From: Sam Kong <sam.s.kong@...>
Date: 2007-02-13T04:55:08+09:00
Subject: Re: What's the correct and fast way to determine if a (gig) number is a perfect square?

Hi Joel,

On Feb 11, 11:37 am, Joel VanderWerf <v...@path.berkeley.edu> wrote:
> Sam Kong wrote:
> > Hello,
>
> > I'm solving a math problem in Ruby.
> > I need to determine if a number is a perfect square.
> > If the number is small, you may do like the following.
>
> > def perfect_square? n
> >   sqrt = n ** 0.5
> >   sqrt - sqrt.to_i == 0
> > end
>
> > But Float number has limitation on precision.
> > Thus the function won't work correctly for big numbers like
> > (123456789123456789).
>
> > How would you solve such a case?
> > It should be fast as well as correct because I will use it repeatedly.
>
> Easy: compare integers rather than floats.
>
> x = 123456789123456789
>
> sqrt = Math::sqrt(x)
> p(x == sqrt.floor**2)

Yes. Your approach is better than mine.
But it gives a wrong answer for big numbers like 55833579873437812.


>
> --
>        vjoel : Joel VanderWerf : path berkeley edu : 510 665 3407

Thanks anyway.

Sam