Colorability of planar graphs. I'm trying to show that every planar simple graph with no cycles of length $\{4,5,6,7,8,9,10,11\}$ is $3$-colourable.
Here is what I've done so far.
Let $S$ be the set of all graphs for which the statement is false.
Choose $G$ from $S$ with $\min V(G)=n$, say ie $G$ is the minimum counterexample.
I showed that $G$ can't have a vertex of degree at most $2$ and that $G$ doesn't have a cut vertex.
Next, I claim that if $H$ is in $S$, then $H$ has a cut vertex or a vertex of degree at most $2$.
I tried to show this by method of discharging by giving a charge of $d(v)-6$ to each vertex and a charge of $2|f|-6$ to each face. So the total charge given to the graph is $-12$.
I'm not sure how to proceed from here though.
 A: Let $H$ be in $S$.
Let $c(f)$: Charge of face $f$
$c(v)$: Charge of face $v$
Case 1: $H$ has a cut vertex. Then we're done.
Case 2: $H$ has no cut vertex
Give a charge of $d(v)-6$ to each vertex and a charge of $2|f|-6$ to each face.
Total charge $=-12$
If possible, $H$ has no vertex of degree at most $2$.
As $|f|\geq3$, therefore $c(f)\geq0$.
Thus, there exists a vertex $v$ st $c(v)<0$ ie $d(v)<6$.
Also, $d(v)\geq3$
Now, let every face of length $\geq12$ contribute a charge of $\frac{3}{2}$ to each of its vertices.
As $v$ is not a cut vertex, therefore it is adjacent to $d(v)$ different faces.
Also, if a face is a triangle, then adjacent faces can't be a triangle(otherwise we'd get a $4$-cycle).
Hence, all other faces have length at least $12$.
As $d(v)\geq 3$ ,it is adjacent to at least $2$ faces of length at least $12$,and hence it gets additional charge at least $3$.
Therefore, $c(v)\geq 0$
Now, $c(f)=2|f|-6-\frac{3}{2}|f|=\frac{|f|}{2}-6\geq0$ whenever $|f|\geq 12$.
Therefore, all vertices and faces now have positive charge(contradicts the fact that total charge is $-12$).
So $H$ has a vertex of degree $\geq 2$.
So claim holds, hence $G$ can't be in $S$(contradiction as G is the minimum counterexample).
