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BREAKING NEWS

Submitted by: uno hu

Indicate which comments you would like to be able to see

SE 8.3
#1 SSTS to UP25.
#2 (3=4)i8-(4=7)h8-g7=(7-1)b7=(1-3)b5=a6 -> -3i6; UP28.
#3 (4=9)i4-i23=g1-(9=26)de1-(6=4)b1 -> -4b4; UP30.
#4 kraken cell (279)c3 -> -7c8; UP81
(9)c3-(9=6)c9-(9=24)h39-(4=7)h8
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(7)c3
||
Y-wing(2)-(2=4)h3-(4=7)h8
09/Apr/17 12:42 AM
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UP25; Kraken (2479)c3 -> [7h8==7c3]-7c8; UP81
2c3-(2=47)h38
4c3-XWing(4bg1, 4cg5)=4h5-(4=7)h8
7c3
9c3-9c9=*Wing[(4=2)h3-(2=*613)hcg9-(3=4)i8]-(4=7)h8
09/Apr/17 1:06 AM
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Here's a path one step shorter using JC's approach with the finned X wing; I have mixed feelings about this style; it does shorten the length of the proof but at the expense of significantly added complexity. JC's path is dynamic; mine is not, with similar trade-offs.
#2 (1)b7=(1-3)b5=g6-(3=1)g9 More...
09/Apr/17 2:14 AM
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pardon me; I found a typo in the kraken. The finned X-wing line was missing a cell address; it should have been (4)c3...
sorry.

(9)c3-(9=6)c9-(9=24)h39-(4=7)h8
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(7)c3
||
(4)c3-c4=finned X-Wing(4b14g14i4)[gi4=b4-b1=g1]-g5=h5-(4=7)h8
||
Y-wing(2)-(2=4)h3-(4=7)h8
09/Apr/17 2:28 AM
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Krakenless path:
lcls to 25
1. (4=7)h8-g7=b7-(7=3)b2-b5=g5-i6=(3)i8 =>-4i8; lcls to 28
2. (4=9)i4-g4=(9-4)g1=(4)b1 =>-4b4; lcls to 30
3. (6)c2=c9-(6=2)h9-(2*=4)h3-(4=7)h8-c8=(679)c239 =>-2c23; lcls to 81
09/Apr/17 2:57 AM
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Colourful photo.
09/Apr/17 4:32 AM
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cenoman, it appears to me that your last step is missing the value of 2 in c3 at that point. (279)c3? I am not seeing how that was eliminated. Thanks, Jim
09/Apr/17 4:44 AM
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Hi Jim,
if you agree with my elimination of 2c2, I don't catch your concern with the elimination of 2c3.
2c3 is eliminated by a sub-chain of the whole chain step #3, namely
(2=4)h3-(4=7)h8-c8=(679)c239 =>-2c3. I have tagged 2h3 with '*' to raise awereness of readers on the use of this More...
09/Apr/17 6:33 AM
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cenoman, in your 3rd step, what you have written implies that the value of 2 in h3 has been removed by that in h9, so (2)h9 is not available to eliminate (2)c3; from what I see, the chain implies (679)c2, but not (79)c3.

(6)c2=c9-(6=2)h9-(2=4)h3-(4=7)h8-(7=1)c8-(1=245)c456-(2)=(79)c23 -> +679c2,+279c3
09/Apr/17 7:00 AM
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One end and/or the other of the implication chain is true, so the result of the aic is the logical union of possibilities: for c2, (6) OR (79) -> (679)c2.

No assumption was made regarding c3, so the result is its initial value, (279) OR (79) -> (279)c3.
09/Apr/17 7:04 AM
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Jim,
you have written yourself the two chains eliminating 2c3 and 2c2
(2=4)h3-(4=7)h8-(7=1)c8-(1=245)c456 =>-2c3
and now the same with a prolongation:
(6)c2=c9-(6=2)h9-(2=4)h3-(4=7)h8-(7=1)c8-(1=245)c456 =>-2c2
You could as well write this
(2=4)h3-(4=7)h8-(7=1)c8-(145=2)c456 More...
09/Apr/17 9:08 AM
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#26 26@h9i7=>i7<>1&hi8g79<>26 1@g789=>i8<>1
3?@i6=>{i8=4=>h8=7,b5=3=>b7=1}=>g7=8=>26@fi7=>ad7= 5=>i6<>3
#29 26@i67=>i234<>26 More...
09/Apr/17 9:12 AM
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You are correct, cenoman. Sorry. Not enough brain food (coffee).
09/Apr/17 10:14 AM
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Variant of cenoman's solution based on (17)b7 -> 1 unique solution :

g2=5, c1=8; UP25

1. 3g5=(3-1)b5=(1-7)b7=HT(781)g789 -> -3g89, -1i78; i8=e9=3, HT(478)g78.h8=1g9; UP28

2. Kite(4bg1, 4i234)-(4=6)b4; b1=4; UP30

3. [7c3=XYWing(269)c39.h9-(2=47)h38]-(7=1)c8; 14 singles; UP45; HP(47)f23; stte
09/Apr/17 5:29 PM
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57:59
17/Apr/17 8:11 PM
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