Re: Surrogate factoring approach, analysis
jstevh_at_msn.com
Date: 01/01/05
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Date: 1 Jan 2005 08:16:18 -0800
B Loggins wrote:
> James,
>
> Some questions, and forgive me if I'm just failing to understand you
-
> I am not a mathematician. It seems that, for this approach to even
get
> off the ground, the number x must be "of the form" that you've given
> (we'll call it "Harris-Factorable"). Are all (interestingly
> factorable) numbers Harris Factorable? Is RSA-2048? If what I say
is
> true, then given a number x, x must be shown to be H-F before this
> method can be used. Hence, we desire an H-Fness test. Is H-Fness
> decidable? Is it NP-Complete? If not, would it take so long to
> determine whether x is H-F that something like the Number Field Sieve
> would give faster results?
With x = n/m where n and m are coprime integers, n must be a factor of
your target number, but it could be a trivial factor.
So, say for RSA-2048, the method I have *will* give you some x, with an
n that is a factor of RSA-2048, but n may just be RSA-2048.
If the theory is correct, then it will work on any non-zero positive
integer, so there wouldn't be "Harris-Factorable" numbers, at all.
If the theory is correct, it would be a complete solution to the
factoring problem, and even a way to determine primeness.
>
> Second, you state that x is "dependent" upon the above pattern. This
> simply means (to me) that x can be put in terms of that pattern. But
> note that the same argument can be given for the original factors.
> Consider a large number x such that x is the product of two large
> primes, p1 and p2, then x = p1*p2 and thus x could be said to "be
> dependent" on the above factors. But this doesn't help to solve the
> problems. How does the H-F pattern help to find primes more quickly?
> Thanks
My method finds a factor of the target number dependent on the factors
of different numbers I call surrogates, as explained in my original
post.
If the set of solutions for x is complete in that you get every integer
factor of your target number, then the factoring problem is solved, and
solutions could be found quite rapidly.
The reason is that factoring some *other* number in order to factor a
number made up of only two primes could be done MUCH faster than
factoring of the original number, and then it's just a matter of
iterating through combinations.
If my approach works, it'd make it possible to factor an RSA number as
fast as it is generated, completely ruining that approach, as my work,
if correct, proves that factoring is NOT a hard problem.
I think it is correct, and that some people simply decided it was hard
because no one they knew of had given a solution and they couldn't find
a solution, which doesn't mean that my finding is necessarily new.
For all we know, expecially since it's so easy, some people might have
figured this out long ago.
James Harris
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