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Submitted by: uno hu

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Kathy! Is that you???
10/Jun/17 1:48 AM
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Basic techniques to UP = 33. Hopefully, this is clear enough. I'm open to suggestions if there is a cleaner or clearer way to present the idea:
(4)a8 = a4 - b6 = HP(14)gh6|(14)h8 - (1)h7 = f7 - (1=7)f1 - h1 = h4 - (7=24)b46 => -4b9 ; UP = 81
10/Jun/17 5:27 AM
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#28 34@ef2=>[f2<>1 & 2@de3=>i3<>2] 17@d8f7=>d8f7<>389=>9@e789=>e13<>9 #33 179@d138=>d6<>9
#34 1?@d8=>h8=4=>a4=4=>g6=4=>g1….9<>1=>d8<>1
VHBC to #81
V:Only cell left in Vertical column for this More...
10/Jun/17 6:27 AM
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Hi rwm
I can't more easily present a path to show b9 <> 4 then you have.
But d8 <> 1 can be easily presented, as shown above.
Cheers Les.
10/Jun/17 6:47 AM
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10/Jun/17 6:49 AM
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lcls to 33
1. Almost turbot fish (4)h6=b6-b9=a8 with (4)g6 as a fin:
(1=4)h8-[TF(4)h6*=b6-b9=a8]=(4-1)g6=(1)g8 =>-1d8; singles to 81
The same logic is in
1. Kraken row (4)bgh6 =>-1d8; singles to 81
10/Jun/17 7:18 AM
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Hi rwm, the idea behind your path is clear.
I would not notate your node HP(14)gh6|(14)h8 - (1)h7 as you did, because you need both the pair (14)gh6 AND the bivalue (14)h8 for the conflict with 1h7.
To me, the symbol | stands for the boolean operator OR while you mean AND. My suggestion:
HP(14)gh6 & (14)h8 - (1)h7
Best regards, cenoman
10/Jun/17 7:37 AM
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1. Unique possibilities to 33.
2. Whether a8=4,h8=1;OR a4=4,gh6=14;h7=26.UP81.
10/Jun/17 8:00 AM
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I got UP to 81.
10/Jun/17 2:16 PM
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2:47 :)
10/Jun/17 5:45 PM
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Hi RWM, I think you need go no further than '-(1)h7' in your chain. Because you have shown that both(4)a8 which results in (1)h8, and (4)a4 which results in -(1)h7, has the same result, namely -(1)h7, which solves the puzzle.

Best Regards, Alfred.
10/Jun/17 5:58 PM
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