1. Secure connectivity of wireless sensor networks Ayalvadi Ganesh 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 K 2 /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 K 2 /P Mean degree of a node is approximately NK 2 /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/r N ) for node pairs at distance x from each other Boolean model: g(x) = 1(x<1) Penrose: Let  Nr N 2 = 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 r N – 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   r 2 K 2 /P If  r 2 K 2 /P = log N + c, then P(graph is disconnected) > e  c /4 If  r 2 K 2 /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  r 2 K 2 /P) N  exp(  r 2 NK 2 /P)  e  c /N Bonferroni inequality: P(there is an isolated node) ≥  i P(i is isolated)   i<j P(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=  r 2 N: expected number of nodes within communication range p=K 2 /P: approximate probability that two nodes share a key

15. Assumptions N,K,P ,K 2 /P  0 nK 2 /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 nK 2 /(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 X 1, X 2, …, X n independent Bernoulli random variables X i ~ Bern(p i ) Y = X 1 +…+X n 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  j keys 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