math.pow(x,y)

**Wojciech Mu³a** · Jun 11 '06, 06:35 PM

Re: math.pow(x,y)

fl1p-fl0p wrote:[color=blue]
> import math
> math.pow(345643 23, 456356)
>
> will give math range error.
>
> how can i force python to process huge integers without math range
> error? Any modules i can use possibly?[/color]

You have to use operator **, i.e. 34564323**45635 6

**Felipe Almeida Lessa** · Jun 11 '06, 06:35 PM

Re: math.pow(x,y)

Em Dom, 2006-06-11 Ã s 11:19 -0700, fl1p-fl0p escreveu:[color=blue]
> import math
> math.pow(345643 23, 456356)
>
> will give math range error.
>
> how can i force python to process huge integers without math range
> error? Any modules i can use possibly?[/color]

34564323**45635 6 ?

--
Felipe.

**Gary Herron** · Jun 11 '06, 07:25 PM

Re: math.pow(x,y)

Wojciech MuÅ‚a wrote:[color=blue]
> fl1p-fl0p wrote:
>[color=green]
>> import math
>> math.pow(345643 23, 456356)
>>
>> will give math range error.
>>
>> how can i force python to process huge integers without math range
>> error? Any modules i can use possibly?
>>[/color]
>
> You have to use operator **, i.e. 34564323**45635 6
>[/color]
That's not very practical. That computation will produce a value with
more than 3.4 million digits. (That is, log10(34564323) *456356 =
3440298.) Python will attempt this, but I was not patient enough to see
if it could calculate an answer today (or even this week).

I doubt that you really *want* all 3.4 million digits. So what is it you
really want? A scientific or engineering result as a floating point
number accurate to some reasonable number of digits? That integer value
modulo some other integer (as used in various cryptology schemes)?

Gary Herron

**Raymond L. Buvel** · Jun 11 '06, 07:35 PM

Re: math.pow(x,y)

Felipe Almeida Lessa wrote:[color=blue]
> Em Dom, 2006-06-11 Ã s 11:19 -0700, fl1p-fl0p escreveu:[color=green]
>> import math
>> math.pow(345643 23, 456356)
>>
>> will give math range error.
>>
>> how can i force python to process huge integers without math range
>> error? Any modules i can use possibly?[/color]
>
> 34564323**45635 6 ?
>[/color]

I just tried this and it is taking an extremely long time even on a fast
machine with 4 Gb of RAM. Killed it after a couple of minutes. This
sort of calculation can be done with extended precision floating point
(as long as you don't need an exact answer). For example (using
defaults on a 64-bit machine),
[color=blue][color=green][color=darkred]
>>> from clnum import mpf
>>> mpf(34564323)** 456356[/color][/color][/color]
mpf('1.39518106 833639480699862 472257296396643 e3440298',36)

compute time is about 160 microseconds.

For more information see

Class Library For Numbers Module

http://calcrpnpy.sourceforge.net/clnumManual.html

For calculations involving large powers, you may still be better off
using logarithms.

**K.S.Sreeram** · Jun 11 '06, 07:45 PM

Re: math.pow(x,y)

Raymond L. Buvel wrote:[color=blue]
> I just tried this and it is taking an extremely long time even on a fast
> machine with 4 Gb of RAM. Killed it after a couple of minutes.[/color]

Thats odd.
34564323**45635 6 completed on my laptop in 28 seconds.
[Python 2.4.3, Celeron-M 1.3GHz, WinXP], and max memory consumption
during the whole process was about 11megs.

What python version are you using?

Regards
Sreeram

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**K.S.Sreeram** · Jun 11 '06, 07:55 PM

Re: math.pow(x,y)

Raymond L. Buvel wrote:[color=blue]
> I just tried this and it is taking an extremely long time even on a fast
> machine with 4 Gb of RAM. Killed it after a couple of minutes.[/color]

You probably tried printing the value.

a = 34564323**45635 6 (takes just 28 seconds)
whereas
b = str(a) takes forever!

Regards
Sreeram

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**Raymond L. Buvel** · Jun 11 '06, 09:35 PM

Re: math.pow(x,y)

K.S.Sreeram wrote:[color=blue]
> Raymond L. Buvel wrote:[color=green]
>> I just tried this and it is taking an extremely long time even on a fast
>> machine with 4 Gb of RAM. Killed it after a couple of minutes.[/color]
>
> Thats odd.
> 34564323**45635 6 completed on my laptop in 28 seconds.
> [Python 2.4.3, Celeron-M 1.3GHz, WinXP], and max memory consumption
> during the whole process was about 11megs.
>
> What python version are you using?
>
> Regards
> Sreeram
>[/color]

Sorry, I tripped over the display problem that Tim Peters has repeatedly
explained on this news group (decimal algorithm is quadratic in the
number of digits) :-(

On a 64-bit 2.2 GHz Opteron running Debian stable, Python 2.3 took 26
seconds and Python 2.4 took 11 seconds.

**casevh@comcast.net** · Jun 11 '06, 11:25 PM

Re: math.pow(x,y)

K.S.Sreeram wrote:[color=blue]
> Raymond L. Buvel wrote:[color=green]
> > I just tried this and it is taking an extremely long time even on a fast
> > machine with 4 Gb of RAM. Killed it after a couple of minutes.[/color]
>
> You probably tried printing the value.
>
> a = 34564323**45635 6 (takes just 28 seconds)
> whereas
> b = str(a) takes forever!
>
> Regards
> Sreeram
>[/color]

If you really want all 3,440,299 digits, use the DecInt module..

http://home.comcast.net/~casevh/

Using Python 2.4.3 on a 1.4Ghz Celeron, a=DecInt(345643 23) ** 456356
takes 20 seconds. astr=str(a) takes just over 1 second.

With DecInt and GMPY, the running times are 6.9 seconds and 0.4 seconds
respectively.

casevh