Zero-One Law for Connectivity in Superposition of Random Key Graphs on Random Geometric Graphs

Random key graph (RKG), also known as uniform random intersection graph, is a random graph defined below. Consider a set with n nodes and another key pool with P n keys; we assume each node randomly chooses K n distinct keys for its key ring; two nodes can establish a secure link between them if they share at least one common key in their key rings. The random key graph is naturally associated with the random key predistribution scheme of Eschenauer and Gligor [1] for wireless sensor networks (WSNs). A WSN is a collection of distributed sensor devices that are able to communicatedwirelessly and supports wide range of applications such as health and environment monitoring, imaging, tracking, and biomedical research; see [2].These applications require all nodes in the network to be within communication range and to be connected with high probability. Some partial results concerning the connectivity of RKGs were given in [3–5]. In [6], Rybarczyk gave asymptotic tight bounds for the thresholds of the connectivity, phase transition, and diameter of the largest connected component in RKGs for all ranges ofK n . With the advent of ad hoc sensor networks, an interesting class of random graphs, namely, random geometric graphs (RGGs), has gained new importance and its properties have been the subject of much study. Here the nodes are randomly distributed in a finite Euclidean space and there is an edge between two nodes if the Euclidean distance between them is below a specified threshold. Much work has been done on graph theoretic properties of such graph, comprehensively summarized in the monograph of [7]. Recently, there is interest in random graphs in which an edge is determined by more than one random property, that is, superposition of different random graphs. The superposition of ER random graphs over RGGs has been of interest for quite some time now. Recent work on such random graphs is in [8, 9] where connectivity properties and the distribution of isolated nodes are analyzed. And the superposition of ER random graphs on RKGs is considered in [10]. Such a graph is constructed as follows: a RKG is first formed based on the key distribution and each edge in this graph is deleted with a specified probability. The superposition of RKGs on RGGs is first studied in [11]. The n nodes are distributed in a finite Euclidean space and each node is assigned a key ring ofK n distinct keys drawn randomly from a pool of P n keys. Two nodes have an edge if and only if they share at least one common key in their key rings and their Euclidean distance is at most r n . Pietro et al. [11] have shown that under the scalingπr n K 2


Introduction
Random key graph (RKG), also known as uniform random intersection graph, is a random graph defined below.Consider a set with  nodes and another key pool with   keys; we assume each node randomly chooses   distinct keys for its key ring; two nodes can establish a secure link between them if they share at least one common key in their key rings.
The random key graph is naturally associated with the random key predistribution scheme of Eschenauer and Gligor [1] for wireless sensor networks (WSNs).A WSN is a collection of distributed sensor devices that are able to communicated wirelessly and supports wide range of applications such as health and environment monitoring, imaging, tracking, and biomedical research; see [2].These applications require all nodes in the network to be within communication range and to be connected with high probability.
Some partial results concerning the connectivity of RKGs were given in [3][4][5].In [6], Rybarczyk gave asymptotic tight bounds for the thresholds of the connectivity, phase transition, and diameter of the largest connected component in RKGs for all ranges of   .
With the advent of ad hoc sensor networks, an interesting class of random graphs, namely, random geometric graphs (RGGs), has gained new importance and its properties have been the subject of much study.Here the nodes are randomly distributed in a finite Euclidean space and there is an edge between two nodes if the Euclidean distance between them is below a specified threshold.Much work has been done on graph theoretic properties of such graph, comprehensively summarized in the monograph of [7].
Recently, there is interest in random graphs in which an edge is determined by more than one random property, that is, superposition of different random graphs.The superposition of ER random graphs over RGGs has been of interest for quite some time now.Recent work on such random graphs is in [8,9] where connectivity properties and the distribution of isolated nodes are analyzed.And the superposition of ER random graphs on RKGs is considered in [10].Such a graph is constructed as follows: a RKG is first formed based on the key distribution and each edge in this graph is deleted with a specified probability.
The superposition of RKGs on RGGs is first studied in [11].The  nodes are distributed in a finite Euclidean space and each node is assigned a key ring of   distinct keys drawn randomly from a pool of   keys.Two nodes have an edge if and only if they share at least one common key in their key rings and their Euclidean distance is at most   .Pietro et al. [11] have shown that under the scaling  2   2  /  = (log /), the one-law that this class of random graphs is connected follows if   > 0 and  > 20.Another notable work is due by Krzywdziński and Rybarczyk [12], where the authors have improved these results and established the one-law for  > 8 without any constraint on   .Recently, Krishnan et al. [13] have shown that for large , this class of random graphs will be connected if   ≥ 2,   and   are selected such that for any  > 0 and 0 <  < 1.They also observed that for large  and 0 <  1 < ∞, the probability that this class of random graphs is disconnected is at least  − 1 /4 if the scaling satisfies The connectivity in the superposition of RKGs on RGGs is still studied in this paper.Assuming that   ≥ , we show that given  2   2  /  = log +  , this class of random graphs is disconnected if   → −∞, and for   → ∞, this class of random graphs is connected.
The rest of the paper is organized as follows.Our main result is presented in Section 2. Namely, the theorem concerning zero-one law for graph connectivity is presented.Section 3 contains technical proof of Theorem 1. Finally, Section 4 discusses prospects of establishing tighter connectivity thresholds in the superposition of RKGs on RGGs.

Main Result
The  nodes are uniformly and independently distributed in R = [0, 1] 2 .Let   ∈ R be the location of point .A key pool with   cryptographic keys is designated for the network of  nodes.Node  randomly chooses a subset   of keys from the key pool with |  | =   .Our interest is in the random graph (  ,   ,   ) with  nodes and edges formed as follows.An edge (, ), 1 ≤  <  ≤ , is present in (  ,   ,   ) if both of the following two conditions are satisfied: where ‖ ⋅ ‖ represents the Euclidean norm.Condition  1 produces a random geometric graph with the transmission range   .Imposing condition  2 on  1 retains the edges of the random geometric graph for which the two nodes share at least one common key.Thus (  ,   ,   ) is a superposition of RKG on RGG.
In the following, to avoid technicalities which obscure the main ideas, we will neglect edge effects resulting due to the fact that  nodes are distributed uniformly and independently over a folded unit square R = [0, 1] 2 with continuous boundary conditions and a node is close to the boundary of R. Throughout the paper, we set  2  =   , where   = (log ) and   = ( (1−)/4 ) for any small 0 <  < 1.The following theorem gives zero-one law for the connectivity of a superposition of RKG on RGG.The first part of the Theorem 1 is proved by using the second moment method, that is, considering the probability of finding at least one isolated node in the network for a random graph (  ,   ,   ).The second part takes a slightly different approach, we assume that R = [0, 1] 2 is entirely coved by circle cells of radius   /2.Connectivity of (  ,   ,   ) is ensured as follows: (1) every circle cell is dense; namely, every circle cell has Θ( 2  ) nodes inside it; (2) the overlapping structure of any two adjacent circle cells has at least one common node; and (3) the nodes in any circle cell form a connected subgraph.

Proof of Theorem 1
Before proceeding, we first introduce some definitions and auxiliary lemmas.For  = 1, 2, . . ., , let   = 1 if node  is isolated in (  ,   ,   ) and  = ∑  =1   .Then,  is exactly the number of isolated nodes in (  ,   ,   ).Let || denote the cardinality of a set  and let (), Var() denote the expectation and variance of random variable , respectively.As a special case of Markov's inequality the first moment method states that and the second moment method (special case of Tschebyscheff 's inequality) states that If  is a binomial distributed random variable,  =  and for any  ≥ 0, then we will use the following variants of Chernoff 's inequality (see [14]): Of course, it is easy to check that under the assumption that lim  → ∞ ( 2  /  ) = 0, where   is the probability that two nodes share at least one common key.Throughout, we make use of the standard bounds valid for all ,  = 1, 2, . . .with  ≤ .Finally, we note the equation , where   is the probability that two nodes share at least one common key in their key rings.By linearity of expectation,  = (∑  =1   ) =   ; hence as   → −∞.The second moment method now implies the result we require, provided that we can show that Var = (() 2 ).Now and so it suffices to show that ( 2 ) = (1 + ( 1))() 2 .Note that where 1, 2 are fixed nodes.Since  → ∞, it therefore suffices to prove that Note that  1  2 take the value 1 exactly when node 1 and node 2 are both isolated.Consider two discs of radius   centered at  1 and  2 ; let  = ‖ 1 − 2 ‖; the cross term (−1)( 1  2 ) is shown to be given by Conditional on the range of , we consider the following three cases.In each case, the conditional joint probability of two nodes being isolated can be obtained from [13].
If     = log  +   for any   → ∞, we have  → 0. Then using the first moment method, we see that the probability Pr( ≥ 1) ≤  → 0 holds; this implies a.a.s.there are no isolated nodes in random graph (  ,   ,   ).The upcoming corollary is immediate from the proof of statement (i) of Theorem 1.
Proof of Statement (ii) of Theorem 1.We consider the unitarea square on R = [0, 1] 2 ; R is divided into square cells of size   ×   , where 1/  is an integer.Let the center of square cell be the center of the circle cell and let the diagonal line of the square cell be the diameter of the circle cell.In this way, R is entirely covered by the circle cells.Also we let   = √ 2  ; this means that two nodes in the same circle cell are within communicating range of each other.

Lemma 3. (i) Every circle cell is dense; specifically, whp every circle cell has Θ(𝑛𝑟
Consequently ) , which implies that every circle cell is dense.
(ii) Now we consider the nodes in the overlapping structure of any two adjacent circle cells.Let   denote the number of nodes in the overlapping structure of any two adjacent circle cells.Clearly   is a binomial random variable with parameters (, ((−2)/4) 2  ).For any 0 <  2 < 1, we also use Chernoff 's inequalities on   : From the two above inequalities we may easily get The above expression implies that   is very likely close to its expectation (( − 2)/4) 2  .So we get our result that whp the overlapping structure of any two adjacent circle cells has at least one common node.Lemma 4. The nodes in any circle cell form a connected subgraph; that is, for any fixed 1 ≤  ≤ 1/ We let  , denote the number of distinct keys in the component of size  in   .Adapting [5], for any  =   , . . ., min{  ,   }, we have From [5], we know that First we use the standard Poissonization technique [7,15] to show the probability of having isolated nodes in any of the circle cells.Denote the circle  by   ; let   = 1 be the event that node  is isolated in   , and let    (V  ) be the intersection of   and the disk centered at position V ∈   with radius   , where node  is at position V .Similar to the discussion in [16], the number of nodes within    (V  ) follows a Poisson distribution with mean    (V  ); and to have an edge with , a node not only has to be within a    (V  ) but also has to share at least a key with node , so the number of nodes neighboring to  follows a Poisson distribution with mean      (V  ).Integrating V over   , the probability that node  is isolated in   is given by The probability that there are no isolated nodes in any of the circle cells is bounded below: The second step and the third step are obtained since Next we prove that every circle cell  contains no component of size , 2 ≤  ≤   /2.The sum term in (30) is evaluated in following three cases based on the size of the component .
First, we analyze the denseness of every circle cell; let   denote the number of nodes in circle cell , 1 ≤  ≤ 1/ 2  .Obviously,   is a binomial random variable with parameters (, (/2) 2  ).Let   denote the event that the circle cell  is not dense, in other words, for any 0 <  1 < 1/2,  1 ≤ , 2  ) nodes in it.