Algorithm Design

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Chapter 8 NP and Computational Intractability
effect is that the committee presents an affirmative decision on the issue
if the number of "Yes" votes is strictly greater than the nurnber of "No"
votes (the "Abstain" votes don’t count for either side), and it delivers a
negative decision otherwise.
Now we have a big table consisting of the vote cast by each committee
member on each issue, and we’d !ike to consider the following definition.
We say that a subset of the member~ M’ c__ M is decisive if, had we looked
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just atlthe votes cast by the members in M’, the committee’s decision
on every issue would have been the same, (In other words, the overall
outcome of the voting among the members in M’ is the same on every
issue as the overall outcome of the v0ting]~ the entire committee.) Such
a subset canbe viewed as a kin d of "inner circle" that reflects the behavior
of the committee as a whole. ~
Here’s the question: Given the votes cast by each member on each
issue, and given a parameter k, we want to know whether there is a decisive
subset consisting of at most k members. We’ll call this an instance
of the Decisive Subset Problem.
Example. Suppose we have four committee members and three issues.
We’re looking for a decisive set of size at most k = 2, and the voting went
as follows.
Issue # rnt /’/12 rrt3 /rt4
Issue 1 Yes Yes Abstain No
Issue 2 Abstain No No Abstain
Issue 3 Yes Abstain Yes Yes
Then the answer to this instance is "Yes," since members ml and m3
constitute a decisive subset.
Prove that Decisive Subset is NP-complete.
Suppose you’re acting as a consultant for the port authority of a small
Pacific Rim nation. They’re currently doing a multi-billion-dollar business
per year, and their revenue is constrained almost entirely by the rate at
which they can unload ships that arrive in the port.
Handling hazardous materials adds additional complexity to what is,
for them, an already complicated task. Suppose a convoy of ships arrives
in the morning and delivers a total of n cannisters, each containing a
different kind of hazardous material. Standing on the dock is a set of m
trucks, each of which can hold up to k containers.
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Exercises
Here are two related problems, which arise from different types of
constraints that might be placed on the handling of hazardous materials.
For each of the two problems, give one of the following two answers:
* A polynomial-time algorithm to solve it; or ....
proof that it is NP-complete.
(a) For each cannister, there is a specified subset of the trucks in ;~hich
it may be safely carried. Is there a way to load all n carmisters -into
the m trucks so that no h-uck is overloaded, and each container goes
in a truck that is allowed to carry it?
In this different Version.of the problem, any cannister can be placed
in any, truck; however, there ~e certain pairs of cannisters that
cannot be placed together in the same truck. (The chemicals they
contain may react e.xplosively if brought into contact.) Is there a
way to load all n cannisters into the m trucks so that no truck is
overloaded, and no two carmisters are placed in the same truck when
they are not supposed to be?
There are many different ways to formalize the intuitive problem of
clustering, where the goal is to divide up a collection of objects into
groups that are "similar" to one another.
First, a natural way to express the input to a clustering problem is via
a set of objects Pl, P2 ..... Pn, with a numerical distance d(p~, PJ) defined on
each pair. (We reqt~e only that d(p~., p~) = 0; that d(p~, Pi) > 0 for distinct
p~ and PJ; and that distances are symmetric: d(p~, PJ) = d(PJ, P~).)
In Section 4.7, earlier in the book, we considered one reasonable
formulation of the clustering problem: Divide the objects into k sets so
as to maximize the minimum distance between any pair of objects in
distinct dusters. This turns out to be solvable by a nice application of
the MinLmum Spanning Tree Problem.
A different but seemingly related way to formalize the clustering
problem would be as follows: Divide the objects into k sets so as to
minimize the maximum distance between any pair of objects in the
same cluster. Note the change. Where the formulation in the previous
paragraph sought clusters so that no two were "close together," this new
formulation seeks clusters so that none of them is too "wide"--that is,
no cluster contains two points at a large distance from each other.
Given the similarities, it’s perhaps surprising that this new formulation
is computationally hard to solve optimally. To be able to think about
this in terms of NP-completeness, let’s write it first as a yes/no decision
problem. Given n objects p~, P2 ..... Pn with distances on them as above,
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