# Sylvy’s Weekly Puzzle #3, Solution (part I)

Cross-reference to puzzle page on my website

I’m really pleased at the interest the third puzzle has generated! Or maybe I just haven’t been able to stop talking about it… 🙂 Anyway, here is the first part of the solution:

The question asked us to investigate the world about which Sonic runs around in a certain special level. Let’s call that world $\mathcal{S}$. I want to describe $\mathcal{S}$ as a topological suface. If you’ve not come across this notion before, you might like to think of a few key examples:

• the (real) plane $\mathbb{R}^{2}$,
• the open unit disk $D^{2}=\{(x,y)\in\mathbb{R}^{2}\;|\;x^{2}+y^{2}<1\}$, and
• the sphere $S^{2}=\{(x,y,z)\in\mathbb{R}^{3}\;|\;x^{2}+y^{2}+z^{2}=1\}$.

Slightly more exotic examples:

• the Möbius band,
• the Torus,
• (variations on the theme…) multi-handled tori, and
• the Klein bottle.

Perhaps now isn’t the time for a complete exposition of topological surfaces (okay, I may have got part-way through drafting one…!), that may come later. For the present I just want to say a few words about the Euler characteristic (which is denoted $\chi$) of a surface, and the formula that goes with it:

$V-E+F=\chi\qquad(\star_{1})$.

The first time I ever heard about this it was in the context of the Platonic solids, as follows.

• Let’s begin by thinking of a cube. A cube has 8 corners (we’re going to call these vertices), 12 edges, and 6 faces. If $V$ denotes the number of vertices, $E$ the number of edges, and $F$ the number of faces; then we have $V-E+F=8-12+6=2$.
• Next, let’s think of a tetrahedron. In this case there are 4 vertices, 6 edges, and 4 faces; thus we have $V-E+F=4-6+4=2$.

(Can you see where this is going?)

• Now let’s think of an octahedron. Here there are 6 vertices, 12 edges, and 8 faces; we have $V-E+F=6-12+8=2$.

You can check that the same formula

$V-E+F=2\qquad(\star_{2})$

holds for the other Platonic solids – but it doesn’t stop there. What about other solids, for example the Archimedean solid the rhombicuboctahedron?

(I picked this one just because it has such a beautiful name.) The formula holds here too: we have $V=24, E=48, F=26$. Therefore $(\star_{2})$ holds for this solid:

$V-E+F=24-48+26=2$.

In fact:

The formula $(\star_{2})$ holds for all polyhedra that are topologically equivalent to the sphere $\mathcal{S}^{2}$.

Although I won’t properly define topological equivalence’ here, let me just say that – roughly speaking – two shapes $A$ and $B$ are topologically equivalent if one can be transformed into the other by continuous shrinking, stretching, bending, twisting, etc. (discontinuous transformations such as tearing or joining are not allowed). Each of the polyhedra discussed above is topologically equivalent to the sphere.

Let’s now think about the surface $\mathcal{S}$ that sonic explores. It is given to us equipped with a vertices/edges/faces structure so we can calculate $V-E+F$. On $\mathcal{S}$, each face has four edges and each vertex is the endpoint of four edges. We don’t seem to know how many faces there are, so let $n$ be the number of faces. Then there are $2n$ edges and $n$ vertices. [Why?] Thus $V-E+F=0$. This shows that $\mathcal{S}$ is not topologically equivalent to a sphere. This is a negative answer to the first part of the puzzle.

On the other hand, $\mathcal{S}$ could be a torus, as shown by the following map for one of these levels:

[Map of a Special Stage in Sonic 3, copyright Sega (fair use of image for educational purposes)]

How does this show that $\mathcal{S}$ could be a Torus? You can see quite easily that the programmers have joined the left edge to the right edge and joined the top edge to the bottom edge. Thus, if Sonic walks off the left edge of the map, he will just pass round to the right edge (without noticing!).  This shape is quite easily seen to be a Torus. This is a positive answer to the second part of the puzzle.

In the third part, I asked about whether $\mathcal{S}$ could be a Klein bottle, and I’ll write a second post with the answer.

[Edit: perhaps it’s unfair to use the map of the level that I found online to argue that $\mathcal{S}$ might be a Torus. In fact the map (plus the description of how the edges are obviously’ joined together) shows that $\mathcal{S}$ definitely is a Torus. If you didn’t have the map of the level then I think it’s reasonable to simply argue that a Torus can admit a vertex/edges/faces structure as seen in the level (i.e. with `square’ faces joined in fours at vertices). We’ll address the question of whether $\mathcal{S}$ must be a Torus in the second part of this answer, when I’ll also discuss the Klein bottle.]