If you guys 'd done your math, you'd know there is an ancient algorithm for approximating numbers by fractions and its called continued fractions.
On 16 December 2011 18:38, Lorenzo Sutton <lorenzofsut...@gmail.com> wrote: > On 16/12/11 16:05, Alexandre Torres Porres wrote: > >> looks like a job for an external >> > Not really answering the OP question but something could be done in Python: > > def find_frac(num): > f = float(num) > last_error = 1000 > best = (0,0) > for i in xrange(1,1001): > for j in xrange(1,i+1): > divide = (float(i) / float (j)) > if divide == f: > return ((i,j),0) > err = abs(divide - f) > if err < last_error: > best = (i,j) > last_error = err > return (best,last_error) > > This would try to find the exact fraction or the one with the smallest > error (trying up to 1000/1000). It would return (numerator, denominator, > error). Guess it would work well at least up to 100 but only for positive > numbers... and... not for numbers < 1.. and surely it's not optimised etc. > etc. :) > > >>> find_frac(2) > ((2, 1), 0) > >>> find_frac(1.5) > ((3, 2), 0) > >>> find_frac(1.**333333333333333333333333333) > ((4, 3), 0) > >>> find_frac(2.4) > ((12, 5), 0) > >>> find_frac(2.8) > ((14, 5), 0) > >>> find_frac(2.987654321) > ((242, 81), 1.234568003383174e-11) > >>> find_frac(50.32) > ((956, 19), 0.004210526315787888) > >>> find_frac(50.322) > ((956, 19), 0.006210526315790332) > >>> find_frac(50.4) > ((252, 5), 0) > >>> find_frac(10.33) > ((971, 94), 0.00021276595744623705) > >>> find_frac(10.**33333333333333333333333333) > ((31, 3), 0) > > Lorenzo. > >> >> >> >> 2011/12/16 i go bananas <hard....@gmail.com <mailto:hard....@gmail.com>> >> >> >> actually, i'm not going to do anything more on this. >> >> i had a look at the articles claude posted, and they went a bit >> far over my head. >> >> my patch will still work for basic things like 1/4 and 7/8, but i >> wouldn't depend on it working for a serious application. As you >> first suggested, it's not so simple, and if you read claude's >> articles, you will see that it isn't. >> >> it's not brain science though, so maybe someone with a bit more >> number understanding can tackle it. >> >> >> >> On Sat, Dec 17, 2011 at 12:51 AM, Alexandre Torres Porres >> <por...@gmail.com <mailto:por...@gmail.com>> wrote: >> >> > i had a go at it >> >> thanks, I kinda had to go too, but no time... :( >> >> > yeah, my patch only works for rational numbers. >> >> you know what, I think I asked this before on this list, >> >> deja'vu >> >> > will have a look at the article / method you posted, claude. >> >> are you going at it too? :) >> >> by the way, I meant something like 1.75 becomes 7/4 and not >> 3/4, but that is easy to adapt on your patch >> >> thanks >> >> cheers >> >> >> >> 2011/12/16 i go bananas <hard....@gmail.com >> <mailto:hard....@gmail.com>> >> >> >> by the way, here is the method i used: >> >> first, convert the decimal part to a fraction in the form >> of n/100000 >> next, find the highest common factor of n and 100000 >> (using the 'division method' like this: >> >> http://easycalculation.com/**what-is-hcf.php<http://easycalculation.com/what-is-hcf.php>) >> >> then just divide n and 100000 by that factor. >> >> actually, that means it's accurate to 6 decimal places, i >> guess. well...whatever :D >> >> >> >> >> >> ______________________________**_________________ >> Pd-list@iem.at mailing list >> UNSUBSCRIBE and account-management -> http://lists.puredata.info/** >> listinfo/pd-list <http://lists.puredata.info/listinfo/pd-list> >> > > > ______________________________**_________________ > Pd-list@iem.at mailing list > UNSUBSCRIBE and account-management -> http://lists.puredata.info/** > listinfo/pd-list <http://lists.puredata.info/listinfo/pd-list> >
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