Introduction to Algorithms
3rd Edition
ISBN: 9780262033848
Author: Thomas H. Cormen, Ronald L. Rivest, Charles E. Leiserson, Clifford Stein
Publisher: MIT Press
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Chapter B.4, Problem 1E
Program Plan Intro
To prove:
The total number of handshaking lemma in terms of undirected graph, can be defined as:
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7. The Bellman-Ford algorithm for single-source shortest paths on a graph G(V,E) as discussed in class has a running time of O|V |3, where |V | is the number of vertices in the given graph. However, when the graph is sparse (i.e., |E| << |V |2), then this running time can be improved to O(|V ||E|). Describe how how this can be done..
8. Let G(V,E) be an undirected graph such that each vertex has an even degree. Design an O(|V |+ |E|) time algorithm to direct the edges of G such that, for each vertex, the outdegree is equal to the indegree.
Please give proper explanation and typed answer only.
Let G be a graph with n vertices. The k-coloring problem is to decide whether the vertices of G can be labeled from the set {1, 2, ..., k} such that for every edge (v,w) in the graph, the labels of v and w are different.
Is the (n-4)-coloring problem in P or in NP? Give a formal proof for your answer. A 'Yes' or 'No' answer is not sufficient to get a non-zero mark on this question.
The total degree of an undirected graph G = (V,E) is the sum of the degrees of all the vertices in V. Prove that if the total degree of G is even then V will contain an even number of vertices with uneven degrees. Handle the vertices with even degrees and the vertices with uneven degrees as members of two disjoint sets.
Chapter B Solutions
Introduction to Algorithms
Ch. B.1 - Prob. 1ECh. B.1 - Prob. 2ECh. B.1 - Prob. 3ECh. B.1 - Prob. 4ECh. B.1 - Prob. 5ECh. B.1 - Prob. 6ECh. B.2 - Prob. 1ECh. B.2 - Prob. 2ECh. B.2 - Prob. 3ECh. B.2 - Prob. 4E
Ch. B.2 - Prob. 5ECh. B.3 - Prob. 1ECh. B.3 - Prob. 2ECh. B.3 - Prob. 3ECh. B.3 - Prob. 4ECh. B.4 - Prob. 1ECh. B.4 - Prob. 2ECh. B.4 - Prob. 3ECh. B.4 - Prob. 4ECh. B.4 - Prob. 5ECh. B.4 - Prob. 6ECh. B.5 - Prob. 1ECh. B.5 - Prob. 2ECh. B.5 - Prob. 3ECh. B.5 - Prob. 4ECh. B.5 - Prob. 5ECh. B.5 - Prob. 6ECh. B.5 - Prob. 7ECh. B - Prob. 1PCh. B - Prob. 2PCh. B - Prob. 3P
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- We have that y*n+1=y*n=y*. Using the given model: yn+1 = byn(1-yn), we find that y* = 1-1/b and y* = 0 are fixed points. I have the function:def main(b,yInit,n): out=np.zeros((n,1)) y = yInit for i in range(n): out[i] = y y = b*(1-y)*y return out The graph model of this would depend both on the value of the initial population (y*) and choice of b. I need to graph this in python using interactive sliders widget [from matplotlib.widgets import slider], where one slider will control b value (x-axis) and another for y value (y-axis). I have the skeleton code for the sliders but have no idea how to graph this equation with it. (y* = between 0 to 1, b = 0 to 4)arrow_forwardTrue or false: let G be an arbitrary connected, undirected graph with a distinct cost c(e) on every edge e. suppose e* is the cheapest edge in G; that is, c(e*) <c(e) for every edge e is not equal to e*. Any minimum spanning tree T of G contains the edge e*arrow_forwardLet G be a directed graph where each edge is colored either red, white, or blue. A walk in G is called a patriotic walk if its sequence of edge colors is red, white, blue, red, white, blue, and soon. Formally ,a walk v0 →v1 →...vk is a patriotic walk if for all 0≤i<k, the edge vi →vi+1 is red if i mod3=0, white if i mod3=1,and blue if i mod3=2. Given a graph G, you wish to find all vertices in G that can be reached from a given vertex v by a patriotic walk. Show that this can be accomplished by efficiently constructing a new graph G′ from G, such that the answer is determined by a single call to DFS in G′. Do not forget to analyze your algorithm.arrow_forward
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