[PD] number to fractions external?

[Jonathan Wilkes]

>________________________________
> 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?
> 
>
>if you had read the thread, you would have seen that claude posted a link to that technique.
>
>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
>>>>   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
>>>>
>>>>   <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
>>>>       3/4, but that is easy to adapt on your patch
>>>>
>>>>       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
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>_______________________________________________
>>>>Pd-list at iem.at mailing list
>>>>UNSUBSCRIBE and account-management ->  http://lists.puredata.info/listinfo/pd-list
>>>>
>>>
>>>
>>>_______________________________________________
>>>Pd-list at iem.at mailing list
>>>UNSUBSCRIBE and account-management -> http://lists.puredata.info/listinfo/pd-list
>>>
>>
>>_______________________________________________
>>Pd-list at iem.at mailing list
>>UNSUBSCRIBE and account-management -> http://lists.puredata.info/listinfo/pd-list
>>
>>
>
>_______________________________________________
>Pd-list at iem.at mailing list
>UNSUBSCRIBE and account-management -> http://lists.puredata.info/listinfo/pd-list
>
>
>