On page 102, we defined the binary relation “connected” on the set of vertices of a directedgraph. Show that this is an equivalence relation(see Exercise 3.29), and conclude that it partitions the vertices into disjoint strongly connected components.

Short Answer

Expert verified

It can be shown that the binary relation “connected” on the set of vertices of a directed graph is an equivalence relation and yes, it partitions the vertices into disjoint strongly connected components.

Step by step solution

01

Explain the Equivalence relation

A relation is said to be in equivalence only if the relation satisfies reflexive, symmetry, and transitive properties.

02

Show that the given relation is the equivalence relation

Consider a set S that has the partitions of an undirected graph. Consider any two vertices x and y in the undirected graph.

From the solution of Exercise 3.29, the binary connected relation of the connected relationship satisfies reflexivity, symmetry, and transitivity. So, it is an equivalence relation.

The strongly connected component is the equivalence class corresponding to this relation.

Thus, it partitions the vertices into disjoint strongly connected components.

Therefore, It is shown that the binary relation “connected” on the set of vertices of a directed graph is an equivalence relation and yes, it partitions the vertices into disjoint strongly connected components.

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Most popular questions from this chapter

In an undirected graph, the degreed(u) of a vertex u is the number of neighbours u is the number of neighbors u has, or equivalently, the number of edges incident upon it. In a directed graph, we distinguish between the indegreedin(u), which is the number of edges into u, and the outdegreedout(u), the number of the edges leaving u.

(a) Show that in an undirected graph, role="math" localid="1658908755010" uevd(u)=2|E|

(b) Use part (a) to show that in an undirected graph, there must be an even number of vertices whose degree is odd.

(c) Does a similar statement hold for the number of vertices with odd indegree in a directed graph?

Perform a depth-first search on the following graph; whenever there’s a choice of vertices, pick the one that is alphabetically first. Classify each edge as a tree edge or back edge, and give the pre and post number of each vertex.

Two paths in a graph are called edge-disjointif they have no edges in common. Show that in any undirected graph, it is possible to pair up the vertices of odd degree and find paths between each such pair so that all these paths are edge-disjoint.

Infinite paths.Let G=(V,E) be a directed graph with a designated “start vertex” sV,asetVGV, a set of “good” vertices, and a set VBV of “bad” vertices. An infinite trace of is an infinite sequence of vertices viV such that (1)v0=s, and (2) for all i0, (vi,vi+1)E. That is, p is an infinite path in G starting at vertex s. Since the setV of vertices is finite, every infinite trace of Gmust visit some vertices infinitely often.

  1. If p is an infinite trace, let Inf(p)V be the set of vertices that occur infinitely often in p. Show that Inf(p) is a subset of a strongly connected component of G.
  2. Describe an algorithm that determines if role="math" G has an infinite trace.
  3. Describe an algorithm that determines if G has an infinite trace that visits some good vertex in VG infinitely often.
  4. Describe an algorithm that determines if role="math" localid="1659627728759" G has an infinite trace that visits some good vertex in VG infinitely often, but visits no bad vertex in VB infinitely often.

Either prove or give a counterexample: if {u,v}is an edge in an undirected graph, and during depth-first search (u)<post (v), then vis an ancestor of uin the DFS tree.

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