I'm looking for an efficient algorithm (at least polynomial in the size of the graph, preferably linear) for the following problem:

Definitions: Given a graph $(V,E)$, with non-negative weights assigned to it's edges $w_E:E \rightarrow \mathbb{R}^+\cup\{0\}$ and non-negative weights assigned to it's vertices $w_V:V \rightarrow \mathbb{R}^+\cup\{0\}$ we define the flow on the edge $e=(v_1,v_2)\in E$ to be the wight of the vertex where the edge starts times the weight of the edge, $f(e=(v_1,v_2)) \equiv w_V(v_1)\cdot w_E(e)$. The inward flow to vertex $v\in V$ is defined by the sum of the flows on all inward edges $f_{in}(v) = \sum_{e\in E\;s.t.\; e=(u,v)}f(e)$ and the outward flow the sum of flows on all the outward edges $f_{out}(v) = \sum_{e\in E\;s.t.\; e=(v,u)}f(e)$.

The Problem Given a finite directed graph $(V,E)$ and an edge weights assignment $w_E$, find a non-trivial weights assignment to the vertices $w_V$ such that the inward flow and the outward flow are equal for each vertex, $\forall v\in V\; f_{in}(v) = f_{out}(v)$. The trivial solution is $w_V\equiv 0$. Notice that I'm only looking for one non-trivial solution (there could be more than one). If there is no non-trivial solution the algorithm should return some error message.

Any ideas? Thank you for your help


I will assume, that all weights are rational (since there is a problem with representing irrational numbers on computer).

You can solve this problem by using linear programming in similar manner to finding maximal flow.

$\forall_{v \in V} w_V$ is constant.

$\forall_{e \in E} w_e$ is variable.

Firstly, for each node we want to be sure that inward flow is equal to outward flow: $$\forall_{v \in V} \sum_{(v_{i}, v)\in E}w_{v_{i}}w_{(v_{i}, v)} = \sum_{(v, v_o)\in E}w_{v}w_{(v, v_o)}$$

Then we want to be sure, that every edge is non negative: $$\forall_{e \in E} w_e \geq 0$$

You can notice, that if there is any non-trivial solution then there are infinitely many, arbitrary large flows (you can multiply weight of every edge by constant $c$ and it will be still valid). So if there is any non-trivial solution, then there is a solution with sum of weights of edges not smaller then 1: $$\sum_{e \in E} w_e \geq 1$$

Now what's left is to check if there is a solution for this linear programming instance.

  • $\begingroup$ Is there a reason for why you switched the roles of $w_V$ and $w_E$? (in the problem I presented $w_V$ were variables and $w_E$ constants). Since it seems like I can use the same schema to solve the original case. $\endgroup$ Sep 26 '19 at 22:13
  • $\begingroup$ Oh my bad, but you can still solve it in the same way $\endgroup$ Sep 26 '19 at 22:23

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