How did you calculated that? The minimum security a spent bitcoin address gives you is 128 bits, how did you get to 80 bits?
A quantum computer can crack those 128 bits within few secs, so we don't count them. 160 bits require 2^80 invocations of RIPEMD-160.
I don't think is correct. First there is no quantum computer in existence thats capable of running Shor's algorithm.
Second, I'm pretty sure the 160 bits of security is from only 1 or 2 rounds of the RIPEMD-160 hash.
True , i also dont believe in quantum theory that much.
No it's 160 bits in total, please join the entropy discussion here and share your thoughts:
https://bitcointalksearch.org/topic/bitcoin-entropy-questions-1523431
So is the RIPEMD160 more secure than SHA256, is more quantum resistant?
Isnt that 32 bits = 160 -128 = 32. If it can crack the SHA256 layer in few secs, than all is left is 32 bits.
Isnt that 32 bits = 160 -128 = 32. If it can crack the SHA256 layer in few secs, than all is left is 32 bits.
SHA256 is more secure (128 bits vs 80 bits).
https://en.wikipedia.org/wiki/Grover%27s_algorithm
How long would it take to crack 80 bits? Would the devs have time to swap the ECDSA into something resistant or would all bitcoin become drained instantly?
You seem a bit confused.
There are no known quantum algorithms that can crack a hash function. ECDSA has 128 bits of security
but if it was somehow completely cracked, spent addresses would have no security and unspent still
have 160 from RIPEMD-160. You don't subtract one from the other, it doesnt work that way.
