I wrote:
From: Brian J. Beesley [mailto:[EMAIL PROTECTED]]
If P-1 does find a factor which is compound, then running P-1 again
with smaller limits will eventually recover a smaller factor. These
extra runs will obviously take less time than the original
Indeed, and with care one
I was wondering if it was common practice (ie: the norm) for P-1
to take the product of two or more factors when giving out a found
factor, if two of more factors are found?
Yes, if both factors are "smooth" enough, they could be found as their
product, rather than individually.
On 12 Jun 00, at 23:20, Eric Hahn wrote:
I was wondering if it was common practice (ie: the norm) for
P-1 to take the product of two or more factors when giving out
a found factor, if two of more factors are found?
I don't know about "common practice", but P-1 uses GCD to find the
factor.
From: Brian J. Beesley [mailto:[EMAIL PROTECTED]]
If P-1 does find a factor which is compound, then running P-1 again
with smaller limits will eventually recover a smaller factor. These
extra runs will obviously take less time than the original
Indeed, and with care one can usually choose
On 14 Jun 00, at 4:49, Paul Leyland wrote:
It's a pity that a similar procedure isn't known for ECM, or at least not
known to me.
Isn't the point that ECM finds numerically small factors much more
easily than it finds numerically large factors? Therefore, if a
"reasonable" amount of trial
From: Brian J. Beesley [mailto:[EMAIL PROTECTED]]
On 14 Jun 00, at 4:49, Paul Leyland wrote:
It's a pity that a similar procedure isn't known for ECM,
or at least not
known to me.
Isn't the point that ECM finds numerically small factors much more
easily than it finds numerically
Greetings all,
I was wondering if it was common practice (ie: the norm) for
P-1 to take the product of two or more factors when giving out
a found factor, if two of more factors are found?
To clarify, I was curious about how P-1 would indicate more
than one factor being found. So, I took
Eric Hahn wrote:
Greetings all,
I was wondering if it was common practice (ie: the norm) for
P-1 to take the product of two or more factors when giving out
a found factor, if two of more factors are found?
Yup, each factor f, for which b^E == 1 (mod f) (b is the base, usually
3, E is
