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Given two MAC schemes $\prod_1 = (keyGen_1, S_1, V_1)$ and $\prod_2=(keyGen_2, S_2, V_2)$.

$\prod_3$ runs $keyGen$ from $\prod_1$ and $\prod_2$, respectively, to obtain $(k_1, k_2)$. $\prod_3$, where $S_3 = ((k_1,k_2), (m_1,m_2))$ then runs $S_1(k_1,m_1)\rightarrow t_1$ and $S_2(k_2,m_2)\rightarrow t_2$. And obtain $t_3 := t_1||t_2$. Would $\prod_3$ be a secure MAC?

And the follow up is when $t_3:= t_1 \oplus t_2 $, would this also be a secure MAC?

My guess is that the concatenation is secure because attackers would have no way of knowing how $t_1$ and $t_2$ is generated. For the second one, my intuition tells me that adversary can somehow swap the messages and produce a valid (m,t) pair?

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    $\begingroup$ Hint: What if $\Pi_1=\Pi_2$? $\endgroup$
    – SEJPM
    Commented Apr 18, 2017 at 16:07
  • $\begingroup$ Hint: for either definition of $\Pi_3$, suppose we had a way to break the security of $\Pi_3$, would that allow us to break the security assumptions of $\Pi_1$? $\endgroup$
    – poncho
    Commented Apr 18, 2017 at 16:18
  • $\begingroup$ $S_3$ has a different input message domain than $S_1$ or $S_2$. How exactly are $m_1$ and $m_2$ created when some message $m$ is to be signed? $\endgroup$
    – Artjom B.
    Commented Apr 18, 2017 at 21:54
  • $\begingroup$ @SEJPM, if $\Pi_1=\Pi_2$, then how it is possible to construct $\Pi_3$ that is ever secure? $\endgroup$
    – Bango
    Commented Apr 22, 2017 at 22:30
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    $\begingroup$ Actually, my previous comments were wrong. I thought that if $\Pi_1=\Pi_2$ then you would sample the very same $k_1=k_2$ (which would lead to $t_3=0$). This is obviously wrong because you call the agorithms independently. $\endgroup$
    – SEJPM
    Commented Apr 23, 2017 at 17:51

1 Answer 1

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In the first case, $\Pi_3$ is not secure.

We can use $(m_1, m_2)$ and $(m_3, m_4)$ to query the adversary $A$.

$t_1||t_2 = A(m_1, m_2)$

$t_3||t_4 = A(m_3, m_4)$

The forged MAC would be $t_1||t_4: Verify(t_1||t_4, (m_1, m_4)) \rightarrow Accept$.

In the second case, $\Pi_3$ is not secure.

We can use $(m_1, m_2)$ and $(m_1, m_3)$ to query the adversary $A$.

$t = t_1 \oplus t_2 = A(m_1, m_2)$

$t' = t_1 \oplus t_3 = A(m_1, m_3)$

$t_f = t \oplus t' = t_2 \oplus t_3$

The forged MAC would be $t_f: Verify(t_f, (m_2, m_3)) \rightarrow Accept$.

Another way to break the scheme is to reverse the message pair. $Verify(t, (m_2, m_1)) \rightarrow Accept$.

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  • $\begingroup$ Nice, another question without any answer off the list. Welcome to cryptography! Bango, if you're still around, could you have a look if this answers your old question as it may help others (even if it may be a bit late for yourself). $\endgroup$
    – Maarten Bodewes
    Commented Apr 28, 2019 at 23:29

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