SudokuSolver Forum

A forum for Sudoku enthusiasts to share puzzles, techniques and software
It is currently Thu Mar 28, 2024 8:30 am

All times are UTC




Post new topic Reply to topic  [ 4 posts ] 
Author Message
 Post subject: Assassin 381
PostPosted: Thu Aug 01, 2019 7:11 pm 
Offline
Grand Master
Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
Attachment:
a381.JPG
a381.JPG [ 61.99 KiB | Viewed 6183 times ]
Assassin 381
Started as a 1.60. Couldn't solve it. Changed it to a 1.15. Had some trouble solving it but ended up with a one-trick pony. Adapted it to another 1.60 puzzle but this time ended up with an Assassin. A couple of really nice steps too. JSudoku needs one chain.
Another puzzle that resists to the end. Enjoy the challenge!
code:
3x3::k:4352:6913:6913:6913:6913:7426:7426:7426:7426:4352:6913:6915:2820:6913:2053:2053:7426:2566:3591:6915:6915:2820:3336:5385:5385:2566:2566:3591:3591:6915:3082:3336:3336:5385:0000:0000:0000:6915:6915:3082:6413:6413:5385:0000:0000:0000:3084:3084:3084:6413:6413:5385:0000:0000:4107:4107:4107:1809:4114:4114:4114:1555:3860:4623:2830:2830:1809:5910:1552:1552:1555:3860:4623:4623:4623:5910:5910:5910:4373:4373:4373:
solution:
Code:
+-------+-------+-------+
| 9 1 2 | 8 5 7 | 4 6 3 |
| 8 7 6 | 2 4 3 | 5 9 1 |
| 3 5 4 | 9 6 1 | 8 7 2 |
+-------+-------+-------+
| 2 9 8 | 5 3 4 | 7 1 6 |
| 4 3 1 | 7 8 6 | 2 5 9 |
| 7 6 5 | 1 9 2 | 3 8 4 |
+-------+-------+-------+
| 5 8 3 | 4 1 9 | 6 2 7 |
| 6 2 9 | 3 7 5 | 1 4 8 |
| 1 4 7 | 6 2 8 | 9 3 5 |
+-------+-------+-------+
Cheers
Ed


Top
 Profile  
Reply with quote  
 Post subject: Re: Assassin 381
PostPosted: Tue Aug 06, 2019 12:29 am 
Offline
Grand Master
Grand Master

Joined: Tue Jun 16, 2009 9:31 pm
Posts: 280
Location: California, out of London
One small chain at the start and a couple of nice and interesting moves later on for me.
Thanks Ed!
Further edits thanks to Andrew. Thanks Andrew!
Assassin 381 WT:
1. Innies n9 = r78c7 = +7(2).
6(2)n9 and 6(2)r8 means r78c7 can only be [61] or [34]
-> r8c6 from (25)
-> 7(2)n8 from {16} or {34}

2. Outies r89 = r7c489 = +13(3)
Since r7c9 is Min 6 -> Max r7c4 = 5
-> 7(2)n8 from [16] or {34}

3. Whichever of (25) is in r8c6 must go in n9 in r9c789
-> Goes in n7 in r7c123
-> 1 not in r7c123
Outies r9 = r8c15 = +13(3) (No 123)
-> 1 in n7 in r9c123
-> 1 in n8 in r7c456
-> 1 in n9 in r8c78
-> 6(2)n9 from {24} or [51]

4. Trying r78c7 = [34] and 6(2)n9 = [51] ...
... puts r8c6 = 2, 7(2)n8 = [16] (HS 1 in r7), 15(2)n9 = [78] (outies r89 = +13(3))
which leaves no solution for 11(2)r8.
Edit: Maybe an easier way to do that is:
Innies r89 = r8c489 = +15(3)
If r8c8 was 1 this puts r8c49 = [68] which leaves no place for 3 in r8.

-> r78c7 = [61]
-> 6(2)n9 = {24}
-> 15(2)n9 = {78}
-> 17(3)n9 = {359}
Also r8c6 = 5

5! Used in many steps below
21(5)r3c6 either contains a 1 in r3c6 or is {23457} with 5 in c7
-> r3c6 is max 7.

6. 17(2)n1 = {89}
-> (89) in r3 only in r3c457
Innies n5 = +8(3) = {125} or {134}
r3c5 from (6789)

For r3c5 from (67) -> r3c47 = {89}
a) r3c5 = 6 -> r6c4 = 1 -> r78c4 = {34} -> 12(2)n5 = {57}
b) r3c5 = 7 -> r6c4 = 2 -> r23c4 = [38] -> 12(2)n5 = {57}

For r3c5 from (89) -> r6c4 from (34) -> Innies n5 = {134} -> 12(2)n5 = {57}

In all cases 12(2)n5 = {57}

-> Innies n5 = {134}
-> 25(4)n5 = {2689}
Also r3c5 from (689)

7. 9 in c4 only in r123c4 (n2)
-> r3c5,r6c4 from [61] or [83]
-> (13) in c4 locked in r678c3
Innies r12 = r2c349 = +9(3) (No 789)
-> 11(2)n4 = [29]

8! 8(2)r2 = [35] or [17]

(A) Consider 8(2)r2 = [35]
-> No 5 in 21(5) -> 21(5) must have 1 in r3c6 (Step 5)
Also -> 10(3) cannot be {235} -> 10(3)n3 must have a 1

(B) Consider 8(2)r2 = [17]
-> 21(5) = [7{2345}]
Cannot have 4 in r3c7 since that would put a second 5 in c6 in r1c6 (IOD n3)
-> r3c7 from {235}
-> 10(3)n3 must have a 1

In both cases 1 is in r2 or r3 in both n2 and n3
-> 1 in r1 in r1c23

-> 27(6)n12 (which cannot have a 9) must have an 8 in n2.
-> HS 8 in n3/r3 -> r3c7 = 8
-> r3c6 = 1
Also -> r3c5 = 6
-> r6c4 = 1
-> 7(2)n8 = {34}
etc.
Easy from here


Top
 Profile  
Reply with quote  
 Post subject: Re: Assassin 381
PostPosted: Tue Aug 06, 2019 8:25 pm 
Offline
Grand Master
Grand Master

Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1893
Location: Lethbridge, Alberta, Canada
Thanks Ed. Another interesting and challenging Assassin!

wellbeback and I used one key point but otherwise solved it very differently.

Thanks Ed for your comments and typo corrections.
Here is my walkthrough for Assassin 381:
Prelims

a) R12C1 = {89}
b) R23C4 = {29/38/47/56}, no 1
c) R2C67 = {17/26/35}, no 4,8,9
d) R45C4 = {39/48/57}, no 1,2,6
e) R78C4 = {16/25/34}, no 7,8,9
f) R78C8 = {15/24}
g) R78C9 = {69/78}
h) R8C23 = {29/38/47/56}, no 1
i) R8C67 = {15/24}
j) 10(3) cage at R2C9 = {127/136/145/235}, no 8,9

1a. Naked pair {89} in R12C1, locked for C1 and N1
1b. 45 rule on R1 4 outies R2C1258 = 28 = {4789/5689}, no 1,2,3, 8,9 locked for R2, clean-up: no 2,3 in R3C4
1c. 45 rule on R12 3 innies R2C349 = 9 = {126/135/234}, no 7, clean-up: no 4 in R3C4

2a. 45 rule on N9 2 innies R78C7 = 7 = {16/34} (cannot be {25} which clashes with R78C8) = [34/61], clean-up: no 1,4 in R8C6
2b. Killer pair 1,4 in R78C7 and R78C8, locked for N9
2c. R78C4 = {16/34} (cannot be {25} which clashes with R8C6), no 2,5
2d. 45 rule on R9 2 outies R8C15 = 13 = [49/58]/{67}, no 1,2,3, no 4,5 in R8C5
2e. Combined cage R8C15 + R8C67 = [49][51]/[58][24]/{67}[24]/{67}[51]
2f. R8C23 = {29/38} (cannot be {47/56} which clash with combined cage)
2g. 16(3) cage at R7C1 = {169/178/259/358/457} (cannot be {268/349} which clash with R8C23, cannot be {367} which clashes with R7C7)

3a. 45 rule on N5 3 innies R4C56 + R6C4 = 8 = {125/134}, 1 locked for N5
3b. 45 rule on N5 1 outie R3C5 = 1 innie R6C4 + 5 -> R3C6 = {6789}, R6C4 {1234}
3c. Combined cage R45C4 = {39/48/57} and R4C56 + R6C4 = {125/134} must contain 5, locked for N5
3d. 45 rule on C4 3 innies R169C4 = 15
3e. Max R6C4 = 4 -> min R19C4 = 11, no 1 in R19C4
3f. Hidden killer triple 7,8,9 in R169C4, R3C4 and R45C4 for C4, R169C4 = 15 cannot contain more than one of 7,8,9, R45C4 contains one of 7,8,9 -> R169C4 must contain one of 7,8,9 and R3C4 = {789}, clean-up: no 5,6 in R2C4

4. 45 rule on N8 3 innies R7C56 + R8C6 = 15 = {159/249/258/267/357} (cannot be {168/348} because R8C6 only contains 2,5, cannot be {456} which clashes with R78C4)

5. 45 rule on R89 2 innies R8C48 = 1 outie R7C9
5a. R7C9 = {6789} -> R8C48 = 6,7,8,9 = [42/34/61/35/62] (cannot be [15] because R78C4 = [61] clashes with R7C9 = 6, cannot be [45] which clashes with R8C67), no 1 in R8C4, clean-up: no 6 in R7C4
5b. R8C48 = 6,7,8 -> R7C9 = {678}, clean-up: no 6 in R8C9
5c. 1 in R8 only in R8C78, locked for N9, clean-up: no 5 in R8C8
5d. 45 rule on R89 3 innies R8C489 = 15 and doesn’t contain 5 = {168/249/267/348}
5e. Consider placements for 1 in R8
R8C7 = 1 => R7C7 = 6 (step 2a) => no 9 in R8C9
or R8C8 = 1 => R8C489 = {168}
-> no 9 in R8C9, clean-up: no 6 in R7C9
[After going through Ed’s walkthrough I realised that if I’d used R7C489 = 13 sooner, to eliminate 6 from R7C9, this short forcing chain would have been avoided.]
5f. R8C489 = {168/267/348}
5g. 3 of {348} must be in R8C4 -> no 4 in R8C4, clean-up: no 3 in R7C4
5h. Naked pair {78} in R78C9, locked for C9 and N9
5i. 16(3) cage at R7C1 (step 2g) = {169/178/259/358/457}
5j. 45 rule on R89 3 outies R7C489 = 13 = {148/247} (cannot be {157} which clashes with 16(3) cage at R6C1), no 5, 4 locked for R8, clean-up: no 1 in R8C8
5k. Naked pair {24} in R78C8, locked for C8 and N9, R8C7 = 1 -> R8C6 = 5, R7C7 = 6 (step 2a), clean-up: no 2,7 in R2C6, no 3 in R2C7, no 8 in R8C5 (step 2d)
5l. Naked triple {359} in 17(3) cage at R9C7, locked for R9
5m. 5 in N7 only in 16(3) cage at R7C1 = {259/358}, no 1,7
5n. 18(4) cage in N7 = {1467} (hidden quad in N7), 1 locked for R9
5o. 2,8 in R9 only in R9C456, locked for N8

6. Consider placement for 1 in R67C4
R6C4 = 1 => R3C5 = 6 (step 3b)
or R7C4 = 1 => R8C4 = 6
-> no 6 in R8C5, clean-up: no 7 in R8C1 (step 2d)
[Ed pointed out that I’d overlooked further eliminations. I’d missed that it also removes 6 from R9C5 and from R1C4, which would have simplified step 7f.]
6a. 7 in N7 only in R9C123, locked for R9

7. 5 in C4 only in R1C4 or in R45C4
7a. R169C4 = 15 (step 3d) cannot contain 5 (cannot be {159/357} because R9C4 only contains even numbers, cannot be {456} which clashes with R78C4, cannot be {258} = [528] which, combined with R3C5 + R6C4 (step 3b) = [72] clashes with R23C4), no 5 in R1C4
7b. 5 in C4 only in R45C4 = {57}, locked for N5, 7 also locked for C4, clean-up: no 4 in R2C4
7c. R4C56 + R6C4 (step 3a) = {134} (only remaining combination), 3,4 locked for N5, clean-up: no 7 in R3C5 (step 3b)
7d. R169C4 = {168/348} (cannot be {249} = [942] which clashes with R3C5 + R6C4 (step 3b) = [94]), no 2,9, 8 locked for C4
7e. R3C4 = 9 -> R2C4 = 2, clean-up: no 6 in R2C6, no 4 in R6C4 (step 3b)
7f. R169C4 = {168/348} = [816/438] (cannot be [618] which clashes with R3C5 + R6C4 (step 3b) = [61], cannot be [834] which clashes with R3C6 + R6C4 (step 3b) = [83]) -> R1C4 = {48}, R9C4 = {68}
7g. Combined cage R169C4 + R3C5 (using step 3b) = [816]6/[438]8 -> 8 in R1C4 + R3C5, locked for N2
7h. 4 in N5 only in R4C56, locked for R4
7i. 6 in C4 only in R89C4, locked for N8
7j. R2C349 (step 1c) contains 2 = {126/234}, no 5

8a. R2C18 = {89} (hidden pair in R2)
8b. 5 in N2 only in R12C5, locked for 27(6) cage at R1C2
8c. R2C1258 (step 1b) = {4789/5689} -> R2C25 = [47/65/74], no 6 in R2C5
8d. 27(6) cage = {124578/134568/234567}
8e. 5,6 of {134568} must be in R2C25, 5,6 of {234567} must be in R2C25 (because 4 of this combination is in R1C4) -> no 6 in R1C235
8f. 6 in R1 only in 29(5) cage at R1C6
8g. 9 in R12 only in R12C1 and 29(5) cage -> 29(5) cage must contain 9 = {15689/25679/34679} (cannot be {24689} which clashes with R12C1 + R1C4, Killer ALS block)
8h. 5 in N1 only in R3C123, locked for R3
8i. 10(3) cage at R2C9 = {127/136}, no 4, 1 locked for N3
8j. 29(5) cage = {25679/34679} (cannot be {15689} = 1{5689} which clashes with R2C67), no 1,8, 7 locked for R1
[Cracked. The rest is fairly straightforward.]
8k. R2C8 = 9 -> R12C1 = [98]
8l. R3C7 = 8 (hidden single in N3), R3C5 = 6 -> R4C56 = 7 = {34}, 3 locked for R4 and N5, R6C4 = 1, R7C4 = 4 -> R8C4 = 3, R1C4 = 8, R9C4 = 6, R78C8 = [24], R8C1 = 6 -> R8C5 =7 (step 2d), R78C9 = [78]
8m. Naked pair {29} in R8C23, 9 locked for N7

9. 45 rule on N3 2 outies R12C6 = 10 = [73] -> R2C7 = 5, R4C56 = [34], R3C6 = 1, R7C56 = [19], R12C5 = [54]
9a. R2C2 = 7 (hidden single in R2)
9b. R1C45 = [85] + R2C25 = [74] = 24 -> R1C23 = 3 = {12}, locked for N1, 2 locked for R1
9c. R2C39 = [61]
9d. R3C89 = [72] (hidden pair in R3)
9e. R3C67 = [18] = 9 -> R456C7 = 12 = {237} (only remaining combination), 3 locked for C7 and N6

10. 2 in C1 only in R456C1, locked for N4
10a. 27(6) at R2C6 = {134568} (only remaining combination), no 7,9, 1,8 locked for N4
10b. R6C4 = 1 -> R6C23 = 11 = [47/65]
10c. R4C2 = 9 (hidden single in N4) -> R34C1 = 5 = [32], R7C1 = 5
10d. Naked pair {47} in R56C1, locked for C1 and N4

and the rest is naked singles.

Rating Comment:
I think I'll give A381 a full 1.5 for steps 7a and 7f, although SudokuSolver might give those steps a lower rating. I also used a couple of short forcing chains so definitely in the 1.5 range.


Top
 Profile  
Reply with quote  
 Post subject: Re: Assassin 381
PostPosted: Sun Aug 11, 2019 6:54 am 
Offline
Grand Master
Grand Master

Joined: Wed Apr 16, 2008 1:16 am
Posts: 1040
Location: Sydney, Australia
Andrew wrote:
Another interesting and challenging Assassin!
Thanks! A perfect Assassin then....what I aim for every time.

I messed up the ending a bit, see note at step 30. The rest of my WT has slight bits of both the other WTs but is mostly entirely different. Love how we can have such variety. Thanks to Andrew for some good brevity suggestions.
a381 WT:
Preliminaries
Cage 17(2) n1 - cells ={89}
Cage 6(2) n9 - cells only uses 1245
Cage 6(2) n89 - cells only uses 1245
Cage 15(2) n9 - cells only uses 6789
Cage 7(2) n8 - cells do not use 789
Cage 8(2) n23 - cells do not use 489
Cage 12(2) n5 - cells do not use 126
Cage 11(2) n7 - cells do not use 1
Cage 11(2) n2 - cells do not use 1
Cage 10(3) n3 - cells do not use 89

1. "45" on n9: 2 innies r78c7 = 7
1a. but {25} blocked by 6(2)n9 = {15/24} = 2 or 5
1b. -> r78c7 = [61/34]
1c. r8c6 = (25)

2. 7(2)n8: {25} blocked by r8c6 = (25)
2a. = {16/34}(no 2,5)

This was the key step for the v1.15 (unpublished) puzzle.
3. "45" on r89: 3 outies r7c489 = 13
3a. the only combinations with 6 are {346} which is blocked by r7c7 = (36)
3b. or {256} which must have 6 in r7c9 but there is no 2,5 for r7c4
3c. -> no 6 in r7c49
3d. no 9 in r8c9, no 1 in r8c4

4. "45" on r9: 2 outies r8c15 = 13 = {49/58/67}(no 1,2,3)

5. 9 in r8 in h13(2) = {49} or in 11(2)n7 = {29}
5a. -> must have 2 or 4
5b. -> [24] blocked from 6(2)r8c6
5c. -> r8c67 = [51], r7c7 = 6 (h7(2))
5d. no 8 in r8c15 (h13(2))
5e. no 6 in 11(2)n7
5f. no 3 in r2c7
5g. no 2,7 in r2c6

6. 6(2)n9 = {24} only: both locked for c8 and n9

7. 15(2)n9 = {78} only: both locked for c9 and n9

8. naked triple {359} in n9: all locked for r9

9. 17(2)n1 in {89}: both locked for c1 and n1
9a. no 4 in r8c5 (h13(2))

Lots of combos here but is a crucial step.
10. 21(5)r3c6: combos with 9 are {12369}, but 1 & 6 are only in r3c6 -> blocked
10a. and {12459}: but it is [1]{2459} only, which blocks all permutations for the 8(2)r2c67 = [17/35/62] -> blocked
10b. -> no 9 in 21(5)

11. 9 in r3 only in r3c45: 9 locked for n2

12. "45" on n5: 1 outie r3c5 - 5 = 1 innie r6c4
12a. r3c5 =(6789), r6c4 = (1234)

13. 9 in r3 in 11(2)r2c4 = [29] or in outie/innie 'cage' n5 = [94]
13a. -> r2c4 = 2 or r6c4 = 4
13b. -> {47} blocked from 11(2)r2c4
13c. {56} blocked from 11(2)r2c4 since {56}[4] in r236c4 clashes with 7(2)r7c4 = 4 or 6
13d. 11(2) = [29]/{38} = 3 or 9

14. {39} blocked from 12(2)r4c4 by 11(2)n2
14a. = {48/57}(no 3,9)

15. hidden single 9 in c4 -> r3c4 = 9, r2c4 = 2
15a. r3c5 + r6c4 = [61/83]
15b. no 6 in r2c6

16. killer pair 1,3 in r6c678: both locked for c4

17. 1 in c4 in r6c4 -> 6 in r3c5 (IODn5=-5), or 1 in c4 in 7(2) = [16]
17a. -> 6 locked: no 6 in r1c4 nor r89c5
17b. no 7 in r8c1 (h13(2))

18. 18(4)n7 must have 4 or 6 for r8c1 = {1467} only: 1,4,7 locked for n7: 1 and 7 for r9

19. 5 & 7 in c4 can't both be in r1c4 -> 12(2)n5 = {57}: both locked for c4 and n5

20. 21(5)r3c6 must have 1 in r1c6 or it is {23457} which must have 7 in r3c6 to avoid {57} clashing with r2c7 = (57)
20a. r3c6 = (17)

Another tricky step
21. "45" on c7: 1 outie r3c6 + 17 = 3 innies r129c7
21a. r3c6 = (17) -> 3 innies = 18 or 24
21b. -> no 9 in r1c7 since r29c7 cannot total 9 or 15
21c. -> r9c7 = 9 (hsingle c7)

22. "45" on c7: 2 outies r23c6 = 1 remaining innie r1c7
22a. r23c6 = [17/31] = 4 or 8 -> r1c7 = (48)
22b. 1 locked in r23c6 for n2 and c6

23. naked pair {48} in r1c47: both locked for r1
23a. r12c1 = [98]

took me a long time to see most of this step. Andrew is brilliant at these!
24. r2c8 = 9 (hsingle n3)
24a. 29(5)r1c6 must have exactly one of 4 or 8 for r1c7 = {15689/23789/34679}
24b. 10(3)n3 = {127/136/145/235} = 1 or 5
24c. -> {15689} blocked from 29(5)
24d. = {23789/34679} (no 1,5)
24e. must have 3 & 7: both locked for r1

25. "45" on r1: 2 remaining outies r2c25 = 11 = {47/56}(no 1,3)

26. 27(6)r1c2: must have both 4 and 5 for n2 and 1 for r1 = {124578} only (no 6)
26a. -> r1c23 = {12}: both locked for n1, 2 for r1
26b. r1c5 = 5
26c. r2c25 = {47} : both locked for r2 and no 4 in r1c4

Not quite fully cracked yet!
27. r1c4 = 8, r3c5 = 6 -> r6c4 = 1 (iodn5=-5)
27a. r12c7 = [45], r2c6 = 3, r1c6 = 7, r3c6 = 1, r2c235 = [764], r2c9 = 1 -> r3c89 = 9 = [72]
27b. r3c8 = 8 (hsingle n3)

28. r3c5 = 6 -> r4c56 = 7 = [34]

29. 7(2)n8 = {34}: 4 locked for n8
29a. r9c4 = 6, r9c56 = 10 -> r8c5 = 7 (cage sum) -> r8c1 = 6 (h13(2)r8c15)

30. r78c9 = [78]
30a. 11(2)n7 = {29}: both locked for n7, 2 for r8

From Andrew's WT, I could have now locked 2 in c1 only in n4, locked for n4, so can then skip straight to the 27(6)r2c3 which only has one combination without 2, making the next steps mostly redundant.

31. [35] blocked from r34c1 by r7c1 = (35)
31a. -> no way to make r34c1 = 8 -> no 6 in r4c2

32. r6c2 = 6 (hsingle n4) -> r6c3 = 5 (cage sum)

33. 27(6)r2c3 must have two of {345} for r3c23 = [6]{12459/13458/23457}
33a. must have 5 -> r3c2 = 5

34. 14(3)r3c1 must have 3 or 4 for r3c1, but can't have both
34a. = {149/239/248}(no 7)
34b. r4c2 = (89)

35. 27(6)r2c3: [65]{1249} blocked by r4c1 = (12)
35a. must have 8 or 9 for n4
35b. = [65]{1348} only (no 2,9)
35c. 1 and 8 both locked for n4

cracked!
Cheers
Ed


Top
 Profile  
Reply with quote  
Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 4 posts ] 

All times are UTC


Who is online

Users browsing this forum: No registered users and 11 guests


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Search for:
Jump to:  
cron
Powered by phpBB® Forum Software © phpBB Group