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16 x 16 broken?

Submitted by: barclay

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 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||(7)c3||Y-wing(2)-(2=4)h3-(4=7)h8 09/Apr/17 12:42 AM |  |
 UP25; Kraken (2479)c3 -> [7h8==7c3]-7c8; UP812c3-(2=47)h384c3-XWing(4bg1, 4cg5)=4h5-(4=7)h87c39c3-9c9=*Wing[(4=2)h3-(2=*613)hcg9-(3=4)i8]-(4=7)h8 09/Apr/17 1:06 AM |  |
 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 |  |
 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||(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 |  |
 Krakenless path:lcls to 251. (4=7)h8-g7=b7-(7=3)b2-b5=g5-i6=(3)i8 =>-4i8; lcls to 282. (4=9)i4-g4=(9-4)g1=(4)b1 =>-4b4; lcls to 303. (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 |  |
 Colourful photo. 09/Apr/17 4:32 AM |  |
 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 |  |
 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 |  |
 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 |  |
 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 |  |
 Jim,you have written yourself the two chains eliminating 2c3 and 2c2(2=4)h3-(4=7)h8-(7=1)c8-(1=245)c456 =>-2c3and now the same with a prolongation:(6)c2=c9-(6=2)h9-(2=4)h3-(4=7)h8-(7=1)c8-(1=245)c456 =>-2c2You could as well write this(2=4)h3-(4=7)h8-(7=1)c8-(145=2)c456 More... 09/Apr/17 9:08 AM |  |
 #26 26@h9i7=>i7<>1&hi8g79<>26 1@g789=>i8<>13?@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 |  |
 You are correct, cenoman. Sorry. Not enough brain food (coffee). 09/Apr/17 10:14 AM |  |
 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 |  |
 57:59 17/Apr/17 8:11 PM |  |

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