0
$\begingroup$

With SHA-2 for hashing, using only the first 128 bits to create a 128 bit hash, what kind of attack would this be vulnerable to and why?

Since 128 bit encryption keys are secure, shouldn't a 128 bit hash also be secure?

Would this be insecure against a brute force attack?

| improve this question | |
$\endgroup$
  • 4
    $\begingroup$ This sounds like a homework question. Think of the birthday problem, for which situation would the birthday problem be an issue? How many bits of security would remain if the birthday problem applies? $\endgroup$ – Maarten Bodewes♦ Apr 13 at 18:35
  • 4
    $\begingroup$ There was a question about What are the consequences of removing a single byte from a sha256 hash?. The same problem with different values. plug and play. $\endgroup$ – kelalaka Apr 13 at 18:46
  • 1
    $\begingroup$ Sometime, birthday attack is considered brute force. But if the problem explicitly asks about brute force attack versus birthday attack, then likely it considers brute force (preimage) attack as: try messages until hitting a given target hash. What is the likelihood of success of that strategy? Is it high enough that it is not doomed in practice? How does it compare to a birthday attack (see my Birthday problem for cryptographic hashing, 101). $\endgroup$ – fgrieu♦ Apr 15 at 4:29
  • 1
    $\begingroup$ The classic reference for implementation of birthday attacks using a practical amount of memory is Paul C. van Oorschot and Michael J. Wiener, Parallel Collision Search with Cryptanalytic Applications, in Journal of Cryptology, 1999. $\endgroup$ – fgrieu♦ Apr 15 at 14:10
  • 2
    $\begingroup$ Birthday attack = collision search. See section 4.1 with "small" = 1. $\endgroup$ – fgrieu♦ Apr 15 at 14:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.