SOLUTIONS MANUAL
All Chapters included
1
, Algorithm Design Solutions Manual By Jon Kleinberg
Eva Tardo
1 Stable Matching
∗ ( ) tend to be more difficult, or to rely on
Note: Exercises denoted with an asterisk
some of the more advanced material.
1. Gale and Shapley published their paper on the stable marriage problem in 1962;
but a version of their algorithm had already been in use for ten years by the
National Resident Matching Program, for the problem of assigning medical
residents to hospitals.
Basically, the situation was the following. There were m hospitals, each with a
certain number of available positions for hiring residents. There were n medical
students graduating in a given year, each interested in joining one of the hospitals.
Each hospital had a ranking of the students in order of preference, and each
student had a ranking of the hospitals in order of preference. We will assume that
there were more students graduating than there were slots available in the m
hospitals.
The interest, naturally, was in finding a way of assigning each student to at most
one hospital, in such a way that all available positions in all hospitals were
filled. (Since we are assuming a surplus of students, there would be some students
who do not get assigned to any hospital.)
We say that an assignment of students to hospitals is stable if neither of the
following situations arises.
• First type of instability: There are students s and
r
s , and a hospital h, so that
– s is assigned to h, and
– sr is assigned to no hospital, and
– h prefers sr to s.
• Second type of instability: There are students s and s , and hospitals h and h ,
r r
so that
– s is assigned to h, and
– sr is assigned to hr, and
– h prefers sr to s, and
– sr prefers h to hr.
So we basically have the stable marriage problem from class, except that (i)
hospitals generally want more than one resident, and (ii) there is a surplus of
medical students.
Show that there is always a stable assignment of students to hospitals, and give an
efficient algorithm to find one. The input size is Θ(mn); ideally, you would like
to find an algorithm with this running time.
2
, Solution. The algorithm is very similar to the one from class. At any point
in time, a student is either “committed” to a hospital or “free.” A hospital
either has available positions, or it is “full.” The algorithm is the following:
While some hospital hi has available positions hi offers a position to
the next student sj on its preference list if sj is free then sj accepts the
offer else (sj is already committed to a hospital hk) if sj prefers hk to
hi then sj remains committed to hk else sj becomes committed to hi
the number of available positions at hk increases by one. the number
of available positions at hi decreases by one.
The algorithm terminates in O(mn) steps because each hospital offers a
positions to a student at most once, and in each iteration, some hospital offers a
position to some student.
Suppose there are pi > 0 positions available at hospital hi. The algorithm
terminates with an assignment in which all available positions are filled,
because any hospital that did not fill all its positions must have offered one to
every student; but then, all these students would be committed to some hospital,
Σ
which contradicts our assumption that
m
i= pi < n.
1
Finally, we want to argue that the assignment is stable. For the first kind of
instability, suppose there are students s and sr, and a hospital h as above. If h
prefers sr to s, then h would have offered a position to sr before it offered one to
s; from then on, sr would have a position at some hospital, and hence would
not be free at the end — a contradiction.
For the second kind of instability, suppose that (hi, sj) is a pair that causes
instability. Then hi must have offered a position to sj, for otherwise it has pi
residents all of whom it prefers to sj. Moreover, sj must have rejected hi in
favor of some hk which he/she preferred; and sj must therefore be committed to
some hl (possibly different from hk) which he/she also prefers to hi.
2. We can think about a different generalization of the stable matching problem, in
which certain man-woman pairs are explicitly forbidden. In the case of employers
and ap- plicants, picture that certain applicants simply lack the necessary
qualifications or degree; and so they cannot be employed at certain companies,
however desirable they may seem. Concretely, we have a set M of n men, a set
W of n women, and a set F ⊆ M × W of pairs who are simply not allowed to
get married. Each man m ranks all the women w for which (m, w) /∈ F , and
each woman wr ranks all the men mr for which (mr, wr) /∈ F .
In this more general setting, we say that a matching S is stable if it does not
exhibit any of the following types of instability.
(i) There are two pairs (m, w) and (mr, wr) in S with the property that m prefers wr
to w, and wr prefers m to mr. (The usual kind of instability.)
∈ S, and a man mr, so that mr is not part of any pair
(ii) There is a pair (m, w)
in the matching,/∈(mr, w) F , and w prefers mr to m. (A single man is
3
, more desirable and not forbidden.)
4
All Chapters included
1
, Algorithm Design Solutions Manual By Jon Kleinberg
Eva Tardo
1 Stable Matching
∗ ( ) tend to be more difficult, or to rely on
Note: Exercises denoted with an asterisk
some of the more advanced material.
1. Gale and Shapley published their paper on the stable marriage problem in 1962;
but a version of their algorithm had already been in use for ten years by the
National Resident Matching Program, for the problem of assigning medical
residents to hospitals.
Basically, the situation was the following. There were m hospitals, each with a
certain number of available positions for hiring residents. There were n medical
students graduating in a given year, each interested in joining one of the hospitals.
Each hospital had a ranking of the students in order of preference, and each
student had a ranking of the hospitals in order of preference. We will assume that
there were more students graduating than there were slots available in the m
hospitals.
The interest, naturally, was in finding a way of assigning each student to at most
one hospital, in such a way that all available positions in all hospitals were
filled. (Since we are assuming a surplus of students, there would be some students
who do not get assigned to any hospital.)
We say that an assignment of students to hospitals is stable if neither of the
following situations arises.
• First type of instability: There are students s and
r
s , and a hospital h, so that
– s is assigned to h, and
– sr is assigned to no hospital, and
– h prefers sr to s.
• Second type of instability: There are students s and s , and hospitals h and h ,
r r
so that
– s is assigned to h, and
– sr is assigned to hr, and
– h prefers sr to s, and
– sr prefers h to hr.
So we basically have the stable marriage problem from class, except that (i)
hospitals generally want more than one resident, and (ii) there is a surplus of
medical students.
Show that there is always a stable assignment of students to hospitals, and give an
efficient algorithm to find one. The input size is Θ(mn); ideally, you would like
to find an algorithm with this running time.
2
, Solution. The algorithm is very similar to the one from class. At any point
in time, a student is either “committed” to a hospital or “free.” A hospital
either has available positions, or it is “full.” The algorithm is the following:
While some hospital hi has available positions hi offers a position to
the next student sj on its preference list if sj is free then sj accepts the
offer else (sj is already committed to a hospital hk) if sj prefers hk to
hi then sj remains committed to hk else sj becomes committed to hi
the number of available positions at hk increases by one. the number
of available positions at hi decreases by one.
The algorithm terminates in O(mn) steps because each hospital offers a
positions to a student at most once, and in each iteration, some hospital offers a
position to some student.
Suppose there are pi > 0 positions available at hospital hi. The algorithm
terminates with an assignment in which all available positions are filled,
because any hospital that did not fill all its positions must have offered one to
every student; but then, all these students would be committed to some hospital,
Σ
which contradicts our assumption that
m
i= pi < n.
1
Finally, we want to argue that the assignment is stable. For the first kind of
instability, suppose there are students s and sr, and a hospital h as above. If h
prefers sr to s, then h would have offered a position to sr before it offered one to
s; from then on, sr would have a position at some hospital, and hence would
not be free at the end — a contradiction.
For the second kind of instability, suppose that (hi, sj) is a pair that causes
instability. Then hi must have offered a position to sj, for otherwise it has pi
residents all of whom it prefers to sj. Moreover, sj must have rejected hi in
favor of some hk which he/she preferred; and sj must therefore be committed to
some hl (possibly different from hk) which he/she also prefers to hi.
2. We can think about a different generalization of the stable matching problem, in
which certain man-woman pairs are explicitly forbidden. In the case of employers
and ap- plicants, picture that certain applicants simply lack the necessary
qualifications or degree; and so they cannot be employed at certain companies,
however desirable they may seem. Concretely, we have a set M of n men, a set
W of n women, and a set F ⊆ M × W of pairs who are simply not allowed to
get married. Each man m ranks all the women w for which (m, w) /∈ F , and
each woman wr ranks all the men mr for which (mr, wr) /∈ F .
In this more general setting, we say that a matching S is stable if it does not
exhibit any of the following types of instability.
(i) There are two pairs (m, w) and (mr, wr) in S with the property that m prefers wr
to w, and wr prefers m to mr. (The usual kind of instability.)
∈ S, and a man mr, so that mr is not part of any pair
(ii) There is a pair (m, w)
in the matching,/∈(mr, w) F , and w prefers mr to m. (A single man is
3
, more desirable and not forbidden.)
4
