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Re: JSH: Upside down situation
Posted:
Mar 15, 2008 6:35 PM
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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
> nT--looking only at positive factors--or 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 cut-off 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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