A Virtual Wiretap Channel for Secure Message Transmission

Abstract

In the Wyner wiretap channel a sender is connected to a receiver and an eavesdropper through two noisy channels. It has been shown that if the noise in the eavesdropper channel is higher than the receiver’s channel, information theoretically secure communication from Alice to Bob, without requiring a shared key, is possible. The approach is particularly attractive noting the rise of quantum computers and possibility of the complete collapse of todays’ cryptographic infrastructure. If the eavesdropper’s channel is noise free however, no secrecy can be obtained. The iJam protocol, proposed by Gollakota and Katabi, is an interactive protocol over noise free channels that uses friendly jamming by the receiver to establish an information theoretically secure shared key between the sender and the receiver. The protocol relies on the Basic iJam Transmission protocol (BiT protocol) that uses properties of OFDM (Orthogonal Frequency-Division Multiplexing) to create uncertainty for Eve (hence noisy view) in receiving the sent information, and use this uncertainty to construct a secure key agreement protocol. The protocol has been implemented and evaluated using extensive experiments that examines the best eavesdropper’s reception strategy. In this paper we develop an abstract model for BiT protocol as a wiretap channel and refer to it as a virtual wiretap channel. We estimate parameters of this virtual wiretap channel, derive the secrecy capacity of this channel, and design a secure message transmission protocol with provable semantic security using the channel. Our analysis and protocol gives a physical layer security protocol, with provable security, that is implementable in practice (BiT protocol has already been implemented).

Appendices

Appendix A: Achievable Transmission Rate Using BiT$^N_{q,\eta }$

For a noise free main channel, the secrecy capacity of BiT$^N_{q,\eta }$ is given by:

$$ C_s(\text {BiT}_{\eta ,q}^N)=-\{\eta \log \eta +(1-\eta )\log \frac{1-\eta }{(2^{Nq}-1)}\}. $$

Figure 4 shows the rate of communication when, the information block length is Nq bits, $q=2,3$ and 4, and $N=64$. The graphs show the achievable rates for $\sigma =128$ semantic security, and $\eta =0.2$ (upper graph) and $\eta =0.4$ (lower graph). The figures show that the achievable secrecy rate and secrecy capacity decreases as $\eta$ grows. This is expected because higher $\eta$ means that the adversary has a better chance of correctly decoding the jammed signal.

Fig. 4.

The secrecy rate and capacity (bits per channel use) for $N=64$ and different values of q for $\eta =0.2$ (upper graph) and $\eta =0.4$ (lower graph).

Appendix B: BiT over Noisy Receiver Channel—An Example

In this section we derive a sufficient relation between $P_b$ and $\eta$ so that the virtual wiretap channel is a stochastically degraded broadcast channel. Following Sect. 3, the transition matrix of the virtual wiretapper channel W for $q=2$ is given by:

$$\mathbf {P}_{\mathsf {W}}= \begin{bmatrix} \eta&\frac{1-\eta }{3}&\frac{1-\eta }{3}&\frac{1-\eta }{3} \\ \frac{1-\eta }{3}&\eta&\frac{1-\eta }{3}&\frac{1-\eta }{3} \\ \frac{1-\eta }{3}&\frac{1-\eta }{3}&\eta&\frac{1-\eta }{3} \\ \frac{1-\eta }{3}&\frac{1-\eta }{3}&\frac{1-\eta }{3}&\eta \end{bmatrix}, $$

where $u=\frac{1-\eta }{3}$, and $v=\eta -\frac{1-\eta }{3}=\frac{4\eta -1}{3}$. Note that the sum of each row is $4u+v=1$. On the other hand, we can compute:

$$ \begin{array}{ll} \mathbf {P}_{\mathsf {M}}^{-1} &{}= \frac{1}{(1-2P_b)^2}\cdot \\ &{}\ \ \begin{pmatrix} (1-P_b)(1-P_b) &{} -P_b(1-P_b) &{} -P_b(1-P_b) &{} P_b^2 \\ -P_b(1-P_b) &{} (1-P_b)(1-P_b) &{} P_b^2 &{} -P_b(1-P_b) \\ -P_b(1-P_b) &{} P_b^2 &{} (1-P_b)(1-P_b) &{} -P_b(1-P_b) \\ P_b^2 &{} -P_b(1-P_b) &{}-P_b(1-P_b)&{} (1-P_b)(1-P_b) \end{pmatrix}. \end{array} $$

Let $a=1-P_b$ and $b=P_b$. The above matrix can be written as:

$$\mathbf {P}_{\mathsf {M}}^{-1} = \frac{1}{(a-b)^2}\cdot \begin{pmatrix} a^2 &{} -ab &{} -ab &{} b^2 \\ -ab &{} a^2 &{} b^2 &{} -ab \\ -ab &{} b^2 &{} a^2 &{} -ab \\ b^2 &{} -ab &{}-ab&{} a^2 \end{pmatrix}. $$

The sum of entries of each row is given by, $\frac{1}{(a-b)^2}(a^2-2ab+b^2)=1$. The following is used to prove the required relation.

Lemma 1

Let there be two matrices

$$ A= \begin{bmatrix} a_{11}&a_{12}&\dots&a_{1n} \\ a_{21}&a_{22}&\dots&a_{2n} \\ \vdots&\vdots&\vdots \\ a_{n1}&a_{n2}&\dots&a_{nn} \end{bmatrix}, B= \begin{bmatrix} b_{11}&b_{12}&\dots&b_{1n} \\ b_{21}&b_{22}&\dots&b_{2n} \\ \vdots&\vdots&\vdots \\ b_{n1}&b_{n2}&\dots&b_{nn} \end{bmatrix}. $$

If $\sum _{j=1}^n a_{ij}=1$ and $\sum _{j=1}^n b_{ij}=1$ for any $i\in [n]$, then $\sum _{j=1}^n (AB)_{ij}=1$, for any $i\in [n]$.

Proof

For any $i\in [n]$,

$$ \begin{array}{ll} \sum _{j=1}^n (AB)_{ij}&{}=\sum _{j=1}^n \left( \sum _{k=1}^n a_{ik}b_{kj} \right) \\ &{}=\sum _{k=1}^n a_{ik}\cdot \left( \sum _{j=1}^n b_{kj} \right) \\ &{}=\sum _{k=1}^n a_{ik}\\ &{}=1. \end{array} $$

   $\square$

Lemma 2

The virtual wiretap channel is a stochastically degraded broadcast channel if $P_b\le \frac{1-\sqrt{\frac{4\eta -1}{3}}}{2}$ and $\eta >\frac{1}{4}$.

Proof

The virtual wiretap channel is a stochastically degraded broadcast channel if there exists a matrix $\mathbf {R}$ such that $\mathbf {P}_{\mathsf {W}}=\mathbf {P}_{\mathsf {M}}\times \mathbf {R}$, and $\mathbf {R}$ is a channel transition matrix; that is, has non-negative entries and each row sums to 1. Using the matrices $\mathbf {P}_{\mathsf {M}}$ and $\mathbf {P}_{\mathsf {W}}$ above, we have:

$$ \begin{array}{ll} \mathbf {R}&{}=\mathbf {P}_{\mathsf {W}}\times \mathbf {P}_{\mathsf {M}}^{-1}\\ &{}=\frac{1}{(a-b)^2}\begin{bmatrix} u(a-b)^2+va^2\ &{} u(a-b)^2-vab\ &{} u(a-b)^2-vab\ &{} u(a-b)^2+vb^2 \\ u(a-b)^2-vab &{} u(a-b)^2+va^2 &{} u(a-b)^2+vb^2 &{} u(a-b)^2-vab \\ u(a-b)^2-vab &{} u(a-b)^2+vb^2 &{} u(a-b)^2+va^2 &{} u(a-b)^2-vab \\ u(a-b)^2+vb^2 &{} u(a-b)^2-vab &{} u(a-b)^2-vab &{} u(a-b)^2+va^2 \\ \end{bmatrix}. \end{array} $$

Using Lemma 1, entries in each row of $\mathbf {R}$ sum to 1.

To ensure entries of $\mathbf {R}$ are all non-negative, we first note that $u(a-b)^2+va^2> 0$ and $u(a-b)^2+vb^2>0$. So the virtual wiretap channel is a stochastically degraded broadcast channel if $u(a-b)^2-vab\ge 0$ and so:

$$ \begin{array}{ll} u(a-b)^2-vab\ge 0 &{}\Leftrightarrow ua^2+ub^2-(2u+v)ab\ge 0\\ &{}\Leftrightarrow ua^2+ub^2-(2u+1-4u)ab\ge 0\\ &{}\Leftrightarrow ua^2+ub^2-(1-2u)ab\ge 0\\ &{}\Leftrightarrow u(a+b)^2-ab\ge 0\\ &{}\Leftrightarrow u-ab\ge 0\\ &{}\Leftrightarrow P_b^2-P_b+u\ge 0,\\ \end{array} $$

where $4u+v=1$ and $a+b=1$ are repeatedly invoked to simplify the expressions. The solution to the above inequality depends on the determinant $1-4u$. When $1-4u>0$, we have

$$ \begin{array}{ll} P_b^2-P_b+u\ge 0&{}\Leftrightarrow \left( P_b-\frac{1-\sqrt{1-4u}}{2}\right) \left( P_b-\frac{1+\sqrt{1-4u}}{2}\right) \ge 0\\ &{}\Leftrightarrow \left( P_b-\frac{1-\sqrt{v}}{2}\right) \left( P_b-\frac{1+\sqrt{v}}{2}\right) \ge 0\\ &{}\Leftrightarrow \left( P_b-\frac{1-\sqrt{\frac{4\eta -1}{3}}}{2}\right) \left( P_b-\frac{1+\sqrt{\frac{4\eta -1}{3}}}{2}\right) \ge 0\\ &{}\Leftrightarrow P_b\le \frac{1-\sqrt{\frac{4\eta -1}{3}}}{2} \text {or}\; P_b\ge \frac{1+\sqrt{\frac{4\eta -1}{3}}}{2}. \end{array} $$

By assumption, $P_b\in [0,\frac{1}{2}]$ and so $P_b\le \frac{1-\sqrt{\frac{4\eta -1}{3}}}{2} = \frac{1}{2} - \sqrt{\frac{4\eta -1}{12}}$.    $\square$

Example 2

Let $P_b=0.1$ and Let $\eta =0.55$. Therefore,

$$ \mathbf {P}_{\mathsf {M}}= \begin{bmatrix} 0.81&0.09&0.09&0.01 \\ 0.09&0.81&0.01&0.09 \\ 0.09&0.01&0.81&0.09 \\ 0.01&0.09&0.09&0.81 \end{bmatrix} $$

and

$$ \mathbf {P}_{\mathsf {W}}= \begin{bmatrix} 0.55&0.15&0.15&0.15 \\ 0.15&0.55&0.15&0.15 \\ 0.15&0.15&0.55&0.15 \\ 0.15&0.15&0.15&0.55 \end{bmatrix}. $$

Therefore

$$ \mathbf {R}= \mathbf {P}_{\mathsf {W}}\times \mathbf {P}_{\mathsf {M}}^{-1}= \begin{bmatrix} 0.66&0.094&0.094&0.156 \\ 0.094&0.66&0.156&0.094 \\ 0.094&0.156&0.66&0.094 \\ 0.156&0.094&0.094&0.66 \end{bmatrix}. $$

$\mathbf {R}$ is the transition probability matrix of a virtual channel that confirms $\mathbf {P}_{\mathsf {W}}$ is degraded with respect to $\mathbf {P}_{\mathsf {M}}$. The secrecy capacity in this example is

$$\begin{aligned} C_s =C_{\mathsf {M}}-C_{\mathsf {W}}=(2-0.7624)-(2-1.1515)=0.3891. \end{aligned}$$