[hobofan]

I mean, technically yes, but it it would be a 128-bit hash function with the security properties of a 64-bit hash function, so it offers little advantage over just using a 64bit hash (which I think was also the point you were trying to make, I think?).

However, that doesn't really address the original question on how much harder cracking 2x64bit hash would be than cracking a single 128bit hash would be.

My best guess there would be that it's really quantify as you start opening up more dimensions besides the number of bits. The gain would mostly come from protection against other properties from one of the algorithms like a potentially hidden backdoor or a undiscovered mathematical weakness. So as long as the strength of the individual hash function holds up, it probably makes sense to diversify between hash functions. E.g. SHA3-256 + BLAKE3-256, probably offers better long-term security properties than using SHA3-512.

[dependenttypes]

> but it it would be a 128-bit hash function with the security properties of a 64-bit hash function

This is not true. Consider two hash functions f and g

    f(x) = md5(x)[0..63]
    g(x) = md5(x)[64..127]

and a third function

   h(x) = f(x) || g(x)

where || is concat

So no, concating multiple smaller hash functions is not any weaker than using a single big one.

[hobofan]

So your point is that if you take the output of the _same_ 128bit hash function twice, split it into 64bit parts and then put it back together, you still have the full properties of the 128bit hash function? Well, no shit.

I have to admit that I'm not the greatest cryptography whiz, but I can't image that this holds up for _independent_ hash functions, where you should be able to more cheaply run a preimage attack against two 64bit hash functions than one 128bit hash function.

[gruez]

>I have to admit that I'm not the greatest cryptography whiz, but I can't image that this holds up for _independent_ hash functions, where you should be able to more cheaply run a preimage attack against two 64bit hash functions than one 128bit hash function.

performance is probably an issue. SHA256 has 256 bits and SHA1 has 160 bits (1.6x more bits), but SHA256 isn't 1.6x slower, it's only 38% slower. benchmarks used: https://www.cryptopp.com/benchmarks.html

Back to the original question of "how secure are 2x 64bit hashes compared to 1x 128 bit hash?", I can't imagine how it could be any more secure, considering that if it were more secure, you could just make your 128 bit hash function be the concatenation of the two 64 bit hash functions. It might be equally secure, but I'm not sure why you'd use it over a properly designed 128 bit hash function.

[dependenttypes]

> where you should be able to more cheaply run a preimage attack against two 64bit hash functions than one 128bit hash function.

Try doing that in my example.

[freemint]

Your hash functions are contrustructed in a way such that the concatenation has the security properties of a 128 bit hash function (because your construction is equivalent md5). I am not sure that results holds for all concatenations of hash functions. Or for SHA concatenations.

I appreciate you making me think deeper about hash functions. Nice construction.