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 Post subject: Assassin 264
PostPosted: Fri Jul 19, 2013 1:34 am 
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 282
Location: California, out of London
I wanted to create one without too many easy IOs. Hope you like it. SudokuSolver gives it a score of 1.60.

Assassin 264

Image

Here is the code. Paste into solver:

Moderator edit to correct code:
3x3::k:4096:4096:5633:5633:2818:3331:3331:3331:2308:5381:5381:1798:5633:2818:2818:4615:3331:2308:5381:1798:1798:6664:4617:4617:4615:4615:4106:5381:2059:6664:6664:5900:4617:4617:4106:4106:2059:2059:6664:5900:5900:5900:6157:4366:4366:2831:2831:6416:6416:5900:6157:6157:4366:5905:2831:4370:4370:6416:6416:6157:3091:3091:5905:2580:5141:4370:3094:3094:4119:3091:5905:5905:2580:5141:5141:5141:3094:4119:4119:2584:2584:


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 Post subject: Re: Assassin 264
PostPosted: Sat Jul 20, 2013 11:32 pm 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Thanks wellbeback for your first Assassin! :D A nice challenging killer. I look forward to more of them.

wellbeback wrote:
I wanted to create one without too many easy IOs.
You managed that. I found quite a few IOs but few were really useful.

Thanks Ed for correcting some typos in my walkthrough.
Here is my walkthrough for Assassin 264:
Prelims

a) R1C12 = {79}
b) R12C9 = {18/27/36/45}, no 9
c) R89C1 = {19/28/37/46}, no 5
d) R9C89 = {19/28/37/46}, no 5
e) 22(3) cage at R1C3 = {589/679}
f) 11(3) cage at R1C5 = {128/137/146/236/245}, no 9
g) 7(3) cage at R2C3 = {124}
h) 8(3) cage at R4C2 = {125/134}
i) 11(3) cage at R6C1 = {128/137/146/236/245}, no 9
j) 13(4) cage at R1C6 = {1237/1246/1345}, no 8,9
k) 26(4) cage at R3C4 = {2789/3689/4589/4679/5678}, no 1

1. Naked pair {79} in R1C12, locked for R1 and N1, clean-up: no 2 in R2C9
1a. 22(3) cage at R1C3 contains 9 -> R2C4 = 9, R1C34 = 13 = {58}, locked for R1, clean-up: no 1,4 in R2C9

2. Naked triple {124} in 7(3) cage at R2C3, locked for N1
2a. 45 rule on N1 1 innie R1C3 = 1 outie R4C1 + 1, R1C3 = {58} -> R4C1 = {47}
2b. 9 in N4 only in R456C3, locked for C3

3. 8(3) cage at R4C2 = {125/134}, 1 locked for N4

4. 45 rule on N3 1 innie R3C9 = 1 outie R1C6 + 5, R1C6 = {1234}, R3C9 = {6789}

5. 45 rule on N7 1 outie R9C4 = 1 innie R7C1 + 2, no 7,8 in R7C1, no 1,2 in R9C4

6. 11(3) cage at R6C1 = {128/137/146/236/245}
6a. 5,6 of {236/245} must be in R6C12 (R6C12 cannot be {23/24} which clashes with 8(3) cage at R4C2), no 5,6 in R7C1, clean-up: no 7,8 in R9C4 (step 5)

7. 45 rule on C12 4 innies R3789C2 = 24, max R3C2 = 4 -> min R789C2 = 20, no 1,2 in R789C2

8. 13(4) cage at R1C6 = {1237/1345} (cannot be {1246} which clashes with R12C9 = [36] because 13(4) cage “sees” R1C9), no 6
8a. 5,7 only in R2C8 -> R2C8 = {57}
8b. 13(4) cage at R1C6 = {1237/1345}, 1,3 locked for R1, clean-up: no 6,8 in R2C9

9. 11(3) cage at R1C5 = {128/146/236/245} (cannot be {137} because R1C5 only contains 2,4,6), no 7
9a. 7 in R2 only in R2C789, locked for N3, clean-up: no 2 in R1C6 (step 4)

10. 45 rule on C89 4 innies R1237C8 = 15 = {1257/1347/1356} (cannot be {1239/1248/2346} because R2C8 only contains 5,7), no 8,9, 1 locked for C8, clean-up: no 9 in R9C9

11. 18(3) cage at R2C7 = {189/279/369/459/468} (cannot be {567} which clashes with R2C8, cannot be {378} which clashes with R2C89, ALS block)
11a. 1,2 of {189/279} must be in R3C8 -> no 1,2 in R23C7

12. 11(3) cage at R1C5 (step 9) = {128/146/236/245}
12a. Hidden killer pair 1,2 in R2C3 and R2C56 for R2, R2C56 cannot contain more than one of 1,2 -> R2C3 = {12}, R2C56 must contain one of 1,2
12b. 4 in N1 only in R3C23, locked for R3

13. 18(3) cage at R2C7 (step 11) = {189/279/369/459/468}
13a. 8 of {468} must be in R3C7, 9 of {189/279/369/459} must be in R3C7 -> R3C7 = {89}
13b. 4 of {459} must be in R2C7 -> no 5 in R2C7

14. 45 rule on C1234 1 outie R7C5 = 2 innies R58C4 + 3
14a. Min R58C4 = 3 -> min R7C5 = 6
14b. Max R58C4 = 6, no 6,7,8 in R58C4

15. 45 rule on C6789 2 innies R25C6 = 1 outie R3C5 + 4, IOU no 4 in R5C6

16. 45 rule on R1234 2 innies R4C25 = 1 outie R5C3 + 3, IOU no 3 in R4C5

17. 45 rule on R6789 2 innies R6C58 = 1 outie R5C7
17a. Min R6C58 = 3 -> min R5C7 = 3
17b. Max R6C58 = 9, no 8,9 in R6C5, no 9 in R6C8

18. 11(3) cage at R6C1 = {128/137/146/236/245}
18a. Consider placements for R4C1 = {47}
R4C1 = 4
or R4C1 = 7 => R1C3 = 8 (step 2a) => 8 in N4 only in 11(3) cage = {128}
-> no 4 in R6C12 + R7C1, clean-up: no 6 in R9C4 (step 5)
18b. 11(3) cage = {128/137/236}, no 5
[Going a bit further]
18c. R4C1 = 4 => 8(3) cage at R4C2 = {125}, locked for N4 => 11(3) cage = {36}2/{37}1
or R4C1 = 7 => R1C3 = 8 (step 2a) => 8 in N4 only in 11(3) cage = {128}, 2 locked for N4
-> no 2 in R456C3, no 3 in R7C1, clean-up: no 5 in R9C4 (step 5)
[And further still]
18d. R4C1 = 4 => R1C3 = 5 (step 2a)
or R4C1 = 7 => R1C3 = 8 (step 2a) => 8 in N4 only in 11(3) cage = {128}, 2 locked for N4 => 8(3) cage = {134} => 5 in N4 only in R456C3
-> 5 in R1C3 + R456C3, locked for C3, 4 in R4C1 + 8(3) cage, locked for N4
18e. R4C1 = 4 => 8(3) cage at R4C2 = {125}, locked for N4 => 11(3) cage = {36}2/{37}1
or R4C1 = 7 => R1C3 = 8 (step 2a) => 8 in N4 only in 11(3) cage = {128}, 2 locked for N4 => 8(3) cage = {134}
-> 3 in 8(3) cage + R6C12, locked for N4
[The above are really just one forcing chain, I’ve split it into sub-steps for clarity; they were also how I saw this step developing.]
18f. 5 in N7 only in R789C2, locked for C2
18g. 5 in 8(3) cage only in R5C1 -> no 2 in R5C1

19. 3 in C3 only in R789C3, locked for N7, clean-up: no 7 in R89C1
19a. R3C789C2 (step 7) = 24 = 2{589}/4{569/578} (cannot be 1{689} which clashes with R89C1, cannot be 2{679} which clashes with R1C2) -> no 1 in R3C2
[I overlooked no 4 in R789C2, but it probably didn’t make any difference to step 21 and it’s eliminated by step 21e.]
19b. 1 in N1 only in R23C3, locked for C3
19c. 1 in N7 only R789C1, locked for C1

20. Deleted. I originally used a somewhat “chainy” locking-out cages step, based on the position of 3 in N7, to eliminate 2 from the 17(3) cage at R8C2. However step 21, based on the position of 4 in N7, is more direct and more powerful.

21. 17(3) cage at R7C2 = {269/278/368/458/467} (cannot be {359} because 5,9 only in R8C2)
21a. 4 in N7 only in 17(3) cage = {458/467} or R89C1 = {46} or 20(4) cage at R8C2 = {458}3/{467}3 -> 17(3) cage = {269/278/458/467} (cannot be {368} which clashes with R89C1 = {46} and with 20(4) cage = {458}3/{467}3, locking-out cages), no 3
[Cracked, but there’s still a fairly long ending.]
21b. R9C3 = 3 (hidden single in N7), R9C4 = 4, R7C1 = 2 (step 5), R6C12 = 9 = {36}, locked for R6 and N4, clean-up: no 4 in 8(3) cage at R4C2, no 6,8 in R8C1, no 8 in R9C1, no 6,7 in R9C89
21c. R5C1 = 5, R45C2 = {12}, locked for C2 -> R3C2 = 4
21d. 1 in C1 only in R89C1 = {19}, locked for C1 and N7 -> R1C12 = [79], R4C1 = 4, R1C3 = 5 (step 2a), R1C4 = 8
21e. R78C3 = {46} (hidden pair in C3), locked for N7, R7C2 = 7 (cage sum)
21f. R9C89 = {28} (only remaining combination, cannot be [91] which clashes with R9C1), locked for R9 and N9 -> R89C2 = [85]

22. 16(3) cage at R8C6 = {169/367} (cannot be {259} because 2,5 only in R8C6), no 2,5
22a. 3 of {367} must be in R8C6 -> no 7 in R8C6
22b. 16(3) cage = {169/367}, CPE no 6 in R9C5
22c. 6 in R9 only in R9C67, locked for 16(3) cage, no 6 in R8C6

23. 2 in N8 only in 12(3) cage at R8C4 = {129/237}, no 5,6
23a. 5 in N8 only in R7C46, locked for R7

24. 12(3) cage at R8C4 (step 23) = {129/237} and 16(3) cage at R8C6 (step 22) = {169/367} form combined 28(6) cage R8C456 + R9C567 = {123679}, 3 locked for R8 and N8
24a. R89C1 and combined cage form grouped X-Wing in 1,9 for R89, no other 1,9 in R8

25. 12(3) cage at R7C7 = {147/156/345}, no 9

26. 45 rule on N9 1 outie R6C9 = 1 innie R9C7 -> R6C9 = {179}, R9C7 = {179}
26a. R9C6 = 6 (hidden single in R9)
26b. 9 in N9 only in R7C9 + R9C7, R6C9 = R9C7 -> 9 in R67C9, locked for C9, clean-up: no 4 in R1C6 (step 4)
26c. R3C7 = 9 (hidden single in N3), R7C9 = 9 (hidden single in N9)
26d. 16(3) cage at R8C6 (step 22) = {169/367}
26e. R9C7 = {17} -> no 1 in R8C6

27. R7C5 = 8 -> R58C4 = 5 (step 14) = {23}, locked for C4

28. 26(4) cage at R3C4 = {5678} (only remaining combination) -> R34C4 = {56}, locked for C4, R45C3 = {78}, locked for C3 -> R6C3 = 9, R67C4 = [71], R6C9 = 1, R9C7 = 1 (step 26), R8C6 = 9 (cage sum), R9C5 = 7, R7C6 = 5 (hidden single in N8), R3C6 = 7 (hidden single in C6)

29. R7C6 = 5 -> 24(4) cage at R5C7 = {4578} (only remaining combination) -> R5C7 = 7, R6C67 = {48}, locked for R6

30. R67C9 = [19] -> R8C89 = 13 = {67}, locked for R8 and N9 -> R78C3 = [64], R8C7 = 5

31. 45 rule on N5 3 remaining innies R4C46 + R6C6 = 15 = {168} (only remaining combination, cannot be {258} which clashes with R6C5, cannot be {348} because R4C4 only contains 5,6, cannot be {456} because no 4,5,6 in R4C6) -> R4C46 = [61], R6C6 = 8

32. R1C6 = 3, R3C9 = 8 (step 4)

33. R34C6 = [71] = 8 -> R2C5 + R3C7 = 10 = [28]

and the rest is naked singles.

Rating Comment:
I'll rate my walkthrough for A264 at least Hard 1.5. I used a fairly heavy forcing chain.


Last edited by Andrew on Tue Jul 30, 2013 11:15 pm, edited 1 time in total.

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 Post subject: Re: Assassin 264
PostPosted: Wed Jul 24, 2013 10:39 am 
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Joined: Wed Apr 16, 2008 1:16 am
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That was a real Assassin wellbeback. Tough one. Thankyou! I spend a lot of time in the wrong area completely, in fact, the very last area to fall when I finally found a way in. I've only had a glance at Andrew's way to see that I did it quite differently. Looks like my steps 6, 7 & 9 are my key ones. Step 6 took the longest to find but the others are pretty tough. I usually like to work with a maximum of 4 combos so feel very chuffed and a bit headachey when there are more than that.

Like Andrew found, it takes a huge number of steps to finally get it out. Going away for a long weekend so don't have time to put them all into a WT.

Start to A264
23 steps:
Preliminaries courtesy of SudokuSolver
Cage 16(2) n1 - cells ={79}
Cage 9(2) n3 - cells do not use 9
Cage 10(2) n7 - cells do not use 5
Cage 10(2) n9 - cells do not use 5
Cage 7(3) n1 - cells ={124}
Cage 8(3) n4 - cells do not use 6789
Cage 22(3) n12 - cells do not use 1234
Cage 11(3) n47 - cells do not use 9
Cage 11(3) n2 - cells do not use 9
Cage 13(4) n23 - cells do not use 89
Cage 26(4) n245 - cells do not use 1

1. 16(2)n1 = {79}: both locked for r1 and n1
1a. no 2 in r2c9

2. 22(3)r1c3 must have two of {568} for r1c34 = {589} only
2a. r2c4 = 9
2b. r1c34 = {58} only: both locked for r1
2c. no 1,4 in r2c9

3. "45" on n1: 1 innie r1c3 - 1 = 1 outie r4c1
3a. r4c1 = (47)

4. 7(3)n1 = {124} only: all locked for n1

5. "45" on c12: 4 innies r3789c2 = 24
5a. must have 1/2/4 for r3c2
5b. but {2679} blocked by r1c2 = (79)
5c. h24(4) = {1689/2589/3489/4569/4578}
5d. can't have two of 1,2,4 -> no 1,2,4 in r789c2

6. r789c3 cannot have more than one of 1,2,4 because of r23c3
6a. r789c1 cannot have more than 2 of 1,2,4
6b. -> r789c3 must have one of 1,2,4 -> Killer triple with r23c3: no 1,2,4 in r456c3
6c. & r789c1 must have two of 1,2,4 -> r7c1 = (124); 10(2) = {19/28/46}(no 3,7)= [6/8/9..]
[edit: Andrew calls this whole step (variable hidden killer triple for 1,2,4)] I'll try and remember that for next time.

7. h24(4)r3789c2: [1]{689} blocked by 10(2)n7 (step 6c)
7a. = {2589/3489/4569/4578}(no 1)
7b. 1 in n1 only in c3: 1 locked for c3

8. "45" in n7: 1 innie r7c1 + 2 = 1 outie r9c4
8a. r9c4 = (346)

9. h24(4)r3789c2 = {2589/3489/4569/4578}(note: can't have two of 3,6,7)
9a. 20(4)r8c2: consider permutations for {3467}:
i. {67}[43]/{367}[4] blocked because h24(4) combo's don't allow two of 3,6,7
ii. {347}[6] blocked by IOD n7 = -2 (step 8)
9b. ie, all permutations for 20(4)r8c2 = {3467} blocked
9c. 20(4) = {2369/2378/2459/2468/2567/3458} = [2/8 in n7..]
9d. 20(4) = [2] in r9c3 or {3458} -> r9c3 = (23458)(no 6,7,9)

10. 10(2)n7: {28} blocked by 20(4)r8c2
10a. = {19/46}(no 2,8)

11. h24(4)r3789c2 = {2589/3489/4569/4578}
11a. but [4]{569} blocked by 10(2)n7 = [6/9]
11b. = {2589/3489/4578}(no 6)
11c. must have 8 -> 8 locked for n7 and c2

12. 17(3)n7: {278} blocked by 20(4)r8c2 = [2/8](step 9c)
12a. {269} blocked by 10(2)n7 = [6/9]
12b. = {359/368/458/467}(no 2)
12c. 7 in {467} must be in r7c2 -> no 7 in r78c3

missing routine clean-up from here
13. 7 in c3 only in r456c3: 7 locked for n4
13a. r4c1 = 4, r1c3 = 5 (IODn1 = -1), r1c4 = 8
13b. 10(2)n7 = {19} only: locked for c1 and n7
13c. r7c1 = 2, r9c4 = 4 (IODn7 = +2)
13d. r1c12 = [79]
13e. r9c3 = 3

14. Naked pair {46} in r78c3: both locked for c3
14a. r78c3 = 10 -> r7c2 = 7 (cage sum)

15. Hidden single 4 in n1 -> r3c2 = 4

16. 26(4)r3c4 = {2789/3689/5678}
16a. must have 8: 8 locked for n4

17. r7c1 = 2 -> r6c12 = 9 = {36} only: both locked for r6 & n4
17a. r5c1 = 5

18. 10(2)n9: {19} blocked by r9c1 = (19)
18a. 10(2)n9 = {28} only: both locked for r9 and n9
18b. r89c2 = [85]

19. "45" on n9: 1 innie r9c7 = 1 outie r6c9
19a. both = (179)

20. 16(3)r8c6: {259} blocked by 2 & 5 only in r8c6
20a. = {169/367}(no 2,5)
20b. [6]{19} blocked by r9c1 = (19): no other way for 6 to be in r8c6 -> no 6 in r8c6
20c. 16(3) must have 6 -> r9c6 = 6
20d. no 7 in r8c6

21. 25(4)r6c3 = {1789/3589}(no 2)
21a. must have 8 -> r7c5 = 8
21b. must have 9 -> r6c3 = 9
21c. no 9 in r9c7 (r6c9=r9c7)

22. "45" on c1234: 2 remaining innies r58c4 = 5 = {23} only: both locked for c4

23. r45c3 = {78} = 15 -> r34c4 = 11 = {56} only: 5 locked for c4

On from there. I found the same as Andrew that need lots more steps to get it to singles. Good luck!
Cheerio
Ed


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 Post subject: Re: Assassin 264
PostPosted: Wed Jul 24, 2013 11:24 pm 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Ed wrote:
I've only had a glance at Andrew's way to see that I did it quite differently.
Yes, you did. After looking at your step 5, I realise that I ought to have got more out of my step 7 but I was trying to avoid using large combinations too early; clearly a mistake in this case.
If I'd:
also eliminated 4 from R789C2, which I would have if I'd written out the valid combinations for 24(4), I would probably have found Ed's steps 6 and 7.
Ed's solving path is definitely more direct than mine. :applause:


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 Post subject: Re: Assassin 264
PostPosted: Fri Jul 26, 2013 2:10 am 
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Posts: 282
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Thanks guys - I went through both your wts (mostly). We certainly work in different ways :)
Your wts had more in common with each other than with mine. I think I had slightly more in common with Andrews.

As both of you pointed out - there was no single clever move which cracked it. (At least that we found). We just had to slog our way through.

Anyway - here's how I did it.

Hidden Text:
1. 7/3@r2c3 = {124}
16/2@r1c1 = {79}
-> 22/3@r1c3 = [{58}9]
-> 21/4@r2c1 = [{368}4] with 4 in r4c1 or [{365}7] with 7 in r4c1


2. Outies - Innies c1 -> r126c2 = r5c1 + 13
-> r12c2 not = +13
-> r12c2 cannot be [76]

If we try r1c3 = 8
puts r4c1 = 7
puts r1c12 = [97]
puts 6 in r23c1

It also puts 8 in n4 in r6c12
i.e., puts r6c12 = {28}

Which leaves no solution for 10/2@r8c1

-> r1c34 = [58]
-> r4c1 = 4 and that part of 21/4@r2c1 in n1 = {368}
-> 8/3@r4c2 = {125}
-> 11/3@r6c1 cannot be {128}
-> 8 and 9 in n4 in r456c4
-> 11/3@r6c1 = [{36}2] or [{37}1]


3.HS 5 in c1 -> r5c1 = 5
-> r45c2 = {12}
-> r23c3 = {12} and r3c2 = 4
-> {12} in n7 in r789c1
-> 10/2@r8c1 = {19} or {28}


4. (34) in n7 in r789c3
Cannot both go in 17/3@r7c2 -> r9c3 from (34)
Innies - outies n7 -> r9c4 = r7c1 + 2
Since r7c1 from (12) -> r9c4 from (34)
-> r9c34 = {34}
-> r89c2 = +13


5. 5 in n7 locked in r789c2
Since (89) in c3 locked in n4 -> 5 cannot go in r7c2
-> r89c2 = {58}
-> 10/2@r8c1 = {19}
-> r7c1 = 2
-> r9c34 = [34]
Also r6c12 = {36} and r456c3 = {789}
-> 17/3@r7c2 = [7{46}]
-> 16/2@r1c1 = [79]


6. 10/2@r9c8 can only be {28}
-> r89c2 = [85] and 8 in n8 in r7c56
-> r9c567 = {67(1|9)}

Innies - Outies n9 -> r6c9 = r9c7
Since 6 already in r6 -> cannot go in r9c7 -> 6 in r9c56

7 cannot go in r8c6 - leaves no place for 7 in r9
Given r9c4 = 4 and (23) already in r9 -> 7 cannot go in r8c45
-> 7 in n8 in r9c56

-> r9c56 = {67} and r9c7 from (19)

Given 8 in n8 is in r7c56 there is no solution for 16/3@r8c6 with 7 in r9c6
-> r9c56 = [76]
-> r8c45 = {23}
-> 16/3@r8c6 = {169} with 6 in r9c6


7. r7c456 = {58(1|9)}
Innies - Outies c1234 -> r7c5 = r58c4 + 3
-> Min r7c5 = 6. Only options for r7c5 are 8 or 9
But since 4 already in c4 and r8c4 from (23)
-> r58c4 cannot = +6
-> r7c5 cannot be 9
-> r7c5 = 8 and r58c4 = {23}


8. r3467c4 = {1567}
26/4@r3c4 cannot contain a 1.
-> 1 in r67c4
-> 25/4@r6c3 can only be [9718]
-> r45c3 = {78} and r34c4 = {56}

Also r7c6 = 5
Also r8c6 = 9
-> r9c7 = 1
-> r6c9 = 1
-> 23/4@r6c9 = [19{67}]
-> 12/3@r7c7 = [{34}5]
-> r78c3 = [64]


9. Innies - Outies n3 -> r3c9 = r1c6+5
But 9 already in c9
-> [r1c6,r3c9] from [16], [27], or [38]
-> 4 in n2 locked in the 11/3@r1c5
-> 11/3@r1c5 = {146} or {245}
-> HS 7 in n2 -> r3c6 = 7
-> [r1c6,r3c9] not [27]

Also - In n2 either 6 in r3c4 or (61) in 11/2@r1c5
Both of those prevent [r1c6,r3c9] = [16]

-> [r1c6,r3c9] = [38]
-> 10/2@r9c8 = [82]


10. r3c45 = [52] or [61]
-> (12) in r3 locked in r3c35
-> 18/3@r2c7 (includes a 9) cannot be {279}
-> 2 in n3 in the part of the 13/4@r1c6 cage in n3
-> 13/3@r1c6 = [3217]
-> r8c89 = [67]


11. 45 rule in n6 -> r456c7 = +19. Also, it must include (78).
-> r456c7 = {478}
-> r7c78 = [34]
-> 9/2@r1c9 = [45]
-> 18/3@r2c7 = [693]
-> 21/4@r2c1 = [8364]
-> r6c12 = [36]
Also -> 11/3@r1c5 = [6{14}]
-> r3c45 = [52]
-> 7/3@r2c3 = [241]

Also, r4c67 (= +9) can only be [18]
-> r45c3 = [78] and r45c2 = [21]
Also r56c7 = [74]
-> r6c6 = 8

Also r4c89 = [53]
-> 17/3@r5c8 = [962]

Just cleanup from here


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