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| JSH... |
 Posted: Tue Nov 03, 2009 4:30 pm |
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Guest
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Over a year ago I realized that one of my math results seemed to lead
to a probabilistic method for finding quadratic residues modulo N, and
while I kind of put that out there, it was a weird year with the
"financial crisis" and I've been in arguments with math people over my
research before, so I mainly figured, hey, if it's right, someone
should notice it!
Around a year later I think that it's quite possible that I DID solve
the factoring problem, but if so, it's a rather easy answer, and for
that reason, no one wants to believe it's true, and if enough people
won't believe, what can you do?
(Implement it? Then what? Get taken to the NSA where MORE
mathematicians can then do who knows what? The place is just a
concentration of you people. Sounds like walking into a lion's den
covered in bloody raw meat.)
I've had lots of math results over the years and repeatedly faced math
people who just don't care if it's correct, but seem confident that no
matter what I find they have the right to ignore it. So ok, if you
wish, ignore a solution to the factoring problem and continue as
you've done, for a year. I've grown so cynical about you people as
I've gone past shock at your behavior.
Too many math people seem to believe that pre-conceived notions about
what a solution should be actually constrain what mathematically is
correct. And WORSE they seem to have strongly held pre-conceived
notions about WHO can solve math problems, and can be vicious to a
point of obscene cruelty with people they deem incapable of solving
certain problems if those people claim they have.
Academic mathematicians on top of that seem to be comfortable with a
brutal view that if a result does not help their career then it's not
only ok to ignore it, but it's their duty to ignore it, if they can.
Now then, if I did not solve the factoring problem then all is ok. If
I did, then the only reason you do not know it a year later is because
of the above. Your society can be cruel to a point of vicious. You
show hatred for outsiders that borders on insane demented blood lust,
and you feel confident in ignoring ANYTHING that doesn't fit your
personal view of what an answer should look like.
The solution I found is absurdly simple. Math people have built
complexity on top of abstruseness in this area with glee and abandon,
and in my opinion don't want a simple solution, refuse to believe a
simple solution to the factoring problem is possible, and cannot be
motivated from their beliefs by an ACTUAL solution.
Ok. Let's see how far that goes as I don't think much will change.
I'm just kind of noting a sad anniversary.
Factoring problem was solved around a year ago, and the world did not
care.
James Harris |
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| JSH... |
| Posted: Tue Nov 03, 2009 6:13 pm |
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Guest
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On Nov 3, 6:30 pm, JSH <jst... at (no spam) gmail.com> wrote:
[quote]Over a year ago I realized that one of my math results seemed to lead
to a probabilistic method for finding quadratic residues modulo N, and
while I kind of put that out there, it was a weird year with the
"financial crisis" and I've been in arguments with math people over my
research before, so I mainly figured, hey, if it's right, someone
should notice it!
Around a year later I think that it's quite possible that I DID solve
the factoring problem, but if so, it's a rather easy answer, and for
that reason, no one wants to believe it's true, and if enough people
won't believe, what can you do?
[/quote]
But that is so strange, so hey, just on an anthropic principle
argument it seems it can't work.
So round and round I go... Maybe part of the reason I mostly put it
out of my mind for a year.
[quote](Implement it? Then what? Get taken to the NSA where MORE
mathematicians can then do who knows what? The place is just a
concentration of you people. Sounds like walking into a lion's den
covered in bloody raw meat.)
[/quote]
The idea is relatively easy algebra, so if it's true my fear is there
is willful blindness going on, and if people are cornered...
Quite simply there is no rulebook for what to do if massively easy
mathematical proofs are just ignored by the experts and people who
supposedly care about such things, as then you're in a bonkers
situation already.
The math looks easy. And I've seen how vicious math people can be.
It's unfortunately quite possible I fear that math society just kind
of didn't want to be bothered with a simple answer to the factoring
problem.
Like, maybe found it inconvenient.
James Harris |
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| Enrico... |
| Posted: Tue Nov 03, 2009 7:24 pm |
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On Nov 3, 9:13 pm, JSH <jst... at (no spam) gmail.com> wrote:
[quote]On Nov 3, 6:30 pm, JSH <jst... at (no spam) gmail.com> wrote:
Over a year ago I realized that one of my math results seemed to lead
to a probabilistic method for finding quadratic residues modulo N, and
while I kind of put that out there, it was a weird year with the
"financial crisis" and I've been in arguments with math people over my
research before, so I mainly figured, hey, if it's right, someone
should notice it!
Around a year later I think that it's quite possible that I DID solve
the factoring problem, but if so, it's a rather easy answer, and for
that reason, no one wants to believe it's true, and if enough people
won't believe, what can you do?
But that is so strange, so hey, just on an anthropic principle
argument it seems it can't work.
So round and round I go... Maybe part of the reason I mostly put it
out of my mind for a year.
(Implement it? Then what? Get taken to the NSA where MORE
mathematicians can then do who knows what? The place is just a
concentration of you people. Sounds like walking into a lion's den
covered in bloody raw meat.)
The idea is relatively easy algebra, so if it's true my fear is there
is willful blindness going on, and if people are cornered...
Quite simply there is no rulebook for what to do if massively easy
mathematical proofs are just ignored by the experts and people who
supposedly care about such things, as then you're in a bonkers
situation already.
The math looks easy. And I've seen how vicious math people can be.
It's unfortunately quite possible I fear that math society just kind
of didn't want to be bothered with a simple answer to the factoring
problem.
Like, maybe found it inconvenient.
James Harris
[/quote]
==============================================
Was that the one where you used T mod p to generate pairs F1 and F2
such that F1 * F2 mod p = T mod p ?
I couldn't resist the temptation to implement.
Minor bug when F1 + F2 mod p = 0 ... Easily fixed.
Works fine. Probabilistic by choice of p.
Small p gives a small number of long lists to search.
Large p gives a large number of short lists to search.
Seems to be an irreducible amount of work needed
for any T that increases with the size of T.
Sample output:
T = 29647477
p = 127
Iterate F1 mod p starting at 1
For each F1, calculate F2 s.t. F1*F2 mod p = T mod p
Factors found at F1 = 8
F1 = 254*A + 135
F2 = 254*B + 27
A = 28
B = 16
F1 = 7247
F2 = 4091
Gotta go.. somebody's pounding on the door ...
Enrico |
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| Dave -Turner... |
| Posted: Wed Nov 04, 2009 3:06 am |
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[quote]Around a year later I think that it's quite possible
that I DID solve the factoring problem
[/quote]
So you've solved the factoring problem but you just don't know it? Is that
by any chance because you can't prove it? |
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| Gordon Burditt... |
| Posted: Wed Nov 04, 2009 4:32 am |
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[quote]Over a year ago I realized that one of my math results seemed to lead
to a probabilistic method for finding quadratic residues modulo N, and
while I kind of put that out there, it was a weird year with the
"financial crisis" and I've been in arguments with math people over my
research before, so I mainly figured, hey, if it's right, someone
should notice it!
[/quote]
You've blown your credibility. If you claim that 2+2=4, no first-grade
math student who recognizes you will believe it, even if you say
you have a proof.
And you still haven't factored 6 yet.
[quote]Around a year later I think that it's quite possible that I DID solve
the factoring problem, but if so, it's a rather easy answer, and for
that reason, no one wants to believe it's true, and if enough people
won't believe, what can you do?
[/quote]
No, it's not that they believe it's an easy answer, it's that they
won't read it because your name is on it. And they believe that any
form of math, even 2+2=4, is too hard for you, because of the number
of false claims and simple errors you've made in the past.
[quote](Implement it? Then what? Get taken to the NSA where MORE
mathematicians can then do who knows what? The place is just a
concentration of you people. Sounds like walking into a lion's den
covered in bloody raw meat.)
[/quote]
I doubt very much they believe you know enough about math to even
be able to semi-intelligently *cheat* on "Are You Smarter Than A
First Trimester Fetus?".
[quote]I've had lots of math results over the years
[/quote]
How many of them are incorrect results for 2+2=?
[quote]and repeatedly faced math
people who just don't care if it's correct, but seem confident that no
matter what I find they have the right to ignore it.
[/quote]
Don't assume that the "you people" you are talking to are some kind
of math elite, just because dead cats are better at math than you
are. However, likely 99% of your readers on USENET completed
elementary-school math.
They always have the right to ignore it. Would you pay any attention
to a claimed proof of Fermat's Last Theorem if it were written by
a dead cat? The dead cat has a much better reputation at math than
you do.
[quote]So ok, if you
wish, ignore a solution to the factoring problem and continue as
you've done, for a year. I've grown so cynical about you people as
I've gone past shock at your behavior.
[/quote]
Your definition of the factoring problem did not include a requirement
of doing it quickly. Trial division as a method of factoring has
been around for a long, long time.
[quote]Too many math people seem to believe that pre-conceived notions about
what a solution should be actually constrain what mathematically is
correct.
[/quote]
You mean that a math proof should not have elementary addition
mistakes in it, like some of yours did?
[quote]And WORSE they seem to have strongly held pre-conceived
notions about WHO can solve math problems,
[/quote]
You might be able to solve 2+2 after a large number of tries, but
I do know it's a waste of my time to bother reading it after the
first few dozen wrong results.
[quote]and can be vicious to a
point of obscene cruelty with people they deem incapable of solving
certain problems if those people claim they have.
[/quote]
Yes, you can expect that when you shoot off your mouth too soon
repeatedly without checking your math first.
[quote]Academic mathematicians on top of that seem to be comfortable with a
brutal view that if a result does not help their career then it's not
only ok to ignore it, but it's their duty to ignore it, if they can.
Now then, if I did not solve the factoring problem then all is ok. If
[/quote]
If you cannot *use* your method to factor 6, what good is it? You've
had a whole year. And it's not like 6 is someone's top-secret RSA
key being used to protect the gold in Fort Knox.
[quote]I did, then the only reason you do not know it a year later is because
of the above. Your society can be cruel to a point of vicious. You
show hatred for outsiders that borders on insane demented blood lust,
and you feel confident in ignoring ANYTHING that doesn't fit your
personal view of what an answer should look like.
[/quote]
If by that you mean that 2+2 should come out as a result in numbers,
not cat pee, you're right. I do expect that the result should be
a number.
[quote]The solution I found is absurdly simple.
[/quote]
The problem as you define it is absurdly simple. If it does not
contain a requirement of factoring *quickly*, the solution (trial
division) was known long ago.
[quote]Math people have built
complexity on top of abstruseness in this area with glee and abandon,
and in my opinion don't want a simple solution, refuse to believe a
simple solution to the factoring problem is possible, and cannot be
motivated from their beliefs by an ACTUAL solution.
[/quote]
When you can't use it to factor 6, why should I believe it works?
[quote]Ok. Let's see how far that goes as I don't think much will change.
I'm just kind of noting a sad anniversary.
Factoring problem was solved around a year ago, and the world did not
care.
[/quote]
The factoring problem as you define it was solved over a thousand
years ago. |
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| rossum... |
| Posted: Wed Nov 04, 2009 5:27 am |
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On Tue, 3 Nov 2009 18:30:13 -0800 (PST), JSH <jstevh at (no spam) gmail.com> wrote:
[quote]Around a year later I think that it's quite possible that I DID solve
the factoring problem, but if so, it's a rather easy answer, and for
that reason, no one wants to believe it's true, and if enough people
won't believe, what can you do?
Despite its name the "Factoring Problem" is not just about finding[/quote]
factors; finding factors is neccessary but not sufficient. The real
factoring problem is finding factors *quickly*. The *quickly* part is
the most important part. We can find factors very easily by using
trial factorisation, but trial factorisation is slow.
You need to test how fast you method runs before you make any claims
about solving the factoring problem. Yes, that does mean implementing
it and running it against an implementation of trial factorisation and
comparing the relative speeds of the two methods.
If, and only if, your method finds factors faster than trial
factorisation you might have something. Then you have to trial your
method against other existing fast methods such as the number field
sieve.
You have more work to do on this James.
rossum |
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| Mark Murray... |
| Posted: Wed Nov 04, 2009 6:28 am |
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Gordon Burditt wrote:
[quote](JSH wrote: added by MM)
Over a year ago I realized that one of my math results seemed to lead
to a probabilistic method for finding quadratic residues modulo N, and
while I kind of put that out there, it was a weird year with the
"financial crisis" and I've been in arguments with math people over my
research before, so I mainly figured, hey, if it's right, someone
should notice it!
[/quote]
Notice the "one of my math results seemed to lead to a probabilistic
method". This is a new tack of his, started in sci.physics, where he
begins with "I may have <done marvellous thing> in blah blah...", and
is soon after presuming the truth of it.
[quote]Your definition of the factoring problem did not include a requirement
of doing it quickly. Trial division as a method of factoring has
been around for a long, long time.
[/quote]
He has now finally admitted (in sci.physics) that "the factoring
problem" involves solving it in (say) polynomial time. He doesn't
seem to get what this means.
[quote]Yes, you can expect that when you shoot off your mouth too soon
repeatedly without checking your math first.
[/quote]
He "can not bring himself to check his work". This was directly said
in his own "MyMathGroup".
More recently, he refused to do so (sci.physics).
[quote]If you cannot *use* your method to factor 6, what good is it? You've
had a whole year. And it's not like 6 is someone's top-secret RSA
key being used to protect the gold in Fort Knox.
[/quote]
About a year ago, he said straight out that he could not do this
(sci.math, I think).
M
--
Mark Murray |
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| JSH... |
| Posted: Wed Nov 04, 2009 2:56 pm |
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On Nov 3, 9:24 pm, Enrico <ungerne... at (no spam) aol.com> wrote:
[quote]On Nov 3, 9:13 pm, JSH <jst... at (no spam) gmail.com> wrote:
On Nov 3, 6:30 pm, JSH <jst... at (no spam) gmail.com> wrote:
Over a year ago I realized that one of my math results seemed to lead
to a probabilistic method for finding quadratic residues modulo N, and
while I kind of put that out there, it was a weird year with the
"financial crisis" and I've been in arguments with math people over my
research before, so I mainly figured, hey, if it's right, someone
should notice it!
Around a year later I think that it's quite possible that I DID solve
the factoring problem, but if so, it's a rather easy answer, and for
that reason, no one wants to believe it's true, and if enough people
won't believe, what can you do?
But that is so strange, so hey, just on an anthropic principle
argument it seems it can't work.
So round and round I go... Maybe part of the reason I mostly put it
out of my mind for a year.
(Implement it? Then what? Get taken to the NSA where MORE
mathematicians can then do who knows what? The place is just a
concentration of you people. Sounds like walking into a lion's den
covered in bloody raw meat.)
The idea is relatively easy algebra, so if it's true my fear is there
is willful blindness going on, and if people are cornered...
Quite simply there is no rulebook for what to do if massively easy
mathematical proofs are just ignored by the experts and people who
supposedly care about such things, as then you're in a bonkers
situation already.
The math looks easy. And I've seen how vicious math people can be.
It's unfortunately quite possible I fear that math society just kind
of didn't want to be bothered with a simple answer to the factoring
problem.
Like, maybe found it inconvenient.
James Harris
==============================================
Was that the one where you used T mod p to generate pairs F1 and F2
such that F1 * F2 mod p = T mod p ?
[/quote]
Nope. I have 2 solutions from surrogate factoring, and one from my
general solution to binary quadratic Diophantine equations.
So 3 possible solutions to the factoring problem.
You're showing an early version of one of the surrogate factoring
approaches.
But your prime p is too small.
[quote]I couldn't resist the temptation to implement.
Minor bug when F1 + F2 mod p = 0 ... Easily fixed.
Works fine. Probabilistic by choice of p.
Small p gives a small number of long lists to search.
Large p gives a large number of short lists to search.
[/quote]
Not from the current surrogate factoring theory.
[quote]Seems to be an irreducible amount of work needed
[/quote]
There are 2 approaches with current surrogate factoring theory, and
you can't be doing any of them, as for one thing, p needs to be quite
large. Your p is way too small.
[quote]for any T that increases with the size of T.
Sample output:
T = 29647477
p = 127
Iterate F1 mod p starting at 1
For each F1, calculate F2 s.t. F1*F2 mod p = T mod p
Factors found at F1 = 8
F1 = 254*A + 135
F2 = 254*B + 27
A = 28
B = 16
F1 = 7247
F2 = 4091
Gotta go.. somebody's pounding on the door ...
Enrico
[/quote]
With surrogate factoring you need a large p, where the limit on the
size of p is that:
p < (f_1 + f_2)/2
where f_1 and f_2 are two positive factors such that f_1*f_2 = T,
where T is your target composite to factor.
IF you didn't even know that, you're not even close to doing the
current surrogate factoring theory.
You may be still using older basic research of mine, and stopped
paying attention as I advanced it.
For others, yes, it does factor, so I discovered a new factoring
method. The issue of course has been, how fast with big numbers? I
refuse to test it with large numbers.
James Harris |
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| Chris McDonald... |
| Posted: Wed Nov 04, 2009 2:58 pm |
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JSH <jstevh at (no spam) gmail.com> writes:
[quote]... I refuse to test it with large numbers.
[/quote]
And this is where you lose all credibility....
--
Chris. |
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| JSH... |
| Posted: Wed Nov 04, 2009 3:02 pm |
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On Nov 4, 1:32 am, gordonb.36... at (no spam) burditt.org (Gordon Burditt) wrote:
[quote]Over a year ago I realized that one of my math results seemed to lead
to a probabilistic method for finding quadratic residues modulo N, and
while I kind of put that out there, it was a weird year with the
"financial crisis" and I've been in arguments with math people over my
research before, so I mainly figured, hey, if it's right, someone
should notice it!
You've blown your credibility. If you claim that 2+2=4, no first-grade
math student who recognizes you will believe it, even if you say
you have a proof.
[/quote]
We'll see.
The answer I found is rather stupid simple, as I found that if you
have a quadratic residue q modulo N, where N is an odd non-unit
natural, then you have some simple congruence relations:
k^2 = q mod N
implies that k = 3^{-1}*(f_1 + f_2) mod N
where f_1 and f_2 are factors of some number T, such that T = f_1*f_2,
where T = 2q mod N, but T does not equal 2q exactly. The "implies" is
there because it's probabilistic and some factorization of T will work
50% of the time.
It's really weirdly simple. And naturally scales as N increases in
size.
It is the method for finding quadratic residues modulo N.
And it has been around for around a year now, as I'm sort of
celebrating a belated birthday for the discovery.
Now you can say I don't have credibility but I figure if it works
SOMEONE in the world is probably using it now.
And may have been using it for months.
James Harris |
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| JSH... |
| Posted: Wed Nov 04, 2009 3:12 pm |
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On Nov 4, 4:58 pm, Chris McDonald <ch... at (no spam) csse.uwa.edu.au> wrote:
[quote]JSH <jst... at (no spam) gmail.com> writes:
... I refuse to test it with large numbers.
And this is where you lose all credibility....
--
Chris.
[/quote]
Fine. I have NO intention of proving this result works personally by
factoring large numbers with it. I'm hoping for some group of
researchers somewhere in the world to prove it. Seems safer...
The result itself again is a weirdly simple relationship from
fundamental algebra that given q, a quadratic residue modulo N, where
N is an odd non-unit Natural, you have that if:
k^2 = q mod N
then (and this is weird but mathematically correct and trivial to
prove):
k = 3^{-1}(f_1 + f_2) mod N
where f_1 and f_2 are factors such that f_1*f_2 = 2q mod N, where
f_1*f_2 does not equal 2q, for roughly 50% of factors f_1 and f_2.
And yes, the 50% is about quadratic residues.
So weirdly enough quadratic residues are more intimately connected
with factoring than mathematicians ever realized, though I do wonder
how they missed such a fascinatingly simple result!
THAT is a fundamental algebraic result that I figured out about a year
ago, which shows you how to solve quadratic residues modulo N.
The factoring problem was solved about a year ago, and you people just
failed to notice because you're pig-headed, arrogant, and also kind of
dumb, unfortunately, despite your pretensions to the contrary.
Else, how did you NOT notice the arrival of such a powerful but simple
result?
The proof is trivial. It's on my math blog. To find it, just go to
Google.
Search on: surrogate factoring
James Harris |
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| JSH... |
| Posted: Wed Nov 04, 2009 6:22 pm |
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On Nov 4, 7:45 pm, gordonb.wf... at (no spam) burditt.org (Gordon Burditt) wrote:
[quote]You've blown your credibility. If you claim that 2+2=4, no first-grade
math student who recognizes you will believe it, even if you say
you have a proof.
We'll see.
Have you admitted yet that "the factoring problem" includes as an
important element the issue of speed of factoring? If not, well,
[/quote]
I always have.
[quote]trial division has been around for a long, long time, much longer
than you have been alive. If you're not going to factor large
numbers, how do you intend to prove that it's fast? "It's freaking
algebra" doesn't cut it. Inverting a trillion x trillion matrix
[/quote]
There is no generally accepted method for solving quadratic residues
modulo N.
I've provided such a method: a world's first.
It just so happens that such a thing also solves the factoring
problem.
So it's a two-fer.
That's how I can drop the argument that I have to factor some large
number first, and then explain why breaking people like you when you
don't BELIEVE you can be broken is dangerous.
Cornered people can be very dangerous.
You put up a challenge you do not believe can be met. If I met it,
alone, there is no telling what someone like you might do.
[quote]is just freaking algebra also, but it's not fast.
The answer I found is rather stupid simple, as I found that if you
have a quadratic residue q modulo N, where N is an odd non-unit
natural, then you have some simple congruence relations:
Have you used it to factor 6 yet? Someone claims that you admitted
earlier that it was impossible. Did you really say that? If not,
[/quote]
The phrase "surrogate factoring" is one I invented for a concept:
factoring one number by instead factoring another.
Under that name I've had lots of failed or near approaches, and one of
those could not factor 15. It might not have factored 6 either but
I'm not sure.
I don't consider even numbers so I doubt I said 6. It was probably 15
and you thought 6 for some reason.
[quote]let's see a demonstration of factoring 6. If so, start tossing out
[/quote]
That's stupid.
[quote]all the proofs that prove you can factor all composites (which, of
course, includes 6), since obviously a proof that you can do something
impossible is broken. And any proof for being able to factor
composites not equal to six will need to use the fact that the
composite isn't 6 somewhere in the proof.
k^2 = q mod N
implies that k = 3^{-1}*(f_1 + f_2) mod N
where f_1 and f_2 are factors of some number T, such that T = f_1*f_2,
where T = 2q mod N, but T does not equal 2q exactly. The "implies" is
there because it's probabilistic and some factorization of T will work
50% of the time.
A proof that your method works is going to need a proof that numbers
that you're supposed to pick actually can be picked. This fails
if you are required to pick an integer between, say, sqrt(2) and
sqrt(3).
[/quote]
The method I've given specifies N, an odd non-unit Natural.
You're babbling out stuff with the assumption that I don't have
anything.
So it's not worth it to you to even care what has already been
presented as you've made up your mind and are just going through the
motions.
I'm only replying to you as I'm looking for something to do while I
try to process why this situation is so impossible.
[quote]Don't assume that the probabilities of different factorizations of
T are independent, so a 50% probability of one working does not say
much for the probability of at least one out of all possible
factorizations working.
It's really weirdly simple. And naturally scales as N increases in
size.
That is not a proof that your method is fast. Do you have any idea
what the O(n) notation means?
[/quote]
Yup.
I don't usually use it.
[quote]It is the method for finding quadratic residues modulo N.
And it has been around for around a year now, as I'm sort of
celebrating a belated birthday for the discovery.
Now you can say I don't have credibility but I figure if it works
SOMEONE in the world is probably using it now.
And may have been using it for months.
So what? I don't buy your doomsday theories that cracking RSA will
cause a world panic. There are lots easier ways of stealing credit
card numbers than cracking SSL used in e-commerce stores.
[/quote]
Unless people like you have watched exploits and rationalized them
away, deciding that they were "human error".
I've seen math people do some weird things to go into denial over my
research.
IT is unfortunately possible that widespread exploits are commonplace
now, and security experts are explaining them all away. If so, then
there really is no way to know what the current security situation is.
For all we know at this point, there is no security currently in place
any more. With hackers worldwide moving into and out of systems, at
will.
And people like you explaining it all away.
James Harris |
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| Gordon Burditt... |
| Posted: Wed Nov 04, 2009 10:45 pm |
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Guest
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[quote]You've blown your credibility. If you claim that 2+2=4, no first-grade
math student who recognizes you will believe it, even if you say
you have a proof.
We'll see.
[/quote]
Have you admitted yet that "the factoring problem" includes as an
important element the issue of speed of factoring? If not, well,
trial division has been around for a long, long time, much longer
than you have been alive. If you're not going to factor large
numbers, how do you intend to prove that it's fast? "It's freaking
algebra" doesn't cut it. Inverting a trillion x trillion matrix
is just freaking algebra also, but it's not fast.
[quote]The answer I found is rather stupid simple, as I found that if you
have a quadratic residue q modulo N, where N is an odd non-unit
natural, then you have some simple congruence relations:
[/quote]
Have you used it to factor 6 yet? Someone claims that you admitted
earlier that it was impossible. Did you really say that? If not,
let's see a demonstration of factoring 6. If so, start tossing out
all the proofs that prove you can factor all composites (which, of
course, includes 6), since obviously a proof that you can do something
impossible is broken. And any proof for being able to factor
composites not equal to six will need to use the fact that the
composite isn't 6 somewhere in the proof.
[quote]k^2 = q mod N
implies that k = 3^{-1}*(f_1 + f_2) mod N
where f_1 and f_2 are factors of some number T, such that T = f_1*f_2,
where T = 2q mod N, but T does not equal 2q exactly. The "implies" is
there because it's probabilistic and some factorization of T will work
50% of the time.
[/quote]
A proof that your method works is going to need a proof that numbers
that you're supposed to pick actually can be picked. This fails
if you are required to pick an integer between, say, sqrt(2) and
sqrt(3).
Don't assume that the probabilities of different factorizations of
T are independent, so a 50% probability of one working does not say
much for the probability of at least one out of all possible
factorizations working.
[quote]It's really weirdly simple. And naturally scales as N increases in
size.
[/quote]
That is not a proof that your method is fast. Do you have any idea
what the O(n) notation means?
[quote]It is the method for finding quadratic residues modulo N.
And it has been around for around a year now, as I'm sort of
celebrating a belated birthday for the discovery.
Now you can say I don't have credibility but I figure if it works
SOMEONE in the world is probably using it now.
And may have been using it for months.
[/quote]
So what? I don't buy your doomsday theories that cracking RSA will
cause a world panic. There are lots easier ways of stealing credit
card numbers than cracking SSL used in e-commerce stores. |
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| Gordon Burditt... |
| Posted: Thu Nov 05, 2009 2:12 am |
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[quote]You've blown your credibility. If you claim that 2+2=4, no first-grade
math student who recognizes you will believe it, even if you say
you have a proof.
We'll see.
Have you admitted yet that "the factoring problem" includes as an
important element the issue of speed of factoring? If not, well,
I always have.
[/quote]
Then how come none of your proofs mention the speed of the algorithm?
[quote]trial division has been around for a long, long time, much longer
than you have been alive. If you're not going to factor large
numbers, how do you intend to prove that it's fast? "It's freaking
algebra" doesn't cut it. Inverting a trillion x trillion matrix
There is no generally accepted method for solving quadratic residues
modulo N.
I've provided such a method: a world's first.
It just so happens that such a thing also solves the factoring
problem.
[/quote]
You haven't proved that it works faster than trial division. For
that matter, you haven't proved that it always works, either. I
think you did prove that *if* it works, it provides a correct answer,
not necessarily faster than trial division. That alone is not much
of an accomplishment.
[quote]So it's a two-fer.
[/quote]
Not without the missing pieces.
[quote]That's how I can drop the argument that I have to factor some large
number first, and then explain why breaking people like you when you
don't BELIEVE you can be broken is dangerous.
[/quote]
If you think that I am an academic in mathematics, or that my job
depends on it, you're very wrong.
[quote]Cornered people can be very dangerous.
You put up a challenge you do not believe can be met. If I met it,
[/quote]
You're right: I don't believe that you have the vaguest idea how
to prove that your algorithm is faster than trial division. Someone
else might be able to prove that, with your method, but YOU can't.
[quote]alone, there is no telling what someone like you might do.
[/quote]
You really have a very overblown sense of your own importance, don't you?
[quote]is just freaking algebra also, but it's not fast.
The answer I found is rather stupid simple, as I found that if you
have a quadratic residue q modulo N, where N is an odd non-unit
natural, then you have some simple congruence relations:
Have you used it to factor 6 yet? Someone claims that you admitted
earlier that it was impossible. Did you really say that? If not,
The phrase "surrogate factoring" is one I invented for a concept:
factoring one number by instead factoring another.
Under that name I've had lots of failed or near approaches, and one of
those could not factor 15. It might not have factored 6 either but
I'm not sure.
I don't consider even numbers so I doubt I said 6. It was probably 15
and you thought 6 for some reason.
[/quote]
That's possible. It has been a long time since that discussion.
[quote]let's see a demonstration of factoring 6. If so, start tossing out
That's stupid.
[/quote]
No, what's stupid is that you don't realize that if you *PROVE* that
your method works for *ALL ODD COMPOSITE NUMBERS*, and it doesn't
work for 15, your proof is invalid, and you have to throw the whole
proof out.
[quote]all the proofs that prove you can factor all composites (which, of
course, includes 6), since obviously a proof that you can do something
impossible is broken. And any proof for being able to factor
composites not equal to six will need to use the fact that the
composite isn't 6 somewhere in the proof.
k^2 = q mod N
implies that k = 3^{-1}*(f_1 + f_2) mod N
where f_1 and f_2 are factors of some number T, such that T = f_1*f_2,
where T = 2q mod N, but T does not equal 2q exactly. The "implies" is
there because it's probabilistic and some factorization of T will work
50% of the time.
A proof that your method works is going to need a proof that numbers
that you're supposed to pick actually can be picked. This fails
if you are required to pick an integer between, say, sqrt(2) and
sqrt(3).
The method I've given specifies N, an odd non-unit Natural.
You're babbling out stuff with the assumption that I don't have
anything.
[/quote]
It's up to you to demonstrate that you do have something. (What
you've shown me is heavily error-ridden). And that includes proving
it's fast, which I don't believe you've ever even attempted.
[quote]So it's not worth it to you to even care what has already been
presented as you've made up your mind and are just going through the
motions.
I'm only replying to you as I'm looking for something to do while I
try to process why this situation is so impossible.
Don't assume that the probabilities of different factorizations of
T are independent, so a 50% probability of one working does not say
much for the probability of at least one out of all possible
factorizations working.
It's really weirdly simple. And naturally scales as N increases in
size.
That is not a proof that your method is fast. Do you have any idea
what the O(n) notation means?
Yup.
I don't usually use it.
[/quote]
If you want credibility, I recommend you learn, especially since
you don't want to demonstrate using your method on large numbers.
[quote]It is the method for finding quadratic residues modulo N.
And it has been around for around a year now, as I'm sort of
celebrating a belated birthday for the discovery.
Now you can say I don't have credibility but I figure if it works
SOMEONE in the world is probably using it now.
And may have been using it for months.
So what? I don't buy your doomsday theories that cracking RSA will
cause a world panic. There are lots easier ways of stealing credit
card numbers than cracking SSL used in e-commerce stores.
Unless people like you have watched exploits and rationalized them
away, deciding that they were "human error".
[/quote]
There are plenty of exploits that don't require cracking SSL, such
as dumpster diving, finding a plaintext file on the server containing
large amounts of credit card info, hacking or social-engineering
the administrator's password, bribing an employee, use of card-number
generators, stealing old backup tapes, etc. These get maybe hundreds
of thousands of cards at a time. These have been going on long
before you came up with your method. (The Egghead breach was at
least 10 years ago, wasn't it?)
Even if you can trivially break SSL, you have to be in a position
where you can tap the traffic to the e-commerce web site (such as
working at their ISP), and that's not very easy, especially if you
are in a different country. And you only get one card at a time
if you break an SSL connection to an ordinary user buying something,
and only if he enters his card number on that connection (he might
just be checking shipment status).
[quote]I've seen math people do some weird things to go into denial over my
research.
[/quote]
You think I'm a "math people"? How wrong you are! I don't even read
sci.math.
[quote]IT is unfortunately possible that widespread exploits are commonplace
now, and security experts are explaining them all away. If so, then
there really is no way to know what the current security situation is.
For all we know at this point, there is no security currently in place
any more.
[/quote]
As I said, if SSL vanished tomorrow, I doubt it would cause worldwide
panic.
[quote]With hackers worldwide moving into and out of systems, at
will.
[/quote]
Very doubtful that it's being done by SSL. Remember, even hackers
can get authorized access to most of these servers using their own
accounts (user certificates are not in use). |
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| Mark Murray... |
| Posted: Thu Nov 05, 2009 3:08 am |
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JSH wrote:
[quote]Have you admitted yet that "the factoring problem" includes as an
important element the issue of speed of factoring? If not, well,
I always have.
[/quote]
Liar.
M
--
Mark Murray |
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