Undirected ST-Connectivity in Log-Space

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Undirected ST-Connectivity in Log-Space

• By Omer Reingold (Weizmann Institute)
• Year 2004
• Presented by Maor Mishkin

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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.

TRUE
FALSE
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ST-Connectivity Problem

• Is called USTCON or Maze Problem.
• The STCON problem is on a Directed Graph.

Exit
t
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ST-Connectivity Problem

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

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t
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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.
• Time O(N) -gt this is also the lower Time bound
  for USTCON problem.
• Space O(N).

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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 2N and put in place j the parity bit
  of the bit array representation of j.

Input101101 Output0
Input100101 Output1
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Log-Space Algorithms

• Definition of a Log-Space algorithm Given an
  input of size N, algorithms Space is O(logN).
• Claim. If the algorithm is Log-Space
  Deterministic gt 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 ij (since
  Time is bigger than Space states)-gt a
  contradiction to algorithm finishing the
  calculation.
• That is, class LOG-SPACE is contained in P.

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Previous Related Work

• STCON J. Savitch (70), Log2 Space a super
  polynomial Time.
• 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.

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Previous Related Work

• 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 Sequence Noam Nisan 92b,
  quasi-polynomial Time in Space Log2.

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Previous Related Work

• USTCON NSW89 improved Savitch (70) to Space
  Log1.5, and not polynomial Time.
• USTCON ATSWZ00 improved the previous one to
  Space Log1.333, but still not polynomial Time
  most Space efficient until this work.

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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.

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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.

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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).

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Powering

• Definition. the kth 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.
• Powering increases the Degree of the Graph and
  will not maintain the Graph as a Constant Degree
  one.

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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 dltltD).
• Each vertex v of G is replaced with a copy Hv
  of H.
• Each of the d vertices of Hv is connected to its
  neighbors in Hv and also to one vertex in Hw,
  where (v,w) is one of the D edges going out of v
  in G.
• The Degree of the Product Graph is d1.

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Expander Graphs

• Definition. For a d-regular Graph G(V,E), VN.
• , ,
• Vertex Expansion Properties give us that the
  Expander Graph will have Logarithmic Diameter.

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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 1970s.
• Widely used for De-Randomization, Error
  Correction, CS theory.

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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).

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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.

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Informal USTCON algorithm

• 3. Now solve USTCON on the resulting Graph that
  has Logarithmic Diameter Constant Degree.

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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 its adjacency matrix
  is symmetric.
• A Graph is D-regular if the sum of entries in a
  row (and column) is D.

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Graph Representation

• Given a D-regular Graph, we can assume that for
  vertex v, the edges are labeled 1D, and we can
  talk about the ith 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).

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Graph Representation

• For a D-regular undirected graph G, let us define
  Rotation Map
• RotG ND --gt ND is defined as
  RotG(v,i) (w,j) if the ith edge incident to v
  leads to w, and this is the jth edge incident to
  w.
• Rotation map will be the input Graph
  representation at this work.

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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 MG of a
  D-regular Graph G, that is the Adjacency Matrix
  of G divided by D.
• Formally -gt

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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 ? iff Mx ?x.
• All one Eignvalue

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Graph Expansion Measurement

• It turns out that the Eigenvalues of MG are at
  most 1.
• We denote by ?(G) the second largest Eigenvalue
  of MG (in the absolute value)
• It is known that ?(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 ?(G)lt ? as an (N,D,?)-graph.

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Solving USTCON given a Constant Degree Expander.

• USTCON in Constant Degree Expanders can be solved
  in Log-Space
• Let ? lt1 be some constant, then there exists an
  O(logDlogN) 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
• If s t are in the same connected component
  this component is an (N,D, ?)-Graph then A
  outputs connected.
• If A outputs connected then s t are indeed in
  the same connected component.

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Solving USTCON given a Constant Degree Expander
(cont.)

• The algorithm A simply enumerates all Dl paths
  of length lO(logN) from s, where the leading
  constant in the big-O notation depends on ?. 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 Dl paths requires O(logDlogN)
  Space. When D is a constant we get O(logN) Space.

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Solving USTCON given a Constant Degree Expander
(cont.)
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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 RotG. The tth power of G
  is the Dt-regular Graph Gt whose rotation map
  is given by
• Where these values are computed via the rule

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Powering Expansion Measure

• Lemma - If G is an (N,D,?)-graph then Gt is an
  (N,Dt, ?t)-graph.
• Proof The normalized adjacency matrix MGt of
  Gt is the tth power of the Normalized Matrix MG
  of G, so all the Eigenvalues also get raised to
  the tth power.
• MGtx MGt-1MGx MGt-1?x
• ? MGt-1 x ?tx

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Two Graph Products

• Reminder - 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 d2 (d2ltltD)

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Zig-Zag Product

• Definition. If G is a D-regular Graph with N
  vertexes with rotation map RotG H is a
  d-regular Graph with D vertexes with rotation map
  RotH, then their Zig-Zag Product G H is defined
  to be the d2-regular graph with ND vertexes
  whose rotation map RotG H is as follows
• RotG H((v,a),(i,j))
• 1. Let(a,i)RotH (a,i)
• 2. Let(w,b)RotG(v,a)
• 3. Let(b,j)RotH(b,j)
• 4. Output((w,b),(j,i))

z
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Expansion Properties of G H
z

• Theorem. (RVW01) If G is an (N,D,?G)-graph H
  is a (D,d, ?H)-graph, then G H is a
• (ND,d2,f(?G, ?H))-graph, where
• We get, that if is a good Expander (?H -gt0)
  then f(?G, ?H)-gt ?G

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Universal Traversal exploration sequence

• 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.

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Issues Summary

• USTCON
• Log-Space Polynomial Time
• Powering
• Replacement Product
• Expander Graphs
• Zig-Zag Product