I'm preparing myself to exam, but I have a lot of troubles with rigorous proofs. This post is very long, but this is because I remind here 2 long definitions.

At the beginning I want to remind the definicion of CBC-MAC.

Let $F$ be a pseudo-random function. We define the CBC-MAC as $(Gen,Mac,Vrfy)$ as follows:

(a) $Gen$ on input $1^n$ outputs a key $(k_1,k_2) \in \lbrace 0,1 \rbrace^n \times \lbrace 0,1 \rbrace^n$ selected uniformly at random.

(b) $Mac$ on input a key $(k_1,k_2) \in \lbrace 0,1 \rbrace^n \times \lbrace 0,1 \rbrace^n$ and message $m \in \lbrace 0,1 \rbrace^*$ does the following.

  • It first unambiguously pads the message $m$ to $m^{pad}=m|1|0^l$, where $l \in \lbrace 0,\ldots,n-1 \rbrace$ and $||m^{pad}|| \equiv 0 \mod n$.

  • Then it splits $m^{pad}$ into $r$ blocks of $n$-bit: $m^{pad}=m_1^{pad}|m_2^{pad}|\ldots|m_r^{pad}$.

  • Let $t_0=0^n$. For $i=1,\ldots,r$, it computes $t_i=F_{k_1} \left( t_{i-1} \oplus m_i^{pad} \right)$.

  • Mac outputs $F_{k_2}(t_r)$.

(c) $Vrfy$ on input key $k_1,k_2 \in \lbrace 0,1 \rbrace^n$, a message $m \in \lbrace 0,1 \rbrace^*$ and tag $t \in \lbrace 0,1 \rbrace^n$ outputs $1$ if $Mac_{k_1,k_2}(m)=t$ and $0$ otherwise.

Now I have some exam from previous year, and I don't know at all how to prove this facts (and, it should be done in rigorous way).

Consider the modification CBC-MAC' of the CBC-MAC where the algorithm $Mac'_{k_1,k_2}(m)$

  • selects $t_0$ uniformly at random,

  • computes $t_1, \ldots, t_r$ as in CBC-MAC,

  • and outputs $\left( t_0,F_{k_2}(t_r) \right)$.

The verification algorithm given tag $(t_0,s)$ performs the same computation as $Mac'_{k_1,k_2}(m)$ using the given value $t_0$ and outputs $1$ if $s=F_{k_2}(t_r)$ and $0$ otherwise.

1) Prove that CBC-MAC' is not strongly existentially unforgeable under an adaptive chosen-message attack (definition is below this exercise).

Consider the simplification CBC-MAC'' of the CBC-MAC where the algorithm $Mac''_{k_1,k_2}(m)$ outputs $t_r$ instead of $F_{k_2}(t_r)$.

2) Prove that CBC-MAC'' is not strongly existentially unforgeable under an adaptive chosen-message attack.

3) Is there a condition on the message space of CBC-MAC'' such that your proof does not hold? Explain your answer.


An efficient MAC is strongly existentially unforgeable under an adaptive chosen-message attack if for all PPT adversaries $\mathcal{A}$ it holds that $Pr[MACforge_{\mathcal{A}}(n)=1]$ is negligible function, where $MACforge_{\mathcal{A}}(n)$ denotes the outcome of the following experiment:

(a) A key $k \leftarrow Gen(1^n) \in \lbrace 0,1 \rbrace^n$ is generated.

(b) The adversary $\mathcal{A}$ is given oracle access to $Mac_k(\cdot)$ and outputs a pair $(m,t)$. Formally: $(m,t) \leftarrow \mathcal{A}^{Mac_k(\cdot)}(1^n)$. Let $Q$ denote the set of query/response-pairs $(m',t')$ for all the queries $m'$ asked by $\mathcal{A}$ during its execution.

(c) The output of the experiment is $1$ if and only if $(m,t) \not\in Q$ and $Vrfy_k(m,t)=1$.

If you could help me with the rigorous proof, I'd be really grateful for your time.

Thanks, Ben.

  • 1
    $\begingroup$ For 1), construct an adversary which creates a slightly different message than one used in the oracle calls before, with the same $t_r$, but different $t_0$. How can $\mathcal A$ change this message? $\endgroup$ – Paŭlo Ebermann Oct 28 '12 at 20:17

1) The adversary queries the oracle (with some randomly chosen message $m$) and gets as a result a message $m=m_1|m_2|...$ and its tag $t=(t_0,F_{k_2}(t_r))$. She then draws $\rho$ uniformly at random in $\{0,1\}^n$ and outputs the message $m=\rho\oplus m_1|m_2|...$ and its (valid) tag $t=(\rho\oplus t_0,F_{k_2}(t_r))$.

2) The adversary queries the oracle (with some randomly chosen message $M$) and gets as a result the padded message $M|1|0^l$ and its tag $t=t_r$. She then queries the oracle with the message $t_r$ and gets the corresponding tag $T$. She eventually outputs the (unpadded) message $M|1|0^l|0^n$ and its valid tag $T$.

3) The message space consists of the set of $\{\mbox{bit_length}(S)|S\}$ where S runs through all possible bit strings of length less than $2^n$.

  • $\begingroup$ In 1), there is no reason to draw $\rho$ randomly, any fixed (non-zero) $\rho$ will do. Also, the $m$ and $M$ could be arbitrary instead of randomly chosen. $\endgroup$ – Paŭlo Ebermann Oct 28 '12 at 21:31
  • 1
    $\begingroup$ It is correct that any $\rho\neq0$ will do. However, there was a reason: the adversary in 1) cannot be distinguished from the real signer and that it is straightforward to show it. (Of course, it might also be the case with other ways to choose $\rho$, but harder to prove.) Note that it was never stated in the question that the adversary has to be indistiguishable from the legitimate signer, but this is a nice property (from an attacker point of view) as it shows that forgeries cannot be detected. $\endgroup$ – bob Oct 29 '12 at 8:41
  • $\begingroup$ @bob, will you please explain 3? $\endgroup$ – Thom Oct 29 '12 at 17:37
  • $\begingroup$ @Avery: the constraint on the message format prevents a valid message from being extended into another valid message by a simple append. $\endgroup$ – bob Oct 29 '12 at 17:49
  • $\begingroup$ @Bob, I still don't understand it. What happens when the message is of length > 2^n? $\endgroup$ – Thom Oct 29 '12 at 22:30

@bob has answered 1. and 2.

For 3., you might try proving that it is secure for a message space that is prefix-free (i.e., there do not exist messages $M,M'$ such that $M$ is a prefix of $M'$ and both $M$ and $M'$ are in the message space). I think this is a sufficient condition, and it is broader than bob's answer. Be warned that the proof of security for this case is not easy.


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