Five Color Theorem

Theorem
Any planar graph $$G \ $$ can be assigned a $k$-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 so $$\frac{2}{3} e \geq f$$.

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

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However, if every vertex has degree greater than $$5$$ as we supposed, $$\sum \text{degrees} \geq 6v$$, 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 $$2$$.



This graph illustrates the case in which the walk from $$y_1$$ to $$y_3$$ can be completed. $$\text{Blue} = 1$$, $$\text{Yellow} = 2$$, $$\text{Red} = 3$$, $$\text{Green} = 4$$, and $$\text{Turquoise} = 5$$. The dotted lines represent edges and vertices that might exist, as this is simply a fairly minimal example graph that matches the conditions.