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roaa
Joined: 18 Apr 2009
Posts: 112
Location: Sweden
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 Posted: Sat May 30, 2015 9:10 am Post subject: May 30 VH |
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| I used two X-wings: One 237 gives r6c8<>2, which opens the second 789 giving r5c6<>9. |
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dongrave
Joined: 06 Mar 2014
Posts: 572
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| Posted: Sat May 30, 2015 1:47 pm Post subject: |
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| Code: |
+----------------+----------------+----------------+
| 9 6 37 | 8 1 27 | 4 23 5 |
| 37 1 4 | 67 5 267 | 238 238 9 |
| 5 2 8 | 9 3 4 | 1 6 7 |
+----------------+----------------+----------------+
| 6 9 23 | 4 7 38 | 28 5 1 |
| 37 8 5 | 1 2 39 | 79 4 6 |
| 1 4 27 | 5 6 89 | 2789 278 3 |
+----------------+----------------+----------------+
| 8 3 9 | 67 4 5 | 67 1 2 |
| 2 57 1 | 3 8 67 | 567 9 4 |
| 4 57 6 | 2 9 1 | 357 37 8 |
+----------------+----------------+----------------+
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I see a few different ways to solve it using two steps but I didn't see any one-steppers. Roaa's two XY-wings solves it - or his 1st XY followed by the a 278 XY - or you can start with the 278 XYZ followed by the 237 XY - but they all require two steps. Anyone out there use just one step? |
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JC Van Hay
Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium
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| Posted: Sat May 30, 2015 2:49 pm Post subject: |
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| Code: | +--------------+------------+----------------+
| 9 6 7-3 | 8 1 27 | 4 (23) 5 |
| 37 1 4 | 67 5 267 | 238 238 9 |
| 5 2 8 | 9 3 4 | 1 6 7 |
+--------------+------------+----------------+
| 6 9 (23) | 4 7 38 | (28) 5 1 |
| 37 8 5 | 1 2 39 | 79 4 6 |
| 1 4 (27) | 5 6 89 | 2789 (278) 3 |
+--------------+------------+----------------+
| 8 3 9 | 67 4 5 | 67 1 2 |
| 2 57 1 | 3 8 67 | 567 9 4 |
| 4 57 6 | 2 9 1 | 357 37 8 |
+--------------+------------+----------------+
[(3=2)r1c8 - 2r6c8=XYWing(278)r6c38,r4c7 - (2=3)r4c3] - (3=7)r1c3; stte |
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rmireland
Joined: 21 Sep 2013
Posts: 33
Location: New Orleans
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| Posted: Sat May 30, 2015 4:25 pm Post subject: |
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It didn't help in the end, but I noticed that r7c7 (67) and r8c6 (67) were locked together by r7c4 (67); that is r7c7 and r8c6 are either both 6 or are both 7.
There is a potential 23-7 XY pivoted in r1c8, but there is no single cell seen by both pincers with a 7 candidate. However, my two locked cells, r7c7 and r8c6, both had a 7 candidate and were both seen by one of my pincers. Therefore if I treat the locked cells as one, neither can be 7, so both are set to 6.
But, even so, I still needed two XYs after that to solve. Darn.
Last edited by rmireland on Sat May 30, 2015 11:57 pm; edited 2 times in total |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sat May 30, 2015 10:42 pm Post subject: |
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| Code: |
+----------+----------+------------+
| 9 6 37 | 8 1 27 | 4 23 5 |
| 37 1 4 | 67 5 267 | 238 238 9 |
| 5 2 8 | 9 3 4 | 1 6 7 |
+----------+----------+------------+
| 6 9 23 | 4 7 38 | 28 5 1 |
| 37 8 5 | 1 2 39 | 79 4 6 |
| 1 4 27 | 5 6 89 | 2789 278 3 |
+----------+----------+------------+
| 8 3 9 | 67 4 5 | 67 1 2 |
| 2 57 1 | 3 8 67 | 567 9 4 |
| 4 57 6 | 2 9 1 | 357 37 8 |
+----------+----------+------------+
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Play this puzzle online at the Daily Sudoku site
I usually try to solve these using the standard VH moves, but I didn't see it, so I used coloring (if that's the right term) on 7.
(7=6)r7c7-(6=7)r7c4-(6=7)r8c6-r12c6=r2c4-r2c1=r5c1=>r5c7<>7 |
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rmireland
Joined: 21 Sep 2013
Posts: 33
Location: New Orleans
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| Posted: Sun May 31, 2015 12:12 am Post subject: |
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Hi Marty,
We've had a real drought of Eureka lately, so I'm glad to see a few sprinkles. However, I don't follow your logic. If I understand it, you are essentially saying "If r7c7 were a 7 then ... r5c7 <> 7" by following your path.
But if r7c7 were 7, then r5c7 <> 7 directly, and the path doesn't seem to add anything. If you ended in a contradiction, then you would have something important.
Don't you need a second path that traces "If r7c7 were a 6 then ... r5c7<>7" or a path that says "If r7c7 were 7 then ... r5c7=7" as a contradiction?
Just wondering,
Rick |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sun May 31, 2015 12:38 am Post subject: |
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Rick,
The first term (7=6)r7c7 says if that cell is NOT 7, then r5c1 IS a 7. R5c7 is toast because it sees both the start and end of the chain. Remember that in a strong inference, as indicated by an = sign, the number to the left of the = sign is always false and the one to the right is true.
Marty |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Sun May 31, 2015 2:03 pm Post subject: |
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| Marty R. wrote: | | Code: |
+----------+----------+------------+
| 9 6 37 | 8 1 27 | 4 23 5 |
| 37 1 4 | 67 5 267 | 238 238 9 |
| 5 2 8 | 9 3 4 | 1 6 7 |
+----------+----------+------------+
| 6 9 23 | 4 7 38 | 28 5 1 |
| 37 8 5 | 1 2 39 | 79 4 6 |
| 1 4 27 | 5 6 89 | 2789 278 3 |
+----------+----------+------------+
| 8 3 9 | 67 4 5 | 67 1 2 |
| 2 57 1 | 3 8 67 | 567 9 4 |
| 4 57 6 | 2 9 1 | 357 37 8 |
+----------+----------+------------+
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Play this puzzle online at the Daily Sudoku site
I usually try to solve these using the standard VH moves, but I didn't see it, so I used coloring (if that's the right term) on 7.
(7=6)r7c7-(6=7)r7c4-(6=7)r8c6-r12c6=r2c4-r2c1=r5c1=>r5c7<>7 |
Marty, I suggest you review your third term, -(6=7)r8c6. It should be -(7=6)r8c6 which does not lead to you deletion. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sun May 31, 2015 4:01 pm Post subject: |
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Thanks Ted, consider it reviewed.   |
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