1. 1 Entropy Waves, The Zigzag Graph Product, and New Constant-Degree Expanders Omer Reingold Salil Vadhan Avi Wigderson Lecturer: Oded Levy

2. 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. 3 The zigzag product G A = (N A, D A, λ A ), G B = (D A, D B, λ B ). We define their zigzag product as a graph which Its vertices are N A × N B (each vertex is represented as a pair (v, u) such that v  N A and u  N B. For each vertex (v, k) we put an edge between a pair (v, k) to (v[k[i]], k[i][j]).

4. 4 Example

5. 5

6. 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. 7 Theorem G = G A Z G B = (N A × D A, D B 2, f(λ A, λ B )). f(λ A’ λ B ) ≤ λ A + λ B + λ B 2. λ A, λ B < 1→ f(λ A, λ B ) < 1. We know that. We aim to show that for each

8. 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.

9. 9 Step 2 is a random walk on G A.Step 3 is a random walk on G B.  uniform within clouds. Step 1 does not change anything.

10. 10 Step 1 is a random walk on G B.Step 2 is a permutation.Step 3 is a random step on a regular graph.  non-uniform within clouds.

11. 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.

12. 12 The relation between and B is The relation between and A will be defined later.

13. 13 Decomposing α

14. 14 Recall that we want to bound Since G B is a regular graph, (1, 1, …, 1) is an eigenvector of B with eigenvalue 1,

15. 15

16. 16 Thus Combining it with previous results our expression becomes Opening the inner product we get

17. 17 Using triangular inequality we get Since à is a permutation matrix, it is unitary thus preserves length Using Cauchy-Schwartz inequality

18. 18 We’ll first bound. Decomposing α  to its components we have By the expansion of G B we get Thus

19. 19 For our analysis we’ll define linear map C such that:

20. 20

21. 21 It is easy to see that In addition

22. 22 Now we’ll bound. We’ll first relate A to Ã. Claim:

23. 23

24. 24 Hence Since Cα is orthogonal to Combining these claims we have

25. 25 pq gets maximum value of ½ (when), thus

26. 26 It remains to show that if both λ 1 and λ 2 are smaller than 1, f(λ 1,λ 2 ) is also smaller than 1. Case I

27. 27 Case II

28. 28 Finally

29. 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 λ 1 and λ 2. If λ 1 <1 and λ 2 <1 then f(λ 1, λ 2 )<1. f(λ 1, λ 2 )≤ λ 1 + λ 2

30. 30 Families of expanders In this section we’ll introduce two families of expanders: Basic construction More efficient construction

31. 31 Basic family construction Let H = (D 4, D, 1/5). We build a family of graph in the following way: G 1 = H 2 G i+1 = G i 2 Z H G i is an infinite family of expanders such that G i = (D 4i, D 2, 2/5)

32. 32 Vertices. Degree. Expansion Bounded by 2/5 since the limit of the series λ n = λ 2 n-1 + 6/25 is 2/5.

33. 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 = (D 8, D, λ). For every t we define

34. 34 For each t ≥ 0, G t = (D 8t, D 2, λ t ) such that λ t = λ + O(λ 2 ). Vertices. Degree. Expansion Let μ t = max{λ i | 1 ≤ i ≤ t}. We have μ t = max{μ t-1, μ t-1 2 + λ + λ 2 }. Solving this yields μ t < λ + O(λ 2 ).

35. 35 Can we do better? The best possible 2 nd 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. 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. 37 Example

38. 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. 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. 40 Claim: A Z’ B = (N A D A, D B 3, λ A +2 λ B 2 ) Let B i a D A ×D A permutation matrix Note that the normalized adjacency matrix corresponding to steps 2-4 is Thus

41. 41 B i is a permutation matrix thus We can now decompose M’α || Substitutting this in the formula we get that

42. 42 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.

43. 43 Using the same techniques from previous results we get the desired result.

44. 44 The affine plane Consider field F q such that q = p t for some prime number t. We define AF q be a F q × F q 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. 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. 46 Lemma: AF q = (q 2, q, q -½ ). Proof : An entry in the square of the normalized adjacency matrix of AF q 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 |L a,b  L c,d | / q 2.

47. 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 I q the identity matrix in size q Denote J q the all one’s matrix in size q

48. 48 Obviously Since eigenvalues of J q are q (one time) and 0 ( q - 1 times), (J q - I q )  J q eigenvalues are q(q – 1), -q and 0. I q  q qI q contribute s q to each eigenvalue. Dividing it by q we get that M 2 eigenvalues are 1, 0 and 1/q, thus the 2 nd largest eigenvalue is q -1, and M’ s 2 nd largest eigenvalue is q -½.

49. 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 = (q 2(i+1), q 2, O(iq -½ ))