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submitting questions

Submitted by: kevin

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Beautiful blues in the jigsaw!
13/Jun/09 12:10 AM
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Now that's pretty! Have a great weekend, everyone!
13/Jun/09 1:15 AM
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maen all! i see why they're called tough! found it doable after i printed it out and got down to eliminating possibilities.
13/Jun/09 1:17 AM
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Kate!! Very very good
13/Jun/09 1:28 AM
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Interesting lighting and perfect flower
13/Jun/09 1:34 AM
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7:52 Hi to all. Lovely colours.
13/Jun/09 1:36 AM
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9:24
13/Jun/09 1:49 AM
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Beautiful reward!
13/Jun/09 4:33 AM
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1. Note pair 17 at c18.Unique possibilities to 43.
2. If g2=9,h1=4=d2,e3=9=a1,d8=5,h8=8,a9=7 and row 8 is devoid of 7.So g2=4 and UPs to 81.
13/Jun/09 6:41 AM
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13:02, hi everyone.
13/Jun/09 8:29 AM
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must have messed up somewhere 31:00
13/Jun/09 11:16 AM
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1) Start at 23,basic techniques to UP=43
2) (7):c8=c1-h1=i3,=>i8<>7
3) (8)d4=g4-(8=4)g6-g2=(4-5)d2=(5)d8,=>d8<>8.UP=81
13/Jun/09 3:54 PM
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Thank-you Alfred and Sotir for posting solutions/proofs. Just for kicks and giggles, here is a rough comparison of the two posted solutions:
Alfred: 2 steps, 10 total Strong Inference Sets, Max depth in any one step: 8
Sotir: 3 steps, 8 total Strong Infernce Sets, Max depth in any one step: More...
13/Jun/09 8:28 PM
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Obviously, I cannot count. Sotir's solution uses a total of 9 SIS, 2+2+5=9.

In both, I do not count SIS considered for an UP. In every case, it takes one SIS for each UP, so that additional count would be the same addition for both solutions.
13/Jun/09 8:37 PM
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Thanks Steve for the analysis.I noticed later I could have done without e3=9 in step 2.After g2=9,a1 becomes 9 without the need of 9 at e3.Regards,Alfred.
14/Jun/09 12:07 AM
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SE=7.2
09/Sep/10 10:46 PM
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