Phase Transitions in Combinatorial Structures

Phase Transitions in Combinatorial Structures Geometric Random Graphs Dr. Béla Bollobás University of Memphis 1

Subcontractors and Collaborators Subcontractors -None Collaborators Paul Balister József Balogh Christian Borgs Jennifer Chayes Martin Haenggi Alexandr Kostochka Kittikoon Naprasit Oliver Riordan Amites Sarkar Alexander Scott Mark Walters 3

Phase Transitions in Combinatorial Structures design of random combinatorial process simulation of combinatorial process estimation of critical parameters heuristic justification of phenomena rigorous bounds for critical parameters New Ideas Model real-world graphs by rigorously defined models of random graph processes. Define geometric random graphs and simulate their behavior Give bounds on the critical parameters of geometric random graphs Impact Construction of novel random combinatorial processes that either model real-life networks or exhibit more complex behavior Experimental and heuristic bounds for critical parameters Models for radio transmission with interference Béla Bollobás, University of Memphis TASKS 1Q 2Q 3Q 4Q 1Q 2Q 3Q 4Q 1Q 2Q 3Q 4Q 1Q 2Q 3Q 4Q design and simulation theoretical bounds write-up of results exchange with integrators FY01 Schedule FY02 FY03 FY04 4 4

Part I. CONNECTEDNESS Joint with Paul Balister, Amites Sarkar and Mark Walters. Suppose n radio transceivers are scattered at random over a certain area. Each radio is able to establish a direct two-way connection with the k radios nearest to it. In addition, messages can be routed via intermediate radios, so that a message can be sent indirectly from radio s to radio t through a series of radios s 1 = s, s 2,..., s l = t, each one having a direct connection to its predecessor.

Precise Formulation Define a random geometric graph G(A, λ, k) as follows. Let P be a Poisson process of intensity λ in a region A, and join every point of P to its k nearest neighbors.

Precise Formulation Define a random geometric graph G(A, λ, k) as follows. Let P be a Poisson process of intensity λ in a region A, and join every point of P to its k nearest neighbors. We would like to know the values of k for which the resulting random geometric graph G(A, λ, k) is likely to be connected.

Equivalent Formulations Two equivalent ways. The first is to fix the area A and let λ. In the second formulation, we instead fix λ = 1 and grow the region A while keeping its shape fixed, so that the expected number of points in A again increases. We use the second: a domain of area n and a Poisson process of intensity 1. The shape is essentially irrelevant, so that we may as well take A = S n, the square of area n (not side n), which ensures that the expected number of points in our region is n. Thus we are interested in the values of k for which G n,k = G(S n, 1, k) is likely to be connected, as n.

Easy Bounds Observe that two essentially trivial arguments give the right order of magnitude for k: there exist constants c 1 and c 2 so that

Easy Bounds Observe that two essentially trivial arguments give the right order of magnitude for k: there exist constants c 1 and c 2 so that if k c 1 log n then the probability that G n,k is connected tends to zero as n, and if k c 2 log n then the probability that G n,k is connected tends to one as n. As usual, we say that an event occurs with high probability (whp) if it occurs with probability tending to one as n.

A trivial upper bound Tessellate the square S n with small squares of area log n. (We use natural logarithms.) Then the probability that a small square contains no points of the process is e log n = n 1 = o( log n n ), so that whp no small square is empty. On the other hand, simple calculations show that every point has at most l = 5πe log n points within distance 5 log n. Implies: for k = l < 42.7 log n whp every point x V (G), contained in a square Q x, is joined to every point in Q x, and also to every point in every adjacent square.

A trivial lower bound For c < 1/8, set k = c log n, define r > 0 by πr 2 = k + 1, and take three concentric discs, D 1, D 3 and D 5, with D j having radius jr, and so area j 2 (k + 1).

A trivial lower bound For c < 1/8, set k = c log n, define r > 0 by πr 2 = k + 1, and take three concentric discs, D 1, D 3 and D 5, with D j having radius jr, and so area j 2 (k + 1). R 1 = D 3 \ D 1 and R 3 = D 5 \ D 3 are rings. Then (i) the probability that we have at least k + 1 nodes in D 1 is at least 1/2; (ii) the probability that every disc centred at a point of R 3 and touching D 1 has at least k + 1 nodes in R 3 is 1 + o(1);

An Illustration to the Lower bound D A A 2 A 1 2r 0 2r 0 r 0 x Figure 0: Lower bound, undirected case.

A Trivial Lower Bound Completed Hence the probability that a given disc D 5 contains a component of order k + 1 is at least n 1+ε. As we have more then n/100 log n > n 1 ε/2 independent choices for such a component, the probability that there is such a component is at least (1 n 1+ε ) n1 ε/2 = 1 + o(1).

The First Problem These trivial arguments show that we should focus attention on the range k = Θ(log n). Indeed, define c l and c u by and c l = sup{c : P(G n, c log n is connected) 0}, c u = inf{c : P(G n, c log n is connected) 1}. We have just shown that 0.125 c l c u 42.7.

GIVE BETTER BOUNDS FOR c l AND c u. The First Problem These trivial arguments show that we should focus attention on the range k = Θ(log n). Indeed, define c l and c u by and c l = sup{c : P(G n, c log n is connected) 0}, c u = inf{c : P(G n, c log n is connected) 1}. We have just shown that 0.125 c l c u 42.7.

Earlier Results By making use of a substantial result of Penrose from 1997, Xue and Kumar (2003?) proved that 0.074 c l c u 5.1774.

Earlier Results By making use of a substantial result of Penrose from 1997, Xue and Kumar (2003?) proved that 0.074 c l c u 5.1774. It seems likely that c l = c u = c,

Earlier Results By making use of a substantial result of Penrose from 1997, Xue and Kumar (2003?) proved that 0.074 c l c u 5.1774. It seems likely that c l = c u = c, and Xue and Kumar conjectured that c = 1. Gilbert s disc model G r (n): join two nodes in S n if their Euclidean distance is at most r. Penrose: if πr 2 = c log n, so that the average degree is c log n, then c = 1 is the critical value:

G r (n) and G n,p Analogous classical result: if in a random graph G = G n,p the average degree is c log n, then if c < 1, whp G is not connected, while if c > 1, whp G is connected. In both cases, the obstruction for connectivity is the existence of isolated vertices, in the sense that whp the graph becomes connected as soon as it has no isolated vertices. In our problem we expressly forbid isolated vertices, indeed, each vertex has degree at least k. Thus the obstruction for connectivity must involve more complicated extremal configurations.

G n,k and G n,k out This motivates the study of the directed case, where, in a Poisson process of intensity 1 in a region S n, we place directed edges pointing away from each point towards its k nearest neighbors. By our construction, in the resulting graph G n,k every vertex has out-degree exactly k. We wish to know how large c should be to guarantee a directed path between any two vertices whp, so that the threshold value of c, if it exists, will be higher than in the undirected case.

Second Strong Connectedness Give bounds on the constants c l and c u for the directed model G n,k corresponding to c l and c u for the undirected model G n,k.

Second Strong Connectedness Give bounds on the constants c l and c u for the directed model G n,k corresponding to c l and c u for the undirected model G n,k. c l = sup{c : whp G n,c log n is not strongly connected}, c u = inf{c : whp G n,c log n is strongly connected}. Trivially, c l c l and c l c u.

The Third Problem Covering For each node x of G n,k, let D x be the disc centred at x just covering the kth nearest node to x. Let D = D(G n,k ) be the union of the discs D x. [Thus if each node broadcasts with a power to reach k other notes (=motes) then D is the domain where at least one broadcast can be heard.] Define c l and c u analogously to c l, etc. Again, ALMOST CERTAINLY, c l = c u = c. HOW LARGE IS c l?

RESULTS Balister, B, Sarkar and Walters

RESULTS Balister, B, Sarkar and Walters For the undirected model G n,k, 0.3043 c l c u 0.5139.

RESULTS Balister, B, Sarkar and Walters For the undirected model G n,k, 0.3043 c l c u 0.5139. For the directed model G n,k, 0.7209 c l c u 0.9967.

RESULTS ctd. COVERING: The constants are at least as close as for the directed model: 0.7209 c l c l c u c u 0.9967.

A sketch of the proofs Lower bound: similar to the easy argument, with changing densities. In the directed case the obstruction is the existence of a vertex of indegree 0. D A A 2 A 1 2r 0 2r 0 r 0 x

Upper Bounds A node P is normal if the smallest circle containing its k nearest neighbors does not intersect the boundary.

Upper Bounds A node P is normal if the smallest circle containing its k nearest neighbors does not intersect the boundary. Assume that all points are normal. This excludes O( n log n) nodes from consideration. Theorem Let c > 1 log 2 1.4427. Then the probability that G = G n,c log n contains a component consisting entirely of normal points tends to zero as n. Proof. Let P be a northernmost node of such a component G. Then P is extreme in the sense that its k = c log n nearest neighbors all lie below it. The probability that a normal node is extreme is 2 k, and so the expected number of extreme normal nodes is at most n2 k = o(1). Thus the probability that there is such a G

The First Pillar Lemma Fix c > 0. Then, there exists a constant c such that the probability that G = G n,c log n contains two components each of (Euclidean) diameter at least c log n tends to zero as n. Based on (i) an isoperimetric inequality, (ii) this simple fact: Fix c > 0. Then there exist c and c + with c < c < c + such that, letting r, R, satisfy πr 2 = c log n and πr 2 = c + log n, then whp every vertex in G n,c log n is joined to every vertex within distance r, and no vertex is joined to a vertex at distance more than R. [The same is true for the directed model G n,c log n.]

The Second Pillar Theorem Let c > 1 0.5139, and set log 7 k = c log n. Then the probability that G n,k is connected tends to one as n.

There is No Small Component H 2 H 1 H 3 A 3 P 3 A 2 P 2 H A P 1 A 1 P 6 A 6 H 6 A 4 P 4 =P 5 H 4 A 5 H 5 Figure 0: The hexagon H

Threshold Our results show that if n = n(k) e 1.94k then lim k P(G n,k is connected) = 1 and if n = n(k) e 3.3k then lim k P(G n,k is connected) = 0.

Threshold Our results show that if n = n(k) e 1.94k then lim k P(G n,k is connected) = 1 and if n = n(k) e 3.3k then lim k P(G n,k is connected) = 0. There is no doubt that there is a constant c, 1.94 < c < 3.3, such that if ε > 0 then for n = n(k) e (c ε)k we have lim k P(G n,k is connected) = 1 and for n = n(k) e (c+ε)k we have lim k P(G n,k is connected) = 0. Although we cannot show the existence of this constant c, let alone determine it, we can prove that the transition from connectedness to disconnectedness is considerably sharper than these relations indicate.

Sharp Threshold The length of the window (going from probability 0.99 of being connected to probability 0.01 of being connected, say) is O(n) rather than n 1+o(1).

Sharp Threshold The length of the window (going from probability 0.99 of being connected to probability 0.01 of being connected, say) is O(n) rather than n 1+o(1). For k 1 and 0 < p < 1, set n k (p) = max{ n P(G n,k is connected) p }. Theorem Let 0 < ε < 1 be fixed. Then, for sufficiently large k, n k (ε) < C(ε)(n k (1 ε) + 1) where C(ε) = 6 ε log ( 1 ε) + 1 2.

Conjecture We cannot show that the threshold in k is sharp to the extent that if G n,k is connected with probability ε > 0, then for k (1 + ε)k the probability that G n,k is connected tends to one as n 1. Nevertheless, we CONJECTURE that, given ε > 0, there is a constant c = c(ε) such that if P(G n,k is connected ) ε then for k = k + c P(G n,k is connected ) 1.

Part II. Random Transmission Joint with Paul Balister and Mark Walters Fix a probability distribution D on measurable regions A R d \ {0}. Construct a random digraph G by placing points {x i } in 2 according to a Poisson process with intensity 1. Independently for each x i, choose regions A xi according to the distribution D. Let the vertices of G be the x i, and let the edges x i x j lie in G if x j A(x i ) = x i + A xi. We wish to know whether the resulting graph has an infinite directed path.

Random Transceiver Networks The nodes x 1, x 2,... are randomly scattered radio transceivers in R 2 ;

Random Transceiver Networks The nodes x 1, x 2,... are randomly scattered radio transceivers in R 2 ; A(x i ) is the area in which the signal strength of x i is sufficiently strong to be received by another transceiver. Individual transceivers may be strongly directional, so we might approximate A xi by, for example, a thin sector of a disk.

Our Result, Loosely Stated We have percolation if the expected area of A xi is slightly more than 1 provided there is not much overlap in the regions A xi (in a sense that we shall make precise later), and the distribution D satisfies some mild boundedness conditions. In the case when the A xi are thin sectors of a disk, these conditions hold if the orientations of the sectors are randomly distributed without too much concentration in a given direction.

Definitions S is the standard Lebesgue measure of S in R d. Fix a distribution on areas D, and define two distributions derived from D. Let D 0 be the distribution of the number N of points of our Poisson process in A where A is distributed according to D. If all the areas A are the same, then D 0 is a Poisson distribution with mean A. Let D c be the distribution on R d given by P Dc (S) = E( S A )/ E( A ) where A is distributed according to D. In other words, D c is the probability distribution of the location of a randomly

Definitions ctd. We require that for any x 1, x 2, if we fix A(x 1 ) and A(x 2 ), then the probability of a randomly chosen neighbor of a randomly chosen neighbor of x 1 being a neighbor of x 2 is at most δ. In addition, we also need the same result if we, rather artificially, replace A xi by A xi. Formally, A and A are distributed according to D. Fixing A, A, and z R d, let X be a random point uniformly distributed in A and let Y be an independent random variable with distribution D c. Then define δ by δ = ess sup A,A sup z max{p(x+y A +z), P( X+Y A +z)} (0)

Example 1 Balls in high dimensions If A is a ball in d dimensions with then we can bound (A + x) A by noting that (A + x) A lies in a hypersphere of volume A (1 α 2 /4) d/2 about the point x/2 when x is α times the radius of A from 0. We can split the integral into a component when α 2 > 0.8, where the integrand is at most A 0.8 d/2 and α 2 0.8 where x is restricted to a volume A 0.8 d/2. Consequently, δ 2(0.8) d/2, and so δ decreases exponentially with d.

Example 2 Random sectors in the plane Take d = 2 and let A be a randomly oriented thin sector of a disk. Assume the sectors are oriented uniformly within some fixed angle θ, and each sector has angle εθ. Then the probability distribution D c has density at most ε/ A at any point. It is then clear that ±X + Y has probability density at most ε/ A anywhere, so δ = sup z,a,a P(±X + Y A + z) ε.

Definitions and Assumptions Define η so that 1 + η is the average number of neighbors of a point in G. (This is also the average volume E A.) Define σ 2 to be the variance of the number of neighbors of a point. ( σ 2 = E A + Var A, so if the volume of A is fixed to be 1 + η then σ 2 = 1 + η.) Assume that all sets A given by our distribution D lie in a ball of radius r 0 about 0. Furthermore, assume that the root mean square distance of a neighbor is at least r m > 0 in any direction, i.e., r 2 m E((Y u)2 ) for any unit vector u, where Y is distributed according to D c.

Main Result There is an absolute constant c > 0 such that if η 1 and δ < cr 9 m r 9 0 η16 σ 32 then G almost surely has an infinite directed path.

Balls in High Dimensions Assume that G consists of the points of a Poisson process with intensity 1 in R d, d 2, and each point is joined to all other points within a ball of volume 1 + η. Then there exists a constant c > 0 such that if η > c(0.9931) d then G almost surely has an infinite component.

Random Sectors in the Plane Assume that G consists of the points of a Poisson process with intensity 1 in 2, and each point is joined to all other points within a sector of a disk of area 1 + η and angle εθ randomly oriented within a fixed angle of θ. Then there exists a constant c > 0 such that if η > cε 1/16 then G almost surely has an infinite directed path.

Part III. FAST TRANSMISSION Joint with Paul Balister, Martin Haenggi and Mark Walters We wish to transmit information from one point, the source s, to another point, the target t, with the following three properties: reliably, i.e., with probability 1 ɛ, quickly, i.e., through few hops, economically, i.e., using little power. Reliability means that s and t lie in the same cluster (component) with high probability. The failure of transmitters would just lower the intensity of the Poisson process.

The Disc Model For each node x, take a disc D x of area a centred at x, and postulate that two nodes x and y of the process can communicate if y D x or, equivalently, x D y. Joining two nodes by an edge if they can communicate, we obtain an infinite geometric random graph G a. Note that the area a is exactly the expected degree of a vertex. This disc model D a was first suggested by Gilbert in 1961, and has been much studied. The bounds that have been proved for the critical area or degree a c are still very bad. Recently, with Balister and Walters I showed that 4.508 < a c < 4.515 with confidence 99.99%.

Directional Transmitters Is it beneficial to use directional transmitters, or ones that broadcast to a non-circular region? We have seen that the answer is YES. We can get an advantage by using directional transmitters: even with very low power there exist points at arbitrarily large distance that can communicate. However, this does not give us any bound on how many hops this will take or how likely this transmission is. Indeed if the power is such that we expect only 1 + η neighbours then there is a significant probability (e (1+η) ) that s has no neighbours at all, and hence definitely can not talk to t.

Directional Transmitters Hence even to hope for reliable transmission with reliability ε the vertex s must expect log(1/ε) neighbours, and similarly for t. Pushing the power of transmission of all nodes up high, we lose all the advantages of the result. We can do much better: s and t are big brothers: they can broadcast much further and they can receive from much further.

First Result Suppose that transmitters are placed in the plane according to a Poisson process of intensity one. Further suppose that all the transmitter apart from s and t broadcast to a randomly oriented sector of angle δ of area 1 + η. Finally suppose that s can communicate with all transmitters within distance R and similarly for t. Then, for any fixed η > 0, provided that R is large enough, the probability that s and t can communicate is arbitrarily close to one independently of the distance from s to t.

Second Result Transmitters: Poisson process of intensity 1. Fix a (small) angle. Suppose that all the transmitters nearer s than t broadcast directionally into a sector of radius r and angle δ oriented randomly inside the sector of angle φ pointed in the st direction, and that all the transmitters nearer t than s receive directionally, in the same way. Finally, s can transmit to any point within R and that t can receive from any point within R. Then, provided that R is large enough independently of the distance from s to t and that the area of a sector, δr 2 /2, is greater than 1, the node s can communicate with the node t with probability arbitrarily close to one. Moreover the number of hops is at most (1 + o(1)) ( 3φ 4r sin(φ/2) ) d(s, t).

Figure R R s φ r δ φ t omnidirectional reception directional transmission directional reception omnidirectional transmission

Conjecture Transmitters: Poisson process of intensity 1. Fix a (small) angle.

Conjecture Transmitters: Poisson process of intensity 1. Fix a (small) angle. Suppose that all the transmitters apart from s and t broadcast to a randomly oriented sector of angle δ and radius r inside a sector of angle φ centred in the st direction. Finally, suppose that s can communicate with all transmitters within distance R and similarly for t. Then, provided that R is large enough, and that the area of a sector δr 2 /2 is greater than one, the probability that s and t can communicate is arbitrarily close to one independently of the distance between s and t. Moreover the number of hops is at most. (1 + o(1)) ( 3φ 4r sin(φ/2) ) d(s, t)