The problem:
Three scorpions are chasing a single ant on the edgegraph of a cube. The scorpions have the same speed ($v$), while the ant is $3$ times as fast ($3v$). They can move in any direction and instantly turn around. Additionally, the scorpions are dumb, meaning that they always go after the ant on the shortest possible path.
The scoprions and the ant can be thought of as a point, they have no volume.
And the edges of the cube can be thought of as a planar graph, with all sidelengths being $1$.
Question: Is there a strategy the ant can use to forever avoid the scorpions? In what positions can the ant accomplish that?
My attempt:
There is definitely a position where the scorpions win, and that's when they can corner the ant:
Also, if the scoprions are $1+ \frac{1}{3}$ edge away from the corner where the ant sits, they can still get into the position drawn above to catch the ant.
Otherwise it gets really complicated and messy. It seems though that if the ant just goes as far away as possible from the scoprions at every moment, they won't be able to catch it. Obviously this would need a mathematical proof though.
I tried to write up the distance function between scorpions and the ant, but since even if $1$ scoprion touches the ant, it's over, we have to take the $\min$ of them.
First of all, this is the area they can move in:
$$C = \{(x_1,x_2,x_3) \in \Bbb{R}^3| \\ |(i_1,i_2,i_3) \in \{1,2,3\}, x_{i_1} \in \{0,1\}, x_{i_2} \in \{0,1\}, x_{i_3} \in [0,1]\}$$
And this is the distance from the ant to the closest scoprion, which we always want to maximize as the ant:
$$d_{\text{closest scorpion}}(x \in C) = \\ = \min (d_{\text{scoprion 1}}(x,s_1),d_{\text{scoprion 2}}(x,s_2),d_{\text{scoprion 3}}(x,s_3))$$
And
$$d_{\text{scoprion i}}(x,s_i) = \text{Metropolis distance}(x,s_i)$$
where
$$\text{Metropolis distance}(x,y) = |x_1-y_1|+|x_2-y_2|+|x_3-y_3|$$
Maximizing $d_{\text{closest scorpion}}$ is tough, because neither $\min$ nor the $\text{Metropolis distance}$ can be differentiated. But if that's done, we could look at the gradient vector to help the ant decide where to go.
I'm kinda stuck at this point, so any help would be appreciated!
The distance you proposed, called the Manhattan distance or $L_{1}$ distance:
$$ d_{L_{1}}(x, y) = \sum_{i=1}^{3} |x_{i} - y_{i}| $$
is not the distance the ant needs, to walk from point $x$ to point $y$ on the cube. Distance the ant needs to walk $d_{G}(x,y)$ is the length of the shortest path between $x$ and $y$, for example if $x=(0.1, 0, 0)$ and $y= (0.2, 0, 1)$: $$ d_{G}(x,y) = 1.3 \neq 1.1 = d_{L_{1}}(x,y). $$
Ant starts on some edge $v_{1}v_{2}$ of the cube or graph $G$, let's say $\delta \geq 0$ away from the vertex $v_{1}$ and $1 - \delta$ from the vertex $v_{2}$. Ant can wait for the scorpions, or it can move to one of the four edges nearby. You can use the line graph $L(G)$ of the graph $G$ drawn in red in the Figure below to derive a state space:
Each red vertex of $L(G)$ is an edge in $G$ and has 4 neighbors. If $d_{G}(S, v_{1}) \leq \frac{\delta}{3}$, for some scorpion $S$, then ant can not cross $v_{1}$ and these two edges (incident to $v_{1}$) are not accessible to the ant and corresponding two vertices are not present in the state space. Similar for vertex $v_{2}$ if $d_{G}(S, v_{2}) \leq \frac{1 - \delta}{3}$.
To show your intuition:
is true, you could try do case analysis of how state space depends on the initial positions. Positions that make state space a tree, means that scorpions win. Otherwise state space has cycles and ant can just walk endlessly in a cycle. (Line graph of a cycle is a cycle, so a cycle in state space corresponds to a cycle on the cube.)