<p>Let <code>g</code> be the number of vertices of valence 4. A closed sphere has <code>G=0</code> and we have the equation <code>V-E+F=2</code>. Now <code>V=x1+g+x2</code> and <code>F=275</code>. </p>

<p>As a first step let us count the edges. If we multiply each face by its valence and sum those up, we get <em>twice</em> the number of edges (as each edge is shared by two faces), so:</p>

<pre><code>(1)   2*E = (274*4 + x3*1)
</code></pre>

<p>Immediately we can see that <code>2</code> divides <code>x3</code>.</p>

<p>Similarly if we multiply each vertex with its valence we get again <em>twice</em> the number of edges, as each edge is shared by two vertices, so:</p>

<pre><code>(2)   2*E = 3*x1 + 4*g + 5*x2
</code></pre>

<p>And again we see that <code>x1</code> and <code>x2</code> must have the same parity.</p>

<p>Plugging the first of this equation (with the very first ones) into <code>V-E+F=2</code> we get</p>

<pre><code>(3)   2 = V-E+F = x1 + g + x2 - (1096+x3)/2 + 275 = x1 + x2 - x3/2 + g + 548
</code></pre>

<p>Equating the two equations we derived above we get </p>

<pre><code>(4)   1096 + x3 = 3*x1 + 4*g + 5*x2
</code></pre>

<p>Now the system of equations <code>(3)&amp;(4)</code> has all the information encoded (excluding the parity information), so it appears you don't already have everything you need to know to determine these four unknowns.</p>

<p>I made a small Haskell program that does some bruteforcing with these conditions: <a href="https://tio.run/##dZDLCoMwEEX3@YpZdOEjsZrUQqHd9BdciqDUV9BG8VGy6L/biSspOhAYZs69E26djU3RtsvCGGhx5gAP0AG4oLF1ocIXhj7BbeDfrmYukBDOylycde9oTua@L4aoGLJX/exmlSMU@FikxC62dEA1p1rQyoav8biz2Kfc8/51CQUCO4X3jMTzLC3SXH5SbjP82BFeHdD73hzhDWmOMePB4Fi0iWMnj4QQFXXtPMlOjSYL8s6kwqYfpJrgBFPWFLBBoFyWHw" rel="nofollow noreferrer" title="Haskell – Try It Online">Try it online!</a></p>

<p>The first solution it does get is <code>(0,732,2564,0) = (x1,x2,x3,g)</code></p>

CC BY-SA 3.0
2018-02-03T07:53:48.710Z
Looks good! However, I am wondering if we can use the available value of valence 4 (vertices = 'g' and faces = '274'), to encode equation 3 & 4. Also, for [x3] we have only 1 face.
Arushi Agarwal
CC BY-SA 3.0
2018-02-03T09:30:44.800Z
Ah right I forgot that g is given in the end. Assumin g is known we have two equations in three variables.  We already included that the valence of g is 4 in equation (2) and the single face with valece x3 in equation (1). I still think one more piece of information is missing.
flawr
CC BY-SA 3.0
2018-02-03T09:59:19.910Z
Ahh, I think No, we have number of edges given in question, let's suppose "q" (see 3rd sentence of question). Still I am unsure how to get final value...
By the way, If 2*number of edges = Faces, then for valence 4 we have 137 (274/2) edges.
Arushi Agarwal
CC BY-SA 3.0
2018-02-03T10:13:27.993Z
Also, I think given that object is sphere and x3 valence has 1 face only, then x3 should be equal to 2, because valence is nothing but number of neighbors (and for 1 face it can have only 2).
Arushi Agarwal
CC BY-SA 3.0
2018-02-03T11:20:20.060Z
A face with only two neighbours isn't really a face, a face needs at least 3 neighbours. And the information that we have a sphere tells us nothing but the genus.
flawr
CC BY-SA 3.0
2018-02-02T23:07:07.800Z
2018-02-02T23:20:08.703Z
2018-02-02T23:20:08.703Z
flawr
<p>Firstly, I would like to tell this is my first question on stackexchange and I am a senior school student. Please bear with me, if I am not able to express myself clearly.</p>

<p>I am trying to solve a question based on euler's formula: V - E + F = 2 (1 - G)</p>

<p>I have been given a closed mesh of a sphere. Number of Edges and Faces and number of Vertices with valence 4 are given.
<a href="https://i.sstatic.net/BEW7L.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/BEW7L.png" alt="Valence with Vertices and Faces"></a></p>

<p>Now, I have to compute V, x_1, x_2 and x_3. The vertices for valence 4 are some how not printed in my book and I assume them to be 137 (274/2) relating to table on the right. I know F = 2 V, but 275 = 2V gives a decimal point number which is not possible !</p>

<p>Pardon for my brevity and correct me if I am not clear :)</p>

CC BY-SA 3.0
2018-02-02T20:44:30.863Z
What is the *valence* of a vertex, edge or face?
flawr
CC BY-SA 3.0
2018-02-02T22:22:00.857Z
The valence is the number of neighbors of a vertex (= number of edges joining in that vertex) or face (= number of edges around it).
Arushi Agarwal
<mesh><polygon>
Topology: Polygonal Meshes, Euler's Formula
Arushi Agarwal
# Topology: Polygonal Meshes, Euler's Formula

(Asked by: username_0 on 2018-02-02)

Firstly, I would like to tell this is my first question on stackexchange and I am a senior school student. Please bear with me, if I am not able to express myself clearly.

I am trying to solve a question based on euler's formula: V - E + F = 2 (1 - G)

I have been given a closed mesh of a sphere. Number of Edges and Faces and number of Vertices with valence 4 are given.
[![Valence with Vertices and Faces](https://i.sstatic.net/BEW7L.png)](https://i.sstatic.net/BEW7L.png)

Now, I have to compute V, x\_1, x\_2 and x\_3. The vertices for valence 4 are some how not printed in my book and I assume them to be 137 (274/2) relating to table on the right. I know F = 2 V, but 275 = 2V gives a decimal point number which is not possible !

Pardon for my brevity and correct me if I am not clear :)

**Tags:** mesh, polygon

> **Comments:**
> **username_1**: What is the *valence* of a vertex, edge or face?  
> **username_0**: The valence is the number of neighbors of a vertex (= number of edges joining in that vertex) or face (= number of edges around it).  

---

## 1 answers

### Answer by username_1 on 2018-02-02. Score: 0 | Accepted answer

Let `g` be the number of vertices of valence 4. A closed sphere has `G=0` and we have the equation `V-E+F=2`. Now `V=x1+g+x2` and `F=275`.

As a first step let us count the edges. If we multiply each face by its valence and sum those up, we get *twice* the number of edges (as each edge is shared by two faces), so:

```
(1)   2*E = (274*4 + x3*1)

```

Immediately we can see that `2` divides `x3`.

Similarly if we multiply each vertex with its valence we get again *twice* the number of edges, as each edge is shared by two vertices, so:

```
(2)   2*E = 3*x1 + 4*g + 5*x2

```

And again we see that `x1` and `x2` must have the same parity.

Plugging the first of this equation (with the very first ones) into `V-E+F=2` we get

```
(3)   2 = V-E+F = x1 + g + x2 - (1096+x3)/2 + 275 = x1 + x2 - x3/2 + g + 548

```

Equating the two equations we derived above we get

```
(4)   1096 + x3 = 3*x1 + 4*g + 5*x2

```

Now the system of equations `(3)&(4)` has all the information encoded (excluding the parity information), so it appears you don't already have everything you need to know to determine these four unknowns.

I made a small Haskell program that does some bruteforcing with these conditions: [Try it online!](https://tio.run/##dZDLCoMwEEX3@YpZdOEjsZrUQqHd9BdciqDUV9BG8VGy6L/biSspOhAYZs69E26djU3RtsvCGGhx5gAP0AG4oLF1ocIXhj7BbeDfrmYukBDOylycde9oTua@L4aoGLJX/exmlSMU@FikxC62dEA1p1rQyoav8biz2Kfc8/51CQUCO4X3jMTzLC3SXH5SbjP82BFeHdD73hzhDWmOMePB4Fi0iWMnj4QQFXXtPMlOjSYL8s6kwqYfpJrgBFPWFLBBoFyWHw "Haskell – Try It Online")

The first solution it does get is `(0,732,2564,0) = (x1,x2,x3,g)`

> **Comments:**
> **username_0**: Looks good! However, I am wondering if we can use the available value of valence 4 (vertices = 'g' and faces = '274'), to encode equation 3 & 4. Also, for [x3] we have only 1 face.  
> **username_1**: Ah right I forgot that g is given in the end. Assumin g is known we have two equations in three variables.  We already included that the valence of g is 4 in equation (2) and the single face with valece x3 in equation (1). I still think one more piece of information is missing.  
> **username_0**: Ahh, I think No, we have number of edges given in question, let's suppose "q" (see 3rd sentence of question). Still I am unsure how to get final value... By the way, If 2*number of edges = Faces, then for valence 4 we have 137 (274/2) edges.  
> **username_0**: Also, I think given that object is sphere and x3 valence has 1 face only, then x3 should be equal to 2, because valence is nothing but number of neighbors (and for 1 face it can have only 2).  
> **username_1**: A face with only two neighbours isn't really a face, a face needs at least 3 neighbours. And the information that we have a sphere tells us nothing but the genus.  

---

