Transcript Lecture 17, March 8

Lecture 17 Path Algebra
 Matrix multiplication of adjacency matrices of directed
graphs give important information about the graphs.
Manipulating these matrices to study graphs is path
algebra.
 With "path algebra" we can solve the following problems:
 Compute the total number of paths between all pairs
of vertices in a directed acyclic graph;
 Solve the "all-pairs-shortest-paths" problem in a
weighted directed graph with no negative cycles;
 Compute the "transitive" closure of a directed graph.
 Think: what is the meaning of M2 where M is an
adjacency matrix of a graph G?
All paths of length r
 Claim. Mr describes paths of length r in G, with entry
i,j denoting number of distinct paths from i to j, here,
M is the adjacency matrix of G.
 Proof. The base case is r = 0, trivial. For the induction
step, assume the claim is true for r' < r; we prove it
for r. Then Mrij = (M · Mr-1)ij
= ∑1 ≤ k ≤ n Mik Mr-1kj
Now any r-step path from i to j must start with a step
to some intermediate vertex k. If there is such a path,
Mik is 1; otherwise it is 0. So adding up Mik Mr-1kj gives
the number of r-step paths.
Computing the matrix powers
 Suppose G is acyclic (no path longer than n-1).
Consider I + M + M2 + ... + Mn-1 : the ij'th entry gives
the total number of distinct paths from i to j.
 Therefore, in O(nω+1) steps, we can compute the total
number of distinct paths between all pairs of vertices.
Here ω denotes the best-known exponent for matrix
multiplication; currently ω = 2.376.
 We can even do better, if n=2k, by first calculating M2,
M4, ..., M2^(k-1), and then calculating
(I+M)(I+M2)(I+M4)...(I+M2^(k-1)) =
I + M + M2 + ... + M2^k–1, where 2k is the least power
of 2 that is ≥ n. This gives an algorithm that runs in
O(nω log n) time, where n is the number of vertices.
Reachability graph
 Given an un-weighted directed graph G = (V,E), and
we want to form the graph G' that has an edge
between u and v if and only if there exists a path (of
any length) in G from u to v.
 Let's first see how to solve it using what we know
from say CS 240. There, we explored depth-first and
breadth-first search algorithms. These algorithms
could find all vertices reachable from a given vertex
in O(|V|+|E|) time. So if we run depth-first search from
every vertex, the total time is O(n(|V|+|E|), which
could be as bad as O(n3) if the graph is dense. Can
we do better than O(n3)?
Transitive closure
 Back to path algebra. Now our matrix M consists of
1's and 0's. How can we find the matrix where there's
a 1 in row i and column j iff there is a length-2 path
connecting vertex i and j? I claim the entry in row i
and column j should be
 OR1 ≤ k ≤ n Mik AND Mkj.
 That is: we just need to use “boolean multiplication
and additions in our matrix computation.
 So the transitive closure of G is given by
M' = I + M + M2 + ... + Mn-1
where +, * are corresponding boolean operations. It
tells if there is a path between any two nodes.
Transitive closure continues ..
 How fast can we compute:
I + M + M2 + ... + Mn-1 ?
 we can multiply 2 Boolean matrices in O(nω)
steps.
 Since, using the “doubling trick”, we have to
do log n Boolean matrix multiplications, this
gives a total cost of O(nω log n) to solve the
transitive closure problem. This is indeed
better than simply running breadth-first or
depth-first search from each vertex.
 Think: what if n is not a power of 2?