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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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 Posted: Mon Oct 11, 2010 3:18 pm Post subject: Free Press Oct 1, 2010 |
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Not yet started ...
| Code: |
Puzzle: FP100110
+-------+-------+-------+
| . . . | 4 6 . | 5 3 . |
| . 2 . | 5 . . | . . . |
| 3 7 . | . . . | . . . |
+-------+-------+-------+
| . . . | . . . | 1 . . |
| 1 . . | 2 . 5 | . . 4 |
| . . 8 | 7 . . | . . . |
+-------+-------+-------+
| . . . | . . . | . 7 8 |
| . . . | . . 2 | . . . |
| . 4 7 | . 5 3 | 2 1 . |
+-------+-------+-------+
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Play this puzzle online at the Daily Sudoku site
Keith |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Mon Oct 11, 2010 10:38 pm Post subject: |
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I'm throwing in the towel.  |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Tue Oct 12, 2010 12:44 am Post subject: |
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I am going to perform basics again, but here is my first result
| Quote: | | xy-wing 36-9 vertex (36)r7c7 and pseudocell (39)box7; r8c79,r9c1<>9 |
Ted |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Tue Oct 12, 2010 1:49 am Post subject: |
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Here's where I'm stuck. If you could explain the pseudo cell, I'd appreciate it.
| Code: |
+------------------+-------------+----------------+
| 89 189 19 | 4 6 7 | 5 3 2 |
| 46 2 46 | 5 3 1 | 8 9 7 |
| 3 7 5 | 9 2 8 | 46 46 1 |
+------------------+-------------+----------------+
| 7 569 2469 | 3 489 469 | 1 2568 569 |
| 1 369 369 | 2 89 5 | 7 68 4 |
| 24569 569 8 | 7 1 469 | 369 256 3569 |
+------------------+-------------+----------------+
| 256 1356 1236 | 16 49 49 | 36 7 8 |
| 689 13689 1369 | 168 7 2 | 3469 456 3569 |
| 689 4 7 | 68 5 3 | 2 1 69 |
+------------------+-------------+----------------+
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Tue Oct 12, 2010 3:02 am Post subject: |
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| Marty R. wrote: | Here's where I'm stuck. If you could explain the pseudo cell, I'd appreciate it.
| Code: |
+------------------+-------------+----------------+
| 89 189 19 | 4 6 7 | 5 3 2 |
| 46 2 46 | 5 3 1 | 8 9 7 |
| 3 7 5 | 9 2 8 | 46 46 1 |
+------------------+-------------+----------------+
| 7 569 2469 | 3 489 469 | 1 2568 569 |
| 1 369 369 | 2 89 5 | 7 68 4 |
| 24569 569 8 | 7 1 469 | 369 256 3569 |
+------------------+-------------+----------------+
| 256 1356 1236 | 16 49 49 | 36 7 8 |
| 689 13689 1369 | 168 7 2 | 3469 456 3569 |
| 689 4 7 | 68 5 3 | 2 1 69 |
+------------------+-------------+----------------+
|
Play this puzzle online at the Daily Sudoku site |
Marty,
Happy to help........
First, you did not finish basics. Column 1 has a triple (689)r189c1 which opens still more basics.
Here is my code after basics,
| Code: | *-----------------------------------------------------------*
| 89 18 19 | 4 6 7 | 5 3 2 |
| 4 2 6 | 5 3 1 | 8 9 7 |
| 3 7 5 | 9 2 8 | 46 46 1 |
|-------------------+-------------------+-------------------|
| 7 59 4 | 3 8 6 | 1 2 59 |
| 1 6 3 | 2 9 5 | 7 8 4 |
| 2 59 8 | 7 1 4 | 369 56 3569 |
|-------------------+-------------------+-------------------|
| 5 13 2 | 16 4 9 | 36 7 8 |
| 689 138 19 | 168 7 2 | 3469 456 3569 |
| 689 4 7 | 68 5 3 | 2 1 69 |
*-----------------------------------------------------------* |
I learned the term xy-wing with pseudocell pincer from Keith. Simply stated, it is two (or more?) cells that form a chain and act as a single cell pincer.
Looking at my code, notice the potential xy-wing with vertex (36)r7c7, and pincer (69)r9c9. What we need is another pincer (39) to complete the xy-wing. Now look at the (13)r7c2 and the (19)r8c3. So, if r7c7=6 then r9c9=9 and if r7c7=3 then r7c2=1 and r8c3<>1=9. Thus the combination (13)r7c2 and (19)r8c3 form a pseudocell and acts as a pincer for the xy-wing. Most of the longer chains I find are extensions of the vertex of a potential xy-wing in this manner until I find a cell with the matching pincer.
Of course, this could be simply described as a four cell xy chain: (9=6)r9c9-(6=3)r7c7-(3=1)r7c2-(1=9)r8c3; r8c79,r9c1<>9.
Hope this helps....
Ted |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Tue Oct 12, 2010 3:40 am Post subject: |
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Really?
I had no real problem with this one.
After basics:
| Code: | +----------------+----------------+----------------+
| 89 18 19 | 4 6 7 | 5 3 2 |
| 4 2 6 | 5 3 1 | 8 9 7 |
| 3 7 5 | 9 2 8 | 46 46 1 |
+----------------+----------------+----------------+
| 7 59 4 | 3 8 6 | 1 2 59 |
| 1 6 3 | 2 9 5 | 7 8 4 |
| 2 59 8 | 7 1 4 | 369 56 3569 |
+----------------+----------------+----------------+
| 5 13 2 | 16 4 9 | 36 7 8 |
| 689 138 19 | 168 7 2 | 3469 456 3569 |
| 689 4 7 | 68 5 3 | 2 1 69 |
+----------------+----------------+----------------+ |
XY-wing 5-69 in B69 solves the puzzle.
(It was not that easy, the basics were tough.)
Keith |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Tue Oct 12, 2010 3:46 am Post subject: |
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| Thanks Ted. I need to get my act together on basics. I don't think I'd ever figure out a pseudo cell except on a UR. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Tue Oct 12, 2010 4:08 am Post subject: |
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It was pretty easy once I played the c1 689 triple that Ted pointed out I missed.
I saw the XY-Wing on 596 but I played the Type 1 UR on 59 first. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Tue Oct 12, 2010 3:23 pm Post subject: |
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Marty,
Their is a 24 pair in B4 that I spotted early on. That avoids the triple in C1.
Keith |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Tue Oct 12, 2010 3:50 pm Post subject: |
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| keith wrote: | Marty,
Their is a 24 pair in B4 that I spotted early on. That avoids the triple in C1.
Keith |
Of course, I would call that a 3569 quad rather than a 24 pair. But I have been missing basics lately and that has caused me problems.
Creeping (galloping?) senility. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Wed Oct 13, 2010 9:14 pm Post subject: |
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Marty,
I found it before I started pencil marks.
I do this scanning thing with each block - look at the solved cells outside the block, that are not solved inside the block. In this case, I found the 24 pair in B4 by noticing the solved 2 and 4 in each of R5 and C2.
(For B4, look at solved cells outside B4 in R456 and C123.)
Keith |
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