37} in R34C1, locked for C1
2p. 2 in R1 only in R1C456, locked for N2
2q. 10(3) cage contains 2 = {127/235}, no 4
2r. R23C4 = {39/48} (cannot be {57} which clashes with 10(3) cage
2s. Combined cage 20(4) at R1C9 + 21(3) at R5C9 = {1469}{57}9/{2357}{48}9, 7 locked for N3
2t. 17(4) cage at R7C6 = {2357/2456} (cannot be {2348} which clashes with R89C9), no 8, 5 locked for C7 and N9
2u. 1 in N9 only in R789C8, locked for C8

3a. 45 rule on N6 3 outies R237C9 = 2 innies R46C7 + 9, R7C9 = 9 -> R23C9 = R46C7
3b. R23C9 contains 6 for N3 = {46/56} = 10,11 -> R46C7 = 10,11 = {46/38}
3c. R56C9 = {57} (cannot be {48} which clashes with R46C7), locked for C9 and N6 -> R4C8 = 9, R4C9 = 1 (cage sum), R1C9 = 3, R35C7 = [12], R3C2 = 5 -> R4C2 = 2, R3C1 = 3 (step 1b) -> R4C1 = 7, clean-up: no 9 in R2C4
3d. Naked pair {28} in R89C9, locked for N9
3e. Naked pair {46} in R23C9, 4 locked for N3
3f. R23C9 = {46} = 10 -> R46C7 = 10 = {46}, locked for C7 and N6
3g. Naked pair {38} in R56C8, locked for C8
3h. R4C89 = [91] = 10 -> R3C6 = 9 (step 2c), clean-up: no 3 in R2C4
3i. Naked pair {48} in R23C4, locked for C4 and N2, clean-up: no 8 in R7C3 (step 1d)
3j. 19(3) cage at R1C1 = {469} (only remaining combination), 4,9 locked for N1, 9 locked for R1
3k. 10(3) cage at R1C4 = {127} (only remaining combination), 1,7 locked for N2, 7 locked for R1 -> R1C78 = [85], R3C5 = 6, R23C9 = [64], R23C4 = [48]
3l. Naked pair {27} in R23C8, 7 locked for C8 and N3
3m. R7C359 (step 1c) = 22 = [589], R5C4 = 7, R56C9 = [57]
3n. R6C1 = 5 (hidden single in C1) -> R5C1 + R6C2 = 12 = [48/84/93]
3o. 45 rule on N8 3 remaining innies R789C4 = 18 = {369} (only remaining combination), locked for C4 and N8 -> R4C4 = 5
3p. 35(6) cage at R3C6 = {345689} (only remaining combination), no 1, 3 locked for N5

4a. 45 rule on N7 1 outie R9C4 = 1 remaining innie R8C3 + 2, R9C4 = {369} -> R8C3 = {147}
4b. 21(4) cage at R7C1 = {1389/2469/2478} (cannot be {1479} which clashes with R8C3, cannot be {2379} because 3,7 only in R7C2, cannot be {3468} which clashes with R15C1, ALS block)
4c. 1 of {1389} must be in R7C1 -> no 1 in R7C2 + R89C1
4d. 4,6,7 of {2469/2478} must be in R7C12 -> no 4,6 in R89C1
4e. 16(3) cage at R7C4 = {169/367} (cannot be [349] which clashes with 21(4) cage at R7C1), no 4, 6 locked for C4
4f. Consider placement for 3 in N7
R7C2 = 3 -> 21(4) cage at R7C1 = {1389}
or 3 in R8C2 + R9C23, locked for 21(4) cage at R8C2 => R9C4 = 9 => R8C3 = 7 => 21(4) cage at R7C1 = {2469}
-> 21(4) cage at R7C1 = {1389/2469}, no 7, 9 locked for C1 and N7, clean-up: no 3 in R6C2 (step 3n)
4g. Naked pair {48} in R5C1 + R6C2, locked for N4
4h. 45 rule on R1234 1 outie R5C6 = 1 innie R4C3 = {36}
4i. 16(3) cage at R7C4 = {367} (cannot be [619] which clashes with 21(4) cage at R7C1) -> R8C3 = 7, R9C4 = 9, R2C3 = 2
4j. R89C1 = [92] (hidden pair in C1) -> R7C12 = {46}, locked for R7 and N7
4k. Naked pair {46} in R17C1, locked for C1

and the rest is naked singles.


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