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Feb 7 VH

 
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Clement



Joined: 24 Apr 2006
Posts: 1111
Location: Dar es Salaam Tanzania
PostPosted: Sun Feb 06, 2011 10:00 pm    Post subject: Feb 7 VH Reply with quote
XY-Wing 25 29 59 pivoted in r8c6 solves it.
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George Woods



Joined: 28 Mar 2006
Posts: 304
Location: Dorset UK
PostPosted: Mon Feb 07, 2011 12:22 pm    Post subject: another way Reply with quote
Having missed the V obvious 6 in box 7 there was no "basic" XY wing for me. But I found an interesting (for me anyway) alternative. Consider removing 5 fom r7c9 Now an XY wing 589 pivot at r7c5 makes r9c9 5, so Either r9c9 is 5 or r7c9 is 5 so r8c8<>5 or r8c7<>5 ......
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tlanglet



Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
PostPosted: Tue Feb 08, 2011 3:17 pm    Post subject: Re: another way Reply with quote
George Woods wrote:
Having missed the V obvious 6 in box 7 there was no "basic" XY wing for me. But I found an interesting (for me anyway) alternative. Consider removing 5 fom r7c9 Now an XY wing 589 pivot at r7c5 makes r9c9 5, so Either r9c9 is 5 or r7c9 is 5 so r8c8<>5 or r8c7<>5 ......

George,

Very nice observation. This may be considered an "almost" pattern where both conditions must lead to a common result in order to actually make a deletion. (Also note that the 5 in r4c9 may also be deleted by your logic which is sufficient to complete the puzzle in one step.)

A variation of your pattern is an "almost" naked pair. Look again at r79c9. If r7c9 did not contain the 9, then we have a naked pair (58) in those two cells which would eliminate the 5 in r8c78. However if r7c9=9 then r7c5=5, then r7c1<>5, the r8c2=5 which also deletes 5 in r8c78. This provides the same result as you found.
In notational form this is described as: ANP(58=9)r79c9-(9=5)r7c5-(5)r7c1=(5)r8c2; r8c78<>5

I also find these patterns interesting.

Ted
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