Connectivity of sparse Bluetooth networks

Consider a random geometric graph defined on $n$ vertices uniformly distributed in the $d$-dimensional unit torus. Two vertices are connected if their distance is less than a"visibility radius"$r_n$. We consider {\sl Bluetooth networks} that are locally sparsified random geometric graphs. Each vertex selects $c$ of its neighbors in the random geometric graph at random and connects only to the selected points. We show that if the visibility radius is at least of the order of $n^{-(1-\delta)/d}$ for some $\delta>0$, then a constant value of $c$ is sufficient for the graph to be connected, with high probability. It suffices to take $c \ge \sqrt{(1+\epsilon)/\delta} + K$ for any positive $\epsilon$ where $K$ is a constant depending on $d$ only. On the other hand, with $c\le \sqrt{(1-\epsilon)/\delta}$, the graph is disconnected, with high probability.


Introduction and results
Consider the following model of random "Bluetooth networks".Let X = X 1 , . . ., X n be independent, uniformly distributed random points in [0, 1] d , and denote the set of these points by X = {X 1 , . . ., X n }.Given a positive number r n > 0 (the so-called visibility radius), define the random geometric graph G n (r n ) with vertex set X in which two vertices X i and X j are connected by an edge if and only if D(X i , X j ) ≤ r n , where is the Euclidean distance on the torus.
It is well known (Penrose [11]) that, for any ǫ > 0, , where v d is the volume of the Euclidean unit ball in R d .
Note that in order to guarantee that a random geometric graph is connected (with high probability), the average degree in the graph needs to be at least of the order of log n, which makes the graph too dense for some applications.To deal with this issue, one may sparsify the graph.This can be done in a distributed way by selecting, for each vertex X i , randomly, and independently a subset of c n edges adjacent to X i , and then considering the subgraph containing the selected edges only.The selection is done without replacement and if a vertex has less than c n neighbors in G n (r n ), then we take all of its neighbors.The obtained random graph model, coined Bluetooth network (or irrigation graph) has been introduced and studied in Crescenzi et al. [5], Dubhashi et al. [6,7], Ferraguto et al. [9], and [12].
Formally, the random Bluetooth graph Γ n = Γ n (r n , c n ) is obtained as a random sub-graph of G n (r n ) as follows.For every vertex X i ∈ X, we pick randomly, without replacement, c n edges, each adjacent to X i in G n (r n ).(If the degree of X i in G n is less than c n , all edges adjacent to X i are kept in Γ n .)We also denote by Γ + n (r n , c n ) the directed graph obtained by placing a directed edge from X i to X j whenever X j is among the c n selected neighbors of X i .
We study connectivity of Γ n (r n , c n ) for large values of n.A property of the graph holds with high probability (whp) when the probability that the property holds converges to one as n → ∞.
When r n > √ d/2, the underlying random geometric graph G n (r n ) is the complete graph and Γ n (r n , 2) becomes the "2-out" random subgraph of the complete graph studied in Fenner and Frieze [8], where it is shown that the graph is connected with high probability.Dubhashi et al. [6] extended this result by showing that when r n = r > 0 is independent of n, Γ n (r, 2) is connected with high probability.When r n → 0 as n → ∞, Crescenzi et al. [5] proved that there exist nected with high probability.Broutin et al. [3] proved that when r n is just above the connectivity threshold for the underlying graph G n (r n ), that is, when r n ∼ γ(log n/n) 1/d for some sufficiently large γ, the connectivity threshold for the irrigation graphs is More precisely, for any ǫ ∈ (0, 1), This result shows that the simple distributed algorithm building the irrigation graph guarantees connectivity with high probability while reducing the average degree to about 2c ⋆ n , which is much less than the average degree of Θ(log n) for the initial random geometric graph.However, the average degree still grows with n, and therefore Γ n (r n , c n ) is not genuinely sparse.
MAIN RESULTS.The main result of this note is that at the price of slightly increasing the visibility radius to r n ∼ n −(1−δ)/d for some δ > 0, a constant number of connections per vertex suffices to achieve connectivity with high probability.The lower bound of Broutin et al. [3] states that for any ǫ ∈ (0, 1) This bound may seem weak since this value of c n is just a constant, independent of n.However, we show here that this bound is essentially tight: We prove that when r n = Ω(n −(1−δ)/d ), for some δ > 0, then Γ n (r n , c n ) is connected whp whenever c n is larger than a certain constant, which depends on δ and d only.
Theorem 1.Let δ ∈ (0, 1), γ > 0, and ǫ ∈ (0, 1) be fixed.Suppose that r n = γ n n −(1−δ)/d where γ n /γ → 1 as n → ∞.There exists a constant c, depending on δ and d only, such that the random Bluetooth graph Γ(r n , c) is connected whp.For x ∈ (0, 1) set where where v d is the volume of the Euclidean ball of radius Furthermore, as δ becomes small, our upper bound for the constant depends on δ as ((1 + ǫ)/δ) 1/2 + O(1), essentially matching the lower bound of Broutin et al. [3] mentioned above.A straightforward combination of the lower bound and Theorem 1 implies the following: Theorem 2. Let c(δ) denote the smallest integer c for which and let c(δ) denote the largest integer c for which Then, for any ǫ > 0, we have, for all δ small enough, Interestingly, the threshold is essentially independent of the value of γ and the dimension d.
This phenomenon was already observed in (1).
REMARKS AND OPEN QUESTIONS.Before we conclude this section, we mention a few questions that might be worth investigating.Theorem 2 above finds the correct asymptotics for the thresholds c(δ), c(δ) when the visibility radius is r n ∼ γn −(1−δ)/d for small δ.One may phrase a related question as follows: given a fixed a constant c, (a budget, in some sense) can one find a threshold function r ⋆ n = r ⋆ n (c) for connectivity?More precisely, r ⋆ n should be such that, for any ǫ > 0, In some sense, Theorem 2 gives the asymptotics of the threshold function for c → ∞, but one would like to know the threshold for fixed values of the budget c.As it was proved in [3] in the case that r n ∼ γ(n −1 log n) 1/d , the main obstacles to connectivity should be isolated (c + 1)cliques.One could also try to prove that this is indeed the case at a finer level: around the threshold, the number of isolated (c + 1)-cliques should be asymptotically Poisson distributed.Thus, one expects that the probability that Γ n (r n , c) is connected should have asymptotics similar to that of classical random graphs [1, Theorem 7.3] or random geometric graphs [11,Theorem 13.11], where the isolated vertices are the main obstacle.Finally, we mention that elsewhere (Broutin et al. [4]) we investigate the birth of the giant component of Γ(r n , c).

Proof sketch
In the course of the proof, we condition on the location of the points and assume that they are sufficiently regularly distributed.The probability that this happens is estimated in the following Lemma 1. Suppose r n ≥ γn −(1−δ)/d for some δ ∈ (0, 1) and γ > 0. Let ǫ > 0 and denote by F the event that for all x ∈ [0, 1] d , Then there exists a constant θ = θ(δ, ǫ) > 0 and a positive integer Proof.The lemma may be proved by standard arguments.It follows, for example, from inequalities for uniform deviations of empirical measures over VC classes such as Theorem 7 in [2].
In the rest of the proof, we fix X = {X 1 , . . ., X n }, and assume that event F holds.Thus, all randomness originates from the random choices of the c neighbors of every vertex.We denote by P c the probability with respect to the random choice of the neighbors only (i.e., conditional given the set X).It suffices to show that if F holds, then with high probability (with respect to P c ), Γ(r n , c) is connected.
SKETCH OF THE PROOF.The general strategy is to prove that, with high probability, from any two points X i , X j , one can find a path that connects X i to X j .To do this, most of the work consists in proving structural properties of the connected component containing a fixed point X 1 .We study the random graph by dividing the c neighbors into four disjoint groups of sizes k 1 , k 2 , k 3 , and 1, thus obtaining four independent sets of edges added in four different phases.The sketch of the proof of Theorem 1 is as follows.We rely on a discretization of the unit cube [0, 1] d into congruent cubes of side length 1/⌈2r n √ d⌉.
• SEARCHING FOR A DENSE CUBE I. (Section 3).In the first phase we start from an arbitrary vertex, say, X 1 , and using only k 1 choices of each vertex, consider the set of the vertices of X which may be reached using paths of at most ℓ ≈ δ 2 log 2 n edges.We show that if this growth process succeeds, there exists a cube in the grid partition that contains a connected component of size at least n min{δ 2 ,1/25}/3 .
• SEARCHING FOR A DENSE CUBE II.(Section 4).In the second phase we show that by adding k 2 new connections to each vertex in the component obtained after the first step, at least one of the grid cells has a positive fraction of its points in a single connected component.
• PROPAGATING THE DENSITY.(Section 5).Once we have a cell containing a constant proportion of points belonging to the same connected component, it is rather easy to propagate this positive density (of a single connected component) to all other cells of the grid by using k 3 new connections per vertex.
• CONNECTIVITY IS UNAVOIDABLE.(Section 6).The previous phases guarantee that from a single vertex X 1 all three phases succeed with probability 1 − o(1/n).So, with high probability, the connected components of every single vertex X i , 1 ≤ i ≤ n, reach in every corner of the space.Then, it is easy to show that with just one additional connection per vertex, any two such components very likely connect, proving that the entire graph is, in fact, connected.
3 First growth process: searching for a dense cube I We first prove that a constant number of edges per vertex suffices to guarantee that there exists a cell that contains at least a polynomial number of points of X that may be reached from X 1 .
Lemma 2. Suppose X = {X 1 , . . ., X n } are such that the event F defined in Lemma 1 occurs.
Let k 1 and r n be defined as in Theorem 1.With probability at least 1 − o(1/n), the connected component of Γ(r n , k 1 ) containing X 1 is such that there exists a cell in the grid partition of [0, 1] d into congruent cubes of side length 1/⌈2 √ d/r n ⌉ that contains at least n min{δ 2 ,1/25}/3 vertices of the component.
Proof.Let ℓ < n be a positive integer specified below.Consider A, the set of vertices of X that can be reached from X 1 using a directed path of length at most ℓ in We first show a lower bound on the size of this connected component: To see this, the key property is that with high probability, the number of new points added in the second generation is at least 2k 1 .First, the k 1 points of the first generation are distinct for they are sampled without replacement.For the second generation, imagine k 2 1 bins, k 1 for each of the k 1 vertices of the first generation, into which we place the points chosen by these vertices.For these k 2  1 bins to contain only j different points that are also different from the points of the first generation, there must exist k 2 1 − j bins that contain only points of the j remaining bins or points from the first generation.There are k 2 1 j ways of choosing these k 2 1 − j bins and for each such bin, the probability that it contains a point either from the remaining j bins or from points of the first generation is at most ) (since, on the event F , each ball of radius r n contains at least nr d n v d (1 − ǫ) points).Thus, the probability that the number of distinct points in the second generation that are distinct and do not belong to the first generation is less than 2k 1 is at most Similarly, assuming that there are at least 2k 1 points in the second generation, the probability that the number of selected neighbors in the third generation not selected before is less than 4k 1 is at most We may continue in this fashion for ℓ − 1 steps, in each step doubling the number of neighbors with high probability.The probability that the ℓ-th generation has less than Now, we choose ℓ = ℓ(δ) = ⌊min{δ 2 , 1/25} log 2 n⌋, and distinguish two cases depending on the value of δ.
Thus, we have shown that if n is sufficiently large, then with probability at least In particular, the points of A all fall in grid cells at most ⌈r n ℓ⌉/(r n /(2 √ d))+ 1 ≤ 1 + 2ℓ √ d away from the cell containing X 1 .Thus, all these vertices fall in a cube of at most This implies that there must exist a cell with at least vertices for n large enough.
By Lemma 2 we see that, with probability at least 1 − o(1/n), after k 1 connections per vertex, there exists a grid cell that contains at least n min{δ 2 ,1/25}/3 vertices of the connected component containing X 1 .Next we show that with new connections, the same cell contains a constant times nr d n vertices in the same connected component (i.e., a linear fraction of all points in the cell).
Consider the cell that contains the largest number of vertices in the connected component containing X 1 after the first k 1 connections, and let N 0 denote the number of such connected vertices in this cell.Then we have seen in Section 3 that with respect to the random choices of the first k 1 connections.Suppose that the event N 0 ≥ n min{δ 2 ,1/25}/3 holds.Now add k 2 fresh connections to each of these N 0 points, resulting in N 1 ≤ k 2 N 0 vertices in the same grid cell that haven't been in the connected component of X 1 so far.If the number N 1 of new vertices in the cell added to the component is less than 2n min{δ 2 ,1/25}/3 , we declare failure, otherwise continue by adding k 2 new connections to these N 1 points.(Note that since these N 1 vertices did not belong to the component of X 1 before the first step, we have not discovered any of their connections and we may use k 2 new connections per vertex.)In this step we add N 2 ≤ k 2 N 1 new vertices in the same grid cell.If N 2 < 4n min{δ 2 ,1/25}/3 , we declare failure, otherwise continue.We repeat adding k 2 connections to all newly discovered vertices until the number of connected vertices in the grid cell ) or else for L steps, requiring in every step i = 1, . . ., L that the number of newly discovered vertices in the same cell be at least 2 i n min{δ 2 ,1/25}/3 .Here L is chosen such that To estimate the probability that the process described above fails, observe first that at step i, the (conditional) probability that a vertex selects a neighbor in the same cell is at least since on the event F , every grid cell has at least n points and the vertex can reach at most (1 + ǫ)nr d n v d points.In the inequality we used the fact that the number of vertices discovered until step i during the process the process stops with success before step i.
Thus, after i − 1 successful steps, the expected number of newly discovered vertices at stage i is at least by the definition of k 2 , where E i denotes conditional expectation given the first growth process and the first i − 1 steps of the second process.Given that the process has not failed up to step i − 1, the conditional probability that it fails at step i is thus where we used the fact that conditionally, N i has a hypergeometric distribution whose moment generating function is dominated by that of the corresponding binomial distribution B(k (see Hoeffding [10]) and we used simple Chernoff bounding for the binomial distribution.
Thus, the probability that the process ever fails is bounded by for all sufficiently large n.

Third growth process: propagating the density
In this third step we show that, by adding a few more connections, the connected component containing X 1 contains a linear fraction of the points in every cell of the grid partition of [0, 1] d into cubes of side length 1/⌈2 √ d/r n ⌉.In order to do so, we start from the cell containing α d nr d n vertices in the connected component of Γ(r n , k 1 + k 2 ) containing X 1 , and "grow" the component cell-by-cell until every cell has the required number of vertices in the same connected component.
We show that a constant number of additional connections per vertex suffices.Thus, the expected number of newly discovered vertices in cell i + 1, conditionally on the fact that the process has not failed earlier, is at least for all n large enough, where the last inequality follows from our choice of k 3 .Thus, the probability of failure in step i is, by a similar argument as in the proof of Lemma 3, bounded by e −η d nr d n , where Hence, the probability that the process ever fails in any step is o(1/n) and this completes the proof.

Final step: proof of Theorem 1
With a giant component densely populating every cell of the grid, it is now easy to show that with at most one extra connection, the entire graph becomes connected, with high probability.The argument is as follows.
Start with growing the component of X 1 as in the growth processes described above.Then, with k 1 + k 2 + k 3 connections per vertex, the component has the property described in Lemma 4, with probability 1 − o(1/n).Now consider points X 2 , X 3 , . . ., X n , one-by-one.If X 2 does not belong to the connected component of X 1 , then grow the same process starting from X 2 until the component of X 2 hits the one of X 1 .If the two components do not connect, then this new component also satisfies the property of Lemma 4, with probability 1 − o(1/n).(Note that until the two components meet, all connections are new in the sense that they have not been revealed in the first process, and therefore these events hold independently.)Now we may continue, taking every X i , i = 3, . . ., n.If X i does not belong to any of the previously grown components, then we grow a new component starting from X i until it hits one of the previous ones, or else until the component has at least 2α d nr d n /(3k 3 ) vertices in every cell of the grid.By the union bound, with probability 1 − o(1), every vertex is contained in such a giant component.
If there is more than one connected component at the end of the process, then we add one more connection to every vertex not belonging to the component of X 1 .Note that each such component has at least n/(3k 3 ) vertices and for every vertex, the probability that it hits the component of X 1 is at least 1/(3k 3 ), so the probability that all new connections miss the component of X 1 is at most (3k 3 ) −n/3k 3 .Thus, by the union bound, with high probability, all components connect to the first one and the entire graph becomes connected.This concludes the proof of Theorem 1.

Lemma 4 .= 2α d nr d n 3k 3 of
Suppose event F defined in Lemma 1 occurs.Let k 1 , k 2 , k 3 , and r n be defined as inTheorem 1.With probability at least1 − o(1/n), the connected component of Γ(r n , k 1 + k 2 + k 3 ) containing X 1 is such that every cell in the grid partition of [0, 1] d into congruent cubes of side length 1/⌈2 √ d/r n ⌉ contains at least 2α d nr d n /(3k 3 ) vertices of the component.Proof.Suppose the event described in Lemma 3 holds so that there exists a cell with α d nr d n vertices in the connected component of X 1 .Label this cell by 1. Next label all the cells from 1 to ⌈2 √ d/r n ⌉ d in such a way that cells i and i + 1 are adjacent (i.e., they share ad − 1-dimensional face), for all i = 1, . . ., ⌈2 √ d/r n ⌉ d − 1.Note that the size of the cells was chosen such that for any x in cell i the ball B(x, r) contains entirely cell i + 1.In particular, every point in cell i + 1 is a potential neighbor of a point in cell i.Let k 3 = ⌈ 4(1 + ǫ)v d /α d ⌉and consider the following process.Select, arbitrarily, M :the already connected vertices in cell 1 and select, one-by-one, k 3 new neighbors for each until M newly connected vertices have been found in cell 2. Declare failure if the number of newly connected vertices in cell 2 is less than M after revealing all possible k 3 M new connections.Otherwise continue and select k 3 new neighbors of the new vertices in cell 2. Visit all cells in a sequential fashion, always stopping the selection when M new vertices are connected in the next cell.If the process succeeds, then X 1 is connected to at least M points in every cell, which is a positive proportion.To analyze the probability of failure, note that for each new connection, the probability of discovering a new vertex in cell i + 1 by a vertex in cell i is at least 2α d nr d n − k 3 M (1 + ǫ)v d nr d n since (assuming the event F ) there are at least 2α d nr d n vertices in cell i + 1, at most (1 + ǫ)v d nr d n within radius r n of the point whose neighbors we select, and we discard at most k 3 M points cell i + 1 that may have been already be chosen by other vertices in cell i.

4 Second growth process: searching for a dense cube II We
now show that we can leverage Lemma 2 and obtain, still using a constant number of extra edges per vertex, a cell that contains a positive density of points of the connected component containing X 1 .
Lemma 3. Suppose event F occurs.Let k 1 , k 2 , and r n be defined as in Theorem 1.With probability at least1 − o(1/n), the connected component of Γ(r n , k 1 + k 2 ) containing X 1 issuch that there exists a cell in the grid partition of [0, 1] d into congruent cubes of side length 1/⌈2 √ d/r n ⌉ that contains at least α d nr d n vertices of the component, where α