4-regular plane graph

There are $\frac{117*4}{2}=234$ edges in $G$ , since each vertex must be connected to $4$ other vertices, and each edge connects $2$ vertices (if we didn't divide by $2$ , we would count each edge twice).

By Euler's formula, $v-e+f=2$ , where $v$ is the number of vertices, $e$ is the number of edges, and $f$ is the number of faces. We plug in our values for $v$ and $e$ , and solving for $f$ , we get our answer to be $\fbox{119}$ .

Let the number of vertices, edges and faces of the planar graph be $v, e, f$ respectively. Note that $v = 117$ and $e = \frac{4 \cdot v}{2} = 2 \cdot 117 = 234$ . Since the graph is planar, by Euler's Formula, $v - e + f = 2 \Rightarrow f = 2 + e - v = 2 + 234 - 117 = 119$ .

To calculate the number of faces, we will use Euler's formula, for a 2D plane: $F+V-E = 2$ .

As G is 4-regular, each vertex is connected to 4 others, for a total of $4 \times 117 = 468$ connections.

However this will count every edge twice (ie. Point A - Point B as well as Point B - Point A), so the total number of edges is:

$E = \frac{468} {2} = 234$

Using Euler's formula with $V=117$ , $E=234$ , we find:

$F+V-E = 2$

$F = 2+E-V$

$F = 2+234-117$

$F = 119$

According to the degree sum formula: $\Sigma deg(v)=2e$ , so $e=\frac{4*117}{2}=234$ .

According to Euler's formula: $v - e + f = 2$ , so $f=2+e-v=119$

According to Euler's Formula V - E + F = 2 where

V = number of vertices

E = number of edges

F = number of faces

We are given the number of vertices so we only need to calculate the number of edges to solve for F.

By the Handshaking lemma ( you need two person for every handshake ) Every 2 degree contributes to 1 extra edge.

So the number of edges in the graph = total degree/2

Since we know that the graph is a 4 regular graph, all vertices has degree four (means four edges coming out of every vertices)

So, total degree = 4 * 117 = 468

Number of edges = 468/2 = 234

By Euler's Formula 117 - 234 + F = 2

Shifting things around...

F = 119

As, it is a planar graph with $117 \text{ vertices (v)}$ , and having $\text{nodes, each having a degree (k) of}$ $4$ . So, using handshaking theorem, for nodes, i.e. $v=\frac{2e}{k}$ , we get $234 \text{ edges (e)}$ . And finally, on applying $\text{ Euler's Formula: } v+e-f=2$ , we get, $\text{No. of Faces (f) }=\fbox{119}.$

Euler's formula for planar graphs states that:

$v-e+f=2$

Where v = # of vertices, e = # of edges, f = # of faces.

We know v = 117, so we just need to find e.

Since the graph is 4-regular, each vertices has 4 edges connected to it. So it can be just a 117-gon with every adjacent vertices and every other vertices connected. The number of edges is then 117*2.

Plug this back into $v-e+f=2 \Rightarrow f = e+2-v \Rightarrow f = 2*117-117+2 = 117+2 = 119$

So 119 is the answer.

We use Euler's Polyhedral Formula: V-E+F=2, where V = #vertices, E = #edges and F = #faces.

Since each vertex is of degree 4, the total number of edges is clearly (4 117)/2=2 117=234. We know V = 117. Hence F=2+E-f=2+234-117=2+117=119.

Euler's Degree Sum Formula

Since Euler's Degree Sum Formula implies that for any $k-regular$ graph with $V$ vertices then there are $\frac{k*V}{2}$ edges.

Therefore, there must be $234$ edges present.

Using Euler's Formula, we have $V-E+F=2$ where $V,E,F$ are the number of vertices, edges and faces respectively.

This gives $F=117+2 = 119$ so there are $119$ faces.

It's an application of Euler's formula for graphs $V - E + F = 2$ where $V$ is the number of vertices, $E$ the number of edges and $F$ the number of faces. We know $V=117$ and $E=\frac{4\times 117}{2}=2\times 117$ . Substituting into the previous formula we obtain, $\boxed{F=119}$ .

$v - e + f = 2$ (vertices - edges + faces = 2) for planar graphs. Each vertex has four edges connecting to it (hence, 4-regular), so there are $2v = 234$ edges. Thus $f = 119$ .

We use a simple rule to generate $4$ -regular planar graph $G_{k}$ with $n$ vertices, where $n$ is of the form : $6 + 3*k$ .

$G_{0}$ is the the graph of the octahedron . It has $6$ vertices and $8$ faces ( we also include the infinite face ).

The outer triangle is oriented upwards.

$G_{k, k \geq 1}$ is obtained by:

• remove the $3$ outer edges of $G_{k-1}$ and obtain $G^{*}_{k-1}$
• put $G^{*}_{k-1}$ inside a triangle that has an opposite orientation than the outer triangle of $G_{k-1}$
• for every vertex of this new triangle, construct $2$ edges that will link the vertex to the $2$ nearest vertices of $G^{*}_{k-1}$

This procedure adds $3$ vertices and $3$ inner faces for every increment of $k$ . So for every vertex, we get one additional face.

It follows that $G_{117}$ has :

$f = 8 + (117-6) = 119$ faces.