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 47:02 01/Oct/17 3:24 AM |  |
 Great design. 01/Oct/17 5:09 AM |  |
 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;ORc9=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 |  |
 #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 |  |
 Another great design feature it has is a self cleaning roof. 01/Oct/17 8:40 AM |  |
 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 |  |
 6:22 With a couple of guesses. 01/Oct/17 7:23 PM |  |
 Hi AlfredThanks 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 |  |
 #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 |  |
 Sorry AlfredWrong double not 39@ce6 but 39@d57Chhers Les 02/Oct/17 12:44 PM |  |

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