Alexander Belopolsky <[EMAIL PROTECTED]> added the comment:
On Fri, Apr 25, 2008 at 4:37 PM, Mark Dickinson <[EMAIL PROTECTED]> wrote:
..
> for i in range(1, p):
> if (i_is_a_nonresidue_modulo_p):
> break
>
> Here p might be a 200-digit prime number, and the situation is that half
> the integers between 1 and p-1 are 'quadratic residues', while the other
> half are 'quadratic nonresidues'; in practice the residues and
> nonresidues are mixed up fairly well, so the first nonresidue shows up
> pretty quickly, but there's no known small upper bound on when the first
> nonresidue appears.
Hmm, AFAIKT there is always at least one non-residue between 1 and p
and therefore you can just write
for i in itertools.count(1):
if (i_is_a_nonresidue_modulo_p):
break
maybe with an additional check for p > 1.
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Tracker <[EMAIL PROTECTED]>
<http://bugs.python.org/issue2690>
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