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Puzzle 10/10/06: A VH

 
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daj95376



Joined: 23 Aug 2008
Posts: 3854
PostPosted: Wed Oct 06, 2010 12:33 am    Post subject: Puzzle 10/10/06: A VH Reply with quote
Code:
 +-----------------------+
 | . . 5 | . 1 . | . . 2 |
 | . 4 . | 7 . 2 | . . . |
 | 1 . . | . 5 4 | . . 6 |
 |-------+-------+-------|
 | . 1 . | 6 . 5 | . 3 . |
 | 8 . 2 | . . 3 | . . . |
 | . 6 3 | 1 2 7 | . 4 . |
 |-------+-------+-------|
 | . . . | . . . | 5 . . |
 | . . . | 2 . 1 | . . 4 |
 | 4 . 6 | . . . | . 2 3 |
 +-----------------------+

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Marty R.



Joined: 12 Feb 2006
Posts: 5770
Location: Rochester, NY, USA
PostPosted: Wed Oct 06, 2010 4:11 am    Post subject: Reply with quote
Quote:
UR (15) pseudo cell 67 forms naked pair
Remote Pairs (79)
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JC Van Hay



Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium
PostPosted: Wed Oct 06, 2010 1:41 pm    Post subject: Reply with quote
Quote:
To prevent DP (15)R25C89 : => r5c9<>15=7
To prevent DP (49)R57C45 : => r7c5<>49=67
BUG+1 : SIS[(9)r13c8,(9)r3c7] : => r2c7<>9
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tlanglet



Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
PostPosted: Wed Oct 06, 2010 10:06 pm    Post subject: Reply with quote
Three steps. The first two were identical to those by JC, and my last step was the same as Marty's last step.

Quote:
Type 1 UR(15)r25c89; r5c9<>15=7
Type 1 UR(49)r57c45; r7c5<>49
RP(79)r7c8 => r3c3; r3c8<>79=8

Ted
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tlanglet



Joined: 17 Oct 2007
Posts: 2468
Location: Northern California Foothills
PostPosted: Wed Oct 06, 2010 10:14 pm    Post subject: Reply with quote
JC Van Hay wrote:
Quote:
To prevent DP (15)R25C89 : => r5c9<>15=7
To prevent DP (49)R57C45 : => r7c5<>49=67
BUG+1 : SIS[(9)r13c8,(9)r3c7] : => r2c7<>9


JC,

I still find myself on shaky ground sometimes trying to determine if I have a valid BUG pattern; this is a example of that situation. I concluded it was not a valid BUG because of the four occurrences of digit 9 , but if it had been valid, I would have labeled it a BUG+3, not BUG+1 since three cells had were not bivalues.

Would you please explain your last step?

Ted
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JC Van Hay



Joined: 13 Jun 2010
Posts: 494
Location: Charleroi, Belgium
PostPosted: Thu Oct 07, 2010 10:10 am    Post subject: Reply with quote
Ted, the last step should have been :
    X Wing (7)C37 : => r3c8<>7
    BUG+1 or BUG+2 (?) : SIS[(9)r1c8,(9)r3c7] (otherwise BUG) : => r2c7,r3c8<>9
However, Kite (7)R8C6 was sufficient after the 1st step !
Regards, JC
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