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Entropy Waves, The Zigzag Graph Product, and New Constant-Degree Expanders

Entropy Waves, The Zigzag Graph Product, and New Constant-Degree Expanders. Omer Reingold Salil Vadhan Avi Wigderson Lecturer: Oded Levy. Introduction. Most random constant degree graphs are good expanders. Most applications need explicit constructions.

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Entropy Waves, The Zigzag Graph Product, and New Constant-Degree Expanders

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  1. Entropy Waves, The Zigzag Graph Product, and NewConstant-Degree Expanders Omer Reingold Salil Vadhan Avi 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 vNA anduNB . • For each vertex (v, k) we put an edge between a pair (v, k) to (v[k[i]], k[i][j]).

  4. Example

  5. 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

  6. Theorem • Theorem • G = GAZ GB = (NA×DA, DB2, 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

  7. 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.

  8.  uniform within clouds. Step 1 does not change anything. Step 2 is a random walk on GA. Step 3 is a random walk on GB.

  9.  non-uniform within clouds. Step 1 is a random walk on GB. Step 2 is a permutation. Step 3 is a random step on a regular graph.

  10. 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.

  11. The relation between and B is • The relation between and A will be defined later.

  12. Decomposing α

  13. Recall that we want to bound • Since GB is a regular graph, (1, 1, …, 1) is an eigenvector of B with eigenvalue 1,

  14. Thus • Combining it with previous results our expression becomes • Opening the inner product we get

  15. Using triangular inequality we get • Using Cauchy-Schwartz inequality • Since à is a permutation matrix, it is unitary thus preserves length

  16. We’ll first bound . • Decomposing α to its components we have • By the expansion of GB we get • Thus

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

  18. It is easy to see that • In addition

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

  20. Hence • Since Cα is orthogonal to • Combining these claims we have

  21. pq gets maximum value of ½ (when ), thus

  22. It remains to show that if both λ1 andλ2 are smaller than 1, f(λ1 ,λ 2) is also smaller than 1. • Case I

  23. Case II

  24. Finally

  25. 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 bothλ1 andλ2. • If λ1<1andλ2<1 then f(λ1, λ2)<1. • f(λ1, λ2)≤ λ1+ λ2

  26. Families of expanders • In this section we’ll introduce two families of expanders: • Basic construction • More efficient construction

  27. Basic family construction • Let H = (D4, D, 1/5). • We build a family of graph in the following way:G1 = H2Gi+1 = Gi2 Z H • Gi is an infinite family of expanders such that Gi = (D4i, D2, 2/5)

  28. Vertices. • Degree. • ExpansionBounded by 2/5 since the limit of the series λn = λ2n-1 + 6/25 is 2/5.

  29. 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

  30. For each t ≥ 0, Gt =(D8t, D2, λt) such thatλt = λ + O(λ2). • Vertices. • Degree. • ExpansionLet μt = max{λi | 1 ≤ i ≤ t}. We have μt = max{μt-1, μt-12 +λ + λ2}. Solving this yieldsμt < λ + O(λ2).

  31. 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.

  32. 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.

  33. Example

  34. 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.

  35. 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.

  36. Claim: A Z’ B = (NADA, DB3, λA+2 λB2) • Let Bi a DA ×DApermutation matrix • Note that the normalized adjacency matrix corresponding to steps 2-4 is • Thus

  37. Bi is a permutation matrix thus • We can now decompose M’α|| • Substitutting this in the formula we get that

  38. 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.

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

  40. 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 .

  41. 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)

  42. 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.

  43. 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 one’s matrix in size q

  44. Obviously • Since eigenvalues of Jq are q (one time) and 0 (q - 1 times), (Jq - Iq)ÄJqeigenvalues are q(q – 1), -q and 0. IqÄ qqIqcontributes q toeacheigenvalue. Dividing it by q we get that M2eigenvalues are 1, 0 and 1/q, thus the 2nd largest eigenvalue is q-1, and M’s2nd largest eigenvalue is q-½ .

  45. Define the following graph family: • (APq)1 = (APq)Ä(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))

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