Relation between radius of a graph and radius of circle?

Question

I was just making a report for my presentation and suddenly this question popped in my mind. Is there any relation between the radius of a graph in graph theory and radius of a circle? I am just curious. I may be wrong too. Thanks for the help.

P.S.

By a graph $G$ we mean a pair of sets $(V,E)$, where $V$ is the set of vertices and $E$ is the set of edges formed by pair of vertices. For $u,v\in V(G)$, the length of a shortest $u$--$v$ path in $G$ is the distance $d(u,v)$ between them. The eccentricity of a vertex $v$ in $G$, denoted by $e(v)$, is the distance between $v$ and a vertex farthest from $v$. The maximum and minimum eccentricity of $G$ is known as diameter $diam(G)$ and radius $rad(G)$ of $G$, respectively.

Answer 1

There's some relationship, at least for "diameter" -- the diameter of a circle is the greatest distance between any two of its points; the diameter of a set is $\sup_{x,y \in S} d(x, y)$ in a metric space, too.

Well, a graph with the edge-length distance is an object for which taking the max distance between points makes sense, and we call that the "diameter."

As for "radius" -- not so much. For a circle, for instance, the "radius", defined as $\min_{x \in S^1} \max_{y \in S^1} d(x, y)$ turns out to always be $2$ (if you measure distance in $\Bbb R^2$, or $\pi$, if you measure distance "within the circle", i.e., along the circular curve. Neither of these bears much relation to the thing we call the radius of the circle in geometry, namely, half the diameter.

Answer 2

There is indeed a very direct connection. But it depends on being very careful about what we mean when we say 'a circle'.

The circle $S^1$, which many people immediately think of when they think of a 'circle', can be defined as the set of points in $\Bbb{R}^2$ that have distance one to the origin. I.e.: $S^1 = \{ (x,y) \in \Bbb{R}^2 \mid d[(x,y),(0,0)]=1 \}$. But notice that in this definition we make reference to a point that exists outside the circle, namely the origin $(0,0)$. And the common definition of the radius of a circle explicitly makes reference to this outside point, too. In other words, most people naturally think of an embedding of a circle, and define the radius in terms of that embedding. But topologically one can have a circle without this embedding, for instance by 'gluing' the ends of the unit interval $[0,1]$ together. In this case, without any 'outside' points to refer to, we need a new way to define the 'radius', whatever that will mean.

Usually when we consider a graph, it is not embedded in any space. So it has the same problem as the topological circle if we want to talk about its extent (radius / diameter): we need a new definition, and preferably one that closely aligns with our intuition. For the diameter it's basically immediate: the diameter of a circle is the maximum distance between any two points on the circle; this definition makes no reference to the embedding and so it is easy to transfer to the context of graphs. What about the radius?

A circle is probably not the best place to start. Instead, consider the closed unit disc $D = \{ (x,y) \in \Bbb{R}^2 \mid x^2 + y^2 \leq 1 \}$. Intuitively we know that the radius of $D$ is one. But it also happens that among all the points in $D$, one is the smallest maximum distance to all points in $D$ that can be obtained. In other words: $$ 1 = \min_{r\in D} \max_{s\in D} d(r,s) \mbox{.}$$ This definition doesn't make explicit reference to the origin, and only requires the distance between two points in $D$ to be defined, so it works nicely for graphs as well.

So what's the connection to the radius of a circle? Well, since a circle is a simple closed curve it has a well-defined 'inside' (the open unit disc). The radius of a circle can be defined as the min-max, taken over all points that are either on the circle or its inside. A nice bonus from this approach is that we can extend the definition of radius to include any simple closed curve, even ones that don't have a well-defined 'center'!