I've casually proved, as application of some ideas that I am developing, a result that might be of interest in itself. I am completely new in this field and then I would like to ask your help to understand: 1) might it be of interest? 2) Is it trivial, in the sense that it can be proved directly? 3) is it well-known?


Let me fix a *planar* and *regular* setting, even if I could state the result in more general settings: let $n\geq1$ be a fixed integer and $P=[-n,n]^2\subseteq\mathbb Z^2$. Let me fix the following notation: given $(x,y)\in P$, I will denote by $A(x,y)$ the set formed by the following *at most* five points: $(x-1,y),(x,y),(x+1,y),(x,y-1),(x,y+1)$, where *at most* means that if one of those points does not belong to $P$, then I will not consider it. 


Now, the situation is the following: for any point $p\in P$, let $\gamma\_p$ a walk starting on $p$ and ending on $p^+$. I suppose that: if a walk hits the boundary, then it ends. In particular, if $p\in\partial P$, then $p^+=p$.


**Definition:** A flow is a family of walks $\gamma\_p$, one for each $p\in P$, such that: for all $p\in P$, whenever $q\in A(p)$, then $q^+\in A(p^+)$.


My result would be: Given a flow of walks, there is at least one walk which does not hit the boundary.


I was thinking that it might be useful to prove that some walks are bounded, but I repeat that I am really new in this field.


Every comment is welcome and also references are appreciated.


Thanks in advance,


Valerio



Walks that cannot hit the boundary
In fact, you can prove that $p^+=p$ for all $p$.


Actually, consider the shortest paths from $(-n,-n)$ to $(n,n)$ and from $(-n,n)$ to $(n,-n)$ passing through $p$. Taking `pluses' of them, you should also obtain the paths of the same lengths connecting the same points. Now from the first path you obtain that the sum of coordinates of $p^+$ is the same as for $p$, and from the second path you see that the difference of coordinates is the same for $p$ and $p^+$. Hence $p=p^+$.


