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PostPosted: Sat Dec 04, 2010 4:53 am 
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Thanks Simon for posting an interesting graphical walkthrough for Ninjutsudoku #1.:
It's a nice concept to use letters in the hidden 45(9) and then placing them gradually in other cells; that simplifies things a lot compared with elimination solving as it gets lots of cells down to 2 or 3 letter candidates (I won't call them pencilmarks ;) ) very quickly.

The treatment of the eight givens as a zero cage is another interesting concept.

This approach, rather than the more conventional elimination solving, seems to make some of the CPEs easier.

I found this puzzle a lot harder to solve:
I found the hidden 45(9) cage, although not at the start; without spotting that I think this puzzle would have been much harder. I used several short forcing chains and a short contradiction move. With hindsight I probably ought to have obtained the results from my first sets of forcing chains using hidden singles and pairs, as in Simon's step 2, although when doing elimination solving they would probably be "clones" and "clone pairs".

My walkthrough is fairly long but many steps come in groups of 2 or 4, because of the symmetry of the cage pattern, in the same way that some of Simon's steps apply in 2 or 4 different positions.
Here is my walkthrough for Ninjutsudoku #1:
Prelims. Eliminations made for givens, including along short diagonals.

1. 1 in C1 only in R89C1, locked for N7

2. 8 in C9 only in R12C9, locked for N3

3. 2 in R1 only in R1C12, locked for N1

4. 7 in R9 only in R9C89, locked for N9

5. 1 in window at R1C3 only in R1C45678, locked for R1

6. 7 in window at R2C1 only in R23456C1, locked for C1

7. 2 in window at R3C9 only in R45678C9, locked for C9

8. 8 in window at R8C2 only in R9C23456, locked for R9

9. Grouped X-Wing for 8 in window at R1C3 and R12C9, no other 8 in R12
9a. 8 in C1 only in R34567C1, locked for window at R2C1, no 8 in R34C2

10. Grouped X-Wing for 2 in R1C12 and window at R2C1, no other 2 in C12
10a. 2 in R9 only in R9C34567, locked for window at R8C2, no 2 in R8C34

11. Grouped X-Wing for 1 in R89C1 and window at R8C2, no other 1 in R89
11a. 1 in C9 only in R34567C9, locked for window at R3C9, no 1 in R67C8

12. Grouped X-Wing for 7 in window at R3C9 and R9C89, no other 7 in C89
12a. 7 in R1 only in R1C34567, locked for window at R1C3, no 7 in R2C67

13. There are eight 45(9) windows -> R1C12 + R12C9 + R5C5 + R89C1 + R9C89 must form a hidden 45(9) window
13a. 2 in R1 only in R1C12, 8 in C9 only in R12C9, 1 in C1 only in R89C1 and 7 in R9 only in R9C89 -> no 1,2,7,8 in R5C5

14. 3,8 in window at R1C3 only in R1C3456 + R2C6, no 3 in R2C4, no 8 in R3C5 using short diagonal

15. 2 in window at R2C1 only in R4567C1 + R4C2, no 2 in R5C3 using short diagonal

16. 7 in window at R3C9 only in R3456C9 + R6C8, no 7 in R5C7 using short diagonal

17. 1,6 in window at R8C2 only in R8C4 + R9C4567, no 1 in R7C5, no 6 in R8C6 using short diagonal

[I ought to have spotted the next two steps earlier. Steps 18a and 19a have since been re-written, after I realised that they were hidden pairs. Also it took me a long time to spot that they eliminated 3,6 from R5C5 although I’d done similar eliminations in step 13.]

18. 6 in N4 only in R456C1 + R4C2, locked for window at R2C1, no 6 in R23C1 + R3C2
18a. R1C12 = {26} (hidden pair for N1), locked for R1 and hidden 45(9) window, no 6 in R12C9 + R5C5 + R9C89
18b. 6 in window at R1C3 only in R2C67, locked for R2
18c. 6 in C9 only in R34568C9, locked for window at R3C9, no 6 in R6C8
18d. 6 in C8 only in R345C8, locked for window at R3C6, no 6 in R3C6 + R4C7 + R5C67
18e. 6 in R9 only in R9C4567, locked for window at R8C2, no 6 in R8C4

19. 3 in N6 only in R456C9 + R6C8, locked for window at R3C9, no 3 in R7C89 + R8C9
19a. R9C89 = {37} (hidden single in N9), locked for R9 and hidden 45(9) window, no 3 in R5C5 + R89C1
19b. 3 in window at R8C2 only in R8C34, locked for R8
19c. 3 in C1 only in R24567C1, locked for window at R2C1, no 3 in R4C2
19d. 3 in C2 only in R567C2, locked for window at R5C2, no 3 in R5C34 + R6C3 + R7C4
19e. 3 in R1 only in R1C3456, locked for window at R1C3, no 3 in R2C6

20. R5C5 = 9 (naked single), locked for hidden 45(9) window, no 9 in R12C9 + R89C1
20a. 9 in R1 only in R1C345678, locked for window at R1C3, no 9 in R2C67
20b. 9 in C1 only in R234567C1, locked for window at R2C1, no 9 in R34C2
20c. 9 in C9 only in R345678C9, locked for window at R3C9, no 9 in R67C8
20d. 9 in R9 only in R9C234567, locked for window at R8C2, no 9 in R8C34

21. R14C2 = {26} (hidden pair in C2)
21a. R69C8 = {37} (hidden pair in C8)

22. Consider the placement of 5 in C1
5 in C1 of window at R2C1 => no 5 in R3C2 => R9C2 = 5 (hidden single in C2)
or 5 in R89C1
-> 5 must be in R89C1 + R9C2, locked for N7
22a. 5 in window at R8C2 only in R9C256, locked for R9
22b. 5 in C3 only in R1234C3, CPE no 5 in R3C5 using short diagonal

23. Consider the placement for 4 in C9
4 in R12C9
or 4 in C9 of window at R3C9 => no 4 in R7C8 => R1C8 = 4 (hidden single in C8)
-> 4 must be in R1C8 + R12C9, locked for N3
23a. 4 in window at R1C3 only in R1C458, locked for R1
23b. 4 in C7 only in R6789C7, CPE no 4 in R7C5 using short diagonal

24. 4 in hidden 45(9) cage only in R2C9 + R89C1, CPE no 4 in R2C1
24a. 5 in hidden 45(9) cage only in R12C9 + R8C1, CPE no 5 in R8C9

25. Consider combinations for R12C9
R12C9 = [84] => R2C6 = 8 (hidden single in R2) => R2C7 = 6 (hidden single in R2)
R12C9 = {58}, locked for N3 => R2C7 = 6
-> R2C7 = 6
25a. 6 in C8 only in R45C8, locked for N6
25b. R8C9 = 6 (hidden single in C9)
25c. 6 in window at R6C5 only in R6C56, locked for R6 and N5

26. Consider combinations for R89C1
R89C1 = {14}, locked for N7 => R8C3 = 3
R89C1 = [51] => R8C4 = 1 (hidden single in R8) => R8C3 = 3 (hidden single in R8)
-> R8C3 = 3
26a. 3 in C2 only in R56C2, locked for N4
26b. R2C1 = 3 (hidden single in C1)
26c. 3 in window at R2C3 only in R4C45, locked for R4 and N5

27. 7,9 in R2 only in R2C345, locked for window at R2C3, no 7,9 in R34C345, CPE no 9 in R1C3 using short diagonal
27a. 2,9 in R8 only in R8C567, locked for window at R6C5, no 2,9 in R67C567, CPE no 9 in R9C7 using short diagonal

28. 4 in N1 only in R2C3 + R3C123, CPE no 4 in R3C4
28a. 5 in N9 only in R7C789 + R8C7, CPE no 5 in R7C6

29. 4 in R3 only in R3C123, locked for N1
29a. 5 in R7 only in R7C789, locked for N9

30. R89C1 = {14}/[51] cannot be [51], here’s how
R89C1 = [51], locked for hidden 45(9) cage => R1C9 = 8, R3C2 = 5 (hidden single in C2), R2C34 = {79}, R1C3 = 7 clashes with R2C34 using short diagonal
30a. -> R89C1 = {14}, locked for C1, N7 and hidden 45(9) cage, no 4 in R2C9
30b. R9C2 = 5 (hidden single in N7), R3C2 = 4
30c. Naked pair {58} in R12C9, locked for C9 and N3
30d. Naked pair {58} in R2C69, locked for R2
30e. Naked pair {79} in R2C34, locked for R2, CPE no 7 in R1C3 using short diagonal
30f. R2C5 = 4
30g. Naked pair {58} in R1C3 + R2C6, locked for window at R1C3, no 5,8 in R1C456
30h. Naked triple {389} in R567C2, locked for window at R5C2, no 8,9 in R5C34 + R6C3 + R7C4
30i. 9 in C2 only in R67C2, CPE no 9 in R7C1 using short diagonal
30j. Naked quad {1379} in R1C456 + R2C4, locked for N2
30k. 1 in window at R2C3 only in R4C345, locked for R4

31. Naked pair {14} in R8C14, locked for R8
31a. Naked pair {29} in R8C67, locked for R8, CPE no 2 in R9C7 using short diagonal
31b. R8C5 = 5
31c. Naked pair {14} in R79C7, locked for C7 and N9 -> R7C8 = 5

32. R1C8 = 4 (hidden single in C8)
32a. Naked triple {169} in R345C8, locked for window at R3C6, no 9 in R4C7, no 1 in R5C6
32b. 9 in C8 only in R34C8, CPE no 9 in R3C9 using short diagonal

33. R7C7 = 4 (hidden single in window at R6C5), R9C7 = 1, R9C1 = 4, R8C1 = 1, R8C4 = 4
33a. Naked quad {2689} in R8C6 + R9C456, locked for N8
33b. 8 in window at R6C5 only in R6C567, locked for R6

34. Naked pair {58} in R13C3, locked for C3 and N1
34a. R3C3 = 5 (hidden single in window at R2C3), R1C3 = 8, R1C9 = 5, R2C9 = 8, R2C6 = 5

35. Consider placements for R3C8
R3C8 = 1 => R3C9 = 7, R4C8 = 9 (hidden single in C8) => R4C9 = 2
R3C8 = 9 => R8C7 = 7 (hidden single in C7) => R7C9 = 2
-> 2 must be in R4C9 + R7C9, locked for C9

36. 2 in R6 only in R6C13, locked for N4 -> R4C2 = 6, R4C3 = 1

and the rest is naked singles.

When I first saw the puzzles in this thread I thought that this might be the easiest one, partly because I've only done a few non-consecutive puzzles and those not for a long time. Maybe I'll try one of those now and see how I get on with it.


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PostPosted: Mon Dec 06, 2010 7:22 am 
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I assume we're not really meant to try NC Bloomdoku #1 so I didn't. Thanks Simon for NC Bloomdoku #2. A fun puzzle and much easier than Ninjutsudoku #1.

It took me a while to spot my key steps 7 and 8. After that it was fairly straightforward.
I suppose that step 21a is my hardest one:
technically a forcing chain although it's so easy that it can scarcely be considered to be one.

Here is my walkthrough for NC Bloomdoku #1. I've given NC eliminations and stated petal placements for those who, like me, don't work with software solvers in editor mode:
Prelims. Eliminate the givens in rows, columns, nonets and petals.

The following are non-consecutive eliminations for the givens.
a) No 2 in R24C4 + R3C35
b) No 2 in R68C6 + R7R57, no 4 in R7C7
c) No 3 in R9C7, no 5 in R8C6 + R9C57
d) No 4 in R46C3 + R5C24, no 6 in R46C3
e) No 5 in R46C7, no 7 in R46C7 + R5C68
f) No 6,8 in R1C35 + R2C4
g) No 7 in R46C1 + R5C2, no 9 in R46C1
h) No 8 in R46C9

1. R5C5 = 7 (hidden single in R5), no 6,8 in R46C5
1a. 7 in upper-left petal only in R234C3, locked for C3
1b. 7 in lower-left petal only in R67C2, locked for C2

2. R5C8 = 4 (hidden single in R5), placed for upper-right petal, no 3,5 in R46C8
2a. 3 in R5 only in R5C24, locked for lower-left petal

3. R8C6 = 7 (hidden single in C6), placed for lower-right petal, no 6 in R8C5, no 8 in R8C57
3a. 1 in N8 only in R789C5, locked for C5

4. 3 in C5 only in R1234C5, CPE no 3 in R2C4
4a. 3 in N1 only in R123C5, locked for C5

5. 5 in N6 only in R46C9, locked for C9

6. 3 in upper-right petal only in R3C78 + R4C7, CPE no 3 in R12C7
6a. 7 in upper-right petal only in R3C78 + R4C8, CPE no 7 in R2C8

7. R5C6 = {12} -> no 1 in R46C6, no 2 in R4C6 (R45C6 and R56C6 cannot be consecutive)
7a. R5C6 = 1 (hidden single in C6), placed for upper-right petal

8. R5C4 = {23} -> no 3 in R4C4, no 2 in R6C4 (R45C4 and R56C4 cannot be consecutive)
8a. R5C4 = 3 (hidden single in C4), R5C2 = 2, placed for lower-left petal, no 1,3 in R4C2, no 4 in R46C4, no 1 in R6C2

9. 4 in N5 only in R46C5, locked for C5
9a. R2C4 = 4 (hidden single in N2), placed for upper-left petal, no 3 in R2C3, no 3,5 in R2C5
9b. R6C5 = 4 (hidden single in N5), placed for lower-right petal, no 5 in R6C6 + R7C5
9c. 4 in lower-left petal only in R7C23, locked for R7 and N7
9d. R8C9 = 4 (hidden single in R8), no 3,5 in R8C8, no 3 in R9C9
9e. R1C7 = 4 (hidden single in N3), no 3 in R1C8, no 5 in R1C68 + R2C7

10. R9C8 = 3 (hidden single in N9), no 2 in R8C8 + R9C79

11. R6C9 = 5 (hidden single in R6), no 6 in R7C9
11a. R6C1 = 3 (hidden single in R6), no 2 in R7C1
11b. 3 in upper-left petal only in R3C35, locked for R3
11c. 3 in N3 only in R12C9, locked for C9 -> no 2 on R12C9 (R12C9 cannot be consecutive)

12. R4C5 = 2 (hidden single in N5), placed for upper-left petal, no 3 in R3C5
12a. R1C5 = 3 (hidden single in C5), no 2 in R1C6
12b. R2C9 = 3 (hidden single in N3), no 2 in R2C8 + R3C9
12c. R3C3 = 3 (hidden single in R3), no 4 in R3C2
12d. R3C1 = 4 (hidden single in R3), no 5 in R2C1 + R3C2
12e. R4C2 = 4 (hidden single in R4)
12f. R7C3 = 4 (hidden single in R7)
12g. R4C7 = 3 (hidden single in R4), no 2 in R3C7
12h. R8C2 = 3 (hidden single in R8), no 2 in R8C13

13. R67C4 must contain 6 (R67C4 cannot be {89} which would be consecutive), locked for C4 and lower-left petal, no 6 in R67C2

14. Naked pair {79} in R4C3 + R6C2, locked for N4 -> R6C3 = 1, placed for lower-left petal, R4C1 = 6
14a. R4C3 = {79} -> no 8 in R4C4
14b. R4C9 = 1 (hidden single in R4)

15. R7C9 = 2 (hidden single in C9), no 1 in R7C8
15a. R8C4 = 2 (hidden single in R8), no 1 in R8C5

16. 5 in R1 only in R1C12, locked for N1
16a. R1C12 cannot be {56} (cannot be consecutive) -> no 6 in R1C2

17. R6C7 = 2 (hidden single in lower-right petal), no 1 in R7C7
17a. Naked pair {78} in R46C8, locked for C8

18. R6C2 = {79} -> no 8 in R7C2
18a. Naked pair {79} in R67C2, locked for C2 and lower-left petal, no 9 in R67C4
18b. Naked pair {68} in R67C4, locked for C4

19. Naked pair {59} in R8C5 + R9C4, locked for N8, no 6 in R9C5 (must be adjacent to one of the 5s)
19a. 6 in N8 only in R7C45, locked for R7

[There’s possibly some UR reason why R3C12 cannot be [46], although I know that it’s risky to use UR when there are cages/windows/petals. Anyway I don’t use UR because I consider it’s not completely solving a puzzle.]

20. 7 in N1 only in R2C13, locked for R2
20a. 8 in R4 only in R4C68, locked for upper-right petal, no 8 in R3C67
20b. 7 in R7 only in R7C12, locked for N7

21. R7C78 = {589} must contain 5 (R7C78 cannot be {89} which would be consecutive), 5 locked for R7 and N9
21a. R7C8 = 9 or R7C7 = {89} (R78C7 cannot be [89] which would be consecutive) -> no 9 in R8C7
21b. R8C7 = 1, placed for lower-right petal
21c. R9C5 = 1 (hidden single in N8)
21d. R7C1 = 1 (hidden single in N7)

22. Naked pair {68} in R7C45, locked for R7
22a. Naked pair {59} in R7C78, locked for R7 and N9 -> R7C2 = 7, R6C2 = 9, R4C3 = 7, R46C8 = [87], R8C8 = 6, no 5 in R7C8
22b. R7C8 = 9, R7C7 = 5, placed for lower-right petal, R8C5 = 9, R8C1 = 5, R8C3 = 8, no 9 in R9C3
22c. R9C2 = 6, R9C3 = 2

and the rest is naked singles.


Last edited by Andrew on Sat Jan 08, 2011 12:42 am, edited 2 times in total.

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PostPosted: Wed Dec 08, 2010 5:42 am 
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I'm still struggling with Torn Bloomdoku #2 so I'll take the liberty of posting a short comment about Torn Bloomdoku #1 now.

Simon wrote:
Apperently, combining with the NC (Non Consecutive) rule, this variant has only a space of 2 solutions! :!:

So to make a valid puzzle, only 1 more constraint is required: R1C1<>1.

That indicates that the other valid puzzle with this petal pattern has R1C1 = 1:
Using numerical symmetry it seems a logical conclusion, although I can't prove it, that Torn Bloomdoku #1 will have R1C1 = 9. That reduces it to, I think, only 698 remaining candidates (various 9s eliminated and a pair of 8s). Good luck to anyone who wants to continue from there. I won't!


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PostPosted: Wed Dec 08, 2010 12:54 pm 
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While NC Bloomdoku #2 is quite easily solvable for novices of NC solving, the same can't be said for NC Torn-Bloomdoku #2.

In a certain old forum there were a few experts (h3lix, JC Bonsai, udosuk, etc) who developed a few advanced techniques on NC solving. Some of those include a move where you look at a certain group of cells (e.g. pair/triple/quad) and determine that they must include certain candidate(s), or else consecutiveness would occur. This is a special case of "NC locked candidates". I don't think it is trial-and-error or forcing chains but other people may disagree.

To solve NC Torn-Bloomdoku #2 you will need to apply the aforementioned technique, not once or twice but many times. But I will still claim it is "human solvable", and I think some other (strong) solvers will agree.

The #1 puzzles for these 2 series aren't really for human solvers, but to test a solving program etc.


As for Andrew's observation:
You are right in the solution for NC Torn-B #1 R1C1=9. The R1C1=1 solution is the one for #2. If you compare the 2 solutions you will notice they are essentially the same, in one solution all cell values are 10 minus the corresponding cell values of the other solution.


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PostPosted: Thu Dec 09, 2010 7:31 am 
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Thanks Simon for your comments about what is required to solve this puzzle without any forcing chains (I don't group those together with trial-and-error, that implies trying something to see if it works while forcing chains make one or more eliminations in cells other than at the start of the hypothesis and stop there; that seems to be a standard distinction between them on the killer forum).

I used several forcing chains, mostly short ones but step 14, which was where I resumed today, is a bit longer (it looks longer than it really is because it includes NC eliminations and hidden single statements).

Here is how I solved NC Torn-Bloomdoku #2:
As I did for NC Bloomdoku #2 I've given NC eliminations and stated petal placements for those who, like me, don't work with software solvers in editor mode.

Prelims. Eliminate the givens in rows, columns, nonets and petals.

The following are non-consecutive eliminations for the givens.
a) No 2,4 in R1C46 + R2C5
b) No 8 in R35C3
c) No 7 in R35C7 + R4C68, no 9 in R35C7
d) No 4 in R46C1, no 6 in R46C1 + R5C2
e) No 3 in R46C9 + R5C8, no 5 in R46C9
f) No 1 in R57C3, no 3 in R57C3 + R6C24
g) No 2 in R57C7
h) No 5,7 in R8C5 + R9C46

[I can see that there is a hidden window for C234 in R189C234. Whether I use it will depend how easily I find progress with this puzzle.]

1. R5C3 = {67} -> no 7 in R5C2, no 6,7 in R5C4 (R5C23 and R5C34 cannot be consecutive)

2. 4 in N4 only in R46C2, locked for C2
2a. 4 in N4 only in R46C2 -> no 3 in R5C2 (R45C2 and R56C2 cannot be consecutive)
2b. 6 in N4 only in R46C2 + R5C3, CPE no 6 in R7C2
2c. 8 in N4 only in R5C2 + R6C12, CPE no 8 in R6C4

3. 5 in N6 only in R46C8, locked for C8
3a. 5 in N6 only in R46C8 -> no 6 in R5C8 (R45C8 and R56C8 cannot be consecutive)
3b. 2 in N6 only in R4C89 + R5C8, CPE no 2 in R4C6
3c. 3 in N6 only in R46C8 + R5C7, CPE no 3 in R3C8

4. 2 in lower-right petal only in R7C56 + R8C567, CPE no 2 in R8C4

5. 3 in N6 only in R46C8 + R5C7 -> no 2 in R5C8 (must be adjacent to one of the 3s)
5a. 2 in N6 only in R4C89, locked for R4
5b. 2 in N6 only in R4C89 -> no 3 in R4C8 (R4C89 cannot be consecutive)
5c. 2 in R5 only in R5C56 -> no 1 in R5C5, no 1,3 in R5C6 (R5C56 cannot be consecutive)
5d. 1 in R5 only in R5C24, locked for lower-left petal, no 1 in R6C24

6. Consider placements for 2 in R5
R5C5 = 2 => no 1 in R5C4 (R5C45 cannot be consecutive)
or R5C6 = 2 => no 3 in R5C7 (R5C67 cannot be consecutive) => R5C4 = 3 (hidden single in R5) => no 1 in R5C4
-> no 1 in R5C4
6a. R5C2 = 1 (hidden single in R5)

7. 8 in R5 only in R5C45, locked for N5
7a. 8 in R5 only in R5C45 -> no 9 in R5C4, no 7,9 in R5C5 (R5C45 cannot be consecutive)
7b. 9 in R5 only in R5C68, locked for upper-right petal, no 9 in R3C68

8. 8 in N4 only in R6C12 -> no 7 in R6C12 (R6C12 cannot be consecutive)

[I’ve just realised that the logic of my comment about Torn Bloomdoku #1, which has the same petal pattern as this puzzle, suggests that R1C1 must be 1 or 9 but, since I can’t prove this I won’t use it. It’s also a plausible thought that, since #1 has R1C1 ≠ 1 then this one may have R1C1 = 1 but that could be bluffing.]

9. 3 in R5 only in R5C47, CPE no 3 in R4C6
9a. 3 in upper-right petal only in R35C7, locked for C7
9b. 3 in lower-right petal only in R678C6, locked for C6
9c. 3 in C6 only in R678C6 -> no 2,4 in R7C6 (cannot be in the same cell as 3 or adjacent to it)

10. Consider placements for R5C3
R5C3 = 6 => R5C7 = 3 => R5C4 = 8, placed for lower-left petal => R6C1 = 8 (hidden single in R6)
or R5C3 = 7 => R4C1 = 3 => R6C1 = 8
-> R6C1 = 8 -> no 7,9 in R7C1 (R67C1 cannot be consecutive)
10a. 3 in N4 only in R4C12, locked for R4
10b. 3 in N4 only in R4C12 -> no 4 in R4C2 (R4C12 cannot be consecutive)
10c. R6C2 = 4 (hidden single in N4), placed for lower-left petal, no 4 in R7C34, no 3,5 in R7C2 (R67C2 cannot be consecutive)
10d. 3 in lower-left petal only in R57C4, locked for C4
10e. 6 in N4 only in R4C2 + R5C3, CPE no 6 in R23C3

11. 4 in R4 only in R4C456 -> no 5 in R4C5 (cannot be in the same cell as 4 or adjacent to it)
11a. 1 in C5 only in R234C5 -> no 2 in R3C5 (cannot be in the same cell as 1 or adjacent to it)

12. 8 in lower-right petal only in R78C56, locked for N8

13. 3 in hidden window R189C234 only in R89C23, locked for N7

14. Consider placements for R5C4
R5C4 = 3 => R5C7 = 6 => R5C3 = 7
or R5C4 = 8 => no 7 in R5C3, no 7,9 in R6C4 => R5C3 = 6, R7C4 = 3 (hidden single in petal) => R7C2 = 9 (hidden single in petal) => R7C3 = 7 (hidden single in petal)
-> 6 in R5C3 + R5C7, locked for R5 and 7 in R5C3 + R7C3, locked for C3 and petal, no 7 in R6C4 + R7C24
14a. R7C2 = {89} -> no 8 in R7C3 (R7C23 cannot be consecutive)
14b. Consider placements for R7C3
R7C3 = {56} => no 5 in R7C4 (R7C34 cannot be consecutive)
R7C3 = 7 => R5C3 = 6 => R5C7 = 3 => R5C4 = 8 => no 9 in R6C4 (R56C4 cannot be consecutive) => R6C4 = 5 => no 5 in R7C4
-> no 5 in R7C4
14c. R57C4 = [39/83] -> no 9 in R6C4 (R56C4 and R67C4 cannot be consecutive)
14d. 9 in lower-left petal only in R7C24, locked for R7

15. R6C4 = {56} -> no 5 in R6C5 (R6C45 cannot be consecutive)
15a. R6C5 = {79} -> no 8 in R5C5 (R56C5 cannot be consecutive)
15b. R5C5 = 2 -> no 3 in R5C4 (R5C45 cannot be consecutive)
15c. R5C4 = 8 -> no 7 in R5C3 (R5C34 cannot be consecutive)
15d. R5C3 = 6, R5C7 = 3, R6C4 = 5, R7C2 = 9, R7C3 = 7, R7C4 = 3, no 8 in R8C23 (R78C2 and R78C3 cannot be consecutive)
15e. 8 in N7 only in R9C23, locked for R9

16. Naked pair {79} in R5C68, locked for upper-right petal, no 7 in R3C68
16a. 5 in upper-left petal only in R23C23, locked for N1
16b. 8 in upper-left petal in R2C23 + R3C2, locked for N1
16c. 8 in R2C23 + R3C2 -> no 7 in R2C2 (cannot be in the same cell as 8 or adjacent to it)

17. Naked triple {679} in R5C8 + R6C89, locked for N6 -> R4C9 = 2, R4C8 = 5, placed for upper-right petal, no 5 in R3C67, no 4,6 in R3C8 (R34C8 cannot be consecutive)
17a. R6C6 = 3 (hidden single in R6)
17b. Naked pair {79} in R5C6 + R6C5, locked for N5, CPE no 7,9 in R8C6

18. 2 in upper-right petal only in R3C678, locked for R3
18a. 2 in upper-left petal only in R2C24, locked for R2
18b. 2 in R2C24 -> no 1,3 in R2C3 (R2C23 and R2C34 cannot be consecutive)
18c. 2 in lower-right petal only in R8C67, locked for R8

19. 3 in C1 only in R234C1 -> no 4 in R3C1 (cannot be in the same cell as 3 or adjacent to it)
[If there’s any similar type of step for 7 in C1 only in R1234C1, for 7 in R134C2 or for 6 in R1234C4, I can’t see it.]

20. R6C5 = {79} -> no 8 in R7C5 (R67C5 cannot be consecutive)
[I ought to have spotted this when I saw step 15a.]
20a. R7C5 = {45} -> no 5 in R7C6 (R7C56 cannot be consecutive), no 4 in R8C5 (R78C5 cannot be consecutive)
20b. R7C6 = 8, R8C5 = 9, R6C5 = 7, R5C6 = 9, R5C8 = 7, no 6 in R6C8 (R56C8 cannot be consecutive)
20c. R6C8 = 9, R6C9 = 6, no 5 in R7C9 (R67C9 cannot be consecutive)
20d. R7C9 = 1, no 2 in R7C8 (R7C89 cannot be consecutive)

21. R7C8 = {46} -> no 5 in R7C7 (R7C78 cannot be consecutive)
21a. Naked pair {46} in R7C78, locked for R7 and N9 -> R7C1 = 2, R7C5 = 5, R8C7 = 2, R8C6 = 4, R7C7 = 6, R7C8 = 4, R3C7 = 4, R9C8 = 3, R8C8 = 8, no 5 in R2C7 (R23C7 cannot be consecutive), no 1 in R8C1 (R78C1 cannot be consecutive), no 7 in R8C9 (R8C89 cannot be consecutive)
21b. R8C1 = 6, R8C9 = 5, R8C2 = 3, R8C3 = 1, R8C4 = 7, R1C3 = 4, R4C12 = [37], R9C1 = 4, no 6,8 in R3C2 (R34C2 cannot be consecutive)

22. R3C2 = 5, R3C3 = 3, R2C3 = 8, R23C5 = [18], R2C8 = 6, R2C2 = 2, R1C2 = 6, R123C4 = [946], no 7 in R1C1 (R1C12 cannot be consecutive), no 7 in R2C79 (R2C78 and R2C89 cannot be consecutive)
22a. R1C1 = 1, R2C7 = 9

and the rest is naked singles.

Maybe after a few more days, to allow other solvers the chance to use your hints, you can post a walkthrough showing how this puzzle should be solved using the non-consecutive techniques referred to in your post. I'm definitely a novice at NC solving so will be interested to see how they should be used.
I may have managed:
to find some of them in my solving path for pairs and triples but not for linear or square quads although at least one step may be a T-shaped quad.


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PostPosted: Sun Dec 12, 2010 12:47 am 
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simon_blow_snow wrote:
NC Torn-Bloomdoku #2.... a few advanced techniques on NC solving. Some of those include a move where you look at a certain group of cells (e.g. pair/triple/quad) and determine that they must include certain candidate(s), or else consecutiveness would occur. This is a special case of "NC locked candidates"
I used closer to the inverse, an extreme sort of "NC blocked candidates". Same as Andrew's steps 6 & 10. Then found some more instead of Andrew's step 14, like a squeeze move. Very enjoyable puzzle. Thanks Simon!

NC Torn-Bloomdoku #2
ALT step 14:
Image shows r456:
Image
Code:
.-------------------------------.-------------------------------.-------------------------------.
| 124679    256789    145678    | 156789    3         156789    | 245679    1246789   1256789   |
| 1234679   235678    134578    | 1245678   15789     12456789  | 245679    12346789  12356789  |
| 1234679   235678    13457     | 1245678   145789    124567    | 23456     12467     12356789  |
:-------------------------------+-------------------------------+-------------------------------:
| 37        367       9         | 14567     147       1456      | 8         256       267       |
| 5         1         67        | 38        28        2679      | 36        79        4         |
| 8         4         2         | 5679      579       35679     | 1         35679     679       |
:-------------------------------+-------------------------------+-------------------------------:
| 1246      789       5678      | 3579      245789    35789     | 45679     12346789  12356789  |
| 124679    2356789   1345678   | 14579     2489      2345789   | 245679    12346789  12356789  |
| 12479     235789    134578    | 1249      6         1249      | 24579     1234789   1235789   |
'-------------------------------.-------------------------------.-------------------------------'


Andrew's step 13 to above.

ALT 14
14. r5c456+r6c5 = [3825]/[82]{5679} ie, r6c5 = 5 or r5c6+r6c5 = {5679}
14a. -> no 6 in r6c6 (NC squeeze with r5c6 and r6c5)

15. r5c345+r6c5 = (67)[385]/[682](579) ie, r6c5 = 5 or r5c3 = 6
15a. -> no 6 in r6c4 (CPE with r5c3 & NC with r6c5)

16. 6 in r6 only in n6: locked for n6
Cracked

Cheers
Ed


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PostPosted: Tue Dec 14, 2010 11:56 am 
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After retrying NC Torn-Bloomdoku #2 and eliminating the unnecessary steps, I found this unbelievably short solving path! 8-)

4-step crack-in for NC Torn-Bloomdoku #2:
Petals: P1=R234C234, P2=R345C678, P3=R567C234, P4=R678C567

0 (NC prelim):
R1C46+R2C5<>{24}
R35C3+R4C24<>{8}
R35C7+R4C68<>{79}
R46C1+R5C2<>{46}
R46C9+R5C8<>{35}
R57C3+R6C24<>{13}
R57C7+R6C68<>{2}
R8C5+R9C46<>{57}

1:
R5C3={6/7} --> R5C24<>{67}
R46C2 must include {4} of N4 [C2] --> R5C2<>{3}
R46C8 must include {5} of N6 [C8] --> R5C8<>{6}
R4C89+R5C8 must include {2} of N6 --> R4C8<>{3}
--> R5C7+R6C8 must include {3} of N6 --> R5C8<>{2}
--> R5C56 must include {2} of R5 [N5] --> R5C56<>{13}

2:
R5C47 must include {3} of R5 [R4C6]
R5C56 must include {2} of R5 --> R5C47<>[13] --> R5C4<>{1}
Hidden single R5: R5C2=1 [N4,P3]
--> R5C45 must include {8} of R5 [N5] --> R5C45<>{79}
R4C12<>[34], must include {6/7}
-->R4C12+R5C3 must include {67} of N4 --> R6C12<>{67}

3:
R67C2<>[43/83] (R5C4) --> R7C2<>{3}
--> R57C4 must include {3} of P3 [C4] --> R6C4<>{4}
--> R4C12 must include {3} of R4 [N4] --> R6C12=[84]
R5C34<>[78], must include {3/6}
--> R5C347 must include {36} of R5 --> R5C6<>{6}

4:
R5C56+R6C5<>[829] --> R5C6+R6C5 must include {5/7}
--> R6C6<>{6} --> R78C7 must include {6} of P4 [C7,N9]

Cracked.


How about this, a one-step-to-break-it-all puzzle (did Ruud call these one-trick-ponies or something)?

So I want to see what Andrew & Ed think:

1. Are steps 1-3 forcing-chain-free? (I think they are.)

2. Should step 4 (first line) be considered as a forcing chain? (I don't think so.)


Last edited by simon_blow_snow on Wed Feb 02, 2011 6:16 pm, edited 1 time in total.

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PostPosted: Wed Dec 22, 2010 11:40 pm 
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About Torn-Bloomdoku 2 simon_blow_snow wrote:
I found this unbelievably short solving path! 8-)
Really good solution Simon! Don't think it's particularly short but have no hesitation in adding my 8-). It's quite similiar to Andrew's with my alternative step though a couple of your steps are simpler. I've described them here.
simon_blow_snow wrote:
a one-step-to-break-it-all puzzle (did Ruud call these one-trick-ponies or something)?
Sure doesn't feel like a O-T-P to me. Lots of NC tricks in there (see link above). You must be very experienced at NC puzzles if all the other NC tricks felt routine. Interesting that we still think about Ruud nearly 3 years after he disappeared! You've presumably been lurking for quite a while!!

As to your questions about forcing chains...I can't comment since I don't even know what the technical definition of one is...I just know I like chains when I've found one!!

Thanks again for a really great puzzle.
Ed


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PostPosted: Sat Jan 08, 2011 12:45 am 
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I've been busy with other things so have only just gone through Simon's and Ed's latest posts in this thread and the linked posts in the techniques forum.

Nice Crack-In by Simon. It shows what a NC expert can achieve, compared with my walkthrough from a NC novice. It included several neat steps; as Ed commented the final lines of steps 2 and 3 were both killer-like, well they were killer pairs even though this isn't a killer sudoku.

Simon wrote:
Are steps 1-3 forcing-chain-free? (I think they are.)
I'd say almost forcing-chain-free. The neat second line of step 2 "R5C56 must include {2} of R5 --> R5C47<>[13] --> R5C4<>{1}" is probably a short forcing chain.

Ed wrote:
As to your questions about forcing chains...I can't comment since I don't even know what the technical definition of one is...I just know I like chains when I've found one!!
I haven't seen a definition either. I agree that it feels good once one has found a forcing chain. I've got nothing against forcing chains; they are preferable to contradiction moves although I find the latter easier to spot.

Simon wrote:
Should step 4 (first line) be considered as a forcing chain? (I don't think so.)
It's permutation analysis, using NC properties, so not a forcing chain.


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