Dodecahedron faces are buttons, vertices have counters that track the use of the buttons As a follow up to this question, I'm trying to teach invariants by creating a game. The idea is to start with a dodecahedron where each of the 20 vertices has a counter on it and each of the 12 faces is a button.
You can click a button to increase the value of each counter on that face by 1, or right-click a button to decrease the value of each counter on that face by 1. Counters are all initially set to 0.
The challenge is to describe all possible sets of values that the counters can take on. Clearly the sum of the counters must be a multiple of 5, but i don't think that this necessary condition is sufficient as there are less faces than vertices.
It's easy enough to create a 20x12 matrix that shows each vertex counter's value in terms of the 12 faces. For a given set of counter values, I think we could create a 20x13 augmented matrix and row reduce to see if there's a combination of buttons that could be pressed to achieve that set - of course we'd need integer value solutions - but I'd like a more general result. Would a 13th column made of constants help any?
A linear algebra solution would be great, but is there a more clever (elementary) way to do this that doesn't involve linear algebra?
Thanks for your help!
 A: Here is a rather involved, but completely elementary, derivation of necessary and sufficient conditions for the possible vertex configurations.
One can inscribe a tetrahedron in a dodecahedron, as shown:
                                       
The animation above shows 5 of these tetrahedra, but there are actually 10 (at each vertex, you can rotate one of the tetrahedra about that vertex to get a second one). Note that every pentagon adds or subtracts $1$ to the sum of the counts on each of these tetrahedra, so we know that the sum of the counts on the tetrahedra are all equal to one another. This lets us rule out many configurations, and shows that if any vertices are nonzero, there must be at least $3$ with the same sign.
One can also combine two of the above tetrahedra to inscribe a cube in the dodecahedron:
                                                                 
The cube doesn't tell us anything the tetrahedra don't, but if we look at its complement $C$, we see that every face has exactly three vertices in $C$, so the vertex sum of $C$ is a multiple of $3$ and is equal to the sum on every congruent copy of $C$. From now on, let's call this a cross, since it's formed from three mutually orthogonal rectangles.
Suppose that we know the vertex counts on a given cross. What can we infer? Well, we know there is some constant sum across any of the tetrahedra; call this sum $k$. Of these tetrahedra, two are in the cube, and eight have three corners in the cross and one in the cube. For each of these eight tetrahedra, we can infer that the value at that corner of the cube is $k$ minus the sum at the other three corners. But then when we consider a tetrahedron within the cube, we know that the sum of its vertices is
$$k-(\text{sum of 3 vertices})+k-(\text{sum of 3 vertices})+k-(\text{sum of 3 vertices})+k-(\text{sum of 3 vertices}) = k$$
But, as it turns out, these four triples of vertices have as their union the entire cross, so we just have $3k=\sum_{v\in\text{cross}}c_v$, where $c_v$is the count at vertex $v$. Since we know this sum is a multiple of three, we will be able to solve for $k$, and thus obtain consistent values across the whole cube.
So this means that a given configuration on the cross determines a unique arrangement on the rest of the dodecahedron; it remains only to find the valid configurations on the cross.
Consider the three intersecting rectangles forming the cross. Label them $A,B,C$. If we do this labeling in the right order, then we can observe that every pentagon either has one vertex in $A$ and two in $B$, one in $B$ and two in $C$, or one in $C$ and two in $A$. Let $s_A, s_B, s_C$ be the sums on each of these rectangles. Then the sum $s_A-2s_B+4s_C$ is always preserved modulo $9$ when we add a pentagon, and so must always be $0$. (Why that particular invariant? We're multiplying the coefficients by $-2$ each time so that things are balanced, so when we wrap back around to $A$ we want for $(-2)^3$ and $1$ to be congruent.)
Now, I claim that any arrangement on the cross which satisfies this mod-$9$ congruence is attainable.
First, let's see that we can get $s_A, s_B,$ and $s_C$ to be whatever we like, subject to our congruence rule. To do this, we add $(1,2,0)$ to each of $(s_A,s_B,s_C)$ until $s_A$ is where we want it, then $(0,1,2)$ until $s_B$ is as desired. This leaves $s_C$ as something congruent mod-$9$ to our target (since otherwise it wouldn't match the congruence), so we just need to be able to add or subtract $9$ from $C$. But it turns out this is possible! Add or subtract $(2,0,1)-2\cdot (1,2,0)+4\cdot(0,1,2) = (0,0,9)$ as desired.
Now, we want to shuffle counts around within each rectangle. But this is also possible! I'll leave as an exercise the fact that it's possible to obtain a $\pm1$ split on any pair of vertices in the same rectangle, without causing any net effect on the other vertices of the cross. (It suffices to show that you can transfer points from one corner of the rectangle to an adjacent one - exchanges along a long side of the rectangle are a little easier to see.)
So we now have an efficient algorithm for determining if a given configuration is possible: pick a cross, check that $s_A-2s_B+4s_C\equiv 0\pmod9$, then verify that the sums on every tetrahedron are equal to $\frac{s_A+s_B+s_C}3$.

Addendum: there's another cute invariant which I didn't end up using in this proof.
Consider "equators" of the dodecahedron: the vertices not on a given face or its polar opposite. Every face contains either $0$ or $3$ of the points on this equator, so the sum on every equator must be a multiple of $3$ (but not necessarily equal to one another). This nicely rules out configurations like having all vertices be at $+1$.
