# Rounds in cryptography

I need to make it clear I know nothing about crypto so in that context I'm hoping to clear up some confusion:

As I understand it a "round" in a cipher is one encryption operation and a cipher like AES has multiple rounds. Did I get this right?

I've seen descriptions of attacks against a cipher say something like it "broke 4 out 8 rounds". In this case, if the plaintext wasn't recovered than how is it determined that any number of rounds were broken?

Thanks

• SHA-1 is broken in the academic sense as we know that the security provided is lower than the 80 bits expected from the algorithm. As yet, not a single collision has been calculated (but we're getting close). So there is a pretty big gap between broken and exploitable. MD5 does have collisions which you can create in milliseconds, but could still be used in e.g. HMAC without too much fuss (but don't). Dec 1, 2014 at 12:27
• It's been a few years since that comment, and now SHA-1 has demonstrated collisions. Crypto is fun!
– forest
Mar 10, 2018 at 10:29

Now we have to mention one important thing: If an algorithm is broken it doesn't mean that someone can decrypt every message encrypted with this algorithm without the key. Nearly every attack on modern algorithms is not about this. There are different kinds of attacks: As example, a known-plaintext attack needs a special amount of plaintexts and corresponding ciphertexts to gain the key. Numbers of $2^{40} = 1\,099\,511\,627\,776$ are low, despite being something like 128 terabyte of encrypted data. A break with this small amount would be considered devastating for AES. For one special attack on AES-256 we still need ca. $2^{100}$ operations - too much in practice, despite being "broken" in the academic sense.