Directed bipartite graphs with weighted edges with uniform input weight

98 Views Asked by At

Here is the set-up:

Let $G = (V(G), E(G))$ be a bipartite graph with $ V_1 \sqcup V_2$ denoting the vertices with respect to 2-colouring of the graph. Consider the induced directed graph $G' = (V(G),A(G))$ with the vertices $V(G)$ and with an arrow of each direction replacing each edge of $G$. Furthermore, we add a weighting $ wt : A(G) \rightarrow \mathbb{R}$ to each arrow such that $wt(u \rightarrow v)*wt(v \rightarrow u) = 1$ for all adjacent $u,v \in V(G)$.

Here is the question:

Does there exist a weighting function such that such that if $u, v \in V_i $ then

$$\sum_{x \in N(u)} wt(x\rightarrow u) = \sum_{x \in N(v) } wt(x\rightarrow v) $$

where $N(v)$ is the set of adjacent vertices to $v$ in $G$ and $V_i \in \{V_1,V_2\}$.

I'm very ignorant of this general area. It seems like this might be a well-known example of some discrete version of some kind of heat equation or harmonic function. Any references to anything related would be extremely useful to me! Thanks!

1

There are 1 best solutions below

0
On

The constraint on the weighting is not very easy to deal with... because it's about the product of the weights, right ? Somehow it doesn't seem natural.

Anyway : sometimes it's impossible (take $V_1 = \{a,b,c\}, V_2 = \{d,e\}, E=\{(ad), (bd), (ce)\}$ : the constraints on $V_1$ say that all weights are equal, and the constraints on $V_2$ say that they are not.), sometimes it will be easy (you have $|V|-2+|E|$ constraints and $2|E|$ variables) so you should have enough degrees of freedom in most instances of the problem).

You could transform this into an instance of semidefinite programming (your constraints are quadratic, defined with a dot product) so this can be solved "efficiently", for example with interior point methods.

Maybe there is a clever way to reformulate this into a classical problem, but I can't seem to find an idea that could make this quadratic constraint appear magically.