Algorithm Design

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Figure 7.29 An instance of
Coverage Expansion.
Chapter 7 Network Flow
18.
ping(s) always reports a path from s to itself.) Since it’s not practical to go
out and inspect every edge of the network, they’d like to determine the
extent of the failure using this monitoring tool, through judicious use of
the ping command.
So here’s the problem you face: Give an algorithm that issues a
sequence of ping commands to various nodes in the network and then
reports the full set of nodes that are not currently reachable from s. You
could do this by pinging every node in the network, of course, but you’d
like to do it using many fewer pings (given the assumption that only
k edges have been deleted). In issuing this sequence, your algorithm is
allowed to decide which node to pIng next based on the outcome of earlier
ping operations.
Give an algorithm that accomplishes this task using only O(k log n)
pings.
We consider the Bipartite Matching Problem on a bipartite gr,aph G =
(V, E). As usual, we say that V is partitioned into sets X and Y, and each
edge has one end in X and the other in Y.
If M is a matching in G, we say that a node y ~ Y is covered by M if y
is an end of one of the edges in M.
(a) Consider the following problem. We are given G and a matching M In
G. For a given number k, we want to decide if there is a matching M’
inG so that
(i) M’ has k more edges than M does, and
(ii) every node y ~ Y that is covered by M is also covered by M’.
We call this the Coverage Expansion Problem, with input G, M, and k.
and we will say that M’ is a solution to the instance.
Give a polynomial-time algorithm that takes an instance of Coverage
Expansion and either returns a solution M’ or reports (correctly)
that there is no solution. (You should include an analysis of the running
time and a brief proof of why it is correct.)
Note: You may wish to also look at part (b) to help in thinking about
this.
Example. Consider Figure 7.29, and suppose M is the matching consisting
of the edge (xl, Y2). Suppose we are asked the above question
with k = !.
Then the answer to ~s instance of Coverage Expansion is yes.
We can let M’ be the matching consisting (for example) of the two
edges (xl, Y2) and (x2, Y4); M’ has one more edge than M, and y2 is still
covered by M’.
19.
Exercises
(b) Give an example of an instance of Coverage Expansion, specified by
G, M, and k, so that the following situation happens.
The instance has a solution; but in any solution M’, the edges of M do
not form a subset of the edges of M’.
(c) Let G be a bipartite graph, and let M be any matching in G. Consider
the following two quantities.
- K1 is the size of the largest matching M’ so that every node y that
is covered by M is also covered by M’.
- K2 is the size of the largest matching M" in G.
Clearly K1 _< K2, since K 2 is obtained by considering all possible matchings
In G.
Prove that in fact K~ = K~; that is, we can obtain a maximum
matching even if we’re constrained to cover all the nodes covered
by our initial matching M.
You’ve periodically helped the medical consulting firm Doctors Without
Weekends on various hospital scheduling issues, and they’ve just come
to you with a new problem. For each of the next n days, the hospital has
determined the number of doctors they want on hand; thus, on day i,
they have a requirement that exactly Pi doctors be present.
There are k doctors, and each is asked to provide a list of days on
which he or she is wiJJing to work. Thus doctor j provides a set L i of days
on which he or she is willing to work.
The system produced by the consulting firm should take these lists
and try to return to each doctor j a list L; with the following properties.
(A) L; is a subset of Li, so that doctor j only works on days he or she finds
acceptable.
(B) If we consider the whole set of lists L~ ..... L~, it causes exactly pi
doctors to be present on day i, for i = 1, 2 .....n.
(a) Describe a polynomial-time algorithm that implements this system.
Specifically, give a polynomial-time algorithm that takes the numbers
Pl, Pa ..... pn, and the lists L 1 ..... L k, and does one of the following
two things.
- Return lists L~, L~ ..... L~ satisfying properties (A) and (B); or
- Report (correctly) that there is no set of lists L~, L’ a ..... L~ that
satisfies both properties (A) and (B).
(b) The hospital finds that the doctors tend to submit lists that are much
too restrictive, and so it often happens that the system reports (correctly,
but unfortunately) that no acceptable set of lists L~, L~ ..... L~
exists.
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