Algorithm Design

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Chapter 7 Network Flow
Now we would like the resulting matching to have as small a cost as
possible. To achieve this, we will search for a cheap augmenting path with
respect to the fo!lowing natural costs. The edges leaving s and entering t will
have cost 0; an edge e oriented from X to Y will have cost ce (as including this
edge in the path means that we add the edge to M); and an edge e oriented
from Y to X will have cost -ce (as including this edge in the path means that
we delete the edge from M). We will use cost(P) to denote the cost of a path
P in G M. The following statement summarizes this construction.
(7.61) Let M be a matching and P be a path in GM from s to t. Let M’ be the
matching obtained [tom M by augmenting along P. Then IM’I = IMI + 1 and
cost(M’) = cost(M) + cost(P).
Given this statement, it is natural to suggest an algorithm to find a
minimum-cost perfect matching: We iterafively find minimum-cost paths in
GM, and use the paths to augment the matchings. But how can we be sure
that the perfect matching we find is of minimum cost?. Or even worse, is this
algorithm even meaningful?. We can only find minimum-cost paths if we know
that the graph GM has no negative cycles.
Analyzing Negative Cycles In fact, understanding the role of negative cycles
in GM is the key to analyzing the algorithm. First consider the case in which M
is a perfect matching. Note that in this case the node s has no leaving edges,
and t has no entering edges in GM (as our matching is perfect), and hence no
cycle in G M contains s or t.
(7.62) Let M be a perfect matching. I[ there is a negative-cost directed cycle
C in G M, then M is not minimum cost.
Proof. To see this, we use the cycle C for augmentation, iust the same way
we used directed paths to obtain larger matchings. Augmenting M along C
involves swapping edges along C in and out of M. The resulting new perfect
matching M’ has cost cos~ (M’) = cost(M) + cost(C); but cost(C) < 0, and hence
M is not of minimum cost. ¯
More importantly, the converse of this statement is true as well; so in fact
a perfect matching M has minimum cost precisely when there is no negative
cycle in G M.
(7.63) Let M be a perfect matching. I[ there are no negative-cost directed
cycles C in G M, then M is a minimum-cost perfect matching.
Proof. Suppose the statement is not true, and let M’ be a perfect matching of
smaller cost. Consider the set of edges in one of M and M’ but not in both.
7.13 A Further Direction: Adding Costs to the Matching Problem
Observe that this set of edges corresponds to a set of node-disjoint directed
cycles in G M. The cost of the set of directed cycles is exactly cost(M’) - cost(M).
Assuming M’ has smaller cost than M, it must be that at least one of these
cycles has negative cost..,
Our plan is thus to iterate through matchings of larger and larger size,
maintaining the property that the graph GM has no negative cycles in any
iteration. In this way, our computation of a minimum-cost path will always
be well defined; and when we terminate with a perfect matching, we can use
(7.63) to conclude that it has minimum cost.
Maintaining Prices on the Nodes It will help to think about a numericalprice
p(v) associated with each node v. These prices will help both in understanding
how the algorithm runs, and they will also help speed up the implementation.
One issue we have to deal with is to maintain the property that the graph
G M has no negative cycles in any iteration. How do we know that after an
augmentation, the new residual graph still has no negative cycles?. The prices
will turn out to serve as a compact proof to show this.
To understand prices, it helps to keep in mind an economic interpretation
of them. For this pm]~ose, consider the following scenario. Assume that the
set X represents people who need to be assigned to do a set of iobs Y. For an
edge e = (x, y), the cost c e is a cost associated with having person x doing job
y. Now we will think of the price p(x) as an extra bonus we pay for person x to
participate in this system, like a "signing bonus." With this in mind, the cost
for assigning person x to do iob y will become p(x) + ce. On the other hand,
we will think of the price p(y) for nodes y ~ Y as a reward, or value gained by
taking care of iob y (no matter which person in X takes care of it). This way
the "net cost" of assign~g person x to do job y becomes p(x) + ce -p(y): this
is the cost of hiring x for a bonus ofp(x), having him do iob y for a cost of c e,
and then cashing in on the reward p(y). We wil! call this the reduced cost of an
edge e = (x, y) and denote it by ~ = p(x) + c e - p(y). However, it is important
to keep in mind that only the costs c e are part of the problem description; the
prices (bonuses and rewards) wi!l be a way to think about our solution.
Specifically, we say that a set of numbers {p(v) : u ~ V} forms a set of
compatible prices with respect to a matching M if
(i) for all unmatched nodes x ~ X we havep(x) = 0 (that is, people not asked
to do any job do not need to be paid);
(if) for all edges e = (x, y) we have p(x) + Ce >_ P(Y) (that is, every edge has
a nonnegative reduced cost); and
(iii) for all edges e = (x, y) ~ M we have p(x) + c e = p(y) (every edge used in
the assignment has a reduced cost of 0).
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