Five Color Theorem

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

Proof
It is obvious the theorem is true for a graph with only one vertex. We will induct on the number of vertices.

Each face of a planar graph is obviously bounded by at least $3$ edges, and each edge bounds at most $2$ faces, so $\dfrac 2 3 e \ge f$.

We first suppose that every vertex of $G$ is incident on $6$ edges or more.

However, if every vertex has degree greater than $5$ as we supposed, $\sum \text{degrees} \ge 6v$, which is a contradiction.

Therefore, $G$ has at least one vertex with at most $5$ edges, which we will call $x$.

Remove that vertex $x$ from $G$ to create another graph, $G'$.

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

If all five colors were not connected to $x$, then we can give $x$ the missing color and thus five-color $G$.

If all five colors were connected to $x$, we examine the five vertices $x$ was adjacent to, and call them $y_1, y_2, y_3, y_4$ and $y_5$ (in order around $x$).

We color $1, 2, 3, 4$ and $5$ respectively (note that "color" is just a way of labeling the vertices of a graph - it does not actually mean you take crayons and color the graph).

We now consider the subgraph $G_{1,3}$ of $G'$ consisting only of vertices colored $1$ and $3$ and the edges that connect vertices of color $1$ to vertices of color $3$.

If there is no walk between $y_1$ and $y_3$ in $G_{1,3}$, then we simply switch colors $1$ and $3$ in the portion of $G_{1,3}$ connected to $y_1$.

Thus, $x$ is no longer adjacent to a vertex of color $1$, so we can color it $1$.

If there is a walk between $y_1$ and $y_3$ in $G_{1,3}$, then we proceed to form $G_{2,4}$ in the same manner.

However, since $G$ is planar and there is a circuit in $G$ that consists of the walk from $y_1$ to $y_3$, $x$, and the edges $xy_1$ and $xy_3$, clearly $y_2$ cannot be connected to $y_4$ within $G_{2,4}$.

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

Thus, $x$ is no longer adjacent to a vertex of color $2$, so we can color it $2$.


 * Five Color Theorem.png

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

$\text{Blue} = 1, \text{Yellow} = 2, \text{Red} = 3, \text{Green} = 4, \text{Turquoise} = 5$

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