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Rocky Mozell
Joined: 28 Jul 2010
Posts: 34
Location: Boston, MA
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 Posted: Thu Jun 12, 2014 2:21 pm Post subject: Jun 12 VH |
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358 xy wing pivot at r1c8.
Where is everyone? |
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blindleader
Joined: 11 Jan 2013
Posts: 7
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| Posted: Thu Jun 12, 2014 4:57 pm Post subject: |
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Two theories.
1 Everyone is busy playing in the summer sun.
2 Everyone thought this puzzle was too easy to comment on. By far the easiest VH I've ever seen here. Simple basics get you so close to the end that the xy-wing just jumps out at you. YMMV  |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Thu Jun 12, 2014 6:52 pm Post subject: |
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| Code: |
+----------+---------+----------+
| 2 1 58 | 6 38 4 | 9 35 7 |
| 9 47 48 | 1 378 5 | 38 6 2 |
| 3 67 568 | 9 78 2 | 18 15 4 |
+----------+---------+----------+
| 7 5 1 | 3 9 6 | 2 4 8 |
| 4 8 3 | 5 2 7 | 6 9 1 |
| 6 9 2 | 8 4 1 | 5 7 3 |
+----------+---------+----------+
| 5 2 67 | 4 16 8 | 137 13 9 |
| 1 3 47 | 2 5 9 | 47 8 6 |
| 8 46 9 | 7 16 3 | 14 2 5 |
+----------+---------+----------+
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Play this puzzle online at the Daily Sudoku site
In addition to the XY-Wing, there's also a BUG+3 for those who are so inclined. 8r2c5=8r3c3=1r7c7=>r7c8<>1. |
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Lee
Joined: 10 Jun 2014
Posts: 24
Location: San Francisco
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| Posted: Sun Jun 15, 2014 3:29 pm Post subject: |
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Marty,
Sorry to be so late in my comments, but we have had internet problems the last few days.
Thanks for pointing out the BUG+3.
It is my understanding that in order to cope with the BUG we find a cell that is driven by forcing chains to the same value by all the extra candidates (the two 8's and the 1 in this case). Cell r7c8 seems to be driven to 3 by one of the 8's and the 1, but is driven to 1 by the other 8 (the 8 in r3c3).
It seems that cell r3c5 is driven to 7 by all the extra candidates, and therefore I would think it is the road to solution rather than cell r7c8. In this case the solution would be the same, but that might just be coincidental.
Am I missing or misunderstanding something. I would appreciate comments. Thanks. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sun Jun 15, 2014 5:22 pm Post subject: |
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| Quote: |
It is my understanding that in order to cope with the BUG we find a cell that is driven by forcing chains to the same value by all the extra candidates (the two 8's and the 1 in this case). Cell r7c8 seems to be driven to 3 by one of the 8's and the 1, but is driven to 1 by the other 8 (the 8 in r3c3). |
R3c3=8-(8=7)r3c5-(7=6)r3c2-(6=4)r9c2-(4=1)r9c7-(1=3)r7c8
R3c3=8-(8=1)r3c7-1r79c7=>r7c8=1
As you can see, r7c8 can be proven to be either 1 or 3 depending on the path you use. However, in a forcing chain it doesn't matter; as long as you can prove the one that suits your needs it can be used. Hope that's clear. If not, give a holler. |
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