1
$\begingroup$

I am looking for the maximum taxicab distance between two points in a 2dim lattice with imposed periodic boundary conditions.

Taxicab Distance in 2 dimensions for two points $(x_1, y_1)$ and $(x_2, y_2)$ is defined as:

$|x_1 - x_2| + |y_1 - y_2| $

Let a 2 dimensional lattice be with size $L \times L$ with periodic boundaries such that: $x=L+1=1$ and $y=L+1=1$, so every point is contained within the lattice.

I guess that the maximum distance is $L$, however I am not entirely sure.

$Edit$:

I have added a photo which illustrates the two points where I think the distance between them would be $L$. (The blue point and the blue X) Points where I think distance might be $L$

$\endgroup$
4
  • $\begingroup$ If by "periodic" you mean that a point moving off the left edge re-appears on the right edge, then I would think the max distance would be $\lfloor L/2\rfloor,$ right? $\endgroup$ Commented Feb 7, 2019 at 16:58
  • $\begingroup$ Yes, that is exactly what i mean. I am thinking about a point at the edge of the square lattice and then another one in the middle of the lattice, so then the distance is $[L/2]$ in $x$ and $[L/2]$ in $y$, so $L$ in total? I am not sure if this is correct. $\endgroup$ Commented Feb 7, 2019 at 17:02
  • $\begingroup$ I have also added a photo illustrating the what I think would be the maximum distance of $L$ in such a setup. $\endgroup$ Commented Feb 7, 2019 at 17:08
  • $\begingroup$ Oh, right. Yes, I think $L$ would be the max. $\endgroup$ Commented Feb 7, 2019 at 17:37

1 Answer 1

Reset to default
0
$\begingroup$

When using periodic boundary conditions on a lattice of this sort, it is helpful to remember that the lattice can, equivalently, just be extended in any direction as far as needed. The arithmetic in a single direction would then be modulo $L,$ so that distance is measured by $$d((x_1, y_1),(x_2, y_2))=((x_2-x_1)(\!\!\!\!\!\!\mod L))+((y_2-y_1)(\!\!\!\!\!\!\mod L)). $$ Plugging in a few points in the image included in the question will convince you that this is correct.

So, taking a single direction under consideration, we see that the maximum distance obtainable in just the $x$ direction would have to be $\lfloor L/2\rfloor,$ where the floor function is necessary to handle division by an odd $L.$ The same is true in the $y$ direction, which makes the maximum distance $$d_{\max}=2\left\lfloor\frac{L}{2}\right\rfloor.$$

$\endgroup$
1
  • 1
    $\begingroup$ Thank you for the answer! $\endgroup$ Commented Feb 7, 2019 at 18:40

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.