Re:How Hash Algorithms Work (2007)
by comicjk on Friday January 13, 2017 @08:48AM
(#13390706)
> For "words" to generate a collision, you'd have to increase the input domain by allowing "words" to mean any sequence of bytes (e.g. bytes of jpg image or audio file).
That is way overkill - the input domain could just be strings of words. Each word of a typical English text adds about one byte of entropy (2^8 states). We get a probable collision by having a number of text states around the square root of the number of hash states (because of the birthday paradox). So, sqrt(10^48) = 10^24 = (10^3)^8 which is about (2^10)^8. So the space of ordinary 10-word strings is big enough to give a sha-1 collision, without invoking inputs like binary files.
That is way overkill - the input domain could just be strings of words. Each word of a typical English text adds about one byte of entropy (2^8 states). We get a probable collision by having a number of text states around the square root of the number of hash states (because of the birthday paradox). So, sqrt(10^48) = 10^24 = (10^3)^8 which is about (2^10)^8. So the space of ordinary 10-word strings is big enough to give a sha-1 collision, without invoking inputs like binary files.