expanders

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ExpandersPresented by Alon Levin

29.10.2006

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Expander - Intuitive Definition

Expander graph is an undirected graph:

Without “bottlenecks” With high connectivity With a “large” minimal cut With “large” edge expansion

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Computational Complexity Theory Economical Robust Networks (Computer

Network, Phone Network) Construction of Hash Functions Error Correcting Codes Extractors Pseudorandom Generators Sorting Networks

Expanders and their applications

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Expander – Definition via Edge Expansion

The edge expansion of a graph G=(V,E) is :

Theorem:

Gn are d-regular.

{ Gn } is an explicit expander family.

0 1 01, 0, { } , ( ) ,n n nd G G

| |,| |

2

min | ( , ) |( )

| |VS V S

E S SG

S

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Expanders – Definition via Spectral Gap

We shall discuss undirected d-regular graphs from now and on

We shall adopt the notion of A = A(G) as G’s adjacency matrix

Since A is a real symmetric nn matrix it has n real eigenvalues:

We will denote = max( |2|, |n|) The spectral gap is defined as d- .

1 2 ... nd

1 1

1 1A d

A d

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Expanders – Definition via Spectral Gap A d-regular graph G is an (n, d, )-expander if

= max { |i|:1<i≤n } = max { |2(G)|, |n(G)| } and d>

If G is an (n, d, )-expander then

2(G)/2d ≤ d- ≤ 2(G)

large expansion ~ large spectral gap!

* We will prove the inequality right after we show a rather helpful characterization of 2(G)

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Reminder – Euclidean Scalar Product and Norm

2

1

2 2 2

1 1 1

2 2

1 1 1

( ) ( )

0 ( 2 ) 4

,

n

i ii

n n n

P P i i i ii i i

n n n

i i i ii i i

P x a x b

a b a b

a b a b

x y x y

2

1 1

, ,n n

i i ii i

x y x y x x x x

Cauchy–Schwarz inequality :

Proof:

,x y x y

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Reminder – Triangle Inequalityx y x y The triangle inequality:

Proof:

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Rayleigh Quotient

The second largest eigenvalue of a real symmetric matrix can be computed this way:

Proof: Let be an orthonormal basis of A’s eigenvectors, where the ith vector corresponds to eigenvalue i(A).

“≤”: If we let then we would get |2| and

|n|, so the maximum is at least .

2 nx v or x v ��

1{ }ni iv

��

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Rayleigh Quotient

(Proof continues):

“≥”: Denote an arbitrary x as and since

a1=0. Now and since that:1

n

i i ii

Ax a v

��

=max{|=max{|i|:1<i≤n}i|:1<i≤n}

1 0x v ��

1

n

i ii

a v

��

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Spectral gap and edge expansion

Now we are ready to prove the following lemma: if G is an (n, d, )-expander then d- ≤ 2(G)

Proof: G=(V,E), |V|=n. Let SV, |S|≤n/2. We shall set a vector x similar to S’s indicator, but so that <x, >=0 and so by the Rayleigh quotient we have

So, we will define:

1

u v

S S| |vx S| |ux S

2,Ax x x

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Spectral gap and edge expansion As we can easily see, x is orthogonal to since

We can also notice the equalities:

;2 2 2 2| | | | (| | | |) | | 2 | || | | |S S n S S S S S S n

2 2 2| | | | | || | | || |x S S S S S S n

1

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Spectral gap and edge expansion By A’s definition, , and since A is

symmetric: ( , )

( )u vu v E

Ax x

1 0 0 1 0 1 0 0 0 1 0 X1

-

Xi

-

xn

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Edges originating in S

Cut edgesCut in half due

to edges double count

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Which implies

since by our assumption , because

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Regular Graph Union

Here we shall prove a simple lemma:

If G is a d-regular graph over the vertex set V, and H is a

d’-regular graph over the same vertices, then

G’ = GH = (V,E(G) E(H)) is a d+d’-regular graph such that (G’) (G) + (H)

Proof: Take x s.t.

Then,

Rayleigh

quotient

AG’=AG+ AH

Since it’s a multigraph (edge set is a multiset)

'1, 1 0, ( ') ,Gx x G A x x

' , , , ( ) ( )G G HA x x A x x A x x G H

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The Final Expander Lemma Let G=(V,E) be an (n, d, )-expander and FE.

Then the probability that a random walk which starts on an edge from F will pass on an edge from F on it’s tth step is

bounded by

Motivation: Showing that a random walk on a good expander (large spectral gap) behaves similarly to independent choice of random edges

F

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The Final Expander Lemma

Proof: x is a vertices distribution vector s.t.

xv=Pr[Our walk starts at vertex v] The probability to reach vertex u at certain point is

, thus

, where x is the current distribution and x’ is

the new one. By the definition of A=A(G) we can write

x’=Ax/d. We denote Ã=A/d, so x’=Ãx. After i steps, the distribution is x’=Ãix.

( , )

Pr[ ]Pr[ ]v u E

v is reached u is reached from v

Pr[u is reached Pr[u is reached from v]=1/dfrom v]=1/dSince G is d-Since G is d-

regularregular( , )

' vu

v u E

xx

d

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The Final Expander Lemma P is the probability that an edge from F will be

traversed on the tth step.

Let yw be the number of edges from F incident on w divided by d.

Then,

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The Final Expander Lemma Calculation of initial x: first step is picking an edge

from F and then one of it’s vertices, so for vertex v it’sdyv is the number of edges

from F incident on v

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The Final Expander Lemma G is d-regular, thus each row in Ã sums to one.

If is the uniform distribution on G, i.e. , then . Since x is a probability function, it can be decomposed into , where

since x and are probability functions:

Then,

x 1vx n

Ax x x x x , 0x x

1 1 1 1t t t tA x A x A x x A x

11 ; 1, 1, Pr[ ] 1 1 0

v V v V

nx x x x start on vn

x

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The Final Expander Lemma When G is d-regular, each row in Ã sums to one, and

thus if then . A more intuitive way of seeing this, other than algebra,

is considering a random walk with uniform distribution on a d-regular graph:

1vx n Ax x

1

n

1

n

1

n

1

n

1

n

1

n

1

d1

d1

d

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Hence,Linearity of

scalar product

1tA x x 2

, ,

, ,

0

x x x x x

x x x x

x

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The Final Expander Lemma2

2 21

1 1n

i

nx

n n n

Cauchy-Schwartz inequality

(Ã)=(A)/d+Rayleigh quotient t-1 times

x x

2

2 2

, ,

, , 2 ,

2 0

x x x x x x x

x x x x x x

x x

x x

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Since xi are positive, Maximum is achieved when all edges incident to v are

in F, and in that case , so:2 | |v

dx

F

1 2 | |,t F

P A x xd

2 2 max max 1v v v v vv V v V v V

x x x x x x

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The Lubotzky-Phillips-Sarnak Expander

Take a prime p, let V=Zp{}. Define 0-1= and connect every vertex x to: x+1 x-1 x-1

It’s a 3-regular graph with <3

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The Margulis/Gaber-Galil Expanders

Take V=ZnZn, so |V|=n2. Given v=(x,y)V connect it to the following vertices: (x+2y,y) (x,2x+y) (x,2x+y+1) (x+2y+1,y)(all operations are done modulo n) This is an 8-regular graph with =52<8, so it’s

spectral gap is about 0.93

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The End

Questions?

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