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Feb 18 VH

 
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Clement



Joined: 24 Apr 2006
Posts: 1111
Location: Dar es Salaam Tanzania

PostPosted: Thu Feb 18, 2010 12:14 am    Post subject: Feb 18 VH Reply with quote

1. XY-Wing <35>-<57>-<37> eliminating 3 in r7c5 leads to
2. XYZ-Wing <36>-<367>-<37> eliminating 3 in r1c9 solving the puzzle.
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Earl



Joined: 30 May 2007
Posts: 677
Location: Victoria, KS

PostPosted: Thu Feb 18, 2010 2:53 am    Post subject: Feb 18 Reply with quote

xy-chain (36-37-37-36) eliminates the 6 in R9C3 and solves the puzzle.

Earl
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Marty R.



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

PostPosted: Thu Feb 18, 2010 6:00 am    Post subject: Reply with quote

After the 375 XY-Wing there was a BUG+2 in row 1. The two possibilities were 36, which combined with a 36 cell to form a naked pair which finished it off.
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arkietech



Joined: 31 Jul 2008
Posts: 1834
Location: Northwest Arkansas USA

PostPosted: Thu Feb 18, 2010 7:42 am    Post subject: Reply with quote

w-wing 63 removing a 6 in r9c3 will do it in one step.
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cgordon



Joined: 04 May 2007
Posts: 769
Location: ontario, canada

PostPosted: Thu Feb 18, 2010 3:02 pm    Post subject: Reply with quote

I used a Type 6 UR (13) to remove <1> from R9C3 and R8C9. Then an ER on <3>.

Which was pretty cool really.
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tlanglet



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

PostPosted: Fri Feb 19, 2010 1:55 pm    Post subject: Reply with quote

cgordon wrote:
I used a Type 6 UR (13) to remove <1> from R9C3 and R8C9. Then an ER on <3>.

Which was pretty cool really.


I also find UR implication very useful. They also seemed to be very productive in providing one step solutions.

In this puzzle, it is the x-wing 1 overlay on the UR13 that allows the removal of digit 1 in r8c9 and r9c3. But if the x-wing 1 was not present, the UR still offers deletions because of the strong inference between the 7 in r8c9 and the 6 in r9c3. One of these two must be true to prevent the deadly pattern: (7)r8c9 = (6)r9c3.

One possible chain using this inference is:
(3=7)r7c2 - (7=3)r2c2 - (3=6)r1c3 - UR13[(6)r9c3 = (7)r8c9]r89c39 - (7=5)r8c1 - (5=3)r8c5; r7c5<>3.

In this specific case, I suspect the end results of both approaches are identical.

Ted
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