Exploring Kangaroo Methods for ECDLP Solutions

Hey everyone, just sharing my research on using kangaroo techniques to tackle ECDLP. Check out Part 1.

Here's the GitHub link: https://github.com/RetiredC/Kang-1. This software showcases different methods for resolving ECDLP with kangaroos.

So the operations needed are roughly K * sqrt(range), where K varies by the method. First method is Classic totally straightforward. You got tame and wild kangaroos, and a collision between them solves the ECDLP.

K is around 2.10 for this approach.

On to Part 2. I’ve got another method for managing kangaroo loops that works for any range and doesn’t increase operations. Only needs a compact list of visited points, and you can code that on GPU too!

That’ll be in Part 3.

Now for Part 3, we got RCKangaroo software for both Windows and Linux. Here’s the link: https://github.com/RetiredC/RCKangaroo. This one features a speedy implementation and advanced loop handling for Nvidia cards.

When's the GPU version dropping?

That's in part #3, so probably not too soon.

Hey all, I struggle with understanding math in C/C++. I use the Fastecdsa Library in Python, then switch to C/C++ for speed. Just wondering about how you define K.

If you can break down that rule of 4 algorithm you mentioned, I’d love to jump in on this convo.

I prefer C++ for speed testing. CUDA is way faster but tricky for research. I only use it at the end.

Added a diagram called 'diagram.jpg' today to help visualize.

Checked the diagram. If X is the pubkey, what’s the elliptic curve process on X when K = 1.15? Thanks!

Look at 'Ec.cpp' for more info. Here’s a link: https://en.bitcoin.it/wiki/Secp256k1.

Can I run this on MacOS?

Only Windows for now, requires Visual Studio with MFC.

Yep, it's old school hardcore C++.

I might be missing something. I get elliptic curve stuff and signature algorithms, but your explanation seems off. Thanks for your work anyway.

I’m proposing a new ECDLP solution needing 1.15*sqrt(N) operations. Check out this paper for methods needing 2*sqrt(N): https://arxiv.org/pdf/1501.07019.