How to prove that a hash function is collision resistant if it's equal to that of a collision resistant hash function?

user104469

10/19/23, 9:21 AM

Given that H is a collision-resistant hash function from 2n-bit strings to n-bit strings.

How do I prove that Hash is collision-resistant if:

$$\text{Hash}(X_1∥X_2∥X_3) := \text{H}(X_1∥H(X_2∥X_3))$$

151

1 + 5

collision-resistance

Maarten Bodewes

10/19/23, 9:52 AM

This looks like homework to me. What have you tried? We only give hints in comments on homework questions, but only if the post shows significant effort to solve the question at hand.

Maarten Bodewes

10/19/23, 11:30 AM

I think you need to argue in the sense of $\text{H}(X)$ is collision resistant, which means that $\text{H}(X |\ \text{F}(Y))$ is collision resistant as long as $F$ doesn't have collisions, i.e. produces the same output.

poncho

10/19/23, 1:15 PM

There are two standard ways to answer this type of question (one positively, one negatively): a) if you are given a collision in $Hash$, you can use that to produce a collision in $H$ (and hence if $H$ is collision resistant, so is $Hash$), and b) here's an artificial $H$ which is collision resistant, but for which $Hash$ is not. One of these two methods works in this case - can you figure out which one it is?

Score:1

Crypto

claudiorlandi

10/20/23, 8:37 AM

Usually we prove these things by contradiction. Assume there is an adversary A that finds collisions for Hash. Can you construct an adversary B that finds collisions for H? If so, B contradicts the fact that H is collision-resistant, therefore A does not exist.

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Elon Musk

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