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herschko
PostPosted: Sat Dec 17, 2011 1:32 am 
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Readers here might be interested in the killer sudoku at

http://www.maa.org/pubs/FOCUSDec11-Jan12_puzzle.html

In this version (new to me), cage edges don't necessarily line up with cell edges, so that one cell may be in up to four cages (though the most is three in this puzzle). Note that numbers may repeat in cages in this puzzle.

In my browser, I don't see the border lines for all eighty-one cells. They should be there.

I confess I haven't tried solving this yet.
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simon_blow_snow
PostPosted: Sat Dec 17, 2011 12:37 pm 
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This puzzle is equivalent to a zero kenken (latin square, repeats allowed) with 3 layers of overlapping + cages.

I solved it without listing candidates, and found the difficulty about par as an easy/medium everyday newspaper puzzle, i.e. only hidden/naked singles needed.
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HATMAN
PostPosted: Sat Dec 17, 2011 3:18 pm 
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I do like the layout here - easy to understand.

By the way Simon I'm just in the Sheraton - using up my points.

Maurice
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Andrew
PostPosted: Mon Jan 30, 2012 8:33 pm 
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Thanks herschko for posting this fun puzzle.

I think a better title would be Overlapping Cages Killer. Maybe kenken, as mentioned by Simon is more appropriate; I'm not familiar with those puzzles.

I set up my own Excel worksheet, using three by three small squares for each cell to accommodate the various overlapping cages. I put the cage totals in each of the cells of the cages, to remind myself how many cells there were for each cage.

Image

To repeat the rules, some cages overlap, repeats are allowed within cages but no repeats in rows and columns; there aren't any nonets.

I think I took longer to set up my diagram than to solve the puzzle. It might well be easier to print out the diagram from the original link and write placements on the printed diagram.

Here is my "paperless" walkthrough for Overlapping Cages Killer:
In this puzzle, cages overlap at several cells. Also repeats are allowed within cages. Normal 1-9 applies for rows and columns, but there are no nonets.

Simon said that it was equivalent to an Easy/Medium newspaper puzzle so I’ve used “paperless” solving.

1. 45 rule on R2 2 innies R2C19 = 3 must contain 1,2 -> 7(2) cage at R2C7 must contain 3,4 -> 13(2) cage at R2C2 cannot be {49} -> R2C23 = [79] (cannot be [88] because no repeats in rows/columns), R2C3 = 6
[Step 1 is probably my hardest step. I saw it before spotting the important feature in the next few steps.]

2. 45 rule on R12 total = 92 -> overlap cell R1C5 = 2, R2C5 = 5, R3C5 = 1

3. 45 rule on R89 total = 98 -> overlap cell R9C5 = 8, R8C5 = 4, R7C5 = 9, R78C8 = [89], R8C7 = 7

4. 45 rule on C12 total = 93 -> overlap cell R5C1 = 3, R5C2 = 4, R5C3 = 1

5. 45 rule on C89 total = 99 -> overlap cell R5C9 = 9, R5C8 = 2, R5C7 = 8

6. 45 rule on R5 3 remaining innies R5C456 = 18, 45 rule on C5 3 remaining innies R456C5 = 16 -> total of 6 innies = 34 -> overlap cell R5C5 = 7 (because of 27(5) cage at R4C5)

7. R78C2 = {12} -> R8C23 = [18], R7C2 = 2

8. R5C7 = 8 -> R67C7 = 12 = [93] (cannot be [66] because no repeats in rows/columns), R7C6 = 5 (cage sum)

9. R7C5 = 9 -> R7C34 = 8 = [71] (cannot be [44] because no repeats in rows/columns), R6C3 = 3 (cage sum)

10. R5C4 = 5, R5C6 = 6 (only remaining places for 5 then 6 in R5)

11. 45 rule on R2 2 innies R2C19 = 3 must contain 1,2 -> 7(2) cage at R2C7 must contain 3,4 -> R2C78 = [43], R3C8 = 6

12. R4C5 = 3, R6C5 = 6 (only remaining places for 3 then 6 in C5), R4C4 = 6 (cage sum), R4C6 = 9 (cage sum), R6C4 = 2 (cage sum), R6C6 = 4 (cage sum)

13. R2C5 = 5 -> R2C46 = 17 = [98]

14. R8C5 = 4 -> R8C46 = 5 = [32]

15. R3C5 = 1 -> R3C67 = 5 = [32], R4C7 = 5 (cage sum)

16. R5C2 = 4 -> R46C2 = 13 = [85]

17. R9C5 = 8 -> R9C67 = 7 = [16]

18. R9C5 = 8 -> R9C34 = 16 = [97] (because 8 already placed in C34)

19. R1C5 = 2 -> R1C67 = 8 = [71]

20. R1C5 = 2 -> R1C34 = 12 = [48] (because 3,6,9 already placed in C4)

21. R4C3 = 2, R3C3 = 5 (only remaining places for 2 then 5 in C3), R3C4 = 4 (cage sum)

22. R5C9 = 9 -> R34C9 = 12 = [84] (because 5,6 already placed in R34)

23. R3C1 = 7 (only remaining cell in R3), R4C1 = 1 (cage sum)

24. R4C8 = 7 (only remaining cell in R4), R6C8 = 1 (cage sum)

25. R5C1 = 3 -> R67C1 = 12 = [84] (because 5 already placed for R67 and 6 for R6)

26. R67C9 = [76] (only remaining cells in R67)

27. R1C1 = 9 (only remaining place for 9 in R1), R1C2 + R2C1 = 8 = [62] (because 5,7 already placed for R2 and C2)

28. R2C9 = 1 (only remaining cell in R2), R1C89 = 8 = [53] (only remaining places for 5 and 3 in R1)

29. R9C8 = 4 (only remaining cell in C8), R89C9 = 7 = [52] (only remaining places for 5 and 2 in C9)

30. R8C1 = 6 (only remaining cell in R8), R9C12 = 8 = [53] (only remaining places for 5 and then 3 in R9)
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