(a)
To prove the breath first search properties in an undirected graph.
(a)
Explanation of Solution
In the undirected graph, breath first search follow properties as:
Suppose in the undirected graph an edge ( u , v ) is consider as forward or back edge then vertex ‘ u’ should occur before vertex ‘ v’ and edges associated with ‘ u’ should be explore before any other edge in breath first search. Thus, the edge ( u , v ) can be considered as the tree edge because there is no forward and backward edge in undirected graph BFS traversal.
In the breath first search, an edge must be a tree if
Therefore for every tree edge ( u, v ) in breath first search,
If vertex ‘ u’ visit before vertex ‘ v’ , and edge ( u, v ) is cross edge. As the properties of tree edge and cross edge, vertex ‘ u’ visit before vertex ‘ v’ that means vertex ‘ v’ must be in the queue. Hence, from the given statement, it will follow the condition for the cross edge ( u, v ):
Therefore,
(b)
To prove the breath first search properties in the directed graph.
(b)
Explanation of Solution
Breath first search properties in the directed graph as follows:
An edge ( u, v ) is not the tree edge if the given graph G is directed. So, the graph will not contain the forward edge ( u, v ).
In the breath first search, an edge must be a tree if
If vertex ‘ u’ visit before vertex ‘ v’ , and edge ( u, v ) is cross edge. As the properties of tree edge and cross edge, vertex ‘ u’ visit before vertex ‘ v’ that means vertex ‘ v’ must be in the queue. Hence, from the given statement, it will follow the condition for the cross edge ( u, v ):
Therefore,
By seeing the above statements, it is clear for all vertices,
In the breath first search, back edge (u, v) where vertex ‘v’ is the ancestor of vertex ‘u’ and from the given vertices if the drawn edge is larger than the other. Hence,
Therefore, for every back edge (u, v),
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Chapter 22 Solutions
Introduction to Algorithms
- Part 2: Random GraphsA tournament T is a complete graph whose edges are all oriented. Given a completegraph on n vertices Kn, we can generate a random tournament by orienting each edgewith probability 12 in each direction.Recall that a Hamiltonian path is a path that visits every vertex exactly once. AHamiltonian path in a directed graph is a path that follows the orientations of thedirected edges (arcs) and visits every vertex exactly once. Some directed graphs havemany Hamiltonian paths.In this part, we give a probabilistic proof of the following theorem:Theorem 1. There is a tournament on n vertices with at least n!2n−1 Hamiltonian paths.For the set up, we will consider a complete graph Kn on n vertices and randomlyorient the edges as described above. A permutation i1i2 ...in of 1,2,...,n representsthe path i1 −i2 −···−in in Kn. We can make the path oriented by flipping a coin andorienting each edge left or right: i1 ←i2 →i3 ←···→in.(a) How many permutations of the vertices…arrow_forwardGiven a graph that is a tree (connected and acyclic). (1) Pick any vertex v. (II) Compute the shortest path from v to every other vertex. Let w be the vertex with the largest shortest path distance. (III) Compute the shortest path from w to every other vertex. Let x be the vertex with the largest shortest path distance. Consider the path p from w to x. Which of the following are true a. p is the longest path in the graph b. p is the shortest path in the graph c. p can be calculated in time linear in the number of edges/vertices a,c a,b a,b,c b.carrow_forwardIn Computer Science a Graph is represented using an adjacency matrix. Ismatrix is a square matrix whose dimension is the total number of vertices.The following example shows the graphical representation of a graph with 5 vertices, its matrixof adjacency, degree of entry and exit of each vertex, that is, the total number ofarrows that enter or leave each vertex (verify in the image) and the loops of the graph, that issay the vertices that connect with themselvesTo program it, use Object Oriented Programming concepts (Classes, objects, attributes, methods), it can be in Java or in Python.-Declare a constant V with value 5-Declare a variable called Graph that is a VxV matrix of integers-Define a MENU procedure with the following textGRAPHS1. Create Graph2.Show Graph3. Adjacency between pairs4.Input degree5.Output degree6.Loops0.exit-Validate MENU so that it receives only valid options (from 0 to 6), otherwise send an error message and repeat the reading-Make the MENU call in the main…arrow_forward
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