2.9. Network Flow

What is network flow?

The network flow problem can be phrased in several ways, the canonical one being: Given a computer network with point to point links of certain capacities, what is the greatest rate at one given machine can send data to another given machine? The problem is more complex that it first sounds because data flows can take several different paths and merge and split.

The problem can be represented in graph theory, with computers corresponding to nodes and the network links to edges. Every edge has a capacity (from the question) and a directed flow (which is part of the answer). The flows may not exceed the corresponding capacities and the inflow and outflow must be equal for every node other than the two nodes in question. The nodes in question are known as the source and the sink.

An edge whose flow equals its capacity is said to be saturated. An important point about netflow flow that is initially not easy to grasp is that because flow is directed, an edge may be saturated when considered in one direction but not the other. I normally store both capacity and flow for each direction. Suppose there is an edge from A to B. For an undirected graph, capacity[A][B] = capacity[B][A], and for a directed graph capacity[B][A] = 0. I always have flow[A][B] equal the directed flow from A to B, so flow[B][A] = -flow[A][B]. The advantage of this approach is that in either direction one is constrained by flow[X][Y] <= flow[Y][X], and saturation is equivalent to equality (so in a directed graph, a zero flow indicates saturation in the reverse direction).

Solution to network flow problem

The standard method for solving the network flow problem is known as the Ford-Fulkerson method, which is basically the following:

  1. Start with no flow along any edge.
  2. Find a path from source to sink that doesn't contain any saturated edges in the direction from source to sink.
  3. Increase the flow of every edge in this path by the same amount. The amount is the largest amount that will not cause one of the capacities to be exceeded.
  4. Go back to step 2 and repeat. When no such path exists, the flow will be maximised. The proof of this is beyond the scope of this page.

Which path?

You may have noticed that the Ford-Fulkerson method is not really an algorithm, because it does not specify how to find the path. Research has found two reasonably simple methods that run quickly in most networks and reasonably fast even in extreme cases. The first is to take the shortest path and the second is to take the path along which the flow could be increased the most. It might seem tempting to use the longest path (since this would seem to "fill up" the network quickly) but in fact there are some networks where this can take arbitrarily long, apart from being difficult to find. Note that the choice of path finding algorithm only affects the speed of the algorithm, not the amount of the final flow.

Directed or undirected?

You might be wondering whether the edges are themselves directed. The answer is: they can be either directed or undirected. It only affects the algorithm in terms of the idea of a saturated edge: an edge is saturated in a particular direction if the flow cannot be increased in that direction, so a directed edge with no flow is actually saturated in the reverse direction.

Node capacities

So far we have only considered capacities in the edges; what about limiting the amount of data that can flow through a particular node? The Ford-Fulkerson method can't be applied directly to this problem, but fortunately it is possible to think of the problem in a different way that allows it to be applied.

Firstly, all undirected edges need to be split into a pair of directed edges, one in each direction. At the end you may finish with flow in both directions, but this can be safely cancelled out.

Using the computer analogy, imagine that all the input data comes in at one point, the output data leaves at another point, and that it must travel across an internal bus between them. The capacity of the node is the capacity of this bus, so just introduce it into the network as another link. The node representing the computer is split into two nodes, one representing the input port and the other the output port, and the bus is the edge between them.

Once the other edges have been fixed up to point to the correct nodes based on their direction, we have an equivalent problem but with node capacities eliminated, which means that we can apply Ford-Fulkerson as usual.

Minimum Cut

The network flow problem gives you another problem for free, known as the minimum cut problem. This problem asks for the minimum total weight of edges that must be removed to disconnect the source from the sink (so that there is no longer a path from the source to the sink). Surprising, the answer is that the total weight of the minimum cut (or mincut) is the same as the maximum network flow.

To find the actual edges, first construct the flow graph using Ford-Fulkerson. Once this is done, there will be a set of nodes S that can be reached from the source by travelling only along edges that are not saturated in the direction of travel. Choosing all edges that connect nodes in S to nodes outside S forms a minimum cut.

The network flow from S to the rest of the graph (call it R) will in fact be the total network flow, since the sink lies in R. However the flow from S to R is equal to the flow in the removed edges, each of which was saturated, so the weight of the mincut will equal the total network flow.

To see this is in fact minimal, consider applying the above argument to some other cut. The difference is that the edges that are removed are not necessarily saturated, so the weight of the cut may be greater than the total flow.


There are many problems that can be solved with network flow that do not look like network flow. One example comes from a USA Computer Olympiad from several years ago: summarised, what is the smallest number of nodes that must fail in a network to possibly prevent two given computers to be able to communicate (the given nodes may not fail)?

The solution is based on the weighted nodes transformation. Remember that this turns every original node into an edge. Give each such edge a weight of 1, and the other edges an infinite weight (in practice, I think a weight of 3 will suffice). Find the minimum cut, which will consist only of these nodal edges, and then choose these corresponding nodes as the ones that must fail.

Network flow also arises in certain limited cases. One such case is that of finding the maximal way to match up two sets. For example, given some boys and some girls at a dance and a list of pairs that are willing to dance with each other, find how many pairs can be matched up (where a person can only be matched up with one person and of the opposite sex).

This doesn't look like network flow because there are no distinguished nodes to use as the source and sink. However one can create source and sink nodes that do not correspond to any attribute of the problem. Represent each person by a node, and if a boy is willing to dance with a girl then place a directed from the boy to the girl of weight 1. Create a new node as the source and connect it to each boy with weight 1, and connect each girl to a sink with weight 1. Now the number of matchings will equal the network flow, and the matchings themselves correspond to the boy-girl edges that are saturated.

Notice that although the network flow problem would allow non-integer flows (such as a boy dancing half with one girl and half with another), an examination of the Ford-Fulkerson algorithm shows that as long as all weights are integers, the flows will also be integers.

Unfortunately this algorithm cannot be easily extended to either the problem of weighted matching (such as a degree of willingness) or the problem where the objects to be matched do not fall into two distinct sets. Both can be solved in polynomial time, but the algorithms are more sophisticated.

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Last updated Sun Nov 28 22:04:38.0000000000 2004. Copyright Bruce Merry (bmerry '@' gmail dot. com).