It can be shown that some form of birthday attack will succeed
against any hash scheme involving the use of cipher block
chaining without a secret key, provided that either the resulting
hash code is small enough or that a larger hash code can be
decomposed into independent subcodes.

Question

Quizplus · Accepted Answer

The birthday attack exploits the mathematical principle of the birthday problem in probability theory, which states that in a set of randomly chosen people, the probability that some pair of them will have the same birthday is quite high. This principle can be applied to hash functions in cryptography, where the "birthday" corresponds to the hash value. If a hash function produces a small enough range of values (making the hash code small) or if a larger hash code can be broken down into smaller, independent parts (subcodes), then it becomes statistically feasible to find two different inputs that produce the same output (a hash collision) with relatively little effort. This vulnerability exists regardless of whether cipher block chaining (CBC) is used without a secret key, as the attack focuses on the properties of the hash function itself rather than the encryption mechanism.

It Can Be Shown That Some Form of Birthday Attack