Secure connectivity of wireless sensor networks

1. Secure connectivity of wireless sensor networks AyalvadiGanesh University of Bristol Joint work with Santhana Krishnan and D. Manjunath

2. Problem statement • N nodes uniformly distributed on unit square • Pool of P cryptographic keys • Each node is assigned K keys at random • Two nodes can communicate if they are within distance r of each other and share a key Q: For what values of N, K, P and r is the communication graph fully connected?

3. Background: Random key graphs Eschenauer and Gligor (2002): Key distribution scheme for wireless sensor networks Yagan and Makowski (2012): Analysis of the full visibility case Theorem: Suppose P=(N). Let K2/P = (log N+N)/N Then, P(connected)  1 if N  + and P(connected)  0 if N  

4. Heuristic explanation • Probability of an edge between two nodes is approximately K2/P • Mean degree of a node is approximately NK2/P = log N + N • Edges are not independent but, if they were: • key graph would be an Erdos-Renyi random graph • has connectivity threshold at mean node degree of log N

5. Background: Random geometric graphs • N nodes uniformly distributed on unit square • Edge probability g(x/rN) for node pairs at distance x from each other • Boolean model: g(x) = 1(x<1) Penrose: Let NrN2 = log N + N P(connected) 1 if N, and 0 if N

6. Generalisations • Mao and Anderson: • Similar model but with Poisson process of nodes on infinite plane. • Same scaling of rN • Under suitable conditions on g, show a threshold between having isolated nodes in unit square, and no components of finite order in unit square

7. Results for geometric key graphs • Mean node degree r2K2/P • If r2K2/P = log N + c, then P(graph is disconnected) > ec/4 • If r2K2/P = c log N and c>1, then P(graph is connected) 1

8. Upper bound on connection probability • Graph is disconnected if there is an isolated node P(node j is isolated)  (1r2K2/P)N  exp( r2NK2/P) ec/N • Bonferroni inequality: P(there is an isolated node) ≥ iP(i is isolated)  i<jP(i and j are isolated)

9. Isolation of pairs of nodes

10. Lower bound on connection probability • Approach for ER graphs • Compute probability that there is a connected component of m nodes isolated from other nm • Take union bound over all ways of choosing m nodes out of n, and over all m between 1 and n/2

11. Approach for geometric key graphs • Tesselate unit square with overlapping squares of side r/2

12. Approach for geometric key graphs • Are there disconnected components of different sizes in the unit square? • Are there “locally” disconnected components of different sizes within the small squares of side r/2, considering only nodes within that square?

13. Big picture of proof • There are no small – size O(1) – components in the unit square disconnected from rest • There are no large – size > 6 – locally disconnected components in any small square • Can also bound the number of nodes in small components within a small square : very few of them • So how might the graph be disconnected?

14. Notation • N: number of nodes in unit square • r: communication radius of a node • P: size of key pool • K: number of keys assigned to each node • n=r2N: expected number of nodes within communication range • p=K2/P: approximate probability that two nodes share a key

15. Assumptions • N,K,P, K2/P0 • nK2/P  c log N for some c>1 • K > 2 log N • Corollary: • Number of nodes in each small square is (log N) • concentrates near its mean value of n/(2) • uniformly over all squares

16. Within a small square • n/(2) nodes, full visibility • Mean degree is nK2/(2P) = c/(2) log N • Even if edges were independent, expect to see local components of size up to 2, somewhere in the unit square Show there are no bigger components, taking edge dependence into account

17. Within a small square • Say there is a connected component of size m isolated from the rest • Say these m nodes have mKj keys between them • Then • j ≥ m1 • None of the other nm nodes in the square has one of these mKj keys

18. Number of keys among m nodes • Assign K distinct keys to first node • Assign subsequent keys randomly with replacement • P(collision at (i+1)th step)  i/P, independent of the past • P(j collisions)  P(≥j collisions)  ?

19. Collision probability bounds • X1, X2, … , Xnindependent Bernoulli random variables • Xi ~ Bern(pi) • Y = X1+…+Xn • Z is Poisson with the same mean as Y • Hoeffding (1956): Z dominates Y in the convex stochastic order

20. Within the big square • Say nodes 1,2,…,m form a connected component isolated from the rest. Then • for some permutation of 1,2,…,m there is an edge between each node and the next • they hold mKjkeys between them, for some j • there is no edge between the remaining Nm nodes and these m

21. Putting the pieces together • Most nodes belong to a giant component • Each small square may contain some nodes that are locally isolated or within small components • These must either be connected to the giant component in a neighbouring cell, or to another small component • Latter is unlikely, doesn’t percolate