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14 13 B E 9 A Apply the cheapest-link algorithm to the graph above. Give your answer as a list of vertices, starting and ending at vertex A. Example: ABCDEA

14 13 B E 9 A Apply the cheapest-link algorithm to the graph above. Give your answer as a list of vertices, starting and ending at vertex A. Example: ABCDEA

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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### Cheapest-Link Algorithm Challenge #### Graph Description: This graph consists of five vertices labeled \(A\), \(B\), \(C\), \(D\), and \(E\). The edges between these vertices are annotated with weights, representing the cost or distance between the vertices. The weights of the edges are as follows: - Between \(A\) and \(B\): 1 - Between \(A\) and \(C\): 10 - Between \(A\) and \(D\): 6 - Between \(A\) and \(E\): 9 - Between \(B\) and \(C\): 3 - Between \(B\) and \(D\): 14 - Between \(B\) and \(E\): 2 - Between \(C\) and \(D\): 5 - Between \(C\) and \(E\): 13 - Between \(D\) and \(E\): 7 #### Task: Apply the cheapest-link algorithm to the graph above. Give your answer as a list of vertices, starting and ending at vertex \(A\). Example format: \(ABCDEA\). #### Explanation of the Cheapest-Link Algorithm: 1. **Sort all edges in the graph in ascending order by weight.** 2. **Select the smallest edge that does not form a cycle or make any vertex have three edges.** 3. **Repeat step 2 until you have a complete Hamiltonian circuit (visits every vertex exactly once and returns to the starting vertex).** #### Example Solution: This is where you would apply the cheapest-link algorithm step by step to find the solution for the given graph. Please follow the above steps and list down the sequence of vertices forming the cheapest Hamiltonian circuit starting and ending at \(A\).
Transcribed Image Text:### Cheapest-Link Algorithm Challenge #### Graph Description: This graph consists of five vertices labeled \(A\), \(B\), \(C\), \(D\), and \(E\). The edges between these vertices are annotated with weights, representing the cost or distance between the vertices. The weights of the edges are as follows: - Between \(A\) and \(B\): 1 - Between \(A\) and \(C\): 10 - Between \(A\) and \(D\): 6 - Between \(A\) and \(E\): 9 - Between \(B\) and \(C\): 3 - Between \(B\) and \(D\): 14 - Between \(B\) and \(E\): 2 - Between \(C\) and \(D\): 5 - Between \(C\) and \(E\): 13 - Between \(D\) and \(E\): 7 #### Task: Apply the cheapest-link algorithm to the graph above. Give your answer as a list of vertices, starting and ending at vertex \(A\). Example format: \(ABCDEA\). #### Explanation of the Cheapest-Link Algorithm: 1. **Sort all edges in the graph in ascending order by weight.** 2. **Select the smallest edge that does not form a cycle or make any vertex have three edges.** 3. **Repeat step 2 until you have a complete Hamiltonian circuit (visits every vertex exactly once and returns to the starting vertex).** #### Example Solution: This is where you would apply the cheapest-link algorithm step by step to find the solution for the given graph. Please follow the above steps and list down the sequence of vertices forming the cheapest Hamiltonian circuit starting and ending at \(A\).
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