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Submitted by: vicky

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Awww, drunk kitty! Good MaeN all!
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whoops forgot to restart the timer after a pause
I guess he fell asleep after a big play with the box
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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...
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there is something called simultaneous equations using matrices, then it's so easy u could cry.
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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).
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lol - the solution I suggested, I mean - not the one suggested from the higher power.
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21:21 all depended on where I went looking for my chains...
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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...
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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'.
'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.

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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' ?
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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'.

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A complete proof for fc-allergic readers :

1) easy fillings to 27 filled
2) locked 8s in gi5 : eliminate def5=8, then easy fillings to 31
3) hidden pair 36 in c17 : eliminate c1=4
4) hidden pair 25 in ab3 : eliminate a3=9
5) colors with 1s : if f7=1 then both g4 and h6 are 1, so More...
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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.
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Apr. 25, 2007. Easy until the end, then lots of 1, 3, 9 triples.
26/Apr/07 8:10 AM
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20:09 - Better
16/Nov/07 1:29 PM
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05/Sep/10 9:59 AM
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