Tony Arcieri 🌹🦀 on Nostr: For anyone worried about 31 (or 64) steps/rounds of SHA-256 being broken, it's ...
For anyone worried about 31 (or 64) steps/rounds of SHA-256 being broken, it's actually not that surprising of a result.
This 2013 paper accomplished something similar, but for a reduced-round SHA-256 with 28 steps: https://eprint.iacr.org/2015/350
It further notes:
> Using a two-block approach we are able to turn a semi-freestart collision into a collision for 31 steps with a complexity of at most 2^65.5
So further breakages at 31 steps are somewhat expected.
This 2013 paper accomplished something similar, but for a reduced-round SHA-256 with 28 steps: https://eprint.iacr.org/2015/350
It further notes:
> Using a two-block approach we are able to turn a semi-freestart collision into a collision for 31 steps with a complexity of at most 2^65.5
So further breakages at 31 steps are somewhat expected.