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HATMAN
 Post subject: Paper Solvable 6
PostPosted: Sun Apr 10, 2011 4:22 pm 
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Grand Master
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Joined: Wed Apr 30, 2008 9:45 pm
Posts: 694
Location: Saudi Arabia
Paper Solvable 6 Decidoku X zero 1

Numbers are zero to nine - I've been trying to sort out a good 10*10 since I've been creating sudoku puzzles.

This one structure is the first one I'm satisfied with. It is very much windoku style as there are 14 hidden extra groups - so 24 in total comapered to Windoku's 18.

Have fun working out the extra groups.

Image
Original link failed; image link replaced by moderator.

I'll move it to variants if you-all insist - but I feel it is just a slightly larger killer.
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wellbeback
 Post subject: Re: Paper Solvable 6
PostPosted: Thu Apr 14, 2011 11:56 pm 
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Grand Master
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Joined: Tue Jun 16, 2009 9:31 pm
Posts: 282
Location: California, out of London
Thanks HATMAN. I had a fun time solving this. Easy to start, slowed down, made a mistake somewhere so had to start again, and got it on my second attempt. I don't know if I got all the hidden groups, but the one I made most use of was
Hidden Text:
r56c15
.
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HATMAN
 Post subject: Re: Paper Solvable 6
PostPosted: Fri Apr 15, 2011 8:46 am 
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Grand Master
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Joined: Wed Apr 30, 2008 9:45 pm
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Location: Saudi Arabia
Glad you enjoyed it.

Note that when all 14 extra groups are in place it is fully symmetrical.
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wellbeback
 Post subject: Re: Paper Solvable 6
PostPosted: Sun Apr 17, 2011 1:58 am 
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Grand Master
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Joined: Tue Jun 16, 2009 9:31 pm
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I think these are the hidden groups. Excuse the crudeness of the drawings...
Hidden Text:
Image

Took some working out!
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Andrew
 Post subject: Re: Paper Solvable 6
PostPosted: Thu May 19, 2011 1:53 am 
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Grand Master
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Joined: Wed Apr 23, 2008 6:04 pm
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Location: Lethbridge, Alberta, Canada
Thanks HATMAN for a very different and challenging puzzle. It was fun once I realised how to get into it.

I tried this puzzle when it first appeared and got stuck; with hindsight after looking at it again this week I'm not sure why I couldn't get it out at the time.

wellbeback's drawings of the hidden groups are neat. I didn't do it that way, I just used Law of Leftovers which I think is equivalent to the hidden groups.

Rather than number rows, columns and groups from 1 to 10, which I thought would be confusing when describing groups of cages I decided to number them unconventionally from 0 to 9 so here is my starting diagram to show my numbering.

Image

I've tried to take a paper solvable approach as far as possible; it's probably best described as insertion solving rather than true paper solvable.
Here is my walkthrough, with a few comments:
To avoid confusion about row and column numbers, I’ve numbered them 0 to 9. Also the tenets (if that’s the right word; they should probably be called decets, using sort-of Latin, but D is used for the diagonals) are numbered T0 to T4 across the top, then T5 to T9 across the bottom, with the L-shaped ones being T2 and T7.

I’ll try to take a paper solvable approach although I don’t know how well I’ll be able to do this since this puzzle is quite a lot different.

Cages totalling 4 or 5 must contain 0, cages totalling more than 17 cannot contain 0 so will be conventional cage totals; the three cages totalling 6 can be {015/024/123}

Each tenet must contain 0 to 9 so the 45 rule can still be used; however that wasn’t needed.

1. 4(3) cage at R0C9 and 6(3) cage at R2C8 total 10 in 5 cells in T4 and one cell outside T4 -> 0 of 4(3) cage must be outside T4 -> R2C7 = 0, placed for D/, R0C9 + R1C8 = {13}, locked for D/
1a. 6(3) cage at R2C8 = {024} (only remaining combination) -> R3C9 = 0, R2C89 = {24}

2. 0 of 5(3) cage at R0C0 must be in T0 -> 6(3) cage at R0C2 = {123}, 5(3) cage at R0C0 = {04}1 -> R2C2 = 1, placed for D\, R0C0 + R1C1 = {04}, locked for D\, R0C3 = 1, R0C9 = 3, R1C8 = 1, R0C2 = 2, R1C2 = 3

3. 23(3) cage at R7C7 = {689}, locked for D\

4. 22(3) cage at R7C2 = {589/679}, 9 locked for D/

5. 21(3) cage at R2C2 = {579/678}
5a. R0C1 + R1C0 = {59/68}

6. 21(3) cage at R6C9 = {579} (cannot be {489/678} which clash with 23(3) cage at R7C7, ALS block), locked for T9 -> R8C8 + R9C9 = {68}, locked for T9 and D\, R7C7 = 9, R7C89 = {57}, locked for R7 and T9, R6C9 = 9

7. 6(3) cage at R8C7 = {024/123}, 2 locked for T9
7a. 0 of {024} must be in R9C6 -> no 4 in R9C6
7b. R8C9 + R9C8 = [13/40]
7c. 0 in T9 only in R9C68, locked for R9

8. 22(3) cage at R7C2 = {589/679}, 9 locked for T5
8a. 21(3) cage at R8C2 = {678}, locked for T5
8b 22(3) cage at R7C2 = {589} (only remaining combination), naked pair {59} in R8C1 + R9C0, locked for D/, R7C2 = 8, placed for D/, naked pair {67} in R89C2, locked for C2 and T5, R9C3 = 8, R9C9 = 6, R8C8 = 8, R89C2 = [67]

9. 6(3) cage at R6C0 = {024/123}, 2 locked for T5
9a. R8C0 + R9C1 = {04/13}
9b. 0 of {04} must be in R8C0 -> no 4 in R8C0

10. 5(3) cage at R7C3 = {014/023}, 0 locked for T6

11. 21(3) cage at R0C6 = {579/678}, 7 locked for T4
11a. 9 of {579} must be in R0C6 -> no 5 in R0C6

12. 22(3) cage at R3C7 = {589/679}, 9 locked for C8
12a. 8 of {589} must be in R3C7 -> no 5 in R3C7

13. R0C6 = 9 (hidden single in T4), 21(3) cage = {579} (only remaining combination) -> R01C7 = {57}, locked for C7 and T4
13a. R0C8 = 6, then R1C9 = 8 (hidden pair in T4)

14. 20(3) cage at R0C5 = {479/569} (cannot be {578} = 8{57} which clashes with R1C7) -> R1C5 = 9, R0C5 + R1C6 = {47/56}
14a. 6 of {56} must be in R1C6 -> no 5 in R1C6

15. R4C8 = 9 (hidden single in C8)

16. R0C1 + R1C0 (step 5a) = {68} (only remaining combination) -> R0C1 = 8, R1C0 = 6
16a. 20(3) cage at R0C5 (step 16) = {479} (only remaining combination) -> R0C5 + R1C6 = {47}, locked for T3, R3C8 = 5, R3C7 = 8 (cage sum), R7C89 = [75]

17. R3C56 = [12] (hidden pair in T3), then R2C56 = {36} (hidden pair in T3), R2C4 = 8 (hidden single in R2), R4C0 = 8 (hidden single in C0), R6C5 = 8 (hidden single in R6), R5C6 = 8 (hidden single in R5)

18. Naked triple {579} in R239C0, 7 only in R23C0, locked for T0 -> no 7 in R2C1
18a. Naked pair {59} in R28C1

19. 6 in T6 only in R6C134, locked for R6
19a. 6 in R7 only in R7C56, 6 in C7 only in R45C7 -> there must be two 6s in R45C7 + R7C56, locked for T7 and T8
19b. 6 on D/ only in R5C4 + R6C3 (cannot be in R4C5 because of step 19a), CPE no 6 in R6C4

20. Remaining candidates on D/ are 4,6,7, no 6 in R4C5 (step 19a) -> R4C5 = {47}
20a. Naked pair {47} in R04C5, locked for C5
20b. 4,7 in T7 only in R4C5679, locked for R4

[HATMAN talked about extra groups; maybe I will be able to get some/all of the same results using Law of Leftovers, which seem to be the way to work out the extra groups.]
21. Law of Leftovers on R0123 4 outies R4C0123 must have the same four candidates as 4 innies R0123C4
21a. R4C8 = 9 -> no 9 in R4C0123 -> no 9 in R0123C4, no 9 in R5C4 (step 4) -> R9C4 = 9 (hidden single in C4), R9C0 = 5, R8C1 = 9, R2C1 = 5

22. No 1 in R0123C4, no 1 in R4C4 (step 2), no 1 in R5C4 (step 1) -> 1 in T2 only in R5C01, locked for R5
22a. No 1 in R0123C4, no 1 in R4C4 (step 2), no 1 in R5C4 (step 1) -> 1 in C4 only in R678C4, locked for T6

23. 2 on D\ only in R4C4 + R5C5 (cannot be in R3C3 + R6C6 which both “see” R3C6), CPE no 2 in R4C579 + R5C0134
23a. 2 in T2 only in R14C4, locked for C4

24. 0 in C5 only in R5789C5 -> no 0 in R4C6, no 0 in R4C4 (step 2) -> 0 in R4 only in R4C123 -> 0 must be in R01C4 (step 21), locked for C4 and T2

25. 5(3) cage at R7C3 (step 10) = {014/023} -> R7C3 = 0, R78C4 = {14}, locked for C4 and T6

26. R4C2 = 0 (hidden single in C2)

27. 0 in T5 only in R68C0, locked for C0 -> R0C0 = 4, R1C1 = 0, R0C5 = 7, R1C6 = 4, R01C7 = [57], R0C4 = 0 (hidden single in R0), R4C5 = 4, placed for D/
27a. R5C4 + R6C3 = {67} (hidden pair on D/)

28. 4 in R3 only in R3C12 (cannot be in R3, step 2), locked for T1
28a. R5C3 = 4 (hidden single in C3), R5C2 = 9 (hidden single in T2), R3C2 = 4 (hidden single in C2), R6C2 = 5 (hidden single in C2),

29. R2C3 = 9 (hidden single in T1, cannot be in R3C3, step 3), R23C0 = [79]

30. R8C6 = 5 (hidden single in T8), R5C5 = 5 (hidden single in C5), placed for D\, R4C4 = 2 (hidden single on D\), R8C3 = 7 (hidden single in R8), R5C4 = 7 (hidden single on D/), R6C3 = 6 (hidden single on D/)
30a. R1C3 = 2, then R1C4 = 5 (hidden singles in R1)

31. R6C1 = 2, then R6C3 = 3 (hidden singles in T6)
31a. R7C0 = 2 (hidden single in T5)

32. R4C3 = 5 (hidden single in C3, cannot be R3C3, step 30), R3C3 = 3 (hidden single in C3), placed for D\
32a. R3C4 = 6 (hidden single in C4), R3C1 = 7 (hidden single in R3)

33. R4C1 = 6 (hidden single in T1), R5C7 = 6 (hidden single in C7), R7C5 = 6 (hidden single in R7), R2C56 = [36]

34. R6C6 = 7 (hidden single on D\)

35. 6(3) cage at R6C0 (step 9) = {024/123}
35a. 3,4 only in R7C1 -> R7C1 = {34}

36. R4C9 = 7 (hidden single in R4)

37. R89C5 = {02} (hidden pair in C5)
37a. Law of Leftovers on R6789 4 outies R5C6789 must have the same four candidates as 4 innies R6789C5, R6789C5 = {0268} -> R5C6789 = {0268} -> R5C89 = [02], R2C89 = [24], R8C9 = 1, R78C4 = [14]

38. R6C0 = 0 (hidden single in R6), R8C0 = 3, R7C1 = 4, R9C1 = 1, R8C7 = 2, R89C5 = [02]

39. R5C0 = 1 (hidden single in C0), R5C1 = 3 (hidden single in R5), R7C6 = 3 (hidden single in R7), R9C6 = 0, R9C8 = 3, R9C7 = 4, R6C8 = 4 (hidden single in C8), R6C7 = 1 (hidden single in R6), R4C6 = 1, R4C7 = 3 (hidden singles in R4)

Solution:
Image
and in text form

4 8 2 1 0 7 9 5 6 3
6 0 3 2 5 9 4 7 1 8
7 5 1 9 8 3 6 0 2 4
9 7 4 3 6 1 2 8 5 0
8 6 0 5 2 4 1 3 9 7
1 3 9 4 7 5 8 6 0 2
0 2 5 6 3 8 7 1 4 9
2 4 8 0 1 6 3 9 7 5
3 9 6 7 4 0 5 2 8 1
5 1 7 8 9 2 0 4 3 6

I hope my solution is correct. It's one reason why I decided to try again; there wasn't a solution posted in this thread.
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