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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.
3
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]).
4
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
7
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
8
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.
11
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
12
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
14
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,
15
16
Thus
)(),(,~
||~
||~
BBAM
Combining it with previous results our expression becomes
~
||||~~
)( BBB
Opening the inner product we get
~~~
||~~~
||~
||||~
,,,,, BBABABAAM
17
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
18
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
19
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
21
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
22
Now we’ll bound . We’ll first relate A to Ã. Claim:
||||~
,A
)1
(~
A
Dvv D
eACAe A
23
A
Dv D
e A1
)1
(~
A
Dv D
eAC A )1
(~
A
Dv D
eA A
)(1~
A
AD
DveAC
24
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
25
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
26
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
27
Case II
223
122
2222
2~2||~
||~
||
~||
~||
~
~||
~||
~
))(1(
)()(
)(
)(),(,
B
A
B
B
BBB
BBA
BBAM
B
A
31
28
Finally
1},min{1),( 2
22
9
)1()1(9
1
B
BAA
BAf
29
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
30
Families of expanders
In this section we’ll introduce two families of expanders:
Basic construction More efficient construction
31
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)
32
Vertices. Degree. Expansion
Bounded by 2/5 since the limit of the series λn = λ2
n-1 + 6/25 is 2/5.
33
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
34
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).
35
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.
36
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.
37
Example
38
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.
39
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.
40
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
41
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
42
~~~
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
43
Using the same techniques from previous results we get the desired result.
22||22|| 2, BBAM
2222 2,
,qpqp
MBBA
222 2,
,BABBA
M
44
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 .
45
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)
46
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.
47
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
48
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 .
49
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))