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Earl
Joined: 30 May 2007
Posts: 677
Location: Victoria, KS
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 Posted: Sat Nov 03, 2007 11:47 am Post subject: Nov 3 DB |
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Here is the Nov 3 DB from the St Louis paper.
Earl
+-------+-------+-------+
| . 8 . | 5 4 . | . 2 . |
| 7 . . | 1 . . | 4 . . |
| . . . | . 6 . | 3 . 1 |
+-------+-------+-------+
| . 2 . | . . 5 | . . 7 |
| 1 . . | . . . | . . 4 |
| 5 . . | 9 . . | . 8 . |
+-------+-------+-------+
| 4 . 9 | . 5 . | . . . |
| . . 3 | . . 6 | . . 9 |
| . 7 . | . 9 1 | . 3 . |
+-------+-------+-------+
[/code]
[url=http://www.dailysudoku |
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Earl
Joined: 30 May 2007
Posts: 677
Location: Victoria, KS
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| Posted: Sat Nov 03, 2007 11:57 am Post subject: Nov 3 DB |
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Here is another try at a better grid.
A rather basic DB: x-wing, x-wing, xy-wing - voila !
Earl
| Code: |
+-------+-------+-------+
| . 8 . | 5 4 . | . 2 . |
| 7 . . | 1 . . | 4 . . |
| . . . | . 6 . | 3 . 1 |
+-------+-------+-------+
| . 2 . | . . 5 | . . 7 |
| 1 . . | . . . | . . 4 |
| 5 . . | 9 . . | . 8 . |
+-------+-------+-------+
| 4 . 9 | . 5 . | . . . |
| . . 3 | . . 6 | . . 9 |
| . 7 . | . 9 1 | . 3 . |
+-------+-------+-------+
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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: Sat Nov 03, 2007 9:58 pm Post subject: |
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| Quote: | | A rather basic DB: x-wing, x-wing, xy-wing - voila ! |
Coloring on 9, then 8, then an XY-Wing. Both my colorings were a rectangular pattern, so I wonder if they were the X-Wings too. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sat Nov 03, 2007 11:12 pm Post subject: |
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I went looking for uniqueness. A type-3 UR and two X-wings get you here:
| Code: | +-------------+-------------+-------------+
| 39 8 1 | 5 4 37 | 79 2 6 |
| 7 39 6 | 1 23 29 | 4 5 8 |
| 2 4 5 | 78 6 789 | 3 79 1 |
+-------------+-------------+-------------+
| 39 2 4 | 36 8 5 | 69 1 7 |
| 1 39 8 | 267 237 27 | 5 69 4 |
| 5 6 7 | 9 1 4 | 2 8 3 |
+-------------+-------------+-------------+
| 4 1 9 | 38 5 38 | 67 67 2 |
| 8 5 3 | 27 27 6 | 1 4 9 |
| 6 7 2 | 4 9 1 | 8 3 5 |
+-------------+-------------+-------------+ |
This is a BUG+3. Either / and R3C6 is <7>, R5C4 is <7>, R5C5 is <2>. The <2> in R5 must be in C56, R5C4 is not <2>. This reduces it to a BUG+1, R3C6 is <7>.
Keith |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sun Nov 04, 2007 12:03 am Post subject: |
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| keith wrote: | I went looking for uniqueness. A type-3 UR and two X-wings get you here:
| Code: | +-------------+-------------+-------------+
| 39 8 1 | 5 4 37 | 79 2 6 |
| 7 39 6 | 1 23 29 | 4 5 8 |
| 2 4 5 | 78 6 789 | 3 79 1 |
+-------------+-------------+-------------+
| 39 2 4 | 36 8 5 | 69 1 7 |
| 1 39 8 | 267 237 27 | 5 69 4 |
| 5 6 7 | 9 1 4 | 2 8 3 |
+-------------+-------------+-------------+
| 4 1 9 | 38 5 38 | 67 67 2 |
| 8 5 3 | 27 27 6 | 1 4 9 |
| 6 7 2 | 4 9 1 | 8 3 5 |
+-------------+-------------+-------------+ |
This is a BUG+3. Either / and R3C6 is <7>, R5C4 is <7>, R5C5 is <2>. The <2> in R5 must be in C56, R5C4 is not <2>. This reduces it to a BUG+1, R3C6 is <7>.
Keith |
Keith, how did you reason that r5c4 is not a 2? Was it the fact that there are only two 2s in column 4, or is there more to it?
Regardless, after removing the 2, you have a 237 in r5c5, but you went directly from removing that 2 to a BUG+1. How did you reduce r5c5 to two candidates? |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sun Nov 04, 2007 1:08 am Post subject: |
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Marty,
Please refer back to the grid I posted.
The deadly pattern is:
| Code: | +-------------+-------------+-------------+
| 39 8 1 | 5 4 37 | 79 2 6 |
| 7 39 6 | 1 23 29 | 4 5 8 |
| 2 4 5 | 78 6 89 | 3 79 1 |
+-------------+-------------+-------------+
| 39 2 4 | 36 8 5 | 69 1 7 |
| 1 39 8 | 26 37 27 | 5 69 4 |
| 5 6 7 | 9 1 4 | 2 8 3 |
+-------------+-------------+-------------+
| 4 1 9 | 38 5 38 | 67 67 2 |
| 8 5 3 | 27 27 6 | 1 4 9 |
| 6 7 2 | 4 9 1 | 8 3 5 |
+-------------+-------------+-------------+ |
To avoid the deadly pattern, (at least) one of the following is true: R3C6 is <7>, R5C4 is <7>, R5C5 is <2>. Either of the <7>'s forces <2> in R5C6, so the <2> in R5 must be in one of C56. R5C4 cannot be <2>.
Eliminating <2> in R5C4 solves <27> in R8, and takes out the <7> in R5C5, leaving a BUG+1.
Keith |
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Marty R.
Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
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| Posted: Sun Nov 04, 2007 4:35 am Post subject: |
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| Thanks Keith. I see the reasoning (I think). You establish the DP, then work back to see what needs to be done to preclude it. I was trying to look at the trivalue cells without looking at the full picture. |
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keith
Joined: 19 Sep 2005
Posts: 3355
Location: near Detroit, Michigan, USA
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| Posted: Sun Nov 04, 2007 1:10 pm Post subject: |
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| Quote: | | You establish the DP, then work back to see what needs to be done to preclude it. |
Marty,
Yes. Establishing the DP (if it exists) is an essential step.
In the BUG DP, each candidate occurs twice in the unsolved cells in every row, column, and box. In this case, look at R3, C4 and C5 to resolve the three-candidate cells. Then, check that the DP condition is correct in B2 and B5.
Keith |
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re'born
Joined: 28 Oct 2007
Posts: 80
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| Posted: Mon Nov 05, 2007 2:17 am Post subject: |
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| keith wrote: |
To avoid the deadly pattern, (at least) one of the following is true: R3C6 is <7>, R5C4 is <7>, R5C5 is <2>. Either of the <7>'s forces <2> in R5C6, so the <2> in R5 must be in one of C56. R5C4 cannot be <2>.
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A way to avoid the BUG+1 situation, is to note the slightly more complicated deduction:
r3c4 -7- [BUG+3:{r3c6, r5c4} =7|2= r5c5] -2- r2c5 -3- r1c6 -7- r3c4, => r3c4<>7, solving the puzzle.
In words, a 7 in r3c4 kills the two 7 BUG candidates, but it also makes r1c6=3, which makes r2c5=2 killing the last BUG candidate (the 2 in r5c5). Therefore, r3c4 can't be 7. |
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