Theoretical Computer Science

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1 Theoretcal Computer Scence 412 (2011) Contents lsts avalable at ScenceDrect Theoretcal Computer Scence journal homepage: Popular matchngs wth varable tem copes Telkepall Kavtha a, Meghana Nasre b, a Tata Insttute of Fundamental Research, Mumba, Inda b Indan Insttute of Scence, Bangalore, Inda a r t c l e n f o a b s t r a c t Artcle hstory: Receved 11 March 2010 Receved n revsed form 3 October 2010 Accepted 27 December 2010 Communcated by P. Spraks Keywords: Bpartte graphs Matchngs One-sded preference lsts NP-hardness We consder the problem of matchng people to tems, where each person ranks a subset of tems n an order of preference, possbly nvolvng tes. There are several notons of optmalty about how to best match a person to an tem; n partcular, popularty s a natural and appealng noton of optmalty. A matchng M s popular f there s no matchng M such that the number of people who prefer M to M exceeds the number who prefer M to M. However, popular matchngs do not always provde an answer to the problem of determnng an optmal matchng snce there are smple nstances that do not admt popular matchngs. Ths motvates the followng extenson of the popular matchngs problem: Gven a graph G = (A B, E) where A s the set of people and B s the set of tems, and a lst c 1,..., c B denotng upper bounds on the number of copes of each tem, does there exst x 1,..., x B such that for each, havng x copes of the -th tem, where 1 x c, enables the resultng graph to admt a popular matchng? In ths paper we show that the above problem s NP-hard. We show that the problem s NP-hard even when each c s 1 or 2. We show a polynomal tme algorthm for a varant of the above problem where the total ncrease n copes s bounded by an nteger k Elsever B.V. All rghts reserved. 1. Introducton In ths paper we consder the problem of matchng people to tems, where each person ranks a subset of tems n an order of preference possbly nvolvng tes, that s, preference lsts are one-sded. Our nput s a bpartte graph G = (A B, E) where A s the set of people and B s the set of tems, and E = E 1 E r s the set of edges, where E t s the set of edges havng rank t. For any a A, we say a prefers tem b to tem b f the rank of edge (a, b) s smaller than the rank of edge (a, b ). The goal s to come up wth an optmal matchng of people to tems. Several notons of optmalty lke rank-maxmalty [8], maxmum-utlty, Pareto-optmalty [2,1,16] have been studed n the lterature for matchngs wth one-sded preferences. But most of these notons use the absolute ranks specfed by people over tems to dstngush between a par of matchngs. One crteron that does not use numercal ranks s the noton of popularty. Let M(a) denote the tem to whch a person a s matched n a matchng M. We say that a person a prefers matchng M to M f () a s matched n M and unmatched n M, or () a s matched n both M and M, and a prefers M(a) to M (a). Work done as part of the DST-MPG partner group Effcent Graph Algorthms at IISc Bangalore. A prelmnary verson of ths work appeared n the 20th Internatonal Symposum on Algorthms and Computaton, ISAAC 09 [11]. Correspondng address: Computer Scence and Automaton Department, Indan Insttute of Scence, Bangalore, , Inda. Tel.: ; fax: E-mal addresses: kavtha@tcs.tfr.res.n (T. Kavtha), meghana@csa.sc.ernet.n, meghanamande@gmal.com (M. Nasre) /$ see front matter 2011 Elsever B.V. All rghts reserved. do: /j.tcs

2 1264 T. Kavtha, M. Nasre / Theoretcal Computer Scence 412 (2011) Defnton 1. M s more popular than M, denoted by M M, f the number of people who prefer M to M s hgher than those that prefer M to M. A matchng M s popular f there s no matchng that s more popular than M. Popular matchngs were frst ntroduced by Gärdenfors [5] n the context of stable matchngs. The noton of popularty s an attractve noton of optmalty snce t s based on relatve rankng rather than the absolute ranks used by any person; also popular matchngs can be consdered stable n the sense that no majorty vote of people can force a mgraton to another matchng. Unfortunately there exst smple nstances that do not admt any popular matchng. Abraham et al. [3] desgned effcent algorthms for determnng f a gven nstance admts a popular matchng and computng one, f t exsts. A smple example that does not admt a popular matchng s the followng: let A = {a 1, a 2, a 3 }, B = {b 1, b 2, b 3 }, and each person prefers b 1 to b 2, and b 2 to b 3. Consder the three symmetrcal matchngs M 1 = {(a 1, b 1 ), (a 2, b 2 ), (a 3, b 3 )}, M 2 = {(a 1, b 3 ), (a 2, b 1 ), (a 3, b 2 )} and M 3 = {(a 1, b 2 ), (a 2, b 3 ), (a 3, b 1 )}. None of these matchngs s popular, snce M 1 M 2, M 2 M 3, and M 3 M 1. In fact, t turns out that ths nstance admts no popular matchng, the problem beng that the more popular than relaton s not transtve. Our focus n ths paper wll be on such nstances that do not admt a popular matchng. Intutvely, the absence of a popular matchng n an nstance s due to a small set of tems beng n too much demand by a large number of people. Suppose that we are allowed to add extra copes or duplcates of tems and that these duplcates are ndstngushable from the orgnal tem. If there were no bounds on the number of copes that we could add for an tem, a straghtforward soluton would be to have as many copes of an tem as the number of people that demand the tem as ther top choce. For example, n the above nstance that does not admt a popular matchng, say we have 2 addtonal copes of b 1 (call them b 1 and b 1 ). Thus, we have 3 copes of b 1 n our nstance now and t s easy to see that the matchng M = {(a1, b 1 ), (a 2, b ), 1 (a 3, b 1 )} s a popular matchng for the new nstance. But our assumpton of allowng arbtrarly many copes of an tem need not be true n practce. Fortunately, popular matchngs do not requre each person to be matched to her top choce tem. In the above nstance t s possble to get a popular matchng by makng just one extra copy of tem b 1 and not duplcatng any other tems. In ths sense, we expect to mprove the stuaton n terms of popularty by havng addtonal copes of some tems wthn the specfed bounds. Such a soluton of makng extra copes s appealng when the tems represent books or DVDs. Say, along wth a DVD b comes a lcense whch allows us to make a specfed number c many copes of that partcular DVD. If we make up to c copes of that DVD, then we do not ncur any addtonal cost. Our goal then s to make the approprate number of copes of every DVD so that the resultng nstance admts a popular matchng. Another relevant scenaro s when the set B represents tranng programs. Any tranng program b can be run for a sngle person, but some tranng programs may be able to accommodate up to c many people. We call ths the capacty of a tranng program and we wsh to fx the capacty of each tranng program so as to enable the resultng nstance to admt a popular matchng. It may seem from the above examples that havng as many copes of a DVD as are allowed or stretchng a tranng program to ts maxmum capacty should be the best thng to do. We show an example n the next secton whch llustrates that havng as many copes as possble does not always help popularty. The problem of fxng copes. Gven a graph G = (A B, E) where B = {b 1,..., b B } and a lst c 1,..., c B of upper bounds on the number of copes, does there exst an x 1,..., x B such that for each {1,..., B }, havng x copes of the -th tem, where 1 x c, enables the resultng graph to admt a popular matchng? We assume that G does not admt a popular matchng. Our problem s to determne f by fxng the copes of each tem approprately, whether the resultng graph admts a popular matchng or not. We now defne a specal case of ths problem whch we call the 1-or-2 copes problem. The 1-or-2 copes problem. In ths case, each c s ether 1 or 2. Note that when all the c s are 1, ths s the standard popular matchng problem. Thus the 1-or-2 copes problem s a generalzaton of the popular matchng problem. Here we have a subset K of tems whch can be duplcated, that s, the tems n K can have up to 2 copes n the resultng nstance, whle the rest of the tems wll have a sngle copy. The problem s to determne f by duplcatng some elements n K we get a graph that admts a popular matchng. The deletng-some-tems problem. When the nput nstance does not admt a popular matchng, another possblty s to delete some tems so that the resultng graph admts a popular matchng. Here we assume that only certan tems: elements of some subset B of B are allowed to be deleted; otherwse there s a trval soluton that says delete all tems, then the empty matchng s popular. The problem s to determne f there exsts a subset T B such that by deletng the tems n T, the resultng graph admts a popular matchng. If the 1-or-2 copes problem s NP-hard, then t s easy to show that the deletng-some-tems problem s also NP-hard. Gven an nstance G = (A B, E) and K B of the 1-or-2 copes problem, we create an nstance of the deletng-sometems problem as follows: our graph s H = (A B, E) where each tem n K has 2 copes. Thus the set of tems B here can be consdered as {b 1,..., b B, b,..., 1 b } k where K = {b 1,..., b k } and b,..., 1 b k are dentcal copes of these tems. Set the subset B of tems that may be deleted to {b,..., 1 b k }. That s, the frst copy of an tem may not be deleted, whle each of the second copes of tems that we created can be deleted. Thus, any effcent algorthm for the deletng-some-tems problem mples an effcent algorthm for the 1-or-2 copes problem Related work Subsequent to the work on popular matchng algorthms n [3], Manlove and Sng [13] generalzed the algorthms of [3] to the capactated case n the context of house allocaton, there tems were called houses and houses had capactes.

3 T. Kavtha, M. Nasre / Theoretcal Computer Scence 412 (2011) Ths problem can be easly solved usng the algorthm n [3] by ncludng c copes of tem b for each. A faster algorthm for determnng f such an nstance admts a popular matchng and computng one f t exsts, was shown n [13] for ths problem. Mestre showed effcent algorthms for the weghted popular matchng problem n [15], here each person s assgned a prorty or weght, and the defnton of popularty takes nto account the prortes of the people. Mahdan [12] consdered the problem of when random graphs (that s, preference lsts are randomly constructed) admt popular matchngs and showed that a popular matchng exsts wth hgh probablty n such graphs, when the number of tems s a factor of α 1.42 larger than the number of people. In order to deal wth the problem of the nput nstance not admttng a popular matchng, the followng extensons of popular matchngs have been consdered so far. Least unpopular matchng: A natural extenson when the graph does not admt a popular matchng s to ask for a least unpopular matchng. The unpopularty margn of a matchng M, call t u(m), s max M people who prefer M to M people who prefer M to M. The least unpopularty margn matchng s that matchng M wth the least value of u(m). McCutchen [14] showed that computng such a matchng s NP-hard. In [7] Huang et al. gave effcent algorthms to compute matchngs wth bounded values of these unpopularty measures n certan graphs. Mxed matchngs: Very recently, Kavtha et al. [10] consdered the problem of computng a probablty dstrbuton over matchngs, also called a mxed matchng, that s popular. It was shown that every nstance admts a popular mxed matchng and a polynomal tme algorthm was gven to compute such a probablty dstrbuton. Although, popular mxed matchngs are guaranteed to exst and are effcently computable, they may not be an acceptable soluton when the soluton needs to be a pure matchng Our contrbutons In ths paper we consder the followng extenson to the popular matchng problem: does havng extra copes of tems wthn certan bounds, yeld a graph that admts a popular matchng? The man results n ths paper are: The above problem s NP-hard. In fact, the 1-or-2 copes problem s NP-hard. As a consequence the problem of deletngsome-tems s NP-hard. Our reducton constructs an nstance of the 1-or-2 copes problem wth preference lsts of length just 2. In fact, each of these preference lsts has a unque top choce tem and there may be at most 2 tems ted as second choce tems. We also show that when preferences lsts have length 2 wth tes allowed only n the frst poston (.e., the second choce tem s unque), then the problem becomes solvable n polynomal tme. We show that the 1-or-2 copes problem remans NP-hard even f preference lsts are derved from a master lst. We consder a varant of the above problem where we only have to mantan an upper bound k on the total number of extra copes of all the tems, rather an upper bound on the number of copes of each tem. That s, here we are gven a graph G = (A B, E) and an nteger k 0 and we have to decde whether there exst 1,..., B where each 1 and k, such that havng + 1 copes of the -th tem, for 1 B, enables the resultng nstance to admt a popular matchng. Note that we add extra copes of the -th tem and n ths case the sum of the extra copes of all tems s bounded by k. We show a polynomal tme algorthm for ths varant. A prelmnary verson of these results can be found n [11]. Organzaton of the paper. Secton 2 outlnes the algorthm from [3] to compute a popular matchng. Secton 3 contans our NP-hardness proofs. Secton 4 shows a polynomal tme algorthm for the varant where only the sum of the extra copes s bounded. 2. Prelmnares We frst revew the algorthmc characterzaton of popular matchngs gven n [3]. As was done n [3], t wll be convenent to add a unque last tem l a at the end of a s preference lst for each person a A. We wll henceforth refer to ths graph as G = (A B, E). In ths way, t can be assumed that every matchng n G s A-complete, snce any unmatched person a can be pared wth l a. A maxmum matchng M n a bpartte graph G 1 = (A B, E 1 ) has the followng mportant propertes: M E 1 defnes a partton of A B nto three dsjont sets: odd (O), even (E), and unreachable (U). A vertex u belongs to the set E (resp. O) f there s an even (resp. odd) length alternatng path n G 1 from an unmatched vertex to u. A vertex u belongs to the set U, that s, t s unreachable, f there s no alternatng path n G 1 from an unmatched vertex to u. The followng lemma, proved n [6], s well-known n matchng theory. We nclude ts proof for completeness. Lemma 1. Let E, O and U be the sets of nodes defned by G 1 and M E 1 above. Then (a) E, O and U are parwse dsjont, and ndependent of the maxmum matchng M E 1. (b) In any maxmum matchng of G 1, every node n O s matched wth a node n E, and every node n U s matched wth another node n U. The sze of a maxmum matchng s O + U /2. (c) No maxmum matchng of G 1 contans an edge between a node n O and a node n O U. Also, G 1 contans no edge between a node n E and a node n E U.

4 1266 T. Kavtha, M. Nasre / Theoretcal Computer Scence 412 (2011) Proof. (a) The set U s dsjont from E and O by defnton. To prove that E s dsjont from O, assume that a node u s reachable by an even length path from a node a and an odd length path from a node b. Note that a b snce G s bpartte. Composng the two paths, we get an augmentng path n G wth respect to M, contradctng the maxmalty of M. To prove that ths partton n ndependent of M, let N be any other maxmum cardnalty matchng n G. M N conssts of alternatng paths and cycles and each of these paths and cycles are of even length. Snce the graph s bpartte, t s mmedate that the cycle have to be of even length. For paths, assume that a path has more edges from N, then such a path s an augmentng path wth respect to M, a contradcton to the maxmalty of M. A smlar argument holds f there are more edges from M. Usng these paths and cycles to swtch from M to N does not alter the odd/even/unreachable status of nodes, hence the partton s ndependent of the maxmum cardnalty matchng. (b) If a matched node u s reachable from a free node by an odd length path wth respect to any maxmum cardnalty matchng, then ts partner s reachable by an even length path. Thus, all edges n any maxmum matchng of G are ether OE or UU edges. Further, any node n U must be matched by a maxmum matchng, for, f not, the node s reachable wth an even length (zero length) path from tself. Also a node n O must be matched by a maxmum matchng snce an odd length alternatng path startng and endng wth a free node s an augmentng path. Thus, the sze of any maxmum matchng s O + U /2. (c) Nodes n E are reachable by an alternatng path from an unmatched node. Such paths end n a matchng edge. If there were an edge between two nodes n E, we could use that to construct an augmentng path, contradctng maxmalty. Fnally, f there were an edge between a node n E and a node n U, such an edge would be a non-matchng edge. We could use that to extend the alternatng path and reach the node n U, a contradcton to the defnton of nodes n U. As every maxmum cardnalty matchng matches all vertces u O U we call these vertces crtcal as opposed to vertces u E whch are called non-crtcal. Usng the above partton, the followng defntons can be made: Defnton 2. For each a A, defne f (a) to be the set of top choce tems for a. Defne s(a) to be the set of a s most-preferred non-crtcal tems n G 1. Theorem 1 ([3]). A matchng M s popular n G ff () M s a maxmum matchng of G 1 = (A B, E 1 ), and () for each person a, M(a) f (a) s(a). The algorthm for solvng the popular matchng problem s now straghtforward: each a A determnes the sets f (a) and s(a). An A-complete matchng that s a maxmum matchng n G 1 and that matches each a to an tem n f (a) s(a) needs to be determned. If no such matchng exsts, then G does not admt a popular matchng. The popular matchng algorthm from [3] s presented n the Appendx. We now present an example to llustrate these defntons as well as to show that havng the maxmum number of copes of an tem need not always help n terms of popularty Illustratve example Consder the followng nstance G = (A B, E), where A = {a 1, a 2, a 3, a 4, a 5 } and B = {f 1, f 2, s 1, s 2, s 3, s 4 } and the preference lsts of people are descrbed n Fg. 1(a). In Fg. 1(b) we have the same set of applcants and preference lsts except that every tem has 2 copes. That s, B = {f 1, f, 1 f 2, f, 2 s 1, s, 1 s 2, s, 2 s 3, s, 3 s 4, s 4 }, where we have explctly ncluded an dentcal copy b of b, for every b B. We refer to ths nstance wth duplcates as the nstance H = (A B, E ). Here and n the rest of the paper, the frst column n the fgure denotes the set of people and the row adjonng t denotes the preference lst of the partcular person. For example, a 1 treats f 1 as ts rank-1 tem, f 2 as ts rank-2 tem and s 1 as ts rank-3 tem accordng to Fg. 1(a). When there are multple tems n the same cell as n case of Fg. 1(b), we say that the tems are ted at that rank. That s, a 1 treats both f 1 and f 1 as ts rank-1 tem accordng to Fg. 1(b). We omt the numberng of the columns as gven by the frst row from the rest of the fgures a 1 f 1 f 2 s 1 a 2 f 1 f 2 s 2 a 3 f 1 f 2 s 3 a 4 f 1 f 2 s a 1 f 1, f 1 f 2, f 2 s 1, s 1 a 2 f 1, f 1 f 2, f 2 s 2, s 2 a 3 f 1, f 1 f 2, f 2 s 3, s 3 a 4 f 1, f 1 f 2, f 2 s 4, s 4 a 5 f 2 (a) a 5 f 2, f 2 (b) Fg. 1. Example showng larger copes do not always help.

5 T. Kavtha, M. Nasre / Theoretcal Computer Scence 412 (2011) Consder the subgraph G 1 = (A B, E 1 ) of G where every person has edges only to her top-choce tem. It s easy to see that every maxmum matchng of G 1 has to match the followng vertces: the tems {f 1, f 2 } and the person a 5. Thus, the tems f 1 and f 2 are crtcal and s 1,..., s 4 are non-crtcal n G 1. Hence s s the most preferred non-crtcal vertex n G 1 for person a, for = 1,..., 4. Applyng Theorem 1 we see that the nstance G admts a popular matchng: for example, {(a 1, f 1 ), (a 5, f 2 ), (a 2, s 2 ), (a 3, s 3 ), (a 4, s 4 )} s a popular matchng. Let us now consder the subgraph H 1 of H where every person has edges to her rank-1 tems. The crtcal vertces n H 1 = (A B, E ) 1 are f 1 and f 1 whle f 2 and f 2 are now non-crtcal n H 1. Thus f 2 and f 2 become the most preferred noncrtcal vertces for a 1,..., a 4 hence any popular matchng has to match each of a 1,..., a 4 to one of {f 1, f, 1 f 2, f 2 }. Also, a 5 has to be matched to f 2 or f 2 (snce a popular matchng s a maxmum matchng on rank-1 edges). Snce there are 5 people and only 4 tems that they can be matched to n any popular matchng, there exsts no popular matchng now. Thus the nstance shown n Fg. 1(b) where each tem has 2 copes does not admt a popular matchng whle the nstance n Fg. 1(a) where each tem has a sngle copy does. 3. The 1-or-2 copes problem Gven a graph G = (A B, E) and a subset K B of tems whch can be duplcated, that s, we can have up to 2 copes of every tem n the set K, the problem s to determne f there exsts a settng of copes of tems as x 1,..., x B where x = 1 for each tem b B \ K and x s ether 1 or 2 for each tem b K such that wth these copes the resultng graph admts a popular matchng. To prove that ths problem s NP-hard, we reduce the problem of monotone 1-n-3 SAT to the 1-or-2 copes problem. Monotone 1-n-3 SAT s a varant of the 3-satsfablty problem (3SAT). Lke 3SAT, the nput nstance s a collecton of clauses, where each clause conssts of exactly three varables and no varable appears n negated form. The monotone 1- n-3 SAT problem s to determne whether there exsts a truth assgnment to the varables so that each clause has exactly one true varable (and thus exactly two false varables). Ths problem s NP-hard [17]; n fact the varant of monotone 1-n-3 SAT where each varable occurs n at most 3 clauses s also NP-hard (refer to [4]). As wth other reductons from the 3SAT problem, our reducton also nvolves desgnng small gadgets whch buld an nstance of the 1-or-2 copes problem. We frst gve an overvew of our reducton and then descrbe these gadgets n detal Overvew of the reducton Let I be an nstance of the monotone 1-n-3 SAT problem wth {X 1, X 2,..., X n } beng the set of varables and {C 1, C 2,..., C m } beng the set of clauses n I. We must construct from I an nstance of the 1-or-2 copes problem G = (A B, E) and a subset K B of tems whch can be duplcated. We want the followng propertes of our nput G, K : () G (wth a sngle copy of each tem) does not admt a popular matchng. () By duplcatng some tems n K we get an nstance that admts a popular matchng ff there exsts an assgnment wth exactly one varable n each clause of I set to true. Wth these observatons we desgn a gadget for every clause n I. Each gadget conssts of a set of people and a set of tems along wth the preference lsts of people. A subset of these tems s nternal to the gadget, that s, such tems appear only on the preference lsts of people wthn the gadget. In addton there wll be tems whch are publc, that s, such tems appear on the preference lsts of people across several gadgets. The set of all publc tems n our nstance G s the set K of tems whch may be duplcated, that s, the tems n K that get duplcated wll have 2 copes n the resultng nstance. Note that the nstance I requres us to decde the true/false status of varables {X 1,..., X n }. Smlarly the nstance G requres us to decde the duplcaton status for tems n K. The non-trvalty of the 1-or-2 copes problem les n the followng: Let b be an tem that s a unque rank-1 tem for exactly one person n G, then b s crtcal n G 1 (the graph where we have only rank-1 edges). Makng an extra copy of b of tem b makes both b and b non-crtcal n the resultng graph restrcted to rank-1 edges. Ths n turn may change the s-tems of people n the resultng graph, and hence the resultng graph may admt a popular matchng. The preference lsts of people n our gadgets therefore ensure that every tem n K s a unque rank-1 tem for exactly one person n G. Publc tems. The set of publc tems n our nstance G s the set K of tems whch may be duplcated. We now descrbe how we derve the set of publc tems from the nstance I. For every occurrence of varable X n I, we have a publc tem n G. That s, f a varable X appears n clause C t we have an tem u t n G. Note that snce I s an nstance of monotone 1-n-3 SAT, no varable appears n negated form n I. We denote by dup(b) the duplcaton status of tem b K : dup(b) = 0 mples that tem b s not duplcated and hence has only 1 copy n the resultng nstance, whle dup(b) = 1 mples that b gets duplcated and hence has 2 copes n the resultng nstance. The value of dup(u t ) s should capture the truth value of varable X appearng n clause C t, that s, assgn varable X n clause C t as false f dup(u t) = 0, whereas assgn varable X n clause C t as true f dup(u t) = 1. For us to make the above translaton of duplcaton status, t has to be the case that for any, all dup(u t ) s have the same value. So f some dup(u t) s set to 1, we wll need to set dup(ul) = 1, for all l where ul K. Thus the set of tems

6 1268 T. Kavtha, M. Nasre / Theoretcal Computer Scence 412 (2011) correspondng to all occurrences of varable X should smultaneously have the same duplcaton status. That s, we want the followng property wth respect to the duplcaton status of tems correspondng to X. ( ) Let u t K, then dup(u t ) = dup(ut ), for all t where u t K. It can be seen that f ths property s satsfed, then the duplcaton status of tem u t can be translated to the truth assgnment for X appearng n clause C t. Further, ths wll be a consstent assgnment, that s, all occurrences of varable X get the same value. Wth ths, we have completely descrbed all the publc tems (elements of K ) n our nstance G. The set K, therefore, conssts of 3m tems as shown below. K =,t {u t : X appears n C t }. We now descrbe our gadgets - one correspondng to each clause and show how all the gadgets ensure that the above constrant s met. (1) 3.2. Gadget correspondng to a clause Let C t = (X 1 X 2 X 3 ) be a clause n I. Correspondng to C t we have a gadget whch we denote by G t. The gadget G t conssts of a set A t = {a t,..., 1 at } 14 of 14 people and a set B t = {p t, 1 pt, 2 pt, 3 qt, 0 qt, 1 qt 2 } of 6 nternal tems. Fg. 2(a) (c) show the preference lsts of the 14 people a t,..., 1 at 14 assocated wth the clause C t. Recall that we ntroduce a last resort tem for each person to ensure that matchngs are always A-complete. The l-tems are these last resort tems. As seen n Fg. 2(a) and (c) the publc tems u t 1 appear on the preference lsts of people shown n these two tables. Apart from these publc tems, people shown n Fg. 2(b) have publc tems u t appearng on ther preference lsts. Here, t 1, t 2, t 3 (each of them t) s such that, X 1, X 2, X 3 appear n C t1, C t2, C t3 respectvely. The role of the applcants a t,..., 6 at 11 s to ensure that the property (*) mentoned above s satsfed. It s easy to see that f some varable X 1 appears n exactly one clause (say C t ) n the nstance I, then we have exactly one publc tem u t 1 n our set K correspondng to varable X 1 and the property (*) s trvally true wth respect to X 1. Hence assume that the varable X 1 appears n at least 2 clauses, then t 1 s such that C t1 denotes the next hgher numbered clause after C t n whch varable X 1 appears. If C t happens to be the hghest numbered clause n whch varable X 1 appears, then let C t1 denote the lowest numbered clause n whch varable X 1 appears. Smlarly let t 2 (resp. t 3 ) denote the next hgher numbered clause after C t n whch varable X 2 (resp. X 3 ) appears. We note that t 1, t 2, t 3 need not be all dstnct. a t 1 p t 1 u t 1, q t, 0 qt 1 l t,1 a t 2 p t 1 u t 2, q t, 0 qt 1 l t,2 a t 3 p t 1 u t 3, q t, 0 qt 1 l t,3 a t 4 p t 1 q t 2 l t,4 a t 5 p t 1 q t 2 l t,5 (a) a t 6 p t 2 u t 1, u t 1 1 l t,6 a t 7 p t 2 u t 2 2 l t,7 a t 8 p t 2 u t 3 3 l t,8 a t 9 p t 3 u t 1, u t 1 1 l t,9 a t 10 p t 3 u t 2 2 l t,10 a t 11 p t 3 u t 3 3 l t,11 (b) Fg. 2. Preference lsts of people correspondng to a clause C t. a t 12 u t 1 l t,12 a t 13 u t 2 l t,13 a t 14 u t 3 l t,14 (c) The preference lsts are desgned such that when each publc tem has a sngle copy, then G t does not admt any popular matchng. Further, any nstance that admts a popular matchng and s obtaned by duplcatng publc tems n G t obeys the followng two propertes: At least one of u t 1 must get duplcated. As seen n Fg. 2(a), a t 4 and at 5 have pt 1 as ther top tem and qt 2 as ther second choce tem as q t 2 s nobody s top choce tem, t follows that qt 2 s the most preferred non-crtcal tem of at 4 and at 5. It s easy to see that n any popular matchng, one of a t, 4 at 5 has to be matched to pt 1 and the other to qt 2. The role of these 2 people s to ensure that a t, 1 at, 2 at 3 always get matched to tems n s(at ), 1 s(at ), 2 s(at 3 ), respectvely. Items u t 1 appear as unque top tems for a t, 12 at 13 and at 14 respectvely (Fg. 2(c)). Further, they do not appear as top choce tems for any other person n G. Thus, wth a sngle copy, all these tems reman crtcal on rank-1 edges. Hence s(a t ) = 1 s(at ) = 2 s(at ) = 3 {qt, 0 qt 1 }. Thus these 3 people cannot be matched to only these 2 tems n any popular matchng, so there exsts no popular matchng when each tem has a sngle copy. Therefore at least one of u t 1 should have 2 copes n the resultng nstance for all of a t, 1 at, 2 at 3 to be matched to tems n s(a t ), 1 s(at ), 2 s(at 3 ), respectvely.

7 T. Kavtha, M. Nasre / Theoretcal Computer Scence 412 (2011) Exactly one of u t 1 can be duplcated. The 6 people a t,..., 6 at 11 ensure that exactly one amongst ut 1 can have 2 copes n any nstance that admts a popular matchng. These 6 people as shown n Fg. 2(b) can be dvded nto two sets S 1 = {a t, 6 at, 7 at } 8 and S 2 = {a t, 9 at, 10 at 11 }. All these people have publc tems as ther second choce tems. Further, the preference lsts of people n gadgets G t1, G t2, G t3 ensure that tems u t also appear as unque top tems for exactly one person. Hence, when each of the publc tems has a sngle copy, then one person from S 1 and one person from S 2 get matched to ther top tem (p t 2 and pt 3 resp.), whereas the rest of the people get matched to ther respectve last resort tems (most preferred non-crtcal tem) n any popular matchng. However, we know that at least one of u t 1 has 2 copes due to a t 1,..., at 5. We wll assume here that copes of the tems u t are unavalable for the people n the gadget G t snce these copes wll be used up by people belongng to the other gadgets. It therefore suffces to focus on the duplcaton status of tems u t 1 wth respect to people n gadget G t. If exactly one among u t 1 (say, u t 1 ) gets duplcated, then both the copes of the tem u t 1 become non-crtcal n the resultng graph restrcted to rank-1 edges. The person a t 12 gets matched to one copy of u t 1 as ts f -tem whereas person a t 1 gets matched to another copy of ut 1 as ts s-tem. Further, a t 6 and at 9 get matched to the respectve top tems (p t 2 and pt 3 ) whle at, 7 at 8 and at, 10 at 11 get matched to ther respectve last resort tems. Assume that any two of u t 1 (say u t 1 ) have 2 copes n the resultng nstance. As n the prevous case, one copy of each of these tems s used up to match people a t 12 and at 13 to ther respectve f -tems. Further the extra copy of one of the 2 tems (say u t 1 ) s matched to people n a t, 1 at, 2 at 3. However, n ths case we also have four people at, 6 at 9 and at, 7 at 10 treatng u t 1 and u t 2 as ther s-tems respectvely. Although one person from each of the above pars can be matched to her top tem, we cannot match one copy of the tem (u t 2 ) amongst the remanng two people. It s easy to see that a smlar case occurs f all 3 of u t 1 are duplcated. Thus the gadget G t ensures that exactly one of the tems u t 1 has 2 copes n the resultng nstance. Note that the gadget descrbed above assumes that every varable appears n at least 2 clauses. For example, by ths assumpton for varable X 1, we have the 2 tems u t 1 and u t 1 1 appearng n the preference lsts of people a t 6 and at 9. If some varable (say X 1 ) appears n exactly one clause (say C t ), then the preference lsts of people a t 6 and at 9 contan only the tem ut 1 as ther second choce tem. Ths change does not affect the above two propertes that the gadget G t ensures Puttng t together The graph G that we construct s the unon of gadgets correspondng to each of the clauses. The sets A and B can be descrbed as below: A = m t=1 A t. B = m t=1 B t K. The set K s the set of all the publc tems as descrbed earler n Eq. (1). The preference lsts of the people are as shown n our gadget. We note that every tem b K s a unque rank-1 tem for exactly one person n A. We now show how all the gadgets co-operate to enforce property ( ) mentoned n Secton 3.1. Lemma 2. Assume that by havng 1 or 2 copes of tems n K there exsts an nstance that admts a popular matchng. In such an nstance f for some, t, dup(u t ) = 1 then dup(ut ) = 1, for all t where u t K. Proof. Recall that the tem u t K because the varable X appears n the clause C t. If C t s the only clause n whch X appears, then we are done, otherwse assume that X appears n another clause say C t1. Wthout loss of generalty let X be the frst varable n every clause that t appears n and t 1 be such that C t s the next hgher numbered clause after C t1 n whch X appears. We frst show that dup(u t) = 1 dup(ut 1 ) = 1. Suppose not. The tem u t appears on the preference lsts of followng people: a t 1 n gadget G t. a t 6 and at 9 n gadget G t. a t 1 6 and at 1 9 n gadget G t 1. Wth dup(u t) = 1, all these people regard ut as ther s-tem. The desgn of our gadgets ensures that no other tem u t can have 2 copes n the resultng nstance. Hence, n any popular matchng n the resultng nstance, whle a t 1 gets matched to ts s-tem u t, each of at, 6 at, 9 at 1, 6 at 1 9 has to be matched to ts f -tems pt, 2 pt, 3 pt 1, 2 pt 1 3 respectvely. Our assumpton that the resultng nstance admts a popular matchng and that dup(u t 1 ) = 0 mples that there exsts another tem u t 1 j K whch has 2 copes, or equvalently dup(u t 1 j ) = 1. Wth dup(u t 1 j ) = 1, the followng people belongng to the gadget G t1 treat u t 1 j as ther s-tem. ether the 3 people a t 1 2, at 1 7, and at 1 10 or the 3 people at 1 3, at 1 8, and at 1 11.

8 1270 T. Kavtha, M. Nasre / Theoretcal Computer Scence 412 (2011) For k = 2 or 3, whle a t 1 k gets matched to her s-tem (whch s some ut 1 j ) n any popular matchng, both a t 1 k+5 and at 1 k+8 are left wthout an tem amongst ther f or s-tems, a contradcton to the lemma hypothess that the resultng nstance admts a popular matchng. Hence we have dup(u t 1 ) = 1. Recall that we started wth an nstance of monotone 1-n-3 SAT where every varable appears n at most 3 clauses. Thus correspondng to the varable X, we have at most one more tem u t 1 K and we need to show that dup(u t 1 ) = 1. We note that a smlar argument as above forbds any nstance where dup(u t ) = dup(ut 1 ) = 1 and dup(u t 1 ) = 0, to admt a popular matchng. Hence we have dup(u t 1 ) = 1. Ths completes our proof. Thus f by havng 1 or 2 copes of each tem n K there exsts an nstance that admts a popular matchng, then the duplcaton status of tems n K always translates to a consstent truth assgnment of the varables n I. We now prove the followng two lemmas whch establsh the correctness of our reducton. Lemma 3. If there exsts an nstance by havng 1 or 2 copes of tems n K such that ths nstance admts a popular matchng, then there exsts a 1-n-3 satsfyng assgnment for I. Proof. We obtan a truth assgnment of varables X 1,..., X n of I from the duplcaton status of tems n K as follows: for X appearng n clause C t, set X to true f dup(u t) = 1, else set X to false. By Lemma 2, ths assgnment s a consstent assgnment. To see that ths s also a 1-n-3 satsfyng assgnment, consder a clause C t = (X 1 X 2 X 3 ) n I. The fact that the resultng nstance admts a popular matchng enforces that n G t, exactly one of u t 1 has ts dup( ) value set to 1. As the above property s true for every clause C t, t follows that we have a truth assgnment to the varables of I such that every clause has exactly one varable set to true. Ths proves the lemma. Lemma 4. If there exsts a 1-n-3 satsfyng assgnment for I, then there exsts an nstance obtaned by duplcatng tems n K whch admts a popular matchng. Proof. Let Truth-val denote a 1-n-3 satsfyng truth assgnment to varables X 1,..., X n of I. Set the dup( ) value of each tem n K =,t {u t : X appears n C t } as follows: dup(u t ) = 0 f Truth-val(X ) = false, else dup(u t ) = 1, t where ut K. Frst, snce Truth-val s a 1-n-3 satsfyng assgnment for I, every clause n I has exactly one varable set to true. Ths mples that for every clause C t = (X 1 X 2 X 3 ), n the gadget G t, exactly one tem among u t 1 has duplcaton status set to 1. Let us assume that Truth-val sets varable X 1 n clause C t to true. Thus the tem u t 1 has 2 copes n our nstance. To prove that the graph as defned by the above duplcaton status admts a popular matchng, we show that there exsts a matchng that s a maxmum cardnalty matchng on rank-1 edges and where each person a A gets matched to an tem n f (a) s(a). We focus on the people belongng to the set A t. Consder the 5 people a t,..., 1 at 5 frst: at 4 and at 5 get matched to pt 1 and qt 2, respectvely whereas at 2 and at 3 get matched to q t 0 and qt 1, respectvely (ther s-tems). Snce we have 2 copes of tem ut 1, both these copes are non-crtcal n the graph on rank-1 edges. Thus, a t 1 can be matched to one copy of ut 1 snce t s one of her s-tems. Consder the 6 people a t,..., 6 at 11 next: we recall that the duplcaton scheme was derved from the 1-n-3 satsfyng assgnment for I and the property (*) mentoned n Secton 3.1 s ensured by our gadgets. Thus, our assumpton that tem u 1 has 2 copes n our nstance mples that each of the tems u t have a sngle copy n our nstance. Thus all these tems reman crtcal on the graph restrcted to rank-1 edges and hence each of the people a t, 7 at, 8 at, 10 at 11 treat ther respectve last resort tems as ther s-tems. We can therefore match a t 6 and at 9 to ther f -tems pt 2 and pt 3 respectvely whle matchng a t, 7 at, 8 at, 10 at 11 to ther unque last-resorts. Ths leaves us wth the 3 people a t, 12 at, 13 at 14 who get matched to ther respectve f -tems ut 1. Ths fnshes our proof that f Truth-val s a 1-n-3 satsfyng assgnment for I, then we have a settng of duplcaton values of tems n K such that the resultng graph admts a popular matchng. We note that all the people n our gadgets have preference lsts of length 2 where tes occur at second choce tems (see Fg. 2). Recall that the last-resort tems were ntroduced by us and hence are not counted. Further, all people except a t, 1 at, 2 at 3 have at most 2 tems ted as ther second choce tems. It s easy to see that we can merge the tems q t, 0 qt 1 nto a sngle tem qt whch now belongs to the set of tems whch can be duplcated. Further, n any nstance that admts a popular matchng, we wll set dup(q t ) = 1, t = 1,..., m. We can now conclude the followng theorem. Theorem 2. The 1-or-2 copes problem s NP-hard for preference lsts of length 2 wth tes of length 2 allowed n second choce tems. Snce the tes n the preference lsts occur only for second choce tems, ths leaves us wth two unresolved cases: The complexty of the 1-or-2 copes problem when preference lsts are strct (that s, no tes are allowed). Our gadgets can be easly modfed to show that ths problem s also NP-hard. The complexty of the 1-or-2 copes problem when preference lsts have length two and tes can occur only at the frst poston. We show that ths case has a polynomal tme algorthm.

9 T. Kavtha, M. Nasre / Theoretcal Computer Scence 412 (2011) Strct lsts We break tes n the above nstance as follows: For each of a t k, k {1, 2, 3}, we frst replace the tems qt 0 and qt 1 by a sngle tem qt whch becomes an element of K (set of those tems that can be duplcated). Further we let tem u t k precede the tem q t n the preference lst of a t k. For any of a t k, k {6,..., 11}, we have two publc tems ted as at k s second choce, call them ut and u t 1. Recall that these two publc tems correspond to the same varable X appearng n two dfferent clauses namely C t and C t1. We break the te n the rank-2 tem for a t k by lettng tem ut precede tem u t 1 such that t < t 1. That s, the tem correspondng to the lower numbered clause precedes the one correspondng to the hgher numbered clause. It s easy to check that for an A-complete matchng to exst, the tems q t for t = 1,..., m need to have 2 copes n the resultng nstance. It s straghtforward to verfy that all our clams hold even wth these strct preference lsts and hence the 1-or-2-copes problem s NP-hard for strct preference lsts. Corollary 1. The 1-or-2 copes problem for strct preference lsts where the longest preference lst has length 3 s NP-hard Master preference lst In ths secton we consder the restrcton of the problem when preference lsts of the people are derved from a master preference lst. A master lst s a rankng of all tems accordng to some global objectve crteron. The master lst may be allowed to contan tes or may be strct. Irvng et al. [9] consdered the stable marrage problem n the presence of master preference lsts and proved that many nterestng varants reman hard under ths master lst model. In the same ven, we consder the 1-or-2 copes problem n the presence of a master lst. In the master lst model, the preference lst of a person s the same as the master lst, except that she can delete all tems that she fnds unacceptable. We show that under ths severe restrcton of master lst also, the 1-or-2 copes problem s NP-hard for strct preferences. For the sake of convenence, we partton the set B as F (f -tems), S (s-tems) and D (duplcable tems). B = F D S where the sets F, D and S are as defned below: F = m t=1 {pt 1, pt 2, pt 3 } S = m t=1 {qt 2 } D = K m t=1 {qt }. Here the set K s as defned by Eq. (1). The tem q t s the replacement for tems q t 0 and qt 1 as done for strct lsts. We note that the set D s the set of tems whch can be duplcated. To get a master lst such that the preference lsts of all people are derved from the master lst, we order elements of F, S and D as follows: Order the tems n F n any arbtrary order n a strct manner. Let F o denote the ordered lst. Order the tems n S n any arbtrary order n a strct manner. Let S o denote the ordered lst. Let D o denote the ordered lst of D as: D o = u t ,..., u l 1 n, u l 2 n, u l 3 n, q 1, q 2,..., q m Here t 1 < t 2 < t 3 and l 1 < l 2 < l 3. It s clear that the lst F o, D o, S o forms a master lst for the strct nstance G constructed above. We therefore conclude the followng corollary. Corollary 2. The 1-or-2 copes problem wth strct preference lsts derved from a master lst s NP-hard Lsts of length 2 wth tes n the top poston only Here we consder the 1-or-2 copes problem for preference lsts of length 2 where the second choce tem s unque. We are gven G = (A B, E) and every a A can gve any number of top choce tems, say f 0, f 1,..., f d, whch are ted as a s most preferred tems and at most a sngle tem, say s, as a s next choce tem. Thus the entre preference lst of a s: (f 0, f 1,..., f d ) followed by s followed by l a. We show there exsts a polynomal tme algorthm for such nstances. Let E, O and U denote the set of even, odd and unreachable vertces respectvely, n G 1 (the graph G restrcted to rank- 1 edges), refer to Secton 2. The set E s the set of non-crtcal vertces, thus only elements of E are canddates for beng elements n s(a), for any a A. Recall that certan tems that are crtcal n G 1 can become non-crtcal on rank-1 edges once they get duplcated. Let G = (A B, E) be the nstance of the above specal case and K be the set of tems that can be duplcated. Assume that G does not admt a popular matchng and there exsts a way of duplcatng tems n K such that the augmented nstance G = (A B, E) admts a popular matchng. Then we have the followng lemma.

10 1272 T. Kavtha, M. Nasre / Theoretcal Computer Scence 412 (2011) Lemma 5. There exsts an augmented nstance H such that every tem that s crtcal n G 1 s also crtcal n H 1 and H admts a popular matchng. Proof. If every tem that s crtcal n G 1 s also crtcal n G1 (recall that G1 s the graph G restrcted to rank-1 edges), then we have H = G and we are done. Suppose not, then there exsts an tem b1 such that b 1 s crtcal n G 1 but wth dup(b 1 ) = 1, both b 1 and ts duplcate b 1 are non-crtcal n G1. Snce b 1 s non-crtcal n G1, deletng b 1 does not change the sze of the maxmum cardnalty matchng n G1. As any popular matchng has to be a maxmum cardnalty matchng n G1, the contrbuton of b 1 s to add an edge (a, b ) 1 of rank-2 n any popular matchng of G. Note that the person a has to be non-crtcal n G1, as all crtcal vertces of G1 are matched along rank-1 edges n any popular matchng of G. The deleton of b 1 makes b 1 crtcal n the resultng graph. Snce a s non-crtcal, all of a s top choce tems have to be crtcal as there s no edge between 2 non-crtcal vertces (Lemma 1, see Secton 2) - thus a treats ts unque last-resort tem l a as ts s-tem once b 1 s deleted. In the resultng graph too, a has a partner to match n the set f (a) s(a). Deletng all such duplcates b from G we get the desred graph H n whch every crtcal tem n G1 s also crtcal n H 1. Further, snce every person a stll has a partner n f (a) s(a), t follows that H admts a popular matchng. The above lemma makes t smple for us to decde whch tems n K should be duplcated - we mantan the nvarant that no crtcal tem n G 1 becomes non-crtcal due to duplcaton. Let O B, E B and U B denote the odd, even and unreachable tems respectvely, n the current graph. Intally, the current graph s G 1. No tem b U B K should be duplcated. Ths s because makng an extra copy b of such an tem b makes both b and b non-crtcal n the resultng graph restrcted to rank-1 edges. We duplcate all tems b E B K. Ths s because makng an extra copy b of such an tem b does not change the crtcalty status of b as well as b n the resultng graph restrcted to rank-1 edges. For b O B we note the followng: There exsts an alternatng path startng from an unmatched person a to an tem b O B. Duplcatng such an tem b creates an augmentng path from a to b; thus no crtcal tem on rank-1 edges turns non-crtcal by ths change. However an tem b O B can now belong to U B due to the above change. Hence we need to update the status of all tems b O B to check whether b belongs to O B or U B. For nstance, let (f 0, f 1 ) be the top choce for 3 people a 1, a 2, a 3. Both f 0 and f 1 are odd, however once f 0 s ntroduced, we wll have 3 people and 3 top tems (f 0, f, 0 f 1), thus all these tems are now unreachable on rank-1 edges. Hence we should not duplcate f 1 now. Our algorthm s presented n Fg. 3. The correctness of the algorthm follows from Lemma Let G 1 denote the graph G on rank 1 edges; that s, G 1 = (A B, E 1 ). 2. Partton vertces n B w.r.t. a maxmum cardnalty matchng M 1 n G 1 as even(e B ), unreachable(u B ), odd(o B ). 3. H 0 = G. 4. For every b E B K do Add an extra copy of b to the graph H Order the tems n O B K as o 1,..., o t. For each = 1 to t do: If o s odd on rank 1 edges n H 1, then set dup(o ) = 1. Ths defnes the graph H, that s, H = H 1 + an extra copy of o f dup(o ) = 1. Else H = H If H t admts an A-complete matchng, then return H t as the graph G. Otherwse output there s no G correspondng to G, K that admts a popular matchng. Fg. 3. Algorthm for a specal case of 1-or-2 copes problem. Theorem 3. There exsts a polynomal tme algorthm for the 1-or-2 copes problem wth preferences lsts of length 2 wth tes occurrng at the top poston only. 4. Bounded total copes problem We now consder the followng problem: suppose we are gven an nteger k and we have to decde whether there exst 1,..., B where each 1 and k, such that havng + 1 copes of the -th tem, for 1 B, enables the resultng nstance to admt a popular matchng. We show that ths problem can be solved n polynomal tme. Let G 1 denote the graph n whch every person adds edges to her f -tems. Every even person n G 1 adds an edge to her s-tem. We call ths graph G, that s where every person a has added edges to tems n f (a) and every even person a n G 1 has added edges to tems n s(a). Let M be any maxmum cardnalty matchng n G. Snce G does not admt a popular matchng, we know that M < A. The followng theorem s useful here.

11 T. Kavtha, M. Nasre / Theoretcal Computer Scence 412 (2011) Theorem 4. Let G = (A B, E) be the gven graph and let nteger k denote an upper bound on the total extra copes of all tems that we can have n the augmented nstance. Then there exsts a new nstance H = (A B, E) wth extra copes of tems that admts a popular matchng ff k A M, where M denotes any maxmum cardnalty matchng n G. Proof. We frst show that k has to be at least A M. We need to show that addng an extra copy of any tem ncreases the sze of the maxmum cardnalty matchng n the resultng graph restrcted to f and s-tems by at most 1. Ths s easy to see snce addng an extra copy of an s-tem b does not change the f /s-status of the tem b and ts duplcate. Thus f b was crtcal n the graph restrcted to f -tems and s-tems, then havng a duplcate b ncreases the sze of the maxmum cardnalty matchng n ths graph by at most 1. Regardng f -tems, note that ths tem s crtcal on rank-1 edges before ts duplcate was ntroduced. Hence the duplcate ntroduced mght make the copy of the tem an s-tem, however the contrbuton to the sze of the maxmum matchng n the graph restrcted to f -tems and s-tems s at most 1. Snce each extra copy of some tem ncreases the sze of a maxmum matchng by at most one, we need to have at least A M extra copes of some tems so that the resultng graph admts a popular matchng. To show the other sde of the mplcaton, let k A M. We wll show that t s possble to construct a new nstance H such that H (the graph H restrcted to f -tems and s-tems) admts an A-complete matchng. Snce G does not admt a popular matchng, there exsts an α A such that α s unmatched n G ; note that such an α s even or non-crtcal n G. Further, the f -tem b for such a person s odd n G. Let H 1 be the same as the graph G, except that we add an extra copy b of b and we match α to b. Note that none of the sets s(a) for a A has changed n H 1 as no crtcal vertex turns non-crtcal after addng the duplcate b. Ths s because the tem b was odd before the duplcaton; so ether b and b are stll odd or they become unreachable now. Thus each person matched n M contnues to reman matched n any maxmum cardnalty matchng n H 1. Further, we have an extra person matched to b. It s easy to see that the same step can be repeated, untl we have an A-complete matchng n some H. Ths process s guaranteed to halt after A M teratons and the resultng graph H A M s ndeed our desred graph H that admts a popular matchng. The algorthm to construct such a graph H from the gven nstance G(A B, E) follows from the proof of suffcency of the above theorem. The algorthm s descrbed n Fg Construct the graph G = (A B, E ) where E = {(a, b) : a A, b f (a) s(a)}. 2. H 0 = G, H 0 = G, Let M denote a maxmum cardnalty matchng n H = Whle M s not A-complete matchng do: Let a be any unmatched person n H. Let b = f (a) [such a b s odd n H snce a s even n H ]. Add an extra copy of b and call the new nstance H +1. Construct the graph H +1 correspondng to H +1 and update M to be a maxmum cardnalty matchng n H +1. = Output the graph H. Fg. 4. Algorthm for bounded total copes problem. The above whle loop runs for A M teratons, each tme addng one more copy of some tem. Note that the same tem mght get chosen n varous teratons and thus ndvdual copes are not necessarly bounded. For example, f the nput had n people wth dentcal preference lsts: top tem s b 1, followed by b 2, and so on, then our algorthm would add n 1 copes of tem b 1. It s also easy to see that M s a popular matchng n the fnal graph. 5. Summary Gven a bpartte graph G = (A B, E) of people and tems where people have preferences over tems, we showed that the problem of decdng f there exsts an (x 1,..., x B ) {1, 2} B such that havng x copes of the -th tems enables the resultng graph to have a popular matchng s NP-hard. We reduced the monotone 1-n-3 SAT problem to an nstance of the above problem. Our reductons constructed nstances where the maxmum length of a preference lst s 2 when preference lsts can have tes and the maxmum length of a preference lst s 3 when preference lsts are strct. We showed that the problem s solvable n polynomal tme when preference lsts have length at most 2 wth a unque second choce tem. Also, the varant of ths problem where the total number of extra copes s bounded rather than a bound on the number of copes of ndvdual tems, s solvable n polynomal tme. Acknowledgements We thank Sourav Chakraborty for dscussons that motvated ths work and Prajakta Nmbhorkar for helpful dscussons. We also thank the anonymous revewers for ther helpful comments.