Hi All,

Not too lazy to research..just wanted to know what to expect.

Using...

Processor AMD Athlon(tm) 64 X2 Dual Core Processor 3600+, 1900 Mhz, 2
Core(s), 2 Logical Processor(s)

...with no-assembly C++ code with Knuth's division algorithm for div/
mod

...roughly, how many keys per second can I expect.

All estimates welcome.

TIA,

-Le Chaud Lapin-

My previous posts note the discovery of what I call fundamental
factoring relations where I find that you can get information from a
difference of squares modulo some prime p, where you can pick that
prime. I'd like to generalize that a bit in this post.

Given

z^2 = y^2 + nT

which is what is commonly called a difference of squares you can solve
for z modulo a given prime p coprime to y, z and nT with the following
relations:

z = (2α)^{-1}(1 + 2α^2)k mod p

and

k^2 = (α^2+1)^{-1}(nT) mod p

where α is found such that the relations hold.

Derivation:

To begin I need to start with an equation containing several unknowns,
where p is the odd prime:

1. x^2 = y^2 + pr_1

Now I introduce α and k, where:

2. 2αx = k + pr_2

New article up at the Tech-Investigator regarding Phishing with
Botnets. You can get it at http://tech-investigator.blogspot.com/

People don't taunt me because I'm wrong. They taunt because I'm
right.

A con man is not a con because he has abilities that would give him
the ability to succeed by not betraying people's trust.

They con because of weakness. If they were strong enough, smart
enough, then they could play within the rules.

Cons do not look at the big picture.

These people can destroy big chunks of civilization not because that's
what they want to do, but because they're not smart enough to realize
that's what their actions can cause.

They just are not smart enough to know better.

The choice in front of you now is not really a choice, but a race.

If you lose this race then you lose civilization as you know it.
Don't think about retirement or savings as those institutions as you
understand them may be gone. Don't think about saving up for your
kids educations or about maybe having your own business someday or
maybe just having time for golf.

This race is the biggest one of your lives. If you lose you lose
everything.

Intriguingly to me, or maybe a curiosity to others my latest factoring
research allows you to find the last digits of the factors of a
product of two primes with absolute certainty and do so for even a
public key size number or a public key itself.

Now what I'm showing is a demonstration to let you know that what I've
discovered is like nothing else found before by doing something that
no one else can do.

Of course, finding the last digits of the factors does not give you
the full factorization, but hey, it's a telling way for me to impress
upon you why what I've found is of interest to REAL mathematicians and
cryptographers so those who act not interested, are fakes.

Let z^2 = y^2 + T,

where at this point, of course, z and y are unknowns but I'm going to
solve for them modulo 5.

You will first find what I call alpha and k, where

k^2 = (a^2 +1)^{-1} T mod 5

so you pick alpha so that the quadratic residue exist.

Now you can solve for z, as

z = k(1+2alpha^2)(2alpha)^{-1) mod 5.

:::@@==>> Earth to earth Ashes to ashes dust to dust <<==@@:::

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Hi there!
Is there any other than author's implementation of Khazad?
I need to implement it on Smart Card but cannot directly port their
version due to lack of memory (huge tables...).
Greetings,
pps

Mahaney 1982 proved that if P != NP then there are no sparse
NP-complete sets. Sparse means that the number of strings of length n
is bounded by n**k for some k:

http://weblog.fortnow.com/2006/04/favorite-theorems-small-sets.html

Consider a block cipher like AES-128, where the key size is equal to
the block size. The known-plaintext cryptanalysis problem is: given a
set of plaintext/ciphertext pairs (P1,C1),(P2,C2),...(PN,CN), find a
set of keys K so that for any k in K, E(P1,k)=C1, E(P2,k)=C2, etc.
Obviously there will almost always be just one such K, but maybe
there's occasionally two or three, etc. Still, one would hope that
for a generalized family of block ciphers (for arbitrary sized blocks)
based on reasonable principles, one can prove some polynomial upper
bound. That means that the set K is sparse under the stated
definition.

We might like to imagine that cryptanalysis is NP-hard because we can
draw a boolean circuit for these ciphers that's a big mess, so it
looks like a general SAT instance. Am I missing something, or does
Mahaney's theorem make this reasoning over-optimistic? Where is the
complexity of cryptanalysis likely to really live?

On 3 Jan, 06:07, JSH gmail.com> wrote:

"JSH" gmail.com> wrote in message
news:469b95e6-69e2-4631-bb58-5e8ba244c68d@l6g2000prm.googlegroups.com...

I have also mused that the way in which the Britland and Yankland
governments attempt to sell us these technologies as uncrackable
is because they _ARE_ crackable by those in the know?

For example, if such schemes are almost impossible to crack because
you'd never have enough time to match up the co-operating prime
numbers, how would you ever have enough time to find them out for yourself
in the first place if you wanted to set up your own link?

Answer ..... because you have to go cap-in-hand to the Britland and Yankland
authorities to be handed out your pair in the first place.