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Q33E

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Discrete Mathematics and its Applications
Found in: Page 803
Discrete Mathematics and its Applications

Discrete Mathematics and its Applications

Book edition 7th
Author(s) Kenneth H. Rosen
Pages 808 pages
ISBN 9780073383095

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Short Answer

Show that if \(G\) is a weighted graph with distinct edgeweights, then for every simple circuit of \(G\), the edge of maximum weight in this circuit does not belong to anyminimum spanning tree of \(G\).

The edge of maximum weight in some circuit of a weighted graph cannot be included in the minimum spanning tree of the graph.

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Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Definition

A minimum spanning tree of a weighted graph \(G\) is a spanning tree of \(G\) with minimal weight.

Given: \(G\) is a connected undirected weighted graph

\(G\) has distinct edge weights

\(C\) is a simple circuit in \(G\)

\(T\) is a minimum spanning tree of \(G\)

Step 2: Proof by contradiction

To proof: The edge of maximum weight in \(C\) does not belong to \(T\)

Let the edge of maximum weight in \(C\)be\({\bf{e = \{ u,v\} }}\)and let \(e\) belong to \(T\).

Let \({\bf{T' = T - \{ e\} }}\). \({\bf{T''}}\) is then a forest with \(2\) connected components.

Let \(f\) be the edge in the path \({\bf{C - \{ e\} }}\) that is used to "jump" between the \(2\) connected components (this edge has to exists, as \(C\) was a circuit). \(f\) is not an edge in \(T\) (because else \(T\) would contain a circuit).

Since \(e\) is the edge in \(C\) with maximum weight and since \(G\) has distinct edge weights, \(e\) has a larger weight than \(f\)

\(w(e){\bf{ > }}w(f)\)

Step 3: Weight of edge

Let \(T''{\bf{ = }}T \cup \{ f\} {\bf{ - }}\{ e\} \), then the weight of \({\bf{T''}}\) is less than or equal to the weight of \(T\) since \(w(e){\bf{ > }}w(f)\)

\(w(T''){\bf{ < }}w(T)\)

Thus \({\bf{T''}}\) is then a spanning tree with a smaller weight than \(T\). However as \(T\) was the minimum spanning tree, there should not exist a spanning tree with a lower weight and thus we have derived a contradiction. This means that the assumption that \(e\) belongs to \(T\) was wrong (and thus \(e\) cannot belong to \(T\)).

Most popular questions for Math Textbooks

Draw the subtree of the tree in Exercise \(4\)that is rooted at

\(\begin{array}{l}{\bf{a}})a.\\b)c.\\c)e.\end{array}\)

Which of these are well-formed formulae over the symbols \(\left\{ {{\bf{x,y,z}}} \right\}\) and the set of binary operators \(\left\{ {{\bf{ \ast , + ,}} \circ } \right\}\)?

  1. \({\bf{ \ast + + xyx}}\)
  2. \( \circ {\bf{xy \ast xz}}\)
  3. \({\bf{ \ast }} \circ {\bf{xz \ast \ast xy}}\)
  4. \({\bf{ \ast + }} \circ {\bf{xx}} \circ {\bf{xxx}}\)

How many vertices does \({{\bf{B}}_{\bf{k}}}\) have? Prove that your answer is correct.

Answer the same questions as listed in Exercise \({\bf{3}}\) for the rooted tree illustrated.

Which connected simple graphs have exactly one spanning tree?
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