Re: JSH: Hammer falls, Pell's Equation used to solve factoring problem

On Feb 19, 9:25 am, rdec...@xxxxxxxxxxxx wrote:
On Feb 19, 10:59 am, rossum <rossu...@xxxxxxxxxxxx> wrote:





On Thu, 19 Feb 2009 07:13:45 -0800 (PST), JSH <jst...@xxxxxxxxx>
wrote:

You don't guess v.  Now if you had read my original post you'd know
that you don't guess v--you calculate v using calculus.

So, if you are correct, one of the calculus minima solutions will give
you a factor as you describe, so?

Mathematics is not subtle.

What is happening with some of you is that you are betraying your
human frailty.

You think the factoring problem is this huge big deal, but the math
does not care, so you ignore the details of the solution to hold on to
your belief.

It's a calculus problem.  You no more pick v than you pick x in the
problem: find the minimum of f(x), where

f(x) = x^2 + 3x + 2

as then f'(x) = 2x + 3, so x = -3/2 at the minimum.

At a point where f'(x) = 0, which may or may not be a minimum; in
general it could also be a maximum or a point of inflection.

You have to show that the value of v you derive from your calculus is
actually a minimum, rather than a maximum or PoI, and that it results
in an integer factor, rather than a rational factor.  Rational factors
are of no use in solving the factorisation problem.

How many minima are there in your formula, how many maxima and how
many points of inflection?  If the proportion of minima is too low
then you will need further work to pick out the minima _quickly_.

How many of the minima result in an integer factor as opposed to a
rational factor?  Again if the proportion is too low you are going to
have to do further work to pick out the integer factors _quickly_.

Did you PICK anything there?  No.

Yes.  You picked a minimum leading to an integer factor.  A curve may
have minima, maxima and points of inflection.  The minima may give
rise to integer factors or not.  Both of those need to be dealt with
explicitly if you are going to avoid making choices.

You need to _show_ us that your method is fast.  Tell us:

  1 what proportion of the points with f'(x) = 0 are minima.

  2 what proportion of the minima give an integer factorisation.

rossum

This talk about minima is a red herring. If you simplify
James' algorithm, as I did in an earlier post, you find
that |r + t|, |r - t| are 2(m - (D - 1))^2 and
2Dm^2, for suitable rational m. It is impossible
to make these simultaneously less than D, and
even restricting the first to be less than D
provides you with no additional information.

For example, with D = 1111 we find that making

2(m - (D - 1))^2 < D

will restrict our choices to m \in [1087, 1133].
If we pick m = 1087 we find

r + t = -2 * 23^2
r - t = -2 * 1087^2 * 1111

and neither gives any useful factors for 1111.
Using t = 1088 gives us

r + t = -2^3 * 11^2
r - t = -2^13 * 17^2 * 1111

and in this case we actually do find a factor
of 1111. This, though, is just what we'd
expect. Since 11 is a factor of 1111, we'd
get the same thing by just trying random
divisors.

As I've been saying for over a week.

Regards,

Rick- Hide quoted text -

- Show quoted text -

===================================================

So, ... no Hammer?

Gosh.


Enrico
.

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