Consider the "infinite chessboard" on the plane. Think of it as the lattice $X\_1:=\mathbb{Z}^2$, and also finer chessboards $X\_n$ corresponding to $\frac{1}{n}\cdot \mathbb{Z}^2$, $n\geq 1$. Given two squares (i.e. vertices) $u,v$ of $X\_n$ one can define the "knight distance" $d\_n(u,v)$ as the minimum number of moves that a "knight" (moving as in the usual game of chess) must do to get from $u$ to $v$, divided by $n$.


Now take two points $a,b$ on the plane $\mathbb{R}^2$, and define their "knight distance at step $n$" to be the minimum of $d\_n(u,v)$ for $u,v\in X\_n$ such that $d\_E(u,a)$ and $d\_E(v,b)$ are minimal (where $d\_E$ is the Euclidean distance on $\mathbb{R}^2$).


Define the "knight distance" on the plane by $d\_K(a,b):=\lim\_{n\to\infty}d\_n(a,b)$.



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> Does it define an actual distance (metric) on $\mathbb{R}^2$? Assuming it does, how does the spheres look like for this metric? Any interesting properties? Is there a self-homeomorphism of $\mathbb{R}^2$ that pullbacks $d\_K$ to $d\_E$?
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(This was a question I happened to ask myself at high school -and never thought really to answer- of which I was reminded of by just reading this MO [thread](https://mathoverflow.net/questions/76623/growing-random-trees-on-a-lattice-rightarrow-voronoi-diagrams) by Joseph O'Rourke. Provided it makes sense, it still looks like a legitimate question to me...)



What is the limit of the "knight" distance on finer and finer chessboards?
Let $(x,y)$ and $(x+2a,y+a)$ be points in space. Then clearly the distance between these two points is $a$. Therefore, the unit ball around 0 must contain the octagon with vertices $(2,1)$, $(1,2)$, $(-1,2)$ and so on.


I argue that this is all it contains. To see this, construct linear invariants showing how far you can get with $k$ knight's moves of length $1/k$. For instance, each knight's move increases $x$ by no more than $2/k$, so if $|x\_1-x\_2|>2$ then $d(x\_1,x\_2)>1$. Similarly, it increases $x+y$ by no more than 3. With the 6 other linear functionals, one can restrict the unit ball to that octagon.


This metric is a kind of taxicab metric. It is not like the Euclidean metric because there are an infinite number of geodesics of length 1 between $(0,0)$ and $(2,0)$.


So, answers: Yes, octagons, I can't think of any other than the properties of the taxicab metric, no.


