BSGS requires a public key (e.g. puzzles 135, 140, 145, 150, 155, 160, or a signed message or TX for all other puzzles).

IDK what you're doing but BSGS is fully deterministic, unless you run multiple smaller BSGS over (random) subintervals, or you tradeoff the 100% chances to find the key with a smaller memory footprint (potentially missing the solution). But before all, you should answer yourself why you want to use BSGS in the first place. For 135 there might not be enough storage on our planet to even finish off iterating through the baby steps, let alone enough fast RAM to store all of that, to run through the giant steps. At each giant step, a lookup must be performed in the table, but if you can't have the table (or it is not a fast table, which means that all and any of it needs to be directly accessible instantly, not using any network, cloud, cables, compression, etc.) you're at a dead end.

then can you explain - random values are generated, and each of them is checked for a match using the public key?

maybe you have a version of kangaroo (or brute force) for GPU and CPU (for comparing addresses and public keys), which would iterate in a row with a given step (example 1+1+2+3+4..etc) ?
to make it easier for me to figure out how to change the step logic. I want to test my theory.

Nikola Tesla lucky number 369
Page 369 appear before and my post was 369 too and about this lucky number 369
After my post 369 Late night, mod shuffle and maybe removed unused post and page appear 367

At most, I will solve puzzle 67 by September this year. Everyone is waiting for my good news.

I'm just leaving a comment here on page 369 - as I was waiting for a long time to see what comes at 369 - I have been so obsessed with this number for a long time.

Guys, while flipping randomly through my 3 notebooks full of OCD-induced EC investigations, I found on one corner the proof on how ECDSA is broken. Unluckily, my cat made confetti out of the rest of the page, and I don't remember squat about how I anded up at that result. All I could recover is that it has to do with inter-dimensional mathematics, specifically fractal theory.

For those unaware, fractals are a cutting edge area of mathematics, since they behave outside of the normal framework of dimensionality (such as points on a line, 2D or 3D shapes). They were developed in the 80s, so they are actually newer than elliptic curves. Since they were impossible to analyze analiticaly due to their immense complexity and recurrent relations, they only made sense once the first computers started to became mainstream in the science community.

In short, there is a way to map Weierstrass curves over to some fractal (of course, not the same one). The fractal can then be looked up and zoomed-in to analyze a particular range. Depending on the intensity of the particular XY (which is simply the fractal recurrent formula landing there), we can derive back private keys. Unfortunately, while my proof seemed to work out OK, the part of the paper that contained the formula for the mapping was lost (but at least the cat had an happy hour).

Note: this doesn't work with addresses.

share the code))) бpo)

or write to me in private messages, I have a couple of questions.

I've been analyzing the puzzle formation algorithm for two years. Just look at this information. After you figure this out, I'm ready for further dialogue.
https://i.postimg.cc/dVHh5k8h/XLS.jpg

It's part of the algorithm, there's another part, but what's the point of it all if the bots take the entire reward?

Dear Sir first try to know your RAM processor unit quantity your device generation and graphics cards then you can check it by making random code like this 
import ecdsa
import time

def generate_key():
    sk = ecdsa.SigningKey.generate(curve=ecdsa.SECP256k1)
    return sk.to_string().hex()

start_time = time.time()
num_keys = 1000  # number of keys to generate for testing
for _ in range(num_keys):
    generate_key()

end_time = time.time()
speed = num_keys (end_time - start_time)
print(f"Generated {num_keys} keys in {end_time - start_time:.2f} seconds.")
print(f"Speed: {speed:.2f} keys per second.")