Odd Number of Cats? A few days ago, a peer jokingly posed a puzzle. In essence, the puzzle states that there are some cats, each cat is in a corner, and each cat sees 3 other cats. The puzzle then asks how many cats there are in total. The obvious solution is that there are $4$ cats...  However, I have long thought about extensions of the problem. I have found polyhedra that allow for $2n$ "cats",  where $n$ is an integer, the "cats" are the vertices, and the edges serve as "hallways" through which cats can see each other.   
This naturally made me curious if the problem is possible with an odd number of vertices. I immediately tried to attack the problem and could not solve it... I even failed to attack a restricted form of the problem, requiring a convex polyhedron with an odd number of vertices with each vertex having 3 connected edges. An example of a failing example would be a square pyramid (as one of the vertices has $4$ edges). However, this definition actually discounts the puzzle's intended solution... If we use my "hallway" interpretation of the problem then the "hallways" between cats in opposite corners form an "X" in the center of the square, creating an additional vertex with four connected edges.  
Realizing how much harder I had made the puzzle, I decided to pursue the restricted case in two dimensions (which the intended solution lies in). Though I conjecture that an odd number of "cats" is impossible in both two and three dimensions, I thought it would be easier to show in two dimensions, especially for closed, planar graphs. However, I have come up short with a proof here as well.  
I realize that my language here is very vague... I lack a decent introduction into topology, and thus my language is incomplete. Should more details be needed I would be glad to elaborate. Nevertheless,  I think the most basic form of my question is as such: does there exist a generalization of the cat puzzle to odd numbers of cats in two dimensions? In three dimensions? I conjecture this impossible. 
 A: You can't have an odd number of cats, not because of geometry, but something more fundamental: the handshaking lemma. Basically, if two cats seeing each other is mutual, then the number of cats that see an odd number of other cats must be even.
A: This is a special case of theorem from graph theory: Given an undirected graph $G$, if $d(g)$ is the degree of each node, then $\sum_{g\in G} d(g)=2E$, where $E$ is the number of edges in the graph. 
In this case, you have $d(g)=3$ for each cat $g$, so you'd want $3|G|=2E$, which means that $|G|$, the number of cats, must be even.
Basically, if $C$ is the number of cats, then $3C$ counts the each pair of cats that can see each other twice, so $3C$ must be even, so $C$ must be even.
This also means that if every cat can see exactly an odd number of cats, then the total number of cats must be even.
A: How about if there was a Triangle of cats. Each cat is a vertex. There are mirrors between the cats, so each cat can see itself.
Graphic Demonstration
Summary: In conclusion, I state that it is possible to see three cats in a triangle of three cats if you use mirrors.
A: No-one's asked what happens if cats being able to see each other isn't mutual: if the "visibility relation" isn't symmetric. In that case, the relation corresponds to a directed graph where every node has in-degree 3. Since every edge going into a given vertex must also come out of some vertex, the sum of the out-degrees must equal the sum of the in-degrees, and if the number of cats is finite then it follows that each cat can be seen by an average of three cats. But other than that, there are no restrictions; it could be that all cats can see the same three cats (including those three cats themselves, perhaps via mirrors).
A: If we assume that "can see" is symmetric (no one-way mirrors, no cats prevented from turning their heads accordingly, etc.), then the situation describes a graph with $n$ vertices (the cats) and a number $e$ of edges, where each vertex has degree $3$. By the hand-shaking lemma, i.e., by counting the vertex-edge-incidences in two different ways. we find $3n=2e$, hence $n$ must be even.
A: Building onto Robin Saunders' idea, it is definitely possible that cat A can see cat B but cat B can't see cat A. It's just like kids playing hide and seek closing their eyes - "if you can't see them they can't see you right?" 
Since cats are not infinitesimally small, nor are they made up of just their eyes and they don't have $360^\circ$ FOV, we can very easily restrict their vision with something like an anti-scratch cone, now of course the question leaves a lot of room for discussion due to how little information there are regarding the restrictions, and how closely one would like to interpret the question to reality is another thing, but I personally really like this type of questions where there are no true answers, nor bounded by no-fun mathematical theories.
A very easy 3 way triangle would look like:
3  →  3
 ↖ ▼ ↙
   3

where '3' are 3 cats and the arrows are the direction they are looking at.
