Question

Question: Circulation problems allow you to have both a lower and an upper bound on the flow through a particular arc. The upper bound I understand (like pipes, there’s only so much stuff that can go through). However, I’m having a difficult time understanding the lower bound idea. What does it mean? Will an algorithm for solving the problem…

• try to make sure every arc with a lower bound will get at least that much flow, failing completely if it can’t find a way?
• simply disregard the arc if the lower bound can’t be met? This would make more sense to me, but would mean there could be arcs with a flow of 0 in the resulting graph, i.e.

Context: I’m trying to find a way to quickly schedule a set of events, which each have a length and a set of possible times they can be scheduled at. I’m trying to reduce this problem to a circulation problem, for which efficient algorithms exist.

I put every event in a directed graph as a node, and supply it with the amount of time slots it should fill. Then I add all the possible times as nodes as well, and finally all the time slots, like this (all arcs point to the right):

 

 

The first two events have a single possible time and a length of 1, and the last event has a length of 4 and two possible times.

Does this graph make sense? More specifically, will the amount of time slots that get ‘filled’ be 2 (only the ‘easy’ ones) or six, like in the picture?

(I’m using a push-relabel algorithm from the LEMON library if that makes any difference.)

Answer 1

Regarding the general circulation problem:

I agree with @Helen; even though it may not be as intuitive to conceive of a practical use of a lower bound, it is a constraint that must be met. I don’t believe you would be able to disregard this constraint, even when that flow is zero.

The flow = 0 case appeals to the more intuitive max flow problem (as pointed out by @KillianDS). In that case, if the flow between a pair of nodes is zero, then they cannot affect the “conservation of flow sum”:

When no lower bound is given then (assuming flows are non-negative) a zero flow cannot influence the result, because

1. It cannot introduce a violation to the constraints
2. It cannot influence the sum (because it adds a zero term).

A practical example of a minimum flow could exist because of some external constraint (an associated problem requires at least X water go through a certain pipe, as pointed out by @Helen). Lower bound constraints could also arise from an equivalent dual problem, which minimizes the flow such that certain edges have lower bound (and finds an optimum equivalent to a maximization problem with an upper bound).

For your specific problem:

It seems like you’re trying to get as many events done in a fixed set of time slots (where no two events can overlap in a time slot).

Consider the sets of time slots that could be assigned to a given event:

E1 — { 9:10 }
E2 — { 9:00 }
E3 — { 9:20, 9:30, 9:40, 9:50 }
E3 — { 9:00, 9:10, 9:20, 9:30 }

So you want to maximize the number of task assignments (i.e. events incident to edges that are turned “on”) s.t. the resulting sets are pairwise disjoint (i.e. none of the assigned time slots overlap).

I believe this is NP-Hard because if you could solve this, you could use it to solve the maximal set packing problem (i.e. maximal set packing reduces to this). Your problem can be solved with integer linear programming, but in practice these problems can also be solved very well with greedy methods / branch and bound.

For instance, in your example problem. event E1 “conflicts” with E3 and E2 conflicts with E3. If E1 is assigned (there is only one option), then there is only one remaining possible assignment of E3 (the later assignment). If this assignment is taken for E3, then there is only one remaining assignment for E2. Furthermore, disjoint subgraphs (sets of events that cannot possibly conflict over resources) can be solved separately.

If it were me, I would start with a very simple greedy solution (assign tasks with fewer possible “slots” first), and then use that as the seed for a branch and bound solver (if the greedy solution found 4 task assignments, then bound if you recursive subtree of assignments cannot exceed 3). You could even squeeze out some extra performance by creating the graph of pairwise intersections between the sets and only informing the adjacent sets when an assignment is made. You can also update your best number of assignments as you continue the branch and bound (I think this is normal), so if you get lucky early, you converge quickly.

I’ve used this same idea to find the smallest set of proteins that would explain a set of identified peptides (protein pieces), and found it to be more than enough for practical problems. It’s a very similar problem.

If you need bleeding edge performance:
When rephrased, integer linear programming can do nearly any variant of this problem that you’d like. Of course, in very bad cases it may be slow (in practice, it’s probably going to work for you, especially if your graph is not very densely connected). If it doesn’t, regular linear programming relaxations approximate the solution to the ILP and are generally quite good for this sort of problem.

Hope this helps.