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Expanders, Universal Graphs and Disjoint Paths

Expanders, Universal Graphs and Disjoint Paths. Noga Alon Joint work with Michael Capalbo. TexPoint fonts used in EMF. Read the TexPoint manual before you delete this box.: A A A A A A A A. Eigenvalues and Expanders. Expanders are constant-degree graphs in which

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Expanders, Universal Graphs and Disjoint Paths

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  1. Expanders, Universal Graphs and Disjoint Paths Noga Alon Joint work with Michael Capalbo TexPoint fonts used in EMF. Read the TexPoint manual before you delete this box.: AAAAAAAA

  2. Eigenvalues and Expanders Expanders are constant-degree graphs in which every set X of at most half the vertices has Ω(|X|) neighbors outside X. For regular graphs (with a loop in each vertex) this is equivalent to a spectral property:

  3. Tanner, A-Milman, A: A regular graph is a “good” A+Chung: The number of edges in any set X

  4. Disjoint Paths The Problem: Given r pairs of distinct vertices s1t1 ,s2t2 … srtr in an expander G on n vertices, find edge-disjoint paths P1, P2,…Pr where Pi is a path from si to ti The larger r is, the harder it is to find the paths Motivation: Communication networks, Distributed memory Computer architecture

  5. History

  6. The new result Moreover: -Each vertex can appear in up to d/3 pairs -The algorithm is online

  7. s1 t2 s2 t1

  8. The result-more precisely: Definition: A d-regular graph G on n vertices is a very strong expander if: -the average degree in any subgraph with at most n/10 vertices is at most d/6 -the average degree in any subgraph with at most n/2 vertices is at most 2d/3 The adversary-Router game on G: in each round i the adversary picks a pair of vertices siti, and the router has to find a path Pi connecting them, keeping all paths edge disjoint.

  9. Theorem: If G is a d-regular very strong expander on n vertices, and then the router can win any r-round adversary- router game on G by a deterministic polynomial time algorithm, assuming the adversary does not choose any vertex as an endpoint si or ti more than d/3 times.

  10. A brief outline of the proof (for disjoint pairs) During the game, call a vertex not yet picked by the adversary available, and an edge not yet used by the router remaining. The router maintains in round i: -A subgraph Hi of G, consisting of remaining edges, in which all degrees are at least 3d/4 +2 -A set of edge disjoint paths Qs consisting of remaining edges, from each available vertex s to Hi.

  11. Two crucial facts: -A subgraph of minimum degree 3d/4 in a very strong d-regular expander is itself a “pretty good” expander

  12. -If G is a very strong d-regular expander on n vertices, Q is a set of edges of G and one keeps removing vertices of G until the minimum degree is at least 3d/4 + 2, then the remaining graph has at least n-15|Q|/d vertices. Indeed, otherwise the set of deleted vertices will span too many edges.

  13. -In each round, the router finds a short path in Hi that augments the paths from si and ti to Hi and provide the path Pi. -She deletes the edges of Pi, updates Hi, and finds new paths Qs using a Network Flow algorithm. si ti

  14. Open

  15. Universal Graphs Definition: H- A family of graphs. G is H- universal if it contains every member of H as a subgraph Example: G= is H-universal for the family of all 2-regular graphs on 7 vertices

  16. Objective: Construct sparse H-universal graphs for interesting families H Motivation: VLSI circuit design

  17. Universal graphs for bounded-degree graphs: H(k,n)=all graphs on n vertices, max-degree ≤ k Question: Estimate the minimum possible number of edges of an H(k,n)-universal graph. A,Capalbo,Kohayakawa,Rödl,Ruciński,Szemerédi (00): Ω(n2-2/k) edges are needed, O(n2-1/k log1/k n) suffice ACKRRS (01):O(n2-2/k log 1+8/k n) edges suffice A, Capalbo (06):O(n2-2/k log 4/k n) edges suffice

  18. New: Theorem: For all k ≥ 3 there is c=c(k) and an explicit H(k,n)-universal G with at most c n2-2/k edges. The proof applies properties of high-girth expanders and provides a deterministic embedding procedure.

  19. The proof for even k is simpler, the one for odd k requires an additional effort: a new graphdecomposition result. Theorem for k=4: The minimum possible number of edges of a graph that contains a copy of every graph on n vertices with maximum degree at most 4 is Θ(n3/2)

  20. The lower bound: Simple counting: there are “many” 4-regular graphs on n vertices, and a graph with m edges cannot contain too many subgraphs with 2n edges The upper bound: Construction using high-girth expanders

  21. The construction: Let a,d be absolute constants, put m=a n1/2, and let F be a d-regular Ramanujan expander of girth at least ⅔ logd-1m. Thus all nontrivial eigenvalues of F are of absolute value at most 2(d-1)1/2. Define G=(V,E), where V=(V(F))2 and (a1,a2) is adjacent to (b1,b2) iff ai and bi are within distance 2 in F for i=1 and/or i=2. Clearly |E|=O(n3/2). Main claim: G is H(4,n)-universal.

  22. A homomorphism from a graph Z to a graph T is a mapping of V(Z) to V(T) such that adjacent vertices in Z are mapped to adjacent vertices in T. Thus there is an injective homomorphism from Z to T iff Z is a subgraph of T. Pn - the path of length n. A homomorphism from Pn to F is a walk on F. The k-th power Tk of a graph T is the graph on V(T) in which two vertices are adjacent iff their distance in T is at most k.

  23. Let H be a graph on n vertices with maximum degree at most 4. By Petersen’s Theorem H can be decomposed into two spanning subgraphs H1,H2, each having max. degree at most 2. There are bijective homomorphisms gi from Hi to Pn2 To embed H in G we define homomorphisms fi from Pn to F so that f(v)=(f1(g1(v)), f2(g2(v)) ) is an injective homomorphism from H to G.This is done by defining each fi as an appropriate non-back-tracking walk on F.

  24. The properties of F: A (simple) walk of length q in F is a sequence W=w0,w1, … ,wq of distinct vertices of F, with wiwi+1 being an edge for all i. If S1, S2, …,Sq are subsets of V(F), then W slips by the sets Siif for all i, wi is not in Si

  25. A vertex w is nice with respect to the sets Si if there are at least m/2 vertices z so that there is a walk w=w0,w1, … ,wq=z of length q that starts at w, slips by the sets Si,and ends at z. A vertex w is very nice with respect to the sets Si if for every set of vertices Q containing at most d/20 – log d neighbors of each vertex, w is nice with respect to the sets Si[ Q.

  26. Lemma: Let F be a high-girth Ramanujan expander on m vertices (as above), and put q=log m/ log 10. Then, for any collection of sets S1,S2, … Sq of vertices of F satisfying |Si| ≤ m/20 for all i, the number of vertices w that are very nice with respect to the sets Si is at least 9m/10. The proof uses the spectral properties of F and the fact it has high girth.

  27. The homomorphism f1 from Pn to F can be any walk in F covering no vertex more than n1/2 times. Given f1, one can define f2deterministically in steps, where is each step it is defined on q consecutive vertices of the path, making sure that the mapping f(v)=(f1(g1(v)), f2(g2(v))) is injective. To do so, the walk has to slip by appropriately defined sets. The difficulty is that these sets change during the process. The notions “nice” and “very nice” help to overcome this difficulty.

  28. A crucial observation: When augmenting the sets Si by the vertices of a walk of length q, every vertex v which was very nice (with respect to the sets Si), becomes nice (with respect to the augmented sets).

  29. The construction of universal graphs for H(k,n) with k>4 even is similar The odd case requires more efforts

  30. A graph is thin if every connected component of it is either a subgraph of a cycle with pendant edges or a graph with max. degree 3 and at most two vertices of degree 3

  31. Fact: Every thin graph can be mapped homomorphically and bijectively to the forth power of a path. Theorem: Let H be a graph of maximum degree k. Then there are k thin spanning subgraphs H1, H2, … ,Hk of H, so that each edge of H lies in two of the graphs Hi.

  32. A universal graph for H(k,n): Let F be a high-girth Ramanujan graph on m=a n1/k vertices. Construct G=(V(G),E(G)) as follows: V(G)=(V(F))k (a1,a2, … ,ak) and (b1,b2, … ,bk) are adjacent iff there are at least two indices i so that ai and bi are within distance 4 in F.

  33. Open: Is there an H(k,n)-universal graph on n vertices with O(n2—2/k) edges ?

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