Dynamic Matching Markets and Voting Paths
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1 Dynamic Matching Markets and Voting Paths David J. Abraham Teikepai Kavitha Abstract We consider a matching market, in which the aim is to maintain a popuar matching between a set o appicants and a set o posts, where each appicant has a preerence ist ranking a subset o posts in some order o preerence. A matching M is popuar i there is no other matching M such that the number o appicants who preer their partners in M to M exceeds the number o appicants who preer their partners in M to M. Popuar matchings M are stabe in the sense that no coaition o appicants can orce a switch to another matching M by requesting a pairwise eection between M and M with one vote per appicant (i.e. an up-or-down vote). The setting here is dynamic: appicants and posts can enter and eave the market, and appicants can aso change their preerences arbitrariy. Ater any change, the matching we have in pace may no onger be popuar, in which case we are required to update it. However, we cannot simpy recompute a popuar matching rom scratch ater every such change. This is because there are instances in which no popuar matching is directy more popuar than the existing non-popuar matching, and hence there woud be no consensus or the appicants to agree to the switch. The aim then is to ind a voting path, which is a sequence o matchings, each more popuar than its predecessor, that ends in a popuar matching. In this paper, we show that, as ong as some popuar matching exists, there exists a 2-step voting path rom any given matching to some popuar matching. Furthermore, given any popuar matching, we show how to ind a shortest-ength such voting path in inear time. 1 Introduction An instance o the popuar matching probem consists o a bipartite graph G = (A P,E), together with a partition E 1 E 2... E r o the edge set E. For exposition purposes, we ca A the set o appicants, P the set o posts, and E i the set o edges with rank i. I (a, p) E i and (a, p ) E j with i < j, we say that a preers p to p. I i = j, then a is indierent between p and p. The ordering o posts adjacent to a is caed a s preerence ist. We say that preerence ists are stricty ordered i no appicant is indierent between any two posts in its preerence ist. A matching M o G is a subset o E, such that no two edges o M share a common endpoint. A node u A P is either unmatched in M, or matched to some node denoted by M(u). We say an appicant a preers matching M to M i (i) a is matched in M and unmatched in M, or (ii) a is matched in both M and M, and a preers M (a) to M(a). Deinition 1 M is more popuar than M, denoted by M M, i the number o appicants preerring M to M is greater than the number o appicants preerring M to M. A matching M is popuar i there is no matching M that is more popuar than M. Computer Science Department, Carnegie Meon University,USA. Emai: dabraham@cs.cmu.edu Indian Institute o Science, Bangaore, India. Emai: kavitha@csa.iisc.ernet.in 1
2 Figure 1 contains an exampe instance in which A = {a 1,a 2,a 3 }, P = {p 1, p 2, p 3 }, and each appicant preers p 1 to p 2, and p 2 to p 3. Consider the three symmetrica matchings M 1 = {(a 1, p 1 ), (a 2, p 2 ), (a 3, p 3 )}, M 2 = {(a 1, p 3 ), (a 2, p 1 ), (a 3, p 2 )} and M 3 = {(a 1, p 2 ), (a 2, p 3 ), (a 3, p 1 )}. None o these matchings is popuar, since M 1 M 2, M 2 M 3, and M 3 M 1. In act, it turns out that this instance admits no popuar matching, the probem being, o course, that the more popuar than reation is not acycic. a 1 : p 1 p 2 p 3 a 2 : p 1 p 2 p 3 a 3 : p 1 p 2 p 3 Figure 1: An instance or which there is no popuar matching. The popuar matching probem is to determine i a given instance admits a popuar matching, and to ind such a matching, i one exists. The irst poynomia-time agorithms or this probem were given in [3]: when preerence ists are stricty ordered, the probem can be soved in O(n+m) time, where n = A P and m = E, and more generay, the probem can be soved in O(m n) time. Note that when E = E 1, a matching is popuar i and ony i it has maximum cardinaity. Hence, the popuar matching probem is at east as hard as the probem o inding a maximum matching in a bipartite graph. 1.1 Probem Deinition In this paper, we consider a matching market in which the aim is to maintain a popuar matching. The setting is dynamic: appicants and posts can enter and eave the matching market, and appicants can change their preerences arbitrariy. More precisey, an instance o the dynamic popuar matching probem consists o an instance G o the popuar matching probem, together with an existing (possiby empty) matching M 0. It turns out that we cannot simpy sove G rom scratch ater each change, since any particuar popuar matching we ind may not be more popuar than M 0, and urthermore, it is possibe that no popuar matching is more popuar than the existing matching M 0. Hence, in genera, there may be no consensus amongst the appicants to move directy rom M 0 to a popuar matching. We show such an exampe beow. Consider the instance in Figure 2 with M 0 = {(a 1, p 5 ), (a 2, p 2 ), (a 3, p 3 ), (a 4, p 1 )}. First note that M 0 is not popuar, since it is ess popuar than M = {(a 1, p 2 ), (a 2, p 3 ), (a 4, p 1 )} (even with a 3 unmatched). We can show using Lemma 2 rom Section 2 that the ony popuar matchings are M = {(a 1, p 1 ), (a 2, p 2 ), (a 3, p 3 ), (a 4, p 4 )} and N = {(a 1, p 1 ), (a 2, p 3 ), (a 3, p 2 ), (a 4, p 4 )}. However, it is cear that neither M nor N is more popuar than M 0. a 1 : p 1 p 2 p 5 a 2 : p 3 p 2 a 3 : p 3 p 2 a 4 : p 1 p 4 Figure 2: Instance motivating voting-path approach. In order to arrive at a popuar matching by consensus, [3] introduced the oowing generaization o the more popuar than reation: 2
3 Deinition 2 A matching M k is reachabe rom M 0 i there is a sequence o matchings M 0,M 1,...,M k, such that each matching is more popuar than its predecessor. Such a sequence is caed a ength-k voting path rom M 0 to M k. Note that the instance above has a ength-2 voting path rom M 0 to a popuar matching, namey M 0,M,N. There is no a priori reason to expect that such a voting path must exist: the more popuar than reation is not acycic, and so perhaps there are some matchings M 0 rom which we cannot avoid cycing. Even i such a path does exist, it may have ength exponentia in the size o G, since there can be an exponentia number o matchings. In this paper, we show the oowing surprising resut. Theorem 1 Let G,M 0 be an instance o the dynamic popuar matching probem, where G admits a popuar matching. Then G admits a voting path o ength at most 2 rom M 0 to some popuar matching. Additionay, given any popuar matching, we can ind a shortest-ength such voting path in ony inear time. Hence, by using the popuar matching agorithms in [3], we can sove the dynamic popuar matching probem in O(m+n) time when preerence ists are stricty ordered, and more generay in O(m n) time 1. This soves the probem o eicienty computing a shortest-ength voting path to a popuar matching, which was posed in [3]. We have aso shown that such paths have ength at most 2, which is better than the bound o 3 previousy caimed. The proo o the ength-3 bound quoted in [3] is unpubished, and ony appies when preerence ists are stricty ordered. This restriction greaty simpiies the more genera probem, which is discussed in this paper. Interestingy, the improvement rom 3 to 2 impies a connection to the amous resut in graph theory that every tournament has a king [14]. The more popuar than reation is a directed graph on an exponentia number o vertices. This graph is not a tournament though, since or any pair o matchings, there is no guarantee that one is more popuar than the other. However, even without these edges, the set o popuar matchings coectivey act as a king, since every unpopuar matching has a voting path o ength at most 2 into this set. 1.2 Reated Previous Work The bipartite matching probem with a graded edge set is we-studied in both economics and computer science, see or exampe [1, 18, 22] and [6, 12, 2]. It modes some important rea-word probems, incuding the aocation o graduates to training positions [10], amiies to government-owned housing [21], and customers to renta DVDs [4, 17]. Gardenors [8] irst introduced the notion o a popuar matching in the context o the stabe marriage probem 2 [7, 9]. O course, the more popuar than concept can be traced back even urther to the Condorcet voting protoco. One drawback o the popuarity criterion is that a popuar matching may not exist. However, in recent work, Mahdian [15] showed that a popuar matching exists with high probabiity, when (i) preerence ists are randomy constructed, and (ii) the number o posts is a sma mutipicative actor arger than o the number o appicants. Other recent work on popuar matching incudes Mestre s [16] generaization o the eicient popuar matching characterization in [3] to the case where appicant votes carry dierent weights. 1 We make the standard assumption in dynamic programming settings that the instance G does not change whie we are computing a voting path. 2 A stabe marriage instance is the same as a popuar matching instance, except that both appicants (i.e. men) and posts (i.e. women) rank each other in order o preerence. 3
4 We remark that the resut in our paper is anaogous to a series o papers [13, 19, 20, 5] on decentraized mechanisms in the stabe matching iterature. The we-known mechanisms or stabe matching, due to Gae/Shapey [7] and Irving [11], require a centra body to coect preerences and dictate the ina matching. Aternativey, in a decentraized setting, a bocking pair (i.e. a man and woman who preer each other to their current partners) wi act ocay by divorcing their current partners and marrying each other. Knuth [13] showed that i the divorced partners aso marry each other, it is possibe or this process to cyce. However, when divorced partners are not required to marry each other, and every bocking pair has some probabiity o acting next, Roth and Vande Vate [19] show by way o a potentia argument that there is aways a path to a stabe matching. In our setting, the anaogue o a bocking pair is a coaition o appicants who (i) preer some matching M to the current matching M, and (ii) have suicient numbers to win a vote between M and M. It is not too diicut to give a potentia argument to prove the existence o voting paths (at east or the restriction to stricty-ordered preerence ists). However, in this paper, we use more poweru techniques rom matching theory, which in addition to proving existence, aso give the surprising ength-2 bound. As with the resut in [19], this means that as ong as every matching more popuar than the current one has some probabiity o an up-or-down vote, then in the imit, a decentraized mechanism wi ead to a popuar matching. 1.3 Organization o the Paper In Section 2, we review the theory o popuar matchings, and then use this to characterize the set o matchings that admit ength-0 voting paths. In Section 3 and 4, we derive the centra emmas o the paper, using these to characterize the set o matchings that admit ength-1 and ength-2 voting paths respectivey. Finay, in Section 5, we concude with an open probem. 2 Preiminaries: Length-0 Voting Paths In this section, we review the agorithmic characterization o popuar matchings given in [3]. We then note that this characterization can be used to determine i an instance G,M 0 o the dynamic popuar matching probem admits a ength-0 voting path to a popuar matching. For exposition purposes, we create a unique stricty-east-preerred post (a) or each appicant a. In this way, we can assume that every appicant is matched, since any unmatched appicant a can be paired with (a). From now on then, matchings are A-perect. Aso, without oss o generaity, we assume that preerence ists contain no gaps, i.e. i a is incident to an edge o rank i, then a is incident to an edge o rank i 1, or a i > 1. Let G 1 = (A P,E 1 ) be the graph containing ony rank-one edges. [3, Lemma 3.1] shows that a matching M is popuar in G ony i M E 1 is a maximum matching o G 1. Maximum matchings have the oowing important properties, which we use throughout the rest o the paper. M E 1 deines a partition o A P into three disjoint sets: a node u A P is even (respectivey odd) i there is an even (respectivey odd) ength aternating path in G 1 (with respect to M E 1 ) rom an unmatched node to u. Simiary, a node u is unreachabe i there is no aternating path rom an unmatched node to u. Denote by E, O and U the sets o even, odd, and unreachabe nodes, respectivey. Lemma 1 (Gaai-Edmonds Decomposition) Let E, O and U be the sets o nodes deined by G 1 and M E 1 above. Then (a) E, O and U are pairwise disjoint, and independent o the maximum matching M E 1. 4
5 (b) In any maximum matching o G 1, every node in O is matched with a node in E, and every node in U is matched with another node in U. The size o a maximum matching is O + U /2. (c) No maximum matching o G 1 contains an edge between a node in O and a node in O U. Aso, G 1 contains no edge between a node in E and a node in E U. Using this node partition, we make the oowing deinitions: or each appicant a, deine (a) to be the set o most-preerred odd/unreachabe posts in a s preerence ist 3. Aso, deine s(a) to be the set o most-preerred even posts in a s preerence ist. We reer to posts in a A (a) as -posts and posts in a A s(a) as s-posts. Note that -posts and s-posts are disjoint, and that s(a) /0 or any a, since (a) is aways even. Aso note that there may be posts in P that are neither -posts nor s-posts. The next emma characterizes the set o a popuar matchings. Lemma 2 ([3]) A matching M is popuar in G i and ony i (i) M E 1 is a maximum matching o G 1 = (A P,E 1 ), and (ii) or each appicant a, M(a) (a) s(a). Using this emma, we can check i G,M 0 admits a ength-0 voting path to a popuar matching: M 0 E 1 is a maximum matching o G 1 i G 1 admits no augmenting path. Aso, given that M 0 E 1 is a maximum matching o G 1, it is trivia to compute the Gaai-Edmonds decomposition and then to check that each appicant a is matched to M(a) (a) s(a). These checks can ceary be perormed in inear time. Henceorth, we assume then that M 0 is not popuar, or otherwise G,M 0 admits a ength-0 voting path, and we are done. We aso assume that G,M 0 admits a popuar matching, or otherwise no voting path can end in a popuar matching. We concude this section by giving the agorithm based on Lemma 2 or soving the popuar matching probem. Popuar-Matching(G = (A P,E)) 1. Construct the graph G = (A P,E ), where E = {(a, p) : a A and p (a) s(a)}. 2. Construct a maximum matching M o G 1 = (A P,E 1 ). (Note that M is aso a matching in G ). 3. Remove any edge in G between a node in O and a node in O U. (No maximum matching o G 1 contains such an edge). 4. Augment M in G unti it is a maximum matching o G. 5. Return M i it is A-perect, otherwise return no popuar matching. 3 Length-1 voting paths In this section, we show that, given any popuar matching o G, the probem o inding a ength-1 voting path rom M 0 to a popuar matching, or proving that no such path exists, can be soved in inear time. First though, we work towards characterizing the set o a popuar matchings that are more popuar than M 0. 3 In [3], (a) is deined as the set o rank-1 posts in a s preerence ist. We ind the deinition above more suitabe. 5
6 Let r a (p) be the rank o edge (a, p) E. Aso, et r a (s(a)) be the rank o any edge (a, p), where p s(a). We deine the signature o a matching M as the 4-tupe ( A M, AM m, A M s, A M ) 4, where: (i) A M = {a A r a (M(a)) = 1, and a is even/unreachabe, i.e. a E U}. (ii) A M m = {a A 1 < r a (M(a)) < r a (s(a))}. (iii) A M s (iv) A M = {a A r a (M(a)) = r a (s(a))}. = {a A r a (M(a)) > r a (s(a))}. Note that an odd appicant a (i.e. one in O) can ony beong to A M s A M, even i r a (M(a)) = 1, or a / A M by deinition, and a / A M m, since r a (s(a)) = 1. Aso note that or any even/unreachabe appicant a (i.e. one in A \ O), r a (s(a)) 1. Hence A M,AM m,a M s, and A M are pairwise disjoint and partition A. This gives us A M + AM m + A M s + A M = A. Finay, note that A M = {a A : M(a) (a)}. Now, et F be the set o -posts - i.e. F = a A (a). The oowing emma characterizes the set o a popuar matchings in terms o their signatures. Lemma 3 A matching M is popuar i and ony i its signature is ( F,0, A F,0). Proo: Suppose M is popuar. Then by Lemma 2, A M m = A M = 0, and so A M + AM s = A. Now, A M F, since every appicant in AM is matched with some post in F. But since M E 1 is a maximum matching o G 1, Lemma 1(b) requires that every post in F is matched with some appicant in A M. Hence, A M = F, AM s = A F, and M has signature ( F,0, A F,0). Conversey, suppose M is a matching with signature ( F, 0, A F, 0). Then A M m = A M = 0, and so or every a A, M(a) (a) s(a). It remains to show that M E 1 is a maximum matching o G 1. We have that M E 1 = A M + {a AM s : a is odd}. Since A M = F, M E 1 = F + {a A : a is odd}. So, M E 1 = {v A P : v is odd} + {p P : p is unreachabe}, and the resut oows rom Lemma 1(b). Lemma 4 For any matching M, A M + AM m F. Proo: For each a A M, M(a) is odd/unreachabe (i.e. beongs to O U), or otherwise, G 1 contains an edge contradicting Lemma 1(c). Aso, or each a A M m, M(a) is odd/unreachabe, since s(a) contains a s most preerred even posts, and by deinition o A M m, a preers M(a) to posts in s(a) (i.e. r a (M(a)) > r a (s(a))). Hence, A M + AM m F. Finay, we come to the main technica emma in this section, which characterizes the set o a popuar matchings that are more popuar than a given matching M. Lemma 5 Let M be a popuar matching. Then M is more popuar than M i and ony i (i) A M + A M m < F, or (ii) A M m > 0, or (iii) A M s > 0. Proo: Let (M,M) be the dierence between the number o appicants who preer M to M, and the number o appicants who preer M to M. That is, (M,M) = [A M m A M s A M ] + A M s [A M A M m] s. 4 The subscripts,m,s, and stand or irst, midde, second, and ast respectivey. 6
7 Now, since M is popuar, by Lemma 3 we have: A M = [A M A M m A M s A M ] = F = [ F A M A M m ] + [A M A M m] [A M A M. Rearranging, we get: [A M A M m] = [A M s A M ] [ F A M A M m ]. s [ ] Hence (M,M) = F A M AM m + A M m + A M s. The theorem oows immediatey, since A M m and A M s are both non-negative, whie F A M AM m 0 by Lemma 4. Given G,M 0 and some popuar matching M o G, we do not need Lemma 5 to determine i M is more popuar than M 0 - instead, we can just count the number o appicants that preer one matching to the other. Suppose, however, that M is not more popuar than M 0 so that A M 0 + A M 0 m = F, A M 0 s = 0, and A M 0 m = 0. Our aim is to use Lemma 5 as a guide in inding a popuar matching that is more popuar than M 0, or proving that no such matching exists. First we remark that A M 0 + A M 0 m = F, or otherwise, any popuar matching, incuding M, is more popuar than M 0 by Lemma 5. It oows that A M 0 m /0 or A M 0 /0, or otherwise, M 0 has signature ( F,0, A F,0), contradicting the assumption rom the previous section that M 0 is not popuar. Suppose A M 0 m /0, so that there is an appicant a A M 0 m s. By deinition, such an appicant is even/unreachabe. I there is a popuar matching that pairs a with a post in (a), then it must be more popuar than M 0 by condition (ii) o Lemma 5. In order to test i there exists such a popuar matching, we proceed in the oowing way. Let G be the subgraph o G deined in Figure 2 ater step 3. So, G contains a edges between appicants and their -posts and s-posts, except those between nodes in O and nodes in O U. Now, modiy G and M by removing a edges between this particuar appicant a and posts in s(a). Ca the resuting structures G a and M a respectivey. Lemma 6 There exists a popuar matching which pairs a with some post in (a) i and ony i G a admits an augmenting path with respect to M a. Proo: Suppose G a admits an augmenting path Q a with respect to Ma. Since M is popuar, the ony unmatched appicant in Ma is a, and so Ma Q a matches a with some post in (a). We want to caim that Ma Q a is popuar. First note that its signature is o the orm (k,0, A k,0) or some k 0, since it is a matching in a subgraph o G, and G ony contains edges between appicants and their -posts and s-posts. Now, Ma Q a matches a posts in F, since every post matched in Ma is aso matched in Ma Q a. Reca that posts in F are incident in G to rank-1 edges ony, and urthermore, odd posts in F are ony adjacent to even appicants, whie unreachabe posts in F are ony adjacent to unreachabe appicants. Hence, k = F, and so by Lemma 3, Ma Q a is popuar, since its signature is ( F,0, A F,0). Conversey, suppose that G a admits no augmenting path with respect to Ma. Then, Ma is a maximum matching in G a, which, since a is unmatched in Ma, means that there is no A-perect matching in G a. But by Lemma 2(b), every popuar matching is an A-perect matching in G. Hence, every popuar matching must contain an edge in G \ G a = {(a, p) : p s(a)}. s ] 7
8 We now make use o the previous emma. Begin by ooking or an augmenting path in G a with respect to Ma by using depth-irst search to construct the Hungarian tree T a rooted at a (see Figure 3). I we ind such a path Q a, then by Lemma 5, M 0,Ma Q a is a ength-1 voting path to a popuar matching. Otherwise, we repeat this process with some other a A M m s, i any. a posts in (a) their partners neighbors not in (a) Figure 3: Exampe o a Hungarian tree T a rooted at a, with matched edges bod Suppose we are successu in inding an augmenting path Q a such that Ma Q a is popuar. We caim that Q a is edge disjoint rom the set o edges in T a, where a a. For suppose otherwise. Then et e be the irst edge in Q a that is aso in T a. Now, G a admits an aternating path rom a through e. However, since Q a does not exist, this path cannot be extended in G a to end in some unmatched post. Hence, Q a must contain an aternating path rom e through the edge matching a with M (a) (which is missing in G a). But M (a) is unmatched in G a, and hence i we join the two aternating paths above, we get an augmenting path Q a (rom a to M (a) through the edge e) in G a. This gives the required contradiction. Since we ony need to examine each edge a constant number o times, it is cear that we can determine in inear time i there is a popuar matching that pairs some appicant in A M 0 m s with one o its -posts. I there is no such appicant, we repeat this procedure with appicants a A M 0, who must by deinition be even/unreachabe. Here, though our aim is to ind a popuar matching that satisies Lemma 5(iii) by pairing a with a post in s(a). It oows that or this to occur, a must be even, since by Lemma 1 and 2(i), every unreachabe appicant is matched by any popuar matching to a post in (a). Suppose we ind such a matching Ma Q a. Since a is even, M (a) is odd, and so any popuar matching must match M (a) aong a rank-1 edge to an even appicant. However, M (a) may be unmatched in Ma Q a, as we removed a edges between a and posts in (a) rom G (incuding (a,m (a))). Hence we may need to augment Ma Q a in G 1. But every odd node M (a) is adjacent to at east one other even appicant aong a rank-1 edge, namey its predecessor in the odd ength aternating path rom a vertex unmatched w.r.t. M E 1 to M (a) in G 1 ). Hence, such an augmentation aways exists. And it is easy to see that here too we ony examine each edge a constant number o times. By Lemmas 5 and 6, it is cear that the above agorithm correcty inds a ength-1 voting path rom M 0 to some popuar matching, or proves that no such path exists. Additionay, given a popuar matching M, we have just shown that the agorithm runs in inear time. 4 Length-2 voting paths In this section, we show that, given any popuar matching M o G, the probem o inding a ength-2 voting path rom M 0 to some popuar matching can be soved in inear time. We wi assume that M 0 admits no shorter such voting path. In particuar, this means that M is not more popuar than M 0, so that A M 0 m /0, or A M 0 /0. Suppose that A M 0 m /0. Let a A M 0 m and et T a be the Hungarian tree associated with a, as described in Section 3. In the oowing emma, we give a suicient condition or the existence o a ength-2 8
9 voting path rom M 0 to M. Lemma 7 Suppose there exists an appicant a T a such that M 0 (a ) / s(a ) and M (a ) s(a ). Then there exists a ength-2 voting path rom M 0 to M. Proo: Our goa is to ind a matching M 1 such that (i) M 1 is more popuar than M 0, and (ii) a A M 1. This ast condition guarantees that M is more popuar than M 1 by Lemma 5(iii), and hence we get the ength-2 voting path M 0,M 1,M. Beore discussing how we construct M 1, we irst need to show that a A M 0 A M 0 m : It is cear that a / A M 0, or otherwise a A M 0 s and M is more popuar than M 0 by Lemma 5(iii) - a contradiction. Suppose then that a A M 0 s. By deinition we have that M 0 (a ) / s(a ), and so it must be the case that M 0 (a ) is odd/unreachabe and beongs to F. But M 0 matches a posts in F to appicants in A M 0 A M 0 m, or otherwise M is more popuar than M 0 by Lemma 5(i) - a contradiction. Hence, a A M 0 A M 0 m. Now we need to show that G a contains no edge between a and (a ): For suppose this is not the case. Then since (a ) is stricty the east preerred post o a, we have that s(a ) = {(a )}. By deinition, no other appicant is adjacent to (a ), and so (a ) is a ea node in T a with parent a. It oows rom the construction o T a that (a ) is unmatched in Ma, and hence Ma admits an augmenting path rom a through a to (a ). This contradicts our assumption that M 0 admits no ength-1 voting path to a popuar matching. Hence, G a contains no edge between a and (a ). Finay, we describe how to construct M 1. Add an edge between a and (a ) to G a. From the argument above, we have that G a admits an augmenting path Q a rom a through a and ending in (a ). Let M 1 = Ma Q a. The signature o M 1 is ( F,0, A F 1,1), by an argument simiar to the one used in the proo o Lemma 6, except that here we have one appicant a A M 1. It remains to show that M 1 is more popuar than M 0. (M 1,M 0 ) = [A M 0 m A M 0 s A M 0 ] A M 1 + A M 0 A M 1 s [A M 0 A M 0 m ] [A M 1 s A M 1 {NB: A M 0 s A M 1 = 0, since a / A M 0 s } [A M 0 m A M 0 s A M 0 ] A M 1 [A M 0 A M 0 m ] [A M 1 s A M 1 ] = [A M 0 A M 0 m A M 0 s A M 0 ] A M 1 A M 0 [A M 1 A M 1 s A M 1 = F A M 0 A M 0 m [A M 1 s A M 1 ] {since A M 1 = F } = A M 0 m A M 0 m [A M 1 s A M 1 ] {by Lemmas 4 and 5(i)} = A M 0 m A M 1 > 0 {since a A M 0 m A M 1 } ] ] A M 0 m [A M 1 s A M 1 ] It is cear that in inear time, we can check i there exists an appicant a T a such that M 0 (a ) / s(a ) and M (a ) s(a ), and i so, we can construct the matching M 1. Suppose there is no such appicant in T a. Our aim then is to ind a dierent popuar matching in which such an appicant exists. We wi ind such a matching by searching or a particuar type o aternating cyce C in G a with respect to M. First though, we make some observations about T a and G a. By construction, posts in T a are discovered aong unmatched edges. Aso, no post p is a ea node in T a, since then p woud be unmatched in M a, and M a woud admit an augmenting path, which contradicts 9
10 our assumption that M 0 admits no ength-1 voting path to a popuar matching. Thereore, posts in T a have degree 2. By construction, an appicant a T a \ {a} is discovered aong a matched edge. I a is even or unreachabe, then a is incident to at east one unmatched chid edge in G a, since (a ) and s(a ) are non-empty and disjoint. I a is odd, then a must aso be incident to at east one unmatched chid edge in G a - since a is odd, G 1 admits an odd-ength augmenting path rom an unmatched vertex to a, and the ast edge e(a ) on this path is unmatched in M. Since every node in T a \ {a} is incident to at east one matching edge and one non-matching edge (w.r.t. the matching M ) in G a, we can buid the oowing aternating path Q. Begin with an edge (a, p) where p (a). Let the successor o any post p in Q be its matched partner M (p). The successor o any even/unreachabe appicant a is any post in (a ) i M (a ) s(a ), or any post in s(a ) i M (a ) (a ). Finay, the successor o any odd appicant a is the post incident to e(a ). Since T a is inite, at some point this aternating path must orm a cyce C by adding a post that is aready in the path. This procedure ceary takes inear time. It is cear that rom Lemma 2 that M C is a popuar matching. Now, i we can show that M C has some appicant a T a \ {a} such that M 0 (a ) / s(a ) and (M C)(a ) s(a ), then we can use M C as the popuar matching in Lemma 7. First, we prove that C contains at east one appicant a such that M (a ) (a ). Since M C matches a with s(a ) by construction, the ina step wi be to show that M 0 (a ) / s(a ). Lemma 8 C contains at east one appicant a such that M (a ) (a ). Proo: Note that the ength o C is at east 4 since the predecessor and successor o each appicant are aways distinct. Aso note that i C contains an even/unreachabe appicant, then its predecessor/successor is an -post, whose partner in M is the required appicant. The ony way that that C may not have contain an -post is i a the appicants in C are odd. We show that this cannot happen. Let a be the irst odd appicant in Q that is in C. Then by construction, C contains a subpath rom a through e(a ) to some post p that is unmatched in M E 1. It oows that p is even, since odd/unreachabe posts are matched in M E 1 by Lemmas 1(b) and 2(i). Since p is matched in M, its partner M (p) must be even (again by Lemmas 1(b) and 2(i)), and so C contains an even appicant. Lemma 9 Suppose there exists no appicant a T a such that M 0 (a ) / s(a ) and M (a ) s(a ). Then or each a T a \ {a}, M 0 (a ) / s(a ) i and ony i M (a ) (a ). Proo: Let a be any appicant in T a \ {a} such that M 0 (a ) / s(a ). Then by the assumption in the statement o the emma, we have M (a ) / s(a ), and so M (a ) (a) F, since M is popuar. Hence, T a \ {a} [A A M 0 s ] is at most the number o -posts in T a, i.e., T a \ {a} [A A M 0 s ] T a F. (1) Now, since there is no augmenting path in T a, we have that T a \ {a} A = T a P. Partitioning A into A \ A M 0 s and A M 0 s and posts in T a into T a F and T a S, where S is the set o a s-posts, we get T a \ {a} [(A \ A M 0 s ) A M 0 s ] = T a F + T a S. Note that no appicant in A M 0 s can be matched by M 0 to an odd/unreachabe post, otherwise A M 0 + A M 0 m < F and M woud have been more popuar than M 0 by Lemma 5(i). Hence each appicant in A M 0 s even posts, that is one o its s-posts, so T a \ {a} A M 0 s has to be matched by M 0 to one o its most preerred T a S. We thus get, T a \ {a} [A \ A M 0 s ] T a F. (2) 10
11 Combining (1) and (2), we have T a \ {a} [A A M 0 s ] = T a F = T a \ {a}. That is, the number o appicants a in T a \ {a} that satisy M (a ) (a ) is equa to the number o appicants in T a \ {a} that satisy M 0 (a ) / s(a ). But each appicant in T a \ {a} that satisies M 0 (a ) / s(a ) has to satisy M (a ) (a ) by the statement o the emma. So the equivaence oows immediatey. By Lemma 8, C contains at east one appicant a such that M (a ) (a ). By Lemma 9, we have that M 0 (a ) / s(a ). Since M C matches a with s(a ), M C satisies the suicient condition in Lemma 7. Hence there exists a ength-2 voting path rom M 0 to the popuar matching M C. As in Section 3, i A M 0 m = /0, then A M 0 is non empty, and we perorm an anaogous procedure on the Hungarian tree associated with some a A M 0. This inishes the proo o Theorem 1 (stated in Section 1). The overa agorithm is presented in Figure 4. Voting-Path(G = (A P,E),M 0 ) i M 0 has signature ( F,0, A F,0) then return M 0 Let M be any popuar matching o G i M is more popuar than M 0 then return M 0,M or each appicant a A M 0 m s and each even appicant a A M 0 Construct the Hungarian tree T a with respect to Ma, incuding ony unmarked edges Mark a edges in T a i T a contains an augmenting path Q a then Augment Ma Q a in G 1 rom M (a) // appies ony when a A M 0 return M 0,Ma Q a Let T a be the irst Hungarian tree constructed or any a A M 0 m s or even a A M 0 i there exists no a T a \ {a} such that M 0 (a ) / s(a ) and M (a ) s(a ) then Construct C in G a as described in Section 4 Let M = M C Let T a be the Hungarian tree associated with a and the new M Add (a,(a )) to T a, and ind the augmenting path Q a in T a Let M 1 = Ma Q a Augment M 1 in G 1 rom M (a) // appies ony when a A M 0 return M 0,M 1,M Figure 4: Linear-time agorithm or inding a shortest-ength voting path 5 Concusions and Open Probems We considered the probem o computing a shortest-ength voting path in a probem instance G,M 0. We showed that i G admits a popuar matching, then there is aways a voting path o ength at most 2 rom M 0 to some popuar matching. Furthermore, we showed that the probem o inding a shortestength voting path rom M 0 to some popuar matching has the same compexity as the probem o computing a popuar matching in G. We concude with an open probem. Suppose we are given an instance G o the popuar matching probem that admits no popuar matching. Rather than return no popuar matching, we want to return a matching that is as popuar as possibe. Since the directed graph H o the more popuar than reation has no sink, we might consider matchings that, at the very east, are members o sink components in the strongy-connected component 11
12 graph o H. Subject to this, a most popuar matching coud be deined in various ways, or exampe one that has the smaest out-degree in H. However, H has size exponentia in G, and so we are interested in the compexity o inding such matchings. Reerences [1] A. Abdukadiroǧu and T. Sönmez. Random seria dictatorship and the core rom random endowments in house aocation probems. Econometrica, 66(3):689 7, [2] D.J. Abraham, K. Cechárová, D.F. Manove, K. Mehhorn. Pareto-optimaity in house aocation probems. In Proceedings o ISAAC 04: the 15th Annua Internationa Symposium on Agorithms and Computation, voume 3341 o Lecture Notes in Computer Science, pages 3-15, Springer- Verag, [3] D.J. Abraham, R.W. Irving, T. Kavitha, and K. Mehhorn. Popuar Matchings. In Proceedings o SODA 05: the 16th Annua ACM-SIAM Symposium on Discrete Agorithms, pages , [4] Amazon.com DVD Renta: see [5] E. Diamantoudi, E. Miyagawa, and L. Xue. Random paths to stabiity in the roommate probem. Games and Economic Behavior, 48(1):18-28, [6] S.P. Fekete, M. Skutea, and G.J. Woeginger. The compexity o economic equiibria or house aocation markets. Inormation Processing Letters, 88: , 2003 [7] D. Gae and L.S. Shapey. Coege admissions and the stabiity o marriage. American Mathematica Monthy, 69:9 15, [8] P. Gardenors. Match Making: assignments based on biatera preerences. Behavioura Sciences, 20: , [9] D. Gusied and R.W. Irving The Stabe Marriage Probem: Structure and Agorithms. MIT Press, [10] A. Hyand and R. Zeckhauser. The eicient aocation o individuas to positions. Journa o Poitica Economy, 87(2): , [11] R.W. Irving. An eicient agorithm or the stabe roommates probem. Journa o Agorithms, 6: , 1985 [12] R.W. Irving, T. Kavitha, K. Mehhorn, D. Michai, and K. Pauch. Rank-maxima matchings. In Proceedings o SODA 04: the 15th ACM-SIAM Symposium on Discrete Agorithms, pages ACM-SIAM, [13] D.E. Knuth. Stabe Marriage and its Reation to Other Combinatoria Probems. CRM Proceedings and Lecture Notes, voume 10, American Mathematica Society. Engish transation o Mariages Stabes, Les Presses de L Université de Montréa, 1976 [14] H. Landau. On dominance reations and the structure o anima societies, III: the condition or secure structure. Buetin o Mathematica Biophysics, 15(2): ,
13 [15] M. Mahdian. Random Popuar Matchings. In submission. [16] J. Mestre. Weighted Popuar Matchings. In submission. [17] Netix DVD Renta: see [18] A.E. Roth and A. Postewaite. Weak versus strong domination in a market with indivisibe goods. Journa o Mathematica Economics, 4: , [19] A.E. Roth and J.H. Vande Vate. Random Paths to Stabiity in Two-Sided Matching. Econometrica, 58: , [20] A. Tamura. Transormation rom arbitrary matchings to stabe matchings. Journa o Combinatoria Theory, Series A, 62(2): , [21] Y. Yuan. Residence exchange wanted: a stabe residence exchange probem. European Journa o Operationa Research, 90: , [22] L. Zhou. On a conjecture by Gae about one-sided matching probems. Journa o Economic Theory, 52(1): ,
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