Grinberg's theorem is a condition used to prove the existence of an Hamilton cycle on a planar graph. It is formulated in this way:
Let $G$ be a finite planar graph with a Hamiltonian cycle $C$, with a fixed planar embedding. Denote by $ƒ_k$ and $g_k$ the number of $k$-gonal faces of the embedding that are inside and outside of $C$, respectively. Then
$$ \sum_{k \geq 3} (k-2)(f_k - g_k) = 0 $$
While i think i understood the definition i do not know how to apply it on a real problem. For instance, in a graph like that:
how can i identify the internal/external faces of an hypothetical Hamilton cycle $C$ if what i want to do is actually find one of it(an Hamilton cycle)? I mean, the theorem should be used(as far as i understood) to prove(or disprove) the existence of an Hamilton cycle, yet the definition implies that i have to find one to use the whole theorem.
Anybody can help me to understand? I'd like to see an example, even a different one from what i brought should be fine.
Of course, before we find a Hamiltonian cycle or even know if one exists, we cannot say which faces are inside faces or outside faces. However, if there is a Hamiltonian cycle, then there is some, unknown to us, partition for which the sum equals $0$.
So the general idea for using the theorem is this: if we prove that no matter how you partition the faces into "inside" and "outside", we cannot make the sum equal to $0$, then there cannot be a Hamiltonian cycle.
(In subtler applications, I can imagine making arguments such as "if all of these faces are inside faces of a Hamiltonian cycle, then this other face cannot be an outside face". But I don't know of any applications where we can't just take "inside" and "outside" to be an arbitrary partition of the faces and get a contradiction.)
In the example you give, I count $21$ faces with $5$ sides, $3$ faces with $8$ sides, and $1$ face with $9$ sides (the external face). So in order to make the sum equal $0$, we must have $$ 3(f_5 - g_5) + 6 (f_8 - g_8) + 7 (f_9 - g_9) = 0 $$ where $f_5 + g_5 = 21$, $f_8 + g_8 = 3$, and $f_9 + g_9 = 1$.
Taking the sum mod $3$, we get $f_9 - g_9 \equiv 0 \pmod 3$, which cannot happen if one of $f_9, g_9$ is $1$ and the other is $0$. So it's impossible to make the sum equal $0$, and therefore there cannot be a Hamiltonian cycle in this graph.
This strategy, by the way, cannot prove the existence of a Hamiltonian cycle. Just because there's an arbitrary partition of the faces into an "inside" category and an "outside" category for which the sum is $0$, doesn't mean there is actually a Hamiltonian cycle that contains all of the "inside" faces and none of the "outside" faces.