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. __________________________________ Tracker <[EMAIL PROTECTED]> <http://bugs.python.org/issue2690> __________________________________ _______________________________________________ Python-bugs-list mailing list Unsubscribe: http://mail.python.org/mailman/options/python-bugs-list/archive%40mail-archive.com