math.pow(x,y)
Tim Peters
tim.peters at gmail.com
Sun Jun 11 21:58:42 CEST 2006
[Wojciech Muła]
>> You have to use operator **, i.e. 34564323**456356
Or the builtin pow() instead of math.pow().
[Gary Herron]
> That's not very practical. That computation will produce a value with
> more than 3.4 million digits.
Yes.
> (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).
On my box it took less than 30 seconds to do
x = 34564323**456356
If you try to _display_ that as a decimal string, it will take
enormously longer. Python uses a power-of-2 base internally, and
conversion to decimal takes time quadratic in the number of digits.
Doing y = hex(x) instead is very fast (small fraction of a second).
> 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)?
For example, if you only want the last 6 digits, pow(34564323, 456356,
1000000) returns 986961 in an eyeblink.
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