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Ed
 Post subject: Assassin 230
PostPosted: Thu Jan 05, 2012 10:30 pm 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
I found this is a bit of a monster so hope you still have some holiday time! To finally crack it I used a move that is pretty unusual. I can't remember using it in any or my WTs. No chains needed [edit: though very close to them. See comments at my WT step 24]. This is one that Jsudoku has very little trouble with (at least with the solving order that I use - the forum won't let me attach the order but send me an email if you'd like it) but SudokuSolver really struggles.


Assassin 230
NOTE: 1-9 cannot repeat on the diagonals
Image

Code: paste into solver:
3x3:d:k:2304:2304:5633:5378:5378:3331:3331:3331:3331:1028:1028:5633:5378:4613:4358:4358:5383:5383:6152:6152:5633:5633:4613:4613:4358:4358:5383:6921:6152:6152:3594:3594:4613:2315:2315:3852:6921:6921:6921:6157:6157:6157:6157:6157:3852:6921:4878:4878:2063:2063:5648:2321:2321:3852:4878:4878:4626:4626:5648:5648:6163:6163:4116:1557:1557:4626:4118:5648:6163:6163:4116:4116:3351:3351:4626:4118:4118:4120:4120:4120:4120:
solution: 
+-------+-------+-------+
| 2 7 9 | 6 8 1 | 3 4 5 |
| 1 3 4 | 7 9 5 | 2 6 8 |
| 5 8 6 | 3 2 4 | 9 1 7 |
+-------+-------+-------+
| 6 4 7 | 9 5 3 | 1 8 2 |
| 8 1 3 | 2 4 6 | 7 5 9 |
| 9 5 2 | 1 7 8 | 6 3 4 |
+-------+-------+-------+
| 3 9 8 | 4 1 7 | 5 2 6 |
| 4 2 1 | 5 6 9 | 8 7 3 |
| 7 6 5 | 8 3 2 | 4 9 1 |
+-------+-------+-------+
Cheers
Ed
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Andrew
 Post subject: Re: Assassin 230
PostPosted: Sat Jan 07, 2012 3:00 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Thanks Ed for another challenging Assassin!

I'll be interested to see how you, or any other solver, managed this puzzle with "No chains needed". I originally used one short one and then added another after finding an error while checking my walkthrough.

Rating Comment:
I'll rate my walkthrough for A230 at Hard 1.5. I used a couple of short forcing chains; also step 13b might be considered to be an implied chain.

Here is my walkthrough for A230:
Prelims

a) R1C12 = {18/27/36/45}, no 9
b) R2C12 = {13}
c) R4C45 = {59/68}
d) R4C78 = {18/27/36/45}, no 9
e) R6C45 = {17/26/35}, no 4,8,9
f) R6C78 = {18/27/36/45}, no 9
g) R8C12 = {15/24}
h) R9C12 = {49/58/67}, no 1,2,3
i) 21(3) cage at R1C4 = {489/579/678}, no 1,2,3
j) 21(3) cage at R2C8 = {489/579/678}, no 1,2,3
l) 13(4) cage at R1C6 = {1237/1246/1345}, no 8,9

1. Naked pair {13} in R2C12, locked for R2 and N1, clean-up: no 6,8 in R1C12

2. 45 rule on R1 3 innies R1C345 = 23 = {689}, locked for R1
2a. 45 rule on R1 1 innie R1C3 = 1 outie R2C4 + 2, no 5,8,9 in R2C4

3. 21(3) cage at R1C4 = {489/678}, 8 locked for R1 and N2,
3a. 4,7 only in R2C4 -> R2C4 = {47}

4. 45 rule on C123 2 outies R37C4 = 7 = {16/25/34}, no 7,8,9

5. 45 rule on N6 2 innies R5C78 = 12 = {39/48/57}, no 1,2,6
5a. 45 rule on N6 3 outies R5C456 = 12 = {129/138/147/246} (cannot be {156} which clashes with R4C45, cannot be {237} which clashes with R6C45, cannot be {345} which clashes with R5C78), no 5

6. 45 rule on N56 2 innies R46C6 = 11 = {29/38/47} (cannot be {56} which clashes with R4C45), no 1,5,6

7. 45 rule on R1234 2 innies R4C19 = 8 = {17/26/35}, no 4,8,9

8. 45 rule on R6789 2 innies R6C19 = 13 = {49/58/67}, no 1,2,3

9. 45 rule on N7 2 innies R7C12 = 1 outie R7C4 + 8, IOU no 8 in R7C12

10. 45 rule on N3 2 outies R12C6 = 6 = [15/24/42], R1C6 = {124}, R2C6 = {245}

11. 45 rule on N9 2 outies R89C6 = 11 = {29/38/47/56}, no 1

12. 3 in R1 only in R1C789, locked for N3
12a. 13(4) cage at R1C6 contains 3 = {1237/1345}
12b. 21(3) cage at R2C8 = {489/678} (cannot be {579} which clashes with 13(4) cage), no 5, 8 locked for N3

13. R12C6 (step 10) = [15/24/42]
13a. Hidden killer pair 6,9 in 17(4) cage at R2C6 and 21(3) cage at R2C8 for N3, 21(3) cage contains one of 6,9 -> 17(4) cage must contain one of 6,9 -> 17(4) cage = {1259/2456} (cannot be {1457} which doesn’t contain 6 or 9), no 7
13b. 5 of {1259} must be in R2C6 (otherwise cannot place 1 for 13(4) cage at R2C6 because of permutations for R12C6)
13c. 4 of {2456} must be in R2C6 (17(4) cage cannot be 2{456} which clashes with 21(3) cage at R2C8), no 2 in R2C6, no 4 in R2C7 + R3C78, clean-up: no 4 in R1C6 (step 10)
13d. 17(4) cage = {1259/2456}, 2 locked for N3
[An alternative way to look at step 13b. 13(4) cage at R1C6 contains 1 so if 17(4) cage also contains 1 then the 1 in the 13(4) cage must be in R1C6 and then R2C6 is 5.
Ed showed me an alternative way
13(4) cage at R1C6 contains one of 2,4
2 in R2C6 “sees” all 2s in N3 apart from in 13(4) cage, in which case there can’t be 4 in the 13(4) cage -> R12C6 cannot be [42].]

14. 45 rule on N23 1 innie R3C4 = 1 outie R4C6, no 1,5,6 in R3C4, no 7,8,9 in R4C6, clean-up: no 2,3,4 in R6C6 (step 6), no 1,2,6 in R7C4 (step 4)

15. R7C12 = R7C4 + 8 (step 9), min R7C4 = 3 -> min R7C12 = 11, no 1 in R7C12
15a. 45 rule on N7 3(2+1) outies R6C23 + R7C4 = 11, min R7C4 = 3 -> max R6C23 = 8, no 8,9 in R6C23

16. 45 rule on N1 2 innies R3C12 = 1 outie R3C4 + 10
16a. Min R3C4 = 2 -> min R3C12 = 12, no 2 in R3C12

17. 45 rule on R789 3 outies R6C236 = 15
17a. 15(3) can only contain one of 7,8,9, R6C6 = {789} -> no 7 in R6C23

18. 45 rule on R123 3 outies R4C236 = 14 = {149/239/248/347} (cannot be {158/167} because R4C6 only contains 2,3,4, cannot be {257} which clashes with R4C19, cannot be {356} which clashes with R4C45), no 5,6

19. 15(3) cage at R4C9 = {159/168/249/267/456} (cannot be {258/348/357} which leave only one combination for the pair of 9(2) cages in N6), no 3, clean-up: no 5 in R4C1 (step 7)

20. 22(4) cage at R1C3 = {2389/2479/2569/3469/3568} (cannot be {2578/3478/4567} which clash with R1C12)
[I originally used incorrect logic to eliminate a further combination, so rather than re-work all my later steps I’ll use a short forcing chain.]
20a. 22(4) cage = {2389/2569/3469/3568}, no 7
or 22(4) cage = {2479} = {279}4 (cannot be {479}2 which clashes with R1C12) => R3C4 = 4, R2C4 = 7, R4C6 = 4 (step 14), R6C6 = 7 (step 6), locked for D\ -> no 7 in R23C3
-> 22(4) cage = {2389/2569/3469/3568}, no 7
20b. 3 of {3469} must be in R3C4 -> no 4 in R3C4, clean-up: no 4 in R4C6 (step 14), no 7 in R6C6 (step 6), no 3 in R7C4 (step 4)
20c. Killer pair 8,9 in R4C45 and R6C6, locked for N5
[Ed pointed out that there’s a much simpler way.
3 in N7 only in R7C12 + R789C3, CPE no 3 in R7C4.
That’s actually been there since the Prelims; maybe that’s why I never spotted it, it’s not a particularly obvious CPE unless one uses a certain artificial aid in SudokuSolver.
The only artificial aids I use are the automatic Sum function in Excel, helpful for working out differences when using the 45 rule, and a table of combinations although I normally do two and three cell combinations from memory.]

21. 4 in N5 only in R5C456, locked for R5, clean-up: no 8 in R5C78 (step 5)
21a. R5C456 (step 5a) = {147/246}, no 3

22. R7C12 = R7C4 + 8 (step 9), min R7C4 = 4 -> min R7C12 = 12, no 2 in R7C12
22a. Hidden killer pair 1,2 in 18(4) cage at R7C3 and R8C12 for N7, R8C12 contains one of 1,2 -> 18(4) cage must contain one of 1,2 in C3
22b. 45 rule on C12 3 outies R456C3 = 12 = {138/147/156/237/246/345} (cannot be {129} which clashes with 18(4) cage), no 9 in R45C3

23. R4C236 (step 18) = {239/248/347} (cannot be {149} because R4C6 only contains 2,3), no 1
23a. R4C236 = {239/347} (cannot be {248} = {48}2 because 24(4) cage at R3C1 = {4578} = {57}{48} clashes with R1C12), no 8, 3 locked for R4, clean-up: no 6 in R4C78, no 5 in R4C9 (step 7)
23b. 9 of {239} must be in R4C2, 3 of {347} must be in R4C6 -> no 2,3 in R4C2
23c. R4C78 = {18/45} (cannot be {27} which clashes with R4C236), no 2,7

24. R4C236 (step 23a) = {239/347} = [473/743/923/932]
24a. 24(4) cage at R3C1 = {2589/2679/3489/4578} (cannot be {3579} = {57}[93] which clashes with R1C12, cannot be {3678} = {68}[73] which clashes with R4C236, cannot be {4569} = {56}[94] which clashes with R4C236)
24b. 2,3,9 of {2589/2679/3489} must be in R4C23 (from permutations for R4C236) -> no 9 in R3C12

25. 9 in N1 only in R123C3, locked for C3
25a. 22(4) cage at R1C3 (step 20a) = {2389/2569/3469}
[I hope Ed won’t mind me using a short forcing chain here, and another in the re-work in step 20a.]
25b. 24(4) cage at R3C1 (step 24a) = {2589/2679/3489/4578}
24(4) cage = {2589/3489/4578}, 8 locked for N1
or 24(4) cage = {2679} => R4C3 = 2
-> 22(4) cage at R1C3 = {2569/3469}, no 8, 6 locked for C3 and N1
25c. 8 in N1 only in R3C12, locked for R3

26. R1C12 = {27} (cannot be {45} which clashes with 22(4) cage at R1C3), locked for R1 and N1 -> R1C6 = 1, R2C6 = 5 (step 10), clean-up: no 6 in R89C6 (step 11)

27. Naked triple {345} in R1C789, locked for N3, clean-up: no 9 in 21(3) cage at R2C8
27a. Naked triple {678} in 21(3) cage, locked for N3

28. 15(3) cage at R4C9 (step 19) = {249/456} (cannot be {159} which clashes with R4C78, cannot be {168} which clashes with 21(3) cage at R2C8, ALS block, cannot be {267} which clashes with R3C9), no 1,7,8, clean-up: no 1,7 in R4C1 (step 7), no 5,6 in R6C1 (step 8)
28a. 1 in C9 only in R789C9, locked for N9

29. Naked pair {26} in R4C19, locked for R4 -> R4C6 = 3, placed for D/, R6C6 = 8 (step 6), placed for D\, R3C4 = 3 (step 14), clean-up: no 5 in R6C45, no 1 in R6C78
29a. Naked pair {59} in R4C45, locked for R4
29b. Naked pair {47} in R4C23, locked for R4, N4 and 24(4) cage at R3C1, no 4 in R3C12 -> R6C1 = 9, R6C9 = 4 (step 8), R1C9 = 5, placed for D/, clean-up: no 5 in R6C78, no 1 in R8C1, no 4,8 in R9C2

30. R3C7 = 9 (hidden single on D/), R2C7 = 2, R3C8 = 1, R4C78 = [18], clean-up: no 3 in R5C8 (step 5), no 7 in R6C8
30a. R2C9 = 8 (hidden single in N3)

31. R123C3 = {469} (hidden triple in N1), locked for C3 -> R4C23 = [47], clean-up: no 2 in R8C1

32. R456C3 (step 22b) = {237} (only remaining combination) -> R56C3 = {23}, locked for C3 and N4 -> R4C19 = [62], R5C9 = 9 (step 28), clean-up: no 3 in R5C7 (step 5), no 7 in R6C7, no 7 in R9C2

33. Naked pair {36} in R6C78, locked for R6 -> R6C3 = 2, R5C3 = 3
33a. R6C2 = 5 (hidden single in R6), clean-up: no 8 in R9C1
33b. R7C3 = 8 (hidden single in R7)

34. Naked triple {15} in R89C3, locked for N7 and18(4) cage at R7C3 -> R7C4 = 4, R2C4 = 7, R2C8 = 6, placed for D/, R3C9 = 7, R8C1 = 4, R8C2 = 2, placed for D/, R9C1 = 7, placed for D/, R9C2 = 6, R7C12 = [39], clean-up: no 7 in R8C6 (step 11)
34a. R89C6 = [92]
34b. Naked triple {136} in R789C6, locked for N9

35. 45 rule on R9 3 innies R9C345 = 16 = {358} (only remaining combination) -> R9C5 = 3, R9C34 = [58], R9C9 = 1, placed for D\, R5C5 = 4, placed for D\, R3C3 = 6

and the rest is naked singles without using the diagonals.
Thanks Ed for some interesting feedback and for correcting typos.
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Ed
 Post subject: Re: Assassin 230
PostPosted: Wed Jan 11, 2012 10:54 pm 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
This is the way I did it. Doesn't feel too bad with an optimised walk-through. But takes 27 steps to get the first placement and many more than that before knowing the way. The hardest ones to find were steps 15, 24 and 27. Step 1 is quite helpful too. I hope I'm not being tricked in step 24 - that sort of step is new to me. Still learning new things about solving killers after all these years!

Start to A230
27 steps:
Prelims
i. 9(2)n1: no 9
ii. 21(3)n2: no 1,2,3
iii. 13(4)r1c6: no 8,9
iv. 4(2)n1 = {13}
v. 21(3)n3: no 1,2,3
vi. 14(2)n5: no 1,2,3,4,7
vii. 9(2) cages in n6: no 9
viii. 8(2)n5: no 4,8,9
ix. 6(2)n7: no 3,6..9
x. 13(2)n7: no 1.2.3


1. 3 in n7 in r7c12 or r789c3 -> no 3 in r7c4 (CPE)

2. "45" on r1: 3 innies r1c345 = 23 = {689} only: all locked for r1
2a. and no 6,8,9 in r3c4 (sees all r1c345: CPE)
2b. no 1,3 in 9(2)n1

3. 21(3) must have two of 6,8,9 for r1c45
3a. = {489/678}(no 5)
3b. must have 8 -> 8 locked for n2
3c. must have 4/7 -> r2c4 = (47)
3d. 8 locked for r1

4. 4(2)n1 = {13}: both locked for r2 and n1

5. 14(2)n5 = {59/68} = [5/6..]
5a. "45" on n5: 2 innies r46c6 = 11 (no 1)
5b. but {56} blocked by 14(2) -> no 5,6

6. "45" on n23: 1 innie r3c4 = 1 outie r4c6
6a. no 1,5,8,9
6b. no 2,3 in r6c6 (h11(2)r46c6)

7. "45" on c123: 2 outies r37c4 = 7 (no 7,8,9)
7a. =[25/34]
7c. r4c6 = (23) (IOD n23 = 0)
7d. r6c6 = (89)(h11(2)r46c6)

8. 4 in n5 only in r5: locked for r5

9. "45" on n6: 2 innies r5c78 = 12 (no 1,2,6)
9a. = {39/57}(no 8) = [3/5..]

10. "45" on n6: 3 outies r5c456 = 12 and must have 4
10a. but {345} blocked by h12(2)n6 (step 9a)
10b. = {147/246}(no 3,5,8,9)

11. "45" on n7: 1 outie r7c4 + 8 = 2 innies r7c12
11a. -> no 8 in r7c12 since it force one of the innies to equal the outie (IOU)
11a. min. r7c4 = 4 -> min. r7c12 = 12 (no 1,2)

12. Hidden killer pair 1,2 in n7: 6(2) has one of 1,2 ->r789c3 must have 1/2 for n7 (but not both)
12a. {1269/1278/3456} all blocked from 18(4)r7c3

13. "45" on c12: 3 outies r456c3 = 12
13a. but {129} blocked by r789c3 (step 12)
13b. no other combos with 9 -> no 9 in r456c3

14. "45" on r789: 3 outies r6c236 = 15
14a. min. r6c6 = 8 -> max. r6c23 = 7 (no 7,8,9)

15. 8(2)n5 = {17/26/35} = [1/2/5, 1/3/6..] (no eliminations yet)
15a. "45" on n56789: 1 innie r4c6 + 4 = 2 outies r6c23
15b. but [2]{15}/[3]{16} clashes with 8(2)n5
15c. -> no 1,6 in r6c23

16. "45" on n1: 3 outies r3c4+r4c23 = 14
16a. max. r3c4 = 3 -> min. r4c23 = 11 (no 1, no 2 in r4c2)

17. "45" on r1234: 2 innies r4c19 = 8 (no 4,8,9)

18. 1 in r4 in h8(2)r4c19 = {17} or in 9(2)n6
18a. -> {27} blocked from 9(2) since it would leave no 1 for r4 (Locking out cages) (no 2,7)

19. 8 in n5 in h11(2)r46c6 = [38] or in 14(2) = {68} =[3/6..] in r4-> {36} blocked from 9(2)r4c7 (no 3,6) (Blocking cages)

20. 9(2)r4c7 = {18/45} = [5/8..]
20a. -> Killer pair 5,8 with 14(2)n5: both locked for r4
20b. no 3 in r4c19 (h8(2))

21. 6 in n1 only in r123c3 or r3c12 -> no 6 in r4c3 (CPE)

22. "45" on r123: 3 outies r4c236 = 14
22a. must have 3 for r4 = {239/347}(no 6)
22b. = [9]{23}/{47}[3]
22c. r4c2 = (479)

23. h12(3)r456c3: {138/156} both blocked since 1,6,8 only in r5c3
23a. = {147/237/246/345}(no 8) = [1->4..]

24. "45" on n4: 4 innies r46c23 = 18
24a. since two cells of the h12(3)r456c3 overlap the 4 innies -> 1 remaining innie r5c3 + 6 = 2 remaining outies r46c2
24b. the only way for r46c2 = 7 with 1 in r5c3 is [43]: but this is blocked by [1-> 4] in h12(3) (step 23a) (Blocking cages?)
24c. ->no 1 in r5c3
[edit: Andrew made an interesting observation about this step.."It's difficult to know what's a chain and what isn't. ...I guess your steps 24c and 27 are in a "grey area", pushing the border of blocking cages but not quite as far as forcing chains or contradiction moves."]

25. 1 in c3 only in n7: 1 locked for n7
25a. -> 6(2)n7 = {24}: both locked for n7 & r8
25b. no 9 in 13(2)n7

26. 22(4)r1c3: {2479/2578} blocked by 9(2)n1 = [4/7,5/7..]
26a. combo's with 2 are {2389/2569} = [2->9..] (no eliminations yet)

27. When hidden 7(2)r37c4 is [25] ->[5]{139} in 18(4)r7c3 blocked by [2-> 9] in 22(4)r1c3 (Blocking cages)
27a. 18(4) must have 4/5 for r7c4 = {1458/1467}(no 3,9)
27b. must have 4 -> r7c4 = 4
27c. -> r3c4 = 3 (h7(2)r37c4)

On from there. Don't forget the diagonals!
Cheers
Ed
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Frank
 Post subject: Re: Assassin 230
PostPosted: Sat Jan 14, 2012 1:45 pm 
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Posts: 47
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Hola from Puerto Vallarta!

Another toughie! But then I guess that is the idea :)

Here is how I cracked it:
I narrowed r7c4 down to {3,4,5}.

3 was easy to dismiss.

5 was somewhat gruelling.

So r7c4 = 4.
And the rest was a pleasant stroll along the Malecon :)

Hasta la vista!
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wellbeback
 Post subject: Re: Assassin 230
PostPosted: Sun Feb 05, 2012 7:03 am 
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Posts: 282
Location: California, out of London
Still haven't worked out how to do this with no chains. Will try a bit more before reading Ed's walkthrough.

The simplest I got following on from Frank's comment was:

Hidden Text:
5 in r7c4 -> r7c12 = +13 (no 3) -> 3 in n7 in r789c3.
Also 5 in r7c4 -> r46c6 = [29] -> r4c45 = {68} and r4c23 = +12.

r4c3 cannot be 9 (leaves no place in n1 for a 9)
Since r4c3 also cannot be 3 -> r4c2 cannot be a 9 -> no place in r4 for a 9.
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