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 Awww, drunk kitty! Good MaeN all! |  |
 GH |  |
 whoops forgot to restart the timer after a pauseI guess he fell asleep after a big play with the box |  |
 If one is to have a discussion about complexity of solving techniques, perhaps non-fc logic needs to be evaluated also. Although the solution I came up with for todays puzzle can be written as an fc, it is somewhat simpler to look at it otherwise. To wit: If h6=1 then both f7 and g7 are 1, More... |  |
 there is something called simultaneous equations using matrices, then it's so easy u could cry. |  |
 The 'solution' suggested above can be performed at 43 filled after using 3 sets depth 1 to get there, I think. A pair elimination, a pointing elim, and a pair elim - all focusing on box a2, or thereabouts.(at 27,28,28 filled, respectively). |  |
 lol - the solution I suggested, I mean - not the one suggested from the higher power. |  |
 21:21 all depended on where I went looking for my chains... |  |
 Hi Steve, interesting comment. 1) In a way the best solution for each solver is the one he found by himself : there's no discussing it (but for trying to create an objective complexity measuring tool)! 2) I must insist on a point about fcs : they are more a writing tool than a More... |  |
 Your argument (at 52 filled, if I understand) is 'If h6=1 then both f7 and g7 are 1, therefor h6 is not 1'.We could also say 'Whether f6=1 or f7=1, giving g4=1, h6 is not 1'.or'Whether g4=1 or g7=1, giving f6=1, h6 is not 1'.or... let me stop here. These are (at least) but three ways of reading the same fc :(f6=1)==(f7=1)--(g7=1)==(g4=1) forbids h6=1. contd... |  |
 Well you may accept or not that your argument is a fc and prefer your way among the others of telling the story, anyway, it needs examining two sets : {f67=1}=possible ones in col f, and {g47=1}=possible ones in col g. So do you agree to give your argument the weigh 'two sets needed' ? More... |  |
 Let's return to my last elimination at 31 filled (after which easy fillings to the end) : (f6=1)==(h6=1)--(g4=1)==(g7=1) forbids f7=1 You can perfectly read it your mood : 'if f7=1 then both g4 and h6 are 1, so f7 is not 1'. ok? |  |
 A complete proof for fc-allergic readers : 1) easy fillings to 27 filled2) locked 8s in gi5 : eliminate def5=8, then easy fillings to 313) hidden pair 36 in c17 : eliminate c1=44) hidden pair 25 in ab3 : eliminate a3=95) colors with 1s : if f7=1 then both g4 and h6 are 1, so More... |  |
 12:23 |  |
 Hi Steve, your more precisely sketched solution gives it now the same complexity than mine (pair elim=2 sets). So it's a draw ? shake hands ! But I didn't mean at all to maintain that my solution was better ! I just wanted to highlight the role of fcs as a unifying vocabulary allowing a precise definition of the complexity of a proof. |  |
 15:49 |  |
 28:29 |  |
 17:01..10/01/06 |  |
 Apr. 25, 2007. Easy until the end, then lots of 1, 3, 9 triples. 26/Apr/07 8:10 AM |  |
 20:09 - Better 16/Nov/07 1:29 PM |  |
 SE=6.6 05/Sep/10 9:59 AM |  |

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