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From: Mark Murray on 15 Jun 2010 13:43
On 15/06/2010 15:18, JSH wrote:
>> So I wasn't lying.
>>
>> Your own Java program is brute force, and its the only actual solution
>> you've offered.
>
> My test program uses a set of primes under 100 to factor T. That
> portion is then brute force.

At which point it cannot proceed.

> But so what?

Firstly, I wasn't lying, as you claimed. Secondly it does not, by any
stretch of the imagination represent a solution. It solves an extremely
limited subset of your equation, inefficiently.

As the relation is vacuously true, there wasn't even a point to doing
THAT.

>> That's my point. Your solution may "solve" the relationship in trivial
>> cases, but it is useless for real-world (such as cryptographic) cases.
>
> Does not follow.
>
> You're making a big deal about my use of "brute force" to factor T,
> when any idiot would know that if someone made a real world
> implementation they could replace that with any factoring method of
> their choice.

And as has been pointed out, this result is so vacuous, that a PRNG
or the digits of Pi can be used to beat a solution out of it.

> It's not the main method. It's a driver of one piece of that method.
>
> The idea is a way to find k, when k^m = q mod N, by factoring, which
> means, guess what?
>
> You need to factor something.
>
> How you do that factoring is your concern.

Back to the trivial relation.

Replacing one inefficient method with another is not significant.

M
--
Mark "No Nickname" Murray
Notable nebbish, extreme generalist.
From: rossum on 15 Jun 2010 18:47
On Tue, 15 Jun 2010 07:18:07 -0700 (PDT), JSH <jstevh(a)gmail.com>
wrote:

>The idea is a way to find k, when k^m = q mod N, by factoring, which
>means, guess what?
>
>You need to factor something.
>
>How you do that factoring is your concern.
Ermmm, James. Why did you call your previous thread "General
factoring result for k^m = q mod N" if this whole thing is nothing to
do with factoring?

rossum

From: JSH on 15 Jun 2010 21:30
On Jun 15, 3:47 pm, rossum <rossu...(a)coldmail.com> wrote:
> On Tue, 15 Jun 2010 07:18:07 -0700 (PDT), JSH <jst...(a)gmail.com>
> wrote:
>
> >The idea is a way to find k, when k^m = q mod N, by factoring, which
> >means, guess what?
>
> >You need to factor something.
>
> >How you do that factoring is your concern.
>
> Ermmm, James.  Why did you call your previous thread "General
> factoring result for k^m = q mod N" if this whole thing is nothing to
> do with factoring?
>
> rossum

Because it's a general method for solving for k, when k^m = q mod N,
where m is a natural number using factoring.


___JSH
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