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Assume there are two keyed hash-functions $H_1(k_1, m)$ and $H_2(k_2, m)$, with $k_1$ and $k_2$ being independently randomly sampled public keys.

The XOR-combiner is defined as $C_\oplus^{H_1, H_2}:=H_1(k_1, m) \oplus H_2 (k_2, m)$.

Assuming at least one of $H_1$ and $H_2$ is collision resistant, is $C_\oplus^{H_1, H_2}$, as defined above, necessarily collision-resistant for any combination of $H_1, H_2$?

Mittelbach (2014) says: "The exclusive-or combiner is also not robust for collision resistance, even assuming independent functions, as a collision on the combiner does not require collisions under both input functions.", but he's talking about Hash-functions without keys. He also does not provide a proof or counterexample which I could attempt to apply for this problem.

Here's my approach which unfortunately did not lead anywhere:

When attempting a reduction proof, I realized that if an adversary $A$ finds a collision against $C_\oplus^{H_1, H_2}$, that means that $H_1(k_1, m_0)\oplus H_2(k_2,m_0)=H_1(k_1, m_1)\oplus H_2(k_2,m_1)$. However, as Mittelbach (2014) stated, this does not necessarily imply that $H_1(k_1, m_0)=H_1(k_1, m_1)$ or $H_2(k_1, m_0)=H_2(k_1, m_1)$, because it is possible that $H_1(k_1, m_0)\oplus H_1(k_1, m_1)=H_2(k_2, m_0)\oplus H_2(k_2, m_1) \neq 0$.

Unfortunately, I couldn't find a counterexample for $H_1, H_2$ for which I could construct an adversary that breaks collision resistance, leaving me stuck in both attempts of proving and disproving the collision-resistance of $C_\oplus^{H_1, H_2}$.

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  • $\begingroup$ What is the origin of this question? The answer is not simple see link.springer.com/article/10.1007/s00145-019-09328-w $\endgroup$
    – kelalaka
    Jan 3 at 15:55
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    $\begingroup$ @kelalaka It's an exercise for an introductary cryptography course... I'll have a look at the article you linked, thank you! $\endgroup$ Jan 3 at 17:39
  • $\begingroup$ If $H_k$ is collision resistant, then $H'_k(x\Vert b) = H_k(x)\Vert b$ is also collision resistant. (left as an exercise to the reader) What happens if you instantiate both hash functions with $H'$? $\endgroup$
    – Maeher
    Jan 3 at 19:44
  • $\begingroup$ I'm assuming $b$ refers to a single bit? Then $C_\oplus^{H_1', H_2'}=H_1(k_1, m)\|b\oplus H_2(k_2, m)\|b=(H_1(k_1, m)\oplus H_2(k_2, m))\|(b\oplus b)=(H_1(k_1, m)\oplus H_2(k_2, m))\|0$. How would you construct an adversary against this without solving the original problem? $\endgroup$ Jan 3 at 23:02
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    $\begingroup$ Well, are $m\Vert 0$ and $m\Vert 1$ for some fixed $m$ the same message? Do they have the same hash value? $\endgroup$
    – Maeher
    Jan 4 at 12:57

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