Showing that no Hamilton Circuit exists

Question

I was confused about a certain concept and I was wondering if I could get some help.

There were three points that were made in my textbook to show that a graph does not contain a Hamilton circuit:

1. A graph with a vertex of degree one cannot have a Hamilton circuit.
2. Moreover, if a vertex in the graph has degree two, then both edges that are incident with this vertex must be part of any Hamilton circuit.
3. A Hamilton circuit cannot contain a smaller circuit within it.

I understand 1 and 3, but I'm a bit confused about 2. I wasn't really sure how this helps. I realize that the definition of a Hamilton circuit is a simple circuit in a graph that traverses over all vertices in the graph. But this is exactly what we are trying to disprove, so how is it helpful to say that both edges must be part of a Hamilton circuit when we don't even know if the graph contains a Hamilton circuit? How is that not circular?

Hopefully you can understand my confused, sleep deprived thought process. If you can't, could somebody at least elaborate on the significance of number 2?

Thanks for the help.

Answer 1

Suppose that $v$ is a vertex of degree $2$. If $G$ has a Hamilton circuit, that circuit must pass through $v$, so it must include both of the edges incident at $v$. In particular, if the configuration

              o----o----o  
              u    v    w

occurs anywhere in $G$, it must be part of any Hamilton circuit in $G$. Such a circuit must be a circuit, so it must include some other path from $u$ to $w$ as well. This can be useful in proving that a graph does not have a Hamilton circuit: if there is no other path between $u$ and $w$ in $G$, then $G$ cannot have a Hamilton circuit.

Answer 2

A Hamilton circuit passes through each vertex of the graph exactly once and forms a closed loop. If $v$ is a vertex and its degree is $2$ we require both edges to be part of the circuit to avoid breaking it. If $(v,w)$ is the only edge in the circuit containing $v$ then $v$ is an endpoint. Therefore, we don't have a closed loop.

Answer 3

A Hamiltonian cycle(also sometimes reffered to as a circuit) is on in which every vertex in the graph $G$ is encountered once and only once.

One necessary condition for the existence of hamiltonian cycles in a graph $G$ is that $\nexists$ a vertex cut $S$ such that The number of components in $G-S$ that is $k(G-S)$ is always less than the cardinality of the vertex cut $|S|$. In other words $k(G-S)\leq |S|$. A simple proof of this suppose $k(G-S) = k$ and $G_1,G_2,...G_k$ are components in $G-S$ then if the graph contains a hamiltonian cycle $C$ then every time C leaves a component $G_i$ it must encounter a vertex in $S$ thus there must be at least $k$ distinct vertices in $S$ which will be on C after vertices in each $G_i$.

Thus we only definitively know that a graph $G$ does not contain a hamiltonian cycle if $\exists S \subseteq V(G)$ such that $|S|<k(G-S)$. Simple example of this would be a Path or a Star where we can delete a vertex and delete more than one component.

In the light of what I have said :

1. A graph with a vertex of degree one cannot have a Hamilton circuit.

   Since we can delete the vertex just before it and generate two components.

2. Moreover, if a vertex in the graph has degree two, then both edges that are incident with this vertex must be part of any Hamilton circuit.

   Since we can imagine the vertex with degree two is joining two different components $G_1$ and $G_2$ where $S={v}$ such that $degree(v)=2$ then both edges of v are used to join different components.

3. A Hamilton circuit cannot contain a smaller circuit within it.

This is by definition since the length of Hamiltonian circuit is len C $=|V(G)|$.

As far as I know this is the only necessary condition for the existence of Hamiltonian Cycle. I am refferring to :

http://www.amazon.com/First-Course-Graph-Theory-Mathematics/dp/0486483681