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Puzzle 10/10/14: C XY

 
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daj95376



Joined: 23 Aug 2008
Posts: 3854

PostPosted: Thu Oct 14, 2010 1:57 am    Post subject: Puzzle 10/10/14: C XY Reply with quote

I'm out of puzzles that solve with a single "wing".

Code:
 +-----------------------+
 | . . . | . 3 . | . . . |
 | . 2 . | . . 7 | . . 9 |
 | . . 4 | . . . | . . 2 |
 |-------+-------+-------|
 | . . . | 5 . . | 2 . . |
 | 1 . . | . 6 2 | . 5 8 |
 | . 4 . | . 7 8 | . . 6 |
 |-------+-------+-------|
 | . . . | 3 . . | 6 . . |
 | . . . | . 1 . | . 2 4 |
 | . 8 1 | . 2 4 | . 7 . |
 +-----------------------+

Play this puzzle online at the Daily Sudoku site
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tlanglet



Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills

PostPosted: Thu Oct 14, 2010 2:48 am    Post subject: Reply with quote

How about an "almost wing" single step.........

Quote:
Axy-wing 3-57 vertex (37)r3c7 with pseudocell (35)r69c7 and fin ((5)r3c7; r1c7,r7c9<>5
[xy-wing=(5)r3c7]-(5=9)r3c5-(9=4)r4c5-(4=1)r4c8-(1=9)r7c8-(9=5)r9c7;

Ted
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daj95376



Joined: 23 Aug 2008
Posts: 3854

PostPosted: Thu Oct 14, 2010 3:54 am    Post subject: Reply with quote

Or 2x XY-Chains using the same cells. No pseudo-cell and no fin cell.

Code:
 (5=9)r3c5-(9=4)r4c5-(4=1)r4c8-(1=9)r7c8-(9=5)r9c7; r3c7<>5
 +--------------------------------------------------------------------------------+
 |  56789   1569    6789    |  2       3       169     |  457     468     57      |
 |  3568    2       368     |  468     45      7       |  1       368     9       |
 |  356789  13569   4       |  689    a59      169     |  37-5    368     2       |
 |--------------------------+--------------------------+--------------------------|
 |  6789    69      6789    |  5      b49      3       |  2      c14      17      |
 |  1       39      379     |  49      6       2       |  47      5       8       |
 |  2       4       5       |  1       7       8       |  39      39      6       |
 |--------------------------+--------------------------+--------------------------|
 |  4       7       2       |  3       8       59      |  6      d19      15      |
 |  3569    3569    369     |  7       1       569     |  8       2       4       |
 |  569     8       1       |  69      2       4       | e59      7       3       |
 +--------------------------------------------------------------------------------+
 # 79 eliminations remain

Code:
 (5=7)r1c9 - (7=3)r3c7 - (3=9)r6c7 - (9=5)r9c7; r1c7,r7c9<>5
 +--------------------------------------------------------------------------------+
 |  56789   1569    6789    |  2       3       169     |  47-5    468    a57      |
 |  3568    2       368     |  468     45      7       |  1       368     9       |
 |  356789  13569   4       |  689     59      169     | b37      368     2       |
 |--------------------------+--------------------------+--------------------------|
 |  6789    69      6789    |  5       49      3       |  2       14      17      |
 |  1       39      379     |  49      6       2       |  47      5       8       |
 |  2       4       5       |  1       7       8       | c39      39      6       |
 |--------------------------+--------------------------+--------------------------|
 |  4       7       2       |  3       8       59      |  6       19      1-5     |
 |  3569    3569    369     |  7       1       569     |  8       2       4       |
 |  569     8       1       |  69      2       4       | d59      7       3       |
 +--------------------------------------------------------------------------------+
 # 78 eliminations remain

The giveaway: All bivalue cells except for the vertex ... and ... if the vertex is true for <5>, then it results in its own elimination along with the desired eliminations.
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peterj



Joined: 26 Mar 2010
Posts: 974
Location: London, UK

PostPosted: Thu Oct 14, 2010 12:18 pm    Post subject: Reply with quote

I went Ted's path initially - but two steps as an xyz-wing to delete 7 from r3c7 and then the short xy-chain to delete 5.

Danny, I think your first xy-chain alone is enough to single-step if you incorporate (5)r2c1=r2c5 and so can eliminate (5)r9c1?
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tlanglet



Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills

PostPosted: Thu Oct 14, 2010 1:11 pm    Post subject: Reply with quote

Very interesting since I initially had also spotted the xyz-wing (457) but it attracted me to the AXY-wing.

Both other observations are great insight into other perspectives of the same arrangement.

Ted
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daj95376



Joined: 23 Aug 2008
Posts: 3854

PostPosted: Thu Oct 14, 2010 4:25 pm    Post subject: Reply with quote

peterj wrote:
Danny, I think your first xy-chain alone is enough to single-step if you incorporate (5)r2c1=r2c5 and so can eliminate (5)r9c1?

Yes. But, I decided to just present the alternate perspective on Ted's solution since he put so much effort into it.

I find it interesting to see what can be found when a solution is reviewed. I'm always doing that to my solver's solutions and often find the alternate perspective to be just as interesting as the original. That's what I was trying to show here.

Of course, you need a solution before you can review it. Thanks Ted! _ Very Happy _
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Marty R.



Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA

PostPosted: Fri Oct 15, 2010 6:23 pm    Post subject: Reply with quote

I played the XYZ-Wing (457), then an XY-Chain, r3c7<>5.
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