Question

a graph is given to the right
a. explain why the graph has at least one euler path
b. use trial and error or fleurys algorithm to find one such path.

a. choose the correct explanation below.
a. it has exactly two even vertices
b. it has exactly two odd vertices
c. it has more than two odd vertices
d. all graphs have at least one euler path

b. write an euler path starting at a.
(use a comma to separate vertices as needed )

Explanation:

Step1: Determine the reason for Euler path existence

According to Euler's theorem, a connected graph has an Euler path if and only if it has exactly 0 or 2 vertices of odd degree. In this graph, vertices \(D\) and \(C\) have odd degrees (count the number of edges incident to each vertex: for \(D\), the number of edges is \(3\); for \(C\), the number of edges is \(3\); for \(A\), \(E\), and \(B\), the number of edges is \(4\), \(4\), and \(4\) respectively). So, it has exactly two odd vertices.

Step2: Find an Euler path starting from \(A\)

Using trial - and - error:
Start at \(A\). From \(A\), go to \(B\) ( \(A - B\) ). From \(B\), go to \(E\) ( \(A - B - E\) ). From \(E\), go to \(C\) ( \(A - B - E - C\) ). From \(C\), since we want to cover all edges, from \(C\), go to \(B\) ( \(A - B - E - C - B\) ). From \(B\), go to \(A\) ( \(A - B - E - C - B - A\) ). From \(A\), go to \(D\) ( \(A - B - E - C - B - A - D\) )

Answer:

a. B. It has exactly two odd vertices
b. \(A - B - E - C - B - A - D\)