Alan Frieze Charalampos (Babis) E. Tsourakakis WAW June 12 WAW '12 1

Transcription

1 Alan Frieze Charalampos (Babis) E. Tsourakakis WAW June 12 WAW '12 1

2 Introduction Degree Distribution Diameter Highest Degrees Eigenvalues Open Problems WAW '12 2

3 Internet Map [lumeta.com] Food Web [Martinez 91] Friendship Network [Moody 01] Protein Interactions [genomebiology.com] WAW '12 3

4 Modelling real-world networks has attracted a lot of attention. Common characteristics include: Skewed degree distributions (e.g., power laws). Large Clustering Coefficients Small diameter A popular model for modeling real-world planar graphs are Random Apollonian Networks. WAW '12 4

5 Construct circles that are tangent to three given circles οn the plane. Apollonius ( BC) WAW '12 5

6 Apollonian Gasket WAW '12 6

7 Higher Dimensional (3d) Apollonian Packing. From now on, we shall discuss the 2d case. WAW '12 7

8 Dual version of Apollonian Packing WAW '12 8

9 Start with a triangle (t=0). Until the network reaches the desired size Pick a face F uniformly at random, insert a new vertex in it and connect it with the three vertices of F WAW '12 9

10 For any t 0 Number of vertices n t =t+3 Number of vertices m t =3t+3 Number of faces F t =2t+1 Note that a RAN is a maximal planar graph since for any planar graph m t 3n t 6 = 3t + 3 WAW '12 10

11 Introduction Degree Distribution Diameter Highest Degrees Eigenvalues Open Problems WAW '12 11

12 Let N k (t)=e[z k (t)]=expected #vertices of degree k at time t. Then: N 3 t + 1 = N 3 t + 1 3N 3(t) 2t+1 N k t + 1 = N k t 1 k 2t+1 + N k 1 t k 1 2t+1 Solving the recurrence results in a power law with slope 3. WAW '12 12

13 Z k (t)=#of vertices of degree k at time t, k 3 b 3 = 2, b 5 4 = 1, b 5 5 = 4, b k = k 6 k(k+1)(k+2) For t sufficiently large E Z k t b k t 3.6 Furthermore, for all possible degrees k Prob Z k t E Z k t 10 tlog(t) = o(1) WAW '12 13

14 Degree Theorem Simulation WAW '12 14

15 Introduction Degree Distribution Diameter Highest Degrees Eigenvalues Open Problems WAW '12 15

16 Depth of a face (recursively): Let α be the initial face, then depth(α)=1. For a face β created by picking face γ depth(β)=depth(γ)+1. e.g., WAW '12 16

17 Note that if k* is the maximum depth of a face at time t, then diam(g t )=O(k*). Let F t (k)=#faces of depth k at time t. Then, E F t k is equal to 1 t 1 <t 2 <..<t k t k j=1 1 2t j k! t j=1 1 2j + 1 t elog t 2k k+1 Therefore by a first moment argument k*=o(log(t)) whp. WAW '12 17

18 WAW '12 18

19 WAW '12 19

20 WAW '12 20

21 WAW '12 21

22 WAW '12 22

23 WAW '12 23

24 Large Deviations for the Weighted Height of an Extended Class of Trees. Algorithmica 2006 Broutin Devroye The depth of the random ternary tree T in probability is ρ/2 log(t) where 1/ρ=η is the unique solution greater than 1 of the equation η-1-log(η)=log(3). Therefore we obtain an upper bound in probability diam G t ρlog (t) WAW '12 24

25 This cannot be used though to get a lower bound: Diameter=2, Depth arbitrarily large WAW '12 25

26 Introduction Degree Distribution Diameter Highest Degrees Eigenvalues Open Problems WAW '12 26

27 Let Δ 1 Δ 2 Δ k be the k highest degrees of the RAN G t where k=o(1). Also let f(t) be a function s.t. f t t +. Then whp t f(t) Δ 1 tf(t) and for i=2,..,k t f(t) Δ i Δ i 1 t f(t) WAW '12 27

28 t 0 = log log (f t ) t 1 = log (f t ) t Break up time in periods Create appropriate supernodes according to their age. Let Xt be the degree of a supernode. Couple RAN process with a simpler process Y such that X t Y t, X t0 = Y t0 = d 0 Upper bound the probability p*(r)=pr Y t = d 0 + r Union bound and k-th moment arguments WAW '12 28

29 Introduction Degree Distribution Diameter Highest Degrees Eigenvalues Open Problems WAW '12 29

30 Let λ 1 λ 2 λ k be the largest k eigenvalues of the adjacency matrix of G t. Then λ i = 1 ± o 1 Δ i whp. Proof comes for free from our previous theorem due to the work of two groups: Chung Lu Vu Mihail Papadimitriou WAW '12 30

31 t 0 = 0 t 1 = t 1/8 t 2 = t 9/16 t. S 1 S 2 S 3.. Star forest consisting of edges between S 1 and S 3 -S 3 where S 3 is the subset of vertices of S 3 with two or more neighbors in S 1. WAW '12 31

32 Lemma: S 3 t 1/6 This lemma allows us to prove that in F... λ i F = 1 o 1 Δ i WAW '12 32

33 Finally we prove that in H=G-F λ 1 Η = o λ k F Proof Sketch First we prove a lemma. For any ε>0 and any f(t) s.t. f t t + the following holds whp: for all s with f t r s then d s r s ε+1 2 r 1 2. s t for all vertices WAW '12 33

34 Consider six induced subgraphs H i =H[S i ] and H ij =H(S i,s j ). The following holds: λ 1 H λ 1 H i 3 i=1 + λ 1 (H i, H j ) i<j Bound each term in the summation using the lemma and the fact that the maximum eigenvalue is bounded by the maximum degree. WAW '12 34

35 Introduction Degree Distribution Diameter Highest Degrees Eigenvalues Open Problems WAW '12 35

36 Conductance Φ is at most t -1/2. Conjecture: Φ= Θ(t -1/2 ) Are RANs Hamiltonian? Conjecture: No Length of the longest path? Conjecture: Θ(n) WAW '12 36

37 Thank you! WAW '12 37