Equilibrium Points of Directed Graphs

[Pauly Man]

Suppose I have a directed graph in Rⁿ. Where the graph is a hypercube, (a square in R², a cube in R³ etc).

Suppose I define an equilibrium point of a directed graph to be a vertex such that I can travel from any adjacent vertex along an edge to that vertex. What is the maximum number of equilibrium points of a directed hypercube in Rⁿ?

As an example in R²:

*****<*****
*          *  
^          ^
*          *
***** >*****

(For some reason the graph isn't formatting properly, hopefully you can imagine that it is supposed to be a square).

The upper left corner is an equilibrium point for the directed hypercube.

I now wish to work out how to find the maximum number of equilibrium points possible in a directed hypercube in Rⁿ. (Any ideas??)

 

[rutwig]

Unfortunately I could not decipher the plotted graph, but in any case you should consider the proyection of the graph on the plane and then argue, as is usually done in graph theory.

 

[damgo]

It's pretty straightforward to show that this is the same as the size of the largest set of unconnected vertices. For that, you can consider the construction procedure of "doubling" the n-dimensional hypercube and connecting each original points with its double. Then using induction you find 2^(n-1) for n dimensions.

There's probably a prettier way to do it, but graph theory isn't my thing.

 

[Pauly Man]

Originally posted by damgo
It's pretty straightforward to show that this is the same as the size of the largest set of unconnected vertices. For that, you can consider the construction procedure of "doubling" the n-dimensional hypercube and connecting each original points with its double. Then using induction you find 2^(n-1) for n dimensions.

There's probably a prettier way to do it, but graph theory isn't my thing.

Thanks damgo. I sat up last night before going to bed and thought about the problem. I came up with this argument (which I'm pleased to see results in the same conclusion you came up with).

There are n2^n-1 edges in a hypercube. If we know that we have one equilibrium point then we know the direction of n edges. So it follows that we don't know the direction of n(2^n-1-1) edges. If we know of another equilibrium point then we know that it cannot share any edges with the previous equilibrium point, and so we know the direction of another n edges. We therefore do not know the direction of n(2^n-1-2) edges.

Continuing on in this fashion we find that for a equilibrium points we have n(2^n-1-a) edges for which we are unsure of the direction.

The maximum number of equilibrium points occurs when we know the direction of every edge, which occurs when:

n(2^n-1-a) = 0
a = 2^n-1

 

[Pauly Man]

Note that I have edited the graph above, so now it hopefully makes sense).