nth root

Paul Rubin http
Sun Feb 1 01:17:07 EST 2009

"Tim Roberts" <t.roberts at cqu.edu.au> writes:
>                     for root in powersp:
>                             nroot = round(bignum**(1.0/root))
>                             if bignum==long(nroot)**root:
>                                                              .........
> which is probably very inefficient, but I can't see anything better.....

Actually that will not be accurate enough to know if bignum is, say, a perfect square.
You will have to do the small-power tests at full precision (or use a fancy algorithm).
Larger powers can be done with floats.

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