<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!
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.  

---

