On 11/12/2013 08:35 AM, scattered wrote: > On Tuesday, November 12, 2013 7:54:53 AM UTC-5, Pubkeybreaker wrote: >> 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. >> >> >> >> >> >> >> >>>> It factored 130642890110987 almost instantly >> >> >> >> Specify: "almost instantly". Exactly how long? > > You seem to need some valium or something. To say that a method is efficient isn't to say that it is the most efficient method in existence. I wrote an inefficient Python implementation of Pollard's rho when I first read about it a few years ago (idle curiosity on my part, this isn't my area of expertise). I just tried it on 130642890110987 and the factorization appeared on my screen even before my pinky left the enter key (even though Python is a "slow" interpreted language). That is "almost instantly" in any reasonable interpretation of that phrase. There is no reason to adopt a scolding tone againt somebody who make the true albeit unnuanced claim that Pollard's rho can efficiently factor numbers of that size. Perhaps your eagerness to slam other posters is why you took seriously an utterly transparent joke the other day. >
ef·fi·cient adjective \i-?fi-sh?nt\
: capable of producing desired results without wasting materials, time, or energy
Is Pollard rho "capable of producing desired results without wasting materials, time, or energy" ?