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Puzzle 10/09/02: C

 
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daj95376



Joined: 23 Aug 2008
Posts: 3854

PostPosted: Thu Sep 02, 2010 3:34 am    Post subject: Puzzle 10/09/02: C Reply with quote

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

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JC Van Hay



Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium

PostPosted: Thu Sep 02, 2010 11:45 am    Post subject: Reply with quote

Quote:
5-SIS M Wing loop (13) : 1C5 3C5 3C8 3R6 1C3 loop : r2c5<>2,r8c479<>3,r6c3<>67
4-SIS AIC : 1C3 7R5 8C1 8C6 : (1)r6c3=(8)r6c6 : => r6c6<>1
XY Wing (36-2), pivot at R6C7 : => r3c9<>2
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JC Van Hay



Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium

PostPosted: Thu Sep 02, 2010 6:52 pm    Post subject: Reply with quote

Second run ...
Quote:
I cannot resist a continuous loop Very Happy ... But as the 2 1st steps in the previous proof have essentially the cell R6C3 as a target, it appears that the elimination of (3)r6c3 is done through a very simple 3-SIS step that conveniently replace them.

An alternative proof :

XY Wing Style : (38)R4C1 8C6 1R6 : (3)r4c1=(1)r6c3 : r6c3<>3
XY Wing (36-2), pivot at R6C7 : => r3c9<>2

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Marty R.



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

PostPosted: Sat Sep 04, 2010 4:35 am    Post subject: Reply with quote

The DP killers of the potential 28 DP in boxes 79 yielded some common eliminations. Following that, the 28 became a Type 1 UR followed by a BUG+2.
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