# Degrees of Vertices III

#### Joystar1977

##### Active member
Let G be a graph with vertex set V = {v1, v2, v3, v4, v5}.

If the degrees of the vertices are 5, 1, 0, 6, 2, respectively, does G have an Eulerian path? Why or why not?

2E= deg v1 + deg v2 + deg v3 + deg v4 + deg v5

2E = 5 + 1 + 0 + 6 + 2

2E = 14

E = 7

Is it correct to say that G does have an Eulerian path because it can happen if the graph has either 0 or 2 vertices with odd degrees? Even an incomplete graph with the degree of 0 at one or more vertices can have an Eulerian path.

Is it correct to say that G doesn't have an Eulerian path because the path is suppose to pass through all edges exactly once?

#### caffeinemachine

##### Well-known member
MHB Math Scholar
Let G be a graph with vertex set V = {v1, v2, v3, v4, v5}.

If the degrees of the vertices are 5, 1, 0, 6, 2, respectively, does G have an Eulerian path? Why or why not?

2E= deg v1 + deg v2 + deg v3 + deg v4 + deg v5

2E = 5 + 1 + 0 + 6 + 2

2E = 14

E = 7

Is it correct to say that G does have an Eulerian path because it can happen if the graph has either 0 or 2 vertices with odd degrees? Even an incomplete graph with the degree of 0 at one or more vertices can have an Eulerian path.
I don't understand your argument here. Can you restate this?

Is it correct to say that G doesn't have an Eulerian path because the path is suppose to pass through all edges exactly once?
This is not an argument which establishes the inexistence of an Eulerian path in the graph.

I think that there is no graph on five vertices which has 5, 1, 0, 6, 2 as the degrees of its vertices. So it doesn't make sense to ask if there is an Eulerian path in such a Graph.

#### Joystar1977

##### Active member
This is a question for one of my problems on my assignment and states the following:

Let G be a graph with vertex set V = {v1, v2, v3, v4, v5}.

If the degrees of the vertices are 5, 1, 0, 6, 2, respectively, how many edges are in G?

I was told that an Eulerian path is a path which passes all edges exactly once. This can happen if the graph has either 0 or 2 vertices with odd degrees. Even an incomplete graph (with degree 0 at one or more vertices) can have an Eulerian path.

My question was that is it correct to say the following:

G does have an Eulerian path because it can take place even if the graph has 0 or many vertices to where you end up with an odd degree?

I don't want to confuse myself any more than what I already am with this problem.

Quote Originally Posted by Joystar1977 View Post
Is it correct to say that G doesn't have an Eulerian path because the path is suppose to pass through all edges exactly once?

I don't understand your argument here. Can you restate this?

This is not an argument which establishes the inexistence of an Eulerian path in the graph.

I think that there is no graph on five vertices which has 5, 1, 0, 6, 2 as the degrees of its vertices. So it doesn't make sense to ask if there is an Eulerian path in such a Graph.

#### caffeinemachine

##### Well-known member
MHB Math Scholar
This is a question for one of my problems on my assignment and states the following:

Let G be a graph with vertex set V = {v1, v2, v3, v4, v5}.

If the degrees of the vertices are 5, 1, 0, 6, 2, respectively, how many edges are in G?

I was told that an Eulerian path is a path which passes all edges exactly once. This can happen if the graph has either 0 or 2 vertices with odd degrees. Even an incomplete graph (with degree 0 at one or more vertices) can have an Eulerian path.

My question was that is it correct to say the following:

G does have an Eulerian path because it can take place even if the graph has 0 or many vertices to where you end up with an odd degree?

I don't want to confuse myself any more than what I already am with this problem.
About the thing marked in red.. An Eulerian path is not a path! An Eulerian path is a trail, in general, which contains each edge exactly once.

Its true that if a connected Graph has zero or two odd vertices then it has an Eulerian Path. But here the problem is that there is no (simple) graph on $5$ vertices which has 5,1,0,6,2 as the degrees of its vertices. So the question in meaningless.

About the thing marked in blue. I don't understand what you are trying to say here. You may need to rephrase that.