Algorithm Design

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Chapter 7 Network Flow
to cross from A to B, and now it does not cross from A’ to B’. But since gx~ > O,
this means that (A’, B’) has smaller capacity than (A, B), again contradicting
our assumption that (A, B) is a minimum cut. So, if both x and y belong to T,
then Uxy ~ A.
Thus we have established the following conclusion, based on the fact that
(A, B) is a minimum cut: uxy ~ A if and only if both x, y ~ T.
Now we just need to work out the minimum-cut capacity c(A, B) in terms
of its constituent edge capacities. By the conclusion in the previous paragraph,
we know that edges crossing from A to B have one of the following two forms:
o edges of the form (v x, t), where x ~ T, and
o edges of the form (s, Uxy), where at least one of x or y does not belong
to r (in other words, {x, y} ~ r).
Thus we have
x~T
x~T
{x,y}~T
x,y~T
Since we know that c(A, B) = g’ < g*, this last inequality implies
and hence
x,y~T
For example, applying the argument in the proof of (7.59) to the instance
in Figure 7.21, we see that the nodes for New York and Toronto ~re on the
source side of the minimum cut, and, as we saw earlier, these two teams
indeed constitute a proof that Boston has been eliminated.
* 7.13 A Further Direction: Adding Costs to the
Matching Problem
Let’s go back to the first problem we discussed in this chapter, Bipartite
Matching. Perfect matchings in a bipartite graph formed a way to model the
problem of pairing one kind of obiect with another--iobs with machines, for
example. But in many settings, there are a large number of possible perfect
matchings on the same set of objects, and we’d like a way to express the idea
that some perfect matchings may be "better" than others.
7.13 A Further Direction: Adding Costs to the Matching Problem
~ The Problem
A natural way to formulate a problem based on this notion is to introduce
costs. It may be that we incur a certain cost to perform a given job on a given
machine, and we’d like to match jobs with machines in a way that minimizes
the total cost. Or there may be n fire trucks that must be sent to n distinct
houses; each house is at a given distance from each fire station, and we’d
like a matching that minimizes the average distance each truck drives to its
associated house. In short, it is very useful to have an algorithm that finds a
perfect matching "of minimum total cost.
Formaliy, we consider a bipartite graph G = (V, E) whose- node set, as
usual, is partitioned as V = X U Y so that every edge e ~ E has one end in X
and the other end in Y. Furthermore, each edge e has a nonnegafive cost c e.
For a matching M, we say that the cost of the matching is the total cost of a!l
edges in M, that is, cost(M) = ~e~v~ Ce" The Minimum-Cost Perfect Matching
Problem assumes that IXI = IYI = n, and the goal is to find a perfect matching
of minimum cost.
~ Designing and Analyzing the Algorithm
We now describe an efficient algorithm to solve this problem, based on the
idea of augmenting paths but adapted to take the costs into account. Thus, the
algorithm will iteratively construct matchings using i edges, for each value of i
from ! to n. We will show that when the algorithm concludes with a matching
of size n, it is a minimum-cost perfect matching. The high-level structure of the
algorithm is quite simple. If we have a minimum-cost matching of size i, then
we seek an augmenting path to produce a matching of size i + 1; and rather
than !ooking for any augmenting path (as was sufficient in the case without
costs), we use the cheapest augmenting path so that the larger matching wil!
also have minimum cost.
Recall the construction of the residual graph used for finding augmenting
paths. Let M be a matching. We add two new nodes s and t to the graph. We
add edges (s, x) for all nodes x ~ X that are unmatched and edges (y, t) for all
nodes y ~ Y that are unmatched. An edge e = (x, y) ~ E is oriented from x to
y if e is not in the matching M and from y to x if e ~ M. We will use G M to
denote this residual graph. Note that all edges going from Y to X are in the
matching M, while the edges going from X to Y are not. Any directed s-t path
P in the graph G M corresponds to a matching one larger than M by swapping
edges along P, that is, the edges in P from X to Y are added to M and all edges
in P that go from Y to X are deleted from M. As before, we will call a path P in
G M an augmenting path, and we say that we augment the matching M using
the path P.
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