# Is SHA-1 collision free on data up to 20 bytes long? [duplicate]

Is SHA-1 collision free on data up to 20 bytes long (lenght of hash / internal state)? That means that every input produce unique output, but you surely know that, i just write it in order my question to be accepted by the site :)

If yes - which other hash functions does it apply to and what is the corresponding maximal length?

## marked as duplicate by CodesInChaos, mikeazoFeb 28 '13 at 20:25

• With the way we typically construct symmetric primitives the fastest way to prove it, is finding a collision with these properties, and that requires $2^{80}$ work. What's clear is that an ideal 160 bit hashfunction does have collisions with length 20 (with overwhelming probability), and we have not the tiniest amount of evidence suggesting that SHA-1 so broken that it doesn't have this property. – CodesInChaos Feb 28 '13 at 16:13
• @mikeazo 1) I used exist in the mathematical sense, where you can show that something exists, without being able to actually construct it. (If the claim that there are collisions is actually true, then there exists a short proof for their existance) 2) Not with probability 1. You need $2^{160}+1$ in the worst case. But it's quite unlikely it will need more than $2^{90}$ operations. – CodesInChaos Feb 28 '13 at 20:23