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 Post subject: Assassin 411
PostPosted: Tue Feb 15, 2022 5:41 am 
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An x puzzle. So 1-9 cannot repeat on either diagonal

Assassin 411

A fun one! Mainly because I saw the key step quickly. Have to keep thinking a long way in. This puzzle started after a post on the Players forum by Mathimagics here who worked out that this cage pattern gives the hardest killers. Presumably, nonsymmetric, disjoint and diagonal cages, and zero cells, were not allowed. I had to add diagonals to get JSudoku to be able to generate any valid puzzle. SudokuSolver generally can't solve them so it is indeed, a super difficult pattern because there are so few "45"s available. But this one finally jumped out at 1.40. JSudoku uses 13 advanced steps!
triple click code:
3x3:d:k:4864:4864:4864:4864:8705:8705:8705:8705:4354:4864:6147:6147:6147:6147:8705:6660:4354:4354:5893:5893:5893:6147:8198:8198:6660:5383:4354:5893:6664:8198:8198:8198:6660:6660:5383:4354:5893:6664:6664:8457:1034:6660:5383:5383:7947:4108:6664:8457:8457:5645:5645:5645:5383:7947:4108:6664:8457:5645:5645:6926:7947:7947:7947:4108:4108:8457:8975:6926:6926:6926:6926:3856:4108:8975:8975:8975:8975:3856:3856:3856:3856:
solution:
+-------+-------+-------+
| 5 3 1 | 2 6 4 | 8 9 7 |
| 8 9 6 | 1 5 7 | 4 3 2 |
| 7 4 2 | 3 9 8 | 5 6 1 |
+-------+-------+-------+
| 9 8 5 | 7 3 2 | 6 1 4 |
| 1 6 3 | 5 4 9 | 7 2 8 |
| 2 7 4 | 8 1 6 | 3 5 9 |
+-------+-------+-------+
| 3 2 9 | 4 8 5 | 1 7 6 |
| 4 1 7 | 6 2 3 | 9 8 5 |
| 6 5 8 | 9 7 1 | 2 4 3 |
+-------+-------+-------+
Cheers
Ed


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 Post subject: Re: Assassin 411
PostPosted: Thu Feb 17, 2022 5:29 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
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Thanks Ed for this new Assassin! As you said, a fun one, it just flowed. Surprised that JSudoku found it so hard; I didn't use any advanced steps.

Here's how I solved Assassin 411:
Prelims

a) 34(5) cage at R1C5 = {46789}
b) 17(5) cage at R1C9 = {12347/12356}, no 8,9
c) 33(5) cage at R5C4 = {36789/45789}, no 1,2
d) 16(5) cage at R6C1 = {12346}
e) 35(5) cage at R8C4 = {56789}
f) 15(5) cage at R8C9 = {12345}
[I overlooked 32(5) cage at R3C5 = {26789/35789/45689}, no 1 so didn’t use it.]

1a. R5C5 = 4, placed for both diagonals
1b. 34(5) cage at R1C5 = {46789}, CPE -> R1C4 = {1235}
1c. 16(5) cage at R6C1 = {12346}, 4 locked for C1
1d. 35(5) cage at R8C4 = {56789}, CPE -> R9C6 = {1234}
1e. 15(5) cage at R8C9 = {12345}, 5 locked for N9
1f. 4 in R9 only in R9C6789, locked for 15(5) cage, no 4 in R8C9
1g. 8,9 in C9 only in R567C9, locked for 31(5) cage at R5C9, no 8,9 in R7C78
1h. 7,8,9 in R9 only in R9C2345 -> R8C4 = {56}
1i. Caged X-Wing for 5 in 35(5) cage and 15(5) cage, no other 5 in R8

2a. 17(5) cage at R1C9 = {12347} (cannot be {12356} which clashes with 15(5) cage at R8C9, ALS block), 4 locked for C9
2b. R567C9 = {689} (hidden triple in C9) = 23 -> R7C78 = 8 = {17}, locked for R7 and N9
2c. Naked quad {2345} in R8C9 + R9C789, 2,3,4 locked for N9 and 15(5) cage -> R9C6 = 1
2d. Naked triple {689} in R7C9 + R8C78, CPE no 6,8,9 in R7C6
2e. 5 in R1 only in R1C1234, locked for 19(5) cage at R1C1, no 5 in R2C1
2f. 19(5) cage = {12358/12457/13456}, no 9
2g. 6,7 of {12457/13456} must be in R2C1 (R1C123 cannot contain {46/47} which clash with 34(5) cage at R1C5, ALS block), no 6,7 in R1C123
2h. 6,9 in R1 only in R1C5678, locked for 34(5) cage, no 6,9 in R2C6
2i. 9 in C1 only in R345C1, locked for 23(5) cage at R3C1, no 9 in R3C23
2j. 1,7 in C9 only in R1234C9, locked for 17(5) cage, no 1,7 in R2C8
2k. 5 in C9 only in R89C9, locked for N9
2l. 1 in R8 only in R8C12, locked for 16(5) cage at R6C1, no 1 in R6C1
2m. 9 in N1 only in R2C23 + R3C1, CPE no 9 in R3C4

3a. 27(5) cage at R7C6 = {23589/23679/24678/34569} (cannot be {24579/34578} because R8C78 must contain two of 6,8,9)
3b. R8C78 = {689} -> no 6,8,9 in R8C56 (or R7C6 but already eliminated there by CPE)
3c. Variable killer pair 7,8 in R8C3 and R8C5678 for R8, R8C3 cannot contain both of 7,8 -> 27(5) cage must contain at least one of 7,8 = {23589/23679/24678}, 2 locked for N8

4a. 19(5) cage (step 2f) = {12358/12457/13456}
4b. Hidden killer triple for 7,8,9 in 19(5) cage and 23(5) cage at R3C1 for C1, 23 (5) cage can only contain two of 7,8,9 -> 19(5) cage must contain one of 7,8 in R12C1 = {12358/12457}, no 6, no 8 in R1C23
4c. 23(5) cage contains 9 and one of 7,8 = {12389/12479}, no 5,6
4d. Two of 7,8,9 must be in R345C1 -> no 7,8 in R3C23
4e. R1C1 = 5 (hidden single in C1), placed for D\
4f. Naked triple {234} in R9C789, 2,3 locked for R9 and N9 -> R9C1 = 6, placed for D/, R8C49 = [65]
4g. Naked pair {89} in R8C78, locked for R8 and N9 -> R7C9 = 6
4h. Naked pair {89} in R56C9, locked for N6
4i. R8C78 = {89} = 17 -> R7C6 + R8C56 = 10 = {235} (only remaining combination) -> R7C6 = 5, R8C78 = {23}, locked for R8 and N8
4j. R8C123 = [417], 1 placed for D/ -> R67C1 = {23}, locked for C1
4k. 1 in R1 only in R1C34, locked for 19(5) cage, no 1 in R2C1
4l. 1 in C1 only in R345C1, locked for 23(5) cage, no 1 in R3C3
4m. 23(5) cage = {12389/12479}, 2 locked for R3 and N1
4n. 19(5) cage = {12358/12457} -> R1C34 = [12]
4o. Naked triple {234} in R1C2 + R3C23, 3,4 locked for N1, 4 locked for C2
4p. R7C4 = 4 (hidden single in N8)
4q. Naked pair {23} in R3C3 + R9C9, locked for D\
4r. Naked pair {23} in R3C3 + R9C9, CPE no 3 in R3C9
4s. 8 in R1 only in R1C5678, locked for 34(5) cage, no 8 in R2C6
4t. 33(5) cage at R5C4 = {36789/45789} -> R6C3 = {46}

5a. 45 rule on R1 2 outies R2C16 = 1 innie R1C9 + 8
5b. R1C9 = {37} -> R2C16 = 11,15 = [74/87], 7 locked for R2
5c. 24(5) cage at R2C2 = {13569} (only possible combination, cannot be {13578} because R2C23 must contain two of 6,8,9) -> R2C23 = {69}, locked for R2, 9 locked for N1, R2C45 + R3C4 = {135}, locked for N2
5d. R45C1 = {19} (hidden pair in C1), 9 locked for N4
5e. 26(5) cage at R4C2 = {23678} (only possible combination, cannot be {23579} = {2357}9 which clashes with R6C1), no 5,9
5f. R46C3 = [54] (hidden pair in N4) -> R9C2 = 5 (hidden single in N7)
5g. 33(5) cage at R5C4 (step 4t) = {45789}, no 3,6, 5 locked for C4 and N5
5h. Naked pair {13} in R23C4, locked for N2, 1 locked for C4 -> R2C5 = 5
5i. R7C7 = 1 (hidden single on D\) -> R7C8 = 7
5j. 7 on D\ only in R4C4 + R6C6, locked for N5
5k. Naked pair {89} in R79C3, locked for C3 and N7
5l. R2C23 = [96], 9 placed for D\ -> R8C78 = [98], 8 placed for D\
5m. R4C34 = [57] = 12 -> R3C56 + R4C5 = 20 = {389} -> R3C56 = {89}, locked for R3 and N2, R4C5 = 3

6a. 17(5) cage at R1C9 = {12347/12356}, 3 locked for N3
6b. R3C1 = 7 -> R3C7 = 5, placed for D/
6c. Naked pair {89} in R6C4 + R7C3, locked for D/ and 33(5) cage at R5C4 -> R5C4 = 5, R4C6 = 2, placed for D/
6d. R3C7 = 5, R4C6 = 2 -> 26(5) cage at R2C7 = {24569} (only remaining combination) -> R2C7 = 4, R4C7 = 6, R5C6 = 9

and the rest is naked singles, without using the diagonals.


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 Post subject: Re: Assassin 411
PostPosted: Tue Feb 22, 2022 8:01 pm 
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 280
Location: California, out of London
Thanks Ed. A fun puzzle indeed. I found it very straightforward, so almost a "Human Solvable" since the programs had difficulty.
Assassin 411 WT:
1. 15(5)r8c9 = {12345}
35(5)r8c4 = {56789}
16(5)r6c1 = {12346}
r5c5 = 4
-> 4 in r9 in r9c678

2. 17(5)r1c9 = {123(47)|(56)}
-> At least two of (123) in c9 in r1234
-> At most one of (123) in r89c9
-> 5 and one of (123) in r89c9
Also whichever of (123) is in r89c9 is also in r2c8
-> 17(5)r1c9 = {12347} with (47) in c9
Whichever of (123) in r89c9 and r2c8 can only be in r9 in 15(5)r8c9
-> r9c9 from (123)
-> r8c9 = 5

3. (HQ) r9c2345 = {5789}
-> r8c4 = 6
-> r9c1 = 6
-> 6 in n9 in r7c789

4. r123489c9 = {123457}
Since 6 in r7c789 -> NP r56c9 = {89}
-> NS r7c9 = 6
-> r7c78 = {17}
-> r9c6 = 1 and r9c789 = {234} with 4 in r9c78
-> r8c78 = {89}
-> 27(5)r7c6 = [5{23}{89}]
-> (HS 4 in n8) r7c4 = 4

5! Outies c1 = r1c234 + r3c23 + r8c2 = +13(6)
Since r13c23 is Min +10(4) -> r1c4 + r8c2 is Max +3(2)
-> (Since 2 already in r8) r8c2 = 1
-> r13c23 is Max +11(4) = {1234} or {1235}
-> 1 in n1 in r13c3
-> 1 in c1 in r45c1
-> r1c3 = 1
-> r1c4 = 2
-> r13c23 = {1234} with 1 in r1c3, 2 in r3c23, 4 in r13c2

6. 34(5)r1c5 = {46789}
-> (HS 5 in r1) r1c1 = 5

7. 7 in n8 only in r9c45
-> (HS 7 in n7) r8c3 = 7
-> (HS 4 in n7) r8c1 = 4
-> r67c1 = {23}

8. 6 in n1 only in r2c23
4 in n2 in r123c6
-> 24(5)r2c2 can only be [{69}{135}]
-> r45c1 = {19}
-> r23c1 = {78}

9. r3c3,r9c9 = {23}
Whichever of (23) is in r9c9 also goes in r2c8 -> r3c3,r2c8 = {23}
-> Neither of (23) in r7c3
-> (HP in n7) r7c12 = {23}

10. Since 32(5)r3c5 cannot contain a 1 -> (HS 1 in D\) r7c7 = 1
-> r7c8 = 7

11. 4 in c3 in r46c3
-> Since r6c1 = r7c2 (2 or 3) -> (IOD n4) r46c3 = +9(2) = {45}
-> 33(5)r5c4 = {45789}
-> (since 4 already in n5) r46c3 = [54]
-> 5 in r56c4
-> r2c5 = 5 and r23c4 = {13}

12. 26(5)r4c2 must be {23678}
Since r137c2 = {234} -> r456c2 = {678}
-> r2c23 = [96]
-> r8c78 = [98]

13. (HS 6 in D\)r6c6 = 6
-> (NS in D\) r4c4 = 7
-> 32(5)r3c5 = [{98}573]
-> r1c78 = [89]
-> r2c1 = 8
-> r1c2 = 3
-> r3c2 = 4
-> r3c3 = 2
-> r9c9 = 3
-> r2c8 = 3
-> r23c4 = [13]
etc.


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 Post subject: Re: Assassin 411
PostPosted: Fri Feb 25, 2022 8:39 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
Fun walkthroughs guys! Andrew's 2a is really well spotted. Really enjoyed wellbeback's step 2. line 5, and start of 11. I found a version of his ! step 5, so of course, loved that too, as well as his 9 (my step 12). I managed to find a way to get his step 5 in at the start of my WT. [Big thanks to Andrew for checking my WT]

Andrew only used one 45 Rule, wellbeback two, I found three!
a411 WT:
Preliminaries
Cage 4(1) n5 - cells ={4}
Last cell in cage 4(1) n5 at r5c5 set to 4
Cage 15(5) n89 - cells ={12345}
Cage 16(5) n47 - cells ={12346}
Cage 34(5) n23 - cells ={46789}
Cage 35(5) n78 - cells ={56789}
Cage 17(5) n36 - cells do not use 89
Cage 33(5) n457 - cells do not use 12
Cage 32(5) n245 - cells do not use 1

1. r5c5 = 4: locked for both Ds

Key step
2. "45" on c1: 6 outies r1c4 + r13c23 + r8c2 = 13
2a. can't have three 1s in those 3 places since that would leave no 1 for c1
2b. min. r13c23 = {1234} = 10
2c. must have 1 since {2345} > 13
2d. -> r1c4 + r8c2 can't be [11]
2e. -> r1c4 + r8c2 = 3 = {12} only: -> no 1,2 in r1c2 or r1c9
2f. -> r13c23 = 10 = {1234}: all locked for n1
2g. -> r1c34 = (12) (hidden pair r1)
2h. 4 locked for c2

3. 5 in r1 only in r1c19 -> no 5 in r9c9

4. 17(5)r1c9 = 123{47/56}
4a. can't have all of 1,2,3 in c9 since r9c9 = (123)
4b. -> r2c8 from (123)
4d. and killer triple 1,2,3 with r9c9: all locked for c9

5. 15(5)r8c9 = {12345} with 1,2,3 only in r9: all locked for r9
5a. -> r9c1 = 6, placed for d/
5b. 35(5)r8c4 must have 6 -> r8c4 = 6
5c. must also have 5: locked for r9
5d. -> 5 in 15(5)r8c9 must be in r8c9
5e. -> r1c1 = 5 (hsingle r1), place for d\

6. 31(5)r5c9 = {16789/34789}(no 2)
6a. 1,2,3 in r2c8 see all 1,2,3 in c9 apart from r9c9 so must repeat there
6b. "45" on c9: 3 outies r2c8 + r7c78 = 1 innie r9c9 + 8
6c. -> r7c78 = 8
6d. = {17} only: locked for r7, n9 and 7 for 31(5)
6e. -> 31(5) = {16789} -> r567c9 = {689}: 6 locked for c9
6f. no 1 in r2c8 (since = r9c9)

7. r9c789 = {234}: all locked for r9 and n9
7a. r9c6 = 1
7b. r6c5 = 1 (hsingle n5)

8. r8c78 = {89}: both locked for r8, n9 and 27(5)
8a. r7c9 = 6
8b. r56c9 = {89}: both locked for n6

9. 1 on d\ in r3c3 or r7c7 -> no 1 r3c7 (Common Peer Elimination CPE)
9a. -> r8c2 = 1 (hsingle d/}
9b. -> r1c4 = 2 (step 2e)
9c. r1c3 = 1
9d. -> r7c7 = 1 (hsingle d\)
9e. r7c8 = 7

10. r8c78 = {89} = 17 -> r7c6 + r8c56 = 10 = [5]{23}: 2 & 3 locked for r8, n8
10a. r8c13 = [47]
10b. r7c4 = 4 (hsingle n8)

11. 24(5)r2c2: must have 6 for n1 = {69}{135}(only valid combination): 6 locked for r2, 9 for r2 and n1; 3,5 for n2

12. naked pair {23} on d\ in r3c3 and r9c9: both locked for d\ and no 2,3 in r3c9 (CPE)
12a. since r2c8 = r9c9 -> r3c3 + r2c8 form a naked pair
12b. -> no 2,3 in r3c78; no 3 in r7c3

13. naked pair {89} in r7c35: both locked for r7

14. naked triple {234} in r137c2: all locked for c2

15. hidden pair {19} in r45c1 for c1: 9 locked for n4

16. 25(5)r4c2 = {23678} only
16a. -> r5c3 = (23)
16b. 6,8 locked for c2, 6 & 8 for n4

17. -> r2c2 = 9 (placed for d\), r2c3 = 6, r9c2 = 5, r8c8 = 8 (placed for d\), r4c4 = 7, r6c6 = 6

18. naked pair {23} in r6c1 + r5c3: both locked for n4

19. r9c34 = {89} -> r9c5 = 7 (cage sum)

3rd "45"
20. "45" on c1234: 1 remaining outie r2c5 = 1 remaining innie r4c3
20a. -> both are 5 (only common digit)

21. 32(5)r3c5 = {89}[573](only permutations)
21a. 8 & 9 locked for r3 and n2

22. r12c6 = {47} -> no 4,7 in r1c78

23. r23c1 = [87]
23a. -> r1c2 = 3 (cage sum)

24. r3c7 = 5 (placed for d/)

25. r6c4 + r7c3 = {89}: both locked for d/ and 33(5)

much easier now
Cheers
Ed


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 Post subject: Re: Assassin 411
PostPosted: Sat Feb 26, 2022 11:08 pm 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
My solving path was very different from those of wellbeback and Ed, because I found a different early breakthrough step. They both found something else in their steps 5 and 2, respectively. In addition to the steps commented on by Ed, I particularly liked the later part of wellbeback's step 2.
Hidden Text:
I had spotted their 45 on C1 but, since it contained 6 cells I hadn't looked at how much their total was.


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