Re: JSH: What are you people?

On Jan 23, 9:23 am, Rotwang <sg...@xxxxxxxxxxxxx> wrote:
On 23 Jan, 15:13, JSH <jst...@xxxxxxxxx> wrote:

On Jan 22, 10:42 pm, Rotwang <sg...@xxxxxxxxxxxxx> wrote:

I believe that this is false.
Your belief is irrelevant.
Then so is yours. But what aren't irrelevant are the results of my
You sound petulant.
And you sound asinine. But I was too polite to point it out.

Belief does not change the results.

I don't know why that is an arguing point with you.

Good grief, you're slow. Look: it's not an arguing point. But because
I'm actually responsible when it comes to information, unlike you, I
am not in the habit of stating things with certainty unless I have
checked them thoroughly. The end result of this is that I will
frequently prefix statements which I have not checked thorougly with
qualifiers such as "I believe", lest people give my statements more
weight than they deserve. That you jump on this as an opportunity to
spout vacuous slogans such as "your belief is irrelevant", as if the
fact that I'm not in the habit of claiming time and time again to be
certain of things which are false is something for which I should be
insulted, makes you look like a ***. Just so you know.



And your program has problems.

program, if I didn't make a mistake (and if I haven't once again
misinterpreted what your claim is supposed to be - could you go to the
trouble of confirming my above guess?). If you don't believe the
results I gave you can run the program yourself. Here is the Python
script:

def Src(Q,R):
yet = 0
yen = 0
tot = 0
for f in Q:
for g in Q:
if g >= f:
for p in R:
for a in range(1,p):
for k in range(0,p):

Why would you loop from 0?

Why not? I don't recall that you specified that k should be non-zero.
But no matter: changing the program to loop from k = 1 and running
again with the aforementioned inputs gives ... exactly the same
results.

That indicates you never bothered to see what the variables actually
mean mathematically.

For instance, f_1, a factor of the target time T times n, as f_1*f_2 =
nT, is given by f_1 = ak, so why would you make k=0?

Maybe because it works? Specifically if p is a factor of T then k = 0,
a = anything will give you a non-trivial factor of T. In fact, in the
cases where I ran the program with the inputs Q and R sharing primes,
the overwhelming majority of correct values for f1 mod p were found in
this manner.

And since you accused me of having not bothered to see what the
variables mean mathematically, I think it only fair to point out that
if you were paying attention you would have realised that including k
= 0 would make no difference to the results I mentioned, since the
only case where k = 0 satisfies k^2 = (a^2 + 1)^(-1)T mod p is the
case where p divides T, which it never did for the inputs I specified.


It's still sloppy.


And why would you loop to p?

Am I looking at that correctly in terms of how the looping goes?

No. It loops to p - 1.



I ask as I don't program in that script.

An easy mistake to make, I think that Python is unusual in this
regard.


Ok, no prob.

What are you thinking here?

That lack of thinking is problematic.

It wastes other people's time.

I would find it very difficult to waste your time any more efficiently
than you waste it yourself.


I'm having fun. I don't think that having fun is considered by MOST
of humanity to be wasting one's time.

The issue is testing out these ideas.

I need help doing so, and some of you feel an ego need to insult, so I
feed that need, by giving you the opportunity and in return I get my
ideas tested out.


if ((a*a+1)*k*k)%p == (f*g)%p:
tot = tot + 1
if (a*k)%p == f%p or (a*k)%p == g%p:
yen = yen + 1
elif (a*k)%p == 1 or (a*k)%p == (f*g)%p:
yet = yet + 1
print yet, yen, tot

The function takes two arguments; the first is a list of primes from
which the factors of T are drawn, and the second a list of primes from
which p is drawn. tot counts the total number of (a,k) pairs found,
yen the number of those which give f1 equal mod p to a non-trivial
factor of T, and yet the number of those which give f1 equal mod p to
a trivial factor of T. I just ran it several times with Q =
[431,677,967,1187,1429,1621] and R = [683], [971] and [1433]. Each
time it actually gave *fewer* f1's which were equal mod p to a factor
of T than you would expect to get by simply picking f1 at random from
the set {1,2, ... p-1}. That's quite a bit less than 75%.

Fix the error I found and then I'll consider the rest.

I wouldn't be surprised if I found more.

And I did.

Care to point them out?


Sloppy stuff, already mentioned.

For you of all people to make this complaint is quite ridiculous. I
notice that you *still* haven't confirmed my guess as to what your
"theorem" is supposed to mean.

It's EASY ALGEBRA.

Pushing the notion that there is this difficult to understand
"theorem" is just wacky.

Yes, that's nice. But you *still* haven't confirmed my guess as to
what you mean by your atrocious technical writing. A simple yes or no
would suffice, and has the added advantage of saving wear and tear on
your caps lock key.


But how hard can it be? Just look over the math.

It is easy algebra as I keep emphasizing.


The proper answer is that the total number j of primes it would take
is less than m such that T/m! < 1.

No, the proper answer is what I said before: the product of the primes
should be greater than sqrt(T). The number you give is actually an
overestimate.

Yup, I prefer an overestimate to shut down people like you trying to
make it seem like there would be a lot of primes.

RSA numbers usually have at least 1024 bits. The smallest number of
distinct primes whose product is larger than 2^1024 is 132. 25% of 132
is 33. That means that, even if your 75% estimate were correct
(*which* *it* *isn't*) you would be attempting to reconstruct a factor
of T from its values mod p_n, with the expectation that 33 of those
values were wrong. So what do you do? Assume that 99 of them are
correct and try every combination of 99 chosen from 132? There are
more than 10^31 of those, and each one only partially narrows down the
candidates for trial factors < sqrt(T). That doesn't sound to me like
a trivial solution to the factoring problem.


It is if you can figure out easily which factors associate with your
target, where I thought my idea was brilliant which is just to shift n
in order to do so but I fiddled with it in a post on my blog and got a
result that was against that theory.

I'm puzzling over it.

trivial factorisations (which you need to do separately for each n
since you will need to find different (a,k) pairs for each n), you
have the problem that you don't know which of f1 mod p_n and f2 mod
p_n is which, nor which of the 25% of your f1 mod p's don't correspond
to a factorisation at all. Though this might still lead to an

Just multiply times 2, or 3 or some other small prime to pull out
valid factorization eliminating concerns about the 25%.

I'm afraid I don't follow this at all. Can you explain in more detail
how this is supposed to work?

If you factor T, with n=1 and a small prime, you can come back and do
the same for 2T or use any other prime as the prime can associate with
only one factor at a time, right?

Er, no? Different values of a and k will give different values for f1
mod p, and there are plenty of them available for a given p and nT.


YES, BUT if you have made n a prime of your CHOICE then you can just
divide it off from the solutions since you have given a factor.

IT IS TRIVIAL!!!

I thought that was it but tried it out with a simple example and the
damn equations associated with a non-rational solution.


But KNOWING ahead of time what one of the factors is you can just use
its modular inverse to remove it from your results and if you get the
same answer as before, guess what?

Pardon? How does this help? If you factor T you're done, but factoring
T is the hard part - you *don't* know ahead of time what one of the
factors is.


You factor nT.

You clearly are not paying attention at all.

z^2 = y^2 + nT

So you can set n=1, get an answer for a small prime, and then set n=2
or some other prime and get another answer and compare since you can
remove the factor you added in with n.

That should work but I tried an example and got the non-rational
solution in two places though I also in the last case got the rational
factorization which didn't show for n=1 and n=2, but finally did for
n=3.

I was using T=185 and p=17.

That's problematic if it's common.

Otherwise that is a nifty idea for finding out which solutions
correspond to the factorizations you want.

That still leaves the problem of which is which between different
primes, as I'm not certain that the f's keep the same factor, though
with my test there wasn't an opportunity to see.

Oh well, it is basic research.


James Harris
.