Finding Stable Matchings that are Robust to Errors in the Input

03/30/2018 ∙ by Tung Mai, et al. ∙ 0

Given an instance A of stable matching, let B be the instance that results after introducing one error from a polynomially large class of errors, and chosen via a discrete probability distribution. We want to find a stable matching for A that maximizes the probability of being stable in B as well. Via new structural properties, related to the lattice of stable matchings, we give a polynomial time algorithm for this problem. To the best of our knowledge, this is the first work to explore the issue of robustness to errors for this problem.

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1 Introduction

Ever since its introduction in the seminal 1962 paper of Gale and Shapley [GS62]

, the stable matching problem has been the subject of intense study from numerous different angles in many fields, including computer science, mathematics, operations research, economics and game theory, e.g., see the books

[Knu97, GI89, Man13]. The very first matching-based market, namely matching medical interns to hospitals, was built around this problem, e.g., see [GI89, Rot16]. Eventually, this led to an entire inter-disciplinary field, namely matching and market design [Rot16]. The stable matching problem and market design were the subject of the 2012 Nobel Prize in Economics, awarded to Roth and Shapley [RS12].

Another topic that has been extensively studied is the design of algorithms that produce robust solutions, e.g., see the books [CE06, BTEGN09]. Yet, there is a paucity of results at the intersection of these two topics. Indeed, we are aware of only two works [ABG16, ABF17]; see Section 1.2 for a detailed description of these works. As explained there, their notion of robustness of a stable matching solution is quite distinct from ours. Our first contribution is a combinatorial polynomial time algorithm for finding a robust stable matching for a special case of allowable errors in the input. Our second contribution is to initiate work on a new structural question, namely finding relationships between the lattices of solutions of two “nearby” instances of stable matching.

We note that a particularly impressive aspect of the stable matching problem is its deep and pristine combinatorial structure. This in turn has led to efficient algorithms for numerous questions studied about this problem, e.g., see the books mentioned above. In studying the new structural question, we have restricted ourselves to “nearby” instances which differ in only one agent’s preference list. Clearly, this is only the tip of the iceberg as far as “nearby” instances go. Moreover, the structural results are so clean and extensive – see also our followup paper [MV18] – that they are likely to find algorithmic applications beyond the problem of finding robust solutions. In particular, with ever more interesting matching-based markets being designed and launched on the Internet [Rot16], these new structural properties could find interesting applications and are worth studying further.

We will introduce our version of the problem of finding robust stable matchings via the following model: Alice has an instance of the stable matching problem, over boys and girls, which she sends it to Bob over a channel that can introduce errors. Let denote the instance received by Bob. Let denote a polynomial sized domain from which errors are introduced by the channel; we will assume that the channel introduces at most one error from . We are also given the discrete probability distribution, over , from which the channel picks one error. In addition, Alice sends to Bob a matching, , of her choice, that is stable for instance . Since consists of only numbers of bits each, as opposed to which requires numbers, Alice is able to send it over an error-free channel. Now Alice wants to pick in such a way that it is stable for instance and has the highest probability of being stable for the instance received by Bob. Hence she picks from the set

We will say that such a matching is robust. We seek a polynomial time algorithm for finding such a matching.

Clearly, the domain of errors, , will have to be well chosen to solve this problem. A natural set of errors is simple swaps, under which the positions of two adjacent boys in a girl’s list, or two adjacent girls in a boy’s list, are interchanged. We will consider a generalization of this class of errors, which we call upward shift. For a girl , assume her preference list in instance is . Move up the position of so ’s list becomes , and let denote the resulting instance. Then we will say that is obtained from by an upward shift. An analogous operation is defined on a boy ’s list. The domain consists of all such upward shifts on each boy’s and each girl’s list. Clearly, is , i.e., it is polynomially bounded. As will be clarified later, the operation of downward shift is much harder to deal with and we leave it as an open problem; see the Remark at the end of Section 4. In the rest of the paper, we will shorten “upward shift” to simply “shift”. Let us also clarify that we will deal with the generalization of stable matching in which incomplete preference lists are allowed.

We next study the set of robust stable matchings for instance under probability distribution . We show that this set forms a sublattice of the lattice of stable matchings for and we give an efficient algorithm for finding a succinct representation for this set. This representation has the property that any member of the set can be efficiently retrieved from it.

Our main theorem is the following.

Theorem 1.

Given an instance of the stable matching problem, with possibly incomplete preference lists, and a probability distribution over the domain of errors defined above, there is an efficient algorithm that finds:

  • A robust stable matching for .

  • A partial order on elements such that its closed sets are isomorphic to the set of robust stable matching for . The latter set forms a lattice which is furthermore a sublattice of the lattice of all stable matchings for .

The main computational step of our algorithm is to find one max-flow in a network with vertices, where is the number of agents of each sex in instance .

1.1 Overview of results and technical ideas

Henceforth, we will assume that we are dealing with a stable matching instance with complete preference lists; we will finally generalize to incomplete lists in Section 4.1.

Let us first summarize some well-known structural facts, e.g., see [GI89]. The set of stable matchings of an instance form a distributive lattice: given two stable matchings and , their meet and join involve taking, for each boy, the optimal or pessimal choice, respectively. It is easy to show that the resulting two matchings are also stable. The extreme matchings of this lattice are called boy optimal and girl-optimal matchings. A deep notion about this lattice is that of a rotation. A rotation, on an ordered list of boy-girl pairs, when applied to a matching in which all these boy-girl pairs are matched to each other, matches each boy to the next girl on the list, closing the list under rotation. The pairs and the order among them are so chosen that the resulting matching is also stable; moreover, a rotation on a subset of these pairs, under any ordering, leads to a matching that is not stable. Hence, a rotation can be viewed as a minimal change to the current matching that results in a stable matching. Rotations help traverse the lattice from the boy-optimal to the girl-optimal matching along all possible paths available.

Birkhoff’s [B37] fundamental theorem for finite, distributive lattices shows that corresponding to each such lattice, , there is a partial order, say , such that the closed sets of are isomorphic to the elements of . It turns out that for a lattice arising from a stable matching instance, the partial order is defined on a set, say , of rotations. Moreover, if is a closed set of , then starting in the lattice from the boy-optimal matching and applying the rotations in in any order consistent with a topological sort of , we will reach the stable matching corresponding to .

Let and be two instances of stable matching over boys and girls, with sets of stable matchings and , respectively, and lattices and , respectively. Then, it is easy to see that the matchings in form a sublattice in each of the two lattices. Next assume that instance results from applying a shift operation, defined above, to instance . Then, we show that is also a sublattice of (Theorem 2). We use this fact crucially to show that there is at most one rotation, , that leads from to and at most one rotation, that leads from to (Theorem 3). Moreover, we can obtain efficiently this pair of rotations for each of the polynomially many instances that result from the polynomially many shifts (Proposition 4).

Let be the partial order for the lattice of instance . It is easy to see that a matching , corresponding to a closed set of , is in iff whenever , .

In Section 4 we give an integer program whose optimal solution is a robust stable matching for the given probability distribution on shifts. The IP has one indicator variable, , corresponding to each rotation in . The constraints of the program ensure that the set of rotations that are set to 0 form a closed set. The rest of the constraints and the objective function ensure that the corresponding matching minimizes the probability that is in . We obtain the LP-relaxation of this IP and then obtain the dual LP. We interpret the latter as solving a maximum circulation problem in a special network. This in turn is solvable as a max-flow problem on a network having vertices and the solution yields an integral optimal solution to the LP, hence yielding an efficient combinatorial algorithm for finding a robust stable matching.

In Section 5, we next study the set of robust stable matchings for instance under probability distribution . We show that this set forms a sublattice of the lattice of stable matchings for (Lemma 13). We give an efficient algorithm for finding a succinct representation for this set. This representation has the property that any member of the set can be efficiently retrieved from it (Lemma 15).

1.2 Related work

Aziz. et. al. [ABG16] considered the problem of finding stable matching under uncertain linear preferences. They proposed three different uncertainty models:

  1. Lottery Model: Each agent has a probability distribution over strict preference lists, independent of other agents.

  2. Compact Indifference Model: Each agent has a single weak preference list in which ties may exist. All linear order extensions of this weak order have equal probability.

  3. Joint Probability Model: A probability distribution over preference profiles is specified.

They showed that finding the matching with highest probability of begin stable is NP-hard for the Compact Indifference Model and the Joint Probability Model. For the very special case that preference lists of one gender are certain and the number of uncertain agents of the other gender are bounded by a constant, they gave a polynomial time algorithm that works for all three models.

The joint probability model is the most powerful and closest to our setting. The main difference is that in their model, there is no base instance, called in our model. The opportunity of finding new structural results arises from our model precisely because we need to consider two “nearby” instances, namely and the instance obtained by executing a shift.

Aziz. et. al. [ABF17] introduced a pairwise probability model in which each agent gives the probability of preferring one agent over another for all possible pairs. They showed that the problem of finding a matching with highest probability of being stable is NP-hard even when no agent has a cycle in his/her certain preferences (i.e., the ones that hold with probability 1).

1.2.1 A matter of nomenclature

Assigning correct nomenclature to a new issue under investigation is clearly critical for ease of comprehension. In this context we wish to mention that very recently, Genc et. al. [GSOS17] defined the notion of an -supermatch as follows: this is a stable matching in which if any pairs break up, then it is possible to match them all off by changing the partners of at most other pairs, so the resulting matching is also stable. They showed that it is NP-hard to decide if there is an -supermatch. They also gave a polynomial time algorithm for a very restricted version of this problem, namely given a stable matching and a number , decide if it is a -supermatch. Observe that since the given instance may have exponentially many stable matchings, this does not yield a polynomial time algorithm even for deciding if there is a stable matching which is a -supermatch for a given .

Genc. et. al. [GSSO17] also went on to defining the notion of the most robust stable matching, namely a -supermatch where is minimum. We would like to point out that “robust” is a misnomer in this situation and that the name “fault-tolerant” is more appropriate. In the literature, the latter is used to describe a system which continues to operate even in the event of failures and the former is used to describe a system which is able to cope with erroneous inputs, e.g., see the following pages from Wikipedia [Wikb, Wika].

2 Preliminaries

2.1 The stable matching problem

The stable matching problem with incomplete preference lists takes as input a set of boys and a set of girls and a bipartite graph , where is a set of edges connecting certain boys to certain girls. Each agent has a preference ranking the subset of opposite sex that are neighbors of in . If cannot be paired to one of the neighbors, it prefers remaining unpaired rather than getting paired to any of the rest of the agents of opposite sex. The notation indicates that girl strictly prefers to in her preference list. Similarly, indicates that the boy strictly prefers to in his list.

A matching is a pairing of boys and girls so it is a maximum matching in . Thus, if is the complete bipartite graph, will be a perfect matching in it. For each pair , is called the partner of in (or -partner), denoted by , and vice versa. For a matching , a pair is said to be blocking if and prefer each other to their partners in ; note that and/or need not be matched under . Matching is stable if there is no blocking pair w.r.t. .

As stated in Section 1.1, from here on we will assume that is a complete bipartite graph. Finally, we will relax this assumption in Section 4 to derive the more general result.

2.2 The lattice of stable matchings

Let and be two stable matchings. We say that dominates , denoted by , if every boy weakly prefers his partner in to (he either has the same partner or prefers his partner in to ). It is well known that the dominance partial order over the set of stable matchings forms a distributive lattice [GI89], with meet and join defined as follows. The meet of and , , is defined to be the matching that results when each boy chooses his more preferred partner from and ; it is easy to show that this matching is also stable. The join of and , , is defined to be the matching that results when each boy chooses his less preferred partner from and ; this matching is also stable. These operations distribute, i.e., given three stable matchings ,

It is easy to see that the lattice must contain a matching, , that dominates all others and a matching that is dominated by all others. is called the boy-optimal matching, since in it, each boy is matched to his most favorite girl among all stable matchings. This is also the girl-pessimal matching. Similarly, is the boy-pessimal or girl-optimal matching.

2.3 Rotations help traverse the lattice

A crucial ingredient needed to understand the structure of stable matchings is the notion of a rotation, which was defined by Irving [Irv85] and studied in detail in [IL86]. A rotation takes matched pairs in a fixed order, say and “cyclically” changes the mates of these agents, as defined below, to arrive at another stable matching. Furthermore, it represents a minimal set of pairings with this property, i.e, if a cyclic change is applied on any subset of these pairs, with any ordering, then the resulting matching has a blocking pair and is not stable. After rotation, the boys’ mates weakly worsen and the girls’ mates weakly improve. One can traverse from to by applying a suitable sequence of rotations, given by any topological sort of the rotation poset (defined below).

Let be a stable matching. For a boy let denote the first girl on ’s list such that strictly prefers to her -partner. Let denote the partner in of girl . A rotation exposed in is an ordered list of pairs such that for each , , is , where is taken modulo . In this paper, we assume that the subscript is taken modulo whenever we mention a rotation. Notice that a rotation is cyclic and the sequence of pairs can be rotated. is defined to be a matching in which each boy not in a pair of stays matched to the same girl and each boy in is matched to . It can be proven that is also a stable matching. The transformation from to is called the elimination of from .

Let be a rotation. For , we say that moves from to , and moves from to . If is either or is strictly between and in ’s list, then we say that moves below . Similarly, moves above if is or between and in ’s list.

2.4 The rotation poset

Let be an instance of stable matching problem and let be the lattice of its stable matchings. Corresponding to lattice , there is a partial order such that the closed subsets of are in one-to-one correspondence with the matchings in ; a closed subset is a subset of the poset such that if an element is in the subset then all of its predecessors are also included. Since lattice arises from a stable matching instance, it turns out that is defined over a set of rotations. This partial order on rotations is called rotation poset. Given a closed subset , the corresponding matching is found by eliminating the rotations starting from according to the topological ordering of the elements in the subset. We say that generates . Let be closed subset generating and . Then , are closed subsets generating and respectively.

Lemma 1 ([Gi89], Lemma 3.2.1).

For any boy and girl , there is at most one rotation that moves to , below , or above . Moreover, if moves to and moves from then .

Lemma 2 ([Gi89], Lemma 3.3.2).

contains rotations and can be computed in polynomial time.

Lemma 3 ([Gi89], Theorem 2.5.4).

Every rotation appears exactly once in any sequence of elimination from to .

3 Structural Results

3.1 The stable matchings in form a sublattice of

Let and be the sets of all stable matchings under instance and respectively. Let . In other words, is the set of stable matchings in that become unstable in . In this section we show that forms a lattice. We first prove a simple observation.

Lemma 4.

Let . The only blocking pair of under instance is .

Proof.

Since , there must be a blocking pair under . Assume is not , we will show that must also be a blocking pair in . Let be the partner of and be the partner of in . Since is a blocking pair in , and . The preference list of remain unchanged from to , so . Next, we consider two cases:

  • If is not , the preference list of does not change. Therefore, , and hence, is also a blocking pair in .

  • If is , for all pairs such that and , we also have . Therefore, is a blocking pair in .

This contradicts the fact that is stable under . ∎

Recall that are boys right above in ’s list such that the position of is shifted up to be above in . From Lemma 4, we can then characterize the set .

Lemma 5.

is the set of all stable matchings in that match to a partner between and in ’s list, and match to a partner below in ’s list.

Proof.

Assume is a stable matching in that contains for and such that . In , prefers to , and hence is a blocking pair. Therefore, is not stable under and .

To prove the other direction, let be a matching in . By Lemma 4, is the only blocking pair of in . For that to happen, and . We will show that for . Assume not, then , and hence, . Therefore, is a blocking pair in , which is a contradiction. ∎

Let be the boy-optimal lattice formed by .

Theorem 2.

The set forms a sublattice of .

Proof.

Assume is not empty. Let and be two matchings in . By Lemma 5, and both match to a partner between and in ’s list, and match to a partner below in ’s list. Since is the matching resulting from having each boy choose the more preferred partner and each girl choose the least preferred partner, also belongs to the set characterized by Lemma 5. A similar argument can be applied to the case of . Therefore forms a sublattice of . ∎

We will denote the lattice formed by as .

3.2 Rotations going into and out of the sublattice

Let be a stable matching in and be a rotation exposed in with respect to instance . If and for a set of stable matchings, we say that goes into . Similarly, if and , we say that goes out of . Let the set of all rotations going into and out of be and , respectively.

Let be the set of possible partners of in any stable matching in , where . Let be a rotation moving to , be the rotation moving below and be a rotation moving from (see 2.3 for definitions). Note that each of and might not exist.

Lemma 6.

can only contain , . can only contain .

Proof.

Consider a rotation . There exists such that . By Lemma 5, matches to a partner between and in ’s list, and matches to a partner below in ’s list. Moreover, either does not contain for any , or contains where , or both. If does not contain for any , then . If contains where , then .

Consider a rotation . There exists such that . Again, by Lemma 5, contains for and where . Since dominates in the boy optimal lattice, must prefer to his partner in . Hence, matches to a partner below in ’s list. Therefore, must not contain for any . It follows that must be . ∎

Lemma 7.

If both and exist then .

Proof.

Assume that and there exists a sequence of rotation eliminations, from to a stable matching in which is exposed, that does not contain . Since moves below , is matched a partner higher than in her list in . Therefore, the partner can only be or a boy higher than in ’s list.

Consider any sequence of rotation eliminations from to . In the sequence, the position of ’s partner can only go higher in her list. Therefore, cannot be exposed in any matching in the sequence. It follows that is not exposed in a sequence of eliminations from to , which is a contradiction by Lemma 3. ∎

Theorem 3.

There is at most one rotation in and at most one rotation in . Moreover, the rotation in must be either or , and the rotation in must be .

Proof.

By Lemma 6, can contain at most 2 rotations, namely and if they are distinct. By Lemma 7, if both of them exist, . Hence, can contain at most one rotation, and it is either or .

Again, by Lemma 6, can contain at most one rotation, namely if it exists. ∎

By Theorem 3, there is at most one rotation coming into and at most one rotation coming out of .

Proposition 4.

and can be computed in polynomial time.

Proof.

Since we can compute efficiently according to Lemma 2, each of , and can be computed efficiently.

First we can check possible partners of and with respect to instance . By Lemma 5, is empty if none of the possible partners of is between and in ’s list or none of the partners of is below in ’s list. It follows that both and do not exist. Hence we may assume that such a case does not happen.

Suppose exists. If exists and , . Otherwise, , and if exists.

Suppose does not exist. If exists, . If exists, . ∎

Lemma 8.

Let be a matching in and be the corresponding closed subset in . If exists, must contain . If exists, must contain . If exists, must not contain .

Proof.

If exists, does not contain for any . Since , by Lemma 5 matches to a boy between and in her list. The set of rotations eliminated from to must include .

If exists, can not be below in . Since is below in , by Lemma 5 the set of rotations eliminated from to must include .

Assume that exists and contains . Since moves up from , can not contain for any . This is a contradiction. ∎

3.3 The rotation poset for the sublattice

From the previous section we know that is a sublattice of . In this section we give the rotation poset that generates all stable matchings in this sublattice.

We may assume that . If exists, let and be the matching generated by . Otherwise, let . Similarly, let be the matching generated by , where , if exists, and otherwise.

Lemma 9.

is the boy-optimal matching in , and is the girl-optimal matching in .

Proof.

Let be a matching in generated by a closed subset . By Lemma 8, if exists, must contain . Since is the minimum set containing , . Therefore, .

To prove that , we show . Assume otherwise, then there exists a rotation such that . It follows that , and hence . Since contains and is a closed subset, must also contain . This is a contradiction by Lemma 8. ∎

Theorem 5.

is the rotation poset generating .

Proof.

Let be a matching in generated by a closed subset . Let . We show that is a closed subset of and eliminating the rotations in starting from according to the topological ordering of the elements gives .

First trivially. Since , does not contain by Lemma 8. Therefore, does not contain , and . It follows that is a closed subset of .

Next observe that we can eliminate rotations in from by eliminating rotations in first and then eliminating rotations in . This can be done because is a closed subset of . Since generates , the lemma follows. ∎

Finally, we observe that the results stated above also follow when we make an upward shift in a boy’s list.

Lemma 10.

Let be an instance of stable matching, and be another instance obtained by introducing a shift in the list of a boy in instance . Then there is at most one rotation, , that leads from to and at most one rotation, that leads from to .

Proof.

Let us switch the roles of boys and girls, and reverse all partial orders in and . Let and be the resulting matching lattice and rotation poset. Let be the rotations leading from to and from to in the this lattice, as guaranteed by Theorem 3. Then and . ∎

4 Efficient Algorithm for Finding a Robust Stable Matching

As stated in Section 1, let be the domain of all possible shifts applied to each boy’s list and each girl’s list in instance and let be a discrete probability distribution on . Pick one shift from under and let be the random variable denoting the resulting instance. As defined in Section 1, a robust stable matching is a stable matching that minimizes the probability that .

For a particular choice of shift from , let denote the resulting stable matching instance and let and denote the rotations going into and out of , respectively. By Proposition 4, and can be computed efficiently for each such . For convenience, we will name the chosen shift also as .

By Lemma 2, can be computed in polynomial time. We add two additional vertices to , a source preceding all other vertices and a sink succeeding all other vertices. For a shift , we may ignore the cases where neither nor exist. In that case, either or . Hence, we may assume that is always a proper non-empty subset of . For a shift such that does not exist, let . Similarly, for a shift such that does not exist, let .

Let be the Hasse diagram of , defined as follows: The Hasse diagram of a poset is a directed graph with a vertex for each element in poset, and an edge from to if and there is no such that . In other words, all precedence relations implied by transitivity are suppressed.

For each , let and . The integer program is as follows:

(IP)
s.t.
Lemma 11.

An optimal solution to (IP) gives a robust stable matching.

Proof.

Let . The set of constraints:

guarantees that is a closed subset.

Notice that if and only if and . This, in turn, happens if and only if the matching generated by is in . Therefore, by minimizing , we can find a closed subset that generates a robust stable matching. ∎

Next, consider the LP-relaxation of this IP:

(LP)
s.t.

Define , and to be the dual variables corresponding to the first three constraints of (LP). Then its dual is:

(DP)
s.t.

We will interpret (DP) as solving a maximum circulation problem in the following network : It has three sets of edges, and . The edges in are of infinite capacity and the flow on is denoted by . contains edges with capacity , and the flow on is denoted by . The goal is to push the maximum amount of flow from to through and then back from to on edge of infinite capacity.

Lemma 12.

(LP) always has an integral optimal solution and there is a combinatorial polynomial time algorithm for solving it.

Proof.

First, remove edge from the above described network and find a maximum flow from to and denote it by . Obtain the residual graph, which clearly will not have any paths from to . Let be the set of vertices reachable from using residual edges. Construct as follows:

Clearly, is integral. Moreover, and satisfy complementarity:

  • for all .

  • for all .

  • for all .

Hence, is an integral optimal solution for (LP). ∎

Remark. Notice that we formulated the IP for robust stable matching as a minimization problem, i.e, minimizing the probability that the matching is in . The reason is that this involves testing only whether crosses the edge . On the other hand, checking whether is in involves testing two edges in general, i.e, that crosses or . The latter would not have led to a linear IP.

Remark. We can now explain why downward shifts are much more difficult to deal with. Observe that the guarantee that there is at most one rotation, , that leads from to and at most one rotation, that leads from to is crucial for forming the IP and showing that its LP-relaxation always has integral an integral solution. For the case of downward shifts, there may be more than one rotation that lead from to and from to . Although we can still formulate an IP for this case, its LP-relaxation is not guaranteed to have an integral solution.

4.1 Extending to incomplete preference lists

Finally, we show how to extend our algorithm to the generalization of stable matching to incomplete preference lists. It is well known that the set of unmatched agents remain unchanged under all stable matchings in this case, e.g., see [GI89]. Let be the given instance and let be obtained by executing an upward shift on one agent’s list. If the set of unmatched agents under is not the same as under , then and we can ignore . Otherwise, the sets of stable matchings in and are defined over the same subset of the agents and all our results carry over.

The first part of Theorem 1 now follows from Lemmas 11 and 12.

5 Succinct Representation for the Sublattice of Robust Stable Matchings

We first prove that the set of robust stable matchings forms a sublattice of the lattice of all stable matchings of the given instance. We then use our combinatorial solution of (DP) to show how to obtain a succinct structure that helps generate all matchings from this sublattice.

Lemma 13.

The set of robust stable matchings to instance under probability distribution forms a sublattice of .

Proof.

Let and be two robust stable matchings and let and be the corresponding closed subsets. It suffices to show that the matchings generated by and are also robust. We say that a closed subset separates and edge in if and .

Divide the edges in into 5 sets as follows:

  1. is the set of edges in from to .

  2. is the set of edges in from to .

  3. is the set of edges in from to .

  4. is the set of edges in from to .

  5. is the set of edges in from to .

Let for each . Since and are robust stable matchings, the objectives obtained by and in (IP) are equal:

Therefore,

We will show that and . Assume without loss of generality that and . Then

In other words, the objective of (IP) obtained by is smaller than the one obtained by . This contradicts the fact that is robust stable matching. Therefore, and as desired.

It follows that and also attain the minimum value of the objective function of (IP). Hence, the matchings generated by and are also robust. ∎

By Lemma 13, the set of robust stable matchings is a sublattice, say , of . By Birkhoff’s Theorem [B37] we know that there is a partial order, say , whose closed sets are isomorphic to . Next, we show how to construct using our combinatorial solution for (DP).

As in Lemma 12, remove edge from network described above and find a maximum flow from to and denote it by . Obtain the residual graph, say . The strongly connected components of are the elements of . Contracting the strongly connected components of yields a DAG , which gives the precedence relations in . To be precise, in if and only if there is a path from to in .

Lemma 14.

The closed sets of partial order correspond exactly to robust stable matchings.

Proof.

Let be a closed set in . There are no edges in the residual graph that go from to . Hence all edges in the cut are fully saturated and therefore this cut minimizes the objective function of (IP). Hence the corresponding matching is a robust stable matching. The reverse direction is straightforward. ∎

Recall that if is the rotation poset for lattice , then the matching corresponding to any closed set in is obtained by starting from the boy-optimal matching in and applying all rotations in in any order consistent with a topological sort of . In [MV18] we show that the corresponding process for finding the matching in corresponding to a closed set of is the following: The elements of are sets of rotations. Let be the union of all these sets. Now starting from the boy-optimal matching in , apply all rotations in in any order consistent with a topological sort of . This yields the matching in corresponding to set . Hence we get.

Lemma 15.

generates the sublattice of robust stable matchings.

The second part of Theorem 1 follows from Lemmas 13 and 15.

6 Discussion

As stated in the Introduction, the two main questions on stable matching considered in this paper are obtaining efficient algorithms for finding solutions that are robust to errors in the input, and the structural question of finding relationships between the lattices of solutions of two “nearby” instances. The current paper and our followup work [MV18] seem to suggest that both these issues are likely to lead to much work in the future. In particular, the structural results are so clean and extensive that they are likely to find algorithmic applications beyond the problem of finding robust solutions. One possible domain of applications that may be able to exploit these structural properties is matching-based markets, particularly as we are seeing ever more interesting such markets being designed and launched on the Internet, e.g., see [Rot16].

At a more detailed level, we have left the open problem of dealing with downward shifts. The domain , for which we have obtained our algorithm, is very restrictive and we need to extend it to a larger domain. Our followup paper [MV18] partially does this, though it works with a weaker notion of “robust”. It seems more should be doable; in particular, what happens if two or more errors are introduced simultaneously?

Beyond these questions, pertaining to the most basic of formulations of stable matching, one can study numerous variants and generalizations, e.g., the stable roommates problem, and matching intern couples to hospitals. Each of these bring their own structural properties and challenges, e.g., see [GI89, Man13].

7 Acknowledgements

We wish to thank David Eppstein, Mike Goodrich and Sandy Irani for interesting discussions that sparked off the question addressed in this paper.

References

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