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 Post subject: Assassin 351
PostPosted: Thu May 31, 2018 7:11 pm 
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Posts: 1040
Location: Sydney, Australia
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Assassin 351
A fun one! SS gives it 1.40 JSudoku has a bug (can't see a CPE) so gets stuck.

As Andrew commented to me after the last puzzle, zero cells have appeared in my puzzles. They are a really good tool to try and stiffen up the middle of a puzzle. I don't like making one-trick ponies
code:
3x3::k:0000:8961:8961:3330:3843:3076:3076:1029:1029:0000:0000:8961:3330:3843:3843:6150:6150:6150:3847:3847:8961:8961:8961:8961:6150:4616:4616:3847:5129:5129:5129:5129:5642:5642:5642:4616:5643:2828:5901:5901:5901:5901:5901:5642:4616:5643:2828:2828:2828:6670:6670:6670:6670:4623:5643:5643:5648:7953:7953:7953:7953:4623:4623:5648:5648:5648:1554:1554:2835:7953:0000:0000:2304:2304:0000:0000:1554:2835:7953:7953:0000:
solution:
Code:
+-------+-------+-------+
| 9 7 2 | 6 5 4 | 8 3 1 |
| 3 1 5 | 7 8 2 | 4 6 9 |
| 6 4 8 | 3 9 1 | 5 7 2 |
+-------+-------+-------+
| 5 9 6 | 4 1 8 | 7 2 3 |
| 8 3 4 | 2 7 9 | 1 5 6 |
| 7 2 1 | 5 6 3 | 9 8 4 |
+-------+-------+-------+
| 1 6 3 | 8 4 7 | 2 9 5 |
| 2 8 9 | 1 3 5 | 6 4 7 |
| 4 5 7 | 9 2 6 | 3 1 8 |
+-------+-------+-------+
Cheers
Ed


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 Post subject: Re: Assassin 351
PostPosted: Tue Jun 05, 2018 8:55 pm 
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Posts: 1893
Location: Lethbridge, Alberta, Canada
Thanks Ed for another nice Assassin.
Yes it is a fun one:
once one finds the interesting steps.

Here's how I solved Assassin 351:
Prelims

a) R12C4 = {49/58/67}, no 1,2,3
b) R1C67 = {39/48/57}, no 1,2,6
c) R1C89 = {13}
d) R89C6 = {29/38/47/56}, no 1
e) R9C12 = {18/27/36/45}, no 9
f) 6(3) cage at R8C4 = {123}
g) 11(4) cage at R5C2 = {1235}
h) 26(4) cage at R6C5 = {2789/3689/4589/4679/5678}, no 1
i) 35(7) cage at R1C2 = {1235789/1245689/1345679/2345678}
j) 31(7) cage at R7C4 = {1234579/1234678}

Steps resulting from Prelims
1a. Naked pair {13} in R1C89, locked for R1 and N3, clean-up: no 9 in R1C67
1b. 6(3) cage at R8C4 = {123}, locked for N8, clean-up: no 8,9 in R89C6
1c. 11(4) cage at R5C2 = {1235}, CPE no 1,2,3,5 in R6C1
1d. 35(7) cage at R1C2 must contain 5, CPE no 5 in R3C12
1e. 31(7) cage at R7C4 = {1234579/1234678}, 1,2,3 locked for N9
1f. 31(7) cage at R7C4 = {1234579/1234678}, CPE no 4,7 in R7C89
1g. Min R7C89 = 11 -> max R6C9 = 7
1h. 8,9 in N8 only in R7C456 + R9C4, CPE no 8,9 in R9C78
[Maybe this is the CPE which JSudoku couldn’t find.]

2. 18(3) cage at R6C9 = {459/468} (cannot be {189/279/369/567} which clash with 31(7) cage at R7C4, cannot be {378} because 3,7 only in R6C9) -> R6C9 = 4
2a. 1 in R6 only in R6C234, locked for 11(4) cage at R5C2, no 1 in R5C2
2b. 26(4) cage at R6C5 = {2789/3689/5678}, 8 locked for R6
2c. 45 rule on N8 4 innies R7C456 + R9C4 = 26 = {4789} (cannot be {5689} which clashes with 18(3) cage, ALS block; alternatively that 5,6,8,9 cannot be the only values in R7C45689), locked for N8
2d. 4,7 of {4789} must be in R7C456 (R7C456 cannot contain both of 8,9 which clashes with 18(3) cage), locked for R7, N8 and 31(7) cage at R7C4
2e. 31(7) cage = {1234579/1234678} contains one of 8,9 which must be in R7C456 -> no 8,9 in R78C7
2f. Killer pair 5,6 in R789C7 + R9C8 and 18(3) cage, locked for N9
2g. R8C8 = 4 (hidden single in N9)
2h. 7 in N9 only in R89C9, locked for C9
2i. Naked pair {56} in R89C6, locked for C6, clean-up: no 7 in R1C7
2j. Killer pair 8,9 in 31(7) cage and 18(3) cage, locked for R7

3. 45 rule on N7 3 innies R7C12 + R9C3 = 14 = {158/167/239/257/356} (cannot be {149/248/347} because 4,7,8,9 only in R9C3), no 4
3a. 4 in R9 only in R9C12 = {45}, locked for R9 and N7 -> R89C6 = [56]
3b. Naked triple {123} in R9C578, locked for R9
3c. R7C12 + R9C3 = {167/239}, no 8
3d. R7C12 = {16/23} -> 22(4) cage at R5C1 = [87]{16}/[89]{23} -> R5C1 = 8, R6C1 = {79}
3e. 6 in R6 only in 26(4) cage at R6C5 = {3689/5678}, no 2
3f. 2 in R6 only in R6C234, locked for 11(4) cage at R5C2 -> no 2 in R5C2

4. 45 rule on N3 3 innies R1C7 + R3C89 = 17 = {278/458/467} (cannot be {269} because R1C7 only contains 4,5,8), no 9
4a. 8 of {278} must be in R1C7, 4 of {458/467} must be in R1C7 -> R1C7 = {48}, clean-up: no 7 in R1C6
4b. 7 of {278/467} must be in R3C8 -> no 2,6 in R3C8
4c. Naked pair {48} in R1C67, locked for R1, clean-up: no 5,9 in R2C4
4d. 45 rule on N3 2 outies R45C9 = 1 innie R1C7 + 1, R1C7 = {48} -> R45C9 = 5,9 = {23/36}/[81], no 5,9, no 1 in R4C9

5. 15(3) cage at R1C5 = {159/168/249/258/267/357} (cannot be {348} which clashes with R1C6, cannot be {456} which clashes with R12C4)
5a. 45 rule on N2 4 innies R1C6 + R3C456 = 17 = {1268/1349/1358/1457/2348/2456} (cannot be {1259/1367/2357} because R1C6 only contains 4,8)
5b. Hidden killer triple 1,2,3 in 15(3) cage and R3C456 for N2, 15(3) cage contains one of 1,2,3 -> R3C456 must contain two of 1,2,3 -> R1C6 + R3C456 must contain two of 1,2,3 -> R1C6 + R3C456 = {1268/1349/1358/2348} (cannot be {1457/2456} which only contain one of 1,2,3), no 7
5c. R3C456 “see” all cells in R1 containing 2,5,6,7,9 except for R1C1 -> R3C456 cannot contain more than one of 2,5,6,7,9 -> R1C6 + R3C456 = {1349/1358/2348} (cannot be {1268} which contains both of 2,6), no 6, 3 locked for R3, N2 and 31(7) cage at R1C2
5d. R3C456 contains one of {259} -> R1C1 must contain one of 2,5,9, no 6,7 in R1C1
5e. 35(7) cage contains 3 so must also contain 7 in R1C23 + R2C3, locked for N1
5f. 15(3) cage at R1C5 = {159/249/258/267} (cannot be {168} which clashes with R1C6 + R3C456)

6. 45 rule on R456 4(2+2) remaining innies R46C1 + R45C9 = 21, R45C9 = 5,9 (step 4d) -> R46C1 = 12,16 = [39/57/79/97] -> R4C1 = {357}

7. 45 rule on R5 3 remaining innies R5C289 = 14 = {239/257/356} (cannot be {167} because R5C2 only contains 3,5), no 1, clean-up: no 8 in R4C9 (step 4d)
7a. 7,9 of {239/257} must be in R5C8 -> no 2 in R5C8
7b. R45C9 (step 4d) = {23/36}, 3 locked for C9 and N6 -> R1C89 = [31]

8. 45 rule on R4 2 innies R4C19 = 1 outie R5C8 + 3, R4C9 cannot equal R5C8 (IOU) -> no 3 in R4C1
8a. R4C19 cannot total 12 -> no 9 in R5C8
8b. R5C289 (step 7) = {257/356}, 5 locked for R5
8c. R46C1 (step 6) = [57/79/97], 7 locked for C1 and N4

9. 15(3) cage at R3C1 = {159/258/456} (cannot be {168/249} because 1,2,4,6,8,9 only in R3C12, cannot be {267} = {26}7 which clashes with 35(7) cage at R1C2), no 7
[I ought to have spotted this step sooner; it would also have eliminated 3 from R4C1. Incorrect second comment deleted.]
9a. R4C1 = 5, R5C2 = 3, R9C12 = [45]
9b. Naked pair {12} in R6C12, locked for R6 and N4 -> R6C4 = 5, clean-up: no 8 in R2C4
9c. R4C9 = 3 (hidden single in C9), R5C8 = 5 (hidden single in R5), R6C1 = 7 (hidden single in C1), clean-up: no 9 in R7C9 (step 2)
9d. R5C289 (step 8b) = {356} -> R5C9 = 6, clean-up: no 8 in R7C8 (step 2)
9e. Naked pair {89} in R6C78, locked for R6 and N6 -> R6C56 = [63]
9f. 6 in N2 only in R12C6 = {67}, locked for C6 and N2
9g. R45C9 = [36] = 9 -> R3C89 = 9 = [72]
9h. R45C9 = 9 -> R1C7 = 8 (step 4d), R1C6 = 4
9i. R1C6 + R3C456 (step 5c) = {1349} (only remaining combination), 1,9 locked for R3, N2 and 35(7) cage at R1C2
9j. R34C1 = [65] -> R3C2 = 4 (cage sum), R3C7 = 5, R3C3 = 8, R2C789 = [469]
9k. R56C1 = [87] = 15 -> R7C12 = 7 = [16]
9l. R7C12 = 7 -> R9C3 = 7 (step 3)
9m. Naked pair {25} in R12C3, locked for C3 and N1 -> R78C3 = [39], R7C7 = 2, R9C8 = 1
9n. R45C8 = [25] = 7 -> R4C67 = 15 = [87]
Thanks Ed for your comments and corrections, mostly for my typos.

With hindsight:
an optimised walkthrough would start
35(7) cage at R1C2 and R3C12 “see” all of each other’s cells -> R3C12 = 10 -> R4C1 = 5
31(7) cage at R7C4 and R7C89 “see” all of each other’s cells -> R7C89 = 14 -> R6C9 = 4
Do these steps make it a 'one trick pony'? ;)

Rating Comment:
I'll rate my solving path for A351 at 1.5; I used a "sees all except" step and also an interactive CPE in step 9. I'd also rate the "with hindsight" start at 1.5.


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 Post subject: Re: Assassin 351
PostPosted: Sun Jun 10, 2018 9:50 pm 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
Andrew wrote:
With hindsight:
an optimised walkthrough would start
35(7) cage at R1C2 and R3C12 “see” all of each other’s cells -> R3C12 = 10 -> R4C1 = 5
31(7) cage at R7C4 and R7C89 “see” all of each other’s cells -> R7C89 = 14 -> R6C9 = 4
Love this! More direct than my step 1 & 3.
Andrew wrote:
Do these steps make it a 'one trick pony'? ;)
Could have but didn't feel like that. Even though the rest is technically very simple (and fun!) in my WT, step 8 took a long time to see (unlike steps 1 & 3) and did lots of steps from Andrew's WT on first solve, but of course, are not in this optimised WT. Also, it takes me 21 steps to get to really simple so have to keep thinking for quite a while.

A351 WT
21 steps:
Preliminaries courtesy of SudokuSolver
Cage 4(2) n3 - cells ={13}
Cage 12(2) n23 - cells do not use 126
Cage 13(2) n2 - cells do not use 123
Cage 9(2) n7 - cells do not use 9
Cage 11(2) n8 - cells do not use 1
Cage 6(3) n8 - cells ={123}
Cage 11(4) n45 - cells ={1235}
Cage 26(4) n56 - cells do not use 1

Note: no routine clean-up done unless stated
1. "45" on n9: 4 outies r6c9 + r7c456 - 4 = 3 innies r8c89 + r9c9
1a. since r7c456 see all of n9 apart from those 3 innies, therefore must exactly repeat numbers in both spots -> 3 outies, r7c456 = 3 innies
1b. -> the IOD of 4 must be r6c9 -> r6c9 = 4
1c. -> r7c89 = 14 = {59/68}(no 1,2,3,7)

2. "45" on n8: 4 innies r7c456 + r9c4 = 28
2a. but {5689} clashes with r7c89 (step 1c)
2b. = {4789} only (no 1,2,3,5,6)
2c. -> 11(2)n8 = {56} only: both locked for c6
2d. no 7 in r1c7

3. "45" on n1, 4 outies r3c456 + r4c1 - 5 = 3 innies r1c1 + r2c12
3a. same deal as n9 -> r4c1 = 5

4. 11(4)r5c2 must have 5 -> r6c4 = 5
4a. naked triple {123} in n4: locked for n4

5. 1 in r6 only in r6c23 -> no 1 in r5c2

6. "45" on r6: 1 innie r6c1 - 4 = 1 outie r5c2
6a. r6c1 = (67)

7. 4(2)n3 = {13}: both locked for r1 and n3
7a. no 9 in 12(2)r1c6

Had a lot of trouble seeing this "45"
8. "45" on n34567! 2 innies r1c7 + r9c3 = 15 = [87] only permutation
8a. r1c6 = 4

9. 13(2)n2 = {67} only: both locked for c4 and n2

10. "45" on n2: 3 innies r3c456 = 13 = {139/238}(no 5)
10a. must have 3: locked for n2 and r3 and no 3 in r2c3

11. 35(7)r1c2 must have 3 = {1235789/1345679/2345678}
11a. must have 7 -> r1c2 = 7
11b. must have 5 which is only in r123c3: locked for n1 and c3
11c. r12c4 = [67]

12. 31(7)r7c4 = {1234579/1234678} -> r7c456 = {479/478}
12a. must have 4 & 7: 4 locked for r7 and n8 and both not in r789c7 + r9c8

13. 4 in r9 only in 9(2)n7 = [45] only permutation
13a. r89c6 = [56]

14. "45" on n7: 2 innies r7c12 = 7 = {16} only: both locked for r7, n7 and no 6 in r56c1
14a. r7c12 = 7 -> r56c1 = 15 = [87] only permutation
14b. -> r5c2 = 3 (IODr6=+4)
14c. r6c23 = {12}: 2 locked for r6

15. r7c89 = 14 = {59} only: locked for r7, n9

16. naked triple {478} in r7c456: 8 locked for r7, n8 and no 8 in r9c8
16a. r9c4 = 9

17. hidden single 4 in r8c8 -> r8c8 = 4, -> r8c9 = 7 (hsingle n9) -> r9c9 = 8 (hsingle n9) -> r8c7 = 6 (hsingle n9)

18. "45" on n3: 2 innies r3c89 = 9 = [72] only permutation
18a. -> r45c9 = 9 = [36] only permutation

19. "45" on r5: 1 innie r5c8 = 5

20. r2c1 = 3 (Hsingle n1)

21. r3c456 = 13 = {139} only: locked for n2, r3 and not in r123c3
21a. r3c1 = 6

easy now.
Cheers
Ed


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 Post subject: Re: Assassin 351
PostPosted: Sun Jun 10, 2018 11:55 pm 
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Thanks Ed! Here is my WT.
As Andrew wrote and Ed quoted:
an optimised walkthrough would start
35(7) cage at R1C2 and R3C12 “see” all of each other’s cells -> R3C12 = 10 -> R4C1 = 5
31(7) cage at R7C4 and R7C89 “see” all of each other’s cells -> R7C89 = 14 -> R6C9 = 4
This exact move is one I always look for and find easy to spot. So my wt does (more or less) start with that.

Assassin 351 WT:
1. r7c89 see all cells in 31(7)
-> they must be the two missing values from 31(7)
-> r7c89 = +14(2) = {59} or {68}
-> r6c9 = 4

2. 6(3)n8 = {123}
-> Whichever of (89) is in r7c89 must go in n8 in in r9c4
-> Whichever of (56) is in r7c89 must go in n8 in r89c6
-> 11(2)n8 = {56}
-> r7c456 = {47(8|9)}
-> r8c8 = 4
-> r89c9 = {7(8|9)}

3. r3c12 see all cells in 35(7)
-> they must be the two missing values from 35(7)
-> r3c12 = +10(2)
-> r4c1 = 5
-> 11(4)r5c2 = [{123}5]
-> Remaining innies r6 -> r6c123 = +10(3) - must be from [7{12}] or [6{13}]

This is the breakthrough move.
4! Max r56c1 = +16(2) -> Min r7c12 = +6(2)
-> (123) in r7 - only one can go in r7c12 -> one goes in r7c3 and one in r7c7
-> One of (56) in r7c12
-> Max r7c12 = +9(2)
Innies n7 -> Min r9c3 = 5
-> 4 in n7 only in r9c12
-> 9(2)n7 = [45]

5. -> 6 in r7c12
-> r7c89 = {59}
-> r7c456 = {478}
-> r9c4 = 9
Also 11(2)n8 = [56]
-> r8c7 = 6
Also r89c9 = {78}

6. Also 22(4)r5c1 can only be [87{16}]
-> r6c23 = {12} and r5c2 = 3
Also innie n7 -> r9c3 = 7
-> r89c9 = [78]

7. Remaining innies r456 -> r45c9 = +9(2) = [36]
-> 26(4)r6 = [63{89}]
Also remaining innie r5 -> r5c8 = 5
-> r7c89 = [95]
-> r6c78 = [98]

8. Also 4(2)n3 = [31]
Also r3c89 = +9(2) can only be [72]
-> NS 9 in c9 -> r2c9 = 9
Also -> Remaining innie n3 -> 12(2)r1 = [48]

9. -> 2 in n2 only in 15(3)
-> Remaining innies n2 = r3c456 = +13(3) = {139}
-> H+10(2)r3c12 can only be [64]
-> r7c12 = [16]
-> Remaining innies n4 r4c23,r5c3 can only be [964]

10. -> r4c45 = {14}
-> r4c678 = [872]
-> NS r5c7 = 1
-> r5c456 = {279}

11. Uncaged cells in n1 can only be [931]
HS 7 in n1 -> r1c2 = 7
HS 8 in r3 -> r3c3 = 8
24(4)n3 can only be [4695]
HS 6 in n2 -> 13(2)r1c4 = [67]
-> 15(3)n2 = [{58}2]
-> r12c3 = [25]
-> r12c5 = [58]
-> r7c456 = [847]
-> r5c456 = [279]
etc.


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 Post subject: Re: Assassin 351
PostPosted: Mon Jun 11, 2018 8:03 pm 
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"Zero cells"? Does that simply refer to the cells that are not part of a cage?


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 Post subject: Re: Assassin 351
PostPosted: Mon Jun 11, 2018 8:09 pm 
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azpaull wrote:
"Zero cells"? Does that simply refer to the cells that are not part of a cage?
Correct! I think the term comes from the "0" in the code for those cells.


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