Find the number of vertices ($|V|$), edges ($|E|$) and faces ($|F|$) of a coherent plane graph with the property that in each vertex descend exactly four faces: three quadrilateral and one triangular.

So from Euler’s formula we know that in our case: $|V| + |E| - |F| = 2$.

We can write that $|F| = x + y$ where $x$ is the number of quadrilateral faces and $y$ is the number of triangular faces.

We know also that:

  • $2$ vertices are incident with every $1$ edge as there is a vertice at both ends of every edge,
  • $4$ edges are incident with every $1$ vertice because every vertice is a point belonging in $4$ edges (since in each vertex descend exactly four faces)

Because of that we can write that: $2|E| = 4|V|$

Also, we know that: $4|V| = 4x + 3y$ (every $1$ vertice belongs in 4 faces) and $2|E| = 4x + 3y$ (every $1$ vertice belongs in 2 edges).

Because of that I have:

  1. $|V| + |E| - a - y = 2$
  2. $2|E| = 4|V|$
  3. $4|V| = 4x + 3y$ or $2|E| = 4x + 3y$

Which are $3$ equations with $4$ variables. I miss one equation to solve the problem and I don't know where to find it.

  • $\begingroup$ Does this answer your question? Planar graph and number of faces of certain degree $\endgroup$ Commented Aug 11, 2023 at 8:27
  • $\begingroup$ @EricNathanStucky The question is the same as the second part of the linked question. The answer for the first part from the link is not complete (as I understand it) - as in my thinking, there are to many variables and to little equations to find $f_3$. As for my original question, and the second part of linked question - it works if it's true that $1$ vertex is incident with $1 f_3$ and it can be written that $v = 3f_3$. I thought that it must be $4v = 4f_4 + 3f_3$ since in every vertex i have $3$ quadrilateral and $1$ triangular face. What do you think? $\endgroup$
    – thefool
    Commented Aug 11, 2023 at 18:24
  • $\begingroup$ I think that both equations hold simultaneously, answering your question :) $\endgroup$ Commented Aug 12, 2023 at 0:48
  • $\begingroup$ For the first part of the other question, you have the equations $$2v = e\\2e = 3f_3 + 4f_4\\v - e + f_3 + f_4 = 2$$ This may seem like three equations in four unknowns, and thus not enough to solve, but if you try to figure out what you can from it anyway, something interesting occurs.... $\endgroup$ Commented Aug 13, 2023 at 5:32
  • $\begingroup$ Consider all the pairs $(\mathbf v, \mathbf f)$ where $\mathbf v$ is a vertex and $\mathbf f$ is a triangular face having $\mathbf v$ as one of its vertices. Since each triangular face has three vertices, there are $3f_3$ such pairs. Since we at told that each vertex has exactly one triangular face incident at it, and thus is part of eactly one such pair, there are $v$ such pairs. Since these two ways of counting the pairs must give the same result, $v = 3f_3$. $\endgroup$ Commented Aug 13, 2023 at 5:43


You must log in to answer this question.

Browse other questions tagged .