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July 4 DB

 
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Earl



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

PostPosted: Sat Jul 04, 2009 3:16 pm    Post subject: July 4 DB Reply with quote

The July 4 DB is a real firecracker. No single bullet here. OK Norm, prove me wrong!

A Complex Solution: I had to use coloring (6), xy-chain, W-wing (15), xyz-wing (167), another xy-chain, and finally an xy-wing to eliminate the 4 in R5C1!

Earl


Code:

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

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: Sat Jul 04, 2009 10:22 pm    Post subject: Reply with quote

I had either a two step or a five step solution depending on how you count a deletion using a pincer transport.
Quote:
W-wing 15 with pincers in r4c4 and r5c2 and strong link 1 in col8 that deletes 5 from r4c1,
then if r4c4=5, r7c4<>5, r7c3=5, r3c3<>5, so r3c2=5; delete 5 r68c2,
then if r5c2=5, r5c9<>5 so r4c9=5; delete 5 r4c5,
then if r4c4=5, r4c9<>5, so r5c9=5; delete 5 r5c1.
After cleanup, xy-wing 34-5 with pivot 34 in r5c1. If r5c9=5, r4c9<>5, so r4c4=5; delete 5 in r6c5 to complete the puzzle.

Ted
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Marty R.



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

PostPosted: Sun Jul 05, 2009 3:51 am    Post subject: Reply with quote

I used a Hidden UR (35), probably useless, ER (6), Mixed W-Wing, W-Wing (15) with pincer transport. At that point I saw no wings and got sick of looking for XY-Chains or more mixed W-Wings, so I finished with looking at the implications of the potential DP (13-17-37) in boxes 789.
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