manhattan to euclidean metric One may define a graph on a square lattice by taking the nodes of the lattice as graph vertices and the bonds of the lattice as edges. Suppose for simplicity that the nodes have integer $(x,y)$ coordinates with unit spacing in both dimensions. 
The Manhattan norm of the vector with ${\bf v}=(x,y)$ is $|| {\bf v }||_m:=|x|+|y|$ and the Euclidean norm is $|| {\bf v }||_e:=\sqrt{x^2+y^2}$.
Obviously, the Manhattan norm is the number of steps one has to take on the graph to join both ends of ${\bf v}$  by the shortest path. In the meantime the euclidean norm is the distance between both ends in the plane where the graph is embedded. 
There is no one-to-one relationship between both norms. But there seems to be one between  $(|| {\bf v }||_m, N_{\bf v})$ and $||{\bf v}||_e$, where $N_{\bf v}$ is the number of shortest paths on the graph to join both ends of ${\bf v}$. This is at least what my computations on finite graphs show. 
Now, my two questions.


*

*is there a proof for this ?

*does it hold with other lattices-graphs?
 A: Your question is, given the manhattan norm of a grid point, and the number of shortest grid-paths connecting the origin to that grid point, can we deduce the Euclidean norm of the grid point?
Let's work out what the number of shortest grid-paths is. Let $v=(x, y)$ be a grid-point, that is, $x,y\in \mathbb Z$. A simple combinatorial argument shows there are ${{|x|+|y|}\choose{|x|}}={{|x|+|y|}\choose{|y|}}$ possible paths from the origin to $v$ (any such path consists of $|x|+|y|$ "steps", and the only choice made for each step is whether it's horizontal or vertical).
So the question is, do the equations:
$$|x|+|y|=A$$
$${{|x|+|y|}\choose{|x|}}=B$$
determine $|x|$ and $|y|$, or at least $|x|^2+|y|^2$? Well, we have ${{A}\choose{|x|}}=B$, which, due to the symmetry of Pascal's triangle, narrows down the possibilities for $|x|$ to at most two possible values: if $a$ is one such value, $A-a$ is the other. By the first equation, the values for $|y|$ are then $A-a$ and $a$ respectively, so these two solutions are just swapping the values of $|x|$ and $|y|$. Since $|x|^2+|y|^2$ is symmetric, it will take on the same value for both solutions, so the above two equations do indeed determine the Euclidean norm of $v$.
I'm not sure exactly what generalization you're implying in your second question.
