Algorithm Design

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Chapter 7 Network Flow
~ Designing and Analyzing the Algorithm
We begin by constructing a flow network that provides an efficient algorithm
for determining whether z has been eliminated. Then, by examining the
minimum cut in this network, we will prove (7.59).
Clearly, if there’s any way for z to end up in first place, we should have
z win all its remaining games. Let’s suppose that this leaves it with m wins.
We now want to carefully a~ocate the wins from all remaining games so that
no other team ends with more than m wins. Allocating wins in this way can
be solved by a maximum-flow computation, via the following basic idea. We
have a source s from which all wins emanate. The i th win can pass through
one of the two teams involved in the i th game. We then impose a capacity
constraint saying that at most m - wx wins can pass through team x.
More concretely, we construct the following flow network G, as shown in
Figure 7.21. First, let S’ = S - {z}, and let g* = ~x,y~s’ gx~ -the total number
of games left between all pairs of teams in S’. We include nodes s and t, a
node vx for each team x ~ S’, and a node Uxy for each pair of teams x, y ~ S’
with a nonzero number of games left to play against each other. We h~ive the
following edges.
o Edges (s, uxy) (wins emanate from s);
o Edges (ux~, Vx) and (ux~, vy) (only x or y can win a game that they play
against each other); and
o Edges (vx, t) (wins are absorbed at t).
Let’s consider what capacities we want to place on these edges. We want
wins to flow from s to uxy at saturation, so we give (s, u~y) a capacity of
We want to ensure that team x cannot win more than m - wx games, so we
,
Balt-Tor
Bait
~ he set T = {NY, Toronto}
roves Boston is
liminated.
Figure 7.21 The flow network for the second example. As the minimum cut indicates,
there is no flow of value g*, and so Boston has been eliminated.
7.12 Baseball Elimination
give the edge (v x, t) a capacity of m - w x. Finally, an edge of the form (Ux~,
should have at least gx3, units of capacity, so that it has the ability to transport
a!l the wins from ux3, on to vx; in fact, our analysis will be the cleanest if we
give it infinite capacity. (We note that the construction stil! works even if this
edge is given only gx7 units of capacity, but the proof of (7.59) will become a
little more complicated.)
Now, if there is a flow of value g*, then it is possible for the outcomes
of all remaining games to yield a situation where no team has more than m
wins; and hence,’if team z wins all its remaining games, it can still achieve at
least a tie for first place. Conversely, if there are outcomes for the remaining
games in which z achieves at least a tie, we can use these outcomes to define
a flow of value g*. For example, in Figure 7.21, which is based on our second
example, the indicated cut shows that the maximum flow has value at most
7, whereas g* = 6 + 1 + 1 = 8.
In summary, we have shown
(7.60) Team z has been eliminated if and only if the maximum flow in G
has value strictly less than g*. Thus we can test in polynomial time whether z
has been eliminated.
Characterizing When a Team Has Been Eliminated
Our network flow construction can also be used to prove (7.59). The idea is that
the Max-Flow Min-Cut Theorem gives a nice "if and only if" characterization
for the existence of flow, and if we interpret this characterization in terms
of our application, we get the comparably nice characterization here. This
illustrates a general way in which one can generate characterization theorems
for problems that are reducible to network flow.
Proof of (7.59). Suppose that z has been eliminated from first place. Then
the maximum s-t flow in G has value g’ < g*; so there is an s-t cut (A, B) of
capacity g’, and (A, B) is a minimum cut. Let T be the set of teams x for which
vx ~ A. We will now prove that T can be used in the "averaging argument" in
(7.59).
First, consider the node u.t:~, and suppose one of x or y is not in T, but
ux3, ~ A. Then the edge (Ux~, vx) would cross from A into B, and hence the
cut (A, B) would have infinite capacity. This contradicts the assumption that
(A, B) is a minimum cut of capacity less than g*. So if one of x or y is not in
T, then ux3, ~ B. On the other hand, suppose both x and y be!ong to T, but
ux3, ~ B. Consider the cut (A’, B’) that we would obtain by adding u~ to the set
A and deleting it from the set B. The capacity of (A’, B’) is simply the capacity
of (A, B), minus the capacity g~ of the edge (s, uxy)--for this edge (s, Uxy) used
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