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Patrick Hamlyn science forum beginner
Joined: 03 May 2005
Posts: 45

Posted: Sun Jul 16, 2006 3:23 am Post subject:
Re: Loose connectivity, factoring and residues



jstevh@msn.com wrote:
Quote:  Next question then is obvious, how do you make the odds high that your
T is the one that is chosen?

This is the equivalent of asking:
How do you factor large numbers easily?
And you're no closer to answering that than before you 'discovered' your
magnificent 'so obvious nobody else saw it before' method. 

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jstevh@msn.com science forum Guru
Joined: 21 Jan 2006
Posts: 951

Posted: Sun Jul 16, 2006 4:13 am Post subject:
Re: Loose connectivity, factoring and residues



Patrick Hamlyn wrote:
Quote:  jstevh@msn.com wrote:
Next question then is obvious, how do you make the odds high that your
T is the one that is chosen?
This is the equivalent of asking:
How do you factor large numbers easily?
And you're no closer to answering that than before you 'discovered' your
magnificent 'so obvious nobody else saw it before' method.

Doesn't sound like a mathematical statement to me.
It sounds like a political one.
Remember, part of my point is that yours is a political society, which
casually uses social forces rather than mathematical tools.
This time period while people attack this method is meant to show just
how dangerous politics can be.
In the big world you can have wars, like the U.S. war with Iraq, and in
the academic world, you can have wars as well, like the political
battles against my ideas.
Just like the U.S. was wrong with something so huge that thousands of
lies have been lost, mathematicians can be wrong with their politics
here.
Difference is, billions of dollars may be lost before this war is over,
and then some of you will begin to comprehend the costs of playing
politics.
I can win one way, or another. You people let the gap continue, and I
gain the momentum I need to quite simply, end the jobs of most
mathematicians worldwide.
I want to clear entire math departments at major universities, and so
you have some context, a university on my list is Princeton.
Before this is over, the Princeton math department will no longer
exist.
I need a lot of momentum to do that, and that takes time and your
refusal to acknowledge important mathematics in an area important to a
lot of people around the world.
James Harris 

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Proginoskes science forum Guru
Joined: 29 Apr 2005
Posts: 2593

Posted: Sun Jul 16, 2006 4:25 am Post subject:
Re: Loose connectivity, factoring and residues



jstevh@msn.com wrote:
Quote:  Patrick Hamlyn wrote:
jstevh@msn.com wrote:
Next question then is obvious, how do you make the odds high that your
T is the one that is chosen?
This is the equivalent of asking:
How do you factor large numbers easily?
And you're no closer to answering that than before you 'discovered' your
magnificent 'so obvious nobody else saw it before' method.
Doesn't sound like a mathematical statement to me.
It sounds like a political one.

Then you need to clean out your ears. It is _purely_ a mathematical
statement.
 Christopher Heckman 

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Proginoskes science forum Guru
Joined: 29 Apr 2005
Posts: 2593

Posted: Sun Jul 16, 2006 4:28 am Post subject:
Re: Loose connectivity, factoring and residues



Tim Peters wrote:
Quote:  [...]
The _hard_ problem, and the one that's useful for factoring, is finding
distinct x and y such that:
x^2 = y^2 (modulo T) [2]
when T is given and /fixed/. [...]

Actually, it isn't too hard. Choose x and A to be arbitrary integers,
and let
y = A T  x.
Then y^2 = (A T  x) = A T^2  2 A T x + x^2 = x^2 (mod T).
 Christopher Heckman 

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jstevh@msn.com science forum Guru
Joined: 21 Jan 2006
Posts: 951

Posted: Sun Jul 16, 2006 4:29 am Post subject:
Re: Loose connectivity, factoring and residues



Tim Peters wrote:
Quote:  [jstevh@msn.com]
One thing that has come up in this thread in my discussion with Tim
Peters is that the mathematics will solve for any T that has the k and
S given, so you have the possibility that when you solve out for x and
y, you find that x^2  y^2 is coprime to YOUR target T, because the
mathematics has solved for another that fits with those same equations
with a different x_res..
A more pertinent explanation is that you _rely_ on:
x = x_res (modulo T) [1]
holding in order to force the critical:
x^2 = y^2 (modulo T) [2]
But, in fact, the congruence class of the x computed by the method has
nothing nonaccidental to do with the x_res you pick. As a result, as the
smallest prime factor of T gets larger, it becomes increasingly unlikely
that you'll manage to find an x_res, S, f_1 and f_2 such that [1] is
actually true, and then [2] is very unlikely to be true too. Then you may
as well pick x and y at random and rely on pure luck (well, that's not quite
accurate: you'd actually be better off picking them at random then).

There are two congruences:
x^2  y^2 = 0 mod T
and
S  2*x_res*k = 0 mod T
and both have to be satisfied, where the method I've outlined involves
factoring
S + k^2
where there is no information about x_res and T, so the mathematics is
free to use any that fit with those two congruences.
That is mathematically significant.
Quote:  However that seems to occur mostly with relatively large y, so
minimizing y, is a practical result of that analysis, which is why
theory does not necessarily mean you immediately have a practical
solution.
It's trivial to find T such that no choice of f_1 and f_2 leads to a
nontrivial factor of T. That becomes increasingly likely as the smallest
prime divisor of T gets larger. All you have to do to learn this is
implement it.
For example, pick T = 79 * 83, which is still tiny. Then at x_res = S = 1,
no factorization of S+k^2 as the product of two integers yields a factor of
T. At x_res=1 S=2, the same. At x_res=1 S=3, also the same.
I got bored then, and figured out a way that works but knowing the
factorization in advance: at x_res=2 S=3, 2 of the 16 possible ways of
factoring S+k^2 = 2689603 as a product of 2 integers lead to x and y that
find T's factors. The only ways that work then are f_1=1561 f_2=1723 and
f_1=1723 f_2=1561.

You are not as smart as you think you are.
I know the reason for your problems and I think I know the resolution.
You're not thinking about
x^2  y^2 = k*T
as the general solution to the congruence, where because of the way I
solve the problem it's also true that
S  2*x*k = k*T
so there must be a size limitation below which S can't fall or you just
can't solve the thing with a natural number k.
The research question is, what is the minimum size for S given the
absolute size of T?
Just guessing I'd say greater than sqrt(T)?
It probably can be less, but it wouldn't help for it to be more.
Quote:  This is most like last year's methods, where you need to introduce search
after search to have a chance of factoring a nontrivial T.

Ah, so you tried to test it to the best of your ability, and couldn't
figure out a way to get it to work, so you felt comfortable saying it
couldn't work.
But if you're wrong, billions of dollars can be the weight that falls
partly on your head.
And I bet you are wrong. I just picked S=1 and x_res =1 because it was
an obvious thing to do with the initial idea, not thinking any serious
researcher would think that written in stone.
There are lots of reasons to think that S was too small, from the
outset.
I guess some may wonder if S can be too big then, so that maybe there
is this narrow ranger of S's that will work.
Think about it.
If you figure out that you were sadly mistaken in attacking this
method, be smart, take a deep breath, and just come back and reply that
you were wrong, or do NOTHING.
Do NOTHING RASH.
Time is the best healer. And in the heat of the moment, wrong
decisions can seem wise.
The situation is just far bigger than you realized. But even worldwide
attention is just a thing.
James Harris 

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Tim Peters science forum Guru
Joined: 30 Apr 2005
Posts: 426

Posted: Sun Jul 16, 2006 5:11 am Post subject:
Re: Loose connectivity, factoring and residues



....
[Tim Peters]
Quote:  This is most like last year's methods, where you need to introduce search
after search to have a chance of factoring a nontrivial T.

[jstevh@msn.com]
Quote:  Ah, so you tried to test it to the best of your ability,

Yes.
Quote:  and couldn't figure out a way to get it to work,

No. I can get it to work well by exploiting that I know the factorization
before I begin. Otherwise, no, it required more gcds to find a factor than
the randomgcd factoring method, under a variety of different schemes for
picking S and x_res values.
Quote:  so you felt comfortable saying it couldn't work.

I'm 100% comfortable saying that no way I tried worked as well as
randomgcd. I'm also 100% comfortable saying that everything I observed
matched the mathematical statements I've made about it. In contrast,
_every_ observed behavior has caught you by surprise so far. Sucks to be
you here, I'm afraid.
Quote:  But if you're wrong, billions of dollars can be the weight that falls
partly on your head.
And I bet you are wrong.

How much?
Quote:  I just picked S=1 and x_res =1 because it was an obvious thing to do
with the initial idea, not thinking any serious researcher would think
that written in stone.

You are so full of s**t, James: a _serious_ researcher would implement it
himself, which leaves you out entirely. And, no, of course I didn't limit
my experiments to S = x_res = 1.
Quote:  There are lots of reasons to think that S was too small, from the
outset.
I guess some may wonder if S can be too big then, so that maybe there
is this narrow ranger of S's that will work.
Think about it.

I already did, thank you.
Quote:  If you figure out that you were sadly mistaken in attacking this
method, be smart, take a deep breath, and just come back and reply that
you were wrong, or do NOTHING.

Enough with the counterfactual hypotheticals, please.
> [more blah blah blah] 

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Tim Peters science forum Guru
Joined: 30 Apr 2005
Posts: 426

Posted: Sun Jul 16, 2006 6:10 am Post subject:
Re: Loose connectivity, factoring and residues



[Tim Peters]
Quote:  [...]
The _hard_ problem, and the one that's useful for factoring, is finding
distinct x and y such that:
x^2 = y^2 (modulo T) [2]
when T is given and /fixed/. [...]

[Proginoskes]
Quote:  Actually, it isn't too hard. Choose x and A to be arbitrary integers,
and let
y = A T  x.
Then y^2 = (A T  x) = A T^2  2 A T x + x^2 = x^2 (mod T).

Thank you Of course the hard problem is finding integer x and y
satisfying [2] when T is given and fixed /and composite/, and where it's
_not_ true that:
x = +/ y (modulo T) [3]
Cases where [2] and [3] hold are indeed trivial to find, but are _usually_
of no use for factoring via [2]. For if
x = y (modulo T) [4]
then
xy = 0 (modulo T)
so gcd(xy, T) = T. Similarly, if
x = y (modulo T)
then
x+y = 0 (modulo T)
so gcd(x+y, T) = T. So at least one of {xy, x+y} is certain to be useless
when [4] holds. Of course "deeper" analysis is possible, and I'll just
sketch one case here (the other is very much alike): suppose [4] holds. We
already showed that xy is useless in this case, but what about x+y? Well,
x = y (modulo T)
implies
x+y = 2*x (modulo T)
so
gcd(x+y, T) =
gcd(2*x, T)
so x+y is useless too unless you were lucky in picking x. In particular, if
T is odd, then
gcd(2*x, T) = gcd(x, T) = gcd(mod(x, T), T)
so "you're lucky" means you have to be lucky enough so that mod(x, T) isn't
1 and divides T. Because that's very unlikely when T has no small prime
factors, cases where [3] holds are usually considered to be wholly
uninteresting.
OTOH, if [2] holds but [3] doesn't, taking gcds of T with x+y and xy _do_
reveal nontrivial factors of T. Exercise for the reader :)
BTW, note that in your example, there was no need to expand (A T x)^2,
since:
A T  x = x (modulo T)
implies:
(A T x)^2 = (x)^2 = x^2 (modulo T) 

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jstevh@msn.com science forum Guru
Joined: 21 Jan 2006
Posts: 951

Posted: Sun Jul 16, 2006 2:40 pm Post subject:
Re: Loose connectivity, factoring and residues



Tim Peters wrote:
Quote:  ...
[Tim Peters]
This is most like last year's methods, where you need to introduce search
after search to have a chance of factoring a nontrivial T.
[jstevh@msn.com]
Ah, so you tried to test it to the best of your ability,
Yes.
and couldn't figure out a way to get it to work,
No. I can get it to work well by exploiting that I know the factorization
before I begin. Otherwise, no, it required more gcds to find a factor than
the randomgcd factoring method, under a variety of different schemes for
picking S and x_res values.

So you failed.
Quote:  so you felt comfortable saying it couldn't work.
I'm 100% comfortable saying that no way I tried worked as well as
randomgcd. I'm also 100% comfortable saying that everything I observed
matched the mathematical statements I've made about it. In contrast,
_every_ observed behavior has caught you by surprise so far. Sucks to be
you here, I'm afraid.

I don't think so, as I like the mathematics, and I don't trust your
abilities over mine.
Quote:  But if you're wrong, billions of dollars can be the weight that falls
partly on your head.
And I bet you are wrong.
How much?

That's an expression.
Quote:  I just picked S=1 and x_res =1 because it was an obvious thing to do
with the initial idea, not thinking any serious researcher would think
that written in stone.
You are so full of s**t, James: a _serious_ researcher would implement it
himself, which leaves you out entirely. And, no, of course I didn't limit
my experiments to S = x_res = 1.

Then I don't know how you screwed up.
Quote:  There are lots of reasons to think that S was too small, from the
outset.
I guess some may wonder if S can be too big then, so that maybe there
is this narrow ranger of S's that will work.
Think about it.
I already did, thank you.

Then there is something else you did wrong, or didn't think of.
Quote:  If you figure out that you were sadly mistaken in attacking this
method, be smart, take a deep breath, and just come back and reply that
you were wrong, or do NOTHING.
Enough with the counterfactual hypotheticals, please.

You claim you worked backwards, using your knowledge of factorizations
to see what would work.
How did S behave then?
If you don't know, then you screwed up.
Was it random in size with respect to the size of T? (Absolute values
here.)
If you don't know, then you need to go back and check.
If you find what I think you'll find, take a deep breath, do nothing
rash, and remember, time is the healer of all wounds.
James Harris 

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Proginoskes science forum Guru
Joined: 29 Apr 2005
Posts: 2593

Posted: Sun Jul 16, 2006 4:56 pm Post subject:
Re: Loose connectivity, factoring and residues



Tim Peters wrote:
Quote:  [Tim Peters]
[...]
The _hard_ problem, and the one that's useful for factoring, is finding
distinct x and y such that:
x^2 = y^2 (modulo T) [2]
when T is given and /fixed/. [...]
[Proginoskes]
Actually, it isn't too hard. Choose x and A to be arbitrary integers,
and let
y = A T  x.
Then y^2 = (A T  x) = A T^2  2 A T x + x^2 = x^2 (mod T).
Thank you Of course the hard problem is finding integer x and y
satisfying [2] when T is given and fixed /and composite/, and where it's
_not_ true that:
x = +/ y (modulo T) [3]
Cases where [2] and [3] hold are indeed trivial to find, but are _usually_
of no use for factoring via [2]. For if
x = y (modulo T) [4]
then
xy = 0 (modulo T)
so gcd(xy, T) = T. Similarly, if
x = y (modulo T)
then
x+y = 0 (modulo T)
so gcd(x+y, T) = T. So at least one of {xy, x+y} is certain to be useless
when [4] holds. Of course "deeper" analysis is possible, and I'll just
sketch one case here (the other is very much alike): suppose [4] holds. We
already showed that xy is useless in this case, but what about x+y? Well,
x = y (modulo T)
implies
x+y = 2*x (modulo T)
so
gcd(x+y, T) =
gcd(2*x, T)
so x+y is useless too unless you were lucky in picking x. In particular, if
T is odd, then
gcd(2*x, T) = gcd(x, T) = gcd(mod(x, T), T)
so "you're lucky" means you have to be lucky enough so that mod(x, T) isn't
1 and divides T. Because that's very unlikely when T has no small prime
factors, cases where [3] holds are usually considered to be wholly
uninteresting.
OTOH, if [2] holds but [3] doesn't, taking gcds of T with x+y and xy _do_
reveal nontrivial factors of T. Exercise for the reader :)
BTW, note that in your example, there was no need to expand (A T x)^2,

Yes, there is, you liar. 8)
Actually, I was parodying the unnecessary complexities that JSH will
throw into his posts, so that he's confused into thinking there's
something to them.
 Christopher Heckman
Quote:  since:
A T  x = x (modulo T)
implies:
(A T x)^2 = (x)^2 = x^2 (modulo T) 


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Tim Peters science forum Guru
Joined: 30 Apr 2005
Posts: 426

Posted: Sun Jul 16, 2006 10:06 pm Post subject:
Re: Loose connectivity, factoring and residues



[added sci.math because there's some mathematical content here,
and I'm done with this method now]
....
[jstevh@msn.com]
Quote:  Ah, so you tried to test it to the best of your ability,

[Tim Peters]
Quote:  Yes.
and couldn't figure out a way to get it to work,
No. I can get it to work well by exploiting that I know the
factorization before I begin. Otherwise, no, it required more
gcds to find a factor than the randomgcd factoring method, under
a variety of different schemes for picking S and x_res values.

[jstevh@msn.com]
Not at all. My _goal_ was to try different strategies and see how they
behaved, and I was successful in meeting that goal. Showing that it worked
well was not a goal; showing that it didn't work well was not a goal;
showing that there was no possible way for it work well was not a goal;
"figuring out a way to get it to work" was not a goal ... /my/ goal was
simply to observe its behavior under a variety of plausible approaches.
Do you understand this? Because I don't have a stake in whether it works
well or not, "works well" and "doesn't work well" are the same to me. The
"doesn't work well" outcome is "failure" to you because you're emotionally
attached to a /different/ outcome. That's /why/ you call "doesn't work
well" failure. I'm not laboring under that burden.
Quote:  so you felt comfortable saying it couldn't work.
I'm 100% comfortable saying that no way I tried worked as well as
randomgcd. I'm also 100% comfortable saying that everything I observed
matched the mathematical statements I've made about it. In contrast,
_every_ observed behavior has caught you by surprise so far. Sucks to
be you here, I'm afraid.
I don't think so, as I like the mathematics, and I don't trust your
abilities over mine.

Obviously not. Indeed, that's why you spend days, weeks, and even months at
times trying to salvage methods after I've given up on them and explained
why. Sucks to be you here ;)
....
Quote:  I just picked S=1 and x_res =1 because it was an obvious thing to do
with the initial idea, not thinking any serious researcher would think
that written in stone.
You are so full of s**t, James: a _serious_ researcher would implement
it himself, which leaves you out entirely. And, no, of course I didn't
limit my experiments to S = x_res = 1.
Then I don't know how you screwed up.

Easy cure: implement it yourself, and when it doesn't work well for you
either, you can figure out how /you/ screwed up instead.
Quote:  There are lots of reasons to think that S was too small, from the
outset.
I guess some may wonder if S can be too big then, so that maybe there
is this narrow ranger of S's that will work.
Think about it.
I already did, thank you.
Then there is something else you did wrong, or didn't think of.

Weren't you bitching just yesterday about unsupported assertions? You don't
have proof, and you don't have empirical evidence. Your only justification
for saying that is wishful thinking.
Quote:  ...
You claim you worked backwards, using your knowledge of factorizations
to see what would work.
How did S behave then?
If you don't know, then you screwed up.

Sorry, but you don't even begin to understand this, and again at least
because you don't /try/ anything yourself. I did the obvious thing to make
it work on the first try: given that I knew T = p*q, I picked:
x = (pq)/2
y = (p+q)/2
Then:
x^2  y^2 = T = 0 (modulo T)
gcd(x+y, T) = gcd(p, T) = p
gcd(xy, T) = gcd(q, T) = q
follow. You can always work backward from that to find some S and x_res and
k and f_1 and f_2 (sheesh ...) "that work", but doing so gives no /usable/
insight into any of them. Your belief that it /must/ is based on nothing
but hope. Here's a complete "backwards" example illustrating exactly what
the snag is (and incidentally proving that it's possible to work backwards
deterministically and efficiently  there's nothing special about the
specific numbers used here):
T = 8023 = 113 * 71 = p*q
From that it follows that these /must/ work:
x = x_res = 21
y = 92
Then the requirement:
S  2*x*k = x^2y^2
reduces to:
S = 42*k  8023 [1]
We also need:
2*S*k = x_res (modulo T)
which reduces to:
2*S*k = 21 (modulo T) [2]
Plugging [1] into [2] to eliminate S, and simplifying modulo T (note that in
general the RHS of [1] is of the form i*k  T, so is congruent to i*k):
4*k^2 = 1 (modulo T)
which is the same as (multiplying both sides by 4's inverse modulo 8023):
k^2 = 4012 (modulo T) [3]
As I told you yesterday (but didn't bother to demonstrate), solving [3] /is/
the modular square root problem. Nobody knows how to solve that efficiently
when T is composite without knowing T's factorization, but I know how to
solve it efficiently when T's factors are known (and that's longwinded, so
I won't explain it here  read a book  or read my earlier posts, where I
did explain much of it).
The four solutions to [3], along with their associated S values from [1],
are (and you can easily /check/ that these satisfy all the congruences from
which I derived them);
k S
 
2065 78707
3698 147293
4325 173627
5958 242213
Almost done. Picking any <k, S> pair from that list "works" in the forward
direction too (do you see why?), although:
a) It required solving a modular square problem to find them; and,
b) You still _also_ need to pick a factorization of S+k^2 as f_1*f_2
that doesn't screw up. Most ways do screw up. This way always
works (and remember that I know x, y, and k in advance here, so
doing this part is completely trivial):
f_1 = x+y+k
f_2 = xy+k
So, as advertised, that's how to find S, x_res, k, f_1 and f_2 that work on
the first try  provided you know T's factorization before you start.
Quote:  Was it random in size with respect to the size of T? (Absolute values
here.)
If you don't know, then you need to go back and check.

See above. For this way of picking x and y,
S = 2*x*k  T
where:
x = (pq)/2
and little useful can be said about k's magnitude beyond that some k that
works always exists in 1 .. floor(T/2).
Quote:  If you find what I think you'll find, take a deep breath, do nothing
rash, and remember, time is the healer of all wounds.

Is that what you thought would happen ? 

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jstevh@msn.com science forum Guru
Joined: 21 Jan 2006
Posts: 951

Posted: Sun Jul 16, 2006 11:47 pm Post subject:
Re: Loose connectivity, factoring and residues



Tim Peters wrote:
Quote:  [added sci.math because there's some mathematical content here,
and I'm done with this method now]
...
[jstevh@msn.com]
Ah, so you tried to test it to the best of your ability,
[Tim Peters]
Yes.
and couldn't figure out a way to get it to work,
No. I can get it to work well by exploiting that I know the
factorization before I begin. Otherwise, no, it required more
gcds to find a factor than the randomgcd factoring method, under
a variety of different schemes for picking S and x_res values.
[jstevh@msn.com]
So you failed.
Not at all. My _goal_ was to try different strategies and see how they
behaved, and I was successful in meeting that goal. Showing that it worked
well was not a goal; showing that it didn't work well was not a goal;
showing that there was no possible way for it work well was not a goal;
"figuring out a way to get it to work" was not a goal ... /my/ goal was
simply to observe its behavior under a variety of plausible approaches.
Do you understand this? Because I don't have a stake in whether it works
well or not, "works well" and "doesn't work well" are the same to me. The
"doesn't work well" outcome is "failure" to you because you're emotionally
attached to a /different/ outcome. That's /why/ you call "doesn't work
well" failure. I'm not laboring under that burden.

The point is that YOU failed. Your inability to find a working
algorithm is not proof of the failure of the approach, but you keep
posting as if it is.
IN this post, I am going to explain mathematically why you are wrong,
and do it VERY quickly, hoping you will get a clue about how YOUR
ability or lack of it, is not a mathematical determinate.
Quote:  so you felt comfortable saying it couldn't work.
I'm 100% comfortable saying that no way I tried worked as well as
randomgcd. I'm also 100% comfortable saying that everything I observed
matched the mathematical statements I've made about it. In contrast,
_every_ observed behavior has caught you by surprise so far. Sucks to
be you here, I'm afraid.
I don't think so, as I like the mathematics, and I don't trust your
abilities over mine.
Obviously not. Indeed, that's why you spend days, weeks, and even months at
times trying to salvage methods after I've given up on them and explained
why. Sucks to be you here ;)

I LIKE discovery. I like playing with ideas, even when they fail,
working out why they fail.
If you try hard enough, you will fail, a lot.
If you think that you gain some points by never failing publicly in a
big way, then fine, that's you.
But I still like the analogy of professional basball players, who go
up to the plate, swinging that bat, and most of the timethey fail.
Quote: 
I just picked S=1 and x_res =1 because it was an obvious thing to do
with the initial idea, not thinking any serious researcher would think
that written in stone.
You are so full of s**t, James: a _serious_ researcher would implement
it himself, which leaves you out entirely. And, no, of course I didn't
limit my experiments to S = x_res = 1.
Then I don't know how you screwed up.
Easy cure: implement it yourself, and when it doesn't work well for you
either, you can figure out how /you/ screwed up instead.

I'll do as I like with this method, but YOU are the one publicly
downplaying it, putting YOUR judgement against it.
So I feel a need to show people that your judgement is flawed.
You put yourself in the position of attacking this approach, and I can
show you are wrong with theory, without ever needing to do my own
example.
I LIKE mathematics. It's logical.
Quote:  There are lots of reasons to think that S was too small, from the
outset.
I guess some may wonder if S can be too big then, so that maybe there
is this narrow ranger of S's that will work.
Think about it.
I already did, thank you.
Then there is something else you did wrong, or didn't think of.
Weren't you bitching just yesterday about unsupported assertions? You don't
have proof, and you don't have empirical evidence. Your only justification
for saying that is wishful thinking.

I've since supported the assertion mathematically, and will do so again
in this post.
Quote:  ...
You claim you worked backwards, using your knowledge of factorizations
to see what would work.
How did S behave then?
If you don't know, then you screwed up.
Sorry, but you don't even begin to understand this, and again at least
because you don't /try/ anything yourself. I did the obvious thing to make
it work on the first try: given that I knew T = p*q, I picked:
x = (pq)/2
y = (p+q)/2
Then:
x^2  y^2 = T = 0 (modulo T)
gcd(x+y, T) = gcd(p, T) = p
gcd(xy, T) = gcd(q, T) = q
follow. You can always work backward from that to find some S and x_res and
k and f_1 and f_2 (sheesh ...) "that work", but doing so gives no /usable/
insight into any of them. Your belief that it /must/ is based on nothing
but hope. Here's a complete "backwards" example illustrating exactly what
the snag is (and incidentally proving that it's possible to work backwards
deterministically and efficiently  there's nothing special about the
specific numbers used here):

You picked a way that made sense TO YOU, but it doesn't consider the
important mathematical points.
Instead, consider a solution for x_res = 1 mod T, which is a PICKED
residue, not just some residue that just comes out of the blue. I
think you're kind of lazy here as well, as it takes a little more
effort to find a particular residue mod T.
Now then, if you find a solution
x^2  y^2 = 0 mod T
where x = 1 mod T, then you will also find that
S = 2*x*k mod T
solves to give
k = S*(2*x)^{1} mod T
so if you solve for k FIRST, with x_res = 1 mod T, then you get the
SAME ANSWER if k is a residue modulo T, understand?
That is an important point.
k = S*(2*x_res)^{1} mod T
will give the same answer with x_res = 1 as you find when you go out
looking for an answer with x = 1 mod T.
So what does that leave for any variation?
It leaves S.
But the S you find works with your searched for solutions must equal
the residue modulo T for any residue that you picked!!!
So, for a picked S, versus your found S, which I'll now call S_found,
you have
S_found = S mod T
and that is crucial.
Do you understand it?
So you can pick an S that is too small, but you are still picking an S
that has the same residue as an S that MUST factor.
That means, mathematically, it must be the case that if you don't find
an answer with a particular S, then there must be an answer with a
multiple of T added or subtracted from that S.
Do you agree or disagree?
Look over my roadmap of discoveries. Why do you think your
mathematical insights are better than mine?
Just because you've seen a lot of my failures?
But could you have ANY of my successes?
James Harris 

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David Moran science forum Guru Wannabe
Joined: 13 May 2005
Posts: 252

Posted: Mon Jul 17, 2006 12:14 am Post subject:
Re: Loose connectivity, factoring and residues



Quote:  Look over my roadmap of discoveries. Why do you think your
mathematical insights are better than mine?

Because it's obvious that you don't know the first thing about the field of
mathematics. I learned how to write a proof in my freshman year of college
and your crap and useless drivel are not proofs. Your biggest problem is
your own ignorance; you have shown yourself to be ignorant about a lot of
things in our field. A degree in physics doesn't equal knowledge of how to
write a math proof.
Dave 

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