Secure connectivity of wireless sensor networks

Ayalvadi GaneshUniversity of Bristol

Joint work with Santhana Krishnan and D. Manjunath

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?

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

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

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

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

Results for geometric key graphs

• Mean node degree r2K2/P • If r2K2/P = log N + c, then

P(graph is disconnected) > ec/4• If r2K2/P = c log N and c>1, then

P(graph is connected) 1•

Upper bound on connection probability

• Graph is disconnected if there is an isolated node

P(node j is isolated) (1r2K2/P)N

exp( r2NK2/P) ec/N• Bonferroni inequality:P(there is an isolated node) ≥i P(i is isolated) i<j P(i and j are isolated)

Isolation of pairs of nodes

Lower bound on connection probability

• Approach for ER graphs– Compute probability that there is a connected

component of m nodes isolated from other nm– Take union bound over all ways of choosing m

nodes out of n, and over all m between 1 and n/2

Approach for geometric key graphs

• Tesselate unit square with overlapping squares of side r/2

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?

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?

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

Assumptions

• N,K,P, K2/P0• 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

Within a small square

• n/(2) nodes, full visibility• Mean degree is nK2/(2P) = 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

Within a small square

• Say there is a connected component of size m isolated from the rest

• Say these m nodes have mKj keys between them

• Then– j ≥ m1– None of the other nm nodes in the square has

one of these mKj keys

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) ?

Collision probability bounds

• X1, X2, … , Xn independent 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

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 mKj keys between them, for some j– there is no edge between the remaining Nm

nodes and these m

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