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Euler’s Polyhedron Formula on Graphs
If a planar embedding of a connected graph $G$ has $n$ vertices, $m$ edges and $f$ regions, then $f + n = m + 2$

Induction Basis $m=0$. Planar embedding of G has only one vertex and one region (the exterior region). Thus, the claim is true.

Hypothesis: The theorem is true for $m = l + 1 $

Choose an edge $e$ of $G$ and examine $G' = G - e$. If $e$ is in a circuit, then $G'$ is connected and by the induction hypothesis we get $f' + n = (m - 1) + 2$, where $f'$ is a number of regions in $G'$. However, closing the circuit with e increases the number of regions by one so $f′ = f − 1$ and the theorem is true. If $G − e$ is disconnected, then it has two planar components, $G_1$ and $G_2$ whose number of vertices, edges and regions are $n_1, n_2, m_1, m_2, f_1$, and $f_2$, respectively. By the Induction Hypothesis, $f_1 + n_1 = m_1 + 2$ and $f_2 + n_2 = m_2 + 2$. While adding $e$, the number of regions becomes $f_1 + f_2 − 1$ ($G1$ and $G2$ share the same exterior region or one exterior region is drawn to be a region of the other component), the number of vertices becomes $n_1 + n_2$ and the number of edges becomes $m_1 + m_2 + 1$. Hence, the claim is true.

The Linear Bound on Number of Edges for Planar Graphs
For a simple connected planar graph $G_n$ where $n ≥ 3$ vertices the following is true
 * $m ≤ 3n − 6$, where $m$ is a number of edges.

Proof. Let sequence $s_1,. . ., s_f$ be the regions of a planar embedding of $G$, then we denote the number of boundary edges of $s_i$ by $r_i \:\: (i = 1, . . . , f)$. In case $f = 1$, $G$ is then a tree and $m = n − 1 ≤ 3n − 6$. Thus, we assume that $f ≥ 2$. Since $G$ is simple, every region has at least $3$ boundary edges and every edge is a boundary edge of one or two regions in the planar embedding, thus:
 * $3k \leq \sum\limits_{i+1}^f r_i \leq 2m$

The linear bound now follows directly from Polyhedron Formula.

The Minimum Degree Bound For A Simple Planar Graph
Let $G$ be a simple planar graph. Then, $G, δ(G) ≤ 5$

Proof
Let us prove by contradiction and consider the counter hypothesis:
 * $G$ is a simple planar graph and $δ(G) ≥ 6$.

Then, (by handshaking lemma) $m ≥ 3n$, where $n$ is the number of vertices and $m$ is the number of edges in $G$. This contradicts The Linear Bound on Number of Edges for Planar Graphs.

every non-trivial path graph is 2-chromatic
A non-trivial path graph Pn is a graph of form: $P = (V,E)\:$ where $V = \{v_1, v_2, .., v_n\}, \:\: n > 1,$ $E = \{e_1, e_2, .., e_k\}, \:\: e_j=\{v_i, v_{i+1}\}$

Consider the following colouring procedure: $L:V \rightarrow \{red, blue\}$ $L(v_i)=$ $\begin{cases} red, & i \:\:is \:\: odd \\ blue, & i \:\:is \:\: even \end{cases}$

$L$ produces a proper colouring since if $i$ is odd, then $i+1$ is even; and if $i$ is even, then $i+1$ is odd. Thus, for any pair $\{v_i, v_{i+1}\}$ both vertices have different colour.

Note, that $\forall n, \: P_n$ has $K_2$ as an induced subgraph, and $\chi(K_n)=n$ by the definition of proper colouring. Hence, $\chi(P_n)\geq \chi(K_2)$.

Every Simple Planar Graph is n-colorable (n>=6)
wlog we can assume that the graph is connected. If it is not connected, we just color each connected component. They never conflict with each other, so we will be done. We will do the induction on the number of vertices. Base case: The simplest connected planar graph consists of a single vertex. Pick a color for that vertex. we are done. Induction step: Assume $k ≥ 1$, and assume that every planar graph with $k$ or fewer vertices can be $n-colored$. Now consider a planar graph with $k + 1$ vertices. From above, we know that the graph has a vertex of degree 5 or fewer (by the minimum degree bond). Remove that vertex (and all edges connected to it). By the induction hypothesis, we can n-color the remaining graph. Put the vertex (and edges) back in. We have a graph with every vertex colored (without conflicts) except for the ”special” one. There are at most 5 colors adjacent, so we have at least one color left. Use an available color for that vertex. We have then n-colored the graph.

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 -- can be tied to circuit_min_3 or by min_polygon (since G is simple, doesn't allow monogons)
 * each edge bounds at most $2$ faces. -- natural

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

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
--> minimum degree bond

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


 * Let this vertex be labeled $x$.  -- rid

- - - case one

According to the minimum degree bond, there is a vertex v in G, d(v) >= 5. On the other hand, according to the induction hypothesis the graph G' (G' = G - v) is 5-colourable. If, in this colouring, the vertices adjacent to v are coloured using at most four colours, then we can colour v in the missing color.

-> 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 $G_{r + 1}$.

- - - - - - - - - - -- - - -- - - --

-- case two --- -

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 $k \le 5$.