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Tough Sudoku for 22/April/2013


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Submitted by: Gath

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Looks like a festive gathering to me!
22/Apr/13 12:19 AM
1. Note h57=26.Unique possibilities to 29.
2. If h5=6,b5=89,a9=8; there are no candidates for c1; because c1 cannot be 89(to avoid the 89 Unresolvable Rectangle at bc15); but then c1=47, creating a 47 UR at ac17. So h5=2. UP81.
22/Apr/13 1:05 AM
SER=7.2 UP=27
1)(6)h7=(47)ac7 or (6)h5-b5=(6)b78 =>ac7<>6 UP=28
2)(8=6)a8-b9=(8-4)e9=(4)h9=(2)i8=(6)h7 => g8,b7<>6 UP=29
3)(-5)g7=(5=3)b7=(3-6)b8=(6)b5-h5=(6)h7 => g7<>6 UP=81
22/Apr/13 1:48 AM
1: UP27
2: (6=89)b15-(8)b89=(8-4)a8=(4-2)f8=f7-(2=6)h7 => h5<>6, UP81
22/Apr/13 2:04 AM
Hmm. Rechecking my proffered solution leaves a doubt that I had left out 4 as a possibility in i8. Anyway, with that possibility still there 4 is not strong in af8, so my chain is probably invalid. Sorry.
22/Apr/13 2:14 AM
Like Jyrki's ... !

#1. 3 Singles; UP25
HP(26)h57; UP27

LC(1,3,5); [FXWing(6b578.h57) :=> -6ac7;] LC(7)
NP(47)ac7; NP(12)ai4; LC(2); HP(24)fi8; UP28; LC(9)

#2. Chain[6] : 2i8=(2-6)h7=6h5-(6=89)b15-8b9=(8-4)e9=4f8 :=> -4i8, -2f8; UP81
22/Apr/13 3:35 AM
Read FXW... instead of [FXW...]
22/Apr/13 3:59 AM
Sure does Vici!
Hello everyone!
22/Apr/13 8:36 AM
26 26@h57=>h5<>48
28 5@c46=>c7<>5 5@h12=>h7<>5
2 9 More...
22/Apr/13 8:43 AM
2 9 More...
should have appeared as
29 More...
22/Apr/13 8:44 AM
My step 1 should read UP27.
Step 2 also works in reverse order:
2. Whether b89=8,b1=9,b5=6;OR a8=8,b789=356,c1=89(to avoid 47UR),c5=6(to avoid 89UR);h5=2.UP81.
22/Apr/13 9:19 AM
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