# Network Flow I. Lecture Overview The Network Flow Problem

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1 Lecture 6 Network Flow I 6. Overview In these next two lectures we are going to talk about an important algorithmic problem called the Network Flow Problem. Network flow is important because it can be used to express a wide variety of different kinds of problems. o, by developing good algorithms for solving network flow, we immediately will get algorithms for solving many other problems as well. In Operations Research there are entire courses devoted to network flow and its variants. opics in today s lecture include: he definition of the network flow problem he basic Ford-Fulkerson algorithm he maxflow-mincut theorem he bipartite matching problem 6. he Network Flow Problem We begin with a definition of the problem. We are given a directed graph G, a start node s, and a sink node t. Each edge e in G has an associated non-negative capacity c(e), where for all non-edges it is implicitly assumed that the capacity is 0. For example, consider the graph in Figure 6. below. Figure 6.: network flow graph. 8

2 6.. HE NEWORK FLOW PROLEM 85 Our goal is to push as much flow as possible from s to t in the graph. he rules are that no edge can have flow exceeding its capacity, and for any vertex except for s and t, the flow in to the vertex must equal the flow out from the vertex. hat is, apacity constraint: On any edge e we have f(e) c(e). Flow conservation: For any vertex v {s,t}, flow in equals flow out: u f(u,v) = u f(v,u). ubject to these constraints, we want to maximize the total flow into t. For instance, imagine we want to route message traffic from the source to the sink, and the capacities tell us how much bandwidth we re allowed on each edge. E.g., in the above graph, what is the maximum flow from s to t? nswer: 5. Using capacity[flow] notation, the positive flow looks as in Figure 6.. Note that the flow can split and rejoin itself. [] [] [] [] [] [] [] Figure 6.: network flow graph with positive flow shown using capacity[flow] notation. How can you see that the above flow was really maximum? Notice, this flow saturates the a c and s b edges, and, if you remove these, you disconnect t from s. In other words, the graph has an s-t cut of size 5 (a set of edges of total capacity 5 such that if you remove them, this disconnects the source from the sink). he point is that any unit of flow going from s to t must take up at least unit of capacity in these pipes. o, we know we re optimal. We just argued that in general, the maximum s-t flow the capacity of the minimum s-t cut. n important property of flows, that we will prove as a byproduct of analyzing an algorithm for finding them, is that the maximum s-t flow is in fact equal to the capacity of the minimum s-t cut. his is called the Maxflow-Mincut heorem. In fact, the algorithm will find a flow of some value k and a cut of capacity k, which will be proofs that both are optimal! o describe the algorithm and analysis, it will help to be a bit more formal about a few of these quantities. efinition 6. n s-t cut is a set of edges whose removal disconnects t from s. Or, formally, a cut is a partition of the vertex set into two pieces and where s and t. (he edges of the cut are then all edges going from to ). efinition 6. he capacity of a cut (,) is the sum of capacities of edges in the cut. Or, in the formal viewpoint, it is the sum of capacities of all edges going from to. (on t include the edges from to.) efinition 6. It will also be mathematically convenient for any edge (u, v) to define f(v, u) = f(u,v). his is called skew-symmetry. (We will think of flowing unit on the edge from u to v as equivalently flowing units on the back-edge from v to u.)

3 6.. HE FOR-FULKERON LGORIHM 86 he skew-symmetry convention makes it especially easy to add two flows together. For instance, if we have one flow with unit on the edge (c,b) and another flow with units on the edge (b,c), then adding them edge by edge does the right thing, resulting in a net flow of unit from b to c. lso, using skew-symmetry, the total flow out of a node will always be the negative of the total flow into a node, so if we wanted we could rewrite the flow conservation condition as u f(u,v) = 0. How can we find a maximum flow and prove it is correct? Here s a very natural strategy: find a path from s to t and push as much flow on it as possible. hen look at the leftover capacities (an important issue will be how exactly we define this, but we will get to it in a minute) and repeat. ontinue until there is no longer any path with capacity left to push any additional flow on. Of course, we need to prove that this works: that we can t somehow end up at a suboptimal solution by making bad choices along the way. his approach, with the correct definition of leftover capacity, is called the Ford-Fulkerson algorithm. 6. he Ford-Fulkerson algorithm he Ford-Fulkerson algorithm is simply the following: while there exists an s t path P of positive residual capacity (defined below), push the maximum possible flow along P. y the way, these paths P are called augmenting paths, because you use them to augment the existing flow. Residual capacity is just the capacity left over given the existing flow, where we will use skewsymmetry to capture the notion that if we push f units of flow on an edge (u,v), this increases our ability to push flow on the back-edge (v,u) by f. efinition 6. Given a flow f in graph G, the residual capacity c f (u,v) is defined as c f (u,v) = c(u, v) f(u, v), where recall that by skew-symmetry we have f(v, u) = f(u, v). For example, given the flow in Figure 6., the edge (s,a) has residual capacity. he back-edge (a,s) has residual capacity, because its original capacity was 0 and we have f(a,s) =. efinition 6.5 Given a flow f in graph G, the residual graph G f is the directed graph with all edges of positive residual capacity, each one labeled by its residual capacity. Note: this may include back-edges of the original graph G. Let s do an example. onsider the graph in Figure 6. and suppose we push two units of flow on the path s b c t. We then end up with the following residual graph: Figure 6.: Residual graph resulting from pushing units of flow along the path s-b-c-t in the graph in Figure 6..

4 6.. HE FOR-FULKERON LGORIHM 87 If we continue running Ford-Fulkerson, we see that in this graph the only path we can use to augment the existing flow is the path s a c b d t. Pushing the maximum units on this path we then get the next residual graph, shown in Figure 6.. t this point there is no Figure 6.: Residual graph resulting from pushing units of flow along the path s-a-c-b-d-t in the graph in Figure 6.. longer a path from s to t so we are done. We can think of Ford-Fulkerson as at each step finding a new flow (along the augmenting path) and adding it to the existing flow, where by adding two flows we mean adding them edge by edge. Notice that the sum of two flows continues to satisfy flow-conservation and skew-symmetry. he definition of residual capacity ensures that the flow found by Ford-Fulkerson is legal (doesn t exceed the capacity constraints in the original graph). We now need to prove that in fact it is maximum. We ll worry about the number of iterations it takes and how to improve that later. Note that one nice property of the residual graph is that it means that at each step we are left with same type of problem we started with. o, to implement Ford-Fulkerson, we can use any black-box path-finding method (e.g., F). heorem 6. he Ford-Fulkerson algorithm finds a maximum flow. Proof: Let s look at the final residual graph. his graph must have s and t disconnected by definition of the algorithm. Let be the component containing s and be the rest. Let c be the capacity of the (,) cut in the original graph so we know we can t do better than c. he claim is that we in fact did find a flow of value c (which therefore implies it is maximum). Here s why: let s look at what happens to the residual capacity of the (,) cut after each iteration of the algorithm. ay in some iteration we found a path with k units of flow. hen, even if the path zig-zagged between and, every time we went from to we added k to the flow from to and subtracted k from the residual capacity of the (,) cut, and every time we went from to we took away k from this flow and added k to the residual capacity of the cut ; moreover, we must have gone from to exactly one more time than we went from to. o, the residual capacity of this cut went down by exactly k. o, the drop in capacity is equal to the increase in flow. ince at the end the residual capacity is zero (remember how we defined and ) this means the total flow is equal to c. o, we ve found a flow of value equal to the capacity of this cut. We know we can t do better, so this must be a max flow, and (,) must be a minimum cut. his is where we use the fact that if we flow k units on the edge (u, v), then in addition to reducing the residual capacity of the (u, v) edge by k we also add k to the residual capacity of the back-edge (v, u).

5 6.. IPRIE MHING 88 Notice that in the above argument we actually proved the nonobvious maxflow-mincut theorem: heorem 6. In any graph G, for any two vertices s and t, the maximum flow from s to t equals the capacity of the minimum (s,t)-cut. We have also proven the integral-flow theorem: if all capacities are integers, then there is a maximum flow in which all flows are integers. his seems obvious, but you ll use it to show something that s not at all obvious on homework 5! In terms of running time, if all capacities are integers and the maximum flow is F, then the algorithm makes at most F iterations (since each iteration pushes at least one more unit of flow from s to t). We can implement each iteration in time O(m+n) using F. o we get the following result. heorem 6. If the given graph G has integer capacities, Ford-Fulkerson runs in time O(F(m + n)) where F is the value of the maximum s-t flow. In the next lecture we will look at methods for reducing the number of iterations the algorithm can take. For now, let s see how we can use an algorithm for the max flow problem to solve other problems as well: that is, how we can reduce other problems to the one we now know how to solve. 6. ipartite Matching ay we wanted to be more sophisticated about assigning groups to homework presentation slots. We could ask each group to list the slots acceptable to them, and then write this as a bipartite graph by drawing an edge between a group and a slot if that slot is acceptable to that group. For example: group slot his is an example of a bipartite graph: a graph with two sides L and R such that all edges go between L and R. matching is a set of edges with no endpoints in common. What we want here in assigning groups to time slots is a perfect matching: a matching that connects every point in L with a point in R. For example, what is a perfect matching in the bipartite graph above? More generally (say there is no perfect matching) we want a maximum matching: a matching with the maximum possible number of edges. We can solve this as follows: ipartite Matching:. et up a fake start node s connected to all vertices in L. onnect all vertices in R to a fake sink node. Orient all edges left-to-right and give each a capacity of.. Find a max flow from s to t using Ford-Fulkerson.

6 6.. IPRIE MHING 89. Output the edges between L and R containing nonzero flow as the desired matching. his finds a legal matching because edges from R to t have capacity, so the flow can t use two edges into the same node, and similarly the edges from s to L have capacity, so you can t have flow on two edges leaving the same node in L. It s a maximum matching because any matching gives you a flow of the same value: just connect s to the heads of those edges and connect the tails of those edges to t. (o if there was a better matching, we wouldn t be at a maximum flow). What about the number of iterations of path-finding? his is at most the number of edges in the matching since each augmenting path gives us one new edge. Let s run the algorithm on the above example. Notice a neat fact: say we start by matching to and to. hese are bad choices, but the augmenting path automatically undoes them as it improves the flow! Matchings come up in many different problems like matching up suppliers to customers, or cellphones to cell-stations when you have overlapping cells. hey are also a basic part of other algorithmic problems.

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