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Earl
Joined: 30 May 2007
Posts: 677
Location: Victoria, KS
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 Posted: Sun Jun 07, 2009 3:21 am Post subject: June 7 VH |
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The old reliable does it again.
Solution (249) xy-wing
Early Earl |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sun Jun 07, 2009 3:25 am Post subject: |
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Alternatively:
X-Wing, Hidden UR and the easiest BUG+2 you'll ever encounter. |
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crunched
Joined: 05 Feb 2008
Posts: 168
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| Posted: Sun Jun 07, 2009 5:05 am Post subject: |
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| Marty R. wrote: | Alternatively:
X-Wing, Hidden UR and the easiest BUG+2 you'll ever encounter. |
The xy wing that Earl discusses did it for me. I am curious to know how the alternative including (hidden UR ?) and (Bug +2 ?) works.
Here is the grid with basics knocked out:
| Code: |
+-----------+-----------+-------+
| 36 36 4 | 8 1 2 | 5 7 9 |
| 9 5 8 | 7 3 6 | 4 1 2 |
| 1 7 2 | 59 49 459 | 8 3 6 |
+-----------+-----------+-------+
| 36 2 356 | 56 8 1 | 9 4 7 |
| 8 9 56 | 4 56 7 | 1 2 3 |
| 4 1 7 | 23 29 39 | 6 5 8 |
+-----------+-----------+-------+
| 5 34 9 | 23 24 8 | 7 6 1 |
| 2 46 1 | 69 7 49 | 3 8 5 |
| 7 8 36 | 1 56 35 | 2 9 4 |
+-----------+-----------+-------+
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Play this puzzle online at the Daily Sudoku site |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sun Jun 07, 2009 5:21 am Post subject: |
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Crunched,
Apparently my Hidden UR was enabled because I didn't get far enough with my basics. It doesn't appear on your grid. In your grid the X-Wing on 6 in rows 59 will reduce the puzzle to a BUG+1. |
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crunched
Joined: 05 Feb 2008
Posts: 168
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| Posted: Sun Jun 07, 2009 5:57 am Post subject: |
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| Marty R. wrote: | Crunched,
Apparently my Hidden UR was enabled because I didn't get far enough with my basics. It doesn't appear on your grid. In your grid the X-Wing on 6 in rows 59 will reduce the puzzle to a BUG+1. |
Okay, thanks Marty. I am just starting to debug BUG solutions.
So the x wing knocks out the 6 in c3, r4. This leaves the 459 cell (rc6, r3) as the only one with 3 digits in the cell. Now the puzzle is ripe for a BUG solution. The 459 is pared down to a 9 because the 459 sees a 49 in another box and the 59 in its own box, and something has to give with all those 9s seeing each other. Therefore, the 9 in the 3 digit cell is the only 9 that can remain.
Do I get this right?
| Code: |
+-----------+-----------+-------+
| 36 36 4 | 8 1 2 | 5 7 9 |
| 9 5 8 | 7 3 6 | 4 1 2 |
| 1 7 2 | 59 49 459 | 8 3 6 |
+-----------+-----------+-------+
| 36 2 356 | 56 8 1 | 9 4 7 |
| 8 9 56 | 4 56 7 | 1 2 3 |
| 4 1 7 | 23 29 39 | 6 5 8 |
+-----------+-----------+-------+
| 5 34 9 | 23 24 8 | 7 6 1 |
| 2 46 1 | 69 7 49 | 3 8 5 |
| 7 8 36 | 1 56 35 | 2 9 4 |
+-----------+-----------+-------+
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Play this puzzle online at the Daily Sudoku site |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Sun Jun 07, 2009 9:36 am Post subject: |
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That was one of these puzzle where everything practically solves itself.
10 minutes into the puzzle, done with basics, only 22 cells to go and almost all bi-value.
Look for xy wings, find a 49 w-wing instead: (4)r3c5=r8c6, which removes 4 from r3c6
done. Not even a second coffee.
NOT funny  |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Sun Jun 07, 2009 1:15 pm Post subject: |
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| Like Marty: an x wing gets rid of a 6 which leaves a BUG+1. |
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tlanglet
Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
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| Posted: Sun Jun 07, 2009 1:22 pm Post subject: |
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I used the BUG+2 as a single step solution.
r4c3 must be 6 or r3c6 must be 9. Both force r3c5=4:
r4c3=6 => r4c4=5 => r3c4=9 => r3c5=4
r3c6=9 => r3c5=4
Ted |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sun Jun 07, 2009 3:50 pm Post subject: |
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| Quote: | Okay, thanks Marty. I am just starting to debug BUG solutions.
So the x wing knocks out the 6 in c3, r4. This leaves the 459 cell (rc6, r3) as the only one with 3 digits in the cell. Now the puzzle is ripe for a BUG solution. The 459 is pared down to a 9 because the 459 sees a 49 in another box and the 59 in its own box, and something has to give with all those 9s seeing each other. Therefore, the 9 in the 3 digit cell is the only 9 that can remain.
Do I get this right? |
Yes, you got it right, I'm just not sure on the reasoning. The conventional reasoning, at least the way I learned it, is to see which number in the trivalue cell appears three times in the row, column and box, which is 9. If you were to remove the 9 you'd be left with the deadly BUG pattern. Thus, the cell is solved with 9 because that is the only way to kill the bug. |
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cgordon
Joined: 04 May 2007
Posts: 769
Location: ontario, canada
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| Posted: Sun Jun 07, 2009 9:01 pm Post subject: |
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| Quote: | r4c3 must be 6 or r3c6 must be 9. Both force r3c5=4:
r4c3=6 => r4c4=5 => r3c4=9 => r3c5=4
r3c6=9 => r3c5=4 |
Isn't this just a "what if" solution - ie - one not based on definitive rules. Sorry - I'm looking for a controversy. |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sun Jun 07, 2009 9:42 pm Post subject: |
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| cgordon wrote: | | Quote: | r4c3 must be 6 or r3c6 must be 9. Both force r3c5=4:
r4c3=6 => r4c4=5 => r3c4=9 => r3c5=4
r3c6=9 => r3c5=4 |
Isn't this just a "what if" solution - ie - one not based on definitive rules. Sorry - I'm looking for a controversy. |
Craig, that wasn't me you quoted, but I do have an opinion. I believe it's a form of trial-and-error, although I use it when presented with the opportunity. It's similar to a Type 3 UR, e.g., 24-24-245-246. Of the last two cells, one must be 5 or the other 6.
A form of pure trial-and-error would be picking a bivalue cell and doing a double-implication chain (DIC), testing for each value to see where it leads. The difference between this and the examples cited in the preceding paragraph is that in those examples the player is led to the DIC as a result of a recognizable pattern, making it more satisfying than pure T&E.
That's not a fact, it's just how I think about such things. |
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nataraj
Joined: 03 Aug 2007
Posts: 1048
Location: near Vienna, Austria
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| Posted: Mon Jun 08, 2009 4:06 pm Post subject: |
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| cgordon wrote: | | Isn't this just a "what if" solution ? |
I was going to say "yes, same as a forcing network starting at any old bi-value cell". But I see a small difference (though not the same one as Marty):
the two candidates in a bi-value cell are mutually exclusive. It is either/or but not both. With BUG+2 (or other UR based strong links) it is an inclusive or: one or the other or both.
Whether one arrives at the DIC by a BUG+2 argument or by looking at a cell with only two candidates, to me makes no big difference. But, from many sodukus I have looked at in the final stages, trying to crack that tough nut, I tend to believe that the UR-based forcing chains often meet pretty soon (like after 2-3 steps), while the forcing chains from random bi-values either run into a new bifurcation very soon (turning the chain into a network) or go on forever before reaching a contradiction.
That - for me - does not make the UR based chains more satisfying (and in fact I avoid them if I can, matter of personal taste) but simply: more efficient. |
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