The Euler Number
There is another approach to identify the configuration space of Narms machine, this approach uses less construction techniques and relies on a certain property of surfaces: the "Euler Number". First, lets look at another property of surfaces. Below are 2 pictures of surfaces, how can we know whether these surfaces are topologically distinct?
The answer is simple, the Sphere consists only of one connected component while the second image contains 2 such components, Since Topologically equivalent surfaces must contain the same number of connected components the surfaces are topologically distinct. By counting the number of connected components of each surface, it is possible to distinguish (topologically) between various surfaces. The advantage of this property is that it is very easily computed, while the disadvantage is that it does not distinguish between many different nonequivalent classes of surfaces. Below are examples of 3 surfaces, each consisting of exactly one connected component, this surfaces are however, topologically distinct as we will conclude later on.
Sphere  Oriented Surface Of Genus 2  Torus 

The Euler Number of a surface is an integer with this properties:
How is the Euler Number defined for the Torus? We divide the torus into regions each equivalent to a disk. Beside the regions which we call faces, we can notice edges which are the borders between 2 neighboring faces and vertexes which are the meeting points of borders of 3 or more faces. We can now compute the Euler Number, denoted by c, for this division by EN=FE+V (F=number of faces, E=number of edges, V=number of vertexes). In this case F=4, E=8, V=4, c=0. It turns out that no matter how we divide the torus into regions the formula FE+V will yield the same result: c=0 
The flat representation of the torus was introduced in the introductory lesson , by identifying opposite sides. The 4 green vertexes represent a unique vertex on the torus. In a similar fashion, other vertexes and edges of equal color represent a unique vertex or edge. 
Invariance of the Euler Number for surfaces:
Starting from 2 divisions of a surface we can always proceed to a new refinement of these 2 divisions, by combining them:  


Division 1  Division 2  refinement of the 2 divisions 

Therefore, in order to prove that the results of computing the Euler Number for different divisions yield equal results, it is suffice to prove this for the case in which one division is a refinement of the other. By sequentially adding edges to the original division it is always possible to achieve the given refinement in a finite number of steps. We will prove that the Euler Numbers computed on these divisions are equal simply by showing equality is preserved in every step:
Additions: 1 Face 3 Edges 2 Points 
Additions: 1 Face 2 Edges 1 Point 
Additions: 1 Face 3 Edges 2 Point 

13+2 = 0 
12+1 = 0 
12+1 = 0 
Therefore the Euler Number is a "Topological invariant" and we can assign to each surface its Euler Number simply by calculating it once on one of its possible divisions.
In order to compute the Euler Number c, of the Sphere we divide the sphere into 3 stripes: F = 3, E =
3, V = 2. We compute: c=33+2=2 

A dissection of the Sphere 
The fact that the Euler Numbers of the Torus and the Sphere are not equal shows that these objects are not topologically equivalent.
We will not go into computing the Euler Number for every surface. An straightforward computation shows that the Euler Number for an oriented surface of genus g is c_{g} = 22g. Thus the Euler Number differs between oriented surfaces.
Suppose we know that the configuration space of the Narms machine is an oriented surface we can find out the configuration space if we could compute the Euler Number for this surface. Since we have shown that the configuration space of the Narms machine is constructed from polygons, these polygons are a division of the configuration space. It would be natural to try to compute the Euler Number simply by computing it for this division. Let us start by computing the Euler Number for the configuration space of the 3Arms Machine. In this case, the configuration space is constructed from triangles, we have previously noticed that each triangle corresponds to a possible bending of the 3 arms. Therefore this construction of the configuration space is made of 2^{3} = 8 such triangles, or F=8. Since each edge of a triangle is common to exactly two triangles:
E =  8·3 2 
= 12 
Each vertex is common to 4 triangles, this triangles consists of all 4 possible bending of two arms and a constant bending of the third arm. We have proved this in the previous page for N>= 4 but the same arguments carry out to the case N = 3 and thus:
V =  8·3 4 
= 6 
Then we compute c=812+6=2, therefore the genus of the surface representing the configuration space is 0 and we have verified that it is a Sphere. By a similar computation, we can find the Euler Number for all Narms Machines. Again, the configuration space is constructed from F=2^{N} polygons, each consist of N edges. Since each edge is common to exactly 2 polygons:
E =  2^{N}N 2 
= 2^{(N1)}N 
Every vertex is common to exactly 4 polygons:
V =  2^{N}N 4 
= 2^{(N2)}N 
The formula for the Euler Number:
c_{N} = 2^{(N2)}(4N)
By the formula: c_{N} = 22g we verify that g = 12^{(N3)}(4N) .
In order to complete the proof we still have to clear two points:
The configuration spaces of Narms machines are surfaces:
Since the configuration space is made by gluing polygons by there common edges it is clear that states contained inside each polygon (these states represent position in which no arm is stretched) has a 2dimensional local property. We have already shown how the polygons are glued (below is an example for N = 4), from this construction you it is clear that the local properties of points located on edges and vertexes is 2dimensional.
The configuration spaces of Narms machines are oriented surfaces:
This paragraph assumes a previous knowledge on "orientability" .
For clarity we shall first examine the 4arms machine's configuration space.
Although we have harnessed our extensive knowledge of the configuration space from the previous pages, in the demonstration, the proof of orientability lies on all of this knowledge. It relies only on the fact that adjacent squares (or polygons in the general case) are mirror images of one another. The definition for the orientation of a given state, we have introduced carries over to the general case. By similar arguments, it follows that this is defines a continuous orientation of the configuration space.
In this lesson we have queried a special family of machines, in the next lesson we will introduce a new family of machines. The investigation of these machines will lead us to a much more general result then we have come up with so far.