Cryptography Stack Exchange is a question and answer site for software developers, mathematicians and others interested in cryptography. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

From what I've found, it is generally accepted a cryptographic hash function like SHA-2 has an evenly, randomly distributed output. Is there a way to test this without running through the entire 2^512 keyspace? Generate a large number of high-quality 512 bit random numbers? What about random 64 character "keyboard ascii" strings?

share|improve this question

Well, for one, SHA-2 (either SHA-256 or SHA-512) doesn't have a 'keyspace'; that's because it doesn't have a key. SHA-2 takes an arbitrary bitstring is input, and generates an output; while there are limits on how long the bitstring can be, those limits are so huge ($2^{64}-1$ bits for SHA-256, $2^{128}-1$ bits for SHA-512), those limits can in practice be ignored.

With that nit aside, your question is 'given arbitrary inputs (that is, inputs that are not specifically selected with the SHA-2 hash in mind), does the SHA-2 hash functions really generate outputs that are evenly distributed?

To test that, you can use any input distribution that strikes your fancy (as long as inputs aren't repeated; that would bias the statistics, and as long as you're not selecting inputs specifically because of what their SHA-2 hashes are). You can certainly consider random 512 bit (or whatever size) bit strings, or random 64 character strings. I would personally advise you to look at "related inputs", that is, inputs that are similar (for example, a 512 bit counter that increments for every hash) -- similar inputs are more likely to stress the SHA-2 function.

Now, one final comment: while you can verify a weakness in SHA-2 with this procedure (although it's extremely unlikely you'll find anything); you cannot verify that SHA-2 is actually well-distributed. Just because the tests you run cannot find anything doesn't mean that a different set of tests wouldn't.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.