Well I finally worked through a lot of the mathematics with my
surrogate factoring idea to the point of getting a good handle on
exactly what it is and how it works--it's a super sieve method.

To understand why it's a sieve, consider that I rely on finding
rational x, y and z, such that

yx^2 + Ax - j^2 = T

and

yz^2 + Ax - j^2 = 0

where A, j, and T are integers, and j^2 + T = M^2, which is the target
integer I'm trying to factor.

What I did was ask a basic question, what rational values for y can
fit?

If you know much mathematics you should know that either quadratic can
be solved by an infinite number of rational y's, and in fact, both can
also be solved by an infinite number, BUT my work shows a unique subset
of that infinity, which is where things get started.

All other rational y's that fit such that all the conditions are held
are just trivial multiples of the base set, and provably, if any x
exists such its numerator has only one prime factor of M, then that
subset of y's will give it.

However, it turns out that the probability as M gets larger that one of
the rationals x's will reveal a prime factor of M is roughly 50%, which
is a fascinating result that has to do with quadratic residues.

So why is it a super sieve, why is it a sieve at all and how does it
relate to quantum methods for factoring?

Well, by putting the two requirements that I do, I make a situation
where the mathematics loops through *combinations* of factors of j and
T, to try and find if ANY of those combinations can be matched with
integers from the full set of integers--yup, it tests out of infinity
checking against ALL integers--to see if x can have a non-trivial
factor of M.

Whether it can or not depends on quadratic residues.

Some of you may have heard of quantum methods for factoring where
somehow simultaneously a huge number of possible factors can be
checked, to factor a number, and that was actually demonstrated with a
factorization of 15.

If you know your math history, you may know that mechanical means of
factoring have been around for some time, but we don't rely on the
mechanical means as it was more important to abstract out the
underlying mathematical reasons.

I have abstracted out *how* quantum methods--basically mechanical
factoring--actually must work mathematically.

You see, there had to be a mathematical basis, which would lead to a
relatively simple way to factor.

Now to many of you I'm just some "crank" who is babbling, but that's
because you lack enough sophisticated mathematical knowledge to
understand how it works, or even to quite get what I'm saying.

That's ok. I can simply demonstrate.

Now it's basically an accident that this information is out on Usenet,
as while I do rely on Usenet to talk out ideas, when I had a
breakthrough with the factoring I went to mathematicians by email
first, and even sent a copy of my paper to Shor.

After I got little back from them I contacted the US Government,
multiple times, and verified receipt and was told my work would be
looked over, but basically I got no response, and clearly they didn't
take me seriously.

When I wrote my program, and was excited to see it factor, and then
disappointed to see it fail, I began to think maybe I'd just screwed
up, and hitting a wall, I put my work in the public domain.

Now there's no way to pull it back, so I'm pushing forward.

The original algorithm in my program, will, my current analysis shows,
factor about 50% of the time, which is astounding.

I get a sense that some of you don't get it, so let's say you take some
RSA challenge number, and calculate j and T, and factor them, and then
run them through the algorithm.

My research indicates you have a 50% chance of factoring the number.

That's NOW with what I've put out there.

Some of you may notice that people who reply to me, chatter a lot, say
a lot of negatives, but don't refute mathematically.

Think about it. If there are people who notice a new factoring
algorithm talked about, and get curious enough to check it out notice
that it factors 50% of the time, do you think they'll talk to any of
you?

I'm talking to you because there are multiple screw-ups and missed
catches at many levels, and it probably will go down in history as one
of the biggest boondoggles by intelligence services...ever.

Even if someone has already broken the RSA challenge numbers and sent
answer to RSA, what makes you think you'd know by now even if that
happened?

You people are out of the loops. As soon as I get a little further,
now that I see myself having broken through, I won't tell you a damn
thing either.

James Harris

From:	Xcott Craver
Subject:	Re: Basically a sieve method, relation to quantum
Date:	Sat, 22 Jan 2005 07:41:17 GMT

wrote:
>
>I get a sense that some of you don't get it, so let's say you take some
>RSA challenge number, and calculate j and T, and factor them, and then
>run them through the algorithm.
>
>My research indicates you have a 50% chance of factoring the number.

If this is true, then why not run your algorithm on ALL of the
challenge numbers, and factor about half of them?

If what you're saying is true, you could factor at least one
challenge number today.

>That's NOW with what I've put out there.

Then just post some big factors, all right?

--X

From:	oðin
Subject:	Re: Basically a sieve method, relation to quantum
Date:	Fri, 21 Jan 2005 15:40:24 -0800

You never explained why you think your foolish factoring algorithm executes
in polynomial time. You have shown some crap code that you say works half of
the time, and everyone else notes is horridly slow. We would be impressed,
in spite of all your embarrassments, if you could at least make an argument
that shows it is scales polynomial time, or at least sub-exponential time.
Are you going to try to do that? If not, nobody cares what you have to
say...

From:	Durk van Veen
Subject:	Re: Basically a sieve method, relation to quantum
Date:	Fri, 21 Jan 2005 20:32:31 -0800

jstevh@msn.com wrote:

>
> The original algorithm in my program, will, my current analysis shows,
> factor about 50% of the time, which is astounding.
>

Wouldn't trial and error division find the factors 100% of the time? Now it
might take a while (maybe longer than the universe takes to reach
heat-death) but the question is whether your algorithm would be faster or
not.

From:	jstevh at msn.com
Subject:	Re: Basically a sieve method, relation to quantum
Date:	22 Jan 2005 08:42:17 -0800

Durk van Veen wrote:
> jstevh@msn.com wrote:
>
>
> >
> > The original algorithm in my program, will, my current analysis
shows,
> > factor about 50% of the time, which is astounding.
> >
>
> Wouldn't trial and error division find the factors 100% of the time?
Now it
> might take a while (maybe longer than the universe takes to reach
> heat-death) but the question is whether your algorithm would be
faster or
> not.

My algorithm is in polynomaial time.

It is significantly faster than what you're suggesting.

I did one test calculation of the number of combinations and found that
a number made up by multiplying the first 1,000 primes together would
give only around 100 million combinations.

That is, with a number that huge, my algorithm would probably loop
through all the combinations on a decent pc in a couple of hours.

In comparison, it could factor an RSA type number, if the theory is
correct, and all the implementation details can be worked out, in
seconds.

Make no mistake, we can talk here as if it's nothing, but if the basic
theory is right, then someone else may already be successfully breaking
Internet encryption as we speak, with some fancy program or programs on
a distributed system that can factor VERY large numbers in a few
seconds, based off the math.

If my research holds up, and I think it will, then the basic algorithm
itself, which is already out there, will factor an RSA type number 50%
of the time, with maybe a few million combinations to iterate through
for any given j that you try.

That's why this work should be taken seriously.

You can talk about it, or downplay it or whatever, but if it's correct
and someone less silly than you people pays attention, they can start
walking through Internet security IMMEDIATELY as in right now, not
later, not in a few months, but RIGHT NOW.

James Harris

From:	Augustus SFX van Dusen
Subject:	Re: Basically a sieve method, relation to quantum
Date:	Fri, 21 Jan 2005 23:57:53 GMT

On Fri, 21 Jan 2005 15:10:41 -0800, jstevh wrote:

> You people are out of the loops. As soon as I get a little further,
> now that I see myself having broken through, I won't tell you a damn
> thing either.

Come, come! You are saying that just to make us happy.

From:	David C. Ullrich
Subject:	Re: Basically a sieve method, relation to quantum
Date:	Sat, 22 Jan 2005 08:11:04 -0600

On 21 Jan 2005 15:10:41 -0800, jstevh@msn.com wrote:

>Well I finally worked through a lot of the mathematics with my
>surrogate factoring idea to the point of getting a good handle on
>exactly what it is and how it works

Now wait a second. For the last few days (weeks?) you've been
abusing people who've been pointing out that it didn't work.
You've been making claims about how it was easy to prove that
it works - then there was this claim that it was easy to prove
that it worked 50% of the time, using quadratic residues...

How could you make all those claims _before_ you finally
thought out exactly how it worked?

Guffaw.

>[...]
>
>Now to many of you I'm just some "crank" who is babbling, but that's
>because you lack enough sophisticated mathematical knowledge to
>understand how it works, or even to quite get what I'm saying.

Huh. I would have thought it was because almost everything you
say turns out to be wrong.

What's truly remarkable is the fact that when someone says
you're wrong they're lying, but then when you finally realize
you _were_ wrong that doesn't seem to change the fact that
the people who said you were wrong were lying. _That's_ the
part about what you're saying that _I_ don't quite get...

>[...]
>You people are out of the loops. As soon as I get a little further,
>now that I see myself having broken through, I won't tell you a damn
>thing either.

Guffaw. When pigs fly.

>
>James Harris

************************

David C. Ullrich

From:	ScottD
Subject:	Re: Basically a sieve method, relation to quantum
Date:	Sat, 22 Jan 2005 06:06:15 GMT

wrote in message news:1106349041.630565.57710@f14g2000cwb.googlegroups.com...
> Well I finally
[snip]
> James Harris

Congrats on the progress. Keep at it.

From:	Gib Bogle
Subject:	Re: Basically a sieve method, relation to quantum
Date:	Sat, 22 Jan 2005 12:14:02 +1300

jstevh@msn.com wrote:
....
> You people are out of the loops. As soon as I get a little further,
> now that I see myself having broken through, I won't tell you a damn
> thing either.

Please tell us! Please!!

From:	jstevh at msn.com
Subject:	Re: Basically a sieve method, relation to quantum
Date:	22 Jan 2005 08:31:18 -0800

Xcott Craver wrote:
> wrote:
> >
> >I get a sense that some of you don't get it, so let's say you take
some
> >RSA challenge number, and calculate j and T, and factor them, and
then
> >run them through the algorithm.
> >
> >My research indicates you have a 50% chance of factoring the number.
>
> If this is true, then why not run your algorithm on ALL of the
> challenge numbers, and factor about half of them?
>
> If what you're saying is true, you could factor at least one
> challenge number today.
>

Conceivably, yes, if I used some other program to factor T, since my
current program just would in all likelihood fail to fully factor it,
as I call it recursively to do all the factorizations, breaking down a
number until it gets to something smaller than 200, at which point it
uses a table of primes!

Remember my method is *surrogate* factoring, which means you have to
factor something!!!

To factor an RSA challenge number I'd need a full factorization of some
number off of it. Now T = M^2 - j^2, where M is the target number, and
j is a number you get to pick.

My analysis is that if you do that with my method there is a
significant chance that you will factor M, if you use a full
decomposition of T, and all combinations of its factors.

Now I know that you people don't believe me, which is actually good, as
it just make it kind of funny, you know?

Like, if you believed me, maybe one of you might do it, but you won't
because you don't, until someone of greater intellectual ability comes
along, or I do it myself.

What I'm doing is getting a handle on the theory so that I don't waste
time searching for a solution, like having to experiment with different
j's to get one that works, without knowing exactly why it works.

Now someone else who only cares about getting the solution might not be
so particular.

I'm about finesse.

You people talk a lot so I'm not worried about you. I can freely talk
about this research without concern that any of you will do anything,
but talk.

I have years of experience dealing with you, from which I can tell what
you will and will not do.

James Harris

From:	Michael Brown
Subject:	Re: Basically a sieve method, relation to quantum
Date:	Sun, 23 Jan 2005 12:53:55 +1100

jstevh@msn.com wrote:
> Xcott Craver wrote:
>> wrote:
>>>
>>> I get a sense that some of you don't get it, so let's say you take
>>> some RSA challenge number, and calculate j and T, and factor them,
>>> and then run them through the algorithm.
>>>
>>> My research indicates you have a 50% chance of factoring the number.
>>
>> If this is true, then why not run your algorithm on ALL of the
>> challenge numbers, and factor about half of them?
>>
>> If what you're saying is true, you could factor at least one
>> challenge number today.
>>
>
> Conceivably, yes, if I used some other program to factor T, since my
> current program just would in all likelihood fail to fully factor it,
> as I call it recursively to do all the factorizations, breaking down a
> number until it gets to something smaller than 200, at which point it
> uses a table of primes!
>
> Remember my method is *surrogate* factoring, which means you have to
> factor something!!!

For what it's worth, I cooked up a couple of ~64-bit primes (was a 104-bit
product, I forgot to set the uppermost bits). They weren't specially chosen
or even well-generated. I just used the rand() function to generate a binary
string. Their product was 17417537469613251264524569748998981537. "factor"
from:
http://www.asahi-net.or.jp/~KC2H-MSM/cn/
factored it in 33.26 seconds (1128714656609730623 * 15431302648208293919).
After a bit of trial and error, it turned out that it could factor the
product squared minus 16 in 3.26 seconds:
303370611505381579716912725262028356146608169452120680027239449463266882353
= 3* 3 * 71 * 307 * 2673397 * 84828952219129 * 8533686513259849 *
799079573776815674841701598797953

How do you get the original primes from the above factorisation? Or does j
have to be specially chosen?

--
Michael Brown
www.emboss.co.nz : OOS/RSI software and more :)
Add michael@ to emboss.co.nz ---+--- My inbox is always open

From:	Michael Brown
Subject:	Re: Basically a sieve method, relation to quantum
Date:	Sun, 23 Jan 2005 15:04:18 +1100

Michael Brown wrote:
[...]
> How do you get the original primes from the above factorisation? Or
> does j have to be specially chosen?

From what I can gather, it's supposed to go something like:
b_i = factors of j^2 ( = 16 in this case)
f_i = factors of T (product squared minus j^2)

b_1 = product of some subset of b_i
b_2 = -j^2 / b1
f_1 = product of some subset of f_i
f_2 = T / f_1

A = some integer (not sure how to calculate this)

Then a possible factor is:
(b_1 f_2 + b_2 f_1 + 2 j^2) / A

Since I'm not sure how to calculate A, I just calculated each possibility
mod each of the original primes to see if it was zero. No such combination
occured (except for the trivial cases b_1 = f_1 = 1 or the prime to be
tested was zero).

Is my interpretation correct?

--
Michael Brown
www.emboss.co.nz : OOS/RSI software and more :)
Add michael@ to emboss.co.nz ---+--- My inbox is always open

From:	Tim Peters
Subject:	Re: Basically a sieve method, relation to quantum
Date:	Sat, 22 Jan 2005 19:25:31 -0500

[jstevh@msn.com]
....
> To factor an RSA challenge number I'd need a full factorization of some
> number off of it. Now T = M^2 - j^2, where M is the target number, and
> j is a number you get to pick.

That's where you often get off track, failing to engage your full abilities.
What you want instead is a method that will factor M given a factorization
of T = M+j, where j is a number you get to pick. For example, pick j=0, and
then your algorithm will run in linear time. Picking j=0 in T=M^2-j^2 is
good too, but then you're left with tedious work to weed out the repeated
factors. Set your goal higher.

From:	David C. Ullrich
Subject:	Re: Basically a sieve method, relation to quantum
Date:	Sun, 23 Jan 2005 08:03:03 -0600

On Sat, 22 Jan 2005 19:25:31 -0500, "Tim Peters"
wrote:

>[jstevh@msn.com]
>...
>> To factor an RSA challenge number I'd need a full factorization of some
>> number off of it. Now T = M^2 - j^2, where M is the target number, and
>> j is a number you get to pick.
>
>That's where you often get off track, failing to engage your full abilities.
>What you want instead is a method that will factor M given a factorization
>of T = M+j, where j is a number you get to pick. For example, pick j=0, and
>then your algorithm will run in linear time.

Damm. I was going to suggest deriving a factorization of M from a
factorization of 2*M, but your idea is much simpler and more elegant.

Oh well, not the first time something like this has happened - I guess
we math guys should leave the actual programming to the pros.

>Picking j=0 in T=M^2-j^2 is
>good too, but then you're left with tedious work to weed out the repeated
>factors. Set your goal higher.
>

************************

David C. Ullrich

From:	Tim Peters
Subject:	Re: Basically a sieve method, relation to quantum
Date:	Sun, 23 Jan 2005 23:49:58 -0500

[jstevh@msn.com]
>>> ....
>>> To factor an RSA challenge number I'd need a full factorization of some
>>> number off of it. Now T = M^2 - j^2, where M is the target number, and
>>> j is a number you get to pick.

[Tim Peters]
>> That's where you often get off track, failing to engage your full
>> abilities. What you want instead is a method that will factor M given a
>> factorization of T = M+j, where j is a number you get to pick. For
>> example, pick j=0, and then your algorithm will run in linear time.

[David C. Ullrich]
> Damm. I was going to suggest deriving a factorization of M from a
> factorization of 2*M, but your idea is much simpler and more elegant.

Indeed, and yours is just the special case of picking j=M (I know James
realizes that, I'm just pointing it out for your benefit). What he may not
have realized yet is that his method is just a special case of picking j =
M^2 - j^2 - M.

I wouldn't call it my idea, though -- it's almost certain I wouldn't have
thought of it without JSH pushing in this direction. Since I'm too stupid
to make it work anyway, James should get all the credit.

Right now he's got this niggling detail where, to factor M in polynomial
time, he at least has to assume that factoring integers with twice as many
digits is cheap. Some posters are making a big deal out of that, as if it
were a fundamental problem. Really! It would be easier to prove that M can
be factored in polynomial time if he got to assume instead that factoring
M+0 takes negligible time. Even the chowderheads on sci.math could
understand that proof. Heck, even I can write a linear-time Python program
for it:

def factor(M):
j = 0
factors = factorize(M+j) # returns list of factors in negligible time
print [hex(p) for p in factors]

The only subtle part is using hex() to display the factors (Python's hex(n)
takes time linear in the number of bits in n; converting to decimal instead
would inject a quadratic-time component).

> Oh well, not the first time something like this has happened - I guess
> we math guys should leave the actual programming to the pros.

Ya, you're way out of your depth here again. I am too. We should charge
admission for visiting this end of the pool.

From:	mensanator at aol.compost
Subject:	Re: Basically a sieve method, relation to quantum
Date:	23 Jan 2005 10:02:57 -0800

Mark Nudelman wrote:
> jstevh@msn.com wrote:
> > Mark Nudelman wrote:
> >> jstevh@msn.com wrote:
> >>> The original algorithm in my program, will, my current analysis
> >>> shows, factor about 50% of the time, which is astounding.
> >>
> >> I'm not sure why that's astounding. There are lots of algorithms
> >> that factor numbers 100% of the time.
> >>
> >
> > Yeah but my algorithm does it in polynomial time.
>
> I must have missed your proof that your algorithm works in polynomial
time.

How about failing in polynomial time? Does that count for anything?

> Obviously this is the crucial point. Could you repeat that argument,
or
> point me to a reference?
>
> --Mark

From:	Mark Nudelman
Subject:	Re: Basically a sieve method, relation to quantum
Date:	Fri, 21 Jan 2005 16:54:49 -0800

jstevh@msn.com wrote:
> The original algorithm in my program, will, my current analysis shows,
> factor about 50% of the time, which is astounding.

I'm not sure why that's astounding. There are lots of algorithms that
factor numbers 100% of the time.

> I get a sense that some of you don't get it, so let's say you take
> some RSA challenge number, and calculate j and T, and factor them,
> and then run them through the algorithm.
>
> My research indicates you have a 50% chance of factoring the number.

Great, so try it on 7 RSA challenge numbers. You should have a probability
of 127/128 of factoring at least one of them.

But this is useful only if your algorithm can factor in polynomial time.

--Mark

From:	David Kastrup
Subject:	Re: Basically a sieve method, relation to quantum
Date:	Sat, 22 Jan 2005 17:46:35 +0100

jstevh@msn.com writes:

> Mark Nudelman wrote:
>> jstevh@msn.com wrote:
>> > The original algorithm in my program, will, my current analysis
>> > shows, factor about 50% of the time, which is astounding.
>> I'm not sure why that's astounding. There are lots of algorithms
>> that factor numbers 100% of the time.
>>
>
> Yeah but my algorithm does it in polynomial time.

Polynomial in what? You don't even know what that means. You are
just spurting buzzwords that you feel sound as impressive as what you
think you are doing.

--
David Kastrup, Kriemhildstr. 15, 44793 Bochum

From:	David Kastrup
Subject:	Re: Basically a sieve method, relation to quantum
Date:	Sun, 23 Jan 2005 20:35:54 +0100

"Nora Baron" writes:

> jstevh@msn.com wrote:
>> Mark Nudelman wrote:
>> > jstevh@msn.com wrote:
>> > > The original algorithm in my program, will, my current analysis
>> > > shows, factor about 50% of the time, which is astounding.
>> > I'm not sure why that's astounding. There are lots of algorithms
>> > that factor numbers 100% of the time.
>> >
>> Yeah but my algorithm does it in polynomial time.
>>
>
> What degree polynomial?

At least a BA, I reckon.

--
David Kastrup, Kriemhildstr. 15, 44793 Bochum

From:	jstevh at msn.com
Subject:	Re: Basically a sieve method, relation to quantum
Date:	22 Jan 2005 08:36:20 -0800

Mark Nudelman wrote:
> jstevh@msn.com wrote:
> > The original algorithm in my program, will, my current analysis
shows,
> > factor about 50% of the time, which is astounding.
>
> I'm not sure why that's astounding. There are lots of algorithms
that
> factor numbers 100% of the time.
>

Yeah but my algorithm does it in polynomial time.

James Harris

From:	Richard Henry
Subject:	Re: Basically a sieve method, relation to quantum
Date:	Sat, 22 Jan 2005 12:45:21 -0800

wrote in message
news:1106411780.564349.169950@c13g2000cwb.googlegroups.com...
> Mark Nudelman wrote:
> > jstevh@msn.com wrote:
> > > The original algorithm in my program, will, my current analysis
> shows,
> > > factor about 50% of the time, which is astounding.
> >
> > I'm not sure why that's astounding. There are lots of algorithms
> that
> > factor numbers 100% of the time.
> >
>
> Yeah but my algorithm does it in polynomial time.

Have you proven that point?

From:	Gib Bogle
Subject:	Re: Basically a sieve method, relation to quantum
Date:	Sun, 23 Jan 2005 10:16:11 +1300

Richard Henry wrote:

> wrote in message
> news:1106411780.564349.169950@c13g2000cwb.googlegroups.com...
>
>>Mark Nudelman wrote:
>>
>>>jstevh@msn.com wrote:
>>>
>>>>The original algorithm in my program, will, my current analysis
>>
>>shows,
>>
>>>>factor about 50% of the time, which is astounding.
>>>
>>>I'm not sure why that's astounding. There are lots of algorithms
>>
>>that
>>
>>>factor numbers 100% of the time.
>>>
>>
>>Yeah but my algorithm does it in polynomial time.
>
>
> Have you proven that point?

The concept of "proof" is a foreign one to JSH.

From:	Tim Smith
Subject:	Re: Basically a sieve method, relation to quantum
Date:	Sat, 22 Jan 2005 21:11:11 GMT

In article , Richard Henry wrote:
>> Yeah but my algorithm does it in polynomial time.
>
> Have you proven that point?

He has, but it is quantum polynomial time, so the proof goes away if you
look for it.

--
--Tim Smith

From:	Mark Nudelman
Subject:	Re: Basically a sieve method, relation to quantum
Date:	Sun, 23 Jan 2005 09:42:55 -0800

jstevh@msn.com wrote:
> Mark Nudelman wrote:
>> jstevh@msn.com wrote:
>>> The original algorithm in my program, will, my current analysis
>>> shows, factor about 50% of the time, which is astounding.
>>
>> I'm not sure why that's astounding. There are lots of algorithms
>> that factor numbers 100% of the time.
>>
>
> Yeah but my algorithm does it in polynomial time.

I must have missed your proof that your algorithm works in polynomial time.
Obviously this is the crucial point. Could you repeat that argument, or
point me to a reference?

--Mark

From:	Xcott Craver
Subject:	Re: Basically a sieve method, relation to quantum
Date:	Sat, 22 Jan 2005 22:24:14 GMT

wrote:
>Mark Nudelman wrote:
>>
>> I'm not sure why that's astounding. There are lots of algorithms
>> that factor numbers 100% of the time.
>
>Yeah but my algorithm does it in polynomial time.

Then, factor an RSA challenge.

If it works 50% of the time, try it on a bunch of challenge
numbers, and give us the factors of at least one.

If you had such an algorithm, it would be incredibly easy to
prove it by doing the above.

>James Harris
--Xcott

From:	Nora Baron
Subject:	Re: Basically a sieve method, relation to quantum
Date:	23 Jan 2005 11:27:11 -0800

jstevh@msn.com wrote:
> Mark Nudelman wrote:
> > jstevh@msn.com wrote:
> > > The original algorithm in my program, will, my current analysis
> shows,
> > > factor about 50% of the time, which is astounding.
> >
> > I'm not sure why that's astounding. There are lots of algorithms
> that
> > factor numbers 100% of the time.
> >
>
> Yeah but my algorithm does it in polynomial time.
>

What degree polynomial?

Nora B.

>
> James Harris

From:	C. Bond
Subject:	Re: Basically a sieve method, relation to quantum
Date:	Sat, 22 Jan 2005 17:31:01 GMT

jstevh@msn.com wrote:

> Xcott Craver wrote:
> > wrote:
> > >
> > >I get a sense that some of you don't get it, so let's say you take
> some
> > >RSA challenge number, and calculate j and T, and factor them, and
> then
> > >run them through the algorithm.
> > >
> > >My research indicates you have a 50% chance of factoring the number.
> >
> > If this is true, then why not run your algorithm on ALL of the
> > challenge numbers, and factor about half of them?
> >
> > If what you're saying is true, you could factor at least one
> > challenge number today.
> >
>
> Conceivably, yes, ...

[snip]

> I have years of experience dealing with you, from which I can tell what
> you will and will not do.

And your posting history reveals "what you will and will not do." Namely,
you *will* launch a self-aggrandizing, irrelevant diatribe and you *will
not* meet the challenge the previous poster laid down.

--
There are two things you must never attempt to prove: the unprovable -- and
the obvious.
--
Democracy: The triumph of popularity over principle.
--
http://www.crbond.com

From:	Chip Eastham
Subject:	Re: Basically a sieve method, relation to quantum
Date:	21 Jan 2005 16:44:08 -0800

o=F0in wrote:

> We would be impressed, in spite of all your embarrassments,
> if you could at least make an argument that shows it is scales
> polynomial time, or at least sub-exponential time.
> Are you going to try to do that? If not, nobody cares what you
> have to say...

I'd be impressed as hell if he can factor something like 50% of
the open RSA challenge problems.

Then again maybe I'm more easily impressed than most, having
been among the 25% of the world's pure mathematicians that he
had fired from their jobs. :wink:

regards, chip