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NonConsecutive sudoku techniques http://www.rcbroughton.co.uk/sudoku/forum/viewtopic.php?f=10&t=855 |
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Author: | Ed [ Mon Dec 20, 2010 11:38 pm ] |
Post subject: | NonConsecutive sudoku techniques |
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. (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. (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. (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. (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 |
Author: | simon_blow_snow [ Tue Dec 21, 2010 5:47 pm ] |
Post subject: | Re: NonConsecutive sudoku techniques |
Ed, thanks for liking the puzzle and the techniques! Ed wrote: 4. (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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