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daj95376
Joined: 23 Aug 2008
Posts: 3854
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 Posted: Sat May 23, 2009 3:00 pm Post subject: Puzzle NR_026 |
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| Code: | +-----------------------+
| 4 . . | . . . | . . . |
| . . 1 | . 5 6 | 7 . . |
| . 9 5 | 8 . . | 2 6 . |
|-------+-------+-------|
| . . 7 | 3 . 9 | . 4 . |
| . 8 . | . 4 . | . . . |
| . 5 . | 1 . . | . . 8 |
|-------+-------+-------|
| . 4 2 | . . . | 5 . . |
| . . 8 | 6 . . | . 2 . |
| . . . | . . 4 | . . 6 |
+-----------------------+
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Play this puzzle online at the Daily Sudoku site |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Sun May 24, 2009 2:21 pm Post subject: |
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I worked for some time to find a solution to this puzzle, and then I noted a UR step that solved the puzzle. However, I believe the move is valid but I do not know how to describe it formally.
First, a simple kite on 9 with hinge in box4 deletes 9 from r9c7. This results in the following code:
| Code: |
*--------------------------------------------------------------------*
| 4 367 36 | 29 29 37 | 8 5 1 |
| 8 2 1 | 4 5 6 | 7 39 39 |
| 37 9 5 | 8 137 137 | 2 6 4 |
|----------------------+----------------------+----------------------|
| 2 16 7 | 3 8 9 | 16 4 5 |
| 139 8 369 | 57 4 57 | 1369 139 2 |
| 39 5 4 | 1 6 2 | 39 7 8 |
|----------------------+----------------------+----------------------|
| 6 4 2 | 79 1379 8 | 5 139 379 |
| 13579 137 8 | 6 1379 1357 | 4 2 379 |
| 13579 137 39 | 2579 12379 4 | 13 8 6 |
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Now the UR 39 in r56c17. In gory detail, either r5c1 must equal 1, r5c7 must equal 1, or r5c7 must equal 6. In all three cases, digit 1 in r9c7 is deleted and the puzzle is solved.
If the UR is viewed as a Type 3, then (I believe) the UR (16) subset pairs with the (16) bivalue in r4c7 to delete digit 1 in r9c7 again solving the puzzle. This is more concise description, but I never fully understood the posting by Asellus regarding how pseudo-cells "see" other cells.
Help is appreciated....
Ted  |
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daj95376
Joined: 23 Aug 2008
Posts: 3854
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| Posted: Sun May 24, 2009 4:34 pm Post subject: |
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| tlanglet wrote: | Now the UR 39 in r56c17. In gory detail, either r5c1 must equal 1, r5c7 must equal 1, or r5c7 must equal 6. In all three cases, digit 1 in r9c7 is deleted and the puzzle is solved.
If the UR is viewed as a Type 3, then (I believe) the UR (16) subset pairs with the (16) bivalue in r4c7 to delete digit 1 in r9c7 again solving the puzzle. This is more concise description, but I never fully understood the posting by Asellus regarding how pseudo-cells "see" other cells.
Help is appreciated....
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Your UR also performs [r5c8]<>1.
FWIW (which isn't much): I like your first description because it's an easy forcing chain.
| Code: | r5c1=1, r4c2=6, r4c7=1 => r5c8,r9c7<>1
r5c1<>1, (39) UR Type 1, r45c7=16 => r5c8,r9c7<>1
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For the pseudo-cell -- [r5c1] and [r5c7] -- to work (for me) the combined <16> effect must be applied in [r5] because this is the only unit they have in common.
===== ===== ===== Side Note
I've been toying with the idea of having my SIN routine check to see if setting a particular candidate true would result in a UR condition. Then I could work backwards to determine how the UR forces the elimination.
Your example would work perfectly with this concept:
| Code: | r9c7=1 r4c7=6 r4c2=1 => (39) UR r56c17 ==> r9c7<>1
r5c8=1 r4c7=6 => (39) UR r56c17 ==> r5c8<>1
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Tue May 26, 2009 12:53 am Post subject: |
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| I needed six moves, including two Type 3 URs where I checked what resulted from all the ways the DP could be broken. I view them as forcing chains, which I'm led to by the UR pattern. In other words, trial and error, one step above something like picking a bivalue cell and checking both values. Even though I use them when I can't find something better, I don't find them very satisfying. |
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storm_norm
Joined: 18 Oct 2007
Posts: 1741
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| Posted: Tue May 26, 2009 1:56 am Post subject: |
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Ted,
excellent 
in order to see this, I am going to attempt to use a partial grid with minimal pencil marks, one step at a time, for better clarity.
we have the UR {3,9} r17c56 with extra candidates. the {1,6} cell in r4c7 and the {1,5} cell in r5c8.
like this
| Code: | +---------+-------+-----------+
| . . . | . . . | 16 . . |
| 139 . . | . . . | 1369 15 . |
| 39 . . | . . . | 39 . . |
+---------+-------+-----------+ |
now, imagine that the naked pair {1,6} r45c7 is false in column 7.
the grid would look like this (in your mind of course). 
| Code: | +---------+-------+---------+
| . . . | . . . | . . . |
| 139 . . | . . . | 39 15 . |
| 39 . . | . . . | 39 . . |
+---------+-------+---------+ |
as one can see, the 3 and 9 must be eliminated from the r5c1 cell in order to avoid the UR (type 1 UR). this then places a 1 in r5c1 and would eliminate the 1 from r5c8. keep that in mind.
---------
now, lets do it in reverse...
imagine the 1 in r5c1 is false...
looks like...
| Code: | +--------+-------+-----------+
| . . . | . . . | 16 . . |
| 39 . . | . . . | 1369 15 . |
| 39 . . | . . . | 39 . . |
+--------+-------+-----------+ |
as one can tell, the 39 can be eliminated from the r5c7 cell to avoid the deadly pattern (type 1 UR)
looks like...
| Code: | +--------+-------+---------+
| . . . | . . . | 16 . . |
| 39 . . | . . . | 16 15 . |
| 39 . . | . . . | 39 . . |
+--------+-------+---------+ |
this leaves the naked pair {16} r45c7. this also (as seen before) eliminates the 1 from r5c8.
----
we now have proved this strong inference. short and about as elegant as one can get. (sounds like some people I know)
UR39[(16)r45c7 = (1)r5c1]; r5c8 <> 1
if you are curious, the chain that eliminates the 1 from r9c7...
(1)r7c8 = (1)r5c8 - UR39[(1)r5c1 = (16)r45c7]; r9c7 <> 1 |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Tue May 26, 2009 1:13 pm Post subject: |
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Danny and Norm,
Thanks for your posts. They were very helpful.
Ted |
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