Expander Graphs and Their Applications (II)basics.sjtu.edu.cn/~chen/teaching/EXPANDER09/exp2.pdf ·...

Expander Graphs and Their Applications (II)

Yijia ChenShanghai Jiaotong University

Last Lecture

Graph Reachability Problem on Good Graphs

For our purpose, let c, d ∈ N, a graph G = (V , E) is (c, d)-good if each v ∈ Vhas degree at most d and for all u, v ∈ V , if there is a path from u to v , then

dist(u, v) ≤ c · log |V |.

TheoremLet c, d ∈ N. For (c, d)-good graphs G = (V , E), the reachability problem canbe solved in space

O(c · log |V | · log d) = O(log |V |),

i.e., in LogSpace.

Graph Expansion

For every S , T ⊆ V we let

E(S , T ) :={~e | ~e is a direction of some e ∈ E with tail in S and head in T

Let S ⊆ V . Then the edge boundary of S is

∂S := E(S , S).

The (edge) expansion ratio of G is

h(G) := min{S||S|≤|V |/2}

|∂S ||S | .

DefinitionLet d ∈ N. A sequence of d-regular graphs {Gi}i∈N of size increasing with i isa family of expander graphs if there exists ε > 0 such that h(Gi ) ≥ ε for all i .

Graph Expansion

∂S := E(S , S).

h(G) := min{S||S|≤|V |/2}

|∂S ||S | .

Graph Expansion

∂S := E(S , S).

h(G) := min{S||S|≤|V |/2}

|∂S ||S | .

Graph Expansion

∂S := E(S , S).

h(G) := min{S||S|≤|V |/2}

|∂S ||S | .

Graph Expansion

∂S := E(S , S).

h(G) := min{S||S|≤|V |/2}

|∂S ||S | .

Graphs as Matrices

The adjacency matrix of an n-vertex graph G , denoted A = A(G), is an n × nmatrix (ai,j)i,j∈[n] where

ai,j = the number of edges in G between the i-th and the j-th vertices.

Note, A is a real and symmetric matrix.

TheoremIf A is real and symmetric, then

I A has exactly n (not necessarily distinct) eigenvalues, i.e., λ1 ≥ λ2 ≥ . . .≥ λn.

I There exists a set of n eigenvectors v1, v2, . . . , vn ∈ Rn, one for eacheigenvalue (i.e., Avi = λivi ), that are orthonormal, i.e., all vi s are of length1 and orthogonal with each other.

Note, v1, . . . , vn form a basis for Rn.

Graphs as Matrices

Graph Spectrum

Let G be a graph and A = A(G). The spectrum of G is the eigenvalues of A,i.e., λ1 ≥ λ2 ≥ . . . ≥ λn.

Let G be a d-regular graph. Then λ1 = d and the corresponding eigenvectorcan be chosen as

v1 =1√n

(1√n

, . . . ,1√n

TheoremLet G be a d-regular graph with spectrum λ1 ≥ λ2 ≥ . . . ≥ λn. Then

d − λ2

2≤ h(G) ≤

√2d(d − λ2).

Note d − λ2 = λ1 − λ2 is the spectral gap of G.

Graph Spectrum

Let G be a graph and A = A(G).

The spectrum of G is the eigenvalues of A,i.e., λ1 ≥ λ2 ≥ . . . ≥ λn.

v1 =1√n

(1√n

, . . . ,1√n

d − λ2

2≤ h(G) ≤

√2d(d − λ2).

Graph Spectrum

v1 =1√n

(1√n

, . . . ,1√n

d − λ2

2≤ h(G) ≤

√2d(d − λ2).

Graph Spectrum

v1 =1√n

(1√n

, . . . ,1√n

d − λ2

2≤ h(G) ≤

√2d(d − λ2).

Graph Spectrum

v1 =1√n

(1√n

, . . . ,1√n

d − λ2

2≤ h(G) ≤

√2d(d − λ2).

Expander Mixing Lemma

We let λ := λ(G) := max{|λ2|, |λn|}.

LemmaLet G = (V , E) be a d-regular graph with n-vertices. Then for all S , T ⊆ V ,∣∣∣∣∣∣E(S , T )

∣∣− d |S ||T |n

∣∣∣∣ ≤ λ√|S ||T |.

We let λ := λ(G) := max{|λ2|, |λn|}.

∣∣− d |S ||T |n

∣∣∣∣ ≤ λ√|S ||T |.

We let λ := λ(G) := max{|λ2|, |λn|}.

∣∣− d |S ||T |n

∣∣∣∣ ≤ λ√|S ||T |.

d-Regular Graphs as Matrices

Lemmaλ1 = d and the corresponding eigenvector is v1 = 1/

Proof.Let λ be any eigenvalue of A = A(G) = (ai,j)i,j∈[n] with correspondingeigenvector v = (e1, . . . , en). We choose an i ∈ [n] such that for every j ∈ [n]

|ei | ≥ |ej |.

By Av = λv , ∑k∈[n]

ai,kek = λei ,

d |ei | = |ei |∑k∈[n]

|ai,k | ≥∑k∈[n]

|ai,k ||ek | ≥

∣∣∣∣∣∣∑k∈[n]

ai,kek

∣∣∣∣∣∣ = |λ||ei |.

|ei | ≥ |ej |.

ai,kek = λei ,

d |ei | = |ei |∑k∈[n]

|ai,k | ≥∑k∈[n]

|ai,k ||ek | ≥

∣∣∣∣∣∣∑k∈[n]

ai,kek

∣∣∣∣∣∣ = |λ||ei |.

|ei | ≥ |ej |.

ai,kek = λei ,

d |ei | = |ei |∑k∈[n]

|ai,k | ≥∑k∈[n]

|ai,k ||ek | ≥

∣∣∣∣∣∣∑k∈[n]

ai,kek

∣∣∣∣∣∣ = |λ||ei |.

d-Regular Graphs as Matrices (cont’d)

LemmaThe graph is connected if and only if λ1 > λ2.

Proof.(⇐) Easy.

(⇒) Let v = (e1, . . . , en) be an eigenvector for d . We choose an i ∈ [n] suchthat for every j ∈ [n]

|ei | ≥ |ej |.

Without loss of generality, we assume ei > 0. Then,

dei = ei

∑k∈[n]

ai,k = ei

∑{i,k}∈E

ai,k ≥∑

{i,k}∈E

ai,kek =∑k∈[n]

ai,kek = dei ,

where {i , k} ∈ E means that there is at least one edge between the i-th andthe k-th vertices.

Therefore, if {i , k} ∈ E , then ek = ei . Then we repeat the above argument onek .

Proof.(⇐) Easy.

|ei | ≥ |ej |.

dei = ei

∑k∈[n]

ai,k = ei

∑{i,k}∈E

ai,k ≥∑

{i,k}∈E

ai,kek =∑k∈[n]

ai,kek = dei ,

Proof.(⇐) Easy.

|ei | ≥ |ej |.

dei = ei

∑k∈[n]

ai,k = ei

∑{i,k}∈E

ai,k ≥∑

{i,k}∈E

ai,kek =∑k∈[n]

ai,kek = dei ,

Proof.(⇐) Easy.

(⇒) Let v = (e1, . . . , en) be an eigenvector for d .

We choose an i ∈ [n] suchthat for every j ∈ [n]

|ei | ≥ |ej |.

dei = ei

∑k∈[n]

ai,k = ei

∑{i,k}∈E

ai,k ≥∑

{i,k}∈E

ai,kek =∑k∈[n]

ai,kek = dei ,

Proof.(⇐) Easy.

|ei | ≥ |ej |.

Without loss of generality, we assume ei > 0.

dei = ei

∑k∈[n]

ai,k = ei

∑{i,k}∈E

ai,k ≥∑

{i,k}∈E

ai,kek =∑k∈[n]

ai,kek = dei ,

Proof.(⇐) Easy.

|ei | ≥ |ej |.

dei = ei

∑k∈[n]

ai,k = ei

∑{i,k}∈E

ai,k ≥∑

{i,k}∈E

ai,kek =∑k∈[n]

ai,kek = dei ,

Proof.(⇐) Easy.

|ei | ≥ |ej |.

dei = ei

∑k∈[n]

ai,k = ei

∑{i,k}∈E

ai,k ≥∑

{i,k}∈E

ai,kek =∑k∈[n]

ai,kek = dei ,

Therefore, if {i , k} ∈ E , then ek = ei .

Then we repeat the above argument onek .

Proof.(⇐) Easy.

|ei | ≥ |ej |.

dei = ei

∑k∈[n]

ai,k = ei

∑{i,k}∈E

ai,k ≥∑

{i,k}∈E

ai,kek =∑k∈[n]

ai,kek = dei ,

LemmaThe graph is bipartite if and only if λ1 = −λn.

Proof.(⇒) Easy.

(⇐) Let v = (e1, . . . , en) be an eigenvector for −d . We choose an i ∈ [n] suchthat for every j ∈ [n]

|ei | ≥ |ej |.

dei = ei

∑k∈[n]

ai,k = ei

∑{i,k}∈E

On the other hand,

dei = −(−dei ) = −∑k∈[n]

ai,kek =∑

{i,k}∈E

ai,k(−ek).

Therefore, if {i , k} ∈ E , then ek = −ei . Then we repeat the above argumenton ek .

Proof.(⇒) Easy.

|ei | ≥ |ej |.

dei = ei

∑k∈[n]

ai,k = ei

∑{i,k}∈E

On the other hand,

dei = −(−dei ) = −∑k∈[n]

ai,kek =∑

{i,k}∈E

ai,k(−ek).

Proof.(⇒) Easy.

|ei | ≥ |ej |.

dei = ei

∑k∈[n]

ai,k = ei

∑{i,k}∈E

On the other hand,

dei = −(−dei ) = −∑k∈[n]

ai,kek =∑

{i,k}∈E

ai,k(−ek).

Proof.(⇒) Easy.

(⇐) Let v = (e1, . . . , en) be an eigenvector for −d .

We choose an i ∈ [n] suchthat for every j ∈ [n]

|ei | ≥ |ej |.

dei = ei

∑k∈[n]

ai,k = ei

∑{i,k}∈E

On the other hand,

dei = −(−dei ) = −∑k∈[n]

ai,kek =∑

{i,k}∈E

ai,k(−ek).

Proof.(⇒) Easy.

|ei | ≥ |ej |.

dei = ei

∑k∈[n]

ai,k = ei

∑{i,k}∈E

On the other hand,

dei = −(−dei ) = −∑k∈[n]

ai,kek =∑

{i,k}∈E

ai,k(−ek).

Proof.(⇒) Easy.

|ei | ≥ |ej |.

dei = ei

∑k∈[n]

ai,k = ei

∑{i,k}∈E

On the other hand,

dei = −(−dei ) = −∑k∈[n]

ai,kek =∑

{i,k}∈E

ai,k(−ek).

Proof.(⇒) Easy.

|ei | ≥ |ej |.

dei = ei

∑k∈[n]

ai,k = ei

∑{i,k}∈E

On the other hand,

dei = −(−dei ) = −∑k∈[n]

ai,kek =∑

{i,k}∈E

ai,k(−ek).

Proof.(⇒) Easy.

|ei | ≥ |ej |.

dei = ei

∑k∈[n]

ai,k = ei

∑{i,k}∈E

On the other hand,

dei = −(−dei ) = −∑k∈[n]

ai,kek =∑

{i,k}∈E

ai,k(−ek).

Therefore, if {i , k} ∈ E , then ek = −ei .

Then we repeat the above argumenton ek .

Proof.(⇒) Easy.

|ei | ≥ |ej |.

dei = ei

∑k∈[n]

ai,k = ei

∑{i,k}∈E

On the other hand,

dei = −(−dei ) = −∑k∈[n]

ai,kek =∑

{i,k}∈E

ai,k(−ek).

Expander Graphs are Almost Good

Recall: A graph G = (V , E) is (c, d)-good if each v ∈ V has degree at most dand for all u, v ∈ V , if there is a path from u to v , then

DefinitionA d-regular graph G on n vertices is called an (n, d)-graph. Moreover, it is an(n, d , α)-graph if λ(G) ≤ αd .

TheoremLet α < 1/2. Then all (n, d , α) graphs are (c, d)-good for some appropriatec ∈ N.

DefinitionA d-regular graph G on n vertices is called an (n, d)-graph.

Moreover, it is an(n, d , α)-graph if λ(G) ≤ αd .

Fix an (n, d , α)- graph G = (V , E). For every v ∈ V and r ∈ N let

B(x , r) :={y ∈ V | dist(x , y) ≤ r

Claim. If |B(x , r)| ≤ n/2, then for some fixed ε > 0∣∣B(x , r + 1)∣∣ ≥ (1 + ε)

∣∣B(x , r)∣∣.

By the Expander Mixing Lemma, for every S ⊆ V

|E(S , S)||S | ≤ d

(|S |n

Then by |E(S , S)|+ |E(S , S)| = d |S |,

|E(S , S)||S | ≥ d

((1− α)− |S |

S has at least |E(S , S)|/d neighbors outside S . The claim follows by takingε := 1/2− α. a

Then for some r = O(log n) and every x ∈ V we have |B(x , r)| > n/2. �

Fix an (n, d , α)- graph G = (V , E).

For every v ∈ V and r ∈ N let

B(x , r) :={y ∈ V | dist(x , y) ≤ r

∣∣B(x , r)∣∣.

|E(S , S)||S | ≤ d

(|S |n

Then by |E(S , S)|+ |E(S , S)| = d |S |,

|E(S , S)||S | ≥ d

((1− α)− |S |

B(x , r) :={y ∈ V | dist(x , y) ≤ r

∣∣B(x , r)∣∣.

|E(S , S)||S | ≤ d

(|S |n

Then by |E(S , S)|+ |E(S , S)| = d |S |,

|E(S , S)||S | ≥ d

((1− α)− |S |

B(x , r) :={y ∈ V | dist(x , y) ≤ r

∣∣B(x , r)∣∣.

|E(S , S)||S | ≤ d

(|S |n

Then by |E(S , S)|+ |E(S , S)| = d |S |,

|E(S , S)||S | ≥ d

((1− α)− |S |

B(x , r) :={y ∈ V | dist(x , y) ≤ r

∣∣B(x , r)∣∣.

|E(S , S)||S | ≤ d

(|S |n

Then by |E(S , S)|+ |E(S , S)| = d |S |,

|E(S , S)||S | ≥ d

((1− α)− |S |

B(x , r) :={y ∈ V | dist(x , y) ≤ r

∣∣B(x , r)∣∣.

|E(S , S)||S | ≤ d

(|S |n

Then by |E(S , S)|+ |E(S , S)| = d |S |,

|E(S , S)||S | ≥ d

((1− α)− |S |

B(x , r) :={y ∈ V | dist(x , y) ≤ r

∣∣B(x , r)∣∣.

|E(S , S)||S | ≤ d

(|S |n

Then by |E(S , S)|+ |E(S , S)| = d |S |,

|E(S , S)||S | ≥ d

((1− α)− |S |

S has at least |E(S , S)|/d neighbors outside S .

The claim follows by takingε := 1/2− α. a

B(x , r) :={y ∈ V | dist(x , y) ≤ r

∣∣B(x , r)∣∣.

|E(S , S)||S | ≤ d

(|S |n

Then by |E(S , S)|+ |E(S , S)| = d |S |,

|E(S , S)||S | ≥ d

((1− α)− |S |

B(x , r) :={y ∈ V | dist(x , y) ≤ r

∣∣B(x , r)∣∣.

|E(S , S)||S | ≤ d

(|S |n

Then by |E(S , S)|+ |E(S , S)| = d |S |,

|E(S , S)||S | ≥ d

((1− α)− |S |

Random Walks on Expander Graphs

Random Walks

A vector p = (p1, . . . , pn) ∈ Rn is a probability (distribution) vector if all pi ≥ 0and

∑i∈[n] pi = 1.

The probability vector for the uniform distribution is

n, . . . ,

DefinitionA random walk on a graph G = (V , E) is a discrete-time stochastic process(X0, X1, . . .) taking values in V . The vertex X0 is sampled from some initialdistribution on V , and Xi+1 is chosen uniformly at random from the neighborsof Xi .

Random Walks

∑i∈[n] pi = 1. The probability vector for the uniform distribution is

n, . . . ,

Random Walks

n, . . . ,

DefinitionA random walk on a graph G = (V , E) is a discrete-time stochastic process(X0, X1, . . .) taking values in V .

The vertex X0 is sampled from some initialdistribution on V , and Xi+1 is chosen uniformly at random from the neighborsof Xi .

Random Walks

n, . . . ,

DefinitionA random walk on a graph G = (V , E) is a discrete-time stochastic process(X0, X1, . . .) taking values in V . The vertex X0 is sampled from some initialdistribution on V ,

and Xi+1 is chosen uniformly at random from the neighborsof Xi .

Random Walks

n, . . . ,

Normalized Adjacency Matrices

Let G be a d-regular graph with adjacency matrix A = A(G). Then itsnormalized adjacency matrix is

I The random walk on G = (V , E) is a Markov Chain with state set V andtransition matrix A.

I A s real, symmetric, and doubly stochastic; i.e. every column and every rowsums up to 1.

I If d = λ1 ≥ . . . ≥ λn are the eigenvalues of A, then 1 = λ1 ≥ . . . ≥ λn

with λi = λi/d for i ∈ [n] are the eigenvalues of A with the sameeigenvectors as of A.

I Sampling a vertex x from some probability distribution p on V and thenmoving to a random neighbor of x is equivalent to sampling a vertex fromthe distribution Ap.

Let G be a d-regular graph with adjacency matrix A = A(G).

Then itsnormalized adjacency matrix is

Normalized Adjacency Matrices (cont’d)

I The matrix At

is the transition matrix of the Markov Chain defined by

random walks of length t, i.e., (At)i,j is the probability a random walk

starting at i is at j after t steps.

I The stationary distribution of the random walk on G is the uniformdistribution, namely, uA =Au = u.

I The matrix At

random walks of length t,

i.e., (At)i,j is the probability a random walk

I The matrix At

The norms of vectors

Let x = (x1, . . . , xn) ∈ Rn. Then:

- ‖x‖1 :=∑

i∈[n] |xi |.

- ‖x‖2 :=√∑

i∈[n] x2i .

- ‖x‖∞ := maxi∈[n] |xi |.

Let x = (x1, . . . , xn) ∈ Rn. Then:

- ‖x‖1 :=∑

i∈[n] |xi |.

- ‖x‖2 :=√∑

i∈[n] x2i .

- ‖x‖∞ := maxi∈[n] |xi |.

Let x = (x1, . . . , xn) ∈ Rn. Then:

- ‖x‖1 :=∑

i∈[n] |xi |.

- ‖x‖2 :=√∑

i∈[n] x2i .

- ‖x‖∞ := maxi∈[n] |xi |.

Let x = (x1, . . . , xn) ∈ Rn. Then:

- ‖x‖1 :=∑

i∈[n] |xi |.

- ‖x‖2 :=√∑

i∈[n] x2i .

- ‖x‖∞ := maxi∈[n] |xi |.

Let x = (x1, . . . , xn) ∈ Rn. Then:

- ‖x‖1 :=∑

i∈[n] |xi |.

- ‖x‖2 :=√∑

i∈[n] x2i .

- ‖x‖∞ := maxi∈[n] |xi |.

Rapid Mixing of Walks

TheoremLet G be an (n, d , α)-graph with normalized adjacency matrix A. Then for anydistribution vector p and any positive integer t:

‖Atp − u‖1 ≤

√n αt

Corollary

(n, d , α)-graphs are (c, d)-good with

c = − 3

2 log α+ ε.

and ε > 0.

‖Atp − u‖1 ≤

√n αt

Corollary

c = − 3

2 log α+ ε.

and ε > 0.

‖Atp − u‖1 ≤

√n αt

Corollary

c = − 3

2 log α+ ε.

and ε > 0.

Rapid Mixing of Walks (cont’d)

‖Atp − u‖2 ≤ ‖p − u‖2α

t ≤ αt .

LemmaFor every probability vector p, ‖Ap − u‖2 ≤ ‖p − u‖2α ≤ α.

‖Atp − u‖2 ≤ ‖p − u‖2α

t ≤ αt .

‖Atp − u‖2 ≤ ‖p − u‖2α

t ≤ αt .

Proof of the Lemma

Obviously, 〈p − u, u〉 = 0, that is, p − u and u are orthogonal. It is easy toshow ‖p − u‖2 ≤ 1. Consequently, for some appropriate (αi )2≤i≤n we have

p − u =∑

2≤i≤n

βivi .

‖Ap − u‖2 = ‖Ap − Au‖2 = ‖A(p − u)‖2

∥∥∥∥∥∥∑

2≤i≤n

λiβivi

∥∥∥∥∥∥2

√√√√⟨ ∑2≤i≤n

λiβivi ,∑

2≤i≤n

λiβivi

√ ∑2≤i≤n

λ2i β

2i 〈vi , vi 〉 ≤ α

√ ∑2≤i≤n

β2i 〈vi , vi 〉

∥∥∥∥∥∥∑

2≤i≤n

∥∥∥∥∥∥2

= α‖p − u‖2 ≤ α.

Proof of the Lemma

Obviously, 〈p − u, u〉 = 0, that is, p − u and u are orthogonal.

It is easy toshow ‖p − u‖2 ≤ 1. Consequently, for some appropriate (αi )2≤i≤n we have

p − u =∑

2≤i≤n

βivi .

‖Ap − u‖2 = ‖Ap − Au‖2 = ‖A(p − u)‖2

∥∥∥∥∥∥∑

2≤i≤n

λiβivi

∥∥∥∥∥∥2

√√√√⟨ ∑2≤i≤n

λiβivi ,∑

2≤i≤n

λiβivi

√ ∑2≤i≤n

λ2i β

2i 〈vi , vi 〉 ≤ α

√ ∑2≤i≤n

β2i 〈vi , vi 〉

∥∥∥∥∥∥∑

2≤i≤n

∥∥∥∥∥∥2

= α‖p − u‖2 ≤ α.

Proof of the Lemma

Obviously, 〈p − u, u〉 = 0, that is, p − u and u are orthogonal. It is easy toshow ‖p − u‖2 ≤ 1.

Consequently, for some appropriate (αi )2≤i≤n we have

p − u =∑

2≤i≤n

βivi .

‖Ap − u‖2 = ‖Ap − Au‖2 = ‖A(p − u)‖2

∥∥∥∥∥∥∑

2≤i≤n

λiβivi

∥∥∥∥∥∥2

√√√√⟨ ∑2≤i≤n

λiβivi ,∑

2≤i≤n

λiβivi

√ ∑2≤i≤n

λ2i β

2i 〈vi , vi 〉 ≤ α

√ ∑2≤i≤n

β2i 〈vi , vi 〉

∥∥∥∥∥∥∑

2≤i≤n

∥∥∥∥∥∥2

= α‖p − u‖2 ≤ α.

Proof of the Lemma

p − u =∑

2≤i≤n

βivi .

‖Ap − u‖2 = ‖Ap − Au‖2 = ‖A(p − u)‖2

∥∥∥∥∥∥∑

2≤i≤n

λiβivi

∥∥∥∥∥∥2

√√√√⟨ ∑2≤i≤n

λiβivi ,∑

2≤i≤n

λiβivi

√ ∑2≤i≤n

λ2i β

2i 〈vi , vi 〉 ≤ α

√ ∑2≤i≤n

β2i 〈vi , vi 〉

∥∥∥∥∥∥∑

2≤i≤n

∥∥∥∥∥∥2

= α‖p − u‖2 ≤ α.

Proof of the Lemma

p − u =∑

2≤i≤n

βivi .

‖Ap − u‖2 = ‖Ap − Au‖2 = ‖A(p − u)‖2

∥∥∥∥∥∥∑

2≤i≤n

λiβivi

∥∥∥∥∥∥2

√√√√⟨ ∑2≤i≤n

λiβivi ,∑

2≤i≤n

λiβivi

√ ∑2≤i≤n

λ2i β

2i 〈vi , vi 〉 ≤ α

√ ∑2≤i≤n

β2i 〈vi , vi 〉

∥∥∥∥∥∥∑

2≤i≤n

∥∥∥∥∥∥2

= α‖p − u‖2 ≤ α.

Expander Graphs and Their Applications (II)basics.sjtu.edu.cn/~chen/teaching/EXPANDER09/exp2.pdf ·...

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