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Pinata
 Post subject: Pinata Killer Sudoku 25
PostPosted: Sun Feb 10, 2013 1:46 pm 
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Joined: Sat Jul 28, 2012 11:05 pm
Posts: 92
Pinata Killer Sudoku 24 Solution:
+-------+-------+-------+
| 9 1 7 | 3 8 6 | 2 5 4 |
| 4 2 6 | 9 7 5 | 8 3 1 |
| 5 8 3 | 2 4 1 | 6 7 9 |
+-------+-------+-------+
| 3 7 2 | 1 6 9 | 5 4 8 |
| 6 9 8 | 5 3 4 | 1 2 7 |
| 1 5 4 | 7 2 8 | 3 9 6 |
+-------+-------+-------+
| 2 6 1 | 4 9 3 | 7 8 5 |
| 8 3 9 | 6 5 7 | 4 1 2 |
| 7 4 5 | 8 1 2 | 9 6 3 |
+-------+-------+-------+

Pinata Killer Sudoku 25
Image

Jsudoku Code 3x3::k:8457:8457:8457:8457:4115:7937:7937:7937:7937:8457:6922:6922:6922:4115:5634:5634:5634:7937:8457:6922:4363:4363:4115:5891:5891:5634:7937:8457:6922:4363:4363:4100:4100:5891:5634:7937:4876:4876:4876:4613:4613:4100:3597:3597:3597:8712:6407:2822:4613:4613:2577:4112:7183:8206:8712:6407:2822:2822:2577:4112:4112:7183:8206:8712:6407:6407:6407:3346:7183:7183:7183:8206:8712:8712:8712:8712:3346:8206:8206:8206:8206:

Sudoku Solver Score: 1.30

This is a Disjoint Group Killer Sudoku, Numbers can't repeat in the following coloured groups as well.

Red Group: R147C147
Dark Green Group: R147C258
Blue Group: R147C369
Yellow Group: R258C147
Pink Group: R258C258
Turquoise Group: R258C369
Dark Grey Group: R369C147
Light Pink Group: R369C258
Faint Green Group: R369C369
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Andrew
PostPosted: Fri Feb 15, 2013 10:11 pm 
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Joined: Wed Apr 23, 2008 6:04 pm
Posts: 1895
Location: Lethbridge, Alberta, Canada
Pinata, you are very good at coming up with different types of killers from time to time! :applause:

This is the first time I've seen a disjoint groups killer, although clearly it's a format which others have used since it's programmed into SudokuSolver.

There's so much information needed in the worksheet diagram for this puzzle:- cages, candidates, nonets and disjoint groups that it was hard to decide which colouring format to use. I decided to colour the cages, as is my normal practice, rather than colour the disjoint groups, and to remember where the disjoint groups are. This may have made my use of the disjoint groups more difficult (I had to re-work a few times when I realised I'd done eliminations in them incorrectly after making placements), but I felt that making the cages clear was more important. While the dashed cage borders are clear in Pinata's diagram, they wouldn't be on my worksheet which has extra cells for candidates. This is a puzzle which is probably much easier for a software solver than for a human solver; so/too many things to keep track of. I'm not saying that I don't want more disjoint groups killers but, please, not too often.

Here is my walkthrough for Pinata Killer #25:
Numbers can’t be repeated in the following Disjoint Groups
R147C147, R147C258, R147C369
R258C147, R258C258, R258C369
R369C147, R369C258, R369C369

Prelims

a) 10(2) cage at R6C6 = {19/28/37/46}, no 5
b) R89C5 = {49/58/67}, no 1,2,3
c) 23(3) cage at R3C6 = {689}
d) 19(3) cage at R5C1 = {289/379/469/478/568}, no 1
e) 11(3) cage at R6C3 = {128/137/146/236/245}, no 9

1. 45 rule on R1234 1 outie R5C6 = 5, placed for R258C369 group
1a. R5C6 = 5 -> R4C56 = 11 = {29/38/47}, no 1,6

2. 45 rule on R5 2 innies R5C45 = 7 = {16/34}
2a. 45 rule on R5 2 outies R6C45 = 11 = {29/38/47}, no 1,6

3. 45 rule on C1234 2 outies R56C5 = 4 = [13], R5C4 = 6 (step 2), R6C4 = 8 (step 2a), clean-up: no 7,9 in R6C6, R7C5 = {68}
6 placed for R258C147 group, 1 placed for R258C258 group, 8 placed for R369C147, 3 placed for R369C258 group

4. 45 rule on R6789 2 remaining innies R46C6 = 9 = [72], R4C5 = 4 (cage sum), R4C4 = 9, R7C5 = 8, clean-up: no 5,9 in R89C5
9 placed for R147C147 group, 4 placed for R147C258 group, 7 placed for R147C369 group, 2 placed for R369C369 group and 8 placed for R147C258 group
4a. 32(7) cage at R6C9 contains 2, locked for N9

5. Naked pair {67} in R89C5, locked for C5 and N8
5a. Naked triple {259} in 16(3) cage at R1C5, locked for N2

6. 23(3) cage at R3C6 = {689} -> R3C7 = 9, placed for R369C147 group

7. 17(4) cage at R3C3 = {1259/1349}, no 6,7,8
7a. 2 of {1259} must be in R4C3 -> no 5 in R4C3
7b. 7 in C4 only in R12C4, CPE no 7 in R2C1

8. 16(3) cage at R6C7 = {169/349/367/457}
8a. 9 of {169} must be in R7C6 -> no 1 in R7C6
8b. 4 of {349} must be in R6C7, 4 of {457} must be in R7C6 -> no 4 in R7C7

9. 31(7) cage at R1C6 = {1234678} (only remaining combination), no 5, 7 locked for N3
9a. 5 in N3 only in 22(5) cage at R2C6, no 5 in R4C8
9b. 5 in R4 only in R4C12, locked for N4, CPE no 5 in R1C2
9c. 5 in C9 only in R6789C9, locked for 32(7) cage at R6C9, no 5 in R9C78
9d. 32(7) cage contains 5 so must also contain 8, locked for N9

10. 22(5) cage at R2C6 contains 5 = {12568/23458}, 31(7) cage at R1C6 (step 9) = {1234678}
10a. 31(7) cage and 22(5) cage both contain 2 which must be in N3 and R4C89, 2 locked for R4 and N6, CPE no 2 in R1C8
[I’ve omitted a similar step for 8. Steps 11 and 11a are a simpler way to make these eliminations.]

11. 14(3) cage at R5C7 = {347} (only remaining combination), locked for R5 and N6
11a. Naked pair {289} in 19(3) cage at R5C1, locked for N4
11b. 9 in R6 only in R6C89, CPE no 9 in R9C8

12. 17(4) cage at R3C3 (step 7) = {1349} (only remaining combination), no 5, 4 locked for R3

13. 16(3) cage at R6C7 (step 8) = {169/367/457} (cannot be {349} because 4,9 only in R7C6)
13a. 7 of {367/457} must be in R7C7 -> no 3,5 in R7C7

14. 33(7) cage at R1C1 contains 2, locked for N1
14a. 7 in C4 and N1 only in 33(7) cage at R1C1 and 27(5) cage at R2C2, 5 in R4 and N1 only in 33(7) cage and 27(5) cage -> both 33(7) cage and 27(5) cage must contain 5,7
14b. 27(5) cage = {15678/34578}, no 9, 8 locked for N1, 5 locked for C2
14c. 33(7) cage = {1235679} (only remaining combination), no 4
14d. 4,8 in R1 only in R1C679, locked for 31(7) cage at R1C6, no 4,8 in R234C9

15. R4C7 = 8 (hidden single in R4), R3C6 = 6
6 placed for R369C369

16. 9 in R6 only in R6C89, 9 in C6 only in R789C6 -> 16(3) cage at R6C7, 28(5) cage at R6C8 and 32(7) cage at R6C9 must all contain 9 -> 16(3) cage (step 13) = {169} (only remaining combination) -> R7C6 = 9, R67C7 = {16}, locked for C7
9 placed for R147C369

17. 11(3) cage at R6C3 = {137/146/245} (cannot be {236} because no 2,3,6 in R6C3)
17a. 6 of {146} must be in R7C3, 4 of {245} must be in R6C3 -> no 4 in R7C3
17b. Killer triple 1,3,4 in R34C3 and 11(3) cage, locked for C3

18. R6C89 = {59} (hidden pair in R6)
18a. 5,9 in R6C89 and N9 -> 28(5) cage at R6C8 and 32(7) cage at R6C9 must both contain 5,9 -> 32(7) cage = {1234589} (only remaining combination), no 6,7
18b. 7 in N9 only in 28(5) cage = {15679/34579}
18c. Consider placement for 1 in 32(7) cage
1 in 1 in R9C6 => 28(5) cage = {34579} (cannot be {15679} which clashes with R7C7)
or 1 in R78C9 + R9C89 => R7C7 = 6 => 28(5) cage = {34579}
-> 28(5) cage = {34579}, no 1,6, 4 locked for R8, CPE no 3 in R8C9

19. R7C7 = 6 (hidden single in N9), R6C7 = 1
1 placed for R369C147 and 6 placed for R147C147
19a. Naked pair {26} in R4C89, locked for R4, CPE no 2,6 in R1C8
19b. 1 in N9 only in R78C9 + R9C89, locked for 32(7) cage at R7C9, no 1 in R9C6
19c. 34(7) cage at R6C1 contains 1, locked for N7

20. Naked pair {34} in R89C6, locked for C6 and N8
20a. Naked pair {18} in R12C6, locked for N2, CPE no 1 in R2C9
20b. 31(7) cage at R1C6 = {1234678}, 3,4 locked for N3

21. 4 in R2 only in 27(5) cage at R2C2 (step 14b) = {34578}, no 1,6
21a. 6 in N1 only in R1C23, locked for R1

22. 4 in C1 only in R679C1, locked for 34(7) cage at R6C1, no 4 in R9C2
22a. 4 in N7 only in R79C1, locked for C1
22b. 6 in C1 only in R69C1, locked for 34(7) cage at R6C1, no 6 in R9C2
22c. 34(7) cage contains 4 so must also contain 7 -> 34(7) cage = {1245679/1345678}
22d. Hidden killer pair 6,7 in R9C123 and R9C5 for R9, R9C5 = {67} -> R9C123 must contain one of 6,7
22e. Killer pair 6,7 in R6C1 and R9C123, locked for 34(7) cage, no 7 in R78C1

[I can’t see any immediate way to reduce the 34(7) cage at R6C1 and 25(5) cage at R6C2 to one combination so I’ll look at the disjoint groups.]
23. 7 in R147C147 only in R1C147, locked for R1
23a. 1 in R147C258 only in R1C28, locked for R1 -> R1C6 = 8, R2C6 = 1, 22(5) cage at R2C5 = {12568}, 8 locked for C8
1 placed for R258C369
23b. 4 in R147C369 only in R17C9, locked for C9
23c. 7 in R369C147 only in R369C1, locked for C1
23d. 8 in R369C369 only in R9C39, locked for R9

[This is the first time I’ve ever done, or even seen, a killer with disjoint groups, so it’s not surprising that I didn’t see this earlier.]
24. 5 in C7 only in R28C7, locked for disjoint group R258C147, no 5 in R2C1 + R8C14

25. 5 in R8 only in R8C78, locked for N9 and 28(5) cage at R6C8 -> R6C8 = 9, R6C9 = 5
9 placed for R369C258, 5 placed for R369C369
25a. R89C9 = {89} (hidden pair in C9), R9C39 = {89} (hidden pair in R9)

26. 5 in R9 only in R9C14, locked for 34(7) cage at R6C9, no 5 in R7C1

27. 5 in R7 only in R7C34 -> 11(3) cage at R6C3 (step 17) = {245} (only remaining combination) -> R6C3 = 4, R7C34 = {25}, locked for R7
4 placed for R369C369
27a. R9C6 = 3, R8C6 = 4
3 placed for R369C369
27b. R3C3 = 1, R4C3 = 3, R3C4 = 4, R3C9 = 7, R4C12 = [15], R3C2 = 8, R2C3 = 7, R2C24 = [43], R1C4 = 7, R5C9 = 3
4 placed for R258C258, 3 placed for R258C147, 4 placed for R369C147 and 5 placed for R147C258

28. R9C7 = 2, R2C7 = 5, R3C8 = 2, R4C8 = 6, R2C8 = 8, R4C9 = 2, R12C9 = [46], R1C78 = [31], R5C78= [47], R7C89 = [31], R7C12 = [47], R6C12 = [76], R9C4 = 5

29. R8C24 = [31] (hidden pair in R8 = 4, R67C2 = [67] = 13 -> R8C3 = 8 (cage sum)

and the rest is naked singles, without using the disjoint groups.

Rating Comment:
I'll rate my walkthrough for Pinata #25 at least 1.5. I used one short forcing chain, which would make it Easy 1.5. It's the fact that using the disjoint groups was such hard work which makes me say at least 1.5. Maybe if I'd used the disjoint groups earlier and more effectively, I might have avoided using the short forcing chain, but I'd still have given the same rating.
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