Explanation Given: The provided graph is: Explanation: Fleury’s algorithm has three steps: Step 1: Start at any vertex. Go along any edge from one vertex to another vertex. Remove this edge from the graph. Step 2: Now on a vertex of the revised graph, choose any edge from this vertex and do not use a cut edge unless you have no other option. Go along the chosen edge. Remove this edge from the graph. Step 3: Repeat Step 2 until all the edges have been used and returned to the vertex at which you started. Here, the path begins at F → B , so the revised graph is: None of the edges BC , BD , BE or BK are the cut edges, so any of the edges can be removed. Remove BE , so the revised graph is: Now the path is F → B → E . Now ED is the cut edge, but this is the only edge remaining for vertex E , so remove it. The revised graph is: The path is F → B → E → D . None of the edges DB, DC or DK is a cut edge, so any edge can be removed. Remove DB , so the revised graph is: And path is F → B → E → D → B . None of the edges BC or BK is a cut edge, so any edge can be removed. Remove BK , so the revised graph is: And path is F → B → E → D → B → K . Now KD is a cut edge, but this is the only edge remaining for vertex K , so remove it. The revised graph is: And the path is F → B → E → D → B → K → D . Now DC is the cut edge, but this is the only edge remaining for vertex D , so remove it. The revised graph is: And the path is F → B → E → D → B → K → D → C . Now CB is a cut edge, but this is the only edge remaining for vertex C , so remove it. The revised graph is: And the path is F → B → E → D → B → K → D → C → B . Now BA is a cut edge, but this is the only edge remaining for vertex B , so remove it. The revised graph is: And the path is F → B → E → D → B → K → D → C → B → A . None of the edges AG, AF and AH are cut edges, so any edge can be removed. Remove AH , so the revised graph is: And the path is F → B → E → D → B → K → D → C → B → A → H

Mathematical Ideas (13th Edition) - Standalone book

13th Edition

ISBN: 9780321977076

Author: Charles D. Miller, Vern E. Heeren, John Hornsby, Christopher Heeren

Publisher: PEARSON

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Chapter 14, Problem 19T

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An Euler circuit for the provided graph using Fleury’s algorithm.

The provided graph is:

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Chapter 14, Problem 18T

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The table below shows the acreage, number of visitors, and total revenue of state parks and recreational areas in Massachusetts, New York, and Vermont in 2010. State Acreage (in thousands) Visitors (in thousands) Revenue (in thousands) Massachusetts 350 35,271 $12,644 New York 1,354 56,322 $85,558 Vermont 69 758 $10,969 Select the three true statements based on the data in the table. A. Vermont had the highest revenue per acre of state parks and recreational areas. B. Vermont had approximately 11 visitors per acre of state parks and recreational areas. C. New York had the highest number of visitors per acre of state parks and recreational areas. D. Massachusetts had approximately 36 visitors per acre of state parks and recreational areas. E. New York had revenue of approximately $63.19 per acre of state parks and recreational areas. F. Massachusetts had revenue of approximately $0.03 per acre of state parks and recreational areas.

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Chapter 14 Solutions

Mathematical Ideas (13th Edition) - Standalone book

Ch. 14.1 - Vertices and Edges In Exercises 1-6, determine how...Ch. 14.1 - Vertices and Edges In Exercises 1-6, determine how...Ch. 14.1 - Vertices and Edges In Exercises 1-6, determine how...Ch. 14.1 - Prob. 4E Ch. 14.1 - Prob. 5E Ch. 14.1 - Prob. 6E Ch. 14.1 - Prob. 7E Ch. 14.1 - Prob. 8E Ch. 14.1 - Prob. 9E Ch. 14.1 - Prob. 10E

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An Euler circuit for the provided graph using Fleury’s algorithm. The provided graph is:

Solution Summary: The author describes Fleury's algorithm, which uses three steps to determine an Euler circuit for the provided graph.

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An Euler circuit for the provided graph using Fleury’s algorithm. The provided graph is:

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An Euler circuit for the provided graph using Fleury’s algorithm.

The provided graph is: