Studying the four color problem, I was analyzing all possible 3-regular planar graphs of 12 faces, with the additional restriction that graphs that have one or more faces with less than 5 edges, are not to be considered.
Note: It counts as a face also the surrounding area (infinite) of the graph
Using a Java program that builds all possible graphs of this kind, I saw that all have an Hamiltonian path, and that this path is very simple to compute using an algorithm I am implementing in Java: Cahit spiral chain algorithm.
The algorithm is:
- Start from an external vertex
- Only to choose the second vertex of the path, move on the external cycle clockwise
- For all the other vertices that define the path, always move left (each vertex has three edges, one is the edge I am coming from, for the other two "left" and "right" are referred to the planar representation of the graph)
- If moving left, you end up on a already visited vertex, move right
Here is the question:
Is this an obvious observations? Is there a basic theorem that implies the existence of an Hamiltonian path for graphs of this kind (3-regular planar graphs of 12 faces …)?
Best Answer
Euler's formula states that
$$V - E + F = 2$$
which holds for any finite, connected, planar graph with $V$ vertices, $E$ edges and $F$ faces. Since each vertex has degree $3$, the total degree is $3V$ and the number of edges is half of this, since each edge contributes $2$ to the total degree. So $$V-\frac{3V}{2}+12 = 2 $$ and therefore: $$ V = 20$$ The total degree is then $3V=60$ and the number of edges is half of this $E=30$. Since each face is made up of at least $5$ edges, each face is made of exactly $5$ edges. The $3$-regular graph with these properties is the Dodecahedron, which is well-known to be Hamiltonian. Indeed, this is the subject of Hamilton's own Icosian game.