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I have to solve this exercise and I really could use some help:

Show that any 2-round key-exchange protocol (that is, where each party sends a single message) satisfying $Definition$ $9.1$ can be converted into a public-key encryption scheme that is CPA-secure.

$Definition$ $9.1$:
A key-exchange protocol II is secure in the presence of an eavesdropper if for every probabilistic polynomial-time adversary $A$ there exists a negligible function negl such that

$Pr [KE^{eav}_{A,\pi} (n) = 1] \le \frac{1}{2}+ negl(n)$

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    $\begingroup$ Hint: Look at DH and ElGamal / ECIES / DHIES. $\endgroup$ – SEJPM May 30 '19 at 8:43
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The methods suggested in the comments are useful in practice, but they base on some specific key exchange protocols. It's useful to proove that a public key encryption scheme can be created from any such $KE^{eav}$-secure protocol.

Notation

Let $\langle s_A, m_A \rangle, \langle s_B, m_B \rangle$ denote the state and transmitted message from Alice and Bob within the protocol respectively. Let $Prot_A(1^n) \rightarrow \langle s_A, m_A \rangle$ denote the process of Alice creating her secret state and the message to be transmitted to bob, $Prot_B(m_A) \rightarrow \langle s_B, m_B \rangle$ be the process of Bob sampling his random state and creating the message based on the random state and Alice's message. We then define the key to be $f(s_A, m_B) = k_A = k = k_B = f(s_B, m_A)$ for $f$ some predefined function.

Construction

Now, we create a public-key encryption scheme as follows:

\begin{align*} Gen&(1^n) \\ & \langle s_A, m_A \rangle \leftarrow Prot_A(1^n) \\ & \langle s_k, p_k \rangle = \langle s_A, m_A \rangle \\ & Output \space \langle s_k, p_k \rangle \\ % % Enc_{p_k}&(m): \\ & \langle s_b, m_b \rangle \leftarrow Prot_B(p_k) \\ & Output \space \langle m_B, f(s_B, p_k) \oplus m \rangle \\ % % Dec_{s_k}&(c): \\ & \langle m_B, c' \rangle = c \\ & k = f(s_k, m_B) \\ & Output \space k \oplus c' \end{align*}

Correctness

Since $f(s_k, m_B) = f(s_A, m_B) = f(s_B, m_A) = f(s_B, p_k)$ the correctness follows from the correctness of a one-time-pad.

Security

We can proceed to prove CPA-security by reduction. Assume there is an adversary $\mathcal{A'}$ which succeeds at breaking the CPA experiment for our public-key encryption scheme $\Pi'$. We construct an adversary $\mathcal{A}$ breaking the KE experiment as follows:

  1. $\mathcal{A}$ receives $m_A, m_B, \hat{k}$ as input
  2. $\mathcal{A}$ runs $\mathcal{A'}(1^n, m_A)$ and receives $m_0, m_1$
  3. $\mathcal{A}$ samples $b \leftarrow \{0,1\}$ uniformly and computes $c = \langle m_B, m_b \oplus \hat{k} \rangle$
  4. $\mathcal{A}$ runs $\mathcal{A}'(c)$ and receives $b'$
  5. $\mathcal{A}$ outputs 0 iff $b = b'$

Now we have 2 cases to consider -- when $\mathcal{A}$ received the true key ($[K=\hat{k}]$) and when it received a uniformly random string ($K \neq \hat{k}]$). But in the first case, the result of the KE experiment will be 1 exactly when $\mathcal{A}'$ succeeds in the CPA experiment. In the other scenario, since $\mathcal{A}$ samples b uniformly, the result of the experiment will be 1 when $b = b'$ with probability $\frac{1}{2}$. Hence:

$Pr[KE^{eav}_{\mathcal{A}, \Pi}(1^n) = 1] = \frac{1}{2}Pr[KE^{eav}_{\mathcal{A}, \Pi}(1^n) = 1 | K = \hat{k}] + \frac{1}{2}Pr[KE^{eav}_{\mathcal{A}, \Pi}(1^n) = 1 | K \neq \hat{k}] = \frac{1}{2}(Pr[Pubk^{CPA}_{\mathcal{A}', \Pi'}(1^n) = 1] + \frac{1}{2}) = \frac{1}{2}(\frac{1}{2} + \frac{1}{p(n)} + \frac{1}{2}) = \frac{1}{2} + \frac{1}{2p(n)}$

This contradicts $\Pi$ being $KE^{eav}$-secure, hence $\mathcal{A}'$ cannot exist and $\Pi'$ is CPA-secure.

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  • $\begingroup$ The notation of tuples is somewhat inconsistent in your construction. Also, the symbols your looking for are probably "\langle"=$\langle$ and "\rangle"=$\rangle$ instead of $<$ and $>$. $\endgroup$ – Maeher Feb 7 '20 at 9:22
  • $\begingroup$ Thanks for the feedback, I fixed the notation. $\endgroup$ – Boyan Hristov Feb 10 '20 at 23:54

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