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Chapter 3: Q27E (page 111)

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.

Short Answer

Expert verified

In any undirected graph. It is possible to pair up the vertices of odd degrees and the paths between each pair can be found so that all these paths are edge-disjoint.

Step by step solution

01

Explain edge-disjoint

In a graph, if there is no common edge that exists between the two paths, then it is called edge-disjoint.

02

Step 2: Show that 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.

Consider any undirected graph that has multiple edges and vertices. In an undirected graph, the number of vertices with an odd degree must be even and it can be divided into two groups.

In all connected branches of the graph, the number of odd-degree vertices is even n . For every odd-degree vertices, there must be a pair of connected vertices (u,v) and the path edge is removed from the graph. Consider a graph containing n-1 pairs of odd-degree vertices, and repeat the process to find each pair. The paths between vertices are disjoint.

Therefore, it has been proved that It is possible to pair up the vertices of odd degrees and the paths between each pair can be found so that all these paths are edge-disjoint.

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Most popular questions from this chapter

You are given a directed graph in which each nodeu∈V, has an associated pricepu which is a positive integer. Define the array cost as follows: for each u∈V,

cost[u] = price of the cheapest node reachable fromu (includingu itself).

For instance, in the graph below (with prices shown for each vertex), the cost values of the nodes A,B,C,D,E,Fare2,1,4,1,4,5,respectively.

Your goal is to design an algorithm that fills in the entire cost array (i.e., for all vertices).

(a) Give a linear-time algorithm that works for directed acyclic graphs.

(b) Extend this to a linear-time algorithm that works for all directed graphs.

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

Design a linear-time algorithm which, given an undirected graph G and a particular edge ein it, determines whetherGhas a cycle containing.

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.

Give a linear-time algorithm for the following task.
Input: A directed acyclic graph G

Does G contain a directed path that touches every vertex exactly once?
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