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Earl
Joined: 30 May 2007 Posts: 677 Location: Victoria, KS
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Posted: Sat Nov 14, 2009 3:39 pm Post subject: Nov 14 DB |
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The Nov 14 DB is a refreshing challenge. Different paths present themselves. Here is a one-stepper. Norm will surely find another.
A Solution: type 4 UR 25 where either solution makes R8C7 <2>. The type I UR 89 does not solve the puzzle.
Earl
Code: |
+-------+-------+-------+
| . . . | 1 . . | . 4 5 |
| . 3 . | . . . | 7 . 9 |
| . . . | . . 3 | . 8 2 |
+-------+-------+-------+
| 3 . 7 | 5 9 . | 4 . . |
| . . . | . 4 . | . . . |
| . . 5 | . 3 6 | 8 . 7 |
+-------+-------+-------+
| 7 4 . | 3 . . | . . . |
| 5 . 6 | . . . | . 3 . |
| 2 9 . | . . 5 | . . . |
+-------+-------+-------+
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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: Sat Nov 14, 2009 6:08 pm Post subject: |
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An xy-chain solution: Code: | *-----------------------------------------------------------*
| 689 7 89 | 1 268 29 | 3 4 5 |
| 168 3 2 | 68 5 4 | 7 16 9 |
| 169 5 4 |c69 7 3 |d16 8 2 |
|-------------------+-------------------+-------------------|
| 3 26 7 | 5 9 8 | 4 126 *16 |
| 89 26 89 | 7 4 1 | 25 256 3 |
| 4 1 5 | 2 3 6 | 8 9 7 |
|-------------------+-------------------+-------------------|
| 7 4 1 | 3 68 29 | 259 25 8-6 |
| 5 8 6 |b49 12 7 | 29 3 a14 |
| 2 9 3 | 468 168 5 |*16 7 148-6 |
*-----------------------------------------------------------*
(6=1)r4c9-(1=4)r8c9-(4=9)r8c4-(9=6)r3c4-(6=1)r3c7-(1=6)r9c7
=> r79c9<>6;
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nataraj
Joined: 03 Aug 2007 Posts: 1048 Location: near Vienna, Austria
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Posted: Sun Nov 15, 2009 12:09 pm Post subject: |
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This one could very well be found in the "one trick pony" section (uniqueness) ...
My approch was slightly different from Earl's:
The UR 26 (r45c28) with strong link on 2 in row 4 means that r5c8 cannot be "6".
This turns the UR 25 (r57c78) into a type 1 and solves r7c7 ("9") and the puzzle. |
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tlanglet
Joined: 17 Oct 2007 Posts: 2468 Location: Northern California Foothills
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Posted: Sun Nov 15, 2009 2:47 pm Post subject: |
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Hi Nataraj, good to see that you still are able to visit occasionally.
I also used the UR26 in r45c28. To prevent the deadly patter, either r4c8=1 or r5c8=5. This strong inference can be used as follows:
(6=1)r4c9 - UR26[(1)r4c8 = (5)r5c8] - (5=2)r5c7 - (2=9)r8c7 - (9=4)r8c4 - (4=1)r8c9 - (1=6)r9c7; r79c9<>6 to complete the puzzle.
(Dan, I just read your post and realize that we used two different chains to form identical pincers. Interesting..........)
Ted |
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keith
Joined: 19 Sep 2005 Posts: 3355 Location: near Detroit, Michigan, USA
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Posted: Sun Nov 15, 2009 4:30 pm Post subject: |
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I had a type 6 UR that put 5's on a diagonal, opening up a useful type 1 UR.
Edit: After basics:
Code: | +----------------+----------------+----------------+
| 689 7 89 | 1 268 29 | 3 4 5 |
| 168 3 2 | 68 5 4 | 7 16 9 |
| 169 5 4 | 69 7 3 | 16 8 2 |
+----------------+----------------+----------------+
| 3 26 7 | 5 9 8 | 4 126# 16 |
| 89 26 89 | 7 4 1 | 25@ 256 3 |
| 4 1 5 | 2 3 6 | 8 9 7 |
+----------------+----------------+----------------+
| 7 4 1 | 3 68 29 | 259 25@ 68 |
| 5 8 6 | 49 12 7 | 29 3 14 |
| 2 9 3 | 468 168 5 | 16 7 1468 |
+----------------+----------------+----------------+ |
I did not notice the 89 UR in B14. Rather, there is a Type 6 UR 25 in B69. It solves the cells @ as 5. This reveals a Type 1 UR 26 in B46 that solves # as 1.
Kind of cool. It is rare to find a Type 6 UR that is of much help.
Keith |
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keith
Joined: 19 Sep 2005 Posts: 3355 Location: near Detroit, Michigan, USA
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Posted: Wed Nov 18, 2009 3:43 pm Post subject: Re: Nov 14 DB |
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Earl wrote: | The Nov 14 DB is a refreshing challenge. Different paths present themselves. Here is a one-stepper. Norm will surely find another.
A Solution: type 4 UR 25 where either solution makes R8C7 <2>. The type I UR 89 does not solve the puzzle.
Earl |
Earl,
I don't see this UR.
Keith |
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