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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 ...)?

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    as far as I understand it, yes, if they are 3-connected.2012-06-17

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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.

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    @Daniel. I verified them with Mathematica and all those graphs are isomorphic!2012-06-22