User:Stixme/Sandbox

aux 1
-- Minimum Degree Bond -- For a simple planar graph G, d(G) <= 5 Consider counter hypothesis: G is a simple planar graph and d(G) >= 6. Then, by handshaking lemma, |G| >= 3|V|. Contradiction.

-- Five-colour by induction -- basis: |G| = 1, G is 1-colourable. hyp: with n <= l, l >= 1; holds for n = l + 1 Acc. to mdb, ex v, d(v) <= 5 && G' = G - v is 5-colourable.

=== =

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 $k \le 5$.