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:

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:

According toMinimum Degree Bound $G_{r + 1}$ 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 clockwise 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 $c_5$.

Let us denote $H_{i,j}$ a subgraph of $G'_r$ induced by the vertices colored with $c_i$ and $c_j$.

Consider $H_{1, 3}$.

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

Thus, $H_{1, 3}$ is disconnected into two components.

We can, then, interchange the colors $c_1$ and $c_3$ in the component that is 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 path between $y_1$ and $y_3$ in $H_{1,3}$.

Including the vertex $x$ in this path we get a circuit $C$.

Since we indexed the vertices $y_1, y_2, y_3, y_4$ and $y_5$ clockwise, exactly one of the vertices $y_2$ and $y_4$ is inside $C$.

Hence, $y_2$ and $y_4$ are in different connected components of $H_{2, 4}$

Then we can switch colors $c_2$ and $c_4$ in the component of $H_{2, 4}$ that is connected to $y_2$.

Now $x$ is no longer adjacent to a vertex of color $c_2$, so we can color it $c_2$.


 * Five Color Theorem.png

This graph illustrates the case in which the path 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 $k \le 5$.

Also known as
This theorem is also known as Heawood’s Theorem, for, although The Five-Color Theorem is more widely used.

The British English spelling of this proof is five colour theorem.

Also see

 * The proof gives a simple (recursive) algorithm for 5-coloring a planar graph, the so-called Heawood's Algorithm.

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.