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Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
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Posted: Thu Aug 30, 2012 4:45 pm Post subject: Vanhegan extreme |
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Another one that I'm stuck on. This is puzzle #6-1645514, rated 2.2.0.1.1.
Code: |
+------------------+------------+-----------------+
| 4 27 2679 | 17 69 3 | 2789 5 1289 |
| 369 37 1 | 5 8 2 | 379 3467 349 |
| 23569 8 235679 | 17 69 4 | 2379 2367 1239 |
+------------------+------------+-----------------+
| 7 6 234 | 8 245 9 | 1 234 2345 |
| 129 5 29 | 3 124 7 | 6 8 24 |
| 123 234 8 | 6 1245 15 | 235 9 7 |
+------------------+------------+-----------------+
| 358 347 3457 | 2 57 6 | 35789 1 3589 |
| 256 1 2567 | 9 3 8 | 4 27 25 |
| 2358 9 2357 | 4 157 15 | 23578 237 6 |
+------------------+------------+-----------------+
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Play this puzzle online at the Daily Sudoku site |
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arkietech
Joined: 31 Jul 2008 Posts: 1834 Location: Northwest Arkansas USA
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Posted: Fri Aug 31, 2012 5:20 pm Post subject: |
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Couldn't break this one. I was hoping someone else could. |
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Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
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Posted: Fri Aug 31, 2012 6:27 pm Post subject: |
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Dan, did you by any chance do something to check if the grid is valid? |
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JC Van Hay
Joined: 13 Jun 2010 Posts: 494 Location: Charleroi, Belgium
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Posted: Fri Aug 31, 2012 6:45 pm Post subject: |
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Looking at the set of bivalues and bilocals, the candidates for the digits 4 and 1 are the "strongest" (see why !).
Furthermore they are "coupled" in r5c5.
Therefore, I suggest, as a hint, to investigate the consequences of assuming either 1r5c5 or 2r5c5 or 4r5c5 to be true.
If you do so, you will save much time in the solving of this puzzle !
Regards, JC. |
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Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
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Posted: Fri Aug 31, 2012 7:18 pm Post subject: |
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I don't know if that is what JC had in mind, but looking at the 124 possibilities in r5c5, I found 4 to be invalid, and that was all that was needed. |
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JC Van Hay
Joined: 13 Jun 2010 Posts: 494 Location: Charleroi, Belgium
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Posted: Sat Sep 01, 2012 7:37 am Post subject: |
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Yes Marty.
With the bilocals on 4s and 1s, one can make 2 chains of length 4 and 2, a "Jellyfish" and a "Kite", respectively.
As they are weakly coupled in r5c5, it is natural to begin the analysis of the puzzle from the cell r5c5 before investigating any other "weaker" patterns.
r5c5=1->solution
Or
r5c5=2->contradiction after a cascade of singles
or
r5c5=4->contradiction after a cascade of singles while r5c9=4->solution
It remains to find one chain to explain -4r5c5. The length of that chain should be the shortest, but this is not necessary.
Example:
Code: | +--------------------------+------------------+---------------------+
| 4 27 2679 | 17 69 3 | 2789 5 1289 |
| 69(3) 7(3) 1 | 5 8 2 | 379 3467 349 |
| 2569(3) 8 2679(35) | 17 69 4 | 2379 2367 1239 |
+--------------------------+------------------+---------------------+
| 7 6 23(4) | 8 245 9 | 1 234 2345 |
| 129 5 29 | 3 12-4 7 | 6 8 (24) |
| 12(3) 2(34) 8 | 6 125(4) 1(5) | 2(35) 9 7 |
+--------------------------+------------------+---------------------+
| 358 (347) 37(45) | 2 (57) 6 | 3789(5) 1 3589 |
| 256 1 267(5) | 9 3 8 | 4 27 (25) |
| 2358 9 237(5) | 4 157 1(5) | 2378(5) 237 6 |
+--------------------------+------------------+---------------------+
Chain[11] : If r5c5=4, then [r5c9=2,r8c9=5=r6c7=r9c6,r7c5=7] and r6c2=4=r7c3;r6c1=3=r7c2,r3c3=5;no 3 in B1 => -4r5c5 |
I hope that this way of writing the chain will seem easier to read than the corresponding "forbidding matrix", where no assumption is done, proving the derived SL 4r5c9=4r6c5 :=> -4r5c5.
Note : the rating of this puzzle is SER=8.4, while the puzzles proposed on this site rarely exceed SER=7.1 !
Regards, JC. |
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