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From: Mark Murray on 1 Jul 2010 10:40
On 07/01/10 14:58, JSH wrote:
>> Spend 5 years getting a decent education in mathematics.
>
> But what about THIS result? What would YOU do with it if it were
> yours?

In no particular order, all some or none of the following:

a) Spend effort on seeing if it could be improved. This would
likely involve becoming familiar with the background material
first.

b) Compare it to the works of predecessors to see what I could
learn from them. Also involves learning the background material
first.

c) Chuck it in the trash as a dead-end.

>> The basic disagreement between you and everyone else is the value of
>> your research. You think it deserves serious consideration, we think
>> it is of very poor quality.
>
> It is my math. Do a search in Google on: mymath

"Yours"?! HAH. You are rehashing (badly) the work of others.

> So who are you anyway? Who is this "we"?
>
> What makes you at all relevant to anyone, especially me?
>
> Why should I give a damn what you think?

No reason whatsoever. Your choice. But if you don't give a damn
what people think, its rather a waste of time asking them.

>> For reference here is a paper doing what you are attempting to
>> achieve.
>>
>> http://www.dtc.umn.edu/~odlyzko/doc/arch/discrete.logs.pdf
>
>
> Oh, Odlyzko. I'm still mad at him for years ago telling me my prime
> counting function wasn't interesting.

So what?

> When is the last time you had a chat with him?

No relevance at all.

>> I do not expect you to understand the maths; from experience even the
>> abstract is beyond you. Do note however the quality of the paper
>> compared to your own work. If you can see it then you will understand.
>
> Now that you've finished strutting and talking down to me, back to the
> CURRENT IDEA.
>
> What do you think should be done with it?

As it stands, and given your likelihood of improving it, I think
it should be trashed.

> Should it be abandoned completely and best forgotten by all?

Ayup. Fat chance, but I can dream.

M
From: big gus on 1 Jul 2010 13:52

"JSH" <jstevh(a)gmail.com> wrote in message
news:cad44d52-e885-409e-aab9-ca869d6f80a4(a)a30g2000yqn.googlegroups.com...
On Jun 30, 10:06 pm, MichaelW <ms...(a)tpg.com.au> wrote:
> On Jul 1, 1:20 pm, JSH <jst...(a)gmail.com> wrote:
>
>
>
>> > So what would you do next if you were me?
>>
>> > James Harris
>>
>> Spend 5 years getting a decent education in mathematics.

>But what about THIS result? What would YOU do with it if it were
>yours?

I would have discarded it long ago and moved on to newer,
more interesting areas.


>> The basic disagreement between you and everyone else is the value of
>> your research. You think it deserves serious consideration, we think
>> it is of very poor quality.

>It is my math. Do a search in Google on: mymath

no need.

>So who are you anyway? Who is this "we"?

We are professional Mathematicians, and you are not.


>What makes you at all relevant to anyone, especially me?

He is correct.


>Why should I give a damn what you think?

Because he is smarter than you are. Dont take that as an insult, it is just
the truth, and you need to deal with the fact that you have great
dificulity with math, you constiantly fail to show proof, fail to factor
numbers when requested, fail to post correct math,..... I guess you are
just not smart in Math. You have not shown me anything to think that you are
over 12 years.

>> For reference here is a paper doing what you are attempting to
>> achieve.
>>
>> http://www.dtc.umn.edu/~odlyzko/doc/arch/discrete.logs.pdf


>Oh, Odlyzko. I'm still mad at him for years ago telling me my prime
>counting function wasn't interesting.

You never knew him.

>When is the last time you had a chat with him?

>> I do not expect you to understand the maths; from experience even the
>> abstract is beyond you. Do note however the quality of the paper
>> compared to your own work. If you can see it then you will understand.

>Now that you've finished strutting and talking down to me, back to the
>CURRENT IDEA.

You are at ground level, so everyone has to talk down to
you, and that is your fault.

>What do you think should be done with it?

change it over to Serriagoat Factoring or something,
something...

>Should it be abandoned completely and best forgotten by all?

Is it useful? has it demonstrated that it works for all numbers ?

no, no.

>James Harris

is jsh troll? yes.


From: MichaelW on 1 Jul 2010 17:10
On Jul 1, 11:58 pm, JSH <jst...(a)gmail.com> wrote:
> On Jun 30, 10:06 pm, MichaelW <ms...(a)tpg.com.au> wrote:
>
> > On Jul 1, 1:20 pm, JSH <jst...(a)gmail.com> wrote:
>
> > > So what would you do next if you were me?
>
> > > James Harris
>
> > Spend 5 years getting a decent education in mathematics.
>
> But what about THIS result?  What would YOU do with it if it were
> yours?
>

I would not publish it on a maths forum as being of any significance.
I would run it through a few thousand test samples, realise that it
was not going anywhere, and bin it.

> > The basic disagreement between you and everyone else is the value of
> > your research. You think it deserves serious consideration, we think
> > it is of very poor quality.
>
> It is my math.  Do a search in Google on: mymath
>
> So who are you anyway?  Who is this "we"?

Basically there are two classes of posters who respond to your
threads. The first is those who simply abuse and never have anything
mathematical to say. The second is those who are prepared to respond
to you at a maths level; this is the "we". There are a lot fewer these
days (Mark, Joshua, myself, a few others).

>
> What makes you at all relevant to anyone, especially me?
>
> Why should I give a damn what you think?
>

Because you asked. *You said* you are brainstorming and want a
response.

> > For reference here is a paper doing what you are attempting to
> > achieve.
>
> >http://www.dtc.umn.edu/~odlyzko/doc/arch/discrete.logs.pdf
>
> Oh, Odlyzko.  I'm still mad at him for years ago telling me my prime
> counting function wasn't interesting.
>
> When is the last time you had a chat with him?

I honestly had no idea and did not know who he is. Remember I am not
even in your country and I certainly don't hang around at academia.
Sounds like a smart guy though.

>
> > I do not expect you to understand the maths; from experience even the
> > abstract is beyond you. Do note however the quality of the paper
> > compared to your own work. If you can see it then you will understand.
>
> Now that you've finished strutting and talking down to me, back to the
> CURRENT IDEA.
>
> What do you think should be done with it?
>
> Should it be abandoned completely and best forgotten by all?

Yes, it should be completely abandoned. I could not be any clearer
without starting to sound like the parrot sketch.

Frankly I have seen several of your "This is the factorisation
breakthrough at last" algorithms over the last couple of years and
compared to them this one is especially poor. As we say here in Oz it
feels like you just phoned it in. Usually it takes me a day to crack
one of your algorithms but this one took less than an hour.

Anyway it appears we are at the personal abuse stage of the JSH thread
cycle where we all stop talking maths and tell each other what we
think is wrong with them. Unless you have more maths to add it is
probably time to bail out. You can have the last word if you want.

Regards, Michael W.
From: Joshua Cranmer on 1 Jul 2010 17:41
On 06/30/2010 11:20 PM, JSH wrote:
> There is only one thing that matters with this approach at this point
> from the basic research side: it directly attacks using a large m by
> directly eliminating a large portion of that m.
>
> So it can reduce an m of arbitrary size to finding 4 factors.
>
> There is NOTHING else that even comes close to that concept out there
> at all.

Okay, introducing a lot of complexity into a problem and then removing a
large portion (but not all of it) is not exactly something to be proud of.

> If it can be figured out how to work it well it simply eliminates the
> advantage of discrete logs.
>
> Blows it away.

Looking at it in a different way: current factoring and discrete log
algorithms are of about the same complexity. So factoring just ONE
number would not make discrete log asymptotically faster. This number
may not necessarily be a semiprime, so it might be a little faster
constant-factor wise, but if you have to factor more than one or two
numbers to solve the discrete log, you will have eroded any lead you
had. If you can't bring down the number of numbers to factor, then this
algorithm will never be able to compete with the most performant algorithms.

> So what would you do next if you were me?

As others have suggested, stop pursuing this avenue, if you are trying
to find a useful discrete log algorithm.

--
Beware of bugs in the above code; I have only proved it correct, not
tried it. -- Donald E. Knuth
From: JSH on 1 Jul 2010 20:23
On Jun 30, 7:01 pm, MichaelW <ms...(a)tpg.com.au> wrote:
> On Jul 1, 9:54 am, JSH <jst...(a)gmail.com> wrote:
>
>
>
>
>
> > I've noted a way to solve for m, when k^m = q mod N, through integer
> > factorization, which is then an approach to solving discrete
> > logarithms in a prior post.  In this post I'll explain when the
> > equations MUST work, where a simple analysis can be done trivially
> > using methods familiar to those who've solved simultaneous equations
> > in regular algebra.
>
> > Here are relevant equations without the complete detail explaining
> > them all of the prior post which should be read for reference:
>
> > Everything follows from use of a simple system of equations:
>
> > f_1 = a_1*k mod N thru f_m = a_m*k mod N
>
> > Two important constraining equations:
>
> > a_1*...*a_m = q mod N
>
> > and
>
> > a_1+...+a_m = m mod N
>
> > Resultant equations:
>
> > f_1*...*f_m = q^2 mod N
>
> > and
>
> > f_1+...+f_m = mk mod N
>
> > (These are arbitrary constraints that I used.  There may be others
> > that are of practical use.)
>
> > Now assume that for some unknown number m-c of the f's that the a's
> > are simply the modular inverse of k, then for that number the f's
> > simply equal 1, which allows me to solve for m with:
>
> > (k-1)*m = (f_1+...+f_c - c) mod N
>
> > If k-1 is coprime to N, you can simply use the modular inverse to get
> > m.  Otherwise you'd to divide off common factors from both sides and
> > then use the modular inverse with what remained.
>
> > All of which was given in my prior post, but notice I can now go back
> > to the constraining equations for the a's with the information that
> > some of the a's have been set to the modular inverse of k:
>
> > a_1*...*a_c*k^{-(m-c)} = q mod N, so a_1*...*a_c= q*k^{-(m-c)}
>
> > and
>
> > a_1+...+a_c + (m-c)k^{-1} = m mod N, so a_1+...+a_c = -(m-c)k^{-1} + m
> > mod N
>
> > which means there are two simultaneous congruence equations with c
> > unknowns:
>
> > a_1*...*a_c= q*k^{-(m-c)}
>
> > and
>
> > a_1+...+a_c = -(m-c)k^{-1} + m mod N
>
> > Using the first to substitute out a_1 into the second and simplifying
> > slightly gives:
>
> > q*k^{-(m-c)}  + a_2^2*a_3*...*a_c + a_2*a_3^2*...*a_c...+
> > a_2*...*a_c^2 = a_2*...*a_c[-(m-c)k^{-1} + m] mod N
>
> > Notice then that with any c-2 variables set arbitrarily, the existence
> > of the remaining one is set if a quadratic residue exists.
>
> > So for instance if c=4, then you can have a_3 and a_4 completely free
> > variables, as long as quadratic residues exist to allow for a_2, which
> > would indicate a 50% probability in that case if N is prime.
>
> > However, you can also further constrain one more of the a's to remove
> > squares, for instance let a_3 = a_2^{-1} mod N, and you have:
>
> > q*k^{-(m-c)}  + a_2*a_4*...*a_c + a_3*a_4*...*a_c...+ a_4*...*a_c^2 =
> > a_4*...*a_c[-(m-c)k^{-1} + m] mod N
>
> > which allows a solution for a_2 to always exist.  Which would leave
> > with c=4, a_4 completely free.
>
> > Assuming human nature will be to look for smaller values to factor to
> > actually use the algorithm, one can assume that
>
> > f_1*...*f_m = q^2 mod N
>
> > will be with smaller values of q^2 mod N, based on human preference,
> > so if a_4 is completely free, and is non-unit, it would likely in many
> > cases mean that f_4 will be 2.
>
> > If a_3 and a_4 are free then one would expect f_3 and f_4 to be 2.
>
> > With c = 4 or greater then the area to consider variability that
> > cannot just be arbitrarily set to a convenience value is with a_1 and
> > a_2 which could make f_1 and f_2 just about any residue mod N.  Worst
> > case they are both near N, so you'd have a size of approximately 4N^2.
>
> > So algorithms based on this method should exit within q^2 mod N less
> > than 4N^2.
>
> > Here's the example given in my prior post, which should make more
> > sense given the information above:
>
> > Solve for m, where:
>
> > 2^m = 13 mod 23
>
> > so k = 2, and f_1*...*f_m = q^2 mod N = 13^2 mod 23 = 8 mod 23.
>
> > And I found a solution with f_1*...*f_m = 54, and
>
> > f_1 = 2, f_2 = 3, f_3 = 3, f_4 = 3
>
> > so c=4, and
>
> > m = (k-1)^{-1}(f_1 +...+ f_c - c) mod N = 2+3+3+3 - 4 = 7
>
> > And 2^7 = 128 = 13 mod 23.
>
> > Notice that 2 is a factor as are small primes.  The method will try to
> > fit small primes if you are using small values which is about human
> > preference.  The algebra tries to give you what you want based on the
> > size of the numbers you are factoring.
>
> > The value of c is dynamically set by the factorization of q^2 mod N.
> > But the results above indicate that the algebra must give you a
> > factorization within that range which will work.
>
> > And that is what's found with a first blush basic analysis.  It's not
> > clear at this time what further information might result from more
> > basic research.  My aim at this point is to answer criticism against
> > this approach.
>
> > Routinely posters reply requesting I demonstrate by breaking current
> > encryption.  Well, if I could do that I wouldn't need to bother
> > posting on newsgroups now would I?
>
> > It's basic research.  Early stages.
>
> > James Harris
>
> Solve one of the following with your algorithm.

Pulling a key equation from the *original post* that starts this
thread, let's see what analysis can be done.

> 2^m = 16 mod 23
A key equation from my original post is:

q*k^{-(m-c)} + a_2^2*a_3*...*a_c + a_2*a_3^2*...*a_c...+
a_2*...*a_c^2 = a_2*...*a_c[-(m-c)k^{-1} + m] mod N

k = 2, q = 16, N = 107, assume c=4 for ease, first blush gives:

16*2^{4-m} + a_2^2*a_3*a_4 + a_2*a_3^2*a_4 + a_2*a_3*a_4^2 =
a_2*a_3*a_4*(-(m-4)2^{-1} + m) mod 107

I can see m=4, so can work backwards to see what should work!

16 + a_2^2*a_3*a_4 + a_2*a_3^2*a_4 + a_2*a_3*a_4^2 = 4a_2*a_3*a_4 mod
107

So:

a_2^2*a_3*a_4 - 3a_2*a_3^2*a_4 + a_2*a_3*a_4^2 = 91 mod 107

Letting a_3 and a_4 be the free variables, I can set a_3 = a_2^{-1}
mod 107:

a_2*a_4 - 3a_3*a_4 + a_4^2 = 91 mod 107

Going for easy again, let a_4 = 1 mod 107, so:

a_2 - 3a_3 + 1 = 91 mod 107

Yuck. So not as easy to escape the quadratic residue as I thought.
Going back then, and letting a_3 = a_4 = 1:

a_2^2 - 3a_2 + a_2 = 91 mod 107

Yeah! Easy completion of the square:

(a_2 - 1)^2 = 92 mod 107

a_2 - 1 = 29 mod 107

is a solution, so lucked out there, but 50% probability of success.

So a_2 = 28 mod 107.

Then f_2 = 28*2 mod 107 = 56 mod 107, f_3 = f_4 = k mod 107 = 2 mod
107.

Oh, going back to my original post:

a_1+...+a_c = -(m-c)k^{-1} + m mod N

so a_1+28+1+1 = 4 mod 107, so a_1 = 81 mod 107, so f_1 = 81*2 mod 107
= 55 mod 107.

And: 55*56*2*2 = 12320. Kind of big.

12320 mod 107 = 15, which isn't right. It should be 42.

One more a_2 available:

a_2 - 1 = -29 mod 107, so a_2 = 79, so f_2 = 51, f_3 = 2, f_4 = 2

a_1+79+1+1 = 4 mod 107, so a_1 = 30, so f_1 = 60.

60*51*2*2 = 12240, and 12240 = 42 mod 107, as required.

Whew! Still a big number though. It would take 114 iterations to
reach it naively.

And that's with two small prime factors as well.

Well one good thing from this exercise, I made a BIG mistake thinking
I could use a free variable to eliminate the quadratic residue. So
that's a big deal.

Useful exercise.


James Harris
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