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 Hello, Nicole! 29/Oct/11 12:12 AM |  |
 WOW Deb! Nicole looks so much like you. Pretty lady! 29/Oct/11 1:12 AM |  |
 1. Note cg5=47;e46=12.Unique possibilities to 24.2. If g5=7,g789=289,h8=6=h4;column c is devoid of 6.So c5=7.UP81. 29/Oct/11 1:41 AM |  |
 site puzzle is FXW, UP81 Alfred 29/Oct/11 3:48 AM |  |
 8.5**** (SE @ 13K x 19FC)000000480000030000058000702009000000030060501406700000000001200000804030000070040 29/Oct/11 3:51 AM |  |
 8.5 ****#1 (3=5697*8)dehai3 - (8=927)i4h45 - c5=b6 - *ab8=(7-4)c8=c3 ; c3<>3// sue-de-coq (69)f7,(1269)d876 ; d1<>2, d9<>1269, e9<>9#2 (2)f1=e2 - f9=c9 - (2=7)c5 - (7=298)h54i4 - i13=(8-1)g1=a1 ; a1<>2// kite(2) ; e2<>2// SSTS, xy-wings... 29/Oct/11 6:10 AM |  |
 your step #1 makes no sense at all to me peterj. 29/Oct/11 7:27 AM |  |
 you can do the deduction with:1. [(4)c3=(4-8)b3=i3-(8=279)i4h45-(7)c5=*(7-4)c8=c3]||(7)c23-(7=3569)adeh3 # -35c3but I simply can't see the logic of your chain. 29/Oct/11 7:30 AM |  |
 only other difference with path I used is// WRing(12)d6c9 instead of SDQ 29/Oct/11 7:50 AM |  |
 Thanks kobold.I could have said:Whether i5=6;OR a5=6=c8;i8=28.UP81. 29/Oct/11 7:59 AM |  |
 also #2(2)e9=c9-(2=7)c5-(7=289)h45i4-(8)g46=(8-1)g1=(1-2)a1=SS(2)a28.ce9 # -2e2 29/Oct/11 7:59 AM |  |
 Kobold, if d3<>3 then there is a ls(569)deh3 locking 7 in a3 and 8 in i3. (My * denotes remember the 7.) The simple chain from 8 and the memory 7 makes (7)ab8=c8 and on to d3<>3. I don't see this as any harder than much you propose and, while I might have made a mistake, "no sense at all" seems a strong criticism. 29/Oct/11 8:41 AM |  |
 Sorry - above d3=3 29/Oct/11 8:46 AM |  |

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