For a Connected Non trivial Graph with exactly $2k$ odd vertices ,the minimum number of trails that decompose it is max $\left\{2k,1\right\} $

Question

I know it is probably duplicate question of

this, but my solution /approach is different and also the point where i am facing problem is different.

Theorem For a Connected Non trivial Graph with exactly $2k$ odd vertices ,the minimum number of trails that decompose it is max $\left\{2k,1\right\} $

Proof is given as -:

A Trail Contributes even degree to every vertex,except that non closed trail contributes odd degree to its endpoint.Therefore,a partition of the edges into trails must have some non closed trail ending at each odd vertex.since each trails has only 2 endpoints,we must use atleast $k$ trails to satisfy $2k$ odd vertices.

I am stuck at point

A Trail Contributes even degree to every vertex,except that non closed trail contributes odd degree to its endpoint

Here is my graph

My trail is $D\rightarrow C\rightarrow A\rightarrow B\rightarrow C$

Here 2 vertex has odd vertex that is $C,D$ and two vertices have odd degree $A,B$

I am stuck here that every trail contributes even degree to each vertex.

Though going through formula ,if i take value of $k$=1,we have 2 vertices of odd degree and we have only $k$ i.e $1$ trail possible.

please help me out ,where i am doing wrong.

Answer 1

Your trail is not closed so the end points can have odd degree.

This argument can be deduced by using the theorem 1.2.26 (looks like you're using Graph Theory by D. B. West)

The theorem states: A graph G is Eulerian iff it has at most one nontrivial component and it's vertices all have even degree.