Algorithm Design

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Chapter 7 Network Flow
Finally, we have to consider how to initialize the algorithm, so as to get it
underway. We initialize M to be the empty set, define p(x) = 0 for all x ~ X,
and define p(y), for y a Y, to be the minimum cost of an edge entering y. Note
that these prices are compatible with respect to M = ¢.
We summarize the algorithm below.
Start with M equal to the empty set
Define p(x)=0 for x~X, and p(y)--- win c e for y~Y
e into y
While M is not a perfect matching
Find a minimum-cost s-[ path P in G M using (7.64) with prices p
Augment along P to produce a new matching M r
Find- a set of compatible prices with respect to M r via (7.65)
Endwhile
The final set of compatible prices yields a proof that GM has no negative
cycles; and by (7.63), this implies that M has minimum cost.
(7.66) The minimum-cost perfect matching can be found in the time required
i
Extensions: An Economic Interpretation of the Prices
To conclude our discussion of the Minimum-Cost Perfect Matching Problem,
we develop the economic interpretation of the prices a bit further. We consider
the following scenario. Assume X is a set of n people each looking to buy a
house, and Y is a set of n houses that they are all considering. Let v(x, y) denote
the value of house y to buyer x. Since each buyer wants one of the houses,
one could argue that the best arrangement would be to find a perfect matching
M that maximizes ~(x,y)~4 v(x, y). We can find such a perfect matching by
using our minimum-cost perfect matching algorithm with costs ce = -v(x, y)
if e = (x, y).
The question we will ask now is this: Can we convince these buyers to
buy the house they are allocated? On her own, each buyer x would want to
buy the house y that has maximum value v(x, y) to her. How can we convince
her to buy instead the house that our matching M al!ocated ~. We will use prices
to change the incentives of the buyers. Suppose we set a price P(y) for each
house y, that is, the person buying the house y must pay P(Y). With these
prices in mind, a buyer will be interested in buying the house with maximum
net value, that is, the house y that maximizes v(x, y) -P(Y). We say that a
Solved Exercises
perfect matching M and house prices P are in equilibrium if, for all edges
(x, y) ~ M and all other houses y’, we have
v(x, y) - P(~) > v(x, y’) - P(y’).
But can we find a perfect matching and a set of prices so as to achieve this
state of affairs, with every buyer ending up happy? In fact, the minimum-cost
perfect matching and an associated set of compatible prices provide exactly
what we’re lookin.g for.
(7,67) LetM be aperfect matchingofminimum cost, where c e = ~v(x, y) for
each edge e ~ (x, y), and let p be a compatible set of prices, Then the matching
M and the set ofprices {P(y) = -p(y):y ~ Y} are in equilibrium,
Proof. Consider an edge e = (x, y) ~ M, and let e’ = (x, y’). Since M and p are
compatible, we have p(x) + c e = p(y) and p(x) + c e, > p(y’). Subtracting these
two inequalities to cance! p(x), and substituting the values of p and c, we get
the desired inequality in the definition of equilibrium. []
Solved Exercises
Solved Exercise 1
Suppose you are given a directed graph G = (V, E), with a positive integer
capacity c e on each edge e, a designated source s ~ V, and a designated sink
t ~ V. You are also given an integer maximum s-t flow in G, defined by a flow
value fe on each edge e.
Now suppose we pick a specific edge e ~ E and increase its capacity by
one unit. Show how to find a maximum flow in the resulting capacitated graph
in time O(m + n), where m is the number of edges in G and n is the number
of nodes.
Solution The point here is that O(m + n) is not enough time to compute a
new maximum flow from scratch, so we need to figure out how to use the flow
f that we are given. Intuitively, even after we add 1 to the capacity of edge e,
the flow f can’t be that far from maximum; after all, we haven’t changed the
network very much.
In fact, it’s not hard to show that the maximum flow value can go up by
at most 1.
(7.68) Consider the flow network G’ obtained by adding 1 to the capacity of
e. The value of the maximum flow in G’ is either v(f) or v(f) + 1.
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