Algorithm Design

4O8
Chapter 7 Network Flow
Why are such prices useful? Intuitively, compatible prices suggest that the
matching is cheap: Along the matched edges reward equals cost, while on
all other edges the reward is no bigger than the cost. For a partial matching,
this may not imply that the matching has the smallest possible cost for its
size (it may be taking care of expensive jobs). However, we claim that if M
is any matching for which there exists a set of compatible prices, then GM
has no negative cycles. For a perfect matching M, this will imply that M is of
minimum cost by (7.63).
To see why G M can have no negative cycles, we extend the definition of
reduced cost to edges in the residual graph by using the same expression
~ =p(v)+ c e -p(w) for any edge e = (v, w). Observe that the definition
of compatible prices implies that all edges in the residual graph GM have
nonnegafive reduced costs. Now, note that for any cycle C, we have
cost(c) = E = E
eEC
since all the terms on the right-hand side corresponding to prices cancel out.
We know that each term on the right-hand side is nonnegafive, and so clearly
cost(C) is nonnegafive.
There is a second, algorithmic reason why it is usefnl to have prices on
the nodes. When you have a graph with negative-cost edges but no negative
cycles, you can compute shortest paths using the Bellman-Ford Algorithm in
O(mn) time. But if the graph in fact has no negative-cost edges, then you can
use Diikstra’s Algorithm instead, which only requires time O(m log n)--almost
a full factor of n faster.
In our case, having the prices around allows us to compute shortest paths
with respect to the nonnegafive reduced costs ~, arriving at an equivalent
answer. Indeed, suppose we use Diikstra’s Algorithm to find the minimum
cost dp,M(U) of a directed path from s to every node u ~ X U Y subje.ct to the
costs ~. Given the minimum costs dp,M(Y) for an unmatched node y ~ Y, the
(nonreduced) cost of the path from s to t through y is dp,M(Y) + P(Y), and so
we find the minimum cost in O(n) additional time. In summary, we have the
following fact.
(7.64) Let M be a matching, and p be compatible prices. We can use one
run of Dijkstra’s Algorithm and O(n ) extra time to find the minimum-cost path
from s to t.
Updating the Node Prices We took advantage of the prices to improve one
iteration of the algorithm. In order to be ready for the next iteration, we need
not only the minimum-cost path (to get the next matching), but also a way to
produce a set of compatible prices with respect to the new matching.
7.13 A Further Direction: Adding Costs to the Matching Problem
Figure 7.22 A matching M (the dark edges), and a residual graph used to increase the
size of the matching.
To get some intuition on how to do this, consider an unmatched node x
with respect to a matching M, and an edge e = (x, y), as shown in Figure 7.22.
If the new matching M’ includes edge e (that is, if e is on the augmenting
path we use to update the matching), then we will want to have the reduced
cost of this edge to be zero. However, the prices p we used with matching M
may result in a reduced cost ~ > 0 -- that is, the assignment of person x to
job y, in our economic interpretation, may not be viewed as cheap enough.
We can arrange the zero reduced cost by either increasing the price p(y)
reward) by ~, or by decreasing the price p(x) by the same amount. To keep
prices nonnegative, we will increase the price p(y). However, node y may be
matched in the matching M to some other node x’ via an edge e’ = (x’, y), as
shown in Figure 7.22. Increasing the reward p(y) decreases the reduced cost
of edge e’ to negative, and hence the prices are no longer compatible. To keep
things compatible, we can increase p(x’) by the same amount. However, this
change might cause problems on other edges. Can we update all prices and
keep the matching and the prices compatible on all edges? Surprisingly, this
can be done quite simply by using the distances from s to all other nodes
computed by Dijkstra’s Algorithm.
(7.6S) Let M be a matching, let p be compatible prices, and let M’ be a
matching obtained by augmenting along the minimum-cost path from s to t.
Then p’ (v) = dp,M(V ) + p(v) is a compatible set of prices for M’.
Proof. To prove compatibility, consider first an edge e =,(x’, y) ~ M. The only
edge entering x’ is the directed edge 0~, x’), and hence dp,M(x’) = dp,M(y) --
~, where ~=p(y)+c e -p(x’), and thus we get the desired equation on
such edges. Next consider edges (x, y) in M’-M. These edges are along the
minimum-cost path from s to t, and hence they satisfy dp,M(y ) = dp,M(X ) + ~
as desired. Finally, we get the required inequality for all other edges since all
edges e = (x,y) ~M must satisfy dp,M(y ) < dp,M(X ) + ~. []
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