# A New Leakage-Resilient IBE Scheme in the Relative Leakage Model

Yu Chen, Song Luo, Zhong Chen

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Yu Chen, Song Luo, Zhong Chen. A New Leakage-Resilient IBE Scheme in the Relative Leakage Model. 23th Data and Applications Security (DBSec), Jul 2011, Richmond, VA, United States. pp.263-270, 10.1007/978-3-642-22348-8_22. hal-01586577

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# A New Leakage-Resilient IBE Scheme in the Relative Leakage Model

Yu Chen*, Song Luo, and Zhong Chen

Information Security Lab, School of EECS, Peking University, Beijing, China
Key Laboratory of High Confidence Software Technologies, Ministry of Education
{chenyu,luosong,chen}@infosec.pku.edu.cn

**Abstract.** We propose the first leakage-resilient Identity-Based Encryp-
tion (IBE) scheme with full domain hash structure. Our scheme is leakage-
resilient in the relative leakage model and the random oracle model under
the decisional bilinear Diffie-Hellman (DBDH) assumption.

**Key words:** identity based encryption, leakage-resilient, relative leak-
age, bilinear Diffie-Hellman assumption

# 1 Introduction

Cryptographic schemes are used to be analyzed in an attack model in
which the internal secret states are completely hidden from the adver-
sary/attacker. However several works [12, 13] indicated that the attack
model fails to capture many attacks in the real world, since the attacker
may obtain some partial information about the secret states via vari-
ous *key leakage attacks*. Therefore it is urgent to design leakage-resilient
cryptographic schemes which remain provably secure in the strengthened
attack model which takes *key leakage attacks* into account.

Recently, the research community pay a lot of attention to construct
IBE schemes with leakage-resilience. Alwen et al. [1] presented three
leakage-resilient IBE schemes from the Gentry IBE [10], the Boneh-Gentry-
Hamburg IBE [4], and Gentry-Peikert-Vaikuntanathan IBE [11], respec-
tively. Among them, the first scheme is secure in the standard model,
while the other two schemes are secure in the random oracle model. Chow
et al. [6] gave three new leakage-resilient IBE schemes from the Boneh-
Boyen IBE [2], the Waters IBE [16], and the Lewko-Waters IBE [14],
respectively. All of them are secure in the standard model.

**Our Contributions.** According to [5], IBE schemes from pairings can be classified into three broad families, the full-domain hash family (e.g.

* Supported by National Natural Science Foundation of China (No.61073156).
Boneh-Franklin IBE [3]), the exponent inversion family (e.g. Gentry-IBE [10]), and the commutative blinding family (e.g. Boneh-Boyen IBE [2]). The existing work [1, 6] have shown that IBE schemes from the exponent inversion family and commutative blinding family can be tailored to be leakage-resilient ones. It is natural to ask if we can strengthen the IBE schemes from the full domain hash family to be leakage-resilient.

We give an affirmative answer to the above question by presenting an IBE scheme with the full domain hash structure based on a variant of Boneh-Franklin IBE [7]. Its leakage-resilient chosen plaintext security can be tightly reduced to the DBDH assumption in the relative leakage model and the random oracle model.

# 2 Preliminaries

**Notations.** $x \stackrel{R}{\leftarrow} S$ denotes that $x$ is picked uniformly at random from the set $S$. We write PPT for probabilistic polynomial time. By $\text{negl}(n)$ we denote a negligible function of $n$. We denote the bit-wise XOR operation by $\oplus$. We denote by $\mathcal{I}$ the identity space and by $\mathcal{SK}$ the private key space.

## 2.1 Bilinear Diffie-Hellman Assumption

The decisional BDH (DBDH) assumption [2,3] is defined via the following game: the challenger runs the bilinear group generator $\text{GroupGen}(1^n)$ to generate $(p, \mathbb{G}, \mathbb{G}_T, e)$, picks four random exponents $x, y, z, w$ from $\mathbb{Z}_p$, then computes $g^x, g^y, g^z, T_0 = e(g, g)^{xyz}$ and $T_1 = e(g, g)^{xyw}$. We denote by $D$ the tuple $(p, \mathbb{G}, \mathbb{G}_T, e, g, g^x, g^y, g^z)$. The challenger picks a random bit $c$ and gives to the adversary $\mathcal{B}$ the challenge instance $(D, T_c)$. We say $\mathcal{B}$ succeeds in solving the DBDH problem if it outputs the right guess $c'$ for $c$ at the end of the game, whose advantage is defined as:

$$|\Pr[c = c'] - 1/2| = |\Pr[\mathcal{B}(D, e(g, g)^{xyz}) = 0] - \Pr[\mathcal{B}(D, e(g, g)^{xyw}) = 0]|$$

**Definition 2.1** The $(t, \epsilon)$-DBDH assumption holds if no $t$-time adversary has at least $\epsilon$ in solving the DBDH problem in $\mathbb{G}$.

## 2.2 Randomness Extractors

The following notions and primitives will be used in our construction. We refer the readers to [1, 15] for a complement knowledge.

For a random variable $X$, we define $\mathbf{H}_\infty(X) = -\log(\max_x \Pr[X = x])$ as its min-entropy. We use the notion of *average min-entropy* [8] which
captures the remaining unpredictability of a random variable X conditioned on another random variable Y, formally defined as

$$ \tilde{\mathbf{H}}_{\infty}(X|Y) = -\log (E_{y \leftarrow Y} [\max_x \Pr[X = x|Y = y]]) $$

where $E_{y \leftarrow Y}$ denotes the expected value over all values of Y.

The average min-entropy measures exactly the optimal probability of guessing X given knowledge of Y. The following lemma was proved in [9] regarding average min-entropy:

**Lemma 1.** For any random variables X, Y, Z, if Y has $2^l$ possible values, then $\tilde{\mathbf{H}}_{\infty}(X|(Y, Z)) \ge \tilde{\mathbf{H}}_{\infty}(X|Z) - l$.

The statistical distance between two random variables X, Y over a finite domain $\Omega$ is defined as

$$ \mathbf{SD}(X, Y) = \frac{1}{2} \sum_{\omega \in \Omega} |\Pr[X = \omega] - \Pr[Y = \omega]| $$

Same as [1, 6, 15], a main tool used in our construction is the strong randomness extractor, which is formally defined as follows to the setting of the average min-entropy.

**Definition 2.2** A polynomial-time function ext: $\mathbb{G} \times \{0, 1\}^\mu \to \{0, 1\}^m$ is an average case $(k, \epsilon)$-strong extractor if for all pairs of random variables $(X, Y)$ such that $X \in \mathbb{G}$ and $\tilde{\mathbf{H}}_{\infty}(X|Y) \ge k$, we have that

$$ \mathbf{SD}((\mathrm{ext}(X, U_\mu), U_\mu, Y)), (U_m, U_\mu, Y)) \le \epsilon $$

where $\mathbb{G}$ is a non-empty set, and $U_\mu, U_m$ are two uniformly distributed random variables over $\{0, 1\}^\mu, \{0, 1\}^m$ respectively.

Dodis et al. [8] proved that any strong extractor is in fact an average-case strong extractor, for a proper setting of the parameters:

**Lemma 2.** For any $\delta > 0$, if ext is a worst case $(m - \log(1/\delta), \epsilon)$-strong extractor, then ext is also an average-case $(m, \epsilon + \delta)$-strong extractor.

As a specific example, they proved the following lemma which essentially gives an explicit construction of an average-case strong extractor:

**Lemma 3.** Let X, Y be two random variables such that $X \in \mathbb{G}$ and $\tilde{\mathbf{H}}_{\infty}(X|Y) \ge k$. Let $\mathcal{H} = \{H: \mathbb{G} \to \{0, 1\}^m\}$ be a family of universal hash functions. If $m \le k - 2\log(1/\epsilon)$ then we have

$$ \mathbf{SD}((H(X), U_s, Y)), (U_m, U_s, Y)) \le \epsilon $$
## 2.3 Leakage Model for IBE Setting

In this paper we use the relative leakage model suitable for the IBE setting. The leakage-resilient chosen plaintext security is defined by the following LeakCPA game, which is refined from the CpaLeak game introduced in [6].

**Setup.** The challenger generates the public parameters mpk and the master secret key msk. It gives mpk to the adversary and keeps msk to itself.

**Phase 1.** The adversary can make one of the following two types of queries to the challenger:

1. **Leak**($I, h_i$) query, where $h_i : SK \rightarrow \{0, 1\}^{l_i}$. The challenger checks if the overall amount leakage will exceed $l$. If not, it responds with $h_i(sk)$. Otherwise it responds with a reject symbol $\perp$.

2. **Reveal**($I$) query, where $I$ is the identity. The challenger responds with the associated private key sk.

**Challenge.** The adversary submits two messages $M_0, M_1$ of equal size and a challenge identity $I^*$, with the restriction that $I^*$ has not been revealed. The challenger picks a random bit $\beta$ and encrypts $M_\beta$ under $I^*$. It sends the resulting ciphertext $C^*$ to the adversary.

**Phase 2.** The same as Phase 1 with the restriction that no leakage queries or reveal queries related to $I^*$ are allowed.

**Guess.** The adversary outputs a bit $\beta'$. We say it succeeds if $\beta = \beta'$. The advantage of an adversary $\mathcal{A}$ on breaking an IBE scheme $\mathcal{E}$ with security parameter $\kappa$ and leakage bound $l$ is defined as $\text{Adv}_{\mathcal{A}, \mathcal{E}}^{\text{CPALeak}}(\kappa, l) = |\text{Pr}[\beta = \beta'] - \frac{1}{2}|$.

**Definition 2.3** An IBE scheme $\mathcal{E}$ is $\ell$-leakage fully secure if for all PPT adversaries $\mathcal{A}$ it holds that $\text{Adv}_{\mathcal{A}, \mathcal{E}}^{\text{CPALeak}}(\kappa, \ell) \le \text{negl}(\kappa)$.

## 3 Our Scheme

Our scheme consists of the following four algorithms:

**Setup.** Run GroupGen($1^\kappa$) → ($p, \mathbb{G}, \mathbb{G}_T, e$), pick $x \stackrel{R}{\leftarrow} \mathbb{Z}_p$, $g_2 \stackrel{R}{\leftarrow} \mathbb{G}^*$, and a cryptographic hash function $H : \{0, 1\}^* \rightarrow \mathbb{G}$. Let $g_1 = g^x$, $\ell = \ell(\kappa)$ be an upper bound on the amount of leakage. Then set an average-case ($\log |\mathbb{G}_T| - \ell, \epsilon_{\text{ext}}$)-strong extractor function $\text{ext} : \mathbb{G}_T \times \{0, 1\}^\mu \rightarrow \{0, 1\}^n$. The message space is $\mathcal{M} \in \{0, 1\}^n$, while mpk = ($g, g_1, g_2$) and msk = $x$.

**KeyGen.** For a given identity $I$, pick $t \stackrel{R}{\leftarrow} \mathbb{Z}_p$, compute $u = H(I)$, and then generate the private key for $I$ as $sk = (d_1, d_2) = (t, (ug_2^{-t})^x)$.
**Encrypt.** To encrypt a message $M$ under identity $I$, pick an exponent $r \stackrel{R}{\leftarrow} \mathbb{Z}_p$ and a seed $s \stackrel{R}{\leftarrow} \{0, 1\}^\mu$ for the extractor function, generate the ciphertext as $C = (c_1, c_2, c_3, c_4) = (g^r, s, e(g_1, g_2)^r, M \oplus \text{ext}(e(u, g_1)^r, s))$.

**Decrypt.** To decrypt a ciphertext $C = (c_1, c_2, c_3, c_4)$ encrypted under $I$ using the associated private key $sk = (d_1, d_2)$ to compute $M = c_4 \oplus \text{ext}(e(c_1, d_2)c_3^{d_1}, c_2)$. It is easy to verify that if the private key matches, we get the right decryption.

## 3.1 Security Analysis

**Theorem 3.1** If the DBDH assumption holds and the extractor's second parameter $\epsilon_{\text{ext}}$ is negligible in $\kappa$, then the proposed scheme is $\ell$-leakage secure, where $\ell = \log |\mathbb{G}_T| - k$ and $k$ is the extractor's first parameter.

To prove the theorem, we organize the proof as a sequence of games, which are defined as follows:

**Game$_{Real}$:** The real CPALeak game.

**Game$_{Final}$:** The real CPALeak game except in the challenge phase the challenger generates the ciphertext as follows:

$$ 
\begin{align*}
z, w &\stackrel{R}{\leftarrow} \mathbb{Z}_p, \beta \stackrel{R}{\leftarrow} \{0, 1\} & W = e(u^*, g_1)^z e(g_1, g_2)^{t^*(w-z)} \\
c_1^* &= g^z & c_2^* \stackrel{R}{\leftarrow} \{0, 1\}^\mu \\
c_3^* &= e(g_1, g_2)^w & c_4^* = M_\beta \oplus \text{ext}(W, c_2^*)
\end{align*}
$$

where $t^*$ is the tag of private key $sk^*$ of the challenge identity $I^*$, $z$ and $w$ are randomly picked from $\mathbb{Z}_p$. The challenge ciphertext is $C^* = (c_1^*, c_2^*, c_3^*, c_4^*)$. Note that if $w \neq z$, then $C^*$ is not a valid ciphertext since it is only decrypted correctly when using the private key with tag $t^*$.

**Lemma 3.2** If there exists a PPT algorithm $\mathcal{A}$ such that $\text{Adv}_{\mathcal{A}, \mathcal{E}}^{\text{GameReal}}$ - $\text{Adv}_{\mathcal{A}, \mathcal{E}}^{\text{GameFinal}} = \epsilon$, then we can build a PPT algorithm $\mathcal{B}$ with advantage $\epsilon$ in breaking the DBDH problem.

*Proof.* Suppose $\mathcal{B}$ is given a DBDH challenge $(p, \mathbb{G}, \mathbb{G}_T, e, g, g^x, g^y, g^z, T)$. We now describe how it interacts with $\mathcal{A}$ in the following game:

**Setup.** $\mathcal{B}$ sets $g_1 = g^x$ (implicitly sets $msk = x$), $g_2 = g^y$, picks a suitable extractor function $\text{ext}$, then gives $\mathcal{A}$ the public parameters $mpk = (p, \mathbb{G}, \mathbb{G}_T, e, g, g_1, g_2, \text{ext})$.

**Hash queries.** For a fresh hash query on $I$, $\mathcal{B}$ picks $a, t \stackrel{R}{\leftarrow} \mathbb{Z}_p$ and responds with $u = g^a g_2^t$.
**KeyGen queries.** For an arbitrary identity $I$, $\mathcal{B}$ computes a private key for it as follows: (1) compute $u = H(I)$; (2) set $d_1 = t$, $d_2 = g_1^a = (ug_2^{-t})^x = (g^a g_2^t g_2^{-t})^x$; (3) return $sk = (d_1, d_2)$.

We note that the keygen queries are always implicitly called by $\mathcal{B}$ when it answers the associated leak queries and reveal queries.

**Phase 1.** To answer the leak queries and reveal queries issued by $\mathcal{A}$, $\mathcal{B}$ creates two lists $L$ and $K$, which are initially empty. $L$ is a list of triples of identities, private keys, and a leakage counter, while $K$ is a list of tuples of identities, private keys.

– **Leak**($I, h_i$) query: $\mathcal{B}$ checks if there is a tuple $\langle I, sk \rangle$ in the existing $K$ list. If it is not $\mathcal{B}$ runs $sk \leftarrow \text{KeyGen}(msk, I)$, inserts the tuple $(I, sk)$ to the $K$ list and the triple $\langle I, sk, 0 \rangle$ to the $L$ list. After this step there must exists a triple $\langle I, sk, num \rangle$ in the $L$ list, $\mathcal{B}$ checks if $num + l_i \le l$. If this is true, it responds with $h_i(sk)$ and sets $num \leftarrow num + l_i$ in $\langle I, sk, num \rangle$. Otherwise $\mathcal{B}$ responds with a reject symbol $\perp$.

– **Reveal**($I$) query: $\mathcal{B}$ checks if there is a tuple $\langle I, sk \rangle$ in the $K$ list. If it is $\mathcal{B}$ responds with $sk$. If it is not $\mathcal{B}$ runs $sk \leftarrow \text{KeyGen}(msk, I)$, inserts the tuple $\langle I, sk \rangle$ to the $K$ list and the triple $\langle I, sk, 0 \rangle$ to the $L$ list, and responds the leak query with $sk$.

Notice that $\mathcal{B}$ can calculate a valid private key for any identity. Therefore, $\mathcal{B}$ is able to answer all the leakage queries **Leak**($I, h_i$) and reveal queries **Reveal**($I$), with the corresponding private key $sk = (d_1, d_2)$.

**Challenge.** $\mathcal{A}$ submits two messages $M_0, M_1$ and an identity $I^*$ on which it want to be challenged to $\mathcal{B}$. $\mathcal{B}$ computes $sk^* = (d_1^*, d_2^*) = (t^*, g_1^{a^*})$, then generates the challenge ciphertext as follows:

$$ 
\begin{align*}
\beta &\leftarrow^R \{0, 1\} & c_1^* &= g^z \\
c_2^* &\leftarrow^R \{0, 1\}^\mu & c_3^* &= T \\
W &= e(c_1^*, d_2^*)(c_3^*)^{d_1^*} &= e(g^z, g_1^{a^*})T^{t^*} & c_4^* &= M_\beta \oplus \text{ext}(W, c_2^*)
\end{align*}
$$

**Phase 2.** The same as Phase 1.

**Guess.** $\mathcal{A}$ outputs a guess $\beta'$. $\mathcal{B}$ returns 0 if $\beta = \beta'$ or 1 if $\beta \neq \beta'$.

We will prove that the advantage of $\mathcal{B}$ in breaking the DBDH problem is $\epsilon$. To see this, notice that if $T = e(g, g)^{xyz}$ the challenge ciphertext is a correct ciphertext according to the original encryption algorithm and thus $\mathcal{A}$ plays the **Game**$_{Real}$. This is because $W = e(g^z, g_1^{a^*})T^{t^*} = e(g^{a^*}, g_1^z)e(g_2^{t^*}, g_1^z) = e(g^{a^*}g_2^{t^*}, g_1^z) = e(u^*, g_1)^z$ as one can easily verify. Thus the probability that $\mathcal{A}$ succeeds in the game is exactly $\frac{1}{2} + \text{Adv}_{\mathcal{A},\mathcal{E}}^{\text{GameReal}}$. Since $\mathcal{B}$ outputs 0 when $\mathcal{A}$ succeeds we get that

$$ \Pr[\mathcal{B}(D, e(g, g)^{xyz}) = 0] = \frac{1}{2} + \text{Adv}_{\mathcal{A},\mathcal{E}}^{\text{GameReal}} $$
On the other hand if $T = e(g, g)^{xyw} = c_3^*$ then $\mathcal{A}$ essentially plays the $\mathbf{Game}_{\mathbf{Final}}$, because $W = e(g^z, g_1^{a^*})T^{t^*} = e(g^{a^*}, g_1^z)e(g_2^{t^*}, g_1^{(w-z)+z}) = e(u^*, g_1)^z e(g_1, g_2)^{t^*(w-z)}$ as one can easily verify. Therefore we have that

$$ \Pr[\mathcal{B}(D, e(g, g)^{xyw}) = 0] = \frac{1}{2} + \mathrm{Adv}_{\mathcal{A}, \mathcal{E}}^{\mathbf{Game}_{\mathbf{Final}}} $$

Combining the above equations we get that the advantage of $\mathcal{B}$ in DBDH is $|\Pr[\mathcal{B}(D, e(g, g)^{xyz}) = 0] - \Pr[\mathcal{B}(D, e(g, g)^{xyw}) = 0]| = \mathrm{Adv}_{\mathcal{A}, \mathcal{E}}^{\mathbf{Game}_{\mathbf{Real}}} - \mathrm{Adv}_{\mathcal{A}, \mathcal{E}}^{\mathbf{Game}_{\mathbf{Final}}} = \epsilon$. Therefore we prove the lemma. $\square$

**Lemma 3.3** For any PPT adversary $\mathcal{A}$ we have $\mathrm{Adv}_{\mathcal{A}, \mathcal{E}}^{\mathbf{Game}_{\mathbf{Final}}} \le 2\epsilon_{\mathrm{ext}}$.

*Proof.* In the $\mathbf{Game}_{\mathbf{Final}}$, it is true that $W = e(u^*, g_1)^z e(g_1, g_2)^{t^*(w-z)}$, where $t^*$ is the tag of the private key for $I^*$. If we assume that the exact private key with tag $t^*$ is perfect hidden from the adversary, then $W$ distributes uniformly at random in $\mathbb{G}_T$, and therefore the challenge ciphertext $C^*$ is totally independent of $M_\beta$ in an PPT adversary $\mathcal{A}$'s view. This is because $w = z \bmod p$ with negligible probability in $\kappa$ and $t^*$ is chosen randomly for $I^*$.

Suppose we denote by $R$ the set of all terms (public parameters, private keys, challenge ciphertext) given to the adversary $\mathcal{A}$ except the leakage, the random seed $c_2^*$, and the part of the challenge ciphertext $c_4^*$, then according to the above argument $\tilde{\mathbf{H}}_\infty(C|R) = \log|\mathbb{G}_T|$. But the attacker has access to at most $\ell$ bits of leakage from the private key, i.e. to a random variable $Y$ with $2^\ell$ values, thus by lemma 1 we know that

$$ \tilde{\mathbf{H}}_\infty(C|(Y, R)) \ge \tilde{\mathbf{H}}_\infty(C|R) - \ell = \log|\mathbb{G}_T| - \ell $$

According to the definition of $(\log|\mathbb{G}_T| - \ell, \epsilon_{\mathrm{ext}})$-strong extractor we have that $\mathbf{SD}((\mathrm{ext}(W, S), S, Y, R), (U_m, S, Y, R)) \le \epsilon_{\mathrm{ext}}$, where $S$ is the random variable for the seed $c_2^* \in \{0, 1\}^\mu$ distributed uniformly at random, $Y$, $R$ are the values of all the random variables known to the adversary: leakage and the rest, respectively. Thus the statistical distance of $c_4^* = M_\beta \oplus \mathrm{ext}(W, c_2^*)$ from the uniform distribution is at most $\epsilon_{\mathrm{ext}}$ for each $\beta$. The statistical distance between the two possible ciphertexts is at most $2\epsilon_{\mathrm{ext}}$ and no adversary (even an unbounded one) can distinguish them with advantage more than this. $\square$

Suppose $\epsilon_{DBDH}$ is the maximum advantage of all PPT adversaries in the DBDH game. Then according to the above lemma, for any PPT adversary $\mathcal{A}$ we have $\mathrm{Adv}_{\mathcal{A}, \mathcal{E}}^{\mathbf{Game}_{\mathbf{Real}}} - \mathrm{Adv}_{\mathcal{A}, \mathcal{E}}^{\mathbf{Game}_{\mathbf{Final}}} \le \epsilon_{DBDH}$. Therefore

$$ \mathrm{Adv}_{\mathcal{A}, \mathcal{E}}^{\mathbf{Game}_{\mathbf{Real}}} \le \mathrm{Adv}_{\mathcal{A}, \mathcal{E}}^{\mathbf{Game}_{\mathbf{Final}}} + \epsilon_{DBDH}(\kappa) \le 2\epsilon_{\mathrm{ext}}(\kappa) + \epsilon_{DBDH}(\kappa) $$
The proposed scheme is leakage-resilient CPA secure if both $\epsilon_{DBDH}(\kappa)$ and $\epsilon_{ext}(\kappa)$ are negligible functions of $\kappa$. This proves the theorem. $\square$

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