0

I have one question. I'm trying to lower complexity when breaking private keys for BTC.

If we have that an address is a hash160 from priv key, then we can assume that we have 2^96 per each address.

If we have, let's say, 10M of active wallets (~2^23.25). Then, 160-23.25=136.75 we can assume that wallets are uniformly distributed on 2^136.75 gaps.

If we start to find those wallets from those assumptions, can it work well?

I mean, supposedly this distribution should follow the Gaussian function, or am I wrong?

normal distribution

Which methods else do I have to lower complexity?

Why does this guy say 2^80? https://bitcoin.stackexchange.com/a/83819/131965 can somebody explain?

1 Answer 1

1

Why does this guy say 2^80? can somebody explain?

That is explained in the comments to the question you linked.

To paraphrase

As numbers become large the number resulting from the birthday paradox approaches the square root. With Bitcoin these numbers are so large that there's nothing to be gained by trying to be more precise.

Original comments reproduced here for convenience

the actual formula for the birthday paradox is complicated, but as the numbers get bigger, taking the square root becomes an increasingly better approximation to it. When talking about cryptography, the numbers are so huge that the difference isn't worth worrying about. –
Mark Jan 20 '19 at 1:17

Technically, the square root of the number of possibilities gives you a 39.3% probability for collisions (1-exp(-1/2)), that's about where you see the probability go from negligable to negligably not. 1/2 sqrt(n) gives you 11.75%. 2 sqrt(n) gives you 86.5%. In cryptography the exact factor is usually ignored, and we just say that you need approximately sqrt(n) elements to see a probable collision. –
Pieter Wuille Jan 20 '19 at 1:17


we can assume that wallets are uniformly distributed on 2^136.75 gaps.

I'm no statistician but, to me, a uniform distribution does not imply fixed length gaps. I suspect the ratio can only inform you of the probability that a search of a smaller space will yield an expected number of results.


Related:

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.