A One-way Secret Key Agreement with Security Against Active Adversaries

TL;DR

This paper introduces an efficient one-way secret key agreement protocol that remains secure even when an active adversary can tamper with the communication, utilizing a specially designed secure MAC.

Contribution

It extends existing OW-SKA protocols to active adversaries by integrating a secure MAC, ensuring robustness and secrecy against tampering.

Findings

Protocol achieves security against active adversaries.

Secrecy of the shared key is formally proven.

The protocol is efficient and practical.

Abstract

In a one-way secret key agreement (OW-SKA) protocol in source model, Alice and Bob have private samples of two correlated variables X and Y that are partially leaked to Eve through Z, and use a single message from Alice to Bob to obtain a secret shared key. We propose an efficient secure OW-SKA when the sent message can be tampered with by an active adversary. The construction follows the approach of an existing OW-SKA with security against passive adversaries, and uses a specially designed secure Message Authentication Code (MAC) that is secure when the key is partially leaked, to achieve security against active adversaries. We prove the secrecy of the established key and robustness of the protocol, and discuss our results.

Equations

$$\displaystyle\tilde{H}_{\nu,\infty}^{\mathsf{fuzz}}(\mathbf{X}|\mathbf{Z})=-\log\Big{(}\underset{\mathbf{z}\leftarrow\mathbf{Z}}{\mathbb{E}}\max_{\mathbf{x}}\sum_{\mathbf{y}:-\log(P_{\mathbf{X}|\mathbf{Y}}(\mathbf{x}|\mathbf{y}))\leq\nu}\mathrm{Pr}[\mathbf{Y}=\mathbf{y}|\mathbf{Z}=\mathbf{z}]\Big{)}.$$

$$-\log (P_{\mathbf{X}|\mathbf{Y}}(\mathbf{x}|\mathbf{y}))\le \nu ⟹\frac{P(\mathbf{x},\mathbf{y})}{P(\mathbf{y})}\ge 2^{-\nu }$$

$$⟹P(\mathbf{y})\le 2^{\nu }P(\mathbf{x},\mathbf{y}).$$

$${\mathsf{max}}_{\mathbf{x}}\sum _{\mathbf{y}:P_{\mathbf{X}|\mathbf{Y}}(\mathbf{x}|\mathbf{y})\ge 2^{-\nu }}P(\mathbf{x},\mathbf{y})$$

$$=\sum _{\mathbf{y}:-P_{\mathbf{X}|\mathbf{Y}}({\mathbf{x}}_{\mathsf{max}}|\mathbf{y})\ge 2^{-\nu }}P({\mathbf{x}}_{\mathsf{max}},\mathbf{y})$$

$$H_{\nu ,\infty }^{\mathsf{fuzz}}(\mathbf{X})$$

$$\displaystyle=-\log\big{(}\mathsf{max}_{\mathbf{x}}\sum_{\mathbf{y}:-\log(P_{\mathbf{X}|\mathbf{Y}}(\mathbf{x}|\mathbf{y}))\leq\nu}\mathrm{Pr}[\mathbf{Y}=\mathbf{y}]\big{)}$$

$$\displaystyle=\log\big{(}\frac{1}{\mathsf{max}_{\mathbf{x}}\sum_{\mathbf{y}:-\log(P_{\mathbf{X}|\mathbf{Y}}(\mathbf{x}|\mathbf{y}))\leq\nu}\mathrm{Pr}[\mathbf{Y}=\mathbf{y}]}\big{)}$$

$$\displaystyle\geq\log\big{(}\frac{1}{\mathsf{max}_{\mathbf{x}}\sum_{\mathbf{y}:-\log(P_{\mathbf{X}|\mathbf{Y}}(\mathbf{x}|\mathbf{y}))\leq\nu}2^{\nu}P(\mathbf{x},\mathbf{y})}\big{)}$$

$$\displaystyle=\log\big{(}\frac{1}{2^{\nu}\mathsf{max}_{\mathbf{x}}\sum_{\mathbf{y}:-\log(P_{\mathbf{X}|\mathbf{Y}}(\mathbf{x}|\mathbf{y}))\leq\nu}P(\mathbf{x},\mathbf{y})}\big{)}$$

$$\displaystyle=\log\big{(}\frac{1}{\mathsf{max}_{\mathbf{x}}\sum_{\mathbf{y}:-\log(P_{\mathbf{X}|\mathbf{Y}}(\mathbf{x}|\mathbf{y}))\leq\nu}P(\mathbf{x},\mathbf{y})}\big{)}-\nu$$

$$\displaystyle=\log\big{(}\frac{1}{\sum_{\mathbf{y}:P_{\mathbf{X}|\mathbf{Y}}(\mathbf{x}_{\mathsf{max}}|\mathbf{y})\geq 2^{-\nu}}P(\mathbf{x}_{\mathsf{max}},\mathbf{y})}\big{)}-\nu\text{ (by equation.~{}\ref{eqn:fuzzyminentropy2})}$$

$$\displaystyle\geq\log\big{(}\frac{1}{P(\mathbf{x}_{\mathsf{max}})}\big{)}-\nu$$

$$\displaystyle\geq\log\big{(}\frac{1}{\mathsf{max}_{\mathbf{x}}P(\mathbf{x})}\big{)}-\nu$$

$$=H_{\infty }(\mathbf{X})-\nu$$

$${\mathsf{max}}_{\mathbf{x}}\sum _{\mathbf{y}:P_{\mathbf{X}|\mathbf{Y}}(\mathbf{x}|\mathbf{y})\ge 2^{-\nu }}\mathrm{Pr}[\mathbf{Y}=\mathbf{y}]$$

$$=\sum _{\mathbf{y}:P_{\mathbf{X}|\mathbf{Y}}({\hat{\mathbf{x}}}_{\max }|\mathbf{y})\ge 2^{-\nu }}\mathrm{Pr}[\mathbf{Y}=\mathbf{y}]$$

$$\ge \sum _{\mathbf{y}:P_{\mathbf{X}|\mathbf{Y}}({\mathbf{x}}_{{\mathbf{y}}_{\max }}|\mathbf{y})\ge 2^{-\nu }}\mathrm{Pr}[\mathbf{Y}=\mathbf{y}]$$

$$H_{\nu ,\infty }^{\mathsf{fuzz}}(\mathbf{X})$$

$$\displaystyle=\log\big{(}\frac{1}{\mathsf{max}_{\mathbf{x}}\sum_{\mathbf{y}:-\log(P_{\mathbf{X}|\mathbf{Y}}(\mathbf{x}|\mathbf{y}))\leq\nu}\mathrm{Pr}[\mathbf{Y}=\mathbf{y}]}\big{)}$$

$$\displaystyle\leq\log\big{(}\frac{1}{\sum_{\mathbf{y}:P_{\mathbf{X}|\mathbf{Y}}(\mathbf{x}_{\mathbf{y}_{\max}}|\mathbf{y})\geq 2^{-\nu}}\mathrm{Pr}[\mathbf{Y}=\mathbf{y}]}\big{)}\text{ (by equation~{}\ref{eqn:fuzzent2})}$$

$$\displaystyle\leq\log\big{(}\frac{1}{P(\mathbf{y}_{\max})}\big{)}$$

$$\displaystyle=\log\big{(}\frac{1}{\max_{\mathbf{y}}P(\mathbf{y})}\big{)}$$

$$=H_{\infty }(\mathbf{Y})$$

$$\mathrm{Pr}[(\mathbf{xyz})\leftarrow \mathsf{gen}(P_{\mathbf{XYZ}}),\mathsf{ver}(\mathbf{y},s,\mathsf{mac}(\mathbf{x},s))=acc\text{ ∣ }\mathbf{Z}=\mathbf{z}]=1\text{vspace}-.3em$$

$$\begin{gathered} \mathrm{Pr}[\mathsf{ver}(\mathbf{y},s^{\prime },t^{\prime })=acc\text{ ∣ }\mathbf{Z}=\mathbf{z}]\le {\delta }_{imp}, \\ \text{vspace}-.3em \end{gathered}$$

$$\begin{gathered} \mathrm{Pr}[\mathsf{ver}(\mathbf{y},s^{\prime },t^{\prime })=acc\text{ ∣ }s,\mathsf{mac}(\mathbf{x},s)=t,\mathbf{Z}=\mathbf{z}]\le {\delta }_{sub}, \\ \text{vspace}-.3em \end{gathered}$$

$$\text{2. Consider the set }\mathcal{R}=\{\mathbf{x}:-\log (P_{\mathbf{X}|\mathbf{Y}}(\mathbf{x}|\mathbf{y}))\le \nu \}$$

$${\tilde{H}}_{\infty }(\mathbf{X}|\mathbf{Z})=-\log \sum _{\mathbf{z}\in {\mathcal{Z}}^n}(P_{\mathbf{Z}}(\mathbf{z})L_{\mathbf{z}})$$

$$=$$

$${\mathcal{E}}_1=\{\mathbf{x}:\mathbf{x}\notin \mathcal{R}\}=\{\mathbf{x}:-\log (P_{\mathbf{X}|\mathbf{Y}}(\mathbf{x}|\mathbf{y}))>\nu \}\text{ and }$$

$${\mathcal{E}}_2=\{\mathbf{x}\in \mathcal{R}:\exists \hat{\mathbf{x}}\in \mathcal{R}\text{ s.t. }h(\mathbf{x},(s^{\prime },s))=h(\hat{\mathbf{x}},(s^{\prime },s)\}.$$

$$\Delta (K,C,\mathbf{Z};U_{\ell },C,\mathbf{Z})$$

$$\displaystyle=\Delta\Big{(}h^{\prime}(\mathbf{X},S^{\prime}),h\left(\mathbf{X},(S^{\prime},S)\right),S^{\prime},S,\mathbf{Z};U_{\ell},h\left(\mathbf{X},(S^{\prime},S)\right),S^{\prime},S,\mathbf{Z}\Big{)}$$

$$\le \frac{1}{2}\sqrt{2^{-{\tilde{H}}_{\infty }(\mathbf{X}|\mathbf{Z},h\left(\mathbf{X},(S^{\prime },S)\right))}\cdot 2^{\ell }}\le \frac{1}{2}\sqrt{2^{-{\tilde{H}}_{\infty }(\mathbf{X}|\mathbf{Z})}\cdot 2^{\ell +t}}$$

$$=\frac{1}{2}\sqrt{2^{-n{\tilde{H}}_{\infty }(X|Z)+\ell +t}}\le \sigma$$

$$\displaystyle t^{\prime}_{f}=\big{[}s_{2f}(y^{\prime}_{2})^{r+2}+{\sum}_{i=1}^{r}s^{\prime}_{if}(y^{\prime}_{2})^{i}\big{]}_{1\cdots t}+(y^{\prime}_{1})^{3}+s_{1f}y^{\prime}_{1},\vspace{-.6em}$$

$$\displaystyle\mathbb{E}_{\mathbf{z}\leftarrow\mathbf{Z}}\Big{[}\mathsf{Pr}_{\mathbf{X}}\big{[}\mathsf{ver}(\mathbf{x},(s^{\prime}_{f},s_{f}),t^{\prime})=acc\text{ $|$ }\mathbf{Z}=\mathbf{z}\big{]}\Big{]}$$

$$\displaystyle=\mathbb{E}_{\mathbf{z}\leftarrow\mathbf{Z}}\Big{[}\mathsf{Pr}_{\mathbf{X}}\big{[}t^{\prime}_{f}=[s_{2f}(y^{\prime}_{2})^{r+2}+{\sum}_{i=1}^{r}s^{\prime}_{if}(y^{\prime}_{2})^{i}]_{1\cdots t}$$

$$\displaystyle\qquad+(y^{\prime}_{1})^{3}+s_{1f}y^{\prime}_{1}\text{ $|$ }\mathbf{Z}=\mathbf{z}\big{]}\Big{]}$$

Full text

I Introduction

Secret key agreement is a fundamental problem in cryptography: Alice wants to share a random string called key with Bob, such that a third party Eve who has access to the communication channel between them, has no information about the key. Information theoretic secret key agreement (SKA) was first proposed by Maurer [14] and Ahlswede [1]. In their model that is referred to as the source model, Alice, Bob and Eve have private samples of random variables (RVs) $X,Y$ and $Z$ with a joint probability distribution $P_{XYZ}$ that is known to all parties. The goal of Alice and Bob is to obtain a common secret key by exchanging messages over a public and error free channel. An important quality parameter of an SKA protocol is the length $\ell$ of the established secret key. In the setting that Alice, Bob and Eve’s variables $\mathbf{X},\mathbf{Y}$ and $\mathbf{Z}$ are $n$ independent realizations of $X,Y$ and $Z$ distributed according to $P_{XYZ}$, the *secret-key rate * is defined as the maximal rate at which Alice and Bob generate a highly secret key where the rate is (informally) given by $\ell/n$.

We consider one-way secret key agreement (OW-SKA) where Alice sends a single message over the public channel to Bob, to arrive at a shared key. OW-SKA problem is important in practice because it avoids interaction between Alice and Bob,

as well as being theoretically interesting because of its relation to circuit polarization and immunization of public-key encryption in complexity theory and cryptography [9].

Adversaries. SKAs were first studied with the assumption that Eve is passive and only observes the communication channel.

Maurer [15] considered a more powerful adversary who can eavesdrop and tamper with the communication. Against such adversaries the protocol is required to establish a secret key when the adversary is passive, and with probability $1-\delta$ Alice *or * Bob must detect the tampering, or a key that is unknown to Eve be established. It was proved [15] that SKA with security against active adversaries exists only when certaing simulatablity conditions are satisfied (see section I-A.)

In the following we use SKAp and SKAa to denote SKA with security against passive and active adversaries, respectively.

Constructions. There are a number of constructions for OW-SKAp that (asymptotically) achieve the highest possible secret-key rate for a given distribution $P_{\mathbf{X}\mathbf{Y}\mathbf{Z}}$ [10, 19, 3, 21]. It was proved (Theorem 9, [15]) that secret-key rate of SKAa is the same as the secret-key rate of SKAp if secure key agreement is possible (simulatability conditions hold). Construction of protocols with security against active adversaries however is less studied. It was also shown [15], through a construction, that it is possible to provide message authentication when Alice, Bob and Eve have private samples of $P_{\mathbf{X}\mathbf{Y}\mathbf{Z}}$. In [18] a concrete construction of a MAC was given, when the key $X$ is partially leaked to Eve through $Z$ for a known $P_{\mathbf{X}\mathbf{Z}}$. These MACs can be used to provide protection against active adversaries in SKAa .

Our Work

We propose an efficient OW-SKAa for the setting of $P_{\mathbf{X}\mathbf{Y}\mathbf{Z}}$. The construction is based on a previous construction of OW-SKAp [21] that achieves the highest secret-key rate for a given distribution, and employs two hash functions, $h$ and $h^{\prime}$ that are used for reconciliation and key extraction, respectively. Security proof of the protocol determines parameters of the hash functions in terms of distribution $P_{\mathbf{X}\mathbf{Y}\mathbf{Z}}$, number of samples, key length and security and reliability parameters of the system. We modify the protocol in two steps.

1. We modify the inputs to $h$ (and so the message sent from Alice to Bob) is modified. Theorem 1 recalculates the parameters of the two hash functions to achieve security against passive adversary, and the key length. This will also give the key length of our final construction when the protocol succeed to establish a key against an active adversary.

2. Noting that one of the inputs to $h$ is Alice’s private sample $\mathbf{x}$, one can see the hash function as effectively a keyed hash function. This MAC however is different from traditional MAC systems that use a shared secret key. In Section II-A we define information theoretically secure MAC in correlated randomness setting of $P_{\mathbf{X}\mathbf{Y}\mathbf{Z}}$, with security against impersonation and substitution attacks, and design a MAC with provable security for the case when $\mathbf{X}=\mathbf{Y}$ and the shared key is partially leaked through $\mathbf{Z}$. Using this MAC, which is a keyed hash function, for the hash function $h$ in our protocol, gives us a secure SKAa . In Theorem 2 we prove robustness of the protocol against active adversaries.

To our knowledge the formal definition of MAC in correlated randomness setting of $P_{\mathbf{X}\mathbf{Y}\mathbf{Z}}$ is new, and our construction of MAC is a new efficient construction with proved concrete security for special case of $\mathbf{X}=\mathbf{Y}$, and so would be of independent interest. Other known constructions of MACs for the general setting of $P_{\mathbf{X}\mathbf{Y}\mathbf{Z}}$ are due to Maurer [15] that uses the theory of typical sequences to prove asymptotic security (proof is not publicly available), and for the case of $\mathbf{X}=\mathbf{Y}$ due to Maurer et al. [18]. In Section IV-A1, we compare our MAC with the MAC due to [18].

I-A Related work

Key establishment with information theoretic setting has been studied in Quantum Key Distribution protocols (QKD) where the correlated random variables of Alice and Bob are obtained using quantum communication [2], as well as fuzzy extractor setting [7] where $\mathbf{X}$ and $\mathbf{Y}$ are samples of the same random source with a bound on the distance between the two. Commonly used distance functions are Hamming distance and set difference.

Key establishment with security against active adversaries was studied in [16, 17, 20, 13]. Feasibility of information theoretic SKAa was formulated by Maurer through simulatability property that is defined as follows: $P_{\mathbf{X}\mathbf{Y}\mathbf{Z}}$ is $\mathbf{X}$-simulatable by Eve if they can send $\mathbf{Z}$ through a simulated channel $P_{\hat{\mathbf{X}}|\mathbf{Z}}$ whose output $\hat{\mathbf{X}}$ has the same joint distribution with $\mathbf{Y}$ as $\mathbf{X}$. One can similarly define $\mathbf{Y}$-simulatability. SKA cannot be constructed if $P_{\mathbf{X}\mathbf{Y}\mathbf{Z}}$ is $\mathbf{X}$-simulatable or $\mathbf{Y}$-simulatable.

Organization. Section II is preliminaries. Section III is the construction of secure OW-SKAp . Section IV is the construction of MAC and its security. Section IV-A2 uses the MAC to construct a secure SKAa . Section V concludes the paper.

II Preliminaries

We use capital letters (e.g., $X$) to denote random variables (RVs), and lower-case letters (e.g., $x$) for their instantiations. Sets are denoted by calligraphic letters, (e.g. $\mathcal{X}$), and the size of $\mathcal{X}$ is denoted by $|\mathcal{X}|$. We denote vectors using boldface letters; for example $\mathbf{X}=(X_{1},\cdots,X_{n})$ is a vector of $n$ RVs, and its realization is given by $\mathbf{x}=(x_{1},\cdots,x_{n})$. $U_{\ell}$ denotes an RV with uniform distribution over $\{0,1\}^{\ell},\ell\in\mathbb{N}$. If $X$ is a discrete RV, we denote its probability mass function (p.m.f) by $\mathrm{P}_{X}(x)=\mathsf{Pr}(X=x)$. The conditional p.m.f. of an RV $X$ given RV $Y$ is denoted as $\mathrm{P}_{X|Y}(x|y)=\mathsf{Pr}(X=x|Y=y)$. For two RVs $X$ and $Y$ defined over the same domain $\mathcal{L}$, the statistical distance between $X$ and $Y$ is given by ${\rm\Delta}(X,Y)=\frac{1}{2}\sum_{v\in{\cal L}}|\Pr[X=v]-\Pr[Y=v]|$. Shannon entropy of an RV $X$ is denoted by $H(X)=-\sum_{x}\mathsf{P}_{X}(x)\log(\mathsf{P}_{X}(x))$.

The min-entropy of a random variable $X$ with p.m.f. $\mathrm{P}_{X}$ is defined as $H_{\infty}(X)=-\log(\max_{x}(\mathrm{P}_{X}({x})))$. The average conditional min-entropy [6] of an RV $X$ given RV $Y$ is $\tilde{H}_{\infty}(X|Y)=-\log(\mathbb{E}_{{y}\leftarrow Y}\max_{x}\mathrm{Pr}({X=x}|{Y=y})).$

We write $[x]_{i\cdots j}$ to denote the block from the $i$th bit to $j$th bit in $x.$ We use universal hash function on the output of weakly random entropy source, together with a random seed, to generate an output that is close to uniformly distributed, as shown by Leftover Hash Lemma [12].

Definition 1** (Universal hash family).**

A family of hash functions $h:\mathcal{X}\times\mathcal{S}\to\mathcal{Y}$ is called a universal hash family if $\forall x_{1},x_{2}\in\mathcal{X}$, $x_{1}\neq x_{2}:$ $\mathrm{Pr}[h(x_{1},s)=h(x_{2},s)]\leq\frac{1}{|\mathcal{Y}|}$, where the probability is over the uniform choices of $s$ from $\mathcal{S}$.

Definition 2** (Strong universal hash family).**

A family of hash functions $h:\mathcal{X}\times\mathcal{S}\rightarrow\mathcal{Y}$ is called a strong universal hash family if $\forall x_{1},x_{2}\in\mathcal{X}$, $x_{1}\neq x_{2}$, and for any $c,d\in\mathcal{Y}:$ $\mathrm{Pr}[h(x_{1},s)=c\wedge h(x_{2},s)=d]=\frac{1}{|\mathcal{Y}|^{2}}$, where the probability is over the uniform choices of $s$ from $\mathcal{S}$.

We use a variant of Leftover Hash Lemma, called Generalized Leftover Hash Lemma [[5], Lemma 2.4] that includes side information about the hash input, to prove security properties of our construction.

Lemma 1** (Generalized Leftover Hash Lemma [5]).**

*Assume a universal hash family $h:\mathcal{X}\times\mathcal{S}\rightarrow\{0,1\}^{\ell}$. Then for any two random variables $A$ and $B$, defined over $\mathcal{X}$ and $\mathcal{Y}$ respectively, applying $h$ on $A$ can extract a uniform random variable of length $\ell$ satisfying:

$\Delta(h(A,S),S,B;U_{\ell},S,B)\leq\frac{1}{2}\sqrt{2^{-\tilde{H}_{\infty}(A|B)}\cdot 2^{\ell}}$, where $S$ is chosen randomly from $\mathcal{S}$.*

We now recall the definition of almost strong universal hash family [22].

Definition 3** (Almost strong universal hash family [22]).**

A family of hash functions $h:\mathcal{X}\times\mathcal{S}\rightarrow\mathcal{Y}$ is called $\epsilon$-almost strong universal hash family if $\forall x_{1},x_{2}\in\mathcal{X}$, $x_{1}\neq x_{2}$, and for any $c,d\in\mathcal{Y}$, it holds that : (a) $\mathrm{Pr}[h(x_{1},s)=c]=\frac{1}{|\mathcal{Y}|}$ and (b) $\mathrm{Pr}[h(x_{1},s)=c\wedge h(x_{2},s)=d]\leq\frac{\epsilon}{|\mathcal{Y}|}$, where the probability is over the uniform choices of $s$ from $\mathcal{S}$.

The notion of fuzzy min-entropy has been introduced in [8] to estimate the guessing probability of a value within distance $t$ of a sample value $x$ of a distribution $P_{\mathbf{X}}$. In [8], Fuller et al. used fuzzy min-entropy to compute length of the extracted key in presence of passive adversaries. We consider active adversaries who try to guess a point around a secret key. We use fuzzy min-entropy to compute the probability that an active adversary can correctly guess that point. We define fuzzy min-entropy of a sample value $\mathbf{x}$ corresponding to a joint distribution $P_{\mathbf{X}\mathbf{Y}}$. The adversary tries to guess $\mathbf{x}_{1}$ such that the inequality: $-\log(P_{\mathbf{X}|\mathbf{Y}}(\mathbf{x}_{1}|\mathbf{y}))\leq\nu$ holds, where $\mathbf{y}$ in the secret key of Bob, and $\nu$ is some predetermined value. That is, the adversary tries to guess $\mathbf{x}_{1}$ such that $P_{\mathbf{X}|\mathbf{Y}}(\mathbf{x}_{1}|\mathbf{y})\geq 2^{-\nu}$. To have the maximum chance that the inequality $P_{\mathbf{X}|\mathbf{Y}}(\mathbf{x}_{1}|\mathbf{y})\geq 2^{-\nu}$ holds, the adversary would choose the point $\mathbf{x}_{1}$ that maximizes the total probability mass of $\mathbf{Y}$ within the set $\{\mathbf{y}:P_{\mathbf{X}|\mathbf{Y}}(\mathbf{x}_{1}|\mathbf{y})\geq 2^{-\nu}\}$. The $\nu$-fuzzy min-entropy [8] of an RV $\mathbf{X}$ with joint distribution $P_{\mathbf{X}\mathbf{Y}}$ is defined as

H_{\nu,\infty}^{\mathsf{fuzz}}(\mathbf{X})=-\log\big{(}\mathsf{max}_{\mathbf{x}}\sum_{\mathbf{y}:-\log(P_{\mathbf{X}|\mathbf{Y}}(\mathbf{x}|\mathbf{y}))\leq\nu}\mathrm{Pr}[\mathbf{Y}=\mathbf{y}]\big{)}.

For a joint distribution $P_{\mathbf{X}\mathbf{Y}\mathbf{Z}}$, the $\nu$-conditional fuzzy min-entropy of $\mathbf{X}$ given $\mathbf{Z}$ is defined as

[TABLE]

The following lemma gives both lower and upper bounds of $H_{\nu,\infty}^{\mathsf{fuzz}}(\mathbf{X})$.

Lemma 2**.**

Let the joint distribution of two RVs $\mathbf{X}$ and $\mathbf{Y}$ be denoted as $P_{\mathbf{X}\mathbf{Y}}$. Then the following properties hold.

(i) $H_{\infty}(\mathbf{X})-\nu\leq H_{\nu,\infty}^{\mathsf{fuzz}}(\mathbf{X}).$

(ii) Let $\max_{\mathbf{y}}P(\mathbf{y})=P(\mathbf{y}_{\max})$ for some point $\mathbf{y}_{\max}$ in the domain of $\mathbf{Y}$, and let there exist a point $\mathbf{x}_{\mathbf{y}_{\max}}$ in the domain of $\mathbf{X}$ such that $-\log(P_{\mathbf{X}|\mathbf{Y}}(\mathbf{x}_{\mathbf{y}_{\max}}|\mathbf{y}_{\max}))\leq\nu$, then $H_{\nu,\infty}^{\mathsf{fuzz}}(\mathbf{X})\leq H_{\infty}(\mathbf{Y})$.

Proof.

(i) Note that,

[TABLE]

Let $\mathsf{max}_{\mathbf{x}}\sum_{\mathbf{y}:-\log(P_{\mathbf{X}|\mathbf{Y}}(\mathbf{x}|\mathbf{y}))\leq\nu}P(\mathbf{x},\mathbf{y})$ occur at a point $\mathbf{x}=\mathbf{x}_{\mathsf{max}}$, then

[TABLE]

Now,

[TABLE]

(ii) Let $\mathsf{max}_{\mathbf{x}}\sum_{\mathbf{y}:-\log(P_{\mathbf{X}|\mathbf{Y}}(\mathbf{x}|\mathbf{y}))\leq\nu}\mathrm{Pr}[\mathbf{Y}=\mathbf{y}]$ occur at a point $\mathbf{x}=\hat{\mathbf{x}}_{\max}$, then

[TABLE]

[TABLE]

∎

As a corollary, we obtain the following lemma.

Lemma 3**.**

Let the joint distribution of three RVs $\mathbf{X}$, $\mathbf{Y}$ and $\mathbf{Z}$ be denoted as $P_{\mathbf{X}\mathbf{Y}\mathbf{Z}}$. Then the following properties hold.

(i) $\tilde{H}_{\infty}(\mathbf{X}|\mathbf{Z})-\nu\leq\tilde{H}_{\nu,\infty}^{\mathsf{fuzz}}(\mathbf{X}|\mathbf{Z}).$

(ii) Let $\max_{\mathbf{y}}P(\mathbf{y})=P(\mathbf{y}_{\max})$ for some point $\mathbf{y}_{\max}$ in the domain of $\mathbf{Y}$, and let there exist a point $\mathbf{x}_{\mathbf{y}_{\max}}$ in the domain of $\mathbf{X}$ such that $-\log(P_{\mathbf{X}|\mathbf{Y}}(\mathbf{x}_{\mathbf{y}_{\max}}|\mathbf{y}_{\max}))\leq\nu$, then $\tilde{H}_{\nu,\infty}^{\mathsf{fuzz}}(\mathbf{X}|\mathbf{Z})\leq\tilde{H}_{\infty}(\mathbf{Y}|\mathbf{Z})$.

II-A One-way secret key agreement (OW-SKA)

One natural setting is that the probabilistic experiment that underlies $(X,Y,Z)$ is repeated $n$ times independently, and Alice, Bob and Eve privately receive realizations of the RVs $\mathbf{X}=(X_{1},\cdots,X_{n})$, $\mathbf{Y}=(Y_{1},\cdots,Y_{n})$ and $\mathbf{Z}=(Z_{1},\cdots,Z_{n})$ respectively, where

$P_{\mathbf{X}\mathbf{Y}\mathbf{Z}}(\mathbf{x},\mathbf{y},\mathbf{z})=P_{\mathbf{X}\mathbf{Y}\mathbf{Z}}(x_{1},\cdots,x_{n},y_{1},\cdots,y_{n},z_{1},\cdots,z_{n})\\ =\prod_{i=1}^{n}P_{XYZ}(x_{i},y_{i},z_{i})$. This setting is considered in Maurer’s satellite scenario where a randomly generated string is received by Alice, Bob and Eve over independent noisy channels  [14, 15]. We note that for all $i=1,\cdots,n,$, the RVs $(X_{i},Y_{i},Z_{i})$ have the underlying distribution $P_{XYZ}$ and $P_{X_{i}Y_{i}Z_{i}}(x,y,z)=P_{XYZ}(x,y,z)$, $\forall(x,y,z)\in{\cal X}\times{\cal Y}\times{\cal Z}$.

Definition 4** (Secure OW-SKA protocol [10]).**

Let $X$ and $Y$ be two RVs over $\mathcal{X}$ and $\mathcal{Y}$ respectively. For shared key length $\ell$, a OW-SKA protocol on $\mathcal{X}\times\mathcal{Y}$ consists of two function families: a (probabilistic) function family {$\tau_{\mathrm{Alice}}:\mathcal{X}^{n}\to\{0,1\}^{\ell}\times\mathcal{C}$} that outputs $k_{A}\in\{0,1\}^{\ell}$ and $c$; and a function family $\{\tau_{\mathrm{Bob}}:\mathcal{Y}^{n}\times\mathcal{C}\to\{0,1\}^{\ell}\}$ that outputs $k_{B}\in\{0,1\}^{\ell}$. A OW-SKA protocol on $\mathcal{X}\times\mathcal{Y}$ is secure for a probability distribution $P_{XYZ}$ over $\mathcal{X}\times\mathcal{Y}\times\mathcal{Z}$ if for $\ell\in\mathbb{N}$, the OW-SKA protocol establishes an $(\epsilon,\sigma)$-secret key $k$ satisfying the following properties:

(i) (reliability) $\mathrm{Pr}(K_{A}=K_{B}=K)\geq 1-\epsilon$

(ii) (security)  $\Delta(K,C,\mathbf{Z};U_{\ell},C,\mathbf{Z})\leq\sigma$,

where $K_{A}$, $K_{B}$, $K$, $C$ are RVs corresponding to $k_{A}$, $k_{B}$, $k$ and $c$ respectively, and the RV $\mathbf{Z}$ is the vector of $n$ instances of the RV $Z$ and is Eve’s side information. The RV $U_{\ell}$ is uniformly distributed over $\{0,1\}^{\ell}$.

Definition 5**.**

(Secure information-theoretic one-time MAC in correlated randomness setting)

Let $P_{\mathbf{X}\mathbf{Y}\mathbf{Z}}$ be a public distribution. An $(|\mathcal{S}|,P_{\mathbf{X}\mathbf{Y}\mathbf{Z}},|\mathcal{T^{\prime}}|,\delta_{mac})$-information-theoretic one-time message authentication code is a triple of algorithms $(\mathsf{gen},\mathsf{mac},\mathsf{ver})$ where $\mathsf{gen}:P_{\mathbf{X}\mathbf{Y}\mathbf{Z}}\to(\mathbf{x}\mathbf{y}\mathbf{z})$ samples $P_{\mathbf{X}\mathbf{Y}\mathbf{Z}}$ and privately gives $\mathbf{x}$ and $\mathbf{y}$ to Alice and Bob, and leaks $\mathbf{z}$ to Eve, $\mathsf{mac}:\mathcal{X}^{n}\times\mathcal{S}\to\mathcal{T^{\prime}}$ is the tag generation algorithm that maps an input message from the message set $\mathcal{S}$ to a tag in the tag set $\mathcal{T^{\prime}}$, using the private input of Alice, $\mathbf{x}$, and $\mathsf{ver}:\mathcal{Y}^{n}\times\mathcal{S}\times\mathcal{T^{\prime}}\to\{acc,rej\}$ takes a message and tag pair, and outputs either accept (acc) or reject (rej) using Bob’s private input $\mathbf{y}$.

*The MAC satisfies correctness and unforgeability properties defined as follows.

Correctness : For any choice of $s\in\mathcal{S}$, we have,*

[TABLE]

$\delta_{ot}$-Unforgeability (one-time unforgeability):* For any $(\mathbf{x}\mathbf{y}\mathbf{z})\leftarrow\mathsf{gen}(P_{\mathbf{X}\mathbf{Y}\mathbf{Z}})$, we consider protection against two types of attacks,

(i) $\delta_{imp}$-impersonation: for any message and tag pair $s^{\prime}\in\mathcal{S}$ and $t^{\prime}\in\mathcal{T^{\prime}}$ chosen by the adversary, the following holds.*

[TABLE]

(ii) $\delta_{sub}$-substitution: for any observed message and tag pair $(s,t)$, for any adversary choice of $s^{\prime}\neq s\in\mathcal{S}$ and $t^{\prime},t\in\mathcal{T^{\prime}}$, the following holds.

[TABLE]

The MAC is called $\delta_{ot}$-one-time unforgeable with $\delta_{ot}=\mathsf{max}\{\delta_{imp},\delta_{sub}\}$, where the probability is taken over the randomness of $P_{\mathbf{X}\mathbf{Y}\mathbf{Z}}$.

A special case of the above definition is when $\mathbf{X}=\mathbf{Y}$ and $P_{\mathbf{X}\mathbf{Y}\mathbf{Z}}=P_{\mathbf{X}\mathbf{Z}}$, where $\mathbf{X}$ is the shared key of Alice and Bob that is partially leaked through $\mathbf{Z}$ to Eve.

Maurer’s construction in [15] is an example of the general case one-time MAC, while the construction in [18] is an example of the latter case $\mathbf{X}=\mathbf{Y}$ with partially leaked $\mathbf{X}$.

In the following we define robustness of a secure SKAa . We follow the definition in [23].

Definition 6** (Robustness of secure OW-SKA protocol).**

A OW-SKA protocol $(\tau_{\mathrm{Alice}},\tau_{\mathrm{Bob}})$ is called $(\epsilon,\sigma)$ OW-SKAa with robustness $\delta$ if for any strategy of an active attacker with access to $Z$ and communicated messages over the public channel, the probability that either Bob rejects the outcome of the protocol or the secret key agreement protocol is successful with reliability parameter $\epsilon$ and key security parameter $\sigma$, is no less than $(1-\delta)$. An $(\epsilon,\sigma)$ OW-SKAa protocol has robustness $\delta$ if for all adversary $\mathcal{D}$, the probability that the following experiment outputs ‘success’ is at most $\delta$: sample $(\mathbf{x},\mathbf{y},\mathbf{z})$ according to the distribution $P_{\mathbf{X}\mathbf{Y}\mathbf{Z}}$; let $(k_{A},c)\leftarrow\tau_{\mathrm{Alice}}(\mathbf{x})$, and let $\tilde{c}\leftarrow\mathcal{D}(\mathbf{z},c)$; output ‘success’ if $\tilde{c}\neq c$ and $\tau_{\mathrm{Bob}}(\mathbf{y},\tilde{c})\neq\perp$.

III Design and analysis of a OW-SKAa protocol

We give the construction of a robust and secure OW-SKAa protocol, and prove its security and robustness. Our construction provides security against active adversary.

III-A *A robust and secure OW-SKAa *

A secure OW-SKA protocol that satisfies Definition 4 provides security against passive adversary in which the adversary observes the message $c$ and tries to learn something about the extracted key $k$. However, it does not say anything about what happens if an adversary can modify a message $c$ as it is sent to Bob over public channel. In particular, the definition does not say anything about the output of $\tau_{\mathrm{Bob}}(\mathbf{y},\tilde{c})$ if $\tilde{c}\neq c$. A robust and secure SKAa satisfies Definition 6 and guarantees that any tampering with $c$ will be either detected by Bob, or does not affect a shared secret key establishment between Alice and Bob. We build on the secure SKAp protocol [21] and modify it to provide security and robustness.

The protocol uses two hash function families: a strong universal hash family $h^{\prime}:\mathcal{X}^{n}\times\mathcal{S^{\prime}}\rightarrow\{0,1\}^{\ell}$ and a universal hash family $h:\mathcal{X}^{n}\times(\mathcal{S}\times\mathcal{S^{\prime}})\rightarrow\{0,1\}^{t}$ that are used to extract the key and construct the protocol message $c$, respectively. $h$ is also an almost strong universal hash family when the probability is taken over $\mathcal{X}^{n}$. The message and key domains are the sets $\mathcal{C}=\{0,1\}^{t}\times\mathcal{S}\times\mathcal{S^{\prime}}$ and $\mathcal{K}=\{0,1\}^{\ell}$, respectively.

Construction 1** (A robust and secure OW-SKAa ).**

The SKAa protocol $\mathsf{OWSKA}_{a}=(\mathsf{OWSKA}_{a}\mathsf{.Gen},\mathsf{OWSKA}_{a}\mathsf{.Alice},\mathsf{OWSKA}_{a}\mathsf{.Bob})$ is given as follows:

The SKAa protocol $\mathsf{OWSKA}_{a}$ has three algorithms, $(\mathsf{OWSKA}_{a}\mathsf{.Gen},\mathsf{OWSKA}_{a}\mathsf{.Alice},\mathsf{OWSKA}_{a}\mathsf{.Bob})$, that are given in Algorithm 1, Algorithm 2 and Algorithm 3, respectively. The parameter $\nu$ in Algorithm 3 relies on the correlation between the random variables $\mathbf{X}$ and $\mathbf{Y}$. Higher correlation between $\mathbf{X}$ and $\mathbf{Y}$ implies smaller value of $\nu$ and smaller number of elements in the set $\mathcal{R}$. The relationship among parameters is given by both Theorem 1 and Theorem 2.

III-B Relation with  [21]

Our construction 1 is inspired by the construction of SKAp in  [21], that is reproduced for completeness in Appendix SKAp protocol of Sharifian et al. [21]. The main difference of the constructions is that the protocol message $c$ in Construction 1 takes the seeds for both hash functions as inputs (in addition to $\mathbf{x}$). In Algorithm 2 of our construction, Alice executes a single hash computation $h(\mathbf{x},(s^{\prime},s))$ and sends the result together with the randomness $(s^{\prime},s)$ to Bob. In Algorithm 3 upon receiving the message, Bob searches the set $\mathcal{R}$ to find a unique $\hat{\mathbf{x}}\in\mathcal{R}$ such that $h(\hat{\mathbf{x}},(s^{\prime},s))=h(\mathbf{x},(s^{\prime},s))$. Robust information reconciliation succeeds if a unique $\hat{\mathbf{x}}$ is found in $\mathcal{R}$, allowing Bob to extract the key $k=h^{\prime}(\hat{\mathbf{x}},s^{\prime})$. In the SKAp protocol in [21], the protocol message is $h(\mathbf{x},s)$ together with the randomness $(s^{\prime},s)$. Bob’s algorithm is similar but uses a different check function: Bob searches the set $\mathcal{R}$ to obtain a unique $\hat{\mathbf{x}}$ such that $h(\hat{\mathbf{x}},s)=h(\mathbf{x},s)$, and is successful if such an $\hat{\mathbf{x}}$ is found. This change to the input of the hash function requires a complete re-evaluation of the protocol parameters including parameters of the hash functions and the set $\cal R$, and security and robustness evaluation of the protocol.

In Theorem 1, we prove that Construction 1 is an SKAp , and provide relationship among parameters and the length of the extracted key, In Theorem 2, we prove the robustness and show that the construction is an SKAa . Combining these two theorems, we conclude that our construction is an $(\epsilon,\sigma)$-OW-SKAa protocol with robustness $\delta$ if the parameters $\ell$, $t$ and $\nu$ are chosen to satisfy both the Theorem 1 and Theorem 2.

The following lemma 4 can be proven using standard properties of independent variables and is given below.

Lemma 4**.**

For any $(X_{1}Z_{1}),\cdots,(X_{n}Z_{n})$ independently and identically distributed RV pairs, each with underlying distribution $P_{XZ}$, it holds that $\tilde{H}_{\infty}({\mathbf{X}}|{\mathbf{Z}})=n\tilde{H}_{\infty}(X|Z)$, where ${\mathbf{X}}=(X_{1},\cdots,X_{n})$ and ${\mathbf{Z}}=(Z_{1},\cdots,Z_{n}).$

Proof.

$L_{\bf z}:=\max_{\bf x}P_{{\bf X}|{\bf Z}}({\bf x}|{\bf z})=\max_{\bf x}\prod_{i=1}^{n}P_{X_{i}|Z_{i}}(x_{i}|z_{i})\\ =\max_{\bf x}\prod_{i=1}^{n}P_{X|Z}(x_{i}|z_{i}).$ Note that ${\bf x}$ goes over ${\cal X}^{n}$. Hence, $L_{\bf z}=\prod_{i=1}^{n}\max_{x_{i}}P_{X|Z}(x_{i}|z_{i}).$ We can define $x_{z}=\arg\max_{x}P_{X|Z}(x|z).$ Hence, $L_{\bf z}=\prod_{i=1}^{n}P_{X|Z}(x_{z_{i}}|z_{i}).$ Since

[TABLE]

This completes the proof. ∎

III-C Security against passive adversaries

Theorem 1** (Secure OW-SKA protocol).**

*Let the parameters $\nu$ and $t$ be chosen such that

$\nu=nH(X|Y)+\sqrt{(}n)\log(|\mathcal{X}|+3)\sqrt{\log(\frac{\sqrt{n}}{(\sqrt{(}n)-1)\epsilon})}$, and

$t\geq nH(X|Y)+\sqrt{(}n)\log(|\mathcal{X}|+3)\sqrt{\log(\frac{\sqrt{n}}{(\sqrt{(}n)-1)\epsilon})}+\log(\frac{\sqrt{(}n)}{\epsilon})$, then the OW-SKAa protocol $\mathsf{OWSKA}_{a}$ given in construction 1 establishes a secret key of length $\ell\leq n\tilde{H}_{\infty}(X|Z)+2\log(\sigma)+2-t$ that is $\epsilon$-correct and $\sigma$-indistinguishable from random (i.e. $(\epsilon,\sigma)$-OW-SKA protocol according to Definition 4).*

Proof.

We need to prove that the construction 1 satisfies Definition 4 for secure OW-SKA protocol. We first prove the reliability of the protocol and then analyze its security.

Reliability. We first determine the value of $\nu$ and $t$ to bound the error probability (i.e. reliability) of the protocol by $\epsilon$, and then we compute the extracted secret key length $\ell$. In algorithm $\mathsf{OWSKA}_{a}\mathsf{.Bob}(\cdot)$ 3, Bob searches the set $\mathcal{R}$ for $\hat{\mathbf{x}}$ and checks whether there is a unique $\hat{\mathbf{x}}$ whose hash value matches with the received hash value $d$. The algorithm succeeds if a unique $\hat{\mathbf{x}}$ is found in the set $\mathcal{R}$ with such property. Hence, the algorithm fails in the following two events: $(i)$ there is no element $\mathbf{x}\in\mathcal{R}$, whose hash value matches with the received hash value $d$ i.e. $\mathbf{x}\notin\mathcal{R}$, $(ii)$ the set $\mathcal{R}$ contains more than one element, whose hash values are the same as the received hash value $d$. Therefore, the probability that Bob fails to reproduce the correct key $\mathbf{x}$ is at most the sum of the probabilities of these two cases. These two cases represent the following two events respectively:

[TABLE]

For any $\epsilon>0$, we choose $\epsilon_{1}>0$ and $\epsilon_{2}>0$ satisfying $\epsilon_{1}+\epsilon_{2}\leq\epsilon$. Let $\delta_{1}$ satisfy the equation

$\epsilon_{1}=2^{\frac{-n{\delta_{1}}^{2}}{2\log^{2}(|\mathcal{X}|+3)}}$ and $\nu=H(\mathbf{X}|\mathbf{Y})+n\delta_{1}$. Then, $\mathsf{Pr}(\mathcal{E}_{1})={\mathsf{Pr}}\left(-\log(P_{\mathbf{X}|\mathbf{Y}}(\mathbf{x}|\mathbf{y}))>H(\mathbf{X}|\mathbf{Y})+n\delta_{1}\right)\leq\epsilon_{1}$ (from [11], Theorem 2). We now proceed to compute an upper bound for $\mathsf{Pr}(\mathcal{E}_{2})$. Since $h(\cdot)$ is a universal hash family, for any $\mathbf{x},\hat{\mathbf{x}}\in\mathcal{R}$, $\mathbf{x}\neq\hat{\mathbf{x}}$, random $s^{\prime}\in\mathcal{S^{\prime}}$ and randomly chosen $s\in\mathcal{S}$, we have $\mathsf{Pr}\left(h(\mathbf{x},(s^{\prime},s))=h(\hat{\mathbf{x}},(s^{\prime},s))\right)\leq 2^{-t}$, where the probability is over the uniform choices of $(s^{\prime},s)$ from $(\mathcal{S}^{\prime}\times\mathcal{S})$. Consequently, $\mathsf{Pr}(\mathcal{E}_{2})\leq|\mathcal{R}|\cdot 2^{-t}$. From equation 9 and considering that the sum of probability of elements of the set $\mathcal{R}$ is less than or equal to 1, we obtain

$\frac{|\mathcal{R}|}{2^{\nu}}\leq\mathsf{Pr}(\mathcal{R})\leq 1\Rightarrow|\mathcal{R}|\leq 2^{\nu}$. Hence, $\mathsf{Pr}(\mathcal{E}_{2})\leq|\mathcal{R}|\cdot 2^{-t}\leq 2^{\nu-t}.$ If we set $t=\nu-\log(\epsilon_{2})$, we obtain $\mathsf{Pr}(\mathcal{E}_{2})\leq\epsilon_{2}$. Thus, if $t=H(\mathbf{X}|\mathbf{Y})+n\delta_{1}-\log(\epsilon_{2})$,

then the probability that Bob fails to reproduce the correct key $\mathbf{x}$ is at most $\mathsf{Pr}(\mathcal{E}_{1})+\mathsf{Pr}(\mathcal{E}_{2})\leq\epsilon_{1}+\epsilon_{2}=\epsilon$. Furthermore, considering that $\mathbf{X},\mathbf{Y}$ are generated due to $n$ independent and identical experiments $P_{X_{i}Y_{i}Z_{i}}(x_{i},y_{i},z_{i})=P_{XYZ}(x_{i},y_{i},z_{i})$ for $1\leq i\leq n$, and $P_{\mathbf{X}\mathbf{Y}\mathbf{Z}}(\mathbf{x},\mathbf{y},\mathbf{z})=\prod_{i=1}^{n}P_{XYZ}(x_{i},y_{i},z_{i})$, we have $H(\mathbf{X}|\mathbf{Y})=nH(X|Y)$. Now, setting

$\epsilon_{1}=(\sqrt{n}-1)\epsilon/\sqrt{n}$ and $\epsilon_{2}=\epsilon/\sqrt{n}$, we have that if

$\nu=nH(X|Y)+\sqrt{n}\log(|\mathcal{X}|+3)\sqrt{\log(\frac{\sqrt{n}}{(\sqrt{n}-1)\epsilon})}$ and

$t\geq nH(X|Y)+\sqrt{n}\log(|\mathcal{X}|+3)\sqrt{\log(\frac{\sqrt{n}}{(\sqrt{n}-1)\epsilon})}+\log(\frac{\sqrt{n}}{\epsilon})$, then $\mathsf{Pr}(\mathcal{E}_{1})+\mathsf{Pr}(\mathcal{E}_{2})\leq\epsilon$. Therefore, we conclude that the construction 1 is $\epsilon$-correct, and the reliability condition of Definition 4 is satisfied.

Security. We now prove that the construction 1 also satisfies the security property of Definition 4. Let the RV $\mathbf{Z}$ correspond to $\mathbf{z}$, the attacker’s initial information. Let $K$, $C$, $S^{\prime}$ and $S$ be the RVs corresponding to the extracted key $k$, the ciphetext $c$, $s^{\prime}$ and $s$ respectively, where $k=h^{\prime}(\mathbf{x},s^{\prime})$ and c=\Big{(}h\big{(}\mathbf{x},(s^{\prime},s)\big{)},s^{\prime},s\Big{)}. The RV $C$ is distributed over $\{0,1\}^{t}$. Since $s^{\prime},s$ are randomly chosen and independent of RV $\mathbf{X}$, from [[5], Lemma 2.2(b)], we obtain \tilde{H}_{\infty}(\mathbf{X}|\mathbf{Z},C)=\tilde{H}_{\infty}\big{(}\mathbf{X}|\mathbf{Z},h\big{(}\mathbf{X},(S^{\prime},S)\big{)}\big{)}\geq\tilde{H}_{\infty}(\mathbf{X}|\mathbf{Z})-t. Therefore, utilizing this expression, from Lemma 1, we have

[TABLE]

The last equality is obtained by applying Lemma 4 that proves $\tilde{H}_{\infty}({\mathbf{X}}|{\mathbf{Z}})=n\tilde{H}_{\infty}(X|Z)$. The last inequality follows due to $\ell\leq n\tilde{H}_{\infty}(X|Z)+2\log(\sigma)+2-t$. Consequently, the security property of Definition 4 for OW-SKA protocol is satisfied. Therefore, construction 1 is an $(\epsilon,\sigma)$-OW-SKA protocol. ∎

IV Robustness of construction 1

We now prove that our construction 1 is a robust OW-SKAa protocol as defined in Definition 6. In order to prove robustness, we consider specific construction of the almost strong universal hash family $h:\mathcal{X}^{n}\times(\mathcal{S}\times\mathcal{S^{\prime}})\rightarrow\{0,1\}^{t}$ described below.

IV-A * A one-time secure MAC in $P_{\mathbf{X}\mathbf{Z}}$ setting*

For the private sample $\mathbf{x}$, we split $\mathbf{x}$ into two strings:

$y^{\prime}_{1}=[\mathbf{x}]_{1\cdots t}$ and $y^{\prime}_{2}=[\mathbf{x}]_{t+1\cdots n}$, where $t\leq n/2$. Observe that $\mathbf{x}=y^{\prime}_{2}\parallel y^{\prime}_{1}$. For a message $m$, we represent it as a sequence $(s^{\prime},s)$, where $s^{\prime}=s^{\prime\prime}_{2}\parallel s^{\prime\prime}_{1}$, $s^{\prime\prime}_{1}\in GF(2^{n})$, $s^{\prime\prime}_{2}\in GF(2^{n})$, $s=s_{2}\parallel s_{1}$, $s_{1}\in GF(2^{t})$ and $s_{2}\in_{R}GF(2^{n-t})$ such that $s_{2}$ is non-zero on the last element. Note that we can always have $s_{2}$ to be non-zero on the last element by suitably appending a 1 to $m$. The verifier checks that the last element of $s_{2}$ is non-zero. We represent $s^{\prime}$, suitably padded with 1s, as a sequence $(s^{\prime}_{r},\cdots,s^{\prime}_{1})$ of elements of $GF(2^{n-t})$, where $r$ is odd. Define h\big{(}\mathbf{x},(s^{\prime},s)\big{)}=h\big{(}\mathbf{x},(s^{\prime},(s_{2},s_{1}))\big{)}=\big{[}s_{2}(y^{\prime}_{2})^{r+2}+{\sum}_{i=1}^{r}s^{\prime}_{i}(y^{\prime}_{2})^{i}\big{]}_{1\cdots t}+(y^{\prime}_{1})^{3}+s_{1}y^{\prime}_{1}. We use this MAC to prove robustness of our construction in later section. Our MAC is inspired by the MAC in [4].

Lemma 5**.**

Let h\big{(}\mathbf{x},(s^{\prime},s)\big{)}=\big{[}s_{2}(y^{\prime}_{2})^{r+2}+{\sum}_{i=1}^{r}s^{\prime}_{i}(y^{\prime}_{2})^{i}\big{]}_{1\cdots t}+(y^{\prime}_{1})^{3}+s_{1}y^{\prime}_{1} as defined in section IV. Let $\mathsf{mac}:=h(\cdot)$. Define $\mathsf{ver}(\mathbf{x},(s^{\prime},s),t)$ s.t. it outputs acc $\text{ if }h(\mathbf{x},(s^{\prime},s))=t$ and rej, otherwise. Then $(\mathsf{gen},h,\mathsf{ver})$ is an $(|\mathcal{S^{\prime}}\times\mathcal{S}|,P_{\mathbf{X}\mathbf{Z}},|\mathcal{T^{\prime}}|,\delta_{mac})$-information-theoretic one-time MAC with $\delta_{imp}=3(r+2)2^{-(t+n\tilde{H}_{\infty}(X|Z)-n)}$, $\delta_{sub}=3(r+2)2^{-(t+n\tilde{H}_{\infty}(X|Z)-n)}$, and $\delta_{mac}=3(r+2)2^{-(t+n\tilde{H}_{\infty}(X|Z)-n)}$, where $\mathcal{T^{\prime}}=\{0,1\}^{t}$.

Proof.

We need to prove that $(\mathsf{gen},h,ver)$ satisfies Definition 5 for the case $\mathbf{x}=\mathbf{y}$. Since $\mathbf{x}=\mathbf{y}$, it is easy to see that the correctness property of Definition 5 is satisfied. We now focus on the unforgeability property. We first compute an adversary’s success probability in an impersonation attack and then compute the adversary’s success probability in an substitution attack.

Impersonation attack. In this attack, an adversary tries to generate a correct authenticated message $(t^{\prime}_{f},s^{\prime}_{f},s_{f})$ such that

[TABLE]

where $s_{f}=s_{2f}\parallel s_{1f}$, $s^{\prime}_{f}=(s^{\prime}_{rf}\parallel\cdots\parallel s^{\prime}_{1f})$, and the last element of $s_{2f}$ is non-zero. This is a non-zero polynomial in two variables $y^{\prime}_{2}$ and $y^{\prime}_{1}$ of degree at most $(r+2)$. The term $s_{2f}(y^{\prime}_{2})^{r+2}+{\sum}_{i=1}^{r}s^{\prime}_{if}(y^{\prime}_{2})^{i}+0^{n-t-t}|((y^{\prime}_{1})^{3}+s_{1f}y^{\prime}_{1})$ takes on each element in $GF(2^{n-t})$ for at most $3(r+2)2^{t}$ times when $y^{\prime}_{2}$ and $y^{\prime}_{1}$ varies. Hence, there are at most $3(r+2)2^{t}(2^{n-t}/2^{t})=3(r+2)2^{n-t}$ values of $(y^{\prime}_{2}\parallel y^{\prime}_{1})$ that satisfies the equation 10. Let $\mathbf{X}$ and $\mathbf{Z}$ denote the RVs corresponding to $\mathbf{x}$ and $\mathbf{z}$ respectively. Note that $\mathbf{x}=(y^{\prime}_{2}\parallel y^{\prime}_{1})$, and each value of $(y^{\prime}_{2}\parallel y^{\prime}_{1})$ (i.e. $\mathbf{x}$) occurs with probability at most $2^{-H_{\infty}(\mathbf{X}|\mathbf{Z}=\mathbf{z})}$. Thus, the probability that an adversary can successfully construct a correctly authenticated message =

[TABLE]

The last equality follows from Lemma 4 that proves $\tilde{H}_{\infty}(\mathbf{X}|\mathbf{Z})=n\tilde{H}_{\infty}(X|Z)$. Therefore, the success probability in an impersonation attack is at most $\delta_{imp}=3(r+2)2^{-(t+n\tilde{H}_{\infty}(X|Z)-n)}$.

Substitution attack. Assume that an adversary is given a correctly authenticated message $c=(t^{\prime},s^{\prime},s)$, where

[TABLE]

Let $T^{\prime}$, $C$, $Y^{\prime}_{1}$ and $Y^{\prime}_{2}$ denote the RVs corresponding to $t^{\prime}$, $c$, $y^{\prime}_{1}$ and $y^{\prime}_{2}$ respectively. After observing the message $c=(t^{\prime},s^{\prime},s)$, the adversary tries to generate a forged message $(t^{\prime}_{f},s^{\prime}_{f},s_{f})$ such that

t^{\prime}_{f}=\big{[}s_{2f}(y^{\prime}_{2})^{r+2}+{\sum}_{i=1}^{r}s^{\prime}_{if}(y^{\prime}_{2})^{i}\big{]}_{1\cdots t}+(y^{\prime}_{1})^{3}+s_{1f}y^{\prime}_{1}, $(t^{\prime}_{f},s^{\prime}_{f},s_{f})\neq(t^{\prime},s^{\prime},s)$, and the last element of $s_{2f}$ is non-zero. Then the expected probability that an adversary can successfully construct a forged message, given any message $c$, is

\mathbb{E}_{(c,\mathbf{z})\leftarrow(C,\mathbf{Z})}\big{[}\mathsf{Pr}_{\mathbf{X}}[t^{\prime}_{f}=\big{[}s_{2f}(y^{\prime}_{2})^{r+2}+{\sum}_{i=1}^{r}s^{\prime}_{if}(y^{\prime}_{2})^{i}\big{]}_{1\cdots t}+(y^{\prime}_{1})^{3}+s_{1f}y^{\prime}_{1}\text{ | }C=c,\mathbf{Z}=\mathbf{z}]\big{]}. Now if $(s^{\prime}_{f},s_{f})=(s^{\prime},s)$, then Bob will reject the message unless $t^{\prime}_{f}=t^{\prime}$. Hence, we only need to focus on the case $(s^{\prime}_{f},s_{f})\neq(s^{\prime},s)$.

Since addition and subtraction correspond to the bit-wise exclusive-or in the corresponding field, we have

[TABLE]

If $(s_{1}=s_{1f})$, then the degree of this polynomial in $y^{\prime}_{2}$ is at most $(r+2)$. Now the term

[(s_{2}-s_{2f})(y^{\prime}_{2})^{r+2}+{\sum}_{i=1}^{r}(s^{\prime}_{i}-s^{\prime}_{if})(y^{\prime}_{2})^{i}\big{]} takes on each element of the field $GF(2^{n-t})$ at most $(r+2)$ times as $y^{\prime}_{2}$ varies. Consequently, there are at most $(r+2)(2^{n-t}/2^{t})=(r+2)2^{n-2t}$ values of $y^{\prime}_{2}$ that satisfies equation IV-A. Equation 12 implies that, for each value of $y^{\prime}_{2}$, there exists at most three values of $y^{\prime}_{1}$ which satisfies the equation. Therefore, there are at most $3(r+2)2^{n-2t}$ values of $(y^{\prime}_{2}\parallel y^{\prime}_{1})$ that satisfies both the equation 12 and equation IV-A. If $(s_{1}\neq s_{1f})$, then representing $y^{\prime}_{1}$ in equation IV-A in terms of $y^{\prime}_{2}$ and substituting it in equation 12, we obtain

$t^{\prime}=[-(s_{1}-s_{1f})^{-3}(s_{2}-s_{2f})^{3}(y^{\prime}_{2})^{3(r+2)}]_{{}_{1\cdots t}}+g(y^{\prime}_{2})$ for some polynomial $g(y^{\prime}_{2})$ of degree at most $3r$. Therefore, there are at most $3(r+2)2^{n-2t}$ values of $y^{\prime}_{2}$ to satisfy this equation. From equation IV-A, we see that, for each value of $y^{\prime}_{2}$, there is a unique $y^{\prime}_{1}$ that satisfies the equation. Therefore, in both the cases, there are at most $3(r+2)2^{n-2t}$ values of $(y^{\prime}_{2}||y^{\prime}_{1})$ that satisfies both the equation 12 and equation IV-A. Note that $\mathbf{x}=y^{\prime}_{2}\parallel y^{\prime}_{1}$. Then each value of $(y^{\prime}_{2}\parallel y^{\prime}_{1})$ (i.e. $\mathbf{x}$) occurs with probability at most $2^{-H_{\infty}(\mathbf{X}|\mathbf{Z}=\mathbf{z},T^{\prime}=t^{\prime})}$, where $\mathbf{Z}$ is the RV corresponding to $\mathbf{z}$, Eve’s initial information. Since $|t^{\prime}|=t$, applying [[5], Lemma 2.2(b)], we obtain $\tilde{H}_{\infty}(\mathbf{X}|\mathbf{Z},T^{\prime})\geq\tilde{H}_{\infty}(\mathbf{X}|\mathbf{Z})-t$.

Therefore, the required expected probability =

[TABLE]

The last equality is obtained by using Lemma 4 that proves $\tilde{H}_{\infty}(\mathbf{X}|\mathbf{Z})=n\tilde{H}_{\infty}(X|Z)$. Therefore, the success probability in a substitution attack is at most

$\delta_{sub}=3(r+2)2^{-(t+n\tilde{H}_{\infty}(X|Z)-n)}$.

Consequently, $(\mathsf{gen},h,ver)$ is an $(|\mathcal{S^{\prime}}\times\mathcal{S}|,P_{\mathbf{X}\mathbf{Z}},|\mathcal{T^{\prime}}|,\delta_{mac})$

-information-theoretic one-time MAC with $\delta_{imp}=3(r+2)2^{-(t+n\tilde{H}_{\infty}(X|Z)-n)}$, $\delta_{sub}=3(r+2)2^{-(t+n\tilde{H}_{\infty}(X|Z)-n)}$, and \delta_{mac}$$=\mathsf{max}\{\delta_{imp},\delta_{sub}\}$$=3(r+2)2^{-(t+n\tilde{H}_{\infty}(X|Z)-n)}. ∎

IV-A1 Comparison with other MAC constructions

Comparing our results with [18], we note that the MAC construction in [ [18], Theorem 3] for $t=n/2$, has success probabilities in *impersonation * and substitution attacks as (roughly) $2^{-(H_{2}(\mathbf{X}|\mathbf{Z}=\mathbf{z})-n/2)/2})\approx 2^{-(n\tilde{H}_{\infty}(X|Z)-n/2)/2}$ and $(3\cdot 2^{-(H_{2}(\mathbf{X}|\mathbf{Z}=\mathbf{z})-n/2)/4})\approx 3\cdot(2^{-(n\tilde{H}_{\infty}(X|Z)-n/2)/4})$, respectively, where $H_{2}(\mathbf{X}):=-\log(\sum_{\mathbf{x}\in\mathcal{X}^{n}}P_{\mathbf{X}}(\mathbf{x})^{2})$, the Rényi entropy of $\mathbf{X}$. In our construction however, the probabilities are significantly less as Lemma 5 shows that the success probabilities of the two attacks are the same, and are at most $3(r+2)2^{-(t+n\tilde{H}_{\infty}(X|Z)-n)}$ which for $t=n/2$, are bounded by $3(r+2)2^{-(n\tilde{H}_{\infty}(X|Z)-n/2)}$.

We note that the construction in  [18] focuses on privacy amplification and assumes $\mathbf{x}=\mathbf{y}$, while our construction uses the MAC for reconcilliation, also. Assuming $\mathbf{x}=\mathbf{y}$ in our protocol, the extracted key length will be roughly $(2n\tilde{H}_{\infty}(X|Z)-n)$, and the extraction is possible as long as $n\tilde{H}_{\infty}(X|Z)>n/2$ i.e. $\tilde{H}_{\infty}(X|Z)>1/2$. These improve the results of Maurer et al.[ [18], Theorem 5] that require $H_{2}(\mathbf{X}|\mathbf{Z}=\mathbf{z})>2n/3$, i.e. roughly $n\tilde{H}_{\infty}(X|Z)>2n/3$, and extracts a key of length roughly $H_{2}(\mathbf{X}|\mathbf{Z}=\mathbf{z})-2n/3$, i.e. approximately $n\tilde{H}_{\infty}(X|Z)-2n/3$.

IV-A2 Robustness analysis

In order to prove robustness, we consider specific constructions of the strong universal hash family $h^{\prime}:\mathcal{X}^{n}\times\mathcal{S^{\prime}}\rightarrow\{0,1\}^{\ell}$ and the universal hash family

$h:\mathcal{X}^{n}\times(\mathcal{S}\times\mathcal{S^{\prime}})\rightarrow\{0,1\}^{t}$ as described below. $h$ is also an almost strong universal hash family when the probability is taken over the randomness of $\mathcal{X}^{n}$. For the private sample $\mathbf{x}$, let $y^{\prime}_{1}=[\mathbf{x}]_{1\cdots t}$ and $y^{\prime}_{2}=[\mathbf{x}]_{t+1\cdots n}$, where $t\leq n/2$. Notice that $\mathbf{x}=y^{\prime}_{2}\parallel y^{\prime}_{1}$. For two random elements $s^{\prime\prime}_{1},s^{\prime\prime}_{2}\in_{R}GF(2^{n})$, let $s^{\prime}=s^{\prime\prime}_{2}\parallel s^{\prime\prime}_{1}$ and h^{\prime}(\mathbf{x},s^{\prime})=h^{\prime}\big{(}\mathbf{x},(s^{\prime\prime}_{2},s^{\prime\prime}_{1})\big{)}=\big{[}s^{\prime\prime}_{2}(\mathbf{x})^{2}+s^{\prime\prime}_{1}\mathbf{x}\big{]}_{1\cdots\ell}. We represent $s^{\prime}$, suitably padded with 1s, as a sequence $(s^{\prime}_{r},\cdots,s^{\prime}_{1})$ of elements of $GF(2^{n-t})$ such that $r$ is odd. For two random elements $s_{1}\in_{R}GF(2^{t})$, $s_{2}\in_{R}GF(2^{n-t})$ such that the last element of $s_{2}$ is non-zero, let $s=s_{2}\parallel s_{1}$ and h\big{(}\mathbf{x},(s^{\prime},s)\big{)}=h\big{(}\mathbf{x},(s^{\prime},(s_{2},s_{1}))\big{)}=\big{[}s_{2}(y^{\prime}_{2})^{r+2}+{\sum}_{i=1}^{r}s^{\prime}_{i}(y^{\prime}_{2})^{i}\big{]}_{1\cdots t}+(y^{\prime}_{1})^{3}+s_{1}y^{\prime}_{1}. While verifying the tag, the receiver checks that the last element of $s_{2}$ is non-zero. Notice that $h$ and $h^{\prime}$ are universal hash family and strong universal hash family respectively. We now prove the following lemma 6 that will be used to prove robustness of our protocol. It is analogous to the Lemma 3.5 of Fuller et al. [8].

Lemma 6**.**

If $A$ is a random variable over a set of at most $2^{b}$ possible values, then $\tilde{H}_{\nu,\infty}^{\mathsf{fuzz}}(\mathbf{X}|A)\geq\tilde{H}_{\nu,\infty}^{\mathsf{fuzz}}(\mathbf{X})-b$.

Proof.

[TABLE]

∎

Theorem 2** (robustness of secure OW-SKA protocol).**

*The robustness of secure OW-SKA protocol as defined in Definition 6 is broken with probability at most

3(r+2)\big{(}2^{-(t-n+\min\{n\tilde{H}_{\infty}(X|Z),\tilde{H}_{\nu,\infty}^{\mathsf{fuzz}}(\mathbf{X}|\mathbf{Z})\})}\big{)}. Hence, if t\geq n+\log\big{(}\frac{3(r+2)}{\delta}\big{)}-\min\{n\tilde{H}_{\infty}(X|Z),\tilde{H}_{\nu,\infty}^{\mathsf{fuzz}}(\mathbf{X}|\mathbf{Z})\}, then the OW-SKAa protocol $\mathsf{OWSKA}_{a}$ given in construction 1 has robustness $\delta$.*

Proof.

We need to prove that the construction 1 satisfies Definition 6. The algorithm $\mathsf{OWSKA}_{a}\mathsf{.Bob}(\cdot)$ 3 successfully outputs an extracted key if there is a unique element $\hat{\mathbf{x}}$ in the set $\mathcal{R}$ such that $h(\hat{\mathbf{x}},s^{\prime},s)$ is equal to the received hash value $d=h(\mathbf{x},s^{\prime},s)$. Note that, if the construction is $(\epsilon,\sigma)$-OW-SKA protocol, $\mathbf{x}$ is in the set $\mathcal{R}$ with probability at least $(1-\epsilon)$. In the robustness experiment as defined in Definition 6, an adversary receives an authenticated message $c$ with $c=(t^{\prime},s^{\prime},s)$, where

[TABLE]

and the last element of $s_{2}$ is non-zero. Define $f(y^{\prime}_{2},(s^{\prime},s))=s_{2}(y^{\prime}_{2})^{r+2}+{\sum}_{i=1}^{r}s^{\prime}_{i}(y^{\prime}_{2})^{i}$. For fixed $s,s^{\prime}$, we denote the RVs corresponding to $t^{\prime}$, $c$, $y^{\prime}_{1}$, $y^{\prime}_{2}$, $\mathbf{x}$ as $T^{\prime}$, $C$, $Y^{\prime}_{1}$, $Y^{\prime}_{2}$, $\mathbf{X}$, respectively and hence the randomness is over ${\mathbf{X}\mathbf{Y}\mathbf{Z}}$ only. After observing the message $c$ corresponding to $\mathbf{x}$, the adversary tries to generate $c^{\prime}$ corresponding to $\mathbf{x}$ or, some $\mathbf{x}_{1}\in\mathcal{R}$ such that $\mathbf{x}_{1}\neq\mathbf{x}$. Let $\delta_{\mathbf{x}}$ and $\delta_{\mathbf{x}_{1}}$ denote the success probabilities of the adversary in generating a $c^{\prime}$ corresponding to the above two cases, respectively. Since Bob’s algorithm looks for a unique element in $\mathcal{R}$, only one of the above two cases succeed and hence, the probability that an adversary can generate a forged message is at most $\mathsf{max}\{\delta_{\mathbf{x}},\delta_{\mathbf{x}_{1}}\}$, and we have the success probability of the adversary bounded as:

[TABLE]

where the expectation is over $P_{\mathbf{X}\mathbf{Y}\mathbf{Z}}$. We first bound $\delta_{\mathbf{x}}$. Let the adversary generate a forged message $(t^{\prime}_{f},s^{\prime}_{f},s_{f})$ corresponding to $\mathbf{x}$ such that

t^{\prime}_{f}=\big{[}s_{2f}(y^{\prime}_{2})^{r+2}+{\sum}_{i=1}^{r}s^{\prime}_{if}(y^{\prime}_{2})^{i}\big{]}_{1\cdots t}+(y^{\prime}_{1})^{3}+s_{1f}y^{\prime}_{1}, $(t^{\prime}_{f},s^{\prime}_{f},s_{f})\neq(t^{\prime},s^{\prime},s)$, and the last element of $s_{2f}$ is non-zero. Thus,

[TABLE]

The expectation is over the distribution $P_{X|C=c,Z=z}$. The inequality 17 follows from the substitution attack part of Lemma 5 (i.e., equation 14).

We now compute the success probability (i.e., $\delta_{\mathbf{x}_{1}}$) that the forged value of $c$ $(t^{\prime}_{f_{\mathbf{x}_{1}}},s^{\prime}_{f_{\mathbf{x}_{1}}},s_{f_{\mathbf{x}_{1}}})$ corresponds to $\mathbf{x}_{1}\neq\mathbf{x}$ such that

[TABLE]

and $(t^{\prime}_{f_{\mathbf{x}_{1}}},s^{\prime}_{f_{\mathbf{x}_{1}}},s_{f_{\mathbf{x}_{1}}})\neq(t^{\prime},s^{\prime},s)$, where $y^{\prime}_{1{\mathbf{x}_{1}}}=[\mathbf{x}_{1}]_{1\cdots t}$, $y^{\prime}_{2{\mathbf{x}_{1}}}=[\mathbf{x}_{1}]_{t+1\cdots n}$, $s_{f_{\mathbf{x}_{1}}}=s_{2{f_{\mathbf{x}_{1}}}}\parallel s_{1{f_{\mathbf{x}_{1}}}}$, $s^{\prime}_{f_{\mathbf{x}_{1}}}=(s^{\prime}_{r{f_{\mathbf{x}_{1}}}}\parallel\cdots\parallel s^{\prime}_{1{f_{\mathbf{x}_{1}}}})$, and the last element of $s_{2{f_{\mathbf{x}_{1}}}}$ is non-zero. Since addition and subtraction correspond to bit-wise exclusive-or, we obtain,

[TABLE]

Since $\mathbf{x}$ can be written as $\mathbf{x}_{1}+e$ for some $e=(e_{2}\parallel e_{1})\in GF(2^{n})$, then $(y^{\prime}_{2}\parallel y^{\prime}_{1})=((y^{\prime}_{2{\mathbf{x}_{1}}}+e_{2})\parallel(y^{\prime}_{1{\mathbf{x}_{1}}}+e_{1}))$. Substituting $(y^{\prime}_{2}\parallel y^{\prime}_{1})=((y^{\prime}_{2{\mathbf{x}_{1}}}+e_{2})\parallel(y^{\prime}_{1{\mathbf{x}_{1}}}+e_{1}))$ in equation IV-A2, we have

[TABLE]

It is a polynomial in two variables $y^{\prime}_{2{\mathbf{x}_{1}}}$ and $y^{\prime}_{1{\mathbf{x}_{1}}}$ of degree at most $(r+2)$. There may be two cases: either $e_{1}\neq 0$ or $e_{1}=0$.

Case 1. Let $e_{1}\neq 0$, then equation 20 implies that there are at most two values of $y^{\prime}_{1\mathbf{x}_{1}}$ for any fixed value of $y^{\prime}_{2\mathbf{x}_{1}}$. From equation 18, we obtain that, for each value of $y^{\prime}_{1\mathbf{x}_{1}}$, there exist at most $(r+2)2^{n-2t}$ values of $y^{\prime}_{2\mathbf{x}_{1}}$ since $s_{2{f_{\mathbf{x}_{1}}}}\neq 0$. Therefore, there are at most $2(r+2)2^{n-2t}$ values of $(y^{\prime}_{2\mathbf{x}_{1}}\parallel y^{\prime}_{1\mathbf{x}_{1}})$ (i.e. $\mathbf{x}_{1}$) that satisfy both the equation 15 and equation 18 (and hence also equation 20).

Case 2. Let $e_{1}=0$. There may be two sub-cases: either $e_{2}\neq 0$ or $e_{2}=0$.

Subcase 2(i). Let $e_{2}\neq 0$, then equation 20 implies that there are at most $(r+2)2^{n-2t}$ values of $y^{\prime}_{2\mathbf{x}_{1}}$ for any fixed value of $y^{\prime}_{1\mathbf{x}_{1}}$ (note that $s_{2}\neq 0$). From equation 18, for each value of $y^{\prime}_{2\mathbf{x}_{1}}$, we see that there exist at most three values of $y^{\prime}_{1\mathbf{x}_{1}}$. Consequently, there are at most $3(r+2)2^{n-2t}$ values of $(y^{\prime}_{2\mathbf{x}_{1}}\parallel y^{\prime}_{1\mathbf{x}_{1}})$ (i.e. $\mathbf{x}_{1}$) that satisfy both the equation 15 and equation 18 (and hence also equation 20).

Subcase 2(ii). Let $e_{2}=0$. Then $y^{\prime}_{2}=y^{\prime}_{2{\mathbf{x}_{1}}}$ and $y^{\prime}_{1}=y^{\prime}_{1{\mathbf{x}_{1}}}$. Now proceeding the same way as substitution attack part of the proof of Lemma 5, we prove that there are at most $3(r+2)2^{n-2t}$ values of $(y^{\prime}_{2\mathbf{x}_{1}}\parallel y^{\prime}_{1\mathbf{x}_{1}})$ (i.e. $\mathbf{x}_{1}$) that satisfy both the equation 15 and equation 18 (and hence also equation 20).

Therefore, in any case, there are at most $3(r+2)2^{n-2t}$ values of $(y^{\prime}_{2{\mathbf{x}_{1}}}\parallel y^{\prime}_{1{\mathbf{x}_{1}}})$ (i.e. $\mathbf{x}_{1}$) that satisfy both the equation 15 and equation 18 (and hence also equation 20).

Let $\mathbf{X}_{1}$, $Y^{\prime}_{2{\mathbf{x}_{1}}}$ and $Y^{\prime}_{1{\mathbf{x}_{1}}}$ be the RVs corresponding to $\mathbf{x}_{1}$, $y^{\prime}_{2{\mathbf{x}_{1}}}$ and $y^{\prime}_{1{\mathbf{x}_{1}}}$ respectively. We assume that the only way to attack the MAC and create a forged ciphertext is guessing the secret key for the MAC. The adversary tries to guess $\mathbf{x}_{1}$ such that the inequality: $-\log(P_{\mathbf{X}|\mathbf{Y}}(\mathbf{x}_{1}|\mathbf{y}))\leq\nu$ holds, where $\mathbf{y}$ in the secret key of Bob. That is, the adversary tries to guess $\mathbf{x}_{1}$ such that $P_{\mathbf{X}|\mathbf{Y}}(\mathbf{x}_{1}|\mathbf{y})\geq 2^{-\nu}$. To have the maximum chance that the inequality $P_{\mathbf{X}|\mathbf{Y}}(\mathbf{x}_{1}|\mathbf{y})\geq 2^{-\nu}$ holds, the adversary would choose the point $\mathbf{x}_{1}$ that maximizes the total probability mass of $\mathbf{Y}$ within the set $\{\mathbf{y}:P_{\mathbf{X}|\mathbf{Y}}(\mathbf{x}_{1}|\mathbf{y})\geq 2^{-\nu}\}$. In addition, the adversary is given $\mathbf{z}$ and the authenticated message $c$. Therefore, an adversary can guess $\mathbf{x}_{1}$ with probability at most $2^{-H_{\nu,\infty}^{\mathsf{fuzz}}(\mathbf{X}|\mathbf{Z}=\mathbf{z},T^{\prime}=t^{\prime})}$ (Lemma 3 proves that guessing $\mathbf{x}_{1}$ given $\mathbf{Z}$ and $C$ in this way is better than guessing Bob’s secret key $\mathbf{y}$ given $\mathbf{Z}$ and $C$). Consequently, each value $\mathbf{x}_{1}$ in $\mathcal{R}$ occurs with probability at most $2^{-H_{\nu,\infty}^{\mathsf{fuzz}}(\mathbf{X}|\mathbf{Z}=\mathbf{z},T^{\prime}=t^{\prime})}$. Since the size of the support of RV $T^{\prime}$ is $2^{t}$, from Lemma 6, we obtain $\tilde{H}_{\nu,\infty}^{\mathsf{fuzz}}(\mathbf{X}|\mathbf{Z},T^{\prime})\geq\tilde{H}_{\nu,\infty}^{\mathsf{fuzz}}(\mathbf{X}|\mathbf{Z})-t$.

Thus,

[TABLE]

Consequently, from equations 16, 17 and 21, we conclude that, after observing a message, the expected probability that an adversary will be able to forge a message is at most

$\max\{3(r+2)2^{-(t+n\tilde{H}_{\infty}(X|Z)-n)},3(r+2)2^{-(t+\tilde{H}_{\nu,\infty}^{\mathsf{fuzz}}(\mathbf{X}|\mathbf{Z})-n)}\}$ =3(r+2)\big{(}2^{-(t-n+\min\{n\tilde{H}_{\infty}(X|Z),\tilde{H}_{\nu,\infty}^{\mathsf{fuzz}}(\mathbf{X}|\mathbf{Z})\})}\big{)} $\leq\delta$ (if t\geq n+\log\big{(}\frac{3(r+2)}{\delta}\big{)}-\min\{n\tilde{H}_{\infty}(X|Z),\tilde{H}_{\nu,\infty}^{\mathsf{fuzz}}(\mathbf{X}|\mathbf{Z})\}). Therefore, if t\geq n+\log\big{(}\frac{3(r+2)}{\delta}\big{)}-\min\{n\tilde{H}_{\infty}(X|Z),\tilde{H}_{\nu,\infty}^{\mathsf{fuzz}}(\mathbf{X}|\mathbf{Z})\}, the OW-SKAa protocol $\mathsf{OWSKA}_{a}$ given in construction 1 has robustness $\delta$. ∎

V Concluding remarks.

We proposed a OW-SKAa in source model, proved its security and robustness, and derived the established key length. To our knowledge there is no explicit construction of OW-SKAa to compare our protocol with. There are numerous OW-SKAp constructions (Section I-A) that can be the basis of OW-SKAa . Our construction in Section III-A is based on the construction in [21], and uses a new MAC construction (Section IV). Interesting directions for future work will be improving efficiency of decoding (which is currently exponential), and proving capacity achieving property of the protocol.

SKAp protocol of Sharifian et al. [21]

We briefly recall the construction of secure SKAp protocol due to Sharifian et al. [21]. The protocol is designed for source model in which Alice, Bob and Eve have $n$ components of the source $(\mathbf{X},\mathbf{Y},\mathbf{Z})$ respectively according to a distribution $P_{\mathbf{X}\mathbf{Y}\mathbf{Z}}(\mathbf{x},\mathbf{y},\mathbf{z})$. The protocol uses two universal hash families: $h:\mathcal{X}^{n}\times\mathcal{S}\to\{0,1\}^{t}$ and $h^{\prime}:\mathcal{X}^{n}\times\mathcal{S^{\prime}}\to\{0,1\}^{\ell}$. The construction is given in Algorithm 4. This construction and its security theorems in [21] are given for the case that $P_{\mathbf{X}\mathbf{Y}\mathbf{Z}}(\mathbf{x},\mathbf{y},\mathbf{z})=\prod_{i=1}^{n}P_{X_{i}Y_{i}Z_{i}}(x_{i},y_{i},z_{i})$ is due to $n$ independent experiments that are not necessarily identical.