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Ed
 Post subject: Assassin 255
PostPosted: Thu Feb 28, 2013 11:02 pm 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Feels easy compared to recent Assassins since I saw the key eliminations quickly. SudokuSolver gives it 1.45.

Assassin 255

Image

code: paste into solver:
3x3::k:1792:1792:6913:6913:6913:6913:8706:8706:4355:4868:4868:4868:5381:8706:6913:8706:4355:4355:4868:4614:5381:5381:8706:8706:8706:3591:4355:4614:4614:5384:1801:1801:6154:5643:3591:3591:3084:3084:5384:3597:2574:6154:5643:2575:2575:2320:1553:5384:3597:2574:6154:5643:4114:1555:2320:1553:5384:3597:5652:6154:5643:4114:1555:3861:3861:3861:5652:5652:5652:4886:4886:4886:3861:5143:5143:5143:5652:3864:3864:3864:4886:
Solution: 
+-------+-------+-------+
| 4 3 5 | 9 8 2 | 7 6 1 |
| 7 2 1 | 6 4 3 | 9 8 5 |
| 9 6 8 | 7 5 1 | 2 4 3 |
+-------+-------+-------+
| 5 7 9 | 1 6 4 | 3 2 8 |
| 8 4 3 | 2 7 5 | 6 1 9 |
| 6 1 2 | 8 3 9 | 5 7 4 |
+-------+-------+-------+
| 3 5 7 | 4 1 6 | 8 9 2 |
| 2 8 4 | 3 9 7 | 1 5 6 |
| 1 9 6 | 5 2 8 | 4 3 7 |
+-------+-------+-------+
Cheerio
Ed
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Frank
 Post subject: Re: Assassin 255
PostPosted: Sat Mar 02, 2013 4:15 pm 
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Joined: Mon Apr 28, 2008 10:58 pm
Posts: 47
Location: Victoria, B.C., Canada
Nice one Ed.

Here is how I did it:
I solved this by considering r3c8, in conjunction with r3c2, r9c4, r9c6.

It can't be 1,2 since it is in a hidden 12(2) cage.
It can't be 7,9 due to the 16(2) cage in c8.
It can't be 3 since this causes a conflict with the 20(3) cage in r9.
It can't be 5 since it is in a hidden 10(2) cage.
6 and 8 take a little bit more work to eliminate.

So r3c8 = 4, and the rest is a walk in the park.

Many thanx Ed - Cheers - Frank
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Andrew
 Post subject: Re: Assassin 255
PostPosted: Sun Mar 03, 2013 2:48 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Thanks Ed for your latest Assassin!

Ed wrote:
Feels easy compared to recent Assassins since I saw the key eliminations quickly.
I think I must have taken quite a long time to make the key eliminations, so it felt like a typical Assassin.

Here is my walkthrough for Assassin 255:
Prelims

a) R1C12 = {16/25/34}, no 7,8,9
b) R4C45 = {16/25/34}, no 7,8,9
c) R5C12 = {39/48/57}, no 1,2,6
d) R56C5 = {19/28/37/46}, no 5
e) R5C89 = {19/28/37/46}, no 5
f) R67C1 = {18/27/36/45}, no 9
g) R67C2 = {15/24}
h) R67C8 = {79}
i) R67C9 = {15/24}
j) 21(3) cage at R2C4 = {489/579/678}, no 1,2,3
k) 20(3) cage at R9C2 = {389/479/569/578}, no 1,2

1. Naked pair {79} in R67C8, locked for C8, clean-up: no 1,3 in R5C9

2. 45 rule on R89 1 outie R7C5 = 1, clean-up: no 6 in R4C4, no 9 in R56C5, no 8 in R6C1, no 5 in R6C2, no 5 in R6C9

3. 34(7) cage at R1C7 must contain 1, CPE no 1 in R3C89

4. 45 rule on N1 3 innies R1C3 + R3C23 = 19 = {289/379/469/478/568}, no 1

5. 45 rule on R9 3 innies R9C159 = 10 = {127/136/145/235}, no 8,9

6. 45 rule on C12 1 innie R9C2 = 2 outies R28C3 + 4
6a. Min R28C3 = 3 -> min R9C2 = 7
6b. Max R28C3 = 5, no 5,6,7,8,9

7. 45 rule on C89 2 innies R19C8 = 1 outie R8C7 + 8, IOU no 8 in R1C8
7a. Max R19C8 = 14 -> max R8C7 = 6

8. 45 rule on N69 2(1+1) outies R3C8 + R9C6 = 12 = {39/48/57}/[66], no 2, no 3,5 in R9C6

9. 45 rule on N23 2 outies R13C3 = 1 innie R3C8 + 9, IOU no 9 in R1C3

10. 45 rule on R123 2 innies R3C28 = 10 = [28/46/64/73], clean-up: no 7 in R9C6 (step 9)
10a. R1C3 + R3C23 (step 4) = {289/379/469/478/568}
10b. 2 of {289} must be in R3C2 -> no 2 in R1C3
10c. 6 of {568} must be in R3C2, 9 in {469} must be in R3C3 -> no 6 in R3C3

11. 45 rule on N5 2 outies R7C46 = 10 = {28/37/46}, no 5,9

12. 45 rule on N58 2 innies R9C46 = 13 = [49/58/76/94], no 3,6,8 in R9C4
[Alternatively these eliminations can be obtained from 45 rule on N69 2(1+1) outies R3C2 + R9C4 = 11, which made me spot the next step …]

13. 45 rule on N47 2 innies R9C23 = 1 outie R3C2 + 9, IOU no 9 in R9C3
Thanks Ed for pointing out my typo. My mind must have still been on the previous step when I typed N58.]

14. R9C46 (step 12) = [49/58/76/94]
14a. 20(3) cage at R9C2 = {389/479/569/578}
14b. 7 of {479/578} must be in R9C4 (otherwise the two remaining cells of the 20(3) cage clash with R9C46, CCC), no 7 in R9C23
14c. 3 of {389} must be in R9C3, 8 of {578} must be in R9C2 -> no 8 in R9C3
14d. 7 of {479} must be in R9C4 -> no 4 in R9C4, clean-up: no 9 in R9C6 (step 12), no 3 in R3C8 (step 8), no 7 in R3C2 (step 10)
14e. 20(3) cage = [839/947/965/857]

15. R1C3 + R3C23 (step 4) = {289/469/478/568} (cannot be {379} because R3C2 only contains 2,4,6), no 3
15a. 9 of {469} must be in R3C3, 4 of {478} must be in R3C2 -> no 4 in R3C3

16. 18(3) cage at R3C2 = {279/369/459/468/567} (cannot be {189/378} because R3C2 only contains 2,4,6), no 1
16a. 2 of {279} must be in R3C2 -> no 2 in R4C12

17. R9C2 = R28C3 + 4 (step 6), R9C2 = {89} -> R28C3 = 4,5 = {13/14/23}
17a. Hidden killer pair 1,2 in R28C3 and 21(4) cage at R4C3 for C3, R28C3 contains one of 1,2 -> 21(4) cage must contain one of 1,2 = {1479/1569/1578/2379/2469/2478/2568} (cannot be {3459/3468/3567} which don’t contain 1 or 2, cannot be {1389} which clashes with R28C3)
17b. Hidden killer pair 7,9 in R1C3 + R3C23 and 21(4) cage for C3, R1C3 + R3C23 cannot contain both of 7,9 -> 21(4) cage must contain at least one of 7,9 = {1479/1569/1578/2379/2469/2478} (cannot be {2568} which doesn’t contain 7 or 9)
17c. Killer triple 1,2,3 in R28C3 and 21(4) cage, locked for C3
[Maybe this step is more accurately done as a combined cage.]

18. 20(3) cage at R9C2 (step 14e) = [947/965/857], no 9 in R9C4, clean-up: no 4 in R9C6 (step 12), no 8 in R3C8 (step 8), no 2 in R3C2 (step 10)
18a. Naked pair {46} in R2C28, locked for R2
18b. 21(3) cage at R2C4 = {489/579/678}
18c. 4,6 of {489/678} must be in R2C4 -> no 8 in R2C4

19. R9C159 (step 5) = {127/136/235} (cannot be {145} which clashes with 20(3) cage at R9C2), no 4

20. 15(3) cage at R9C6 = {168/258/348} (cannot be {159/267/357/456} which clash with 20(3) cage at R9C2, cannot be {249} because R9C6 only contains 6,8), no 7,9

21. R9C2 = 9 (hidden single in R9), R9C34 = 11 = [47/65], clean-up: no 3 in R5C1
21a. 9 in R7 only in R7C78, locked for N9

22. 18(3) cage at R3C2 (step 16) = {369/459/468/567}
22a. 3 of {369} must be in R4C2 -> no 3 in R4C1

23. R9C2 = R28C3 + 4 (step 6), R9C2 = 9 -> R28C3 = 5 = {14/23}
23a. 21(4) cage at R4C3 (step 17b) = {1479/1569/1578/2379} (cannot be {2469/2478} which clash with R28C3)

24. R9C34 (step 21) = [47/65], R9C46 (step 12) = [58/76] -> R9C346 = [476/658], 6 locked for R9

25. R9C159 (step 19) = {127/235}, 2 locked for R9
25a. 15(3) cage at R9C6 (step 20) = {168/348}, no 5
25b. 15(3) cage = 6{18}/8{34}

26. 19(4) cage at R8C7 = {1567/2467/3457} (cannot be {1378/1468/2368/2458} which clash with 15(3) cage at R9C6), no 8, 7 locked for C9 and N9 -> R67C8 = [79], clean-up: no 3 in R5C5, no 3 in R5C8, no 2 in R7C1
26a. Killer pair 1,4 in 19(4) cage and 15(3) cage at R9C6, locked for N9, clean-up: no 2 in R6C9

27. 7 in N3 only in R123C7, locked for 34(7) cage at R1C7, no 7 in R2C5 + R3C56
27a. 34(7) cage contains 7 so must also contain 4, CPE no 4 in R2C89

28. 14(3) cage at R3C8 = {149/248/356} (cannot be {158/239} because R3C8 only contains 4,6)
28a. R3C8 = {46} -> no 4,6 in R4C89
28b. 9 of {149} must be in R4C9 -> no 1 in R4C9

29. R7C46 (step 11) = {28/37/46}, R9C46 (step 12) = [58/76] -> combined cage R79C46 = {28}[76]/{37}[58]/{46}[58], 8 locked for N8
[Cracked.]

30. 8 in R8 only in R8C12, locked for N7, clean-up: no 1 in R6C1
30a. 15(4) cage = {1248} (only remaining combination), locked for N7, 4 also locked for R8 -> R7C2 = 5, R6C2= 1, R67C9 = [42], R9C3 = 6, R9C4 = 5 (cage sum), R9C6 = 8, R3C8 = 4 (step 8), R3C2 = 6, clean-up: no 1,2 in R1C1, no 2 in R4C5, no 7 in R5C1, no 6 in R5C5, no 6,8 in R5C8, no 6 in R5C9, no 3,5 in R6C1

31. Naked pair {37} in R7C13, locked for R7
31a. Naked pair {46} in R7C46, locked for R7 and N8 -> R7C7 = 8

32. R9C6 = 8 -> R9C78 = 7 = [43]

33. R3C8 = 4 -> R4C89 = 10 = [19/28]
33a. Naked pair {12} in R45C8, locked for C8 and N6
33b. Naked pair {89} in R45C9, locked for C9 and N6

34. Naked triple {356} in R456C7, locked for C7 -> R8C7 = 1, R9C9 = 7, R9C5 = 2, R9C1 = 1, clean-up: no 8 in R56C5

35. R2C3 = 1 (hidden single in N1), R1C9 = 1 (hidden single in N3), R3C6 = 1 (hidden single in N2)

36. 34(7) cage contains 4 (step 27a) -> R2C5 = 4, R5C5 = 7, R6C5 = 3, R8C5 = 9

37. 45 rule on N3 1 remaining outie R3C5 = 5, R1C8 = 6, R3C9 = 3, R2C89 = [85], R1C5 = 8, R4C5 = 6, R4C4 = 1, R4C8 = 2, R5C89 = [19], R4C9 = 8, clean-up: no 5 in R5C1, no 3 in R5C2

38. Naked pair {48} in R5C12, locked for R5 and N4 -> R5C4 = 2, R5C6 = 5, R5C3 = 3, R4C2 = 7, R4C1 = 5 (cage sum), R7C3 = 7, R7C1 = 3, R1C1 = 4, R1C2 = 3

and the rest is naked singles.

Rating Comment:
I'll rate my walkthrough for A255 at Hard 1.25. No steps were quite at that level but taken together, plus the difficulty in finding some steps, makes this rating feel appropriate.
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Ed
 Post subject: Re: Assassin 255
PostPosted: Thu Mar 07, 2013 9:14 am 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Glad you found this one a good challenge guys. Here's how I got started. Used all the same key eliminations as Andrew's walkthrough but all of them a different way. Worked in the diagonally opposite area to Frank so he found a different way in. :)
18 steps to crack it:
1. 16(2)r6c8 = {79} only: both locked for c8

2. "45" on r89: single outie r7c5 = 1

3. "45" on n5: 2 outies r7c46 = 10 = {28/37/46}(no 5,9)

4. "45" on n69: 2 outies r3c8+r9c6 = 12 = [39]/{48/66}/[57](no 1,2: no 3,5 in r9c6)

5. "45" on n58: 2 innies r9c46 = 13 = {49}/[58]/{67}(no 2,3: no 8 in r9c4)

6. "45" on r123: 2 innies r3c28 = 10 = [28/73]/{46}(no 5, no 1,3,8,9 in r3c2)
6a. no 7 in r9c6 (12(2)r3c8+r9c6)
6b. no 6 in r9c4 (h13(2)r9c46)

7. "45" on n58: 2 outies r9c23 = 1 innie r9c6 + 7
7a. no 7 in r9c23 since 7 would force the other outie to equal the 1 innie r9c6 which is not possible in the same row (IOU)(Andrew's step 14b. saw this as CCC)

8. 20(3)r9c2 = {389/479/569/578}(no 1,2)
8a. 7 in {479} must be in r9c4 -> no 4 in r9c4
8b. no 9 in r9c6 (h13(2)r9c46)
8c. no 3 in r3c8 (12(2)r3c8+r9c6)
8d. no 7 in r3c2 (h10(2)r3c28)

9. "45" on n58: 2 outies r9c78 = 1 innie r9c4 + 2
9a. no 2 in r9c78 since 2 would force the other outie to equal the 1 innie r9c4 which is not possible in the same row (IOU)

10. 15(3)r9c6: {159} blocked since no 1,5,9 in r9c6
10a. no other combo with 9 possible -> no 9 in r9c7

11. "45" on r9: 3 innies r9c159 = 10 (no 8,9)

The crucial steps
12. "45" on c12: 2 outies r28c3 + 4 = 1 innie r9c2
12a. = {13}[8]/{14/23}[9] = [8->3..]
12b. r9c2 = (89): r28c3 = {1234}

13. 9 in r9 only in 20(3)r9c2 = {389/479/569}
13a. but [839] blocked by [8->3] in IODc12 (step 12a)
13b. -> 20(3) = {479/569}(no 3,8)
13c. r9c2 = 9
13d. r9c34 = 11 = [47/65]
13e. no 4 in r9c6 (h13(2)r9c46)
13f. no 8 in r3c8 (12(2)r2c8+r9c6)
13g. no 2 in r3c2 (h10(2)r3c28)
(Andrew got the same elimination of no 3 in r9c3 in his step 17 by a couple of Hidden killer pairs then Killer triple so my move is a shortcut way......)

14. 9 in r7 only in n9: 9 locked for n9

15. 8 in r9 only in 15(3)r9c6 = {168/348}(no 5,7) = [1/3,1/4,3/6/8 in n9....]

16. 6(2)r6c9 = [15]/{24}(no 3,6,7,8,9: no 5 in r6c9)
16a. 19(4)n9: {2458} blocked by r7c9 = (245)
16b. {1378/1468} blocked by r9c78 (Killer halves, step 15)
16c. {2368} blocked by r9c78 (Killer thirds, step 15)
16d. = {1567/2467/3457}(no 8)
16e. Must have 7: 7 locked for n9 (don't need to fix 9 in r7c8 to crack it)

Found this next step took the longest to find.
17. 9 in r8 must be in 22(5)n8. Must also have 1 -> other three cells = 12
17a. Min. any two of others = {23} = 5 -> max. 3rd other = 7: no 8 in r8c456

18. 8 in r8 only in 15(4)n7 = {1248} only: all locked for n7

cracked.
Cheerio
Ed
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