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garytorborg
Joined: 19 Jan 2011
Posts: 28
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 Posted: Mon Jan 31, 2011 9:50 pm Post subject: Another Fiendish Puzzle |
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Thanks to all who helped me last time. I have since successfully solved several "fiendish" puzzles from the program I use, but have once again run into a brick wall.
After the basics, three x-wings (!), and an xy-wing, I have arrived at this position and am now stuck.
| Code: |
|-----|-----|-----|-----|-----|-----|-----|-----|-----|
| 89 | 5 | 4 | 128 | 189 | 7 | 3 | 6 | 12 |
| 1 | 689 | 3 | 5 | 689 | 289 | 4 | 29 | 7 |
| 7 | 2 | 69 | 3 | 169 | 4 | 5 | 8 | 19 |
|-----|-----|-----|-----|-----|-----|-----|-----|-----|
| 246 | 69 | 8 | 46 | 7 | 3 | 1 | 5 | 29 |
| 234 | 7 | 1 | 9 | 48 | 5 | 6 | 24 | 38 |
| 346 | 369 | 5 |1468 | 2 | 18 | 7 | 49 | 38 |
|-----|-----|-----|-----|-----|-----|-----|-----|-----|
| 89 | 1 | 29 | 7 | 5 | 6 | 28 | 3 | 4 |
| 5 | 38 | 7 | 248 | 348 | 28 | 9 | 1 | 6 |
| 36 | 4 | 26 | 18 |1389 | 189 | 28 | 7 | 5 |
|-----|-----|-----|-----|-----|-----|-----|-----|-----|
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Include things like doubles, triples, and quads in the "basics". Anybody have an idea? Thanks in advance. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Mon Jan 31, 2011 10:23 pm Post subject: |
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| There's a W-Wing on the 69 cells in boxes 14, linked by the 9s in c9. That should make a big dent, maybe solve it. |
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Mon Jan 31, 2011 10:24 pm Post subject: |
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You need a new type of wing - a w-wing!
Look at the two (69) bivalues in b1 and b4 - notice the aligned strong link on (9) in c9... |
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garytorborg
Joined: 19 Jan 2011
Posts: 28
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| Posted: Mon Jan 31, 2011 10:31 pm Post subject: |
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| As far as I can see, the W-wing you describe only results in a single elimination: the 9 in r2c2, leaving 68. Any other further clues? |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Mon Jan 31, 2011 10:35 pm Post subject: |
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If you are familiar with XY-Chains, then here is a 4-cell XY-Chain that cracks the puzzle.
| Code: | (4=6)r4c4 - (6=9)r4c2 - (9=2)r4c9 - (2=4)r5c8 => r5c5<>4
+--------------------------------------------------------------+
| 89 5 4 | 128 189 7 | 3 6 12 |
| 1 689 3 | 5 689 289 | 4 29 7 |
| 7 2 69 | 3 169 4 | 5 8 19 |
|--------------------+--------------------+--------------------|
| 246 b69 8 | a46 7 3 | 1 5 c29 |
| 234 7 1 | 9 8-4 5 | 6 d24 38 |
| 346 369 5 | 1468 2 18 | 7 49 38 |
|--------------------+--------------------+--------------------|
| 89 1 29 | 7 5 6 | 28 3 4 |
| 5 38 7 | 248 348 28 | 9 1 6 |
| 36 4 26 | 18 1389 189 | 28 7 5 |
+--------------------------------------------------------------+
# 55 eliminations remain
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peterj
Joined: 26 Mar 2010
Posts: 974
Location: London, UK
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| Posted: Mon Jan 31, 2011 10:48 pm Post subject: |
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| garytorborg wrote: | | As far as I can see, the W-wing you describe only results in a single elimination: the 9 in r2c2, leaving 68. Any other further clues? |
You have made the wrong elimination - it should be a r2c2<>6!
If r3c3 is not a 6 it must be a 9, so r3c9 is not a 9...
So r4c9 must be a 9 and hence r4c2 must be a 6.
Whatever one end of this chain must be a 6 and so r2c2<>6
Also you can eliminate the 9 from r2c2 because of locked 9s pointing in b4...
I think that cracks it. |
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