On Mar 31, 2:49=C2=A0pm,

On 31 Mar, 19:03, "Douglas Eagleson" <eaglesondoug...@yahoo.com>
wrote:

On Mar 31, 12:57 pm,

On 31 Mar, 17:23, "Douglas Eagleson" <eaglesondoug...@yahoo.com>
wrote:

On Mar 31, 11:20 am,

On 31 Mar, 17:01, "Douglas Eagleson" <eaglesondoug...@yahoo.com>
wrote:

On Mar 31, 10:58 am,

On 31 Mar, 16:10, "Douglas Eagleson" <eaglesondoug...@yahoo=

wrote:

On Mar 29, 4:52 pm, "Douglas Eagleson" <eaglesondoug...@y=

wrote:

On Mar 28, 9:36 pm, "Douglas Eagleson" <eaglesondoug...=

wrote:

On Mar 28, 9:35 pm, "Douglas Eagleson" <eaglesondoug.=

wrote:

A for as the logcial loopie.

I was trying to find a new programming language and=

language called ?

If you are familiar with it you know of the questio=

terminating loops based on the act of the called fi=

A simple fifth order language was developed to allo=

NO other language can express this mathematical ser=

Be careful out there.

Note: =C2=A0A field loop controls!

When a prime is reached a loop exit can occur!

Fast executioner =C2=A0programs.

I forgot to clarify the definition of field.

A cause as a set.

All function field as assignment to the domain defines.

A field as category in domain!- Hide quoted text -

- Show quoted text -

Here is the crack for the old obsolete RSA using a field =

statement:

For(a){

A =C2=A0 =C2=A0//as a product

}

Effil is supposed to answer RSA like this. =C2=A0Little a=

closes a loop. =C2=A0And if "A" is a prime product, well =

also.

Please try it out. "a" as the first solution stops the lo=

single slight control flipper;

A- must be defined abstract number. =C2=A0SO declare the =

of "a"

And the logic controls as all field!- D=C3=B6lj citerad t=

- Visa citerad text -

But if it did not halt you did not find the answer to the R=

challenge, or did you?

JT- Hide quoted text -

- Show quoted text -

I am trying to get Eiffel now. =C2=A0I got upset at the Java =

recently and tried to study the various alternatives.

Eiffel as detailed appears to allow this crack. =C2=A0I have =

Eiffel yet. As a matter of fact I cannot afford it.

So whoever has it needs to try it. I have therefor not tried =

- Visa citerad text -

Visual Eiffel

The Open Source Version is free. Details are described in our S=

License , here are the essentials: You can use Visual Eiffel fo=

as long you either

use the product only for educational or purely private, non com=

purposes and/or
publish the sources of any created products that you distribute=

an GPL (Open Source License)

JT- Hide quoted text -

- Show quoted text -

I will download it and give it a whirl. =C2=A0Thanks.- D=C3=B6lj =

- Visa citerad text

You are not Douglas, who gave you eiffel?
I have a feeling you already have eiffel.

JT- Hide quoted text -

- Show quoted text -

No I just downloaded it. =C2=A0The last time i saw it I had no time..=

file was 30 megbyte. =C2=A0Today I let it download.- D=C3=B6lj citera=

- Visa citerad text -

How long would it take to factor RSA 704 with your method, using your
current CPU?

Here you have it
74037563479561712828046796097429573142593188889231
28908493623263897276503402826627689199641962511784
39958943305021275853701189680982867331732731089309
00552505116877063299072396380786710086096962537934
650563796359

Did you read James Harris latest surrogate factoring method how does
it relate to your idea?

Quote: James
Harris.................................................................=

Abstracting to the gist of surrogate factoring I realized it says that
for every factorization of a composite T, with difference of squares

x2 =E2=89=A1 y2 mod T
which of course is the same as, (x-y)(x+y) =E2=89=A1 0 mod T, and is fo=

reason a factorization, there is an alternate factorization of an
integer S, where

S =E2=89=A1 (=CE=B1 + 1)k2 mod T

and alpha and k are defined by

=CE=B1k =E2=89=A1 2x mod T

as then trivially you can multiply both sides of that relation by k,
add back to the original difference of squares, add k2 to both sides
and find

(x+k)2 =E2=89=A1 y2 + (=CE=B1 + 1)k2 mod T

explaining the second factorization, as now you have a factorization
of this alternate number using the same x, y and this
=C2=A0extra variable k:

(x+k - y)(x+k+y) =E2=89=A1 (=CE=B1 + 1)k2 mod T

So for EVERY composite factorization of T using difference of square
that algebra tosses a factorization of ANOTHER composite (=CE=B1 + 1)k2=

T back at you!!!

It's just been there all the time. There has just been another
factorization out there all along with difference of squares and I
call the second factorization the surrogate factorization.

But hey, that goes both ways! I like to use T as the target, and S as
the surrogate, but what if you reverse the equations so that you pick
some surrogate and factor it to see how easily you can factor the
target T?

Equations then become

x2 =E2=89=A1 y2 mod S and =CE=B1k =3D 2x mod S gives you

(x+k)2 =E2=89=A1 y2 + (=CE=B1 + 1)k2 mod S

and now, of course, you'd want to choose
(=CE=B1 + 1)k2 mod S =E2=89=A1 0 mod T

and you can go explicit to start solving everything, so I have

x2 =E2=89=A1 y2 + aS

=CE=B1k =3D 2x + u1S, so =CE=B1 =3D (2x + u1S)/k, and ((2x + u1S)/k + 1=

+u1k)S =E2=89=A1 0 mod T, so

2xk + 2u1Sk + k2 + aS =E2=89=A1 0 mod T

so you can collect to get
k2 + 2(x + u1S)k + u2S =E2=89=A1 0 mod T and now complete the square to=

(k + x + u1S)2 =E2=89=A1 (x + u1S)2 + aS mod T

and now you have a fairly trivial way to solve for k, as u1 =C2=A0and aS
are integers of your choice, so you can, intriguingly enough, simply
CHOOSE two quadratic residues of T, and get their difference to
determine aS, and in that way determine x and y. =C2=A0Then it's just a
matter of selecting u1 such that you get the first residue.

Let r1 and r2 be residues modulo T, and further let
aS =3D r22 - r12 so once you pick residues you can determine aS and then
find x and y, and then let

x + u1S =E2=89=A1 r1 mod T

And you are free to pick u1 so now you can just solve for

u1 =E2=89=A1 (r1 - x)(S)-1 mod T

and finally you can get k with
=C2=A0k + x + u1S =E2=89=A1 r2 mod T

What makes this reversal fascinating to me is that it turns the
factoring problem on its head, where instead of searching for
quadratic residues modulo T, like most major factoring methods work to
do to solve the congruence of squares, you just pick pick them. You
pick quadratic residues of your target T, and use a surrogate
factorization to get a difference of squares with your target, so then
you should have a 50% chance of non-trivially factoring your target
with that choice,

(x+k)2 =E2=89=A1 y2 + (=CE=B1 + 1)k2 mod S
as then also you will have an alternate factorization, the surrogate
factorization

(x+k)2 =E2=89=A1 y2 mod T

as shown above with trivial algebra.

Quote: end James Harris

So will you implement your factorization method and James in Eiffel
Douglas?
How long would it take solve RSA 704 with your and James method
JT ;D- Hide quoted text -

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I will try the eiffel crack method. Harris has changed the function
that causes a field. So the easy way was the triangular relation. =C2=A0A=

the expoentential is slower.

I will try. =C2=A0It will take a few days. =C2=A0Hopefully Eiffel can do =

numbers.- D=C3=B6lj citerad text -

- Visa citerad text -