Create your own MD5 collisions

Question

I\'m doing a presentation on MD5 collisions and I\'d like to give people any idea how likely a collision is.

It would be good to have two blocks of text which hash t

Answer 1

The whole point of such hashes is that collisions are extremely unlikely. You're not going to generate one by chance--your machine will almost certainly die of old age before you succeed. The whole point of using a hash would go away if you could reasonably generate collisions!

Answer 2

These following two different 128 byte sequences hash to the same:

MD5 Hash: 79054025255fb1a26e4bc422aef54eb4

The differences below are highlighted (bold). Sorry it's kind of hard to see.

d131dd02c5e6eec4693d9a0698aff95c 2fcab58712467eab4004583eb8fb7f89 
55ad340609f4b30283e488832571415a 085125e8f7cdc99fd91dbdf280373c5b 
d8823e3156348f5bae6dacd436c919c6 dd53e2b487da03fd02396306d248cda0 
e99f33420f577ee8ce54b67080a80d1e c69821bcb6a8839396f9652b6ff72a70

and

d131dd02c5e6eec4693d9a0698aff95c 2fcab50712467eab4004583eb8fb7f89 
55ad340609f4b30283e4888325f1415a 085125e8f7cdc99fd91dbd7280373c5b 
d8823e3156348f5bae6dacd436c919c6 dd53e23487da03fd02396306d248cda0 
e99f33420f577ee8ce54b67080280d1e c69821bcb6a8839396f965ab6ff72a70

The visualization of the collision/block1 (Source: Links.Org)

The visualization of the collision/block2 (Source: Links.Org)

Answer 3

I would take a look at Hashcash. With an effective hash algorithm, like md5, the time to calculate a collision to exponential with the number of bits. What Hashcash does is calculates partial collisions. That is, a match of say the lower 16 bits of the hash. To get the lower 16 bits to match, one would have to try hashing 2^15 different combinations on average. If you know how long it takes to come up with a 16, 24, or 32 bit collision, then you can easily calculate out the time for higher numbers of bits.

Answer 4

It's hard to do it with just text files, AFAIK. You can get some collisions, but having them also be from just [a-zA-Z] is not easy (yet).

On the other hand, if you just want two "meaningful"-looking files with the same hash, you can do it with something like, say, PostScript: have different binary blobs causing the collision, and use a conditional expression to display different output accordingly.

See e.g. this problem (the H2 part) and solution. For example, this PS file and this one have the same MD5sum but they are both well-formed PostScript files that have entirely different text in them when you open them.

Answer 5

If you're talking about how likely a straightforward collision is - one where there is no deliberate attempt to cause one - then you're going to be disappointed: You'd need to generate on average 2^64 plaintexts before you can expect to see a collision, and that's substantially more than you're going to be able to do in a reasonable (or really, even an _un_reasonable) time.

If you're looking to demonstrate the difficulty of deliberately crafting a collision, other answers have already demonstrated that. The extra constraint of requiring the strings to be entirely textual makes even those approaches largely impractical, though.