does anyone run ECCp?

[hirschY]

Just wondering if anyone here runs ECCp?
If so, who are ya!
If not, can I post links for those interested?

[lurqa_MU]

What is ECCp?

Regards,
lurqa

[hirschY]

Here is the long winded description
for a shorter verion, just go here http://extremedc.com/eccp/

We are coordinating a distributed effort to solve Certicom's ECCp-109 challenge. The challenge is to solve a particular elliptic curve discrete logarithm problem.
The Elliptic Curve Discrete Logarithm Problem (ECDLP) is the basis for a powerful cryptosystem. The very rudimentary idea of our particular problem is the following:
We have a curve, C, of the form:
y2 = x3 +ax +b

For some constants 'a' and 'b'. We're also given some fixed large prime 'p'. A "point on the curve" is a pair of integers (u,v) that satisfy this equation modulo 'p'. This means, simply, that 'p' divides v2-u3-au-b. Now the leap of faith: There is a method by which we can "add" two points on the curve to get another point on the curve. We call it addition, but it looks nothing like what we normally think of as addition. Just think of it as a rule that tells us how to obtain a third point on the curve from two given points. If you know a little algebra, I'll tell you that these points together with this operation form an abelian group.
Here's the point of the challenge: Certicom has chosen a point, 'P', on this curve and a very big integer 'k'. They then computed Q := kP. This means that they added 'P' to itself 'k' times and called the result 'Q'. But there is a clever way to do this with only log(k) operations. So they can choose a REALLY big 'k', and still compute kP. Now, we know 'P' and 'Q' and our mission is to find 'k'. But 'k' is far too big to simply start trying k=1,k=2,k=3,... so we need to do it in a smarter way. Here are the exact challenge parameters for ECCp-109:
p =564538252084441556247016902735257
y2 = x3 +321094768129147601892514872825668x +430782315140218274262276694323197
P = (97339010987059066523156133908935, 149670372846169285760682371978898)
Q = (44646769697405861057630861884284, 522968098895785888047540374779097)


Why are we doing this and why are you going to participate?

Information security is intimitely linked with current computational power. Solving challenges like this help to give an idea of what current computational power really is. Of course, it's a trivial matter to shop around and see how many MIPS you can get for your dollar. But it's another matter entirely to gauge the power of internet distributed computing. Certicom (and other crypto companies like RSA Security Inc. ) understand this and issue these challenges both to show their confidence in the systems they distribute and to help gauge what can be considered secure with current algorithms and computer resources. Here is Certicom's answer to why they've issued these challenges.

You will participate because you, too, are interested in such things. You also don't want to let your machine go to waste while you're not using it. Oh yeah, and there will be a cash prize. Certicom is offering USD $10,000 for the solution to this particular problem. But before you go nuts spending it, let me describe how this money will be distributed: The solution will be found by exactly two machines. What will happen is we'll have a bunch of machines running around computing points on a particular elliptic curve. When two machines have computed the same point, we'll have the solution, with very high probability. Each of the two people responsible for the final solution will get USD $1000, and the remaining USD $8000 will go to the Free Software Foundation which, among other things, helps to support the GNU Project. If you're not familiar with the GNU project, I encourage you to go check it out. They have loads of great software for free. Indeed, I used much of their software in writing the code for this challenge, so it only seems right to give something back. It's actually quite remarkable that the only pieces of non-free software I needed to develop this project was the Windows operating system to TEST the Windows screen saver and the SEAU program from the good folks over at Gammadyne to create the self-installing archives (This is an awesome program and is available in a shareware version).

For more information on the client software and how to help http://www.team-tnt.net and http://www.extremedc.com/eccp

[High_Jumbllama]

[Pace]

Me thinks I'll continue helping the human race with genome research

P.

[Guru]

Join Team Weazy http://www.team-tnt.net and get in on the fun! For more information on the client software and how to join the team http://www.extremedc.com/eccp

Why in the h*ll should we join you when we - as in MURC - can create our own team???

( And pass your team in process... )

ECCP

Genome

[hirschY]

sorry, didnt realize that part was there, I just cut and pasted

I changed the wording, is that better? Didnt mean to sound that way