
Re: JSH: Upside down situation
Posted:
Mar 15, 2008 6:35 PM


JSH wrote: > On Mar 15, 5:49 am, Vend<ven...@virgilio.it> wrote: >> On 15 Mar, 02:01, JSH<jst...@gmail.com> wrote: >> <snip> >> >>> Oh, so why might you get "misfactors" searching k modulo p? >> What do you mean by "searching k modulo p"? >> >> <snip> > > Given z^2 = y^2 + nT, where T is a target composite to factor, I've > shown that > > z = (1 + 2a^2)k/(2a) > > where > > k^2 = (1 + a^2)^{1}(nT) mod p > > so you find k by finding 'a' such that k exists for a given odd prime > p, but there is a rule on p that it be less than the smaller factor of > nTlooking only at positive factorsor that p minus the smaller > factor be less than that factor. > > So, say you pick a prime p approximately sqrt(nT) hoping that your > smaller factor isn't too small, and then you try some 'a', like a=1, > to see if k exists, if it doesn't you use another prime. If it does, > then you have k modulo p. > > Further research shows that k is near a value I call k_0 which is > given by finding k_0 such that > > abs(nT  (1+a^2)k_^2) > > is a MINIMUM, so you have to find an integral quadratic minima. > > Full surrogate factoring theory then says that k_0 is within k_0/2p > steps from a k that will factor your target i.e. give you z = (1 + > 2a^2)k/(2a) such that > > z^2 = y^2 + nT. > > Since p increases in size as T increases in size, it is crucial to > search for k modulo p, as otherwise you have to get really lucky with > your k_0, and in fact, searching by just incrementing k by 1 means you > have approximately k steps, versus k/2p steps. A big difference. > > To figure out how big, just run some examples, where you have a > factorization of T so you have z, and can calculate k, and then see > how many steps you'd have versus using a prime p approximately equal > to your smaller prime factor so that you can get k/2p. > > Oh, so a big issue is guessing at how small p has to be versus your > wish to make p as large as possible, which is probably where most of > the problems with implementing this practically step in, as otherwise, > on paper, it is a solution to the factoring problem. > > There is a preference for z divisible by 3, so n can be used to force > nT mod 3 = 2. > > The subject of this thread is a gesture of my frustration at being > able to find cutting edge mathematics that is just ignored by the bulk > of math society which cheers its own, however. > > Outsiders are just cutoff with no options but forcing the issue as > I'm attempting to do with the factoring problem, as mathematicians > ONLY will support each other, without regard to the research, which I > call an upside down situation where they fight for each other and give > "gold medals" to their colleagues but will simply ignore wins by
``Upside Down" video, Diana Ross:
http://www.youtube.com/watch?v=ovcjA18JF6s
(platinum eh?)
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