Algorithm Design

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Chapter 8 NP and Computational Intractability
21.
and a botmd B, we define the Low-Diameter Clustering Problem as fol-
lows: Can the objects be partitioned into k sets, so that no two points in
the same set are at a distance greater than B from each other?
Prove that Low-Diameter Clustering is N-P-complete.
After a few too many days immersed in the popular entrepreneurial selfhelp
book Mine Your Own Business, you’ve come to the realization that
you need to upgrade your office computing system. This, however, leads
to some tricky problems.
In configuring your new system, there are k components that must
be selected: the operating system, the text editing software, the .e-mail
program, and so forth; each is a separate component. For the ]th component
of the system, you have a set A i of options; and a configuration of
the system consists of a selection of one element from each of the sets
of options A1, A2 .....
Now the trouble arises because certain pairs of options from different
sets may not be compatible. We say that option x~
form an incompatible pair ff a single system cannot contain them both.
(For example, Linux (as an option for the operating system) and I~crosoft
Word (as an option for the text-editing software) form an incompatible
pair.) We say that a configuration of the system is fully compatible if it
Xk ~ Ak such that none of the pairs
consists of elements xl ~ A1, x2 ~ A2 ....
(x~, x i) is an incompatible pair.
We can now define the Fully Compatible Configuration (FCC) Problem.
Aninstance of FCC consists of disjoint sets of options A~, A2 ..... Ak, and a
set P of incompatible pairs (x, y), where x and y are elements of different
sets of options. The problem is to decide whether there exists a fully
compatible configuration: a selection of an element from eachoption set
so that no pair of selected elements belongs to the set P.
Example. Suppose k = 3, and the sets A1, A2, A3 denote options for the
operating system, the text-editing software, and the e-mail program,
respectively. We have
AI= {Linux, Windows NT},
A2= {emacs, Word},
A3= {Outlook, Eudora, rmail},
with the set of incompatible pairs equal to
p _-- {(Linux, Word), (Linux, Outlook), 6qord, rmail)}.
22.
Exercises
Then the answer to the decision problem in this instance of FCC is
yes--for example, the choices Linux ~ A1, emacs ~ A2, rmail ~ A 3 is a fully
compatible configuration according to the definitions above.
Prove that Fully Compatible Configuration is NP-complete.
Suppose that someone gives you a black-box algorithm A that takes an
undirected graph G = (V, E), and a number k, and behaves as follows.
o If G is no.t connected, it.simply returns "G is not connected."
° If G is connected and has an independent set of size at least k, it
returns "yes." -~
¯ If G is connected and does not have an independent set 6f size at
least k, it returns "no." ~
Suppose that the algorithm A runs in time polynomial in the size of G
and k.
Show how, using co!Is to A, you could then solve the Independent Set
Problem in polynomial time: Given an arbitrary undirected graph G, and
a number k, does G contain an independent set of size at least k?
23. Given a set of finite binary strings S = {s 1 ..... Sk} , we say that a string
~ is a concatenation over S if it is equal to s~s~2.. ¯ s~, for some indices
i~ ~ {! ..... k}.
A friend of yours is considering the following problem: Given two
sets of finite binary strings, A -- {a 1 ..... am} and B = {b~ ..... ha}, does
there exist any string u so that u is both a concatenation over A and a
concatenation over B?
Your friend announces, "At least the problem is in N~P, since I would
just have to exhibit such a string u in order to prove the answer is yes."
You point out (politely, of course) that this is a completely inadequate
explanation; how do we know that the shortest such string u doesn’t
have length exponential in the size of the input, in which case it would
not be a polynomial-size certificate?
However, it turns out that this claim can be turned into a proof of
membership in X~P. Specifically, prove the following statement.
If there is a string u that is a concatenation over boti~ A and B, then there
is such a string whose length is bounded by a polynomial in the sum of the
lengths of the strings in A U B.
24. Let G = (V, E) be a bipartite graph; suppose its nodes are partitioned into
sets X and Y so that each edge has one end in X and the other in y. We
define an (a, b)-skeleton of G to be a set of edges E’ _c E so that at most
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