[PD] number to fractions external?
Mathieu Bouchard
matju at artengine.ca
Sat Dec 17 21:29:00 CET 2011
Le 2011-12-17 à 03:44:00, i go bananas a écrit :
> how do you get from a continued fraction in the form like this:
> [0;1,5,2,2]
> to a fraction in the form like this:
> 27/32
> this patch gets as far as that [0; 1,5,2,2] form. but i'm still not sure how to get further
keep track of the last two fractions and feed that to something like expr.
e.g.
$f1 = element of the list
$f2 = last numerator
$f3 = last denominator
$f4 = next-to-last numerator
$f5 = next-to-last denominator
the number before the semicolon should go with the values 1 0 0 1 to start
the algorithm (yes, you have to pretend that a denominator is zero, but
don't worry)
then it would be something like :
[expr $f1*$f2+$f4;
$f1*$f3+$f5;
$f2;
$f3]
and then you take the four outputs and feed them back into expr together
with the next number in the continued fraction...
the advantage of this, is that you can go left-to-right.
With the more obvious method, you have to go right-to-left, because the
innermost terms (inside many parentheses) are at the right.
The left-to-right formula is something I learned in Number Theory course
in fév.2003, but naturally, I had to look it up in Wikipédia in order to
remember, as I don't use this very often...
Let's try the formula with [0;1,5,2,2] :
0 1 0 0 1 gives 0 1 1 0
1 0 1 1 0 gives 1 1 0 1
5 1 1 0 1 gives 5 6 1 1
2 5 6 1 1 gives 11 13 5 6
2 11 13 5 6 gives 27 32 11 13
looks like it works ! 27 32 appear in the 2nd and 1st outlets.
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| Mathieu BOUCHARD ----- téléphone : +1.514.383.3801 ----- Montréal, QC
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