Five Color Theorem

Theorem
A planar graph $G$ can be assigned a proper vertex $k$-coloring such that $k \le 5$.

Proof
Principle of Mathematical Induction on the number of vertices:

Let $G_n$ be a planar graph with $n$ vertices.

We have that:
 * each face of $G_n$ is obviously bounded by at least $3$ edges
 * each edge bounds at most $2$ faces.

Thus:
 * $\dfrac 2 3 e \ge f$

For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition:
 * $G_n$ can be assigned a proper vertex $k$-coloring such that $k \le 5$.

Basis for the Induction
$P \left({r}\right)$ is trivially true for $1 \le r \le 5$, as there are no more than $5$ vertices to be colored.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({r}\right)$ is true, where $r \ge 5$, then it logically follows that $P \left({r + 1}\right)$ is true.

So this is our induction hypothesis:
 * $G_r$ can be assigned a proper vertex $k$-coloring such that $k \le 5$.

Then we need to show:
 * $G_{r + 1}$ can be assigned a proper vertex $k$-coloring such that $k \le 5$.

Induction Step
This is our induction step:

Suppose that every vertex of $G_{r + 1}$ is incident on $6$ edges or more.

Let:
 * $V$ denote the number of vertices of $G_{r + 1}$
 * $E$ denote the number of edges of $G_{r + 1}$
 * $F$ denote the number of faces of $G_{r + 1}$.

Then:

where $\deg \left({v}\right)$ denotes the degree of vertex $v$.

However, if every vertex has degree greater than $5$ as we supposed:
 * $\displaystyle \sum_{v \mathop \in G_{r + 1} } \deg \left({v}\right) \ge 6 V$

which is a contradiction.

Therefore, $G_{r + 1}$ has at least one vertex with at most $5$ edges.


 * Let this vertex be labeled $x$.

Remove vertex $x$ from $G_{r + 1}$ to create another graph, $G'_r$.

By the induction hypothesis, $G'_r$ is five-colorable.

Suppose all five colors were not connected to $x$.

Then we can give $x$ the missing color and thus five-color $G_{r + 1}$.

Suppose all five colors are connected to $x$.

Then examine the five vertices $x$ was adjacent to.

Call them $y_1, y_2, y_3, y_4$ and $y_5$ (in order around $x$).

Let $y_1, y_2, y_3, y_4$ and $y_5$ be colored respectively by colors $c_1, c_2, c_3, c_4$ and $c_5$.

Consider the subgraph $G_{1, 3}$ of $G'_r$ consisting only of:
 * the vertices colored $c_1$ and $c_3$
 * the edges that connect vertices of color $c_1$ to vertices of color $c_3$.

Suppose there exists no walk between $y_1$ and $y_3$ in $G_{1,3}$.

Then exchange colors $c_1$ and $c_3$ in the portion of $G_{1,3}$ connected to $y_1$.

Thus $x$ is no longer adjacent to a vertex of color $c_1$, so $x$ can be colored $c_1$.

Suppose there exists a walk between $y_1$ and $y_3$ in $G_{1,3}$.

Then the subgraph $G_{2, 4}$ of $G'_r$ is formed in the same manner as $G_{1, 3}$.

We have that $G_{r + 1}$ is planar.

Consider the circuit in $G_{r + 1}$ that consists of:
 * the walk from $y_1$ to $y_3$
 * $x$
 * the edges $x y_1$ and $x y_3$.

Clearly $y_2$ cannot be connected to $y_4$ within $G_{2, 4}$.

Thus, we can switch colors $c_2$ and $c_4$ in the portion of $G_{2, 4}$ connected to $y_2$.

Thus, $x$ is no longer adjacent to a vertex of color $c_2$

Thus we can color it $c_2$.


 * Five Color Theorem.png

This graph illustrates the case in which the walk from $y_1$ to $y_3$ can be completed.


 * $\text{Blue} = c_1, \text{Yellow} = c_2, \text{Red} = c_3, \text{Green} = c_4, \text{Turquoise} = c_5$.

The dotted lines represent edges and vertices that might exist, as this is simply a fairly minimal example graph that matches the conditions.

So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * For all $n \in \N_{>0}$, $G_n$ can be assigned a proper vertex $k$-coloring such that $k \le 5$.

Historical Note
It was shown in 1976 by and  that four colors suffice.

Their proof relies heavily on computers and for the moment is not to be found on.

Hence the Five Color Theorem is not the strongest result possible.