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Finding second-preimage to this use of weak hash

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ashidc

$h'(x) := h(a_1 \parallel x \parallel b_1) \parallel h(a_2 \parallel x \parallel b_2) \parallel h(a_3 \parallel x \parallel b_3) \parallel \dots \parallel h(a_k \parallel x \parallel b_k)$

$a_i$ and $b_i$ are known prefixes and suffixes.

If $h$ is MD2 or MD4, how much work would it take to find a second-preimage to $h'$?

Just to be clear: The output of $h'$ is relatively long, $k$ times longer than the output of $h$.

Answers to the same question but with an even weaker cryptographic-hash, may also be of interest. But not as weak as a checksum.

Naively I'd thought that if it requires $2^n$ work to find second-preimage of $h$ then it require $2^{kn}$ work to find second-preimage to $h'$. Since you need to find a specific $x$ that satisfied second-preimage to all $h(a_i \parallel x \parallel b_i)$. But maybe there is a faster way.

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md4
kelalaka avatar
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kelalaka
What are you trying to achieve? Why do you need improve MD2 or MD4? Why don't you just use modern hashes?
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