An alternative proof for Theorem 10.1.3 has the following outline. Suppose G is a connected graph in which every vertex has even degree. Suppose the path C: v1e1v2e2v3…envn+1 has maximum length in G. That is, C has at least as many vertices and edges as any other path in G. First derive a contradiction from the assumption that . Next let H be the subgraph of G that contains all the vertices and edges in C. Then derive a contraction from the assumption that . Show that H contains every vertex of G, and show that H contains every edge of G.
To derive the contradiction from the assumption that and then derive the contraction from the assumption that to show that contains every vertex and every edge of .
Suppose is a connected graph in which every vertex has even degree. Suppose the path has maximum length in . Let be the sub graph of that contains all the vertices and edges in .
It is given that is a connected graph in which every vertex has even degree and the path has maximum length in that means has at least as many vertices and edges as any other path in .
Now by using the method of contradiction let us suppose that there is no cycle in the graph that means if and are the starting and ending vertex.
Step 1. It is given that the every vertex in has even degree. This means the graph should have a closed walk with starting and ending vertices same. The graph will contain a Euler Circuit.
Step 2. By Euler’s theorem if all the vertices of a graph has the even degree, then that graph is known as the Euler’s graph.
Step 3. In our assumption means there will be no path like
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