21: Algorithms
Lecture Maximum Flows and Minimum Cuts [Fa'10]
Col. Hogan: One of these wires disconnects the fuse, the other one fires the bomb. Which one would you cut, Shultz? Sgt. Schultz: Don't ask me, this is a decision for an officer. Col. Hogan: All right. Which wire, Colonel Klink? Col. Klink: This one. [points to the white wire] Col. Hogan: You're sure? Col. Klink: Yes. [Hogan cuts the black wire; the bomb stops ticking] Col. Klink: If you knew which wire it was, why did you ask me? Col. Hogan: I wasn't sure which was the right one, but I was certain you'd pick the wrong one. -- "A Klink, a Bomb, and a Short Fuse", Hogan's Heroes (1966)
Harris and Ross's map of the Warsaw Pact rail network
Figure 2 From Harris and Ross [1955]: Schematic diagram of the railway network of the Western Soviet Union and Eastern European countries, with a maximum flow of value 163,000 tons from Russia to Eastern Europe, and a cut of capacity 163,000 tons indicated as `The bottleneck'.
21
Maximum Flows and Minimum Cuts
In the mid-1950s, Air Force researchersThe E. Harris and F. S. Ross published a classified report studying T. max-flow min-cut theorem the rail network that In the RAND Report of 19 November 1954,its satellite countries(next Eastern Europe. The network linked the Soviet Union to Ford and Fulkerson [1954] gave in to defining the flow problem and suggesting the for max-flow was modeled as a graph maximumfor undirected graphs, saying thatsimplex methodflow it) the is equal tomin- edges, representing with 44 vertices, representing geographic regions, andthe 105 cut theorem the maximum value minimum in the rail network. Each edge was given constructive, links between those regionscapacity of a cut separating source and terminal. Their proof is nota weight, representing the rate but for planar graphs, with source and sink on the outer boundary, they give a polynomialat which material could constructive method. In a report of 26 May 1955,the next. Essentially by trial and error, they time, be shipped from one region to Robacker [1955a] showed that the max-flow min-cut theorem of stuff that could be moved from Menger's determined both the maximum amountcan be derived also from the vertex-disjoint version ofRussia into Europe, as well as theorem. the cheapest way to disrupt the directed case, Ford and Fulkerson [1955] observed that less abstract terms, blowing up train As for the network by removing links (or in the max-flow min-cut theorem tracks), which they called holds Dantzig directed on integer solutions for the transportationthe figure at the top of the page) `thealso for [1951a] graphs. Dantzig and Fulkerson [1955] showed, by extending bottleneck'. Their results (including problem to the the results of were only declassified in 1999.1 This one of the first recorded applications of 25 maximum flow and minimum cut problems. For the both problems, the input is a directed graph G = (V, E), along with special vertices s and t called the source and target. As in the previous lectures, I will use u v to denote the directed edge from vertex u
1 Both the map and the story were taken from Alexander Schrijver's fascinating survey `On the history of combinatorial optimization (till 1960)'.
c Copyright 2011 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/). Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms for the most recent revision.
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Algorithms
Lecture 21: Maximum Flows and Minimum Cuts [Fa'10]
to vertex v. Intuitively, the maximum flow problem asks for the largest amount of material that can be transported from one vertex to another; the minimum cut problem asks for the minimum damage needed to separate two vertices.
21.1
Flows
0
An (s, t )-flow (or just a flow if the source and target are clear from context) is a function f : E that satisfies the following conservation constraint at every vertex v except possibly s and t: f (u v) =
u w
f (v w).
In English, the total flow into v is equal to the total flow out of v. To keep the notation simple, we define f (u v) = 0 if there is no edge u v in the graph. The value of the flow f , denoted | f |, is the total net flow out of the source vertex s: | f | :=
w
f (s w) -
u
f (u s).
It's not hard to prove that | f | is also equal to the total net flow into the target vertex t, as follows. To simplify notation, let f (v) denote the total net flow out of any vertex v: f (v) :=
u
f (u v) -
w
f (v w).
The conservation constraint implies that f (v) = 0 or every vertex v except s and t, so f (v) = f (s) + f (t).
v
On the other hand, any flow that leaves one vertex must enter another vertex, so we must have v f (v) = 0. It follows immediately that | f | = f (s) = - f (t). Now suppose we have another function c : E 0 that assigns a non-negative capacity c(e) to each edge e. We say that a flow f is feasible (with respect to c) if f (e) c(e) for every edge e. Most of the time we will consider only flows that are feasible with respect to some fixed capacity function c. We say that a flow f saturates edge e if f (e) = c(e), and avoids edge e if f (e) = 0. The maximum flow problem is to compute a feasible (s, t)-flow in a given directed graph, with a given capacity function, whose value is as large as possible.
0/5 10/20 0/15 s 0/10 10/10 10/10 5/10 5/20 t 5/15
An (s, t)-flow with value 10. Each edge is labeled with its flow/capacity.
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Algorithms
Lecture 21: Maximum Flows and Minimum Cuts [Fa'10]
21.2
Cuts
An (s , t )-cut (or just cut if the source and target are clear from context) is a partition of the vertices into disjoint subsets S and T --meaning S T = V and S T = --where s S and t T . If we have a capacity function c : E 0 , the capacity of a cut is the sum of the capacities of the edges that start in S and end in T : S, T :=
v S w T
c(v w ).
(Again, if v w is not an edge in the graph, we assume c(v w) = 0.) Notice that the definition is asymmetric; edges that start in T and end in S are unimportant. The minimum cut problem is to compute an (s, t)-cut whose capacity is as large as possible.
5 20 s 10 10 10 15 10 15
t
20
An (s, t)-cut with capacity 15. Each edge is labeled with its capacity.
Intuitively, the minimum cut is the cheapest way to disrupt all flow from s to t. Indeed, it is not hard to show that the value of any feasible (s, t )-flow is at most the capacity of any (s, t )-cut. Choose your favorite flow f and your favorite cut (S, T ), and then follow the bouncing inequalities: |f | =
w
f (s w) -
u
f (u s) f (u v)
u
by definition
=
vS w
f (v w) - f (v w) -
vS wT uT
by the conservation constraint
=
vS wT
f (u v)
removing duplicate edges since f (u v) 0 since f (u v) c(v w) by definition
f (v w) c(v w)
vS wT
= S, T
Our derivation actually implies the following stronger observation: | f | = S, T if and only if f saturates every edge from S to T and avoids every edge from T to S. Moreover, if we have a flow f and a cut (S, T ) that satisfies this equality condition, must f be a maximum flow, and (S, T ) must be a minimum cut.
21.3
The Max-Flow Min-Cut Theorem
Surprisingly, for any weighted directed graph, there is always a flow f and a cut (S, T ) that satisfy the equality condition. This is the famous max-flow min-cut theorem: 3
Algorithms
Lecture 21: Maximum Flows and Minimum Cuts [Fa'10]
The value of the maximum flow is equal to the capacity of the minimum cut. The rest of this section gives a proof of this theorem; we will eventually turn this proof into an algorithm. Fix a graph G, vertices s and t, and a capacity function c : E 0 . The proof will be easier if we assume that the capacity function is reduced: For any vertices u and v, either c(u v) = 0 or c(v u) = 0, or equivalently, if an edge appears in G, then its reversal does not. This assumption is easy to enforce. Whenever an edge u v and its reversal v u are both the graph, replace the edge u v with a path u x v of length two, where x is a new vertex and c(u x) = c(x v) = c(u v). The modified graph has the same maximum flow value and minimum cut capacity as the original graph.
Enforcing the one-direction assumption.
Let f be a feasible flow. We define a new capacity function c f : V V capacity, as follows: c(u v) - f (u v) if u v E c f (u v) = f (v u) 0 if v u E . otherwise
, called the residual
Since f 0 and f c, the residual capacities are always non-negative. It is possible to have c f (u v) > 0 even if u v is not an edge in the original graph G. Thus, we define the residual graph G f = (V, E f ), where E f is the set of edges whose residual capacity is positive. Notice that the residual capacities are not necessarily reduced; it is quite possible to have both c f (u v) > 0 and c f (v u) > 0.
0/5 10/20 0/15 s 0/10 10/10 10/10 5/10 5/20 t s 10 10 5/15 10 10 10 15 5 5 5 15 5 t 5 10
A flow f in a weighted graph G and the corresponding residual graph G f .
Suppose there is no path from the source s to the target t in the residual graph G f . Let S be the set of vertices that are reachable from s in G f , and let T = V \ S. The partition (S, T ) is clearly an (s, t)-cut. For every vertex u S and v T , we have c f (u v) = (c(u v) - f (u v)) + f (v u) = 0, which implies that c(u v) - f (u v) = 0 and f (v u) = 0. In other words, our flow f saturates every edge from S to T and avoids every edge from T to S. It follows that | f | = S, T . Moreover, f is a maximum flow and (S, T ) is a minimum cut. On the other hand, suppose there is a path s = v0 v1 vr = t in G f . We refer to v0 v1 vr as an augmenting path. Let F = mini c f (vi vi+1 ) denote the maximum amount of flow that we can
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Algorithms
Lecture 21: Maximum Flows and Minimum Cuts [Fa'10]
5 10 10 s 10 10 10 15 5 5 5 15 5/10 5 t s 5/10 10 10/20
5/5 5/15 0/15 0/10 t 10/20 10/10
An augmenting path in G f with value F = 5 and the augmented flow f .
push through the augmenting path in G f . We define a new flow function f : E f (u v) + F f (u v) = f (u v) - F f (u v) if u v is in the augmenting path if v u is in the augmenting path otherwise
as follows:
To prove that the flow f is feasible with respect to the original capacities c, we need to verify that f 0 and f c. Consider an edge u v in G. If u v is in the augmenting path, then f (u v) > f (u v) 0 and f (u v) = f (u v) + F f (u v) + c f (u v) = f (u v) + c(u v) - f (u v) = c(u v) by definition of f by definition of F by definition of c f Duh.
On the other hand, if the reversal v u is in the augmenting path, then f (u v) < f (u v) c(u v), which implies that f (u v) = f (u v) - F f (u v) - c f (v u) = f (u v) - f (u v) =0 by definition of f by definition of F by definition of c f Duh.
Finally, we observe that (without loss of generality) only the first edge in the augmenting path leaves s, so | f | = | f | + F > 0. In other words, f is not a maximum flow. This completes the proof!
Exercises
1. Let (S, T ) and (S , T ) be minimum (s, t)-cuts in some flow network G. Prove that (S S , T T ) and (S S , T T ) are also minimum (s, t)-cuts in G. 2. Suppose (S, T ) is the unique minimum (s, t)-cut in some flow network. Prove that (S, T ) is also a minimum (x, y)-cut for all vertices x S and y T .
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Algorithms
Lecture 21: Maximum Flows and Minimum Cuts [Fa'10]
3. Cuts are sometimes defined as subsets of the edges of the graph, instead of as partitions of its vertices. In this problem, you will prove that these two definitions are almost equivalent. We say that a subset X of (directed) edges separates s and t if every directed path from s to t contains at least one (directed) edge in X . For any subset S of vertices, let S denote the set of directed edges leaving S; that is, S := {u v | u S, v S}. (a) Prove that if (S, T ) is an (s, t)-cut, then S separates s and t. (b) Let X be an arbitrary subset of edges that separates s and t. Prove that there is an (s, t)-cut (S, T ) such that S X . (c) Let X be a minimal subset of edges that separates s and t. Prove that there is an (s, t)-cut (S, T ) such that S = X . 4. A flow f is acyclic if the subgraph of directed edges with positive flow contains no directed cycles. (a) Prove that for any flow f , there is an acyclic flow with the same value as f . (In particular, this implies that some maximum flow is acyclic.) (b) A path flow assigns positive values only to the edges of one simple directed path from s to t. Prove that every acyclic flow can be written as the sum of O(E) path flows. (c) Describe a flow in a directed graph that cannot be written as the sum of path flows. (d) A cycle flow assigns positive values only to the edges of one simple directed cycle. Prove that every flow can be written as the sum of O(E) path flows and cycle flows. (e) Prove that every flow with value 0 can be written as the sum of O(E) cycle flows. (Zero-value flows are also called circulations.) 5. Suppose instead of capacities, we consider networks where each edge u v has a non-negative demand d(u v). Now an (s, t)-flow f is feasible if and only if f (u v) d(u v) for every edge u v. (Feasible flow values can now be arbitrarily large.) A natural problem in this setting is to find a feasible (s, t)-flow of minimum value. (a) Describe an efficient algorithm to compute a feasible (s, t)-flow, given the graph, the demand function, and the vertices s and t as input. [Hint: Find a flow that is non-zero everywhere, and then scale it up to make it feasible.] (b) Suppose you have access to a subroutine MAXFLOW that computes maximum flows in networks with edge capacities. Describe an efficient algorithm to compute a minimum flow in a given network with edge demands; your algorithm should call MAXFLOW exactly once. (c) State and prove an analogue of the max-flow min-cut theorem for this setting. (Do minimum flows correspond to maximum cuts?)
c Copyright 2011 Jeff Erickson. Released under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License (http://creativecommons.org/licenses/by-nc-sa/3.0/). Free distribution is strongly encouraged; commercial distribution is expressly forbidden. See http://www.cs.uiuc.edu/~jeffe/teaching/algorithms for the most recent revision.
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