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Ed
 Post subject: Assassin 259
PostPosted: Thu May 02, 2013 8:31 pm 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
5 years here! Seems to have flown. To celebrate, here's a really interesting one. Hard start and end, easy middle. SudokuSolver gives it a score of 1.40. Before it was a zero killer it had an easy ending. Much more interesting like this. Thanks HATMAN for introducing this way of making Assassins.

Assassin 259
Image

code: paste into solver:
3x3::k:3840:3840:4097:4097:4098:4098:0000:0000:0000:3840:8709:4097:1798:1799:0000:0000:0000:0000:8709:5897:8709:1798:1799:4362:0000:0000:0000:8709:5897:8709:5897:4362:4362:0000:1548:1548:8709:5897:5897:5897:4365:9230:9230:9230:7183:4112:4112:3601:4365:4365:9230:7183:9230:7183:4114:2067:3601:4365:3331:3331:7183:9230:7183:4114:2067:4619:4619:4619:3331:4100:7183:0000:4114:4114:3592:3592:3592:4100:4100:0000:0000:
solution: 
+-------+-------+-------+
| 1 8 3 | 6 9 7 | 5 4 2 |
| 6 5 7 | 4 2 1 | 8 3 9 |
| 9 4 2 | 3 5 8 | 1 6 7 |
+-------+-------+-------+
| 4 3 6 | 8 7 2 | 9 1 5 |
| 8 2 1 | 5 3 9 | 6 7 4 |
| 7 9 5 | 1 6 4 | 3 2 8 |
+-------+-------+-------+
| 5 1 9 | 7 4 3 | 2 8 6 |
| 2 7 8 | 9 1 6 | 4 5 3 |
| 3 6 4 | 2 8 5 | 7 9 1 |
+-------+-------+-------+
Cheerio
Ed
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Andrew
 Post subject: Re: Assassin 259
PostPosted: Sat May 04, 2013 2:06 am 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Ed wrote:
To celebrate, here's a really interesting one. Hard start and end, easy middle.
Yes, definitely an interesting one. Thanks, Ed!

I found the start fairly easy, but the end was definitely hard.

Here is my walkthrough for Assassin 259:
Prelims

a) R1C56 = {79}
b) R23C4 = {16/25/34}, no 7,8,9
c) R23C5 = {16/25/34}, no 7,8,9
d) R4C89 = {15/24}
e) R6C12 = {79}
f) R67C3 = {59/68}
g) R78C2 = {17/26/35}, no 4,8,9
h) 23(6) cage at R3C2 = {123458/123467}, no 9

Steps resulting from Prelims
1a. Naked pair {79} in R1C56, locked for R1 and N2
1b. Naked pair {79} in R6C12, locked for R1 and N4, clean-up: no 5 in R7C3
1c. 9 in C4 only in R789C4, locked for N8
1d. 34(6) cage at R2C2 must contain 9, locked for N1

2. 45 rule on N7 3 innies R789C3 = 21 = {489} (only remaining combination, cannot contain 7 because two remaining cells cannot total 14 which clashes with R67C3, CCC), locked for C3 and N7
2a. 7 in C1 only in R23C3, locked for N1
2b. 8 in N4 only in R45C12, CPE no 8 in R2C2
[I realised later that Ed would do this step as
45 rule on N7 2 innies R89C3 = 1 outie R6C3 + 7, IOU no 7 in R89C3
then 45 rule on N7 3 innies R789C3 = 21 = {489} (only remaining combination), …
Where there’s a CCC there’s usually an IOU. The 45 for the CCC is often easier to spot, as in this case, but if one finds the innies-outies the IOU is then easier to use.]

3. 45 rule on N14 3 outies R145C4 = 1 innie R6C3 + 14
3a. R6C3 = {56} -> R145C4 = 19,20 = {478/568/578}, no 1,2,3, 8 locked for C4
3b. R145C4 = {478/568} (cannot be {578} because 23(6) cage at R3C2 cannot contain 7 and one of 5,8) = 19 -> R6C3 = 5, R7C3 = 9
3c. 4 of {478} must be in R45C4 (because 23(6) cage at R3C2 cannot contain both of 7,8), no 4 in R1C4
3d. 6 of {568} must be in R1C4 (because 23(6) cage at R3C2 cannot contain 6 and one of 5,8), no 6 in R45C4

4. 16(3) cage at R1C3 = {178/268/367} (cannot be {358} because 5,8 only in R1C4), no 5
4a. 7 of {178/367} must be in R2C3 -> no 1,3 in R2C3

5. 34(6) cage at R2C2 = {136789/145789/235789/245689/345679}
5a. All combinations contain 7 in R3C3 except for {245689} -> R3C3 = {267}

6. 45 rule on N78 2 innies R7C4 + R9C6 = 12 = [48/57/75]
6a. R145C4 (step 3b) = {478/568}
6b. R145C4 + R7C4 must contain 5, locked for C4, clean-up: no 2 in R23C4

7. 23(6) cage at R3C2 = {123458/123467} -> R5C3 = {123} (R345C2 cannot be {123} which clashes with R78C2)
7a. Killer triple 1,2,3 in R345C2 and R78C2, locked for C2
7b. 6 in C3 only in R1234C3, CPE no 6 in R2C2

8. 9 in N8 only in R89C4
8a. 45 rule on N7 4 remaining outies R89C45 = 20 = {1289/1379/1469/2369} (cannot be {2459} which clashes with R7C4 + R9C6), no 5
8b. 9 in N8 must be in 18(3) cage at R8C3 = [891] (cannot be [495] because no 5 in R8C5) or in 14(3) cage at R9C3 = [491] -> 1 in R89C5, locked for C5 and N8, clean-up: no 6 in R23C5
8c. 18(3) cage at R8C3 = {189/378/468} (cannot be {279/369} because R8C3 only contains 4,8), no 2
8d. 14(3) cage at R9C3 = {149/248/347} (cannot be {167/239} because R9C4 only contains 4,8), no 6, 4 locked for R9
8e. R89C45 contains 1 (step 8b) = {1289/1379/1469}

[I’ve changed the order of the next three steps, for simplification.]

9. Hidden killer triple 1,2,3 in R23C4, R6C4 and R89C4 for C4, R23C4 contains one of 1,3, R89C4 cannot contain more than one of 2,3 (step 8e) -> R6C4 = {123}, R89C4 must contain one of 2,3 = {1289/1379}, no 4,6
9a. R89C4 contains one of 2,3 -> no 2,3 in R89C5

10. 18(3) cage at R8C3 (step 8c) = {189/378} -> R8C3 = 8, R9C3 = 4
10a. 3,9 only in R8C4 -> R8C4 = {39}
10b. 14(3) cage at R9C3 = {149/248/347}
10c. 3 of {347} must be in R9C4 -> no 7 in R9C4
10d. 7 in C4 only in R457C4, CPE no 7 in R5C5

11. 6 in N8 only in 13(3) cage at R7C5 = {256/346}, no 7,8
11a. 8 in N8 only in R9C56, locked for R9

12. R145C4 (step 3b) = {478/568} contains one of 5,7 in R45C4
12a. 17(4) cage at R5C5 = {1259/1349/1358/1367} (cannot be {1268} because R7C4 only contains 4,5,7, cannot be {1457/2357} which clash with R45C4, cannot be {2348} because 17(4) cage “sees” R45C4, which cannot contain both of 5,7, cannot be {2456} because 17(4) cage “sees” R45C4 and 23(6) cage at R3C2 cannot contain both of 7,8)
12a. 17(4) cage at R5C5 = {1259/1349/1358/1367} -> R6C4 = 1, clean-up: no 6 in R23C4
12b. R7C4 = {457} -> no 4,5 in R56C5
12c. 9 of {1259} must be in R5C5 -> no 2 in R5C5

13. Naked pair {34} in R23C4, locked for C4 and N2 -> R8C4 = 9, R8C5 = 1 (cage sum), R9C4 = 2, R9C5 = 8 (cage sum), clean-up: no 7 in R7C2
13a. Naked pair {25} in R23C5, locked for C5 and N2
13b. Naked pair {57} in R7C4 + R9C6, locked for N8

14. R1C4 = 6 (hidden single in C4), R12C3 = 10 = [37]
14a. Naked pair {18} in R23C6, locked for C6
14b. 8 in R6 only in R6C789, locked for N6

15. R145C4 (step 3b) = {568} (only remaining combination) -> R45C4 = {58}, locked for C4, N5 and 23(6) cage at R3C2 -> R7C4 = 7
15a. R67C4 = [17] = 8 -> R56C5 = 9 = {36} (only remaining combination), locked for C5 and N5 -> R7C5 = 4
15b. R9C6 = 5 -> R89C7 = 11 = [29/47]

16. R45C4 = {58} -> 23(6) cage at R3C2 = {123458} (only remaining combination), no 6
16a. 6 in C2 only in R789C2, locked for N7

17. 6 in C3 only in R34C3, locked for 34(7) cage at R2C2, no 6 in R345C1
17a. R2C1 = 6 (hidden single in C1), R3C3 = 2, R45C3 = [61], R3C2 = 4, R23C4 = [43], R23C5 = [25]
17b. Naked pair {23} in R45C2, locked for C2 and N4, clean-up: no 5,6 in R78C2
17c. R78C2 = [17], R9C12 = [36], R6C12 = [79], R2C2 = 5, R1C12 = [18], R3C1 = 9

18. 17(3) cage at R3C6 = {179/278}, no 4, 7 locked for R4 and N5
18a. 4 in N5 only in R56C6, locked for 36(6) cage at R5C6, no 4 in R5C78 + R6C8
18b. 36(6) cage at R5C6 = {246789/345789}, 7,9 locked for R5, 8 locked for C8

19. Killer triple 4,5,8 in R4C1, R4C6 and R4C89, locked for R4

20. 28(6) cage at R5C9 = {234568}, R8C7 = {24} -> at least one of 2,4 in 28(6) cage must be in N6
20a. R4C89 = {15} (only remaining combination, cannot be {24} because 28(6) cage contains at least one of 2,4 in N6), locked for R4 and N6
20b. 4 in N6 only in R5C9 + R6C79, locked for 28(6) cage, no 4 in R8C8
20c. 28(6) cage = {234568}, 5 locked for N9
20d. 36(6) cage at R5C6 (step 18b) = {246789} (only remaining combination), no 3
20e. 28(6) cage and 36(6) cage both contain 6, locked for N69
20f. R1C8 = 4 (hidden single in C8)
20g. 1 in C7 only in R123C7, locked for N3

21. 28(6) cage at R5C9 = {234568} -> R4C7 + R8C9 must contain the other 3 for N69
21a. R4C7 = 3
or R8C9 = 3 => R6C7 = 3
-> 3 in R46C7, locked for C7 and N6

22. 36(6) cage at R5C6 (step 20d) = {246789}
22a. R5C7 = {67} (R567C8 cannot be {678} which clashes with R3C8)
22b. Killer pair 6,7 in R3C8 and R567C8, locked for C8
22c. R8C6 = 6 (hidden single in R8), R7C6 = 3

[I looked for a long time at 45 rule on N12456789 5(1+1+3) innies R2C6 + R4C7 + R8C9 + R9C89 = 23
R4C7 + R8C9 contain one 3, R9C89 contain 1 for R9 (this doesn’t stop R2C6 + R4C7 also containing 1)
but couldn’t get anything useful from it, at least at this stage.]

23. 28(6) cage at R5C9 = {234568}
23a. Consider placements for R1C79 = {25}
R1C7 = 2
or R1C9 = 2 => 2 in R67C7 + R8C8, CPE no 2 in R8C7
-> no 2 in R8C7
[Cracked …]

24. R8C7 = 4, R9C7 = 7, R5C7 = 6, R5C8 = 7, then R5C6 = 9 (hidden singles in R5), R56C5 = [36]
24a. R4C56 = [72], R3C6 = 8 (cage sum)

25. R6C7 = 3 (hidden single in R6), R8C9 = 3 (hidden single in N9)

and the rest is naked singles.

Rating Comment:
Maybe a touch high, but I'll rate my walkthrough for A259 at Hard 1.5. The final breakthrough was hard to find.
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Ed
 Post subject: Re: Assassin 259
PostPosted: Thu May 09, 2013 8:45 pm 
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Joined: Wed Apr 16, 2008 1:16 am
Posts: 1044
Location: Sydney, Australia
Andrew wrote:
I found the start fairly easy, but the end was definitely hard.
Your step 3 was my way to start as well - but took me a lot longer to find. Then I found a quick way to make big inroads. Great way to finally crack it Andrew! I've pinched your step 23 - so much better than my contradiction move.

Assassin 259 walkthrough
30 steps:
1. 14(2)r6c3 = {59/68}(no 1,2,3,4,7)
1a. "45" on n7: 1 outie r6c3 + 7 = 2 innies r89c3
1b. -> no 7 in r89c3 since a 7 would force the one remaining innie to equal the one outie which is impossible in the same column (IOU: As so often happens, Andrew saw this as CCC).

2. "45" on n7: 3 innies r789c3 = 21 = {489} only: all locked for n7 & c3

3. 16(2)n2 = {79} only: both locked for r1 and n2

4. 23(6)r3c2 = {123458/123467}(no 9) = [5/6 but not both;5/7 but not both)
4a. max. any two cells = 13
4b. "45" on n14: 1 innie r6c3 + 14 = 3 outies r145c4
4c. r6c3 = (56) -> r145c4 = 19/20
4d. max. r45c4 = 13 -> min. r1c4 = 6

5. r145c4 = 19/20: but [8]{56/57} blocked by no {56/57} combos in 23(6)r3c2 (step 4.)
5a. -> r145c4 = [6]{58}/[8]{47}(no 1,2,3; no 6 in r45c4)
5b. ie r145c4 = 19 -> r6c3 = 5
5c. r7c3 = 9
5d. r145c4 must have 8: 8 locked for c4

From here, Andrew and I veer apart.
6. 23(6)r3c2 = {123458/123467} and r45c4 = {58/47}(step 5a)
6a. 5 & 8 in {12358} must be in r45c4 -> no 5 or 8 elsewhere in 23(6)
6b. no 5 nor 8 in r345c2
6c. 7 in {123467} must be in r45c4 -> no 7 in r345c2 nor r5c3

7. 16(2)n4 = {79} only: both locked for r6 and n4
7a. 8 in n4 only in r45c1 in 34(6)r2c2: 8 locked for c1 and 34(6) cage
7b. no 8 in r2c2

8. Hidden single 8 in c2: r1c2 = 8
8a. r1c4 = 6 -> h19(3)r145c4 = [6]{58} only: 5 locked for c4 and both 5&8 locked for n5
8b. 23(6)r3c2 = {123458} only: must have 4 -> 4 locked for c2

9. r1c4 = 6 -> r12c3 = 10 = [37] only permutation

10. 23(6)r3c2 = {123458}: must have 3 which is only in r45c2: 3 locked for c2 and n4

11. 16(2)n4 = {79} only: both locked for n4 and r6
11a. 34(6)r2c2 = {245689} only (no 1)
11b. must have 5: 5 locked for n1

12. r1c2 = 8 -> r12c1 = 7 = [16] only permutation
12a. r345c3 = [261], r3c2 = 4
12b. 7(2)r2c4 = [43] only permutation
12d. 7(2)r2c5 = [25] only permutation, r3c1 = 9, r2c2 = 5, r6c12 = [79]

13. r789c1 = {235} = 10 -> r9c2 = 6 (cage sum)

14. 17(4)r5c5: {1349} blocked by no 3,4,9 in r67c4
14a. = {1367} only -> r56c5 = {36} only: both locked for c5 and n5
14b. r67c4 = [17] only permutation
14c. 8(2)r7c2 = [17] only permutation

15. 18(3)r8c3 must have 4 or 8 for r7c3 = [891] only permutation, r9c345 = [428], r7c5 = 4
15a. r7c5 = 4 -> r78c6 = 9 = {36} only: locked for c6, r9c6 = 5, r9c1 = 3

16. r9c6 = 5 -> r89c7 = 11 = [29/47](no 1,3,6)

17. 17(3)r3c6 must have 1 or 8 for r3c6 = {179/278}(no 4)
17a. must have 7: 7 locked for r4 and n5

18. 4 in n5 only in r56c6 in 36(6)r5c6: 4 locked for that cage
18a. no 4 in r5c78 nor r6c8

19. 28(6)r5c9 = {234568} only (no 7,9)
19a. must have both 2 & 4 -> to avoid clashing with r8c7 = (24) must have at least one of 2 or 4 in n6 -> {24} blocked from 6(2)n6
19b. 6(2)n6 = {15} only: both locked for r4 & n6
19c. r45c4 = [85], r45c1 = [48]
19d. 28(6)r5c9 must have 5 which is only in n9: 5 locked for n9

20. 36(6)r5c6 = {246789} only (no 3)

21. 3 in n9 only in 28(6)r5c9 or r8c9: r56c9 see all of these -> no 3 in r56c9

22. 36(6)r5c6 and 28(6)r5c9 must both have 6: 6 locked for n69
22a. no 6 in r8c9

23. 4 in n6 only in 28(6)r5c9 -> no 4 in r8c8
23a. hidden single 4 in c8 -> r1c8 = 4

Andrew's breakthrough step 23a works from this spot. His way is much better than mine which was a contradiction move around n6. I'll stick with Andrew's way.
24. 2 in n3 is in r1c79: 2 in r1c7 -> no 2 in r8c7: or 2 in r1c9 -> 2 which must be in 28(6)r5c9 only in r67c7 or r8c8, r8c7 "sees" all those cells -> no 2 in r8c7
24a. -> no 2 in r8c7 -> r89c7 = [47]

25. Hidden single 7 in n6 -> r5c8 = 7, r3c9 = 7 (another hidden single)

26. 28(6) = {234568}: must have both 2 & 3 -> to avoid clashing with r8c9, must have 2 or 3 in r6c7: ie, r8c9 = r6c7 = (23)(Clone)

27. 2 in c8 only in r678c8 -> r6c7 & r8c9 cannot be [22] since they see all 2s in c8
27a. -> r6c7 = 3, r8c9 = 3, hidden single 3 in n3 -> r2c8 = 3

28. Hidden single 9 in c8 -> r9c8 = 9, r9c9 = 1, r4c89 = [15]

29. Hidden single 5 in c8 -> r8c8 = 5, r1c9 = 2, r1c7 = 5

30. 28(6)r5c9 must have 2 -> r7c7 = 2, r4c7 = 9, r4c56 = [72] -> r3c6 = 8 (cage sum)

singles now.
Cheerio
Ed
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