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Submitted by: Gath

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Unfortunately, this is the beach (which I'm standing on) where a man was attacked by a shark recently while early morning swimming. His leg was badly mauled but a very brave woman swam in and brought him back to shore. It was quite a rare incident for this particular beach.
26/Jun/08 12:11 AM
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Another very nice picture Anne. As in Greece, nice waters!!!
26/Jun/08 12:53 AM
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Looks pretty calm... And lonely.
26/Jun/08 1:14 AM
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Those beautiful blues wouldn't give you the blues, but a shark certainly would.
A big to all the heros of the world.
26/Jun/08 1:30 AM
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The water is beautiful and I am always ready for a boat ride. What a terrible thing to happen to that man. That woman was very brave and he was fortunate to have her near.
26/Jun/08 2:21 AM
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21:01
26/Jun/08 2:43 AM
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Good afternoon to all! Another nice picture Anne. I assumed the man who was attacked survived?
26/Jun/08 3:03 AM
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Wow, what a story. I salute the courage of that woman, and all those "everyday" heroes who reach out to help others.
26/Jun/08 3:08 AM
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A path:

1)start at 22, basic techniques to UP=30
2)XYZ: (56)bc9=(9)c9-(9=5)c6,=>b4<>5. UP=32
3)fc:(9=8)g5-(8=5)h4-(5=9)a4-(9)a8=(9)c9,=>g9<>9.UP=33
(note nt:ahi8=569)
4)fc:(8=3)g1-(3=9)g7-(9)i8=(9)a8-(9)c9=(9-5)c6=(5)c1
=>c1<>8.UP=34
5)fc:((5 More...
26/Jun/08 4:54 AM
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Anne, the variations in the color of the water are stunning!
Bless that woman who put her life on the line to save another life!
26/Jun/08 6:04 AM
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A path to solve this one
Start at 22, basic techniques UP30
Note : pairs 18 at b78, 47 at a5b6, 12 at i25
1) (9)c9=(9)c6-(9=hp56)ab4-(5=8)h4-(8=9)g5
=> g9<>9, UP31
2) Color on 9’s : c9=a8 – a4=e4 => e9<>9, UP81

Can’t follow sotir’s step 2…
26/Jun/08 6:24 AM
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Germany def. Turkey : 3-2, come to final match EURO2008. Congrats...!
26/Jun/08 7:04 AM
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Yes he did, Greg! Thanks for a wonderful photo Anne!
26/Jun/08 7:59 AM
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Love your Sand Dollar Kathy!
26/Jun/08 8:00 AM
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Another nice picture Anne great part of Oz.
26/Jun/08 8:45 AM
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sotir, step 2 is xyz. the z is the common element to all three cells in this case 5. but it only affects cells that have the three cells in common. if c9 is 6 or 9 b 4 is not 5 but if c9 is 5 then b4 must be 5.in this case c78 would not be 5.
26/Jun/08 10:50 AM
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27:30
26/Jun/08 11:13 AM
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Hi Chiefmsb,

Your explanation for step 2 is very clear.
In fact step 2 is not necessary.In this case for step 3 and 5 we have:
3)(9=8)g5-(8=5)h4-(5)ab4=(5-9)c6=(9)c9,=>g9<>9.UP=31
5)(5)d1=(5)c1-(5)c6=(5)ab4-(5= 8)h4-(8)g5=(8)g1,=>d1<>8.UP=36
Thanks.
26/Jun/08 1:02 PM
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19.03
Anne, your beach shots are sending me sea-green with envy!
26/Jun/08 5:18 PM
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20...mmmm, havent been here for a while..think waves in bay of bengal are more active...
cheers
26/Jun/08 10:57 PM
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43
27/Jun/08 2:55 AM
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sotir,
I believe there is still a problem with your step 4, I think that the strong link (9)i8=(9)a8 that you used is not valid since 9 is still a possible candidate in e8. Without this chain the remainder of the proof doesn't work.
Regards,
Jeff
27/Jun/08 11:58 AM
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