Undirected ST-Connectivity in Log-Space By Omer Reingold (Weizmann Institute) Year 2004 Presented by Maor Mishkin

ST-Connectivity Problem  Given an Undirected Graph and given two Vertices in the Graph, s and t, We need to return “ true ” if s and t are connected and to return “ false ” if they are not connected. s t TRUEFALSE

ST-Connectivity Problem USTCON Maze  Is called USTCON or Maze Problem. STCON  The STCON problem is on a Directed Graph. Entrance Exitt s

ST-Connectivity Problem  Related problem (Cook late 70 ’ s)- Universal Traversal Problem, where we need to also return the path, which connects s and t – Not the focus of this lecture. s t

Basic Search Algorithms  Breadth First Search (BFS) and Depth First Search (DFS) solve the problem in directed graphs (denoted STCON), therefore will solve in Undirected also.  Let N be the input Graph size. USTCON  Time O(N) -> this is also the lower Time bound for USTCON problem.  Space O(N).

Space vs. Time  Algorithms have central resources of computation, Space & Time.  Space is the work memory – not including Input memory.  Trade-Off example: Given a bit array, calculate the parity bit (XOR over all the bits) in Time O(1).  Algorithm: Pre-Calculation - Create an array in the size of 2^N and put in place j the parity bit of the bit array representation of j. Input: Output:0 Input: Output:1

Log-Space Algorithms  Definition of a Log-Space algorithm: Given an input of size N, algorithm ’ s Space is O(logN).  Claim. If the algorithm is Log-Space & Deterministic => Time is polynomial.  Proof. Lets assume that it is not polynomial Time and is Log-Space, therefore will have different Space states. By looking on Space at time i (Si), we know that we will have Si & Sj that Si = Sj and i^=j (since #Time is bigger than #Space states)-> a contradiction to algorithm finishing the calculation.  That is, class LOG-SPACE is contained in P.

Previous Related Work  STCON  STCON – J. Savitch (70), Log^2 Space & a super polynomial Time.  USTCON  USTCON – R. Aleliunas (79), Randomized Log-Space. Use a pointer to current vertex and a counter. Randomly start finding a path from s to t, stop when counter hits a limit. * the Random walk has a one side error.

Previous Related Work  USTCON  USTCON – Massive work to solve the problem without Randomization, but still pseudo-randomized algorithms [AKS87, BNS89, Nis92b, INW94] – We will fully remove the Randomization factor.  Universal Traversal  Universal Traversal Sequence – Noam Nisan 92b, quasi- polynomial Time in Space Log^2.

Previous Related Work  USTCON  USTCON – [NSW89] improved Savitch (70) to Space Log^1.5, and not polynomial Time.  USTCON  USTCON – [ATSWZ00] improved the previous one to Space Log^1.333, but still not polynomial Time – most Space efficient until this work.

Approach  To solve the connectivity problem between s & t, improve the connectivity of every connected component in the Graph, that is: Transform the input Graph into a Graph, which has Logarithmic Diameter (with the same connected components). We also make it a Constant Degree one. s t

Approach  Since the Graph will have Logarithmic Diameter, we can build all Logarithmic length paths, starting from s, and to see if one visits t.  Since the Degree is Constant and does not depend on N, the number of such paths is (polynomial). We later explain how to execute this in Log-Space.

Open issues  How can we enumerate paths on a Graph that is Constant Degree & Logarithmic Diameter Graph in Log- Space (relatively easy).  How can we Transform the input Graph into a Constant Degree & Logarithmic Diameter Graph (the main issue).

Powering  Definition. the k ’ th Power of G contains an edge between two vertices v & w, iff there exists a path of length k from v to w in G.  Repeatedly squaring the graph logarithmic number of times will turn G into a Logarithmic Diameter Graph. not  Powering increases the Degree of the Graph and will not maintain the Graph as a Constant Degree one.

Decreasing Degrees  Replacement Product - an operation with two Graphs, a D-regular Graph G with N vertices and a d-regular Graph H on D vertices (with d<<D).  Each vertex v of G is replaced with a “ copy ” H v of H.  Each of the d vertices of H v is connected to its neighbors in H v and also to one vertex in H w, where (v,w) is one of the D edges going out of v in G.  The Degree of the Product Graph is d+1.

Expander Graphs  Definition. For a d-regular Graph G(V,E), |V|=N. ,,  Vertex Expansion Properties give us that the Expander Graph will have Logarithmic Diameter.  SSSN  1

Expander Graphs  We can turn a Graph into an Expander by Squaring it Logarithmic Times.  Algebraic Expansion Properties will give us a way to measure the Graph Expansion Properties.  Introduced in the 1970 ’ s.  Widely used for De-Randomization, Error Correction, CS theory.

Not Damaging Logarithmic Diameter  It turns out that if H is a “ good enough ” Expander, the expansion properties of the Replacement Product are not worse by much than those of the original Graph.  Formal statements to this effect were proved by Reingold, Vadhan & Wigderson [RVW01] for the Replacement Product and for the Zig-Zag Product (to be described later).

Informal USTCON algorithm 1. First turn the input Graph into a constant-degree, regular Graph with each connected component being non-bipartite (not Replacement Product). 2. The main transformation turns each connected component of the Graph, in a logarithmic number of phases, into an Expander (a logarithmic diameter) 2.1. Each phase starts by raising the current graph to some constant power and then reducing the degree back via a Replacement or Zig-Zag product, using a constant size Expander.

Informal USTCON algorithm 3. Now solve USTCON on the resulting Graph that has Logarithmic Diameter & Constant Degree.

Graph Representation  Adjacency Matrix - A way to represent a Graph G(V,E) – not the input graph – will be used for theoretic discussion only.  At entry (v,u) will have a non-negative integer that equals to the number of edges that go from vertex v to vertex u.  A Graph is undirected iff it ’ s adjacency matrix is symmetric.  A Graph is D-regular if the sum of entries in a row (and column) is D.

Graph Representation  Given a D-regular Graph, we can assume that for vertex v, the edges are labeled 1 … D, and we can talk about the i ’ th neighbor of v.  When taking a step from v to w, it may be useful to keep track of the edges traversed to get to w (rather then just remembering that we are now at w).

Graph Representation  For a D-regular undirected graph G, let us define Rotation Map: Rot G : [N]*[D] --> [N]*[D] is defined as Rot G (v,i) = (w,j) if the i ’ th edge incident to v leads to w, and this is the j ’ th edge incident to w.  Rotation map will be the input Graph representation at this work.

Measure of Graph Expansion  We would like to make sure that our iterations will give us a “ good ” Expander, therefore we would like to measure our Graph Expansion.  Expansion Properties can be calculated. we will look at the Normalized Adjacency Matrix M G of a D-regular Graph G, that is the Adjacency Matrix of G divided by D.  Formally ->

Eigenvalues and Eigenvectors  Eigenvalue - The factor by which a linear transformation multiplies one of its Eigenvectors.  Eigenvector – For matrix M, vector x is an Eigenvector with Eigenvalue λ iff Mx= λ*x.  All one Eignvalue

Graph Expansion Measurement  It turns out that the Eigenvalues of M G are at most 1.  We denote by λ(G) the second largest Eigenvalue of M G (in the absolute value)  It is known that λ(G) is a good measure of the Expansion property of G.[alpha vs. lambda]  We refer to a D-regular undirected Graph G with N vertices such that λ(G)< λ as an (N,D,λ)-graph.

USTCON Solving USTCON given a Constant Degree Expander.  USTCON  USTCON in Constant Degree Expanders can be solved in Log-Space:  Let λ <1 be some constant, then there exists an O(logD*logN) Space algorithm A, such that when a D-regular undirected Graph G with N vertices is given to A as an input the following holds: 1. If s & t are in the same connected component & this component is an (N ’,D, λ )- Graph then A outputs ‘ connected ’. 2. If A outputs connected then s & t are indeed in the same connected component.

USTCON Solving USTCON given a Constant Degree Expander (cont.)  The algorithm A simply enumerates all D^l paths of length l=O(logN) from s, where the leading constant in the big-O notation depends on λ. The algorithm A outputs ‘ connected ’ iff at least one of these paths encounter t.  Following any path from s with length l requires O(logN) Space.  Enumerating all D^l paths requires O(logD*logN) Space. When D is a constant we get O(logN) Space.

USTCON Solving USTCON given a Constant Degree Expander (cont.) t logd logN s

Powering & Expansion Measure  Our main transformation will take a graph and transform each one of its connected components into a Constant Degree Expander. If we ignore the constant degree requirement, a simple way of amplifying the Graph is by Powering.  Let G be a D-regular multi-graph with N vertexes, given by rotation map Rot G. The t ’ th power of G is the D^t-regular Graph G^t whose rotation map is given by Where these values are computed via the rule:

Powering & Expansion Measure Lemma - If G is an (N,D,λ)-graph then G^t is an (N,D^t, λ^t)-graph. Proof – The normalized adjacency matrix M G^t of G^t is the t ’ th power of the Normalized Matrix M G of G, so all the Eigenvalues also get raised to the t ’ th power. M G t x = M G t-1 M G x = M G t-1 λx= λ M G t-1 x = λ t x

Two Graph Products Replacement Product  Reminder - Replacement Product  Zig-Zag Product Replacement Product  Zig-Zag Product of G & H correspond to a subset of the paths of length three in the Replacement Product of these Graphs.  The degree of the output graph is d^2 (d^2<<D)

Zig-Zag Product Zig-Zag Product  Definition. If G is a D-regular Graph with N vertexes with rotation map Rot G & H is a d- regular Graph with D vertexes with rotation map Rot H, then their Zig-Zag Product G H is defined to be the d^2-regular graph with N*D vertexes whose rotation map Rot G H is as follows:  Rot G H ((v,a),(i,j)): 1. Let(a ’,i ’ )=Rot H (a,i) 2. Let(w,b ’ )=Rot G (v,a ’ ) 3. Let(b,j ’ )=Rot H (b ’,j) 4. Output((w,b),(j ’,i ’ )) z z z

Expansion Properties of G H  Theorem. ([RVW01]) If G is an (N,D,λ G )-graph & H is a (D,d, λ H )- graph, then G H is a (N*D,d^2,f(λ G, λ H ))-graph, where:  We get, that if is a “ good ” Expander (λ H ->0) then f(λ G, λ H )-> λ G z z

Universal Traversal Universal Traversal & exploration sequence Universal Traversal  The mentioned Algorithm also solves Universal Traversal (i.e. finding the path from s to t if such a path exist).  Every edge in the logarithmic long path of the final G H Graph is a sequence in G (original input) & can be followed by the “ Rotation Graph Labeling ” in the Zig-Zag Product. z

Issues Summary  USTCON  Log-Space & Polynomial Time  Powering  Replacement Product  Expander Graphs  Zig-Zag Product