On Monday, November 11, 2013 11:03:00 PM UTC-5, Kermit Rose wrote: > On Monday, November 4, 2013 12:49:47 PM UTC-5, Michael F. Stemper wrote: > On 11/04/2013 11:35 AM, me wrote: > > > tell me what you think? > > > http://www.davesinvoice.com/papers/factorization2.pdf > > > > Interesting idea. How about using it to factor 130642890110987? >>> Factor(130642890110987) [58789, 2222233583, 134, 'Pollard Rho, x^2 + 1, First factor check'] Pollard Rho is a very efficient means to factor numbers of this size.
No, it is NOT. Please tell us how long the computation took you.
(1) It requires double length arithmetic. i.e. multi-precision arithmetic that is twice the length of the number being factored. (2) It is more efficient than trial division but still runs in exponential time. O(N^1/4) (3) Both SQUFOF and ECM are more efficient (i.e. faster on average) for numbers this size. (4) If you know in advance that it is the product of two large primes, MPQS will be even more efficient.
My NFS code factors number this size (and slightly larger/smaller) by the billions. I tried Pollard Rho in the past. It is slow. I first run ECM, then if it fails, MPQS.