Coupling of Scale-Free and Classical Random Graphs

Transcription

1 Coupling of Scale-Free and Classical Random Graphs April 18, 2007

2 Introduction Consider a graph where we delete some nodes and look at the size of the largest component remaining. Just how robust are scale free graphs to random failures

3 Introduction Consider a graph where we delete some nodes and look at the size of the largest component remaining. Just how robust are scale free graphs to random failures to deliberate attacks

4 Introduction Consider a graph where we delete some nodes and look at the size of the largest component remaining. Just how robust are scale free graphs to random failures to deliberate attacks should the adversary have infinite time?

5 Introduction If we take the adversary out of the picture we can phrase the question as.

6 Introduction If we take the adversary out of the picture we can phrase the question as. How many vertices do we need to remove so that the graph breaks up into small pieces?

7 The Model Consider the Preferential Attachment Model. Assume m is fixed, G[k] G (k) m. We create G[k + 1] by adding a new vertex k + 1 with m edges (k + 1, t 1 ),..., (k + 1, t m ) where d G[k],i(j) Pr[t i = j] = 2mk + 2i 1

8 Main Result Theorem 1 There exist c > 0,m 0 s.t. for m m 0, whp G (n) m the following holds (a) Every induced subgraph of size 10n log(m) m contains a component of size at least 2n log m m (b) The graph contains an independent set of size cn log(m) m

9 Proof Technique Proof by coupling We create two graphs (G 1, G 2 ) s.t. G 1 G (n) m, G 2 G(n, p). But G 1 and G 2 are not independent but heavily correlated.

10 What is coupling? Let X D X, Y D Y. A coupling is a joint distribution D X,Y s.t. the marginals are correct, i.e. D X, = D X and D,Y = D Y.

11 What is coupling? Let X D X, Y D Y. A coupling is a joint distribution D X,Y s.t. the marginals are correct, i.e. D X, = D X and D,Y = D Y. Suppose X, Y are uniform [0, 1] and we let Y = 1 X then both marginals are still uniform but the joint distribution is far from independent.

12 Chernoff Bounds Let X Bin(n, p) then ( e δ Pr[X > (δ + 1)np] < (1 + δ) (1+δ) ( e δ Pr[X < (1 δ)np] < (1 δ) (1 δ) ) np ) np

13 The Coupling Theorem 2 Fix η < 1/2, there exist constants A, c > 0 s.t. for fixed m we can construct G 1 G (n) m, G 2 G(n, ηm/n) s.t. whp e(g 2 \ G 1 ) Ae cm n

14 The Coupling Sidenote We can create G(n, p) one node at a time G[1] contains a single node

15 The Coupling Sidenote We can create G(n, p) one node at a time G[1] contains a single node Given G[k]

16 The Coupling Sidenote We can create G(n, p) one node at a time G[1] contains a single node Given G[k] generate Y Bin(k, p)

17 The Coupling Sidenote We can create G(n, p) one node at a time G[1] contains a single node Given G[k] generate Y Bin(k, p) let S be a random subset of [k] of size Y

18 The Coupling Sidenote We can create G(n, p) one node at a time G[1] contains a single node Given G[k] generate Y Bin(k, p) let S be a random subset of [k] of size Y create edges between k and S

19 The Coupling Proof We grow G 1, G 2 one node at a time. Start with G 1 [k 0 ] G (n) m, G 2 [k 0 ] G(k 0, ηm/n), where k 0 grows slowly with n. Let t 1,..., t m be the vertices picked when growing G 1 [k]. For j k + 1 we have Pr(t i = j) = d G[k],i(j) 2mk + 2i 1 m 2km + 2m = 1 2k + 2

20 The Coupling Proof 1 2k k 1 k k + 1 Construct s 1,..., s m [k], s.t. s i s are independent Pr(s i = j) = 1 2k+2 for j and s i = j implies t i = j.

21 The Coupling Proof Let X = {i s i }, then X Bin(m, Let Y Bin(k, ηm/n), then E[X] = mk mk (1 + ɛ)η 2k + 2 n k 2k+2 ). = (1 + ɛ)e[y ] So Pr[X < Y ] Ae c m for some constants A, c > 0.

22 The Coupling Proof All the s i s are distinct w.p 1 O( m2 k ), given this and X the set S 1 = {s i s i } is a random subset of [k] of size X.

23 The Coupling Proof All the s i s are distinct w.p 1 O( m2 k ), given this and X the set S 1 = {s i s i } is a random subset of [k] of size X. If Y X we pick a random subset S 2 of size Y from S 1.

24 The Coupling Proof All the s i s are distinct w.p 1 O( m2 k ), given this and X the set S 1 = {s i s i } is a random subset of [k] of size X. If Y X we pick a random subset S 2 of size Y from S 1. Otherwise we pick S 2 a random subset of size Y from [k], this happens w.p. Ae c m + o(1).

25 The Coupling Proof Set D k+1 = #of edges added to G 2 and not G 1. Pr[D k+1 > 0] = Ae c m + o(1), and D k+1 Y so E[D k+1 ] Ame c m

26 The Coupling Proof The number of edges in G 2 \ G 2 is at most ( ) k0 2 + n k=k 0 +1 Assuming k 0 = n 1/4 we see that the sum has expected value at most Ane c m and is concentrated around its mean. D k

27 The Coupling Proof The number of edges in G 2 \ G 2 is at most ( ) k0 2 + n k=k 0 +1 Assuming k 0 = n 1/4 we see that the sum has expected value at most Ane c m and is concentrated around its mean. QED D k

28 Proof of Theorem 1(a) Let G 1, G 2 be as described in Theorem 2. Suppose G 1 has an induced subgraph on a set V, with V = 10n log(m) m where every component has size less than 2n log(m) m. Then we can partition V into two set V 1, V 2 s.t. V 1, V 2 4n log(m) m and G 1 has no V 1 -V 2 edges.

29 Proof of Theorem 1(a) Let x = Ae c m n, if the coupling works then G 2 has at most x, V 1 -V 2 edges. The number of V 1 -V 2 edges in G 2 is a binomial with mean µ = V 1 V 2 ηm/n 24η(log(m)) 2 n/m Pick m large enough s.t. x < µ/100

30 Proof of Theorem 1(a) By Chernoff bounds the probability that we have fewer than x edges crossing V 1, V 2 is at most e 11µ/12. Now ( ) n Pr[G 2 has at most x V 1 V 2 edges] 2 V e 22η(log(m))2 n/m V

31 Proof of Theorem 1(a) By Chernoff bounds the probability that we have fewer than x edges crossing V 1, V 2 is at most e 11µ/12. Now ( ) n Pr[G 2 has at most x V 1 V 2 edges] 2 V e 22η(log(m))2 n/m V ( ) en V 2 V e 22η(log(m))2 n/m V

32 Proof of Theorem 1(a) By Chernoff bounds the probability that we have fewer than x edges crossing V 1, V 2 is at most e 11µ/12. Now ( ) n Pr[G 2 has at most x V 1 V 2 edges] 2 V e 22η(log(m))2 n/m V ( ) en V 2 V e 22η(log(m))2 n/m V ( ) 2em V e 22η(log(m))2 n/m 10 log(m)

33 Proof of Theorem 1(a) By Chernoff bounds the probability that we have fewer than x edges crossing V 1, V 2 is at most e 11µ/12. Now ( ) n Pr[G 2 has at most x V 1 V 2 edges] 2 V e 22η(log(m))2 n/m V ( ) en V 2 V e 22η(log(m))2 n/m V ( ) 2em V e 22η(log(m))2 n/m 10 log(m) e (10 22η)(log(m))2 n/m

34 Proof of Theorem 1(a) By Chernoff bounds the probability that we have fewer than x edges crossing V 1, V 2 is at most e 11µ/12. Now ( ) n Pr[G 2 has at most x V 1 V 2 edges] 2 V e 22η(log(m))2 n/m V ( ) en V 2 V e 22η(log(m))2 n/m V ( ) 2em V e 22η(log(m))2 n/m 10 log(m) e (10 22η)(log(m))2 n/m QED

35 Overview of the proof The idea of the proof was this

36 Overview of the proof The idea of the proof was this We create G 1, G 2 simultaneously s.t.

37 Overview of the proof The idea of the proof was this We create G 1, G 2 simultaneously s.t. a bad event in G 1 implies w.h.p another bad event in G 2

38 Overview of the proof The idea of the proof was this We create G 1, G 2 simultaneously s.t. a bad event in G 1 implies w.h.p another bad event in G 2 which is easy to show is rare in G 2.

39 Overview of the proof The idea of the proof was this We create G 1, G 2 simultaneously s.t. a bad event in G 1 implies w.h.p another bad event in G 2 which is easy to show is rare in G 2. But we are working in the joint probability space so this implies that the bad event is rare in G 1.

40 Another Coupling Theorem 3 Let ɛ > 0 be given, there is a constant C s.t. for fixed m we can couple G 1, G 2 s.t. G 1 G (n) m, G 2 G(n, Cm/n) s.t. whp G 2 contains G 1 \ V for a set of vertices V s.t. {i V i ɛn} ɛn/m.

41 Another Coupling Proof of Theorem 3 We start with G 1 [k 0 ] G (k 0) m and G 2 [k 0 ] G(k 0, Cm/n) independent, and here k 0 = ɛn. A vertex is bad at time k if it has degree at leas Am in G 1 [k].

42 Another Coupling Proof of Theorem 3 We need the following result from Bollobás et al.: Vertex i has degree d + m in G (n) m with probability ( ) d + m 1 o(n 1 ) + (1 + o(1)) ( i m 1 n )m/2 (1 i n )d i.e. for i ɛn it is exponentially small in d for large d If we choose A large enough the expected number of bad vertices is at most e Bm n so whp there are at most ɛn/m bad vertices.

43 Another Coupling Proof of Theorem 3 Let t 1,..., t m be defined as in the previous proof. Let V k be the set of vertices j, s.t. k 0 j k that are good. Now for j V k and Pr[t i = j] (Am + m)/(2mk) A/k Aɛ 1 /n Pr[t i / V k ] k 0 m/(2mn) = ɛ/2 so Pr[t i = j r t i / {j 1,..., j r 1 }] 2Aɛ 2 /n = p

44 Another Coupling Proof of Theorem 3 Let T i be a random subset of V k s.t. every vertex is picked independently with probability p. Then we can couple T i and t i s.t. t i T i if t i is good. p j 1 j 2 j 3... j r Pr[t i = j r t i / {j 1,..., j r 1 }]

45 Another Coupling Proof of Theorem 3 Let S 1 be the set of good vertices adjacent to k + 1 in G 1. Let S 2 = T 1 T 2... T m. Then S 2 is a random subset of V k where every vertex is picked w.p. 1 (1 p) m mp Cm/n for some large C. This completes the proof since we ve constructed S 1 s.t. S 1 S 2.

46 Proof of Theorem 1(b) We re almost there For a constant a > 1 we know that G n, a n has an independent set of size log(a) a n when a. Now couple G 1 and G 2 as in Theorem 3, with ɛ = 1 2 and assume C > 2. Look at the subgraph of G 2 of vertices that come in after n 2. It has distribution G n 2, Cm n G n 2, Cm/2 n/2 so it has an independent set of size n log(m) Cm. Removing the bad vertices from this set we see that this set is also independent in G 1. So G 1 has an independent set of size n log(m) Cm n 2m. By picking m large enough we get the result.

47 The End Thank You

48 The End Thank You Questions?