# [PD] number to fractions external?

Mike Moser-Booth mmoserbooth at gmail.com
Sat Dec 17 03:30:18 CET 2011

```I just gave this a go, and here's what I have so far based on the
Wikipedia link Claude gave. Send the decimal through the left inlet,
and it outputs the numerator and denominator as a list. The argument
is the maximum value the denominator is allowed to be, which keeps it
from going crazy trying to figure out irrational numbers and also
seems to make up for some floating-points errors. You can change the
max with the right inlet as well. Right now it defaults to 100, but
that may be too low. Higher values=more accurate, but potentially more
computation.

I haven't implemented the rules for decrementing the last value of the
continuous fractions, so it's not perfect. But it does give 355/113
for pi. :-)

.mmb

On Fri, Dec 16, 2011 at 2:46 PM, Jonathan Wilkes <jancsika at yahoo.com> wrote:
>>________________________________
>> From: i go bananas <hard.off at gmail.com>
>>To: Ludwig Maes <ludwig.maes at gmail.com>
>>Cc: pd-list at iem.at
>>Sent: Friday, December 16, 2011 1:16 PM
>>Subject: Re: [PD] number to fractions external?
>>
>>
>>
>>now go and make a PD patch that does it, mr smart guy.
>
>
> Wow, how much cpu does that take in Python?  I tried this approach in the form
>
> of an abstraction, with a nested until, and worst case it can take as much as a quarter of
>
> a second to compute with the constants provided below.
>
> (Pentium Dual-core 2.6gHz in WinXP with 0.43 nightly build)
>
> -Jonathan
>
>
>>
>>
>>
>>
>>On Sat, Dec 17, 2011 at 3:00 AM, Ludwig Maes <ludwig.maes at gmail.com> wrote:
>>
>>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 <lorenzofsutton at 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.off at gmail.com <mailto:hard.off at 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
>>>>>
>>>>>   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
>>>>>
>>>>>   <porres at gmail.com <mailto:porres at 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
>>>>>
>>>>>       thanks
>>>>>
>>>>>       cheers
>>>>>
>>>>>
>>>>>
>>>>>       2011/12/16 i go bananas <hard.off at gmail.com
>>>>>
>        <mailto:hard.off at 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 )
>>>>>
>>>>>           then just divide n and 100000 by that factor.
>>>>>
>>>>>           actually, that means it's accurate to 6 decimal places, i
>>>>>           guess.  well...whatever :D
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>
>>>>>
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>>>>>
>>>>
>>>>
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>>
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>
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--
Mike Moser-Booth
mmoserbooth at gmail.com
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