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Clement
Joined: 24 Apr 2006 Posts: 1111 Location: Dar es Salaam Tanzania
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Posted: Fri Jun 10, 2011 11:35 pm Post subject: Jun 11 VH |
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I used three steps: A kite, x-wing and BUG+1.
1) A kite in 9 originating from BOX 1 sets r7c9<>9. This opens
2) An X-Wing in 9 in cols 3 and 7 setting r3c1<>9 which is followed by
3) BUG+1; r8c1=5. |
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prakash
Joined: 02 Jan 2008 Posts: 67 Location: New Jersey, USA
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Posted: Sat Jun 11, 2011 1:16 am Post subject: |
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I think the kite may not be needed. The X wing followed by the BUG+1 solves the puzzle. For those who don't like BUG+1, there is an XYZ wing with 175 pivoted in r8c1 which solves the puzzle as well.
PJ |
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Marty R.
Joined: 12 Feb 2006 Posts: 5770 Location: Rochester, NY, USA
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Posted: Sat Jun 11, 2011 1:25 am Post subject: |
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There was also at least one XY-Wing available as the second move after the X-Wing. |
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Clement
Joined: 24 Apr 2006 Posts: 1111 Location: Dar es Salaam Tanzania
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Posted: Sat Jun 11, 2011 10:25 am Post subject: Jun 11 VH |
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Prakash,
Yes, you are quite right, the kite was not needed. The x-wing was there from the beginning. It is sometimes difficult to see these things while they are in front of your eyes. |
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George Woods
Joined: 28 Mar 2006 Posts: 304 Location: Dorset UK
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Posted: Sat Jun 11, 2011 10:41 am Post subject: |
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Code: |
+-------------+---------------+----------------+
| 139 138 5 | 4 7 36 | 2 368 1389 |
| 6 4 2 | 8 135 9 | 35 7 135 |
| 139 7 89 | 2 1356 36 | 3569 3568 4 |
+-------------+---------------+----------------+
| 4 358 78 | 1 9 2367 | 356 356 235 |
| 357 9 1 | 356 356 2367 | 8 4 235 |
| 2 35 6 | 35 8 4 | 1 9 7 |
+-------------+---------------+----------------+
| 8 2 479 | 36 346 5 | 379 1 39 |
| 157 15 47 | 9 34 8 | 357 2 6 |
| 59 6 3 | 7 2 1 | 4 58 589 |
+-------------+---------------+----------------+
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Play this puzzle online at the Daily Sudoku site
From this postion, or one very like it. I used the 15 59 and implied 19 in BoxA since put a 3 anywhere in BOx A and the 19 in col1 is revealed leading to r8c2 being 1. What is this type of xy wing called? |
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peterj
Joined: 26 Mar 2010 Posts: 974 Location: London, UK
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Posted: Sat Jun 11, 2011 2:34 pm Post subject: |
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George, if I understand your move correctly it is an xy-wing but using an almost-locked-set, in this case an almost-naked-pair 13, as one of the pincers.
The logic is...
Either a (13) pair exists in r13c1 ...
or
there is a 9 in one of r13c1, hence the pivot r9c1=5 and the other pincer r8c2=1.
In both cases, r1c2<>1 and so r8c2=1.
In eureka it would be...
Code: | anp(13=9)r13c1 - (9=5)r9c1 - (5=1)r8c2 ; r1c2<>1 |
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