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PostPosted: Mon Dec 20, 2010 11:38 pm 
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NonConsecutive (NC) sudoku is a really interesting variation on traditional sudoku with interesting ways of locking and blocking candidates. I'm a novice at this variant but really enjoy them....and have only ever solved them with another constraint thrown in eg with killers.

Here are a couple of really nice NonConsective (NC) locking and blocking techniques used in Torn-Bloomdoku #2 here. These come from Andrew's walkthrough here, my alternative step here and most from Simon's Crack-in here. I have taken the liberty of expressing the steps in my expansive stye. Any mistakes are mine.

1.
Image
(This is from Andrew's step 6 and Simon's step 2 line 2.)

No 1 in r5c4 because of the 3s and 2s in r5. Like this.
a. r5c47 cannot be [13] because 2 is locked at r5c56 (NC)
b. r5c47 cannot be [16] because 3 is locked at one of those two cells
c. -> no 1 in r5c4

2.
Image
(This is from Simon's step 2 lines 5 & 6.)

This one is a really nice way hidden pair which feels very killer like.
a. r4c12 cannot be [34](NC)
b. -> must have 6 or 7
c. which forms a hidden pair with r5c3
d. -> no 6 or 7 in r6c12

3.
Image
(This is from Simon's step 3, line 1).

This one combines NC with a "petal" extra group to make an important elimination (no 3 in r7c2).
a. r67c2 cannot be [43](NC)
b. r67c2 cannot be [83] since that would clash with r5c4 = (38) which is in the same petal.
c. -> no 3 in r7c2

4.
Image
(This is from my alt step and Simon's step 4 line 1.)

This was a tricky step for me to find because quite a few candidates are involved. Andrew used a different way to make this elimination (his step 14).
a. r5c56+r6c5 = [2]{579}: ie r5c6+r6c5 must have at least one of 5 or 7
b. or 5c56+r6c5 = [82][5] (r56c5 can't be [87/89](NC) [corrected after following post]; and 2 is locked at r5c56 for n5)
c. both options for r5c6+r6c5 have 5 or 7 -> no 6 in r6c6 (NC)

Cheers
Ed


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PostPosted: Tue Dec 21, 2010 5:47 pm 
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Joined: Thu Oct 07, 2010 3:21 pm
Posts: 170
Ed, thanks for liking the puzzle and the techniques! :santa:

Ed wrote:
4.
Image
(This is from my alt step and Simon's step 4 line 1.)

This was a tricky step for me to find because quite a few candidates are involved. Andrew used a different way to make this elimination (his step 14).
a. r5c56+r6c5 = [2]{579}: ie r5c6+r6c5 must have at least one of 5 or 7
b. or 5c56+r6c5 = [82](57): ie must have 5 or 7
c. both options for r5c6+r6c5 have 5 or 7 -> no 6 in r6c6 (NC)


Case 4b should be:

b. or r5c56+r6c5 = [825] because r56c5 <> [87/89] (NC rule)


Or better yet, this is how I would present this step:

r5c56+r6c5 <> [829] (NC)
--> r5c56+r6c5 = {??5/??7} with {5/7} in r5c6+r6c5
--> r6c6 <> {6} (NC)


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