Given an hash H that is not collision resistant, for example 80-bit digest, if we use the following double hashing scheme: H(SHA2-256(x)).
Does this scheme increase the collision resistance?
Cryptography Stack Exchange is a question and answer site for software developers, mathematicians and others interested in cryptography. It only takes a minute to sign up.
Sign up to join this communityDoes this scheme increase the collision resistance?
No (at least, not if we assume that $H$ is not collision resistant because of the limited output size); the standard birthday attack (where you compute $2^{40}$ hashes, and look for a common value) still works. The only thing the initial SHA-256 does is make evaluating the hash a bit more expensive - it doesn't frustrate the attack in any other way.
And, in case you're wondering, we know how to search for such a collision without a memory that can store $2^{40}$ hashes...