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A UNIFYING APPROACH TO THE STRUCTURES OF THE STABLE MATCHING PROBLEMS

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Computers Math. Applic. Vol. 22, No. 6, pp. 13-27, 1991 6997-4943/ 91 s3.69 + 0.00 Printed in Great Britain. All rights reserved Copyright@ 1991 Pergamon Prem plc

A UNIFYING APPROACH TO THE STRUCTURES

OF THE STABLE MATCHING PROBLEMS

YUANO-CHEH HSUEH

Department of Computer and Information Science, National Chiao Tnng University Hsinchu, Taiwan 36950 R.O.C.

(Received July 1989)

Abstract-It is well-known that the structure of the set of stable marriages of a stable marriage instance can be represented as a ilnite distributive lattice and, conversely, every finite distributive lattice is a set of stable marriages for some stable marriage instance. Recently, lrvmg [12] and Gnsfield [9] propose some representations of the set of all stable assignments for a given solvable instance of the stable roommates problem. In this paper, we will give a nnifying approach to the structures of the stable marriage problem and the stable roommates problem. To achieve this purpose, we fintt study the duality in the structure of a stable marriage instance, then transform every stable roommates instance into a corresponding stable marriage instance and obtain the structure of the stable roommates instance directly from that of the corresponding stable marriage instance. The main results of this paper are: (1) There is a oneone correspondence between the set of stable marriages for a stable marriage instance and the set of feasible words of some Faigle geometry; (2) There is a one-one correspondence between the set of stable assignments for a stable roommates instance and the set of basic words of some Faigle geometry.

1. INTRODUCTION

An instance of size n of the stable marriage problem consists of n men and n women, where each of the n men and the n women ranks the members of the opposite sex in order of preference. A complete matching of the men and women is called a marriage. A marriage it4 is unstable, if there is a man and a woman who are not married to each other in M, but who both prefer each other to their partners in M. A marriage that is not unstable is called stable. It is well-known that there is a stable marriage for any instance of the stable marriage problem [l]. It is also well-known that the structure of the set of stable marriages can be represented as’s finite distributive lattice. Conversely, it is shown [2,3] that every finite distributive lattice is a set of stable marriages for some instance of the stable marriage problem.

There is a closely related problem to the stable marriage problem, called the stable roommates problem. An instance of size n of the stable roommates problem consists of a set of 2n people, where each person in the set ranks the 2n - 1 others in order of preference. A pairing of the 2n people into n disjoint pairs is called an assignment. An assignment (Y is called unstable if there are two persons who are not paired together in CK, but they prefer each other to their respective mates in Q. A stable assignment is one which is not unstable. An instance of the stable roommates problem is called solvable if there is at least one stable assignment. Contrary to the case of the stable marriage problem, there are unsolvable instances of the stable roommates problem.

Recently, Irving [4] and Gusfield [5] give some “small,” “ implicit” representations of the set of all stable assignments for a given solvable instance of the stable roommates problem: the poset II’ on the set of “rotations,” the poset II and the undirected graph G on the set of “nonsingleton rotations.” An interesting result (Theorem 5.3., Gusfield [9]) says that there is a one-one correspondence between the maximal independent sets in G and the set of stable assignments. Furthermore, every maximal independent set in G has the same cardinality (an alternative version of Lemma 5.6. in [5]). If we let

M = (GJ)

Typeset by &S-lj$ 13

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14 Y.-C. HSUEH

be the system such that Z is the collection of all independent sets in G, including the empty set, then the system M is indeed a “matroid” [6] like structure, referred to as “Faigle geometry” [7,8], on the poset II. This observation motivates our study on the combinatorial structures of the stable matching problems. In addition, we observe that the poset H is a member of the class of “self-dual posets.”

The dual poset of a poset P = (5’; 5) is the poset Pd = (S; 2). If P is order-isomorphic to Pd, i.e., if there is a bijective function 6 : z H x6 from S into itself such that for all elements &, $I E S

(1) x 5 y, if and only if, y6 5 x6, (2) (x6)6 = 2,

then P is called a self-dual poset. Such a function 6 is called a dual assignment on P. Note that there may be many non-isomorphic dual assignments on a given self-dual poset. For example, consider the poset P whose diagram is given by

d

b Let

a6 = c, b6 = d, c6 = a, d6 = b,

a6’ =

d, b6’ =c, c 6’ = b, d6’ = a.

Then it is easy to see that both 6 and 6’ are dual assignments on P. However, since a 5 a6 and a 11 a6’ (a is incomparable to a6’), we have that 6 and 6’ are non-isomorphic. This motivates the following definition:

DEFINITION 1 .l. A self-dualized poset Pa = (S;l,S) isastructureconsistingofaset S, apartial order 5 on S, and a dual assignment 6 on the poset (S; I). For each x E S, the element x6 is called the dual element of x in Pa.

The purpose of this paper is to obtain a unifying combinatorial structure, called Faigle geometry (we follow the terminology used in Korte and Lovbz [8]), for both the stable marriage and the stable roommates problems. To achieve this purpose, we first study the duality in the structure of the stable marriage problem, then transform every instance of the stable roommates problem into a corresponding instance of the stable marriage problem. The structure of the stable roommates problem can be obtained from that of the stable marriage problem directly by duality.

Given a stable roommates instance RI of size n. Let S be the set of the given 2n preference lists. Then the instance RI can be transformed into an instance of size 2n of the stable marriage problem by the following:

(1) Add person i to the end of the list of himself, i = 1,. . . ,2n. Let S’ be the resulting set of lists.

(2) Let MS and WS be two identical copies of S’.

(3) Let MI be the instance of the stable marriage problem with MS and WS as the sets of male and female preference lists, respectively.

DEFINITION 1.2. The instance MI is called the instance of the stable marriage problem coffe- sponding to RI. The rotation poset of MI is also called the rotation poset of RI.

It should be noted that Definition 1.2. is valid for both solvable and unsolvable instances of the stable roommates problem, and the rotation poset of a stable roommates instance under this definition is different from that given in Irving [4] or Gusfield [5].

In this paper, we will show that the rotation poset of a given instance of the stable roommates problem, together with some dual assignment on it, is a self-dualized poset and, conversely, every finite self-dualized poset is an instance of the stable roommates problem. Moreover, we will show

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Structures of the stable matching problems 15 that there is a one-one correspondence between the stable marriages of an instance I of the stable marriage problem and the feasible words of a Faigle geometry on the rotation poset of I, and there is a one-one correspondence between the stable assignments of an instance RI of the stable roommates problem and the basic words of a Faigle geometry on the rotation poset of RI.

2. DEFINITIONS AND ALGORITHMS FOR

THE STABLE MARRIAGE PROBLEM

Given an instance of size n of the stable marriage problem, there is a fundamental “proposal- rejection” algorithm [l] which finds a stable marriage of the given instance, called the male optimal matiage (or the female pessimal marriage). Recall that a marriage is a complete matching of the n men and n women. Henceforth, we will denote a marriage by the notation

{man i/woman ji; i = 1,2,. . . ,?J}.

DEFINITION 2.1. We say that woman j accepts the proposal from man i if she removes from her list each man k ranked below man i on her list and, at the same time, is removed from man k’s list.

GALE-SHAPLEY ALGORITHM.

Input: A set of n maZepreference lists and n female-preference lists. Step 1. Every man proposes to the first woman in his current list. Step 2. Every woman who receives proposals accepts the best proposal.

Step 3. If every woman has a proposal, then STOP; otherwise, GO TO Step 1.

The output of this algorithm is a set of 2n sublists of the original preference lists, and the male optimal marriage is the set of the pairs

(man i/Woman ji; i=l,...,n},

where woman ji is the first on man i’s list. The set of all lists of this output possesses several interesting properties [9,10]:

;z;

Every list is nonempty.

Woman j is first on msn i’s list, if and only if man i is last on woman j’s list.

Man i is on woman j’s list, if and only if woman j prefers man i to the last man on her list. Woman j is on man i’s list, if and only if man i is on hers.

DEFINITION 2.2. Given an instance of size n of the stable marriage problem. A table is a set of 2n Jists, each of which is a sublist of the original preference list, such that the above properties (Tl) - (T4) are satisfied.

Obviously, the output of the Gale-Shapley algorithm is a table. We will call this table the male optimal table (or the female pessimal table).

LEMMA 2.3. For any table T of an instance of size n of the stable marriage problem, if we pair each man with the first woman on his list in T, then the resulting matching is a stable marriage. PROOF. Let woman ji be the first on man i’s list, i = 1,2,. . . , n. For any pair (man i, woman j), where j # ji, let man ij be the last on woman j’s list. If woman j prefers man i to man ii, then, by properties (T3) and (T4), man i is on woman j’s list and vice versa. However, since woman ji is first on man i’s list, man i does not prefer woman j to woman ji. This concludes that the matching, by pairing each man i with woman ji, is a stable marriage. I DEFINITION 2.4. Given a table T. The stable marriage obtained as in Lemma 2.3 is called the stable mam’age corresponding to T.

LEMMA 2.5. Given a stable marriage it4 of an instance of size n of the stable marriage problem. There exists a table T such that M is the stable marriage corresponding to T.

PROOF. Let M = (man i/WOmaIl ji; i= 1,2,..., n) and let T be the set of lists obtained from the original preference lists by letting woman ji accept the proposal from man i. We claim that T is a table and, hence, is the desired table.

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16 Y.-C. HSUEH

After woman ji accepts the proposal from man i, properties (Tl), (T3), (T4) and the property that man i is on the last of woman ji’s list follow directly from Definition 2.1. It remains for us to show that woman ji is the first on man i’s list. Suppose woman jk is the first on man i’s list and Ic # i. Again, by Definition 2.1, woman jk must prefer man i to man h. Since man i prefers woman jl: to woman ji, the given marriage M is unstable. This yields a contradiction. Hence, T

is a table. I

DEFINITION 2.6. Given an instance of the stable marriage problem. A cyclic sequence

R

of man/woman pairs

(man &/Woman jk; h = 0, I,. . . , T - 1)

is called a rotation if there exists a table T such that woman jk is the first and woman jk+, mod r is the second on man &‘s list in T for k = 0, 1, . . . , P. The rotation R is said to be exposed in T. The notion of rotations for the stable marriage problem has been studied in detail in Irving and Leather [lo]. In Gusfield [9], a rotation-elimination algorithm is proposed to find all rotations of a stable marriage instance of size n in O(na) time. To help understand thii algorithm, we need the following definition and results.

DEFINITION 2.7. Let R = (man &/WOmaII jk;k = O,l,. . . , r - 1) be a rotation exposed in a table T. If each woman jk accepts the proposd from man &_I mod r, where k = 0, 1, . . . , r - 1, then the rotation R is said to be eliminated from T.

LEMMA 2.8. Let R be a rotation exposed in a table T. Let T’ be the set of lists obtained by eliminating R from T. Then, T’ is a table.

PROOF. It is sufficient to show that T’ possesses property (T2). Observe that man i is removed from a list in T, if and only if he is ranked below man i&l mod r on woman jk’s liit in T for some k. Moreover, woman j is removed from a list in T if j = jl: for some k. Let man i be such that i # ik for any k = 0, 1, . . . , r - 1, and let woman j be the first on man i’s list in T. Since the matching, by pairing each man with the first woman on his list in T, is a stable marriage, we have that j # jk for any k. Thus T’ inherits the property that man i is the last on woman j’s list and woman j is the first on man i’s list. As for man ik, k = 0, 1, . . . , r - 1, by Definition 2.1., it is clear that woman jk is the first on the list of man ik-1 mod t and he is the last on woman

jk’s list in T’. I

In notation, the table T’ will be denoted ss T\R. Observe that the proof of Lemma 2.8. also implies the following result.

COROLLARY 2.9. Let R and R’ be two distinct rotations exposed in a table T. Then R’ is also a rotation exposed in the table T\R.

It is shown (Lemma 4.6. in [lo]) that each table can be obtained from the male optimal table by a sequence of zero or more rotation eliminations. We are now in a position to describe the rotation-elimination algorithm.

ROTATION-ELIMINATION ALGORITHM. Input: The male optimal table. Step 1. Let T be the current table.

Step 2. If there are no rotations exposed in T, then STOP; otherwise, GO TO next step. Step 3. Find a rotation R exposed in T.

Step 4. Eliminate R from T; GO TO Step 1.

This algorithm outputs all rotations of a given stable marriage instance. Gusfield [S] also uses this algorithm to find all stable pairs, which are the pairs appearing in at least one stable marriage. An earlier version of this algorithm is proposed in McVitie and Wilson [ll] and is used to find all stable marriages.

3. THE LATTICE OF STABLE MARRIAGES AND THE ROTATION POSET

Let S be the set of stable marriages of a given stable marriage instance of size n. For any two stable marriages

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Structurea of the stable matching problems 17

and

M2 = {man i/woman ji; i = 1,2, . . . , fl}, let

and

ki = ji if man i prefers Woman ji to woman ji j: otherwise,

k; = ji ifki=ji, ji ifki=ji. Then the two sets of man/woman pairs

MS = {man i/woman ki; i= 1,2,...,n}

and

M4 = {man i/woman ki; i= 1,2,...,n} are stable marriages [15]. Define

MIA& =Ms and MlVMz=Mq.

Then, the algebra

1: = (S;A,V,O,l)

is a distributive lattice [15], where the least element 0 is the male optimal marriage and the greatest element 1 is the female optimal marriage. Note that the female optimal marriage can be obtained from the Gale-Shapley algorithm by reversing the roles of men and women. Dually, if we define

MI A’ M2 = M4 and Ml V’ M2 = Ma, then we have the dual lattice

Cd

=

(S;

Al, v’, Ol,

I’) = (S; v, A, I, 0). Let & the set of all rotations of a given stable marriage instance. Let

R1 = (manik/woman j,; k=O,l,...,r-1)

and

R2 = (manii/woman ji; h=O,l,...,s-1) be two distinct rotations in E.

DEFINITION 3.1. Rotation RI is said to eqlicidy precede Rz, if and only if there exist k and h (0 5 k 5 T - 1, 1 5 h 5 B - l), such that woman j,+l ,,,d r prefers man ik to man ii and maII ii prefers Womb jk+l mod r tCI WOm?UI ji.

Now, define a binary relation “2” on & as below:

R 5 R’ if and only if there are rotations RI = R, a,. . . , & = R’ such that &-I explicitly precedes Ri for each i = 2,. . . , k.

It is easy to see that the binary relation 5 is a partial order on & and the structure f3 = (E, <) is a poset.

DEFINITION 3.2. The poset B = (&; 5) .is called the rotation poset of the given instance.

DEFINITION 3.3. Let P = (S; 5) be a poset. A subset H of S is called hereditary ifh E H and x _< h imply x E H for all x E S.

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18 Y.-C. HSUEH

For any finite poset P = (5’; <), let Hered denote the set of all hereditary subsets of S. It is a fundamental theorem in lattice theory [12-141 that the lattice

C(Hered(?)) = (Hered( fl, U, 4, S)

is a distributive lattice. Conversely, let 3 be the poset of all nontrivial join-irreducible elements of a finite distributive lattice L under the partial order from L, then

t Y t(Hered(,7)).

In other words, if t is the distributive lattice of stable marriages and t3 is the rotation poset of a given instance, then

t Y t(Hered(B)).

The explicit meaning of this isomorphism [lo] can be rephrased as:

If M is the corresponding stable marriage of the hereditary subset 3-1 of &, then M

is the stable marriage corresponding to the table obtained by eliminating all rotations in H. Given a rotation R in E, let 7i(R) denote the hereditary subset and ‘Hd(R) denote the dual hereditary subset of Z generated by R. That is,

and

LEMMA 3.4. For any R E E, the difference subset & - ?id(R) is bereditary.

PROOF. Let Q E & - 7fd(R) and Q’ < Q. If Q’ E Xd(R), then R 5 Q’ 5 Q, a contradiction.

Hence, Q’ E e - ‘Hd(R) and & - Yd(R) is hereditary. I

DEFINITION 3.5. Tbe hereditary subset 8 - ‘Iid is called the dual-ezclusive hereditary subset of & generated by R and is denoted by W(R).

REMARK 3.6. Given a rotation R in E. Let T be the resulting table by eliminating all rotations in W(R) starting from the male optimal table. Since R 5 Q for any rotation Q not in ‘H’(R), we have that R is the only rotation exposed in T.

LEMMA 3.7. For any R E I, tbe subset ‘H(R) is ‘oin-irreducible in t(Hered(f?)). J Conversely, if X is a nontrivial join-irreducible element in Z(Hered(J?)), then ‘H = %(R) for some R E E. PROOF. Let 3c1 and ‘Hz be two hereditary subsets of f, such that ‘HI U ‘HZ = X(R). Then R E ‘HI or R E %2. If R E ‘HI, then X1 = ‘H(R); if R E ‘H2, then N2 = N(R). Hence, H(R) is join-irreducible. Conversely, let 7f # 4 be join-irreducible in L(Hered(B)). We claim that ‘H has

a greatest element R E E and hence X = X(R).

Suppose ‘H does not have a greatest element. Let RI, Ra, . . . , RI be all the maximal elements of%. Let3-1 1 = ‘H(R1) and ‘Hz = N(R2) U s. - U 3t(&). It is clear that ‘HI U ‘Hz = ‘H. Since 3t is join-irreducible, either 311 = X or N2 = ‘H. If X1 = X, then & 5 R for all i E (2,. . . ,k}; if 3-12 = ‘Ii, then R 5 & for some i E (2,. . . , k}. Either case will result in a contradiction. I

Similarly, we have:

LEMMA 3.8. For any R in E, the subset W(R) is meet-irreducible in t(Hered(B)). ConverseJy, if 7f is a nontriviaJ meet-irreducible element in c(Hered(B)), then X = W(R) for some R in E.

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Structures of the stable matching problems 19

4. DUALITY IN THE STABLE MARRIAGE PROBLEM

Note that the previous definitions for tables and rotations in Section 2 are male oriented. Henceforth, we will call them male-oriented 2ables and mole-oriented rotations, respectively. If we reverse the roles of men and women in those definitions, we have the definitions for female-

oriented tables and female-oriented rotations. Moreover, after this reversal, Gusfield’s algorithm finds the set of all female-oriented rotations. As for stable marriages, there is no such distinction. That is, the marriage {man i/woman ji; i = 1, . . . , n} is exactly the same as the marriage {woman ji /man i; i= l,...,n}.

Let 3 be the set of female-oriented rotations. Applying the same reversal to the definition of the partial order on 8, we have a partial 5’ on 3. We call the poset B’ = (3; 5’) the female-oriented rotation poset of the given instance. It is known [15] that t(Hered(B’)) % td.

LEMMA 4.1. Let B = (8; 5) be the male-oriented rotation poset and B’ = (fi 5’) the female- oriented rotation paset of a given stable marriage instance. Then B’ Y Bd. In particular, there is a one-one correspondence between male-oriented rotations in & and femaleoriented rotations in 3.

PROOF. Let Bd = (E; 2) be the dual poset of the male-oriented rotation poset f3 = (6; 5). Since Ld Y L(Hered(Bd)) and a finite distributive lattice is uniquely characterized by the poset of its nontrivial joint-irreducible elements up to isomorphism [14], we have

B’ = (3,s’) E Bd = (8, 2). I

Given any rotation & (male- or female-oriented), let X(Q) be the hereditary subset and W(Q) the dual-exclusive hereditary subset generated by Q. Then, let M(Q) and W(Q) be the respec- tive stable marriages corresponding to a(Q) and W(Q).

For any male-oriented rotation R in d, since M(R) is join-irreducible in L, M(R) is meet- irreducible in Ld. Hence, R corresponds to a female-oriented rotation R’ in 3 such that

M(R)

=

Me(R’). The following lemma gives an explicit form of such correspondence.

LEMMA 4.2. Let R=(man ik/WOmm jk ; k = 0, 1, . . . , P - 1) be a male-oriented rotation in E. If R’ is a female-oriented rotation in 3 such that M(R) = Me(R’), then

R’ = (woman jk/marI i&l mod r; k = O,l,. . . ,P - 1).

PROOF. Let T’ be the female-oriented table corresponding to M’(R’). We claim that (woman jk/man ik-1 mod r; k = 0, 1, . . . , r - 1) is a rotation exposed in T’.

First, since the marriage M(R) contains the pairs man ik_r mod ?/woman jk, where k = 0, 1,. . . , r - 1, we have that, in T’, man it-1 mod r is the first on woman jk’s list and woman jk

is the last on man ik-i mod ?‘s list for each k. Next, since man ik prefers woman jk to woman

jk+l mod ?, by properties (T3) and (T4), man ik is on woman jk’s list and vice versa for each k. If there is an h,O 5 h 5 r - 1, such that man ih is not the second on woman jh’s list, let man i be the second on woman jh’s list, then woman j,, prefers man i to man ih. Let woman j be the partner of man i in Me(R’). Since, in T’, woman j is the last on man i’s list and, by property (T4), woman j, is also on man i’s list, we must have that man i prefers woman j, to woman j. In summary, man i and woman jh are not married to each other in Me(R’) but they prefer each other to their partners in Me(R’). That is, the marriage M”(R’) is unstable, a contradiction. Thus, man ik is the second on woman jk’S list for each k and then (woman jk/man ik-1 mod r; k = O,l,. . . , P - 1) is a rotation exposed in T’.

Since, by Remark 36, R’ is the only rotation exposed in T’, we conclude that

R’ = (woman jk/man ik_1 mod r; k = 0, 1,. . . , r - 1). I

Dually, we have:

LEMMA 4.3. Let R’ = (woman jk/maII ik; k = 0, 1, . . . , P - 1) be a female-oriented rotation in 3. IfR is a male-oriented rotation in & such that M(R’) = M”(R), then

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20 Y.-C. HSUEH COROLLARY 4.4. For any R in & and R’ in 3,

M(R) = W(R’) if and only if M(R’) = Me(R).

PROOF. Assume R = (man &/woman j,; k = 0, 1, . . . , r - 1) and M(R) = MC(R). By Lemma 4.2., we have

R’=(womanjk/mani,_i modr;k=O,l ,..., r-l).

If & is a male-oriented rotation in & such that M(R’) = W(Q), then, by Lemma 4.3., Q = (man ik_1 ,,,od ,/WOmail j,-, mod ?; k = 0, 1, . . . , P - 1)

=(mani,/womanjrt; k=O,l,...,r-1) =R

Hence, M(R) = W(R). The “if” part then follows by duality. I

DEFINITION 4.5. A pair (R, R’) of male- and female-oriented rotations with M(R) = Me(R’) is called a dual pair of rotations. In notation, we write

Rd =R’ and (R’)d=R.

For any rotation Q, the rotation Qd is called the dual rotation of Q in opposite sex orientation. REMARK 4.6. From Corollary 4.4., it follows that (Qd)d = Q for any rotation Q.

LEMMA 4.7. For any two rotations RI and Rz in &,

PROOF. If R1 5 R2 in B, then M(R1) 5 M(R2) in C. Since M(Rl) = Me(@) and M(R2) = iW(R$), we have W(R$) 5’ MC(@) in Ld. H ence, @ 5’ Rf in B’. The proof of the “if” part

is similar. I

The duality in the stable marriage problem plays a central role in studying the structure of the rotation poset of a stable roommates instance. This is the main subject of the next section.

5. ROTATION POSETS OF THE STABLE ROOMMATES PROBLEM

In Section 1, we transform a stable roommates problem instance HI into a stable marriage problem instance MI and call MI the instance of the stable marriage problem corresponding to RI. The rotation posets (male- and female-oriented) of MI are also called the rotation posets of RI. To start exploiting the structure of these rotation posets, we make the following observation. OBSERVATION 5.1. Let c and 3 be the sets of male- and female-oriented rotations of RI, respec- tively. Since the two sets MS and WS of preference lists are identical up to sex reversal, these two sets ,? and 3 are also identical in the following sense:

If R=(man &/woman jk, k = O,l,. . . ,r - 1) is a male-oriented rotation in E, then R’=(woman ik/man ja, k = 0, 1, . . . , r - 1) is a female-oriented rotation in 3, and vice versa. Such property will be called the equal right property of rotations.

Henceforth, for the sake of convenience, we will make use of the following notations and ter- minology:

(1) A male-oriented rotation R = (man &/woman j,, k = 0, 1, . . . , r - 1) in & will be simply written 88

R = (ir/jk, k = O,l,. . . ,r - 1). (2) An assignment Q of the instance HI is denoted as a permutation

1 2 2n

CY= . . .

jl j2 j2,

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Structures of the stable matching problems 21 (3) Let M be a stable marriage of the instance MI. If M is of the form

M={mani/womanji; i=1,...,2n},

then the mapping UM: i H ji is a permutation of (1,. , . ,2n}, called the permutation corwponding to M.

(4) A permutation u of { 1,. . . ,2n} is called a feasible permutation of the instance RI if u is the permutation corresponding to a stable marriage of MI. If, in addition, u is an assignment of RI, then it is called a feasible assignment.

LEMMA 5.2. An assignment u of an instance RI of the stable roommates problem is stable if and only if it is feasible.

PROOF. It is trivial that Q is feasible if it is stable. Conversely, assume LY is feasible but is not stable. Then there exist two persons, say person i and person j, such that person i prefers person j to person a(i) and person j prefers person i to person a(j). Then, in the correspond- ing stable marriage problem instance MI, there exist man i and woman j such that man i prefers woman j to woman a(i) and woman j prefers man i to man a(j). That is, the marriage {man k/woman cr(k); k = 1,. . . ,2n} is unstable, a contradiction. Hence, a is stable if it is

feasible. I

Let R = (ib/jk; k = 0, 1, . . . , T - 1) be a rotation in &. Recall that the dual rotation Rd of R in the opposite sex orientation is a female-oriented rotation in 3 of the form Rd = (woman jk/man ik_1 mod r; k = O,l,... , P - 1). By the equal right property, the male- oriented rotation (j/i&l mod r; k = O,l,. . . , r - 1) is also in &. Now, we reach a position to establish the following theorem.

THEOREM 5.3. Given a stable roommates instance RI. Let 6 be the function from & into itself definedby:foranyR=(ik/jk; k=O,l,...,r-l),

6(R) = (jk/ik_l mod r; k = 0, 1, . . . , P - 1). Then, the structure B6 = (E; <,6) is a self-dualized poset.

PROOF. From Remark 4.6., Lemma 4.7. and the equal right property, it is easy to see that the function 6 : R H 6(R) is a dual assignment on (t; 5). Hence, B6 is self-dualized. I DEFINITION 5.4. For any rotation R in E, the rotation 6(R) given in Theorem 5.3. is called the dual rotation of R in same sex orientation.

REMARK 5.5. If there is no danger of confusion, the rotation S(R) will be simply called the dual rotation of R, and will be written as R*. Also, the male-oriented rotation poset B = (E; 5) will be simply called the rotation poset of a given stable roommates instance.

LEMMA 5.6. Let ‘?i be a hereditary subset of & and M be the stable marriage corresponding to 3-1. If for some rotation R, both R and R6 are in ‘Ii, then the permutation UM is not a feasible assignment.

PROOF. Assume R = (ik/jk, k = O,l,. . . , r- 1) and R6 be both in ‘X. Let T be the male-oriented table corresponding to ‘H. If man io /woman j is a pair in M, since R has been eliminated, woman j is ranked below woman jo in man io’s list. Similarly, since R6 has been eliminated, woman io cannot be paired with any man ranked below man jo. Thus, man j/woman io is not in M. 1

Dually, we have:

LEMMA 5.7. Let 3t be a hereditary subset of E, and Jet M be its corresponding stable marriage. For any rotation R in &, if 7f does not contain R and R6, then tbe permutation UM is not a feasible assignment.

We summarize the above results as in the next theorem.

THEOREM 5.8. Given an instance RI of the stable roommates problem. Let & be the set of all male-oriented rotations of RI. Let 3-1 be a hereditary subset of & and M be the stable marriage

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22 Y.-C. HSUEH

corresponding to ‘Ft. Then the permutation u,w is a feasible assignment of RI if and only if for each R in &, either R or Rb is in 3c but not both.

DEFINITION 5.9. A rotation R in & is called self-dual if R = R!.

COROLLARY 5.10. An instance of the stable roommates problem is unsolvable if and only if there is a self-dual rotation.

LEMMA 5.11. Any self-dual rotation is of odd length.

PROOF. Let R = (ib/jk; k = O,l,. . . ,P - 1) be a self-dual rotation of length r. Since R is self-dual, that is, (jkljk-1 mod t; k = O,l,...,T- 1) = (it/jk; k = O,l,. . . ,P - l), we have

{in; k = 0, 1,. . . , P - 1) = {jb; k = 0, 1, . . . , P - 1). Let js = i, for some m > 0. Then, observe that j, = &+I, . . . ,jr-,,+l = &_I. On the other hand, since im/jm = jc/ir-1, we have

i,_l=j,.Therefore,m=r-m-landr=2m+l. I

For the reason of completeness, we establish the converse of Theorem 5.3. at the end of this section.

THEOREM 5.12. Let Pa = (S; 5,6) be a finite self-dualized poset. Then the poset P = (S; 5) is the rotation poset of a stable roommates instance.

PROOF. See Appendix. I

6. COMBINATORIAL STRUCTURES OF THE STABLE MATCHING PROBLEMS

“Greedoids” are combinatorial structures introduced by Korte and Lov& [7,8] as a structural framework for the greedy algorithm. These structures generalize the well-known combinatorial structures “matroids” by extending the independence axioms of matroids from set systems to languages. There are other combinatorial structures, called Faigle geometries, which extend the concept of matroids on finite sets to posets [16,17]. We briefly introduce them in the following. DEFINITION 6.1. Let S be a finite set. A word (Y on S is a finite sequence of elements of S, and is shortly written as the form Q = 2122.. . a+, where xi’s are elements of S. The number r is called the lengih of o and is usually denoted as Ial. The collection of all possible words on S is denoted by S’ .

DEFINITION 6.2. Let L be a subset of S*. The pair (S; L) is called a language on S.

DEFINITION 6.3. A word o is called simple if no element in a is repeated. A language (S; L) is called simple if any word in L is simple.

DEFINITION 6.4. A language (S; L) is called henditary if it satisfies: g;; 4 E L; if a E L and CY = & tben ,L? E L.

DEFINITION 6.5. Let (S;L) b e a simple hereditary language on S. Any word in L is called a feasible word. Maximal feasible words are called basic words. An element 2 in S is called an

isthmus of L ifit belongs to every basic word.

DEFINITION 6.6. A simple hereditary language 0 = (S; L) is called a gnzedoid, if in addition it satisfies:

((5’3) if o, P E L and Ial > IPI,

then there is an element x E Q such that @x is in L.

Note that property (G3) means that every feasible word can be extended to a basic word and every basic word is of the same length.

DEFINITION 6.7. The length of any basic word of a greedoid Q is called the rank of $7.

Let 0 = (S; L) be a greedoid. For any word Q, let h denote the underlying set of elements in o. Then define a binary relation E on L by:

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Structu- of the stable matching problems 23 It is easy to see that g is an equivalence relation on L. Let 1 be the set of equivalence classes induced by Y. Obviously, the structure f$ = (S; z) is a greedoid.

DEFINITION 6.8. The greedoid c = (S; f;) is called the quodient grcedoid of B relative to the equivalence relation Y.

DEFINITION 6.9. Let P = ($5) be a finite poset. For any subset A of S, a simple word a = 21x2.. . x,. is called a linear extension of A if zi = A and xi 5 xj implies i 2 j for any 1 < i,j < P.

LEMMA 6.10. Let P = (S; 5) be a finite poset and cy,p,y be words on S such that (Y = /37. If (Y is a linear extension of &‘, then p is a linear extension of 6. If, in addition, & is a hereditary subset of S, then so is p.

PROOF. Let p = x122...+ and 7 = zr+i .‘.x~+~. If cr is a linear extension of &, then, in particular, ti 5 xj implies i 5 j for 1 5 i, j 5 r. Hence, p is a linear extension of p. Moreover, if & is hereditary, then for any xj E $ C & and y 5 xj, we have 21 E &. That is, g = x) for some 1 2 k 5 T + s. Since LY is a linear extension, y = zk 5 zj implies k _< j. Hence, y E ,8 and B is

hereditary. I

DEFINITION 6.11. Let P = (S;<) b e a finite poset and (S; L) be a greedoid on S. Then the structure r = (S; 5, L) is a faigle geometry if:

(F4) if Ly E L, then (Y is a linear extension of&;

(FS) for alI hereditary subsets HI and HZ with Hi c Hz, if x E Hi is an isthmus of L II H,’ then x is an isthmus of L rl Hf.

REMARK 6.12.. Ifr = (S; 5, L) is a FGgle geometry and (S; 1) is the quotient greedoid of (S; L), then it is easy to see that T = (S; 5, L) is also a Faigle geometry on (S; I). This geometry r will be called the quotient geometry of F.

EXAMPLE 6.13. Let (S; 5) be a poset with the following diagram x3

1% x4

x1 0 x2

Let LI = {~,~~,x~x~}. Then (S; L ) 1 is a greedoid. However, consider the hereditary subsets HI = ($1) and Hz = { x1, x2}, since xl is an isthmus of L1 n H,’ = LI but is not an isthmus of LI f~ Hf = {d}, the structure (S; 5, LI) is not a Faigle geometry.

If we let L2 = (4, x~,z~,xiz2,x~z~}, then it is easy to see that the structure (S;r, Lz) is a Faigle geometry.

THEOREM 6.14. Let P = (S; s) be a finite poset and let (S; L) be a simple hereditary language on S such that the basic words are the linear extensions of S. Then (S; 5, L) is a Faigle geometry on P.

PROOF. From Lemma 6.11., it is easy to verify that properties (Hl), (H2) and (F4) hold. Observe that any element in a hereditary subset H is in all linear extensions of H. That is, every element in H is an isthmus of L rl H’. Particularly, property (F5) also holds. It remains for us to show property (G3) is satisfied. Let cr,P E L with loI > [/?I. Let x be a minimal element of 6 - p. We claim that px is in L. First, since /3 is hereditary-and x is not in /3, we have that x is not < y for any y in p. Hence, ,Bx is a linear extension of p U i. Next, let z E ($ U 2) and y 5 z.

Casel. .zEP.

Since p is hereditary, we have y E @ c (p U 2). Case2. x=x.

Since 5 is hereditary, y E &. If y 4,8, since x is minimal in & - p, we have y = x. Hence, either y E p or y = x.

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24

DEFINITION 6.15. Let as in Theorem 6.14. is S in [7,8]).

Y.-C. HSUEH

P = (S; 5) be a finite poset. The FaigJe geometry (S; 5, L) on P obtained caJJed the complete Faigle geometry on P (is called the p<wet greedoid OLI COROLLARY 6.16. Let (S; 5, L) be the complete Faigle geometry on the finite poset P = (S; 5). Then there is one-one correspondence between the feasible words of ,! and the hereditary subsets OfS.

PROOF. The mapping by sending 6 to h is one to one from ,? onto Hered( 1 Therefore, the quotient geometry (S; 5, z) is a combinatorial aspect of the distributive lattice

L = (Hered(P);n,U,4,S).

Corresponding to the stable marriage problem, we have the following theorem.

THEOREM 6.17. Given an instance I of the stable marriage problem. Let t? = (8; 5) be the rotation poset of I, and Jet 3 = (E; 5, L) be th e complete FaigJe geometry on B. Then there is a one-to-one correspondence between the stable marriages of I and the feasible wor& of the quotient geometry F = (E; <,I) on B.

DEFINITION 6.18. Given a se&dualized poset Pa = (S; 5,5). Let 3 = (S; 5, L) be the complete FaigJe geometry on (S; 2). A word (Y in L is called dual-exclusive if for any z in S, x and x6 cannot appear in (Y at the same time.

It should be clear that a dual-exclusive word does not contain any self-dual element. Let Le = {(Y E L : Q is dual-exclusive}.

LEMMA 6.19. The structure 3” = (S; 5, L’) is a FaigJe geometry.

PROOF. It is enough for us to verify tha_t property (G3) holds. Let cr,p be in Lc with Ial > IPI. Let j” = (x6 : x E /I) and assume & c (p U fi”). Since Q is dual-exclusive,

I4 I (IPI + IP6W = IPL

we have a contradiction. Hence, the set A = & - (PUfi”) is nonempty. Choose a minimal element x in A. Obviously, /?x is dual-exclusive and is a linear extension of fi U i. To complete the proof, we have to show that p U 5 is hereditary. Let z E (3 U 5) and y 5 z.

Casel. *Ej.

Since j is hereditary, z E fi implies y E p. Case2. 2=x.

Since & is hereditary, we have y E &. If y # (@ U ,L@), then the minimality of x implies y = z. If y E (p u p6), then we must have y E 6; otherwise, y6 E j and ~~~y~imply&~audx~~~.

Therefore, the set p U 5 is hereditary. I

REMARK 6.20. The rank of 3e is less than or equal to ISl/2. If the rank of 3’ is less than ISl/2, then there must be an element x E S with x = x6, i.e., x is a self-dual element.

DEFINITION 6.21. The FaigJe geometry Te is called the dual-exclusive geometry on (S; 5). Corresponding to the stable roommates problem, we have:

THEOREM 6.22. Given an instance RI of the stable roommates problem. Let B = (C; 5) be the rotation poset of RI and 3e = (8; 5, Le) be the dual-exclusive geometry on B. Then:

(1) The instance RI is soJvabJe if and only the rank of 3e is equal to lEl/2.

(2) If RI is solvable, then there is a one-t&one correspondence between the stable assignments of RI and the basic words of the quotient geometry p = (E; 5, ze) of 3”.

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Structures of the stable matching problems 25

7. CONCLUSION

Given a stable marriage instance of size n. Its rotation poset can be constructed [9] in 0(n2) time. Hence, the rotation poset of a stable roommates instance of size n can be constructed in 0(n2) time as well. It should be noted that the “singleton rotations” mentioned in Irving [4,18] are the rotations R with R _< R6, and the “nonsingleton rotations” are the rotations R with RllR6. Moreover, a path from the root to a leaf in the execution tree D defined in [5] is a basic word in the Faigle geometry 3e and a path set is a basic word in the quotient geometry Te. Therefore, the Faigle geometry 3e can be served as the universal structure on the rotation-elimination algorithm of the stable roommates problem.

Finally, since the greedy algorithm for some structure on finite poset works if and only if this structure is a Faigle geometry [16], the greedy algorithm might work for some optimization problems of the stable matching problems.

REFERENCES

1. D. Gale and L. Shapley, College admissions and the stability of marriage, Amer. Murh. Monlhlly 69 (1962).

2. C. Blair, Every finite distributive lattice is a set of stable matchings, J. Comb. Theory, Setie~ A 37,353-356

(1984).

3. D. Gusfield, R. Irving, P. Leather and M. Saks, Every finite distributive lattice is a set of stable matchings for a smaU stable marriage, J. Comb. Theory, Series A 44, 304-309 (1987).

4. R. Irving, On the stable mom-mates problem, Res. Report CSC/86/R5 , Department of Computing Science, University of Glasgow, (1986).

5. D. Gusfield, The structure of the stable roommate problem: efficient representation and enumeration of all

stable assignments, SIAM J. Comput. 1’7, 742-769 (1988). 6. D. WeIsh, Matroid Theory, A.P., (1976).

7. B. Korte and L. Lov&z, Posets, matroids, and greedoids, in matroid theory, Coil. M&h. Sot. J. Bolyoi 40, 239-265 (1982).

8. B. Korte and L. Lovbz, Greedoids-A structural framework for the greedy algorithm, P~o~~JJ in Combi- natorial Optimization, A.P., pp. 221-243, (1984).

9. D. Gusfield, Three fast algorithms for four problems in stable marriage, SIAM J. Comput. 16, 111-128 (1987).

10. R. Irving and P. Leather, The complexity of counting stable marriages, SIAM J. Comprt. 15, 655-667 (1986).

11. D. McVitie and L.B. Wilson, The stable marriage problem, CACM 14, 486-492 (1971). 12. G. Birkhoff, Lattice Theory, AMS, Providence, R.I., (1979).

13. P. Crawley and R. Dilworth, Algebraic Theory of Latticer, Prentice-Hall Inc., Englewood Cliffs, N.J., (1973). 14. G. Griitzer, General Lattice Theory, A.P. (1978).

15. D. Knuth, Mariagea Stables, Les Presses de L’Universit~ de Montreal, (1976).

16. U. Faigle, The greedy algorithm for partially ordered sets, Dircrete Math. 28,153-159 (1979).

17. U. Faigle, Geometries on partially ordered sets, J. Comb. Theory, Series B 28, 26-51 (1989). 18. R. Irving, An efficient algorithm for the stable room-mates problem, J. Aigorithmr 8, 577-595 (1985).

APPENDIX

EVERY FINITE SELF-DUALIZED POSET IS THE

ROTATION POSET OF AN INSTANCE OF THE STABLE ROOMMATES PROBLEM

Let P6 = (S; I, 6) be a finite self-dualized poset. Label the elements of s so that

s = {Il,l;,21,~;, . . . ,lk,t;,Ok+l=l;+1’. . . vQ+#=~f+sl and

either si 5 zf or ZillZ” in P,foreach i=l,..., k.

Foreachi=l , . . . , k, we associate it with four persons, person 4(i - 1) + j, j = 1,2,3,4, and construct a portion of the preference lists as follows:

iirst current-last

position position person 4(i-1)+1 4(i-1)+3 . . . 4(i-1)+4 4(i-1)+2 4(i-1)+4 . . . 4(i-1)+3 4(i-1)+3 4(i-1)+2 . . . 4(i-l)+l 4(i-1)+4 4(i-1)+1 . . . 4(i-1)+2

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26 Y.-C. HSUEH

For&t=1 ,._.,s, we aseociate it with three p ersons,pezson3(t-1)+4k+j, j=1,2,3audcoustructa portion of the preference lists as follows:

I

first current-last

DoBition DOSitiOll

P-n 3(t-1)+4k+1 3(t-1)+4k+2 . . . 3(t1)+4k+3 3(t-1)+4k+2 3(t-1)+4k+3 . . . 3(e1)+4k+1 3(t-1)+4k+3 3(t-1)+4k+1 . . . 3@-1)+&+2 Then consider the following cases:

case 1.

case 2.

case 3.

Case 4.

Case 5.

Zi 5 Xf for some i, l<i<k.

Place 4(i- 1) +4 in any position between the iirst and curmnt-last poaitionson thelist ofpemon4(i-1)+3 and place 4(i - 1) + 3 in any position between the first and current-la& pceitioxu on the Et of perma~ 4(i - 1) + 4.

li IZj forsomei#j, 1 s i,j 5 k.

Place 4(i-1)+3inanyposition between thefh-st andcurrent-last poaitionsonthelist ofpemon4(j-l)+l and place 4(j - 1) + 1 in any position between the f&et and curre&laatpoeitionsonthelistofpereon 4(i - 1) + 3.

~i<~~ffarsOmei#j, 15 i,j < k.

Place4(i-1)+3inanypoeitionbetwecn the iirst audcmrent-la& poaitioneon thelist ofpemon4(j-1)+3 and place 4(j - 1) + 3 in any position between the fhxt and cunwnt-lastpomitionsonthelistofperson 4(i - 1) + 3.

oi 5 zk+t for some i and t, 1 5 i 5 k and 1 5 t 5 S.

Place4(i-1)+3inanyplacebetweenthefirstsndcurrent-laetp~t~Monthe~tdpsrroa3(t-l)+4lr+l

and place 3(t - 1) + 4k + 1 in any position between the tit and curre.nMaat ponitiono ou the list of person 4(i - 1) + 3.

rk+t 5 xi for some i and t, lli<kandl<t<s.

Place 3(t - 1) + 4k + 1 in any poeition between the first and current-last poeitione on the list of person 4(i - 1) + 1 and place 4(i - 1) + 1 in any position between the &at and cunwnt-last poeitionm on the list ofpexson3(t-1)+4k+l.

To complete the preference lists, place any missing entries after the cmTent-la6tpoaitiononeachli&inany order.1lsisodd,thenwejainpaaon3s+4k+ltothegroup,plrrce3s+4k+lintheicutpaiti~onthe~t

ofperson jforeach j=l,..., 3s + 4k, and fill up the list of person 3s + 4k + 1 arbitrcuily.

Let RI denote the constructed instance of the roommates problem. We claim that the rotation poaet of RI is isomorphic to the self-dualized poeet Ps.

Applying the Gale-Shapley algorithm to the corresponding stable ma&age instance MI, we obtain the following male-optimal table:

first cmTe&lMt

position pcmition

male lists: man1 f 1 . . . 11

3s+4k f3.+4k *** hr+lk 3s+4k+l 3s+4k+1 (if II is odd)

iht current-last

position position female lists: woman 1 f 1 . . . 11

where

3s+4k h4k -** h4*

3s+4k+l 3s+4k+1 (ifs is odd)

(l)ifj=4(i-l)+l, l~i~k,thenfj=4(i-1)+3~lj=4(i-l)+~ (2)ifj=4(i-1)+2, llilk,then jj=4(i-1)+4~lj=4(i-l)+3;

(3) if j = 4(i - 1) + 3, 1 5 i 5 k, then _fj = 4(i - 1) + 2 and lj = 4(i - 1) + 1; (4) if j = 4(i - 1) + 4, 1 < i 5 k, then fj = 4(i - 1) + 1 and Ij = 4(i - 1) i- 2; (5)ifj=3(t-1)+4k+l, 1<t<s,thenjj=3(t-l)t4k+2~lj=3(t-l)+4k+3;

(6) if j = 3(t - 1) + 4k + 2, 1 < t 5 8, then jj =3(t-1)+4k+3dlj=3(t-1)+4k+1; (7)ifj=3(t-I)+4k+3, l~t~s,thenjj=3(t-l)+4k+l~d~j=3(t-l)+~+2~

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Structures of the stable matching problems 27 Observe that the rotationa of the inetance MI are:

J& =(4(i - 1) + 1/4(i - 1) + 3,4(i - 1) + 2/4(i - 1) + 4)

Rf =(4(i - 1) + 3/4(i - 1) f 2,4(i - 1) + 4/4(i - 1) + 1)~ i= 1 ,..., k,

and

Rt =(3(t - 1) + 4k + 1/3(t - 1) + 4k + 2, 3(t - 1) + 4k + 2/3(t - 1) + 4k + 3,3(t - 1) + 4k + 3/3(t - 1) + 4k + 1,

=R: t=1 ( . . . ,6.

Furthermore, observe that the mapping zi w Ri ia 8~ isomorphit~~ from the Po& P6 = (S; <,6) into the rotatim poset B* = (E; <,6) of MI. For instance, if xi 5 sj, 1 5 i # j 5 k, then on the list of pereon 4(j - 1) + 1 before the elimination of Ri, 4(i - 1) + 3 is sitting between 4(j - 1) + 3 and 4(j - 1) + 4. Thus, Z& < Rj. The other wzes aresimilar. Wes ummarize the above result as the following theorem.

THEOREM. Let Ps be a self-dualized paset with 2k non-selfdual elements and 8 self-dual elements. Then there is an instance RI of size 4k + 3s if 8 is even and 4k + 3s + 1 if B is odd of the stable roommates problem such that Pa is order-isomorphic to the rotation poset of RI.

EXAMPLE. Let

Then a corresponding instance of the stable roommates problem is one with the following prefexence lists.

o/

Ee”~f~ZE lists arbitrarily)

The rotations of the above instance are:

and

RI = (l/3,2/4),

R: = (3/%4/l),

R2 = (s/7,6/9), R26 = (7/6,8/5),

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