Short Answer

Suppose that \({\bf{G}}\) is a connected multi graph with \({\bf{2k}}\) vertices of odd degree. Show that there exist \({\bf{k}}\) sub graphs that have \({\bf{G}}\) as their union, where each of these subgraphs has an Euler path and where no two of these subgraphs have an edge in common. (Hint: Add \({\bf{k}}\) edges to the graph connecting pairs of vertices of odd degree and use an Euler circuit in this larger graph.)
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Here,\(k\)sub graphs that have \(G\)as their union, where each of these sub graphs has an Euler path and where no two of these sub graphs have an edge in common.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Determine the sub graphs

Let us take a connected multi graph\(G\)with \(2k\)vertices of odd degree and go in the reverse direction for the proof, initially it starts pairing the vertices of odd degree and go on adding an extra edge joining the vertices in each pair, i.e. a total of \(k\) edges are added.

It will obtain a multi graph which will have all vertices even degree which satisfies the condition for being a Euler circuit.

Step 2: Conclusion

Hence, there exist \(k\)sub graphs that have \(G\)as their union, where each of these sub graphs has an Euler path and where no two of these sub graphs have an edge in common

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Discrete Mathematics and its Applications

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