Andy's Math/CS page

Wednesday, September 28, 2011

A geometric-graphs offering

Introduction

After a night of board games, I found myself thinking about the peculiarities of movement on discrete versions of the plane. This suggested a number of questions. As they would likely suffer neglect at my hands, I'm posting them here for others to enjoy---any ideas or references are very welcome.

The basic structure that got me thinking was the 8-neighborhood (or Moore) graph:


This graph (which we'll denote by $G_{8N}$) describes how a chess king moves on an infinite chessboard. It's often convenient for game designers, but there's something... wrong about it: it distorts distances in the plane.

To make this formal, let $G$ be a (finite or countably infinite) undirected graph. Let $d_G(u, v)$ denote the shortest-path distance in $G$ between vertices $u, v$.

We say $F: V(G) \rightarrow \mathbf{R}^2$ is an embedding of $G$ if $|| F(u) - F(v)||_2 \geq d_G(u, v)$ for all distinct vertices $u, v$. (This is just a normalization convention.) Define the distortion of $F$ as the maximum (supremum) of
\[ \frac{|| F(u) - F(v) ||_2}{d_G(u, v)} , \quad{} u \neq v . \]

The study of low-distortion embeddings (which can be pursued in a more general setting) has been a highly-active TCS research topic, largely due to its role in designing efficient approximation algorithms for NP-hard problems. My initial focus here will be on embeddings for periodic and highly-symmetric graphs like $G_{8N}$.

As an example, look at the usual embedding of the 8-neighborhood graph into the plane. This has a distortion of $\sqrt{2}$, witnessed by points along a diagonal.

Warm-up: Show that $\sqrt{2}$ is the minimum distortion of any embedding of the 8-neighborhood graph $G_{8N}$.

Symmetries and Distortion

Now a basic observation here is that, when we started with a graph with a high degree of inherent symmetry, we found that its optimal (distortion-minimizing) embedding was also highly-symmetric. I would like to ask whether this is always the case.

For background, an automorphism of $G$ is a bijective mapping $\phi$ from $V(G)$ to itself, such that $(u, v) \in E(G) \Leftrightarrow (\phi(u), \phi(v)) \in E(G)$.
Let's say that a graph $G$ has 2D symmetry if there's an embedding $F$ of $V(G)$ into the plane, and linearly independent vectors $\mathbf{p}, \mathbf{q} \in \mathbf{R}^2$, such that a translation of the plane by $\mathbf{p}$ or by $\mathbf{q}$ induces an automorphism of $G$ (in the obvious way). In this case we also say the embedding $F$ has 2D symmetry.

So for example, with the usual embedding of $G_{8N}$, we can take $\mathbf{p} = (1, 0), \mathbf{q} = (0, 1)$.

Question 1: Suppose $G$ has 2D symmetry. Does this imply that there is a distortion-minimizing embedding of $G$ with 2D symmetry?

Say that $G$ is transitive if "all points look the same:" there's an automorphism of $G$ mapping any vertex $u$ to any other desired vertex $v$. Similarly, say that an embedding $F$ of $G$ is transitive, if translating the plane by any vector of form $(F(u) - F(v))$ induces an automorphism of $G$. (The usual embedding of $G_{8N}$ is transitive.)

Question 2: Suppose $G$ has 2D symmetry and is transitive. Is there a distortion-minimizing, transitive embedding of $G$ with 2D symmetry?

Question 3: Suppose $G$ has 2D symmetry, and is presented to us (in the natural way, by a finite description of a repeating "cell"). What is the complexity of determining the minimum distortion of any embedding of $G$? What about the case where $G$ is also transitive?

It seems clear that the answers to Questions 1 and 2 are highly relevant to Question 3.

Graph Metrics and Movement

I want to shift focus to another type of question suggested by $G_{8N}$. Let's back up a bit, and think about the familiar 4-neighborhood graph $G_{4N}$. It's not hard to see that the minimum-distortion embedding of $G_{4N}$ also has distortion $\sqrt{2}$. (You have to blow up the grid by a $\sqrt{2}$ factor to make the diagonals long enough.) Yet $G_{4N}$ seems considerably more natural as a discrete representation of movement in the plane somehow. Why?

I think the answer is that, with the usual embedding of $G_{4N}$, the graph distances $d_G(u, v)$ correspond to actual Euclidean travel-distances, under the restricted form of paths in which we confine ourselves to the line segments between vertices. (You can see why this metric is sometimes called "taxicab geometry.") By contrast, the usual embedding of $G_{8N}$ doesn't have this interpretation.

However, consider the following system of paths connecting points in the plane:


If we restrict ourselves to these paths, and if we make those squiggles the right length, then shortest Euclidean travel-distances actually do correspond to distances in the graph $G_{8N}$! This is so, even if we're allowed to switch paths at the crossover points.
So $G_{8N}$ is not totally weird as a discrete model of movement in the plane; it just corresponds to an odder restriction of movement.

More generally, say that a graph $G$, with nonnegative edge-weights ("lengths"), is an obstructed-plane graph, if there is an embedding of $G$ into $\mathbf{R}^2$ along with a set of "obstructions" (just a point-set in $\mathbf{R}^2$), such that shortest paths in $G$ correspond to shortest obstruction-avoiding paths in $\mathbf{R}^2$.

Question 4: What is the complexity of deciding whether a given graph (finite, say) is an obstructed-plane graph?

It simplifies things a bit to realize that, in trying to find an obstructed-plane realization of a graph $G$, the obstructions may as well be all of the plane except the intended shortest paths between all pairs of points. Using this observation, we can at least show that our problem is in NP. Is it NP-complete?

Any planar graph, with arbitrary nonnegative edge-weights, is clearly an obstructed-plane graph. But we've seen that $G_{8N}$, a non-planar graph, is also an obstructed-plane graph. (Quick---prove that $G_{8N}$ is non-planar!) The essence of the problem is to find systems of paths in the plane which, though they may cross, do not introduce any undesired "short-cuts" between vertices.

Now suppose we draw $G$ in the plane, along with a collection of "intended" shortest paths between each vertices. (That is, we will obstruct the rest of the plane, and hope that these paths are indeed shortest in what remains.) We expect that the intended $u$-$v$ path is of Euclidean length $d_G(u, v)$.

A simple observation is that in order to avoid short-cuts, all 4-tuples of distinct vertices $u, u', v, v'$ must obey the following property:

$\bullet$ If the "intended path" from $u$ to $v$ intersects the intended path from $u'$ to $v'$, then
\[ d_G(u, v) + d_G(u', v') \geq d_G(u, u') + d_G(v, v') \]
and
\[ d_G(u, v) + d_G(u', v') \geq d_G(u, v') + d_G(u', v) . \]

Question 5: Is the necessary condition above also sufficient?

In a narrow sense, the answer to Question 5 is No: it's possible to draw a graph in this way and still introduce undesired short-cuts. My real question is whether, from a graph drawing with the property above, we can lengthen and contract the lengths of segments, without changing the topological structure of the drawing, in order to get the desired obstructed-plane realization.

It may be foolish to hope for such a simple condition to be sufficient. Also, an affirmative answer to Question 5 wouldn't seem to imply any new complexity upper bound for our problem (except perhaps to speed up the NP verification a bit). I ask only because I find the question interesting, and wasn't able to cook up any counterexamples in my brief attempt.

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4 Comments:

  • if you made the appropriate 'hills' on the game board, couldn't you get rid of the squiggly bits and still make it work out? might make an interesting looking board.

    By Anonymous pat, at 3:07 PM  

  • Yes, hills would also work! Good point.

    Another interpretation of the 8-neighborhood metric: imagine you're a pious member of a religion that has four gods, corresponding to the four cardinal directions. Then whenever you're traveling N, S, E, or W, you have to make the appropriate displays of respect, and this slows you down relative to traveling NW, SW, etc.

    By Blogger Andy D, at 5:17 PM  

  • nice to see you back on the blogging beat!

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