Tough Sudoku for 2/February/2006

Awww, drunk kitty! Good MaeN all!

GH

whoops forgot to restart the timer after a pause
I 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, therefor h6 is not 1. This is easily seen upon inspection, but evaluates to an fc of equivalent length and sets as gb's solution. Nevertheless, it seems easier and simpler to look at it as described above.

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 proving one. I went on them by realizing there was many ways of justifying a given elimination.

Oh God, comment window will be too small once more... contd...

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' ?

Incidentally, you did not detail how you came to 52 filled. I'm sure it needed other eliminations. If you want to compare fc and 'nonfc' proofs, you have to choose an objective way of counting them.

contd...

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 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 f7 is not 1. Then easy fillings to the end.

Is it really such important to rely on various vocabularies to describe the same reality that a single tool can handle?

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.

20:09 - Better

SE=6.6

19:05.