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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

Give an efficient algorithm that takes as input a directed acyclic graph G=V,E, and two vertices s,tV, and outputs the number of different directed paths from S to t in G.

Suppose a CS curriculum consists of n courses, all of them mandatory. The prerequisite graph G has a node for each course, and an edge from course v to course w if and only if v is a prerequisite for w. Find an algorithm that works directly with this graph representation, and computes the minimum number of semesters necessary to complete the curriculum (assume that a student can take any number of courses in one semester). The running time of your algorithm should be linear.

Biconnected componentsLet G=(V,E) be an undirected graph. For any two edgese,e'E,, we’ll saye:e'if eithere=e'or there is a (simple) cycle containing both e and e'.

a. Show that : is an equivalence relation (recall Exercise 3.29) on the edges.

The equivalence classes into which this relation partitions the edges are called the biconnected components of G . A bridge is an edge which is in a biconnected component all by itself.

A separating vertexis a vertex whose removal disconnects the graph.

b. Partition the edges of the graph below into biconnected components, and identify the bridges and separating vertices.

Not only do biconnected components partition the edges of the graph, they almost partition the vertices in the following sense.

c. Associate with each biconnected component all the vertices that are endpoints of its edges. Show that the vertices corresponding to two different biconnected components are either disjoint or intersect in a single separating vertex.

d. Collapse each biconnected component into a single meta-node, and retain individual nodes for each separating vertex. (So there are edges between each component node and its separating vertices.) Show that the resulting graph is a tree.

DFS can be used to identify the biconnected components, bridges, and separating vertices of a graph in linear time.

e. Show that the root of the DFS tree is a separating vertex if and only if it has more than one child in the tree.

f. Show that a non-root vertex v of the DFS tree is a separating vertex if and only if it has a child v' none of whose descendants (including itself) has a back edge to a proper ancestor of v .

g. For each vertex u define:

Iowu=minpreuprew

Where (v,w) is a back edge for some descendant v of u.

(h) Show how to compute all separating vertices, bridges, and biconnected components of a graph in linear time.

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.

For each node in an undirected graph, let twodegreeube the sum of the degrees of’s neighbors. Show how to compute the entire array of two degree. values in linear time, given a graph in adjacency list format

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