Size of giant component in Random Geometric Graphs -G. Ganesan Indian Statistical Institute, Delhi.

Random Geometric Graph n nodes uniformly distributed in S = [-½, ½]2 Two nodes u and v connected by an edge if d(u, v) < rn Resulting graph  random geometric graph (RGG)

Figure 1: Random Geometric Graph

Radius of Connectivity  

Giant Component Regime   Intermediate, Has giant component Sparse, no Giant Comp. Dense, fully connected r n

Intermediate range  

Theorem 1 (Ganesan, 2012)  

Proof Sketch of Theorem 1 Divide S into small squares {Si}i each of size Choose 4 < ∆ < 5 so that nodes in adjacent squares are joined by an edge. Say that Si is occupied if it contains at least one node and vacant otherwise.

Also, divide the unit square S into ``horizontal” rect. each of size where Fix the bottom rectangle R R

Say that a sequence of occupied squares L = (S1,…,St) form a occupied left-right crossing of R if: Si and Si+1 share an edge for each i. S1 intersects left edge of R. St intersects right edge of R. S St S1

1

Probability of left-right crossing Let LR(R) denote the event that R has an occupied left-right crossing… Lemma (Ganesan, 2012): We have that for some positive constant δ. The constant delta here is independent of the choices of m1 and m2…this is the key…Hence we can fix m1 sufficiently large…to get arbitrarily high probabilities….

if M is sufficiently large. Thus we have that if M is sufficiently large. 1

What is the advantage of identifying occupied left-right crossings.. Ans: We get a path of edges from left to right…

1

1

Thus

What is the probability that each rectangle has such a path? Ans:

Number of rectangles is The number of rectangles is less than since by our choice of rn, we have

Thus… The event that each rectangle has an occupied left-right crossing occurs with prob. And we then get a network of paths from left to right…

1

Thus

Perform an analogous procedure vertically…

1

Thus

Why is this useful? We have obtained a connected “backbone” of paths in G… We have essentially “trapped” isolated components of G in boxes of size…

Isolated component

So, we know that if backbone occurs… then all components not attached to backbone are “boxable” ,i.e., can be fitted in a box…

Let X denote the sum of size of all components of G that are boxable… for some positive constants (1) Recall that backbone occurs with prob..

Thus Note: Whenever backbone occurs, X denotes total sum of sizes of components not attached to the backbone… Therefore

We need to prove the estimate (1) regarding sum of sizes of boxable components… i.e. to prove that Recall: X sum of size of all components of G that are boxable

Proof of (1) How to compute the size of a component that can be fitted in a size box? Main idea…count the number of vacant squares attached to the component…

Observation from figure… “Boxable” components have a circuit of vacant squares attached to them… For any square Si (proved using standard binomial estimates)

We therefore cannot have large boxable components… Because, such components have a lot of vacant squares ``attached” to them…

More precise computation Let S0 be the square containing the origin… Define C0 to be the maximal connected set of occupied squares containing S0… We say that C0 is the cluster containing S0…

S0 C0

S0

We count the number of vacant squares Vs attached to C0 Suppose C0 contains k squares and the outermost boundary ∂C0 of C0 contains L edges…(thick line in fig.) C0 is empty if S0 is vacant…

The following hold: (1) The number of distinct vacant squares attached to ∂C0 is at least Vs > L/8… (2) Since ∂C0 contains k squares in its interior, we must have

(3) Also, the ``last edge” of ∂C0 can cut the X axis at at most L distinct points… (4) And ∂C0 has a self-avoiding path of L-1 steps. (5) The number of choices of ∂C0 is therefore at most L.4L-1

Last edge S0

Thus (1)-(5) implies

Thus…. And if N(C0) denotes the number of nodes is C0, we also have

Recall C0 is the occupied cluster containing S0… Define Ci for each square Si Same conclusion holds…

Use Markov’s inequality to get… for θ sufficiently small… But is the sum of size of all boxable components… And we are done…

Main summary of steps (1) We constructed a backbone of paths with high prob..… (2) We deduced that each component not attached to the backbone is “boxable” (3) We showed that the sum of sizes of all “boxable” components is less than with high prob…

References M. Franceschetti, O. Dousse, D. N. C. Tse and P. Thiran. (2007). Closing Gap in the Capacity of Wireless Networks via Percolation Theory. IEEE Trans. Inform. Theory, 53, 1009–1018. G. Ganesan. (2012). Size of the giant component in a random geometric graph. Accepted for publ. Ann. Inst. Henri Poincare. A. Gopalan, S. Banerjee, A. K. Das and S. Shakottai. (2011). Random Mobility and the Spread of Infection. Proc. IEEE Infocomm, pp. 999–1007.

P. Gupta and P. R. Kumar. (1998). Critical Power for Asymptotic Connectivity in Wireless Networks. Stoch. Proc. and Appl., 2203–2214. M. Penrose. (2003). Random Geometric Graphs. Oxford.