On Fri, Dec 27, 2013 at 1:59 AM, S. Dale Morrey <sdalemor...@gmail.com> wrote: > So here's the problem... > > I'm exploring the strength of the SHA256 algorithm. > Specifically I'm looking for the possibility of a hash collision. >
Man, talk about a lot of unhelpful answers. You don't need an answer to your question, you need to rethink what you're doing entirely. Hash collisions are always possible when the set of outputs is smaller than the set of inputs. You don't need to run any programs to know this. Furthermore, the algorithm is fully deterministic, so the probability of collisions is fixed for any fixed input parameters. But considering the fact that the algorithm was designed to make hash collisions as unlikely as possible, you should realize that the collision probability is going to be vanishingly tiny. And even if, by some cosmically unlikely chance, your 24 million random strings happen to have a pair of distinct strings that hash to the same value--you still have not really explored the strength of SHA256. The strength is a probabilistic thing, and finding a collision through sheer luck does not affect the probability at all. The strength of a hash function is determined by the evenness of its distribution of outputs for given inputs. A perfect hash maps its inputs to the full range of possible outputs completely evenly, with no output more likely than any other. For a hash function with 256 bits of output, the number of output buckets is 2^256, which is an astronomically huge number. I'll leave it to you to calculate the probability of generating a collision with 24 million samples given the assumption of a perfectly even distribution, but I think you'll find it's not worth trying. If you are unfamiliar with the math behind the test you're trying to run, you should take a look at Wikipedia's article on the Birthday Paradox (http://en.wikipedia.org/wiki/Birthday_Paradox) for an introduction. Don't just read it; run the numbers. Figure out how much RAM, CPU time, and hard drive space you'd need to have a reasonable chance of finding a collision (assuming SHA256 has a perfectly even output distribution). It will be a much better use of your time than what you're doing now, and it might even lead you to some ideas that could actually approach the problem of measuring the strength of SHA256 instead of just wasting electricity. --Levi /* PLUG: http://plug.org, #utah on irc.freenode.net Unsubscribe: http://plug.org/mailman/options/plug Don't fear the penguin. */
