You are given tree T=(V,E) along with a designated root node rV. The parent of any node Vr, denoted p(V), is defined to be the node adjacent to v in the path from r to v . By convention, p(r)=r. For k>1, define pk(v)pk-1(pv)andp1(v)=p(v)(so pk(v)is the k th ancestor of v ). Each vertex v of the tree has an associated non-negative integer label l(v). Given a linear-time algorithm to update the labels of all the vertices T according to the following rule: lnew(v)=l(plvv).

Short Answer

Expert verified

Depth-first search algorithm can be used to update the labels of all vertices T according to the rule lnewv=lplvv.

Step by step solution

01

Explain Depth-first search

Depth-first search is the linear time algorithm that gives information about a graph. A graph is considered in the form of an adjacency list. Neighbors of the vertices can be found by exploring the vertices and marking the visited vertices.

02

Give a linear-time algorithm to update the labels of all the vertices T  

Consider the binary tree T =(v,E) with the designated root node rV. The parent of any node vr, denoted p (v), is the node adjacent to v in the path from r to v . By convention p (r)=r .

For k>1 ,define pkv=pk-1pvand p1v=pv. Each vertex v of the tree has an associated non-negative integer label lv.

Consider the following algorithm,that updates the labels of all vertices Taccording to the rule lnewv=lplvv.

Input: TV,E

Output: Inewv

Procedure:

perform DFS onTV,E

Declare DFS on stack

During traversal

if node v is pushed onto the stack

update Inewv=1stack.atmaxstack,size-1v,1

at stack.atI

take out the vertex at ith position

set bottom element to be in the first position

return Inewv

The above algorithm considers the stack to perform DFS. During traversal if node is pushed onto the stack, then the label of the vertex at stack is updated. Then the vertex is moved to the bottom and the bottom element is moved to the first position.

Therefore, Depth-first search algorithm has been used to update the labels of all vertices T according to the rule Inewv=IpIvv.

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Most popular questions from this chapter

The reverse of a directed graph G = (V,E) is another directed graphGR=(V,ER) on the same vertex set, but with all edges reversed that is,ER={(v,u):(u,v)E} . Give a linear-time algorithm for computing the reverse of a graph in adjacency list format.

As in the previous problem, you are given a binary tree T=(V,E) with designated root node. In addition, there is an array x[.]with a value for each node in V Define a new array z[.]as follows: for each uV,

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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.
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