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Submitted by: uno hu

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47:02
01/Oct/17 3:24 AM
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Great design.
01/Oct/17 5:09 AM
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1. Note de2=69.Unique possibilities to 33.
2. Whether c4=4=f5;OR c46=39,b8=3=g7,a87=79,d7=5=e1;f1=8.UP38.
3. Whether b8=3=g7,a87=79,d7=5;OR
c9=3,c2=2,b2=4,b4=1=g6,g4=7,d6=3=i4=h3,a3=1;d1=1.UP46.
4. Whether h8=3=c9;OR h8=2;c2=2.UP81.
01/Oct/17 6:29 AM
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#29 69@de2=>de2<>7 3@a13=>a69<>3 6@ef4=>de6<>6 #31 5@de1=>i1<>5
#32 8?@e1=>{d1=5,39@ce6}=>d6=7=>39@d57=>e2=9=>e6=3=>c9=3=>c6=9=>c4=4=>69@ef4= >d5={}=>e1<>8
#39 4@a89=>b8<>4 More...
01/Oct/17 8:36 AM
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Another great design feature it has is a self cleaning roof.
01/Oct/17 8:40 AM
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Hi Les. In your chain starting with 8@e1, when e6 becomes 3, you give c9=3 as the result. I think you forgot to state that first,d9=1, before c9 becomes 3.

Best Regards, Alfred.
01/Oct/17 4:13 PM
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6:22 With a couple of guesses.
01/Oct/17 7:23 PM
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Hi Alfred
Thanks for your scrutiny of my pathway.
e6=3=>c9=3 because of the double 39@ce6=>d9<>3 which means c9 is the only cell left in the top row for a 3.
I did notice a few other unrelated emissions however, so I will repost my path.
Thanks Alfred.
02/Oct/17 10:59 AM
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#29 69@de2=>de2<>57 3@a13=>a69<>3 6@ef4=>de6<>6 #31 5@de1=>ai1<>5
#32 8?@e1=>{d1=5,39@ce6}=>d6=7=>39@d57=>e2=9=>e6=3=>c9=3=>c6=9=>c4=4=>69@ef4= >d5={}=>e1<>8
#39 More...
02/Oct/17 11:04 AM
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Sorry Alfred
Wrong double not 39@ce6 but 39@d57
Chhers Les
02/Oct/17 12:44 PM
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