# Various proofs in graph theory

I'm struggling on some of these practice problems in my text book. More accurately, I'm struggling in graph theory as a whole. I was hoping to get some insight.

1. Show that when any edge is removed from $K_5$, the resulting sub-graph is planar.

I know that

For any planar graph with $V\ge3$, $E\le3V-6$.

Thus $K_5$ would not be a planar graph since it has $10$ edges and $E\le3\cdot5-6=9$. Removing one edge would make it $9$ edges, which fulfils the requirement for any planar graph. However, since the converse is not necessarily true, this isn't a valid proof.

Is it possible to say this?

If there exists a planar graph $G_1$ with $5$ vertices and $9$ edges, then $K_5$ with one edge removed can be redrawn to be planar since it is isomorphic to $G_1$.

I spent a lot of time trying to graph a planar graph with 5 vertices and 9 edges with no luck.

1. Consider the $n$-dimensional cube $Q_n$. $Q_n$ is bipartite by considering the partition $\{V_1,V_2\}$ where $V_1$ denotes the set of vertices that have an odd number of 1s and $V_2$ denotes the set of vertices that have an even number of 1s. Suppose $n$ is even. Explain why there is no Hamiltonian path in $Q_n$ that starts at the vertex 00...0 and ends at the vertex 11...1. (Hamiltonian path: A path (alternating between vertex and edge with no vertex repeated) that includes every vertex.)

I know there is no Hamiltonian path in $Q_n$ where $n$ is even with the requirements. In $Q_2$, it is a square. 00 and 11 are in opposite corners, so to reach both 01 and 10 and start on 00 or 11 then end on the other you would need to double back. The vertices 00...0 and 11...1 must be in the same partite set since they include an even number of 1s. I'm lost on how to continue.

1. State the converse of the 4-color theorem: If $G$ is planar, then $\chi(G)\le4$. Is the converse true? $\chi(G)$ is the chromatic number of a graph $G$, which is the minimum value of $n$ for which an $n$-coloring of $G$ exists.

The converse should be:

If a graph $G$ has $\chi(G)\le4$, then $G$ is planar.

I know this isn't true. I know $\chi(K_{3,3})=2$ is a counterexample, but why does $K_{3,3}$ only require 2 colors?

1. Let $G$ be a graph with exactly one cycle. Prove that $\chi(G)\le3$.

I know a cycle starts and ends on the same vertex. I'm confused on why $\chi(G)\le3$ rather than $\chi(G)\le2$. I'm not entirely sure where to begin with the proof either.

• You would do better by dividing your question into several other questions. – user320254 Feb 10 '17 at 17:26
• Thanks for the suggestion Peter. I will do that for future posts. I will also look into how to make text easier to read. – Jason N. Feb 10 '17 at 20:27

(1) Note that all edges are equivalent in $K_5$, so we only need to find one planar graph with five vertices and nine edges. Here it is: (2) Any Hamiltonian path in $Q_n$ must alternate the two colours, and since $Q_n$ has an even number of vertices any Hamiltonian path must have its endpoints on vertices of opposite colours. However, because (as you have pointed out) the vertices corresponding to 000...0 and 111...1 are of the same colour if $n$ is even, no Hamiltonian path between them can exist.
(3) All complete bipartite graphs $K_{m,n}$ only need two colours: each bipartition receives one colour. The colouring is proper because the subgraph induced by each bipartition is empty.
(4) Any graph with exactly one cycle can be represented as that cycle with trees branching off. If the cycle is of odd length, it needs at least three colours, hence why $\chi(G)\le3$ is stated instead of $\chi(G)\le2$. Regardless of how the cycle is coloured, each branching tree can then be coloured with only two colours (the colour of the vertex on the cycle and another colour), which does not change the fact that $\chi(G)\le3$ in all cases.