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Submitted by: Gath

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Now that is something special Sue. Thanks so much
30/Jul/08 12:02 AM
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You're up late, fii.
30/Jul/08 12:31 AM
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A perfect place. Well done, Sue! Have a wonderful day/night, everyone.
30/Jul/08 12:37 AM
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Sue, this photo is incredible! Beautiful spot, and excellent photography. Thanks!
30/Jul/08 1:07 AM
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26:16 Maen! Gorgeous picture, Sue!
30/Jul/08 1:14 AM
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Wow, Sue! Beautiful picture - - I love how you 'framed' your focal point with the foreground. Awesome effect!!
30/Jul/08 1:21 AM
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Composition & Color - Great Sue!!
30/Jul/08 2:29 AM
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great picture Sue!!!! We love our great lakes!
30/Jul/08 3:26 AM
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I've been there MANY times, usually in horseback.
30/Jul/08 3:27 AM
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27:37
30/Jul/08 4:42 AM
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Good afternoon to all! A super picture Sue! Thank you!!
30/Jul/08 5:17 AM
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UP:c6=2, c5=9.
8 can appear in c3 or d2 in the lower middle box.
If f2=8, pairs 67 at f79. Hence f5=5, f4=4, f6=9, pairs 13 at e89, e6=9. There would thus be two 9s at e6,f6. Hence f2 cannot be 8, and so c2=8. The rest follows for all the blanks.
30/Jul/08 6:47 AM
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Well done, Sue. The water looks inviting!
30/Jul/08 7:29 AM
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What a beautiful window with a fabulous view! Thanks Sue!
30/Jul/08 9:41 AM
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Beautiful! Thank you, Sue!
30/Jul/08 9:51 AM
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Beautiful Sue...looks like a great place for a relaxing day.
30/Jul/08 11:33 AM
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35:42
30/Jul/08 11:40 AM
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Rugram, very nice solution.

Does anyone know how to write Rugram's elimination of the 8 at f2 as an AIC (chain)? Or maybe it would be a nice loop that places the 8 at d3? I've been trying for about 15 minutes without making much progress and I would like to see it.

Thanks.
30/Jul/08 1:39 PM
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Thanks, Jeff! I don't much understand the notation used here for the proofs, and so am somewhat descriptive in posting proofs. Sorry for that!
30/Jul/08 1:57 PM
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Hi Jeff,
(8=hp67)f89-(67=hp13)e89-(13=9)e6-(9)f6=(9-8)f2=(8)d3
=> d8<>8 => f9=8
Hope to help.
30/Jul/08 2:45 PM
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Oops... maybe shorter:
(8=hp67)f89-(67=hp13)e89-(13=9)e6-(9)f6=(9)f2 => f2<>8
30/Jul/08 2:52 PM
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Thanks Sue, would love to visit on horseback Heidi.
30/Jul/08 4:43 PM
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Hi Jeff,
I studied more and see that above chain can’t work if go from Right to Left. I had questioned Experts on other Site and I’m afraid this deduction can’t present as AIC (must be AAIC)
30/Jul/08 5:13 PM
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12.17 Fantastic colours.
30/Jul/08 8:31 PM
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Sorry for the typos in my solution.
1. The possibilities for 8 should be at d3 or f2 (not c3/d2 as I wrote).
2. What follows should be: d3=8, not c2=8.
Thanks, Alfred from Sydney, for pointing out these typos.
31/Jul/08 12:04 AM
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