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| Author: | JanVerbeke [ Wed Oct 21, 2009 12:58 pm ] |
| Post subject: | New technique? |
I possibly found a new technique, but since I’m new to this community I would like to submit it here for verification. Take as an example this Sudoku (Tazuku Sudoku Extreme 9 no. 31) 700090004001000800000207000008102400900000005007809300000905000009000100500030006 in this state of solution: .-------------------------------.-------------------------------.-------------------------------. | 7 236 2356 | 36 9 8 | 256 1 4 | | 236 9 1 | 5 4 36 | 8 367 237 | | 8 4 356 | 2 1 7 | 569 369 39 | :-------------------------------+-------------------------------+-------------------------------: | 36 356 8 | 1 56 2 | 4 79 79 | | 9 1 26 | 34 7 34 | 26 8 5 | | 4 256 7 | 8 56 9 | 3 26 1 | :-------------------------------+-------------------------------+-------------------------------: | 1 36 346 | 9 28 5 | 7 34 28 | | 23 7 9 | 46 28 46 | 1 5 238 | | 5 8 24 | 7 3 1 | 29 249 6 | '-------------------------------.-------------------------------.-------------------------------' Looking at the connected pairs for candidate 2 you will find two clusters: • Cluster 1 = R1C2 – R6C2 – R5C3 – R5C7 – R6C8 – R9C8 • Cluster 2 = R1C7 – R2C9 – R2C1 – R8C1 – R9C3 We give the first cell in a cluster a positive parity, the second a negative and so on. Let’s have a look now to the candidates 2 that are not part of either clusters. The candidate in cell R8C9 sees both clusters, namely cluster 1 in R9C8 and cluster 2 in R2C9. Both have a positive parity. Should R8C9 contain a 2 then both clusters will have the negative parity (meaning that all cells with negative parity in the cluster will get a 2). This also has as a consequence that both R5C3 and R9C3 would get a 2, two cells in the same column. And both R1C2 and R2C1 would get a 2, two cells in the same box. This implies that R8C9 cannot contain a 2. The same reasoning holds for R7C9 where we can eliminate the 2 too. With this technique clusters of the same candidate are linked, which is not the case with normal coloring techniques. Also, 3D Medusa coloring is different in that clusters of different candidates are linked. Thanks for the feedback! |
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| Author: | Afmob [ Wed Oct 21, 2009 5:31 pm ] |
| Post subject: | |
Hi JanVerbeke, welcome to the forum! Your reasoning is right though R9C8 and R2C9 have negative parity so overall both clusters would have positive parity if one of R78C9 has 2. This technique might be useful if those two clusters aren't connected but in this case they are connected (R5C7 = 2 forces R2C9 = 2, HS @ N3), so it can be replaced by just using coloring without finding a contradiction. Here's how: (R1C2 - R5C3 and R2C9 means: R1C2 = 2 forces R5C3 = 2 and R2C9 = 2 via Hidden Single) Chain 1) R1C2 - R5C3 and R2C9 - R6C8 and R8C1 - R7C5 and R9C3 Chain 2) R6C2 - R5C7 - R2C9 - R1C3 and R9C8 - R8C1 - R7C5 Conclusion: R2C9 = 2, R8C1 = 2 and R7C5 = 2 which also explains why R78C9 <> 2. |
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| Author: | JanVerbeke [ Wed Oct 21, 2009 6:58 pm ] |
| Post subject: | Re: New technique? |
Hello Afmob, Thanks, that's clear to me now. I was suspecting this, but my only source was the Sudoku Solving Guide of SudoCue Ruud. Chains are only built with connected pairs, not by e.g. hidden singles. But it looks quite natural to do this and in fact, I could use it again in the sudoku I was currently solving! Any advise on guides/sites that complement the Sudoku Solving Guide well? Jan |
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| Author: | Børge [ Wed Oct 21, 2009 7:03 pm ] |
| Post subject: | Re: New technique? |
JanVerbeke wrote: Any advise on guides/sites that complement the Sudoku Solving Guide well? www.sudopedia.orgSudoku Players' Forums |
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