Re: Solving the factoring problem

On Jan 26, 1:41 pm, Vend <ven...@xxxxxxxxxxx> wrote:
On Jan 26, 5:00 pm, JSH <jst...@xxxxxxxxx> wrote:



On Jan 26, 1:35 am, Vend <ven...@xxxxxxxxxxx> wrote:

On 26 Gen, 01:46, JSH <jst...@xxxxxxxxx> wrote:

On Jan 25, 4:35 pm, JSH <jst...@xxxxxxxxx> wrote:

On Jan 25, 10:53 am, Vend <ven...@xxxxxxxxxxx> wrote:

On 25 Gen, 07:13, JSH <jst...@xxxxxxxxx> wrote:

Previously I've noted the easy derivation of some remarkably simple
congruences that give f_1 mod p and f_2 mod p, 75% of the time, when
f_1*f_2 = nT, where T is a composite you wish to factor, p is a prime
of your choice and n is a non-zero integer of your choice:

How can they be true 75% of time? As far as I know, mathematical
relations are either true or false.

For 75% of the combinations of integer factors f_1 and f_2, the
congruences will be true, as in you will get integer solutions.

For the remaining 25% some of the solutions are non-rational.

What do you mean by "75% of the combinations of integer factors"?
It's impossible to sample integers at random with uniform probability.

But you can choose p at random, and depending on what p you choose you
can get a lot of different answers.

So you are part of the randomization.

Please explain which variables are chosen at random and from which
probability distribution and which variables are under an universal
quantificatior since it's not clear.


You choose T as that's the target composite to factor, and you choose
other variables in order to factor T, but your choice is not exactly
predictable, right? As you're a human being and you can be quirky.

The best choice if possible for p is a prime p such that p>sqrt(T),
but the minimal prime that satisfies is what you SHOULD want unless
you want something else.

But the pressure is on me to give a method that can factor RSA public
keys--and do it fast and easy--so I've put more in there which is all
about factoring RSA public keys or numbers on that scale or bigger.

Then it's not feasible to do a search for a's that will work with very
big primes and n=1 or something small, so instead I've explained a way
to use small primes, where you can loop through to find every possible
'a' that can work, which gives you k, so that is just set.

However, in trying to factor you need to be sure you have a non-
trivial factorization, so I talk about using n=1, and then n=2, and
n=3, and then n=5, and using more primes in succession so that you can
isolate out a non-trivial factorization.

Now then, for any particular choice of p and n, the factoring
congruences will give you solutions on average for about 75% of them.

So like if you have some composite that has 4 unique factorizations,
on average, you'll get an answer for 3 of them from the factoring
congruences, where the trivial factorization is included.

But you may get more and you may get less as it's a probability.

And that explains everything you need to know about how probability is
a factor in making things random.

It's fun and easy math.


James Harris
.

Relevant Pages

  • Re: Solving the factoring problem
    ... probability distribution and which variables are under an universal ... You choose T as that's the target composite to factor, ... about factoring RSA public keys or numbers on that scale or bigger. ... congruences will give you solutions on average for about 75% of them. ...
    (sci.crypt)
  • Re: I was right, surrogate factoring proof
    ... primes divide n -- at most I'll have to look through sqrt/ ... But don't forget that's included in the cost of the algorithm! ... if surrogate factoring ever becomes ... two values are congruent modulo p, and hopefully not congruent modulo n/p. ...
    (sci.math)
  • Re: I was right, surrogate factoring proof
    ... primes divide n -- at most I'll have to look through sqrt/ ... But don't forget that's included in the cost of the algorithm! ... if surrogate factoring ever becomes ... two values are congruent modulo p, and hopefully not congruent modulo n/p. ...
    (sci.crypt)
  • Re: JSH: Contradictory behavior, issue of math fraud
    ... But if the idea turns out to be a brilliant one which means factoring ... No one has said math journals routinely publish ... current methods in speed for counting primes. ... about being a genius. ...
    (sci.math)
  • Re: JSH: Now were golden!
    ... factoring problem, and I want to definitely thank Enrico and would ... which the negative of the quadratic residue is a quadratic residue and ... they are the primes for which the negative of their quadratic residue ... It's a duty. ...
    (sci.crypt)