Find an invariant quantity 
All the vertices, except one (say $v$) of a $12$-gon are marked $+1,$ and $v$ is marked $-1.$ At each step, we can choose $3$ adjacent vertices and change their signs.
Is it possible to have every vertex marked $+1$ except for one adjacent to $v?$

I have tried to prove that it is impossible by searching invariant, and also tried to prove that it is possible, but didn’t succeed.
 A: Hint: color the vertices red, green, and blue in a repeating pattern, so there are four vertices of each color spaced equally around the dodecagon. Let $R$ be the number of red vertices which are $+1$, similarly for $G$ and $B$. How does each move affect $R, B$ and $G$? What are $R,B,G$ initially? What would these quantities become if you succeeded?
A: My solution is a bit of a cheat. It’s a non-elementary solution (the core idea is essentially some linear algebra) shown in an elementary light.
Assume this is possible. Then, composing this with a “flip” centered at the neighbor of $v$, we find a configuration where just one vertex sign was changed (the other neighbor of said neighbor).
So the question is instead – can we, with the allowed operations, reverse a single sign?
Were this possible, then we could reach any configuration from any starting configuration by applying iteratively authorized moves.
Now, there are $2^{12}$ possible configurations, so this means that there are exactly $2^{12}$ possible sequences of moves doing different things each.
Given that all the “elementary moves” commute and are involutions, any sequence of moves produces the same result as some $f_{i_1}\ldots f_{i_p}$ where $1 \leq i_1 < \ldots < i_p \leq 12$ and $f_k$ is the move flipping the sign of vertex $k$ and its neighbors. Such sequences are called the R-sequences.
Again, there must be $2^{12}$ $R$-sequences producing different results, and there are exactly $2^{12}$ $R$-sequences. So two different $R$-sequences must produce different results.
However, the distinct $R$-sequences $f_1f_2f_4f_5f_7f_9f_{10}f_{11}$ and the empty sequence produce the same result and we get a contradiction.
