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I am currently trying to solve the following problem (2.18) from the book "Introduction to Modern Cryptography (3rd edition)" by Katz and Lindell:

Let $\epsilon > 0$ be a constant. Say an encryption scheme is $\epsilon$-perfectly secret if for every adversary $A$ it holds that $Pr[PrivK^{eav}_{A,\Pi}] \leq 1/2 + \epsilon$.

Consider a variant of the one-time pad where $M = \{0, 1\}^l$ and the key is chosen uniformly from an arbitrary set $K \subseteq \{0, 1\}^l$ with $|K| = (1-\epsilon) \cdot 2^l$. Encryption and decryption are otherwise the same.

Prove that this scheme is $\epsilon$-perfectly secret.

I've tried the following:

$Pr[PrivK^{eav}_{A, \Pi} = 1]$ = $1/2 \cdot Pr[PrivK^{eav}_{A,\Pi} | b = 0]$ + $1/2 \cdot Pr[PrivK^{eav}_{A,\Pi} | b = 1]$

Assuming that the adversary is deterministic we can fix $m_0$ and $m_1$ generated by $A$. In addition, let $C_{m}$ be the set of possible ciphertexts derived from any given $m \in M$.

Then

$Pr[PrivK^{eav}_{A,\Pi} | b = 0]$ = $\sum_{c \in C_{m_0}} Pr[Priv^{eav}_{A, \Pi} = 1 | C = c] \cdot Pr[C = c]$

= $1/|K| \cdot \sum_{c \in C_{m_0}} Pr[Priv^{eav}_{A, \Pi} = 1 | C = c]$

For any $c \in C_{m_0}$, let $M(c)$ be the set of messages that can be encrypted to $c$. Therefore

$Pr[Priv^{eav}_{A, \Pi} = 1 | C = c] = 1 \cdot Pr[m_1 \notin M(c)] + 1/2 \cdot Pr[m_1 \in M(c)]$

Because if $m_1$ is not in $M(c)$, $A$ knows for sure that $m_0$ was encrypted. Otherwise, $A$ can only give a random guess.

Now, what is $Pr[m_1 \notin M(c)]$? In the best case, $A$ selects $m_0, m_1$ such that $|C_{m_0} \setminus C_{m_1}| = |C_{m_1} \setminus C_{m_0}| = 2^l - |K|$

As a consequence, $Pr[m_1 \notin M(c)] = Pr[c \in C_{m_0} \setminus C_{m_1}] \leq \frac{2^l - |K|}{|C_{m_0}|} = \frac{\epsilon}{1 - \epsilon}$

Thus, $Pr[Priv^{eav}_{A, \Pi} = 1 | C = c] = 1/2 + 1/2 \cdot Pr[m_1 \notin M(c)] \leq 1/2 + \frac{\epsilon}{2 \cdot (1 - \epsilon)}$

This implies that $Pr[PrivK^{eav}_{A,\Pi} | b = 0] \leq 1/2 + \frac{\epsilon}{2 \cdot (1 - \epsilon)}$.

Since the same holds for $Pr[PrivK^{eav}_{A,\Pi} | b = 1]$, $Pr[PrivK^{eav}_{A,\Pi}] \leq 1/2 + \frac{\epsilon}{2 \cdot (1 - \epsilon)}$

Can someone point me out where my mistake in this reasoning is?

Thanks in advance.

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  • $\begingroup$ Since $|C_{m_1} \setminus C_{m_0}| <= |K|$, the argument above works only for $\epsilon \leq 1/2$ $\endgroup$ – eee3 Apr 19 at 21:53
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There is no mistake.

Just prefix your work with the observation that any scheme is secure for $\epsilon\ge 1/2$ because all probabilities are less than or equal to 1.

Then consider the case $0<\epsilon<1/2$. Repeat your argument, then note that $2(1-\epsilon)>1$ and so $\epsilon/2(1-\epsilon)<\epsilon$.

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  • $\begingroup$ Thanks for the analysis, I was indeed missing this important point. $\endgroup$ – eee3 Apr 19 at 21:04

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