1

Entropy Waves, The Zigzag Graph Product, and NewConstant-Degree Expanders

Omer ReingoldSalil VadhanAvi Wigderson

Lecturer: Oded Levy

2

Introduction

Most random constant degree graphs are good expanders.

Most applications need explicit constructions.

This lecture deals with the Zigzag product which is used in order to explicitly build expanders.

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The zigzag product

GA = (NA, DA, λA), GB = (DA, DB, λB). We define their zigzag product as a

graph which Its vertices are NA × NB (each vertex is

represented as a pair (v, u) such that vNA and uNB .

For each vertex (v, k) we put an edge between a pair (v, k) to (v[k[i]], k[i][j]).

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Example

5

6

We can separate it to three steps: Move from one vertex to another

vertex at the same cloud. Jump between clouds. Move from one vertex to another

vertex at the same cloud

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Theorem

Theorem G = GA Z GB = (NA× DA, DB

2, f(λA , λB)).

f(λA’λB) ≤ λA + λB + λB 2.

λA, λB < 1→ f(λA , λB) < 1. We know that . We aim to show that for each AA DN 1

),(,

,BAf

M

,

,max

1

_ M

ADAN

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Why does it work?

Given a distribution vector π over G’s vertices, we distinguish between two cases of π:

uniform within clouds. non-uniform within clouds.

We aim to show that after one step in G, π becomes more uniform.

9Step 2 is a random walk on GA.Step 3 is a random walk on GB.

uniform within clouds.

Step 1 does not change anything.

10Step 1 is a random walk on GB. Step 2 is a permutation.Step 3 is a random step on a regular

graph.

non-uniform within clouds.

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Proof Let M be G’s normalized adjacency

matrix. We’ll decompose M into three matrices,

corresponding to three steps as defined before. Then

is a normalized matrix where we connect vertices within each cloud.

is a normalized matrix where we connect clouds.

~

B

~~~

BABM

~

A

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The relation between and B is

AA

A

NN

N

B

B

B

B

BB

00

00

00

00

00

1~

The relation between and A will be defined later.

~

B

~

A

13

A

AAAA

A

AAA

A

AAA

A

AA

A

A

A

A

A

AA

A

A

A

A

A

AA

A

AA

A

AA

A

A

A

A

A

A

A

A

A

A

A

A

D

iiNDDN

D

iiNDkN

D

iiNDN

D

iivDDv

D

iivDkv

D

iivDv

D

iiDD

D

iiDk

D

iiD

D

iiND

D

iiND

D

iiND

D

iivD

D

iivD

D

iivD

D

iiD

D

iiD

D

iiD

1,

1,

1,

1,

1,

11,

1,

1,

1,

1,

1,

11,

1,1

1,1

1,1

1,1

1,1

11,1

1,

1

1,

1

1,

1

1,

1

1,

1

1,

1

1,1

1

1,1

1

1,1

1

AA N

v

N

v

1

||

||

||1

AA

A

A

A

A

DN

kN

N

Dv

kv

v

D

k

,

,

1,

,

,

1,

,1

,1

1,1

||

Decomposing α

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Recall that we want to bound

||

||

||1

||

||

||1

||

||

||1

||~

000

00

00

000

AAA N

v

N

v

N

v

B

B

B

B

B

B

)(),(,,, ||~

||~~~~~~~~

BBABBABABM

Since GB is a regular graph, (1, 1, …, 1) is an eigenvector of B with eigenvalue 1,

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16

Thus

)(),(,~

||~

||~

BBAM

Combining it with previous results our expression becomes

~

||||~~

)( BBB

Opening the inner product we get

~~~

||~~~

||~

||||~

,,,,, BBABABAAM

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Using triangular inequality we get

2~~|||||| 2,, BBAM

Since à is a permutation matrix, it is unitary thus preserves length

~~~

||~~~

||~

||||~

,,,,, BBABABAAM

2~||

~~||||||

~

,, BBBAM

Using Cauchy-Schwartz inequality

~~~

||~~~

||~

||||~

,,, BBABABAAM

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We’ll first bound . Decomposing α to its components we

have

~

B

vv

vvv

v BeeBB )(~~

BB~

By the expansion of GB we get vBvB

Thus

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For our analysis we’ll define linear map C such that:

A

AAA

AAA

N

D

kkN

D

kkv

D

kk

DNN

v

DNN

v

C

C

C

CC

1

1

1

1

0,

0,

0,1

11

)(

)(

)(

)(

20

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It is easy to see that

A

A

A

A

A

A

A

A

AA

D

kkN

D

kkN

D

kkv

D

kkv

D

kk

D

kk

DAD DC

1,

1,

1,

1,

1,1

1,1

1|| /1

In addition

A

AA

AA

NDN

DNC 1

1

1

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Now we’ll bound . We’ll first relate A to Ã. Claim:

||||~

,A

)1

(~

A

Dvv D

eACAe A

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A

Dv D

e A1

)1

(~

A

Dv D

eAC A )1

(~

A

Dv D

eA A

)(1~

A

AD

DveAC

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Hence CACCACCAA

AAAA DDDD ,,1,, 1||~

1||~

1||||~

||||111||||~

,,,,, ADDADD A

AD

A

AD

A

A

ACCCCCACA

Since Cα is orthogonal to

Combining these claims we have22||2|| 2, BBAM

AN1

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pq gets maximum value of ½ (when ),thus

0

0.1

0.2

0.3

0.4

0.5

0.6

0 0.2 0.4 0.6 0.8 1

1,, 22

||

qppq

222,

,2 qpqp BBA

M

2,

,

BBA

M

22qp

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It remains to show that if both λ1 and λ2

are smaller than 1, f(λ1 ,λ 2) is also smaller than 1.

Case I

2

2

2

2

2

)1(

)27(

))1(79(

)1()1(2

2,

91

9

9

9122

2

3122

312

A

B

A

B

A

A

AA

AAA

BBAM

B

A

31

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Case II

223

122

2222

2~2||~

||~

||

~||

~||

~

~||

~||

~

))(1(

)()(

)(

)(),(,

B

A

B

B

BBB

BBA

BBAM

B

A

31

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Finally

1},min{1),( 2

22

9

)1()1(9

1

B

BAA

BAf

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We can show that we the bound may be reduced to

f(λ, 0)= f(0, λ)=0 f(λ, 1)= f(1, λ)=1 f is strictly increasing in both λ1 and λ2.

If λ1<1 and λ2<1 then f(λ1, λ2)<1. f(λ1, λ2)≤ λ1+ λ2

222212

21 4)1()1(),( BBABABAf

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Families of expanders

In this section we’ll introduce two families of expanders:

Basic construction More efficient construction

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Basic family construction

Let H = (D4, D, 1/5). We build a family of graph in the

following way:G1 = H2

Gi+1 = Gi2 Z H

Gi is an infinite family of expanders such that Gi = (D4i, D2, 2/5)

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Vertices. Degree. Expansion

Bounded by 2/5 since the limit of the series λn = λ2

n-1 + 6/25 is 2/5.

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More efficient construction We use tensoring in order to make the

construction in more efficient by reducing the depth of the recursion.

Let H = (D8, D, λ). For every t we define

HZGGG

HHG

HG

ttt 22

21

)(21

21

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For each t ≥ 0, Gt = (D8t, D2, λt) such thatλt = λ + O(λ2).

Vertices. Degree. Expansion

Let μt = max{λi | 1 ≤ i ≤ t}. We have μt = max{μt-1, μt-1

2 + λ + λ2}. Solving this yields μt < λ + O(λ2).

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Can we do better?

The best possible 2nd largest eigenvalues of infinite families of graph is 2(D – 1)½ / D.

Ramanujan graphs meet this bound.

In this section we’ll try to get closer to this bound.

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Derandomizing the Zigzag product

In order to improve the bound we’ll make two steps within the zig part and two steps within the zag part.

In order to save on the degree, we’ll correlate the second part of the zig part and the first part of the zag part.

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Example

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We’ll map each step within each cloud to a step on other cloud.

Given a step in the initial cloud and a target cloud, the permutation will return a step in the target cloud.

We’ll use a matrix to represent this permutation.

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Each edge on the improved Zigzag product can be separated into six steps:

1. Move one step within the first cloud.2. Move one step within the first cloud.3. Jump between clouds.4. Move one step within the second cloud

according to the step made in step 2.5. Move one step within the second cloud.

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Claim: A Z’ B = (NADA, DB3, λA+2 λB

2)

Let Bi a DA ×DA permutation matrix

i

t

iiD BABMA

~~~1'

Note that the normalized adjacency matrix corresponding to steps 2-4 is

A

AAi

D

i

iDiN BBBIB1

~1

~~

Thus~~

' BMBM

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Bi is a permutation matrix thus||||

~||||

~

tBB

We can now decompose M’α||

||~~

||~~

1||~~~

1||' ABABBABMi

iDi

tiiD AA

Substitutting this in the formula we get that

,,,, |||||| MMMM

,',',',~~

||~~

||||~~

BMBBMBBMBM

~~

||~

||||~

,',',', BBMMBMBM

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~~~

2||~

||||~

,',,, BBMBAAM

~~

||~~~

||||~~~

,',,, BBMABBABBM

All the eigenvalues of M’ are smaller than 1 since they are eigenvalues of an normalized adjacency matrix of an undirected regular graph. Thus M’ does not increase the length of any vector.

2~~2||||||

~

2,, BBAM

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Using the same techniques from previous results we get the desired result.

22||22|| 2, BBAM

2222 2,

,qpqp

MBBA

222 2,

,BABBA

M

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The affine plane

Consider field Fq such that q = pt for some prime number t.

We define AFq be a Fq × Fq graph such that each vertex is a pair. We put an edge between (a, b) and (c, d) if and only if ac = b + d, i.e. we connect for all the points on the line y = ax - b .

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Example

(4, 4) (4, 3) (4, 2) (4, 1)

(1, 4)

(1, 3)

(1, 2)

(1, 1)

(2, 1) (2, 2) (2, 3) (2, 4)

(3, 1)

(3, 2)

(3, 3)

(3, 4)

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Lemma: AFq = (q2, q, q-½). Proof:

An entry in the square of the normalized adjacency matrix of AFq in row (a, b) and column (c, d) holds the number of common neighbors of (a, b) and (c, d). Since each vertex neighbor is a line, the common neighbors of (a, b) and (c, d) is |La,bLc,d| / q2.

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We distinguish between 3 cases: a ≠ c

The two lines intersect in exactly one point. a = c and b ≠ d

The two lines does not intersect. a = c and b = d

The two lines intersect in exactly q points. Denote Iq the identity matrix in size q Denote Jq the all one’s matrix in size q

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Obviously

222 )(1

q

JIJqII

qIJJJ

JqIJJ

JJqIJ

JJJqI

qM qqqqq

qqqq

qqqq

qqqq

qqqq

Since eigenvalues of Jq are q (one time) and 0 (q - 1 times), (Jq - Iq) Jq eigenvalues are q(q – 1), -q and 0. Iq

qqIq contributes q to each eigenvalue. Dividing it by q we get that M2 eigenvalues are 1, 0 and 1/q, thus the 2nd largest eigenvalue is q-1, and M’s 2nd largest eigenvalue is q-½ .

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Define the following graph family: (APq)1 = (APq)APq)

(APq)i+1 = (APq)i Z (APq)

Combining it with the previous theorem we get that (APq)i = (q2(i+1), q2, O(iq-½))