Here is a paper that claims to find SHA-1 collisions in 2^52 crypto operations. And optimally secure hash would take 2^80 operations. 2^52 time is still large, but it is getting into cluster and botnet range.
2^80 is if you can use a birthday attack. You can’t use a birthday attack for this, so the difficulty is the full 2^160 bits. Although, if you were trying to crack any one of 1 million (2^20) transactions, you could do a partial birthday attack 2^160/2^20 = 2^140.
Bitcoin Addresses are the only place where 160-bit hash is used. Everything else is SHA-256. They’re calculated as:
bitcoinaddress = RIPEMD-160(SHA-256(publickey))
Correct me if I’m wrong (please, and I’ll gladly eat crow) but I think it would be hard to use an analytical attack on RIPEMD-160 in this case. An analytical attack prescribes a certain range or pattern of inputs to try that will greatly increase your chance of finding a collision. Here, you don’t have that kind of control over RIPEMD-160’s input, because the input is the output of SHA-256. If an analytical attack helps you find an input to RIPEMD-160 that produces a collision, what are you going to do with it? You still have to get SHA-256 to output that value, so you would still have to break SHA-256 too.
For brute force, RIPEMD-160(SHA-256(x)) is no stronger than RIPEMD-160 alone. But for analytical attack, it seems like you must analytical attack both RIPEMD-160 and SHA-256. If I’m wrong, then the strength is the same as RIPEMD-160 and the SHA-256 only serves as one round of key strengthening.
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