Entropy Waves, The Zigzag Graph Product, and New ConstantDegree Expanders

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

• Omer Reingold
• Salil Vadhan
• Avi Wigderson
• Lecturer Oded Levy

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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, ?A), GB (DA, DB, ?B).
• 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 v?NA and
  u?NB .
• For each vertex (v, k) we put an edge between a
  pair (v, k) to (vki, kij).

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Example
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• 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, DB2, f(?A , ?B)).
• f(?A?B) ?A ?B ?B 2.
• ?A, ?B lt 1? f(?A , ?B) lt 1.
• We know that .
• We aim to show that for each

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Why does it work?

• Given a distribution vector p over Gs vertices,
  we distinguish between two cases of p
• ? uniform within clouds.
• ? non-uniform within clouds.
• We aim to show that after one step in G, p
  becomes more uniform.

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? uniform within clouds.
Step 1 does not change anything.
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? uniform within clouds.
Step 2 is a random walk on GA.
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? uniform within clouds.
Step 3 is a random walk on GB.
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? non-uniform within clouds.
Step 1 is a random walk on GB.
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? non-uniform within clouds.
Step 2 is a permutation.
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? non-uniform within clouds.
Step 3 is a random step on a regular graph.
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Proof

• Let M be Gs normalized adjacency matrix.
• Well 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.

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Decomposing a
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• The relation between and B is

• The relation between and A will be defined
  later.

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is symmetric

• Recall that we want to bound

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

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• Thus

• Combining it with previous results our expression
  becomes

• Opening the inner product we get

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• Using triangular inequality we get

• Using Cauchy-Schwartz inequality

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

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• Well first bound .
• Decomposing a? to its components we have

• By the expansion of GB we get

• Thus

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• For our analysis well define linear map C such
  that

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• Now well bound .
• Well first relate A to Ã.
• Claim

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• ev is a basis, thus this claim holds for all
  vectors .
• Substituting ß Ca yieldsAC a CÃ a .

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• Hence

• Since , . Thus

• Combining these claims we have

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• Well mark by p and q
• By Pythagoras theorem we have .
• Thus

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pq gets maximum value of ½ (when ), thus we
showed that for every
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• It remains to show that if both ?1 and ?2 are
  smaller than 1, f(?1 ,? 2) is also smaller than
  1.
• Case I

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• Case II

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• We have
• Finally

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Summary

• We started from vector a orthogonal to the
  uniform distribution.
• We decomposed it to its parallel and
  perpendicular components.
• We showed that we can shrink both the parallel
  and the perpendicular components individually.

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• We decreased the length of the perpendicular
  component in the Zig part.
• The second step did not increase the length of
  the parallel component, but the third step surly
  decreased it.
• The zigzag graph decreased the length of a.

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• We can show that we the bound may be reduced to
  
• f(?, 0) f(0, ?)0
• f(?, 1) f(1, ?)1
• f is strictly increasing in both ?1 and ?2.
• If ?1lt1 and ?2lt1 then f(?1, ?2)lt1.
• f(?1, ?2) ?1 ?2

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Families of expanders

• In this section well 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
  wayG1 H2Gi1 Gi2 Z H
• Gi is an infinite family of expanders such that
  Gi (D4i, D2, 2/5)

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• Vertices.
• Degree.
• ExpansionBounded by 2/5 since the limit of the
  series ?n ?2n-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, ?). For every t we define

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• For each t 0, Gt (D8t, D2, ?t) such that?t
  ? O(?2).
• Vertices.
• Degree.
• ExpansionLet µt max?i 1 i t. We have
  µt maxµt-1, µt-12 ? ?2. Solving this
  yields µt lt ? O(?2).

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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 well try to get closer to this
  bound.

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Derandomizing the Zigzag product

• In order to improve the bound well make two
  steps within the zig part and two steps within
  the zag part.
• In order to save on the degree, well 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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• Well 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.
• Well use a matrix to represent this permutation.

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• Each edge on the improved Zigzag product can be
  separated into six steps
• Move one step within the first cloud.
• Move one step within the first cloud.
• Jump between clouds.
• Move one step within the second cloud according
  to the step made in step 2.
• Move one step within the second cloud.

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• Claim A Z B (NADA, DB3, ?A2 ?B2)
• Let Bi a DA DA permutation matrix

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

• Thus

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• Bi is a permutation matrix thus

• We can now decompose Ma

• Substitutting this in the formula we get that

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• 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.

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

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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
(2, 1)
(2, 2)
(2, 3)
(2, 4)
(1, 4)
(3, 1)
(1, 3)
(3, 2)
(1, 2)
(3, 3)
(1, 1)
(3, 4)
(4, 4)
(4, 3)
(4, 2)
(4, 1)
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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,b?Lc,d / q2.

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• We distinguish between 3 cases
• a ? cThe two lines intersect in exactly one
  point.
• a c and b ? dThe two lines does not intersect.
• a c and b dThe two lines intersect in
  exactly q points.
• Denote Iq the identity matrix in size q
• Denote Jq the all ones matrix in size q

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• Obviously

• Since eigenvalues of Jq are q (one time) and 0 (q
  - 1 times), (Jq - Iq)Ä Jq eigenvalues are q(q
  1), -q and 0. IqÄ 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 Ms 2nd largest
  eigenvalue is q-½ .

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• Define the following graph family
• (APq)1 (APq)Ä(APq)
• (APq)i1 (APq)i Z (APq)
• Combining it with the previous theorem we get
  that (APq)i (q2(i1), q2, O(iq-½))