Randomized gossip algorithms under attack

Abstract

Recently, gossip-based algorithms have received significant attention for data aggregation in distributed environments. The main advantage of gossip-based algorithms is their robustness in dynamic and fault-prone environments with unintentional faults such as link failure and channel noise. However, the robustness of such algorithms in hostile environments with intentional faults has remained unexplored. In this paper, we call attention to the risks which may be caused by the use of gossip algorithms in hostile environments, i.e., when some malicious nodes collude to skew aggregation results by violating the normal execution of the protocol. We first introduce a model of hostile environment and then examine the behavior of randomized gossip algorithms in this model using probabilistic analysis. Our model of hostile environment is general enough to cover a wide range of attacks. However, to achieve stronger results, we focus our analysis on fully connected networks and some powerful attacks. Our analysis shows that in the presence of malicious nodes, after some initial steps, randomized gossip algorithms reach a point at which the lengthening of gossiping is harmful, i.e., the average accuracy of the estimates of the aggregate value begins to decrease strictly.

Appendices

Appendix A: Proof of Theorem 3

Proof

We use conditional expectation to prove the theorem. Let the event $\mathcal {A}$ occur at $t+1$, $i$ and $j$ be the participants of the gossip update, and $x(t)=x=(x_1,\ldots ,x_n)$. Then,

$$\begin{aligned}&ASE(t\!+\!1|i\in \mathcal {H},j\!\in \! \mathcal {H},x(t)=x) \!=\! \frac{1}{n}\sum _{k=1}^{n}\left( x_k^{t+1}\!-\!\overline{x}(0)\right) ^2\\&\quad = ASE(t) +\frac{1}{n}\left( 2 \left( \frac{x_i+x_j}{2}-\overline{x}(0)\right) ^2 -\left( x_i-\overline{x}(0)\right) ^2\right. \\&\quad \quad \quad \quad \quad \qquad \quad \quad \quad \quad -\left. \left( x_j-\overline{x}(0)\right) ^2\right) . \end{aligned}$$

Then, by taking the expectation over all $i,j\in \mathcal {H}$, and applying straightforward calculations, we have

$$\begin{aligned} E[\textit{ASE}(t\!+\!1)|\mathcal {A},x(t)=x]&= \left( 1\!-\!\frac{1}{n}\right) ASE(t)+\frac{1}{n}S(t). \end{aligned}$$

Taking the expectation over $x$ yields

$$\begin{aligned} E[\textit{ASE}(t+1)|\mathcal {A}]= \left( 1-\frac{1}{n}\right) E[\textit{ASE}(t)]+\frac{1}{n}E[S(t)].\nonumber \\ \end{aligned}$$

(30)

Now, suppose the event $\mathcal {B}$ occurs at $t+1$, $i\in \mathcal {H}$ and $j\in \mathcal {M}$ be the participants of the gossip update, $x_j^t=x_m$, and $x(t)=x=(x_1,\ldots ,x_n)$. Then,

$$\begin{aligned}&ASE\left( t+1|i\in \mathcal {H},x_j^t=x_m,x(t)=x\right) \\&\quad = \frac{1}{n}\sum _{k=1}^{n}(x_k^{t+1}-\overline{x}(0))^2 \\&\quad = ASE(t)+\frac{1}{n} \left( -\left( x_i-\overline{x}(0)\right) ^2+ \left( \frac{x_i+x_m}{2}-\overline{x}(0)\right) ^2\right) \\&\quad = ASE(t)-\frac{3}{4n} \left( x_i-\overline{x}(0)\right) ^2+\frac{1}{4n} \left( x_m-\overline{x}(0)\right) ^2 \\&\quad \quad +\frac{1}{2n} \left( x_m-\overline{x}(0)\right) \left( x_i-\overline{x}(0)\right) . \end{aligned}$$

On the other hand,

$$\begin{aligned} E\left[ \left( x_m-\overline{x}(0)\right) ^2\right] = \sigma _m^2+\left( \mu _m-\overline{x}(0)\right) ^2. \end{aligned}$$

Then, taking the expectation over $x_m$ yields

$$\begin{aligned}&ASE(t+1|i\in \mathcal {H},j\in \mathcal {M},x(t)=x)\\&\quad =ASE(t)-\frac{3}{4n}\left( x_i-\overline{x}(0)\right) ^2+\frac{1}{4n} \!\left( \sigma _m^2+\left( \mu _m-\overline{x}(0)\right) ^2\!\right) \\&\qquad +\frac{1}{2n}\left( \mu _m-\overline{x}(0)\right) \left( x_i-\overline{x}(0)\right) . \end{aligned}$$

Taking the expectation over all $i\in \mathcal {H}$ and $j\in \mathcal {M}$, and then taking the expectation over $x$, implies

$$\begin{aligned} E[\textit{ASE}(t+1)|\mathcal {B}]&= \left( 1-\frac{3}{4n}\right) E[\textit{ASE}(t)] \nonumber \\&+\frac{1}{4n}\left( \sigma _m^2+\left( \mu _m-\overline{x}(0)\right) ^2\right) \nonumber \\&+\frac{1}{2n} \left( \mu _m-\overline{x} (0)\right) \left( E[\overline{x}(t) ]-\overline{x}(0)\right) .\nonumber \\ \end{aligned}$$

(31)

If the event $\mathcal {C}$ occurs, $\textit{ASE}(t)$ remains unchanged, and then

$$\begin{aligned} E[\textit{ASE}(t+1)|\mathcal {C}]=E[\textit{ASE}(t)]. \end{aligned}$$

(32)

Putting together all the pieces yields:

$$\begin{aligned} E[\textit{ASE}(t+1)]&= +\Pr (\mathcal {A})E[\textit{ASE}(t+1)|\mathcal {A}]\\&+\Pr (\mathcal {B})E[\textit{ASE}(t+1)|\mathcal {B}] \\&+\Pr (\mathcal {C})E[\textit{ASE}(t+1)|\mathcal {C}] \end{aligned}$$

Then, using (11), (30), (31), and (32), we have

$$\begin{aligned} E[\textit{ASE}(t+1)]&= \left( 1-\frac{1+\frac{p}{2}}{N}\right) E[\textit{ASE}(t)] \\&+\frac{1-p}{N} E[S(t)]+\frac{p}{2N} \sigma _m^2 \\&+\frac{p}{N} \left( \frac{3}{2}-c^t \right) \left( \mu _m-\overline{x}(0)\right) ^2. \end{aligned}$$

To obtain $E[S(t+1)]$, we should first compute $E[S(t+1)|\mathcal {A}]$, $E[S(t+1)|\mathcal {B}]$, and $E[S(t+1)|\mathcal {C}]$. Due to its similarity to $E[\textit{ASE}(t+1)]$, we present only the results and omit the proofs.

$$\begin{aligned} E[S(t+1)|\mathcal {A}]&= E[S(t)],\end{aligned}$$

(33)

$$\begin{aligned} E\left[ S(t+1)|\mathcal {B}\right]&= \left( 1-\frac{1}{n}\right) E[S(t)]+\frac{1}{4n^2} E[\textit{ASE}(t)] \nonumber \\&+\frac{1}{n}(1-c^t )\left( \mu _m-\overline{x}(0)\right) ^2+\frac{1}{4n^2} \sigma _m^2\nonumber \\&-\frac{1}{2n^2}\left( \frac{1}{2}-c^t\right) \left( \mu _m-\overline{x}(0)\right) ^2, \end{aligned}$$

(34)

$$\begin{aligned} E[S(t+1)|\mathcal {C}]=E[S(t)]. \end{aligned}$$

(35)

On the other hand,

$$\begin{aligned} E[S(t+1)]&= +\Pr (\mathcal {A})E[S(t+1)|\mathcal {A}]\\&+\Pr (\mathcal {B})E[S(t+1)|\mathcal {B}] \\&+\Pr (\mathcal {C})E[S(t+1)|\mathcal {C}] \end{aligned}$$

Thus, using (33), (34), and (35), we have

$$\begin{aligned} E[S(t+1) ]&= \left( 1-\frac{2p}{N}\right) E[S(t)]+\frac{p}{2nN}E[\textit{ASE}(t)] \\&+\frac{2p}{N}\left( 1-c^t-\frac{1}{4n}+\frac{1}{2n}c^t\right) \left( \mu _m-\overline{x}(0)\right) ^2 \\&+\frac{p}{2nN}\sigma _m^2. \end{aligned}$$

This relation completes the proof. $\square$

Appendix B: Proof of Lemma 4

Proof

Let the event $\mathcal {B}$ occur at $t+1$, $i\in \mathcal {H}$ and $j\in \mathcal {M}$ be the participants of the gossip update, $x_j^t=x_m$, and $x(t)=x=(x_1,\ldots ,x_n)$. Then,

$$\begin{aligned}&S\left( t\!+\!1|i\in \mathcal {H},x_j^t\!=\!x_m,x(t)\!=\!x\right) \!=\! \left( \frac{1}{n}\sum _{k=1}^{n}x_k^{t+1}\!-\!\overline{x}(0)\right) ^2\\&\quad = \left( \frac{1}{n}\sum _{k=1}^{n}x_k -\frac{x_i}{n}+\frac{x_i+x_m}{2n}-\overline{x}(0)\right) ^2 \\&\quad = \left( \overline{x}(t) -\overline{x}(0)+\frac{x_m-x_i}{2n}\right) ^2 \\&\quad = \left( \overline{x}(t) -\overline{x}(0)\right) ^2+ 2\left( \overline{x}(t) -\overline{x}(0)\right) \left( \frac{x_m-x_i}{2n}\right) \\&\qquad +\left( \frac{x_m-x_i}{2n}\right) ^2 \\&\quad \ge S(t)+\frac{1}{n} \left( \overline{x}(t)-\overline{x}(0)\right) \left( x_m-x_i\right) . \end{aligned}$$

Taking the expectation over $x_m$, we have

$$\begin{aligned}&E\left[ S(t+1)|i\in \mathcal {H},\; j\in \mathcal {M},x(t)=x\right] \\&\quad \ge S(t)+\frac{1}{n}\left( \overline{x}(t)-\overline{x}(0)\right) \left( \mu _m-x_i\right) \end{aligned}$$

Taking the expectation over all $i\in \mathcal {H}$ and $j\in \mathcal {M}$, we have

$$\begin{aligned}&E[S(t+1)|\mathcal {B},x(t)=x]\\&\quad \ge S(t)+\frac{1}{n} \left( \overline{x}(t)-\overline{x}(0)\right) \left( \mu _m-\overline{x}(t)\right) . \end{aligned}$$

Taking the expectation over $x$, and using (11), we have

$$\begin{aligned}&E[S(t+1)|\mathcal {B}]\\&\quad \ge E[S(t)]+ \frac{1}{n} \left( E[\overline{x}(t)]-\overline{x}(0)\right) \left( \mu _m-E[\overline{x}(t)]\right) \\&\quad =E[S(t)]\!+\!\frac{1}{n}(1\!-\!c^t )c^t\left( \mu _m\!-\!\overline{x}(0)\right) ^2 \end{aligned}$$

Hence, $E[S(t+1)|\mathcal {B}]>E[S(t)]$. On the other hand, due to (33) and (35), if the event $\mathcal {A}$ or the event $\mathcal {C}$ occurs at $t+1$, then $E[S(t)]$ remains unchanged. Therefore, $E[S(t+1)]>E[S(t)]$ for any $t\ge 0$. $\square$