On learning cluster coefficient of private networks

Abstract

Enabling accurate analysis of social network data while preserving differential privacy has been challenging since graph features such as clustering coefficient or modularity often have high sensitivity, which is different from traditional aggregate functions (e.g., count and sum) on tabular data. In this paper, we treat a graph statistics as a function f and develop a divide and conquer approach to enforce differential privacy. The basic procedure of this approach is to first decompose the target computation f into several less complex unit computations $f_1,\ldots,f_m$ connected by basic mathematical operations (e.g., addition, subtraction, multiplication, division), then perturb the output of each f _i with Laplace noise derived from its own sensitivity value and the distributed privacy threshold $\epsilon_i,$ and finally combine those perturbed f _i as the perturbed output of computation f. We examine how various operations affect the accuracy of complex computations. When unit computations have large global sensitivity values, we enforce the differential privacy by calibrating noise based on the smooth sensitivity, rather than the global sensitivity. By doing this, we achieve the strict differential privacy guarantee with smaller magnitude noise. We illustrate our approach using clustering coefficient, which is a popular statistics used in social network analysis. Empirical evaluations on five real social networks and various synthetic graphs generated from three random graph models show that the developed divide and conquer approach outperforms the direct approach.

A Proof

Proof of Lemma 1

$$\begin{aligned} & E(u\cdot(a+Lap(0,\, a')) + v\cdot(b+Lap(0,b')))\\ &\quad =E(u\cdot a+v\cdot b) + u\cdot E( Lap(0,a')) +v\cdot E(Lap(0,b')) \end{aligned}$$

Since E(Lap(0, a′)) = 0 and E(Lap(0, b′)) = 0, we have

$$E(u\cdot(a+Lap(0,a'))+v\cdot(b+Lap(0,b'))) =E(u\cdot a+v\cdot b)$$

Proof of Lemma 2

$$\begin{aligned} E((a+Lap(0,a'))\cdot(b+Lap(0,b'))) &=E(a\cdot b+b\cdot Lap(0,a') \\ & \quad+a\cdot Lap(0,b')+Lap(0,a')\cdot Lap(0,b')) \\ &=E(a\cdot b)+b\cdot E(Lap(0,a')) \\ &\quad +a\cdot E(Lap(0,b'))+E(Lap(0,a')\cdot Lap(0,b')) \\ \end{aligned}$$

E(Lap(0, a′)) = E(Lap(0, b′)) = 0; besides, $E(Lap(0,a')\cdot Lap(0,b'))=E(Lap(0,a'))\cdot E(Lap(0,b'))$ since a, b are independently perturbed with Laplace noise; hence

$$E((a+Lap(0,a'))\cdot(b+Lap(0,b'))) =E(a\cdot b)$$

Proof of Lemma 3

$$\begin{aligned} E\left(\frac{a+Lap(0,a')}{b+Lap(0,b')}\right) &=E\left(\frac{a}{b}\right)+E\left(\frac{a+Lap(0,a')}{b+Lap(0,b')}-\frac{a}{b}\right) \end{aligned}$$

Since E(Lap(0,a′)) = 0 and E(0, Lap(b′)) = 0, we have

$$\begin{aligned} E\left(\frac{a+Lap(0,a')}{b+Lap(0,b')}-\frac{a}{b}\right) & = E\left(\frac{(a+Lap(0,a')b-a(b+Lap(0,b')))}{b\cdot(b+Lap(0,b'))}\right) \\ & = E\left(\frac{b\cdot Lap(0,a')}{b\cdot(b+Lap(0,b'))}\right) -E\left(\frac{a\cdot Lap(0,b')}{b\cdot(b+Lap(0,b'))}\right)\\ &=0 \end{aligned}$$

Proof of Result 1

To derive $LS_{C_{i}}^{(s)}(G),$ we first consider the case for s = 0, i.e., $LS_{C_{i}}(G).$

Let G and G′ respectively denote the original graph G and its neighbor graph by deleting an edge from G. For a given node i: let $N_{\Updelta}(i)$ and N ₃(i) denote the attributes of i in G, while $N'_{\Updelta}(i)$ and N′₃(i) denote the same attributes in G′. By definition, we have $0\leq N_{\Updelta}(i) \leq N_{3}(i) = 2/(d_{i}(d_{i}-1)),$ when deleting an edge from $G, N_{\Updelta}(i)$ would be decreased by at most d _i  − 1; while N ₃(i) would be decreased by exactly d _i  − 1. Therefore,

$$\begin{aligned} LS_{C_{i}}(G) &=\frac{N_{\Updelta}(i)}{d_{i}(d_{i}-1)/2}-\frac{N'_{\Updelta}(i)}{d_{i}(d_{i}-1)/2-(d_{i}-1)} \end{aligned}$$

(18)

when $0\leq N_{\Updelta}(i)\leq d_{i}-1,$ we have

$$\begin{aligned} LS_{C_{i}}(G) &\leq \frac{N_{\Updelta}(i)}{d_{i}(d_{i}-1)/2}\leq \frac{d_{i}-1}{d_{i}(d_{i}-1)/2}=2/d_{i} \end{aligned}$$

when $d_{i}-1 \leq N_{\Updelta}(i)\leq d_{i}(d_{i}-1)/2 ,$ we have

$$\begin{aligned} LS_{C_{i}}(G) &\leq \frac{N_{\Updelta}(i)}{d_{i}(d_{i}-1)/2}-\frac{N_{\Updelta}(i)-(d_{i}-1)}{d_{i}(d_{i}-1)/2-(d_{i}-1)}\\ &= \frac{2}{d_{i}-2}-\frac{4 N_{\Updelta}(i)}{d_{i}(d_{i}-1)(d_{i}-2)} \\ &\leq \frac{2}{d_{i}-2}-\frac{4(d_{i}-1)}{d_{i}(d_{i}-1)(d_{i}-2)}=2/d_{i} \end{aligned}$$

So that $LS_{C_{i}}(G)= \frac{2}{d_{i}}.$

In general case, for s > 0, we have (Eq. 7),

$$LS_{C_{i}}^{(s)}(G)=\max_{G'\in D: d(G,G')\leq s} {LS_{C_{i}}(G')}=\frac{2}{d_{i}-s}$$

for d _i  − s > 2; and $LS_{C_{i}}^{(s)}(G)=GS_{C_{i}}(G)=1$ otherwise.

Proof of Lemma 4

Since E(Lap(0,a′)) = 0 and $E(Lap^{2}(0,a'))=a^{'2}, \,\, \tilde{a}=a+Lap(0,a'),$ therefore

$$\begin{aligned} &E((u_1\cdot \tilde{a}+v_1)(u_2\cdot \tilde{a}+v_2)) \\ &\quad=E((u_1\cdot a+v_1)(u_2\cdot a+v_2)) \\ &\qquad +(u_1\cdot(u_2\cdot a+v_2)+u_2\cdot(u_1\cdot a+v_1))E(Lap(0,a^{\prime}))\\ &\qquad +u_1\cdot u_2\cdot E(Lap^2(0,a^{\prime})) \\ &\quad=E((u_1\cdot a+v_1)(u_2\cdot a+v_2))+u_1\cdot u_2\cdot a^{\prime 2} \end{aligned}$$

Proof of Result 3

We first consider the situation of s = 0,

$$\begin{aligned} LS_{C}(G)&=\max_{a_{ij}=1} \bigg\{ \frac{2}{d_{i}}+\frac{2}{d_{j}}+\sum_{a_{ik}a_{jk}=1} \frac{1}{d_{k}(d_{k}-1)/2} \bigg\} \\ &\leq \max_{a_{ij}=1}\left\{\frac{2}{d_{i}}+\frac{2}{d_{j}}+\frac{2(d_{\max}-1)}{d_{k}(d_{k}-1)}\right\} \\ &\leq 2\left(\frac{1}{2}+\frac{1}{2}+\frac{2(d_{\max}-1)}{2*(2-1)}\right) =d_{\max} \end{aligned}$$

LS ^(s) _C (G) = LS _C (G) + s for s ≤ b _ij because we may add one edge to complete a half-built triangle involving edge (i, j) which makes the sensitivity increased by at most one; meanwhile, $LS_{C}^{(s)}(G)=LS_{C}(G)+\lfloor\frac{s+b_{ij}}{2}\rfloor$ for s > b _ij because we have to add two edges to form a triangle to make the sensitivity increased by one, after completing all the b _ij half-built triangles involving edge (i, j). Besides, LS ^(s) _C (G) ≤ GS _C (G) = n − 1. So we have Eq. (17).

Proof of Result 4.

In addition to the Proof of Result 2 given in Nissim et al. (2007), $LS_{N_{\Updelta}}^{(s)}(G)=3\cdot\max_{i\neq j;\,j\in[n]} {c_{ij}(s)}$ because each of the two entries corresponding to vertex i and j will be decreased by at most $\max_{i\neq j;\,j\in[n]} {c_{ij}(s)} ,$ when edge (i, j) is deleted from G. Besides, there are $\max_{i\neq j;\,j\in[n]} {c_{ij}(s)}$ other entries whose values will be decreased by one, corresponding to the neighbors in common by vertex i and j.

For N ₃, we first consider the situation of s = 0,

$$\begin{aligned} &LS_{N_{3}}(G)= \max_{a_{ij}=1} \left\{ \frac{d_{i}(d_{i}-1)}{2}-\frac{(d_{i}-1)(d_{i}-2)}{2} + \right. \left.\frac{d_{j}(d_{j}-1)}{2}-\frac{(d_{j}-1)(d_{j}-2)}{2} \right\}\\ &\leq \max_{a_{ij}=1} \{d_{i}-1+d_{j}-1 \} \leq d_{\max}+d_{\rm secondmax}-2 \end{aligned}$$

In general case, $LS_{N_{3}}^{(s)}(G)=LS_{N_{3}}(G)+s$ for s ≤ b _ij and $LS_{N_{3}}^{(s)}(G)=LS_{N_3}(G)+\lfloor\frac{s+b_{ij}}{2}\rfloor$ for s > b _ij which are similar to those of LS ^(s) _C (G), and $LS_{N_{3}}^{(s)}(G)\leq GS_{N_{3}}(G)=2n-4$. So we have the form for $LS_{N_{3}}^{(s)}(G).$