Let G be a graph. Suppose that a depth-first search traversal of G starting at an...
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Let G be a graph. Suppose that a depth-first search traversal of G starting at an arbitrary vertex results in a depth-first search forest that is a path (that is to say, each level of the tree contains only one vertex). Does this imply that G is a complete graph? Know how to perform a depth-first search traversal of a graph and how to construct a depth-first search forest, and you should do fine. Let G be a graph. Suppose that a depth-first search traversal of G starting at an arbitrary vertex results in a depth-first search forest that is a path (that is to say, each level of the tree contains only one vertex). Does this imply that G is a complete graph? Know how to perform a depth-first search traversal of a graph and how to construct a depth-first search forest, and you should do fine.
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No the fact that a depthfirst search DFS traversal of a graph results in a DFS forest that is a path ... View the full answer
Related Book For
Data Structures and Algorithms in Java
ISBN: 978-1118771334
6th edition
Authors: Michael T. Goodrich, Roberto Tamassia, Michael H. Goldwasser
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