To represent the graph’s acyclic topological nature.

Question

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Answer 1

Question

Chapter 24, Problem 1P

(a)

Program Plan Intro

To represent the graph’s acyclic topological nature.

(a)

Expert Solution

Explanation of Solution

Given Information: A graph G=(V,E) that is having an arbitrary linear order of vertices v1,v2....,v|V| and edges set partitioned into Ef∪Eb . Where Ef={(vi,vj)∈E:i<j} and Eb={(vi,vj)∈E:i>j} . This graph contains no self-loop therefore every edges is in either Ef or Eb .

Explanation:

As mentioned, graph does not contained any self-loop therefore; every edge in graph Gf will go from higher index vertices to lower indexes vertices. It means, it will never come back to same index where it was started and vice−versa .

It means Gf is acyclic and its vertices (v1,v2,.......,v|V|) are sorted with topologically. As in same case, vertices (v|V|,.......,v2,v1) for graph Gb are sorted topologically.

(b)

Program Plan Intro

To represent the Yen’s improvement in Bellman-Ford algorithm.

(b)

Expert Solution

Explanation of Solution

Given Information: Implementation of Bellman- Ford algorithm and a graph with relaxing edges of Ef and Eb .

Explanation:

Bellman- Ford algorithm is simpler than Dijkstra’s algorithm and worked very well with distributed systems. In the first step, it calculate the shortest path that is having at-most one edge in bottom-up approach then it will go for at most 2-edges and so-on up to ith iterations. This process proved that there is no negative weight cycle.

In Bellman- Ford algorithm and previous part a declare that the graph’s edges Ef are going in increasing order and Eb are going in decreasing order. It means any sequence {ki}i of vertices’ length can change from |V| to ||V|/2| times and it initially expected that vertices will increase their indexes before it run through Ef and Eb . For this change, vertices will increase |v+1| with respect to represent the sources in ordering of the vertices |vk2| .

(c)

Program Plan Intro

To calculate the effects of running time complexity for Bellman-Ford algorithm.

(c)

Expert Solution

Explanation of Solution

Given Information: Implementation of Bellman- Ford algorithm and a scheme for evaluating the time complexity.

Explanation:

It cannot improve the asymptotic running time of Bellman-Ford algorithm becausehaving a co-efficient of 1 to 12 .It will drop the runtime of the algorithm.

The final runtime for the algorithm will be Ο(EV) .

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Answer 2

Question

Chapter 24, Problem 1P

(a)

Program Plan Intro

To represent the graph’s acyclic topological nature.

(a)

Expert Solution

Explanation of Solution

Given Information: A graph G=(V,E) that is having an arbitrary linear order of vertices v1,v2....,v|V| and edges set partitioned into Ef∪Eb . Where Ef={(vi,vj)∈E:i<j} and Eb={(vi,vj)∈E:i>j} . This graph contains no self-loop therefore every edges is in either Ef or Eb .

Explanation:

As mentioned, graph does not contained any self-loop therefore; every edge in graph Gf will go from higher index vertices to lower indexes vertices. It means, it will never come back to same index where it was started and vice−versa .

It means Gf is acyclic and its vertices (v1,v2,.......,v|V|) are sorted with topologically. As in same case, vertices (v|V|,.......,v2,v1) for graph Gb are sorted topologically.

(b)

Program Plan Intro

To represent the Yen’s improvement in Bellman-Ford algorithm.

(b)

Expert Solution

Explanation of Solution

Given Information: Implementation of Bellman- Ford algorithm and a graph with relaxing edges of Ef and Eb .

Explanation:

Bellman- Ford algorithm is simpler than Dijkstra’s algorithm and worked very well with distributed systems. In the first step, it calculate the shortest path that is having at-most one edge in bottom-up approach then it will go for at most 2-edges and so-on up to ith iterations. This process proved that there is no negative weight cycle.

In Bellman- Ford algorithm and previous part a declare that the graph’s edges Ef are going in increasing order and Eb are going in decreasing order. It means any sequence {ki}i of vertices’ length can change from |V| to ||V|/2| times and it initially expected that vertices will increase their indexes before it run through Ef and Eb . For this change, vertices will increase |v+1| with respect to represent the sources in ordering of the vertices |vk2| .

(c)

Program Plan Intro

To calculate the effects of running time complexity for Bellman-Ford algorithm.

(c)

Expert Solution

Explanation of Solution

Given Information: Implementation of Bellman- Ford algorithm and a scheme for evaluating the time complexity.

Explanation:

It cannot improve the asymptotic running time of Bellman-Ford algorithm becausehaving a co-efficient of 1 to 12 .It will drop the runtime of the algorithm.

The final runtime for the algorithm will be Ο(EV) .

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Answer 3

Question

Chapter 24, Problem 1P

(a)

Program Plan Intro

To represent the graph’s acyclic topological nature.

(a)

Expert Solution

Explanation of Solution

Given Information: A graph G=(V,E) that is having an arbitrary linear order of vertices v1,v2....,v|V| and edges set partitioned into Ef∪Eb . Where Ef={(vi,vj)∈E:i<j} and Eb={(vi,vj)∈E:i>j} . This graph contains no self-loop therefore every edges is in either Ef or Eb .

Explanation:

As mentioned, graph does not contained any self-loop therefore; every edge in graph Gf will go from higher index vertices to lower indexes vertices. It means, it will never come back to same index where it was started and vice−versa .

It means Gf is acyclic and its vertices (v1,v2,.......,v|V|) are sorted with topologically. As in same case, vertices (v|V|,.......,v2,v1) for graph Gb are sorted topologically.

(b)

Program Plan Intro

To represent the Yen’s improvement in Bellman-Ford algorithm.

(b)

Expert Solution

Explanation of Solution

Given Information: Implementation of Bellman- Ford algorithm and a graph with relaxing edges of Ef and Eb .

Explanation:

Bellman- Ford algorithm is simpler than Dijkstra’s algorithm and worked very well with distributed systems. In the first step, it calculate the shortest path that is having at-most one edge in bottom-up approach then it will go for at most 2-edges and so-on up to ith iterations. This process proved that there is no negative weight cycle.

In Bellman- Ford algorithm and previous part a declare that the graph’s edges Ef are going in increasing order and Eb are going in decreasing order. It means any sequence {ki}i of vertices’ length can change from |V| to ||V|/2| times and it initially expected that vertices will increase their indexes before it run through Ef and Eb . For this change, vertices will increase |v+1| with respect to represent the sources in ordering of the vertices |vk2| .

(c)

Program Plan Intro

To calculate the effects of running time complexity for Bellman-Ford algorithm.

(c)

Expert Solution

Explanation of Solution

Given Information: Implementation of Bellman- Ford algorithm and a scheme for evaluating the time complexity.

Explanation:

It cannot improve the asymptotic running time of Bellman-Ford algorithm becausehaving a co-efficient of 1 to 12 .It will drop the runtime of the algorithm.

The final runtime for the algorithm will be Ο(EV) .

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 24, Problem 1P

(a)

Program Plan Intro

To represent the graph’s acyclic topological nature.

(a)

Expert Solution

Explanation of Solution

Given Information: A graph G=(V,E) that is having an arbitrary linear order of vertices v1,v2....,v|V| and edges set partitioned into Ef∪Eb . Where Ef={(vi,vj)∈E:i<j} and Eb={(vi,vj)∈E:i>j} . This graph contains no self-loop therefore every edges is in either Ef or Eb .

Explanation:

As mentioned, graph does not contained any self-loop therefore; every edge in graph Gf will go from higher index vertices to lower indexes vertices. It means, it will never come back to same index where it was started and vice−versa .

It means Gf is acyclic and its vertices (v1,v2,.......,v|V|) are sorted with topologically. As in same case, vertices (v|V|,.......,v2,v1) for graph Gb are sorted topologically.

(b)

Program Plan Intro

To represent the Yen’s improvement in Bellman-Ford algorithm.

(b)

Expert Solution

Explanation of Solution

Given Information: Implementation of Bellman- Ford algorithm and a graph with relaxing edges of Ef and Eb .

Explanation:

Bellman- Ford algorithm is simpler than Dijkstra’s algorithm and worked very well with distributed systems. In the first step, it calculate the shortest path that is having at-most one edge in bottom-up approach then it will go for at most 2-edges and so-on up to ith iterations. This process proved that there is no negative weight cycle.

In Bellman- Ford algorithm and previous part a declare that the graph’s edges Ef are going in increasing order and Eb are going in decreasing order. It means any sequence {ki}i of vertices’ length can change from |V| to ||V|/2| times and it initially expected that vertices will increase their indexes before it run through Ef and Eb . For this change, vertices will increase |v+1| with respect to represent the sources in ordering of the vertices |vk2| .

(c)

Program Plan Intro

To calculate the effects of running time complexity for Bellman-Ford algorithm.

(c)

Expert Solution

Explanation of Solution

Given Information: Implementation of Bellman- Ford algorithm and a scheme for evaluating the time complexity.

Explanation:

It cannot improve the asymptotic running time of Bellman-Ford algorithm becausehaving a co-efficient of 1 to 12 .It will drop the runtime of the algorithm.

The final runtime for the algorithm will be Ο(EV) .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 24, Problem 1P

(a)

Program Plan Intro

To represent the graph’s acyclic topological nature.

(a)

Expert Solution

Explanation of Solution

Given Information: A graph G=(V,E) that is having an arbitrary linear order of vertices v1,v2....,v|V| and edges set partitioned into Ef∪Eb . Where Ef={(vi,vj)∈E:i<j} and Eb={(vi,vj)∈E:i>j} . This graph contains no self-loop therefore every edges is in either Ef or Eb .

Explanation:

As mentioned, graph does not contained any self-loop therefore; every edge in graph Gf will go from higher index vertices to lower indexes vertices. It means, it will never come back to same index where it was started and vice−versa .

It means Gf is acyclic and its vertices (v1,v2,.......,v|V|) are sorted with topologically. As in same case, vertices (v|V|,.......,v2,v1) for graph Gb are sorted topologically.

(b)

Program Plan Intro

To represent the Yen’s improvement in Bellman-Ford algorithm.

(b)

Expert Solution

Explanation of Solution

Given Information: Implementation of Bellman- Ford algorithm and a graph with relaxing edges of Ef and Eb .

Explanation:

Bellman- Ford algorithm is simpler than Dijkstra’s algorithm and worked very well with distributed systems. In the first step, it calculate the shortest path that is having at-most one edge in bottom-up approach then it will go for at most 2-edges and so-on up to ith iterations. This process proved that there is no negative weight cycle.

In Bellman- Ford algorithm and previous part a declare that the graph’s edges Ef are going in increasing order and Eb are going in decreasing order. It means any sequence {ki}i of vertices’ length can change from |V| to ||V|/2| times and it initially expected that vertices will increase their indexes before it run through Ef and Eb . For this change, vertices will increase |v+1| with respect to represent the sources in ordering of the vertices |vk2| .

(c)

Program Plan Intro

To calculate the effects of running time complexity for Bellman-Ford algorithm.

(c)

Expert Solution

Explanation of Solution

Given Information: Implementation of Bellman- Ford algorithm and a scheme for evaluating the time complexity.

Explanation:

It cannot improve the asymptotic running time of Bellman-Ford algorithm becausehaving a co-efficient of 1 to 12 .It will drop the runtime of the algorithm.

The final runtime for the algorithm will be Ο(EV) .

Answer 6

Chapter 24, Problem 1P

(a)

Program Plan Intro

To represent the graph’s acyclic topological nature.

(a)

Expert Solution

Explanation of Solution

Given Information: A graph G=(V,E) that is having an arbitrary linear order of vertices v1,v2....,v|V| and edges set partitioned into Ef∪Eb . Where Ef={(vi,vj)∈E:i<j} and Eb={(vi,vj)∈E:i>j} . This graph contains no self-loop therefore every edges is in either Ef or Eb .

Explanation:

As mentioned, graph does not contained any self-loop therefore; every edge in graph Gf will go from higher index vertices to lower indexes vertices. It means, it will never come back to same index where it was started and vice−versa .

It means Gf is acyclic and its vertices (v1,v2,.......,v|V|) are sorted with topologically. As in same case, vertices (v|V|,.......,v2,v1) for graph Gb are sorted topologically.

Answer 7

Chapter 24, Problem 1P

(a)

Program Plan Intro

To represent the graph’s acyclic topological nature.

Answer 8

(a)

Expert Solution

Explanation of Solution

Given Information: A graph G=(V,E) that is having an arbitrary linear order of vertices v1,v2....,v|V| and edges set partitioned into Ef∪Eb . Where Ef={(vi,vj)∈E:i<j} and Eb={(vi,vj)∈E:i>j} . This graph contains no self-loop therefore every edges is in either Ef or Eb .

Explanation:

As mentioned, graph does not contained any self-loop therefore; every edge in graph Gf will go from higher index vertices to lower indexes vertices. It means, it will never come back to same index where it was started and vice−versa .

It means Gf is acyclic and its vertices (v1,v2,.......,v|V|) are sorted with topologically. As in same case, vertices (v|V|,.......,v2,v1) for graph Gb are sorted topologically.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Program Plan Intro

To represent the Yen’s improvement in Bellman-Ford algorithm.

(b)

Expert Solution

Explanation of Solution

Given Information: Implementation of Bellman- Ford algorithm and a graph with relaxing edges of Ef and Eb .

Explanation:

Bellman- Ford algorithm is simpler than Dijkstra’s algorithm and worked very well with distributed systems. In the first step, it calculate the shortest path that is having at-most one edge in bottom-up approach then it will go for at most 2-edges and so-on up to ith iterations. This process proved that there is no negative weight cycle.

In Bellman- Ford algorithm and previous part a declare that the graph’s edges Ef are going in increasing order and Eb are going in decreasing order. It means any sequence {ki}i of vertices’ length can change from |V| to ||V|/2| times and it initially expected that vertices will increase their indexes before it run through Ef and Eb . For this change, vertices will increase |v+1| with respect to represent the sources in ordering of the vertices |vk2| .

Answer 13

(b)

Program Plan Intro

To represent the Yen’s improvement in Bellman-Ford algorithm.

Answer 14

(b)

Expert Solution

Explanation of Solution

Given Information: Implementation of Bellman- Ford algorithm and a graph with relaxing edges of Ef and Eb .

Explanation:

Bellman- Ford algorithm is simpler than Dijkstra’s algorithm and worked very well with distributed systems. In the first step, it calculate the shortest path that is having at-most one edge in bottom-up approach then it will go for at most 2-edges and so-on up to ith iterations. This process proved that there is no negative weight cycle.

In Bellman- Ford algorithm and previous part a declare that the graph’s edges Ef are going in increasing order and Eb are going in decreasing order. It means any sequence {ki}i of vertices’ length can change from |V| to ||V|/2| times and it initially expected that vertices will increase their indexes before it run through Ef and Eb . For this change, vertices will increase |v+1| with respect to represent the sources in ordering of the vertices |vk2| .

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

Program Plan Intro

To calculate the effects of running time complexity for Bellman-Ford algorithm.

(c)

Expert Solution

Explanation of Solution

Given Information: Implementation of Bellman- Ford algorithm and a scheme for evaluating the time complexity.

Explanation:

It cannot improve the asymptotic running time of Bellman-Ford algorithm becausehaving a co-efficient of 1 to 12 .It will drop the runtime of the algorithm.

The final runtime for the algorithm will be Ο(EV) .

Answer 19

(c)

Program Plan Intro

To calculate the effects of running time complexity for Bellman-Ford algorithm.

Answer 20

(c)

Expert Solution

Explanation of Solution

Given Information: Implementation of Bellman- Ford algorithm and a scheme for evaluating the time complexity.

Explanation:

It cannot improve the asymptotic running time of Bellman-Ford algorithm becausehaving a co-efficient of 1 to 12 .It will drop the runtime of the algorithm.

The final runtime for the algorithm will be Ο(EV) .