To show that the updation procedure of transitive closer G * ( V , E *) of graph G ( V, E ) will take O ( V 2 ) time, while an edge is added to the graph G .

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Answer 1

Question

Chapter 25, Problem 1P

(a)

Program Plan Intro

To show that the updation procedure of transitive closer G * ( V , E *) of graph G ( V, E ) will take O ( V² ) time, while an edge is added to the graph G .

(a)

Expert Solution

Explanation of Solution

Here, in the graph G , if the updation of transitive closure takes O ( V² ) time. To understand this scenario, suppose, addition of an edge ( x₁ , x₂ ) is performed in the graph G . Now, take the set of vertices ( u, v ) in order to get the path from u to x₁ and x₂ to v . If there is a possibility of having an edge that contains vertices in such manner like ( u , x₁ ) and ( x₂ , v ).

Therefore, perform addition of an edge ( u, v ) in transitive closure only if the closure holds the edges in ( u, x₁ ) and ( x₂, v ) manner. So, the consideration of pair required only once and the total run time of this procedure will be O ( V² ).

(b)

Program Plan Intro

To give an example that the update of transitive closer will take O ( V² ) time, when a new edge ‘ e’ is added to the graph G .

(b)

Expert Solution

Explanation of Solution

Consider the condition where there aretwo strongly connected components with sizes | V |/2 and there is no common edge between them. The transitive closure can be computed by adding these two connected components of the graph.

Now, perform addition of a single edge between two connected components that provide connectivity between two separate components of the graph. Here, it is clearly visible that the total no. of edges will be increased by | V |/4 and the total no of edges will be [| V |/2 + | V |/4]. So, every time while adding a new edge required a constant time at least.

Therefore, the update of transitive closer will take O ( V² ) time, when a new edge ‘ e’ is added to the graph G .

(c)

Program Plan Intro

To define an algorithm that takes O ( V³ ) time to update the transitive closure of graph G , when an edge is inserted in to the graph.

(c)

Expert Solution

Explanation of Solution

Consider a set of vertices in the graphG , and there is a path between every pair of vertices. Now, while performing an addition of an edge ( u, v ), look at the ancestor of vertex u and add it there. The reason of looking at the ancestor of an edge is to explore the branches of the tree. This procedure will also applicable for the vertex v. through this, it will be very easy to get the already added edges and will only consider those edges in n times mostly.

Since, the edges in the tree required O ( V² ) time for consideration. Therefore, the total time taken by the algorithm will be O ( V³ ).

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Answer 2

Question

Chapter 25, Problem 1P

(a)

Program Plan Intro

To show that the updation procedure of transitive closer G * ( V , E *) of graph G ( V, E ) will take O ( V² ) time, while an edge is added to the graph G .

(a)

Expert Solution

Explanation of Solution

Here, in the graph G , if the updation of transitive closure takes O ( V² ) time. To understand this scenario, suppose, addition of an edge ( x₁ , x₂ ) is performed in the graph G . Now, take the set of vertices ( u, v ) in order to get the path from u to x₁ and x₂ to v . If there is a possibility of having an edge that contains vertices in such manner like ( u , x₁ ) and ( x₂ , v ).

Therefore, perform addition of an edge ( u, v ) in transitive closure only if the closure holds the edges in ( u, x₁ ) and ( x₂, v ) manner. So, the consideration of pair required only once and the total run time of this procedure will be O ( V² ).

(b)

Program Plan Intro

To give an example that the update of transitive closer will take O ( V² ) time, when a new edge ‘ e’ is added to the graph G .

(b)

Expert Solution

Explanation of Solution

Consider the condition where there aretwo strongly connected components with sizes | V |/2 and there is no common edge between them. The transitive closure can be computed by adding these two connected components of the graph.

Now, perform addition of a single edge between two connected components that provide connectivity between two separate components of the graph. Here, it is clearly visible that the total no. of edges will be increased by | V |/4 and the total no of edges will be [| V |/2 + | V |/4]. So, every time while adding a new edge required a constant time at least.

Therefore, the update of transitive closer will take O ( V² ) time, when a new edge ‘ e’ is added to the graph G .

(c)

Program Plan Intro

To define an algorithm that takes O ( V³ ) time to update the transitive closure of graph G , when an edge is inserted in to the graph.

(c)

Expert Solution

Explanation of Solution

Consider a set of vertices in the graphG , and there is a path between every pair of vertices. Now, while performing an addition of an edge ( u, v ), look at the ancestor of vertex u and add it there. The reason of looking at the ancestor of an edge is to explore the branches of the tree. This procedure will also applicable for the vertex v. through this, it will be very easy to get the already added edges and will only consider those edges in n times mostly.

Since, the edges in the tree required O ( V² ) time for consideration. Therefore, the total time taken by the algorithm will be O ( V³ ).

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Answer 3

Question

Chapter 25, Problem 1P

(a)

Program Plan Intro

To show that the updation procedure of transitive closer G * ( V , E *) of graph G ( V, E ) will take O ( V² ) time, while an edge is added to the graph G .

(a)

Expert Solution

Explanation of Solution

Here, in the graph G , if the updation of transitive closure takes O ( V² ) time. To understand this scenario, suppose, addition of an edge ( x₁ , x₂ ) is performed in the graph G . Now, take the set of vertices ( u, v ) in order to get the path from u to x₁ and x₂ to v . If there is a possibility of having an edge that contains vertices in such manner like ( u , x₁ ) and ( x₂ , v ).

Therefore, perform addition of an edge ( u, v ) in transitive closure only if the closure holds the edges in ( u, x₁ ) and ( x₂, v ) manner. So, the consideration of pair required only once and the total run time of this procedure will be O ( V² ).

(b)

Program Plan Intro

To give an example that the update of transitive closer will take O ( V² ) time, when a new edge ‘ e’ is added to the graph G .

(b)

Expert Solution

Explanation of Solution

Consider the condition where there aretwo strongly connected components with sizes | V |/2 and there is no common edge between them. The transitive closure can be computed by adding these two connected components of the graph.

Now, perform addition of a single edge between two connected components that provide connectivity between two separate components of the graph. Here, it is clearly visible that the total no. of edges will be increased by | V |/4 and the total no of edges will be [| V |/2 + | V |/4]. So, every time while adding a new edge required a constant time at least.

Therefore, the update of transitive closer will take O ( V² ) time, when a new edge ‘ e’ is added to the graph G .

(c)

Program Plan Intro

To define an algorithm that takes O ( V³ ) time to update the transitive closure of graph G , when an edge is inserted in to the graph.

(c)

Expert Solution

Explanation of Solution

Consider a set of vertices in the graphG , and there is a path between every pair of vertices. Now, while performing an addition of an edge ( u, v ), look at the ancestor of vertex u and add it there. The reason of looking at the ancestor of an edge is to explore the branches of the tree. This procedure will also applicable for the vertex v. through this, it will be very easy to get the already added edges and will only consider those edges in n times mostly.

Since, the edges in the tree required O ( V² ) time for consideration. Therefore, the total time taken by the algorithm will be O ( V³ ).

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Answer 4

Question

Chapter 25, Problem 1P

(a)

Program Plan Intro

To show that the updation procedure of transitive closer G * ( V , E *) of graph G ( V, E ) will take O ( V² ) time, while an edge is added to the graph G .

(a)

Expert Solution

Explanation of Solution

Here, in the graph G , if the updation of transitive closure takes O ( V² ) time. To understand this scenario, suppose, addition of an edge ( x₁ , x₂ ) is performed in the graph G . Now, take the set of vertices ( u, v ) in order to get the path from u to x₁ and x₂ to v . If there is a possibility of having an edge that contains vertices in such manner like ( u , x₁ ) and ( x₂ , v ).

Therefore, perform addition of an edge ( u, v ) in transitive closure only if the closure holds the edges in ( u, x₁ ) and ( x₂, v ) manner. So, the consideration of pair required only once and the total run time of this procedure will be O ( V² ).

(b)

Program Plan Intro

To give an example that the update of transitive closer will take O ( V² ) time, when a new edge ‘ e’ is added to the graph G .

(b)

Expert Solution

Explanation of Solution

Consider the condition where there aretwo strongly connected components with sizes | V |/2 and there is no common edge between them. The transitive closure can be computed by adding these two connected components of the graph.

Now, perform addition of a single edge between two connected components that provide connectivity between two separate components of the graph. Here, it is clearly visible that the total no. of edges will be increased by | V |/4 and the total no of edges will be [| V |/2 + | V |/4]. So, every time while adding a new edge required a constant time at least.

Therefore, the update of transitive closer will take O ( V² ) time, when a new edge ‘ e’ is added to the graph G .

(c)

Program Plan Intro

To define an algorithm that takes O ( V³ ) time to update the transitive closure of graph G , when an edge is inserted in to the graph.

(c)

Expert Solution

Explanation of Solution

Consider a set of vertices in the graphG , and there is a path between every pair of vertices. Now, while performing an addition of an edge ( u, v ), look at the ancestor of vertex u and add it there. The reason of looking at the ancestor of an edge is to explore the branches of the tree. This procedure will also applicable for the vertex v. through this, it will be very easy to get the already added edges and will only consider those edges in n times mostly.

Since, the edges in the tree required O ( V² ) time for consideration. Therefore, the total time taken by the algorithm will be O ( V³ ).

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Answer 5

Chapter 25, Problem 1P

(a)

Program Plan Intro

To show that the updation procedure of transitive closer G * ( V , E *) of graph G ( V, E ) will take O ( V² ) time, while an edge is added to the graph G .

(a)

Expert Solution

Explanation of Solution

Here, in the graph G , if the updation of transitive closure takes O ( V² ) time. To understand this scenario, suppose, addition of an edge ( x₁ , x₂ ) is performed in the graph G . Now, take the set of vertices ( u, v ) in order to get the path from u to x₁ and x₂ to v . If there is a possibility of having an edge that contains vertices in such manner like ( u , x₁ ) and ( x₂ , v ).

Therefore, perform addition of an edge ( u, v ) in transitive closure only if the closure holds the edges in ( u, x₁ ) and ( x₂, v ) manner. So, the consideration of pair required only once and the total run time of this procedure will be O ( V² ).

(b)

Program Plan Intro

To give an example that the update of transitive closer will take O ( V² ) time, when a new edge ‘ e’ is added to the graph G .

(b)

Expert Solution

Explanation of Solution

Consider the condition where there aretwo strongly connected components with sizes | V |/2 and there is no common edge between them. The transitive closure can be computed by adding these two connected components of the graph.

Now, perform addition of a single edge between two connected components that provide connectivity between two separate components of the graph. Here, it is clearly visible that the total no. of edges will be increased by | V |/4 and the total no of edges will be [| V |/2 + | V |/4]. So, every time while adding a new edge required a constant time at least.

Therefore, the update of transitive closer will take O ( V² ) time, when a new edge ‘ e’ is added to the graph G .

(c)

Program Plan Intro

To define an algorithm that takes O ( V³ ) time to update the transitive closure of graph G , when an edge is inserted in to the graph.

(c)

Expert Solution

Explanation of Solution

Consider a set of vertices in the graphG , and there is a path between every pair of vertices. Now, while performing an addition of an edge ( u, v ), look at the ancestor of vertex u and add it there. The reason of looking at the ancestor of an edge is to explore the branches of the tree. This procedure will also applicable for the vertex v. through this, it will be very easy to get the already added edges and will only consider those edges in n times mostly.

Since, the edges in the tree required O ( V² ) time for consideration. Therefore, the total time taken by the algorithm will be O ( V³ ).

Answer 6

Chapter 25, Problem 1P

(a)

Program Plan Intro

To show that the updation procedure of transitive closer G * ( V , E *) of graph G ( V, E ) will take O ( V² ) time, while an edge is added to the graph G .

(a)

Expert Solution

Explanation of Solution

Here, in the graph G , if the updation of transitive closure takes O ( V² ) time. To understand this scenario, suppose, addition of an edge ( x₁ , x₂ ) is performed in the graph G . Now, take the set of vertices ( u, v ) in order to get the path from u to x₁ and x₂ to v . If there is a possibility of having an edge that contains vertices in such manner like ( u , x₁ ) and ( x₂ , v ).

Therefore, perform addition of an edge ( u, v ) in transitive closure only if the closure holds the edges in ( u, x₁ ) and ( x₂, v ) manner. So, the consideration of pair required only once and the total run time of this procedure will be O ( V² ).

Answer 7

Chapter 25, Problem 1P

(a)

Program Plan Intro

To show that the updation procedure of transitive closer G * ( V , E *) of graph G ( V, E ) will take O ( V² ) time, while an edge is added to the graph G .

Answer 8

(a)

Expert Solution

Explanation of Solution

Here, in the graph G , if the updation of transitive closure takes O ( V² ) time. To understand this scenario, suppose, addition of an edge ( x₁ , x₂ ) is performed in the graph G . Now, take the set of vertices ( u, v ) in order to get the path from u to x₁ and x₂ to v . If there is a possibility of having an edge that contains vertices in such manner like ( u , x₁ ) and ( x₂ , v ).

Therefore, perform addition of an edge ( u, v ) in transitive closure only if the closure holds the edges in ( u, x₁ ) and ( x₂, v ) manner. So, the consideration of pair required only once and the total run time of this procedure will be O ( V² ).

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Program Plan Intro

To give an example that the update of transitive closer will take O ( V² ) time, when a new edge ‘ e’ is added to the graph G .

(b)

Expert Solution

Explanation of Solution

Consider the condition where there aretwo strongly connected components with sizes | V |/2 and there is no common edge between them. The transitive closure can be computed by adding these two connected components of the graph.

Now, perform addition of a single edge between two connected components that provide connectivity between two separate components of the graph. Here, it is clearly visible that the total no. of edges will be increased by | V |/4 and the total no of edges will be [| V |/2 + | V |/4]. So, every time while adding a new edge required a constant time at least.

Therefore, the update of transitive closer will take O ( V² ) time, when a new edge ‘ e’ is added to the graph G .

Answer 13

(b)

Program Plan Intro

To give an example that the update of transitive closer will take O ( V² ) time, when a new edge ‘ e’ is added to the graph G .

Answer 14

(b)

Expert Solution

Explanation of Solution

Consider the condition where there aretwo strongly connected components with sizes | V |/2 and there is no common edge between them. The transitive closure can be computed by adding these two connected components of the graph.

Now, perform addition of a single edge between two connected components that provide connectivity between two separate components of the graph. Here, it is clearly visible that the total no. of edges will be increased by | V |/4 and the total no of edges will be [| V |/2 + | V |/4]. So, every time while adding a new edge required a constant time at least.

Therefore, the update of transitive closer will take O ( V² ) time, when a new edge ‘ e’ is added to the graph G .

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

Program Plan Intro

To define an algorithm that takes O ( V³ ) time to update the transitive closure of graph G , when an edge is inserted in to the graph.

(c)

Expert Solution

Explanation of Solution

Consider a set of vertices in the graphG , and there is a path between every pair of vertices. Now, while performing an addition of an edge ( u, v ), look at the ancestor of vertex u and add it there. The reason of looking at the ancestor of an edge is to explore the branches of the tree. This procedure will also applicable for the vertex v. through this, it will be very easy to get the already added edges and will only consider those edges in n times mostly.

Since, the edges in the tree required O ( V² ) time for consideration. Therefore, the total time taken by the algorithm will be O ( V³ ).

Answer 19

(c)

Program Plan Intro

To define an algorithm that takes O ( V³ ) time to update the transitive closure of graph G , when an edge is inserted in to the graph.

Answer 20

(c)

Expert Solution

Explanation of Solution

Consider a set of vertices in the graphG , and there is a path between every pair of vertices. Now, while performing an addition of an edge ( u, v ), look at the ancestor of vertex u and add it there. The reason of looking at the ancestor of an edge is to explore the branches of the tree. This procedure will also applicable for the vertex v. through this, it will be very easy to get the already added edges and will only consider those edges in n times mostly.

Since, the edges in the tree required O ( V² ) time for consideration. Therefore, the total time taken by the algorithm will be O ( V³ ).

Answer 21

(c)

Expert Solution

Answer 22

(c)

Expert Solution

Answer 23

Expert Solution

Answer 24

Ch. 25.1 - Prob. 1E Ch. 25.1 - Prob. 2E Ch. 25.1 - Prob. 3E Ch. 25.1 - Prob. 4E Ch. 25.1 - Prob. 5E Ch. 25.1 - Prob. 6E Ch. 25.1 - Prob. 7E Ch. 25.1 - Prob. 8E Ch. 25.1 - Prob. 9E Ch. 25.1 - Prob. 10E

To show that the updation procedure of transitive closer G * ( V , E *) of graph G ( V, E ) will take O ( V 2 ) time, while an edge is added to the graph G .