Suppose we have a tag game, with an unbounded (2D) map, where the player possesses a character in the map represented by it's (x,y) position, that can, at any frame, move 1 unit left, right, up or down, or stay still. Suppose there are other characters in the game (taggers) that possess the same moving capabilities, that spawn in a box left to the player, and the game can only end if the player gets tagged and dies (that is, if its position is equal to that of any of the taggers).
Now, if the goal is to survive as long as possible, I suppose it is possible by just moving right at each turn. The monsters will never catch me if I do this contantly, therefore allowing me to survive for as long as I like.
What I want, is to formalize this in mathematical terms somehow, but I literally don't know how to. Something that seems so clear, intuitive and logical, is also so hard to prove using that same logic, at least to me. I don't know where to start whether to prove if it is (not) possible to survive indefinitely, nor do I know to do the same to the particular case of bolting right at every instance.
If someone would like to have a crack at it, I would be very grateful.
Let $d(p-q)$ be the Manhattan distance between two position $p$ and $q$, that is: $$d(p-q)=|p_x-q_x|+|p_y-q_y|$$ The goal of the tagger is to let $p(n) = q(n)$ after some $n$ moves a.k.a. $d(p(n)-q(n))=0$. WLOG, suppose that your position is $p$. Your strategy can be described as follows: $$\begin{cases} p_x(n+1)=p_x(n) +1 \text{, if } p_x(n) \geq q_x(n)\\ p_x(n+1)=p_x(n) -1 \text{, if } p_x(n) < q_x(n) \end{cases}$$
To prove that this strategy allows you to survive indefinitely is to prove that $d(p(n),q(n))>0$ for all $n$ no matter what $q(n)$ is (assuming that $p(0) \neq q(0)$, ofc). This can be done by showing that $d(p(n)-q(n))$ doesn't decrease: $$d(p(n+1),q(n+1))=|p_x(n+1)-q_x(n+1)|+|p_y(n+1)-q_y(n+1)|$$ If the opponent decided to take a vertical move, then $q_x(n+1) = q_x(n)$ and $$ \begin{align} d(p(n+1),q(n+1))&=|p_x(n)-q_x(n)| + 1 + |p_y(n+1)-q_y(n+1)|\\ &\geq |p_x(n)-q_x(n)| + 1 + |p_y(n)-q_y(n)|-1\\ &= |p_x(n)-q_x(n)| + |p_y(n)-q_y(n)|\\ &= d(p_x(n),q_x(n)) \end{align} $$ If the opponent decided to take a horizontal move, then $q_y(n+1) = q_y(n)$ and $|p_x(n+1)-q_x(n+1)| \geq |p_x(n)-q_x(n)|$ hence $$\begin{align} d(p(n+1),q(n+1)) &\geq |p_x(n)-q_x(n)| + 1 + |p_y(n)-q_y(n)|\\ &= d(p(n),q(n)) \end{align} \\$$