Re: Easy test of surrogate factoring
From: Matt Gutting (tchrmatt_at_yahoo.com)
Date: 02/18/05
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Date: Fri, 18 Feb 2005 09:40:03 -0500
jstevh@msn.com wrote:
> Paul Rubin wrote:
>
>>Here's an even easier test of surrogate factoring: I gave you a list
>>of fifty 20-digit composite numbers a couple weeks ago and you
>
> haven't
>
>>factored a single one. I think I know what test score to assign,
>>based on that result.
>
>
> No, it's not a fair test. I'll explain why, again, and I'll also
> explain again, what the test is, and why my test is an easy one.
>
[snip]
> If it's not random, then some way exists to figure out how to get all
> the integer Az's without knowing M.
>
> Those are the two possibilities.
>
That's true, James. And if you look and listen, I believe you'll find that
everyone here agrees that the second possibility is the case.
> Now posters working hard to discount this research--apparently deciding
> that it's Harris mathematics or JSH algebra--are avoiding the actual
> issue from what I've seen in looking at posts in this thread.
>
> The reason is simple, the mathematics doesn't work for them either way,
> if their intent is to discount it.
>
> That's what happens with major math results, human beings working hard
> to dismiss them have to ignore facts.
>
> Now my test is simple: calculate Ax, by taking an M with known factors,
> and *look* at the prime factors of the denominator of Ax, and you will
> find that they are the same prime factors as T has.
>
> That proves that there are rules governing the value of the rationals
> Ax's that results from the integer Az's.
>
> And then, algorithms *can* be developed.
>
> Now, if mathematicians were what they claimed then it wouldn't be
> required that I actually produce a working program that can factor RSA
> numbers, as that's an unreasonable request. It's an unresaonable
> request as if I were at that point I wouldn't need to convince anyone,
> as I could just demonstrate.
>
> Worse, my ability or lack of ability to write such a program does not
> disprove the mathematics!
>
> And in this case that means someone else might, or may already have.
>
> But mathematicians are NOT what they claim, and in this case they are
> setting other people up to be responsible for their failures, if things
> go badly.
>
> Now I'll work at building that program to factor an RSA number, but
> while time passes, it's you who may suffer the consequences, in a world
> where hackers may now have the upper hand.
Heck, James, *I* can write a program to factor an RSA number. How about this?
int M;
M = //insert the RSA number here
for (int i=2; i<=sqrt(M);i++) {
if (M % i = 0) {
cout << i << " is a factor of " << M;
}
}
This will produce *every* factor of M less than sqrt(M) (and the only reason
it doesn't stop after just one factor is because I don't know enough C++).
Of course, it will take quite a while. This is precisely what others have been
saying: not that your math can't produce a workable algorithm, but that it's
(at the very best) no more efficient than the least efficient algorithms
currently existing. I think it's great that you've (apparently) arrived at
a new algorithm, but it's not important if it's not an improvement over
what's already available.
Matt
(excess newsgroup stripped)
>
> And mathematicians are de facto protecting them.
>
>
> James Harris
>
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