Covering Pairs in Directed Acyclic Graphs

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1 Advance Access publcaton on 5 November 2014 c The Brtsh Computer Socety All rghts reserved. For Permssons, please emal: ournals.permssons@oup.com do: /comnl/bxu116 Coverng Pars n Drected Acyclc Graphs Nko Beerenwnkel 1, Stefano Beretta 2, Paola Bonzzon 3, Rccardo Dond 4 and Yur Prola 3, 1 Department of Bosystems Scence and Engneerng, ETH Zurch, Basel, Swtzerland 2 Isttuto d Tecnologe Bomedche, Consglo Nazonale delle Rcerche, Segrate, Italy 3 Dp. d Informatca Sstemstca e Comuncazone, Unv. degl Stud d Mlano-Bcocca, Mlan, Italy 4 Dp. d Scenze Umane e Socal, Unv. degl Stud d Bergamo, Bergamo, Italy Correspondng author: prola@dsco.unmb.t The Mnmum Path Cover (MnPC) problem on drected acyclc graphs (DAGs) s a classcal problem n graph theory that provdes a clear and smple mathematcal formulaton for several applcatons n computatonal bology. In ths paper, we study the computatonal complexty of three constraned varants of MnPC motvated by the recent ntroducton of Next-Generaton Sequencng technologes. The frst varant (MnRPC), gven a DAG and a set of pars of vertces, asks for a mnmum-cardnalty set of (not necessarly dsont) paths such that both vertces of each par belong to the same path. For ths problem, we establsh a sharp tractablty borderlne dependng on the overlappng degree of the nstance, a natural parameter n some applcatons of the problem. The second varant we consder (MnPCRP), gven a DAG and a set of pars of vertces, asks for a mnmum-cardnalty set of (not necessarly dsont) paths coverng all the vertces of the graph and such that both vertces of each par belong to the same path. For ths problem, we show that, whle t s NP-hard to compute f there exsts a soluton consstng of at most three paths, t s possble to decde n polynomal tme whether a soluton consstng of at most two paths exsts. The thrd varant (MaxRPSP), gven a DAG and a set of pars of vertces, asks for a sngle path contanng the maxmum number of the gven pars of vertces. We show that MaxRPSP s W[1]-hard when parameterzed by the number of covered pars and we gve a fxed-parameter algorthm when the parameter s the maxmum overlappng degree. Keywords: mnmum path cover; sequence reconstructon; pared-end reads; computatonal complexty Receved 19 March 2014; revsed 29 August 2014 Handlng edtor: Ian Stewart 1. INTRODUCTION The Mnmum Path Cover (MnPC) problem s a well-known problem n graph theory. Gven a drected acyclc graph (DAG), MnPC asks for a mnmum-cardnalty set of paths such that each vertex of G belongs to at least one path of. The problem can be solved n polynomal tme wth an algorthm based on a proof of the well-known Dlworth s theorem for partally ordered sets, whch allows to relate the sze of a mnmum path cover to that of a maxmum matchng n a bpartte graph obtaned from the nput DAG [2]. The MnPC problem has mportant applcatons n several felds rangng from bonformatcs [3 5] to software testng A prelmnary verson of ths paper appeared n [1]. [6, 7]. In partcular, n bonformatcs, manly thanks to the advent of the Next-Generaton Sequencng technques, the MnPC problem s wdely appled to the reconstructon of nucleotde sequences startng from a large set of ther short fragments (called short reads) [3, 4]. More precsely, each fragment s represented by a sngle vertex and two vertces are connected f the algnments of the correspondng reads on the genomc sequence overlap. In [4], the paths on such a graph represent putatve RNA transcrpts and a mnmum-cardnalty set of paths coverng all the vertces represents a set of proten soforms whch lkely orgnated the observed reads. Instead, n [3], the paths of the graph represent the genomes of putatve vral quasspeces and a mnmum-cardnalty set of paths coverng the whole graph represents the lkely structure Secton A: Computer Scence Theory, Methods and Tools Downloaded from

2 1674 N. Beerenwnkel et al. of a vral populaton. In these applcatons, the am s that of reconstructng complete (hence, as long as possble) sequences that possbly share some substrngs. Hence, t s often assumed that paths start from a source vertex, end at a snk vertex and possbly share some vertces. In the rest of the paper, we wll mplctly make ths assumpton. Recently, the avalablty of new knds of data have motvated the defnton of new constraned varants of graph problems n dfferent felds, such as, for example, n the context of socal network analyss [8, 9]. Reconstructng sequences va Mnmum Path Cover s partcularly effectve on relatvely small regons of the sequences to reconstruct, snce on long regons there s not enough nformaton for establshng f two dstant fragments were orgnated from the same sequence. In order to solve (or lessen the mpact of) ths ssue, a partcular knd of reads, called pared-end reads, could consderably help. In fact, pared-end reads are pars of reads obtaned from the same sequence at a fxed dstance, typcally larger than the fragments length. Hence, n a vald path cover, for each pared-end read, there must exst at least a path whch contans the two vertces representng the assocated reads. However, the most wdely used methods for sequence reconstructon n bonformatcs [4, 10] do not take fully advantage of the constrants mposed by pared-end reads durng the reconstructon process and they only use them to valdate (or dscard) the reconstructed sequences. Recent approaches are tryng to ncorporate the new addtonal constrants carred by pared-end or longer sequences nto the problem formulaton, n order to explot t for the reconstructon of the sequences [5, 11, 12]. For example, n CLASS [11], complete sequences are frst exhaustvely enumerated and then a smallest set of them satsfyng all the constrants derved from pared-end or long reads s selected usng a greedy set-cover approxmaton algorthm. Clearly, ths method s both computatonally ntensve (snce t requres a nearly exhaustve enumeraton of the transcrpts) and also approxmate (snce t employs the approxmaton algorthm for set cover). As a consequence, both ts applcablty to real large datasets and the accuracy of ts results are lmted. BRANCH [5] overcomes these lmts by consderng only constrants derved from long reads (or prevously found transcrpts), whch are modeled as subpath constrants (.e. subpaths that must be contaned n some path of the soluton). For ths scenaro, the authors present a polynomal-tme algorthm that reduces the constraned path cover problem to MnPC by contractng the subpath constrants n sngle vertces. However, by not consderng pared-end reads, the accuracy of BRANCH degrades wth the length of the reconstructed sequences (unless there are many long subpath constrants, obvously). Recently, Rzz et al. [12] extended the reducton used by BRANCH n order to correctly handle also the cases where a constrant s a subpath of another constrant. Moreover, they also begn a prelmnary study of the computatonal complexty of the MnPC problem when constrants derved from pared-end reads are ntroduced. In ths paper, we present a systematc study of the computatonal complexty of some varants of the MnPC problem where constrants dervng from pared-end reads are ntroduced. Smlar constraned varants have been also studed n the past by Ntafos and Hakm n the context of software testng [7]. More precsely, n that context, each procedure to be tested s modeled by a graph where vertces correspond to sngle nstructons and two vertces are connected f the correspondng nstructons are executed sequentally. The test of the procedure should check each nstructon at least once, hence a mnmum path cover of the graph represents a mnmum set of executon flows that allows one to test all the nstructons. Moreover, snce there are pars of vertces that a feasble soluton must nclude, Ntafos and Hakm proposed and formalzed the concept of requred pars. In partcular, one of the problems they ntroduced s the Mnmum Requred Pars Cover (MnRPC) problem where, gven a DAG and a set of requred pars, the goal s to compute a mnmum set of paths coverng all the requred pars,.e. a mnmum set of paths such that, for each requred par, at least a path contans both vertces of the par. It s easy to see that the concept of requred pars ntroduced by Ntafos and Hakm correctly models the constrants dervng from pared-end reads n the sequence reconstructon problems we presented before. However, note that MnRPC asks for a soluton that covers only the requred pars, whle, n the sequence reconstructon problems, we are nterested n solutons that also cover all the vertces. For ths reason, we consder a varant of the MnPC problem, called Mnmum Path Cover wth Requred Pars (MnPCRP), that, gven a DAG and a set of requred pars, asks for a mnmum set of paths coverng all the vertces and all the requred pars. Clearly, MnPCRP s closely related to MnRPC. In fact, as we show n Secton 2, the same reducton used n [7] to prove the NPhardness of MnRPC can be appled to our problem, leadng to ts ntractablty. In ths paper, we contnue the analyss of [7] by studyng the complexty of path coverng problems wth requred pars. More precsely, we study how the complexty of these problems s nfluenced by two parameters relevant for the sequence reconstructon applcatons n bonformatcs: () the mnmum number of paths coverng all the vertces and all the requred pars and () the maxmum overlappng degree (defned later). In the bonformatcs applcatons we dscussed, the frst parameter the number of coverng paths s often small, thus an algorthm exponental n the sze of the soluton could be of nterest. The second parameter we consder n ths paper, the maxmum overlappng degree, can be nformally defned as follows. Two requred pars overlap when there exsts a path that connects the vertces of the pars, and the path cannot be splt n two dsont subpaths that separately connect the vertces of the two pars. Then, the overlappng degree of Secton A: Computer Scence Theory, Methods and Tools Downloaded from

3 Coverng Pars n Drected Acyclc Graphs 1675 a requred par s the number of requred pars that overlap wth t. In the sequence reconstructon applcatons, as the dstance between two pared-end reads s fxed, the maxmum overlappng degree s small compared wth the number of vertces, hence t s a natural parameter for nvestgatng the computatonal complexty of the problem. Frst, we nvestgate how the computatonal complexty of MnRPC s nfluenced by the maxmum overlappng degree. We show that MnRPC s APX-hard (hence also NP-hard) when the maxmum overlappng degree s bounded by 1, whle t s polynomal tme solvable when the maxmum overlappng degree s 0. Note that MnPCRP s already NP-hard f the maxmum overlappng degree s 0. In fact, ths can be easly obtaned by modfyng the reducton presented n [7] to hold also for restrcted nstances of MnPCRP wth no overlappng requred pars. Then, we nvestgate how the computatonal complexty of MnPCRP s nfluenced by the number of paths that compose a soluton. We prove that t s NP-complete to decde f there exsts a soluton of MnPCRP consstng of at most three paths (va a reducton from the 3-Colorng problem). We complement ths result by gvng a polynomal-tme algorthm for computng a soluton wth at most two paths, thus establshng a sharp tractablty borderlne for MnPCRP when parameterzed by the sze of the soluton. These results sgnfcantly mprove the hardness result that Ntafos and Hakm [7] presented for MnRPC (and that holds also for MnPCRP), where the soluton contans a number of paths whch s polynomal n the sze of the nput. Some of these results have been ndependently obtaned by Rzz et al. [12]. A natural heurstc approach for solvng MnPCRP s the one whch computes a soluton by teratvely addng a path that covers a maxmum set of requred pars not yet covered by a path of the soluton. Ths approach leads to a natural combnatoral problem, the Maxmum Requred Pars wth Sngle Path (MaxRPSP) problem, that, gven a DAG and a set of requred pars, asks for a path that covers the maxmum number of requred pars. We nvestgate the complexty of MaxRPSP and we show that t s not only NP-hard, but also W[1]-hard when the parameter s the number of covered requred pars. Ths result shows that t s unlkely that the problem s fxed-parameter tractable when parameterzed by the number of requred pars covered by a sngle path. We refer the reader to [13, 14] for an n-depth presentaton of the theory of fxedparameter complexty. We consder also the MaxRPSP problem parameterzed by the maxmum overlappng degree and, dfferently from MnPCRP, we gve a fxed-parameter algorthm for ths case. Ths postve result shows a gap between the complexty of MaxRPSP and the complexty of MnPCRP when parameterzed by the maxmum overlappng degree. The rest of the paper s organzed as follows. Frst, n Secton 2 we gve some prelmnary notons and we ntroduce the formal defntons of the problems we are nterested n. In Secton 3, we nvestgate how the computatonal complexty of MnRPC s nfluenced by the maxmum overlappng degree, whle n Secton 4, we nvestgate the computatonal complexty of MnPCRP when the soluton conssts of a constant number of paths, and n Secton 5, we nvestgate the computatonal complexty of MaxRPSP. We conclude n Secton 6 by presentng some fnal remarks and some open problems. 2. PRELIMINARIES In ths secton, we ntroduce the basc notons used n the rest of the paper and we formally defne the three combnatoral problems we are nterested n. We denote an undrected graph as G = (V, E), where V s the set of vertces and E s the set of (undrected) edges, and a drected graph (or dgraph)asd = (N, A), where N s the set of vertces and A s the set of (drected) arcs. We denote an edge of G = (V, E) as {v, u} E, where v, u V. Moreover, we denote an arc of D = (N, A) as (v, u) A, where v, u N. Gven a drected graph D = (N, A), apath π from vertex v to vertex u, denoted as vu-path, s a sequence of vertces v 1,...,v n such that (v,v +1 ) A, v = v 1 and u = v n.we say that a vertex v belongs to a path π = v 1,...,v n, denoted as v π, fv = v for some 1 n. Gven a path π = v 1,...,v n, we say that a path π = v,v +1,...,v 1,v, wth 1 n, s a subpath of π. Gven a set N N of vertces, a path π covers N f every vertex of N belongs to π. In the paper, we consder a set R of pars of vertces n N. We denote each par as [v,v ], to avod ambguty wth the notatons of edges and arcs. Now, we are able to defne the combnatoral problems we are nterested n. Problem 1. Mnmum Requred Pars Cover (MnRPC) Input: a DAG D = (N, A), a source s N, asnkt N and asetr ={[v x,v y ] v x,v y N,v x v y } of requred pars. Output: a mnmum cardnalty set = {π 1,...,π n } of drected st-paths such that every requred par [v x,v y ] R belongs to at least one st-path π,.e. v x, v y belongs to π. Problem 2. Mnmum Path Cover wth Requred Pars (MnPCRP) Input: a DAG D = (N, A), a source s N, asnkt N and asetr ={[v x,v y ] v x,v y N,v x v y } of requred pars. Output: a mnmum cardnalty set = {π 1,...,π n } of drected st-paths such that every vertex v N belongs to at least one st-path π and every requred par [v x,v y ] R belongs to at least one st-path π,.e. v x, v y belongs to π. Problem 3. Maxmum Requred Pars wth Sngle Path (MaxRPSP) Input: a DAG D = (N, A), a source s N, asnkt N and asetr ={[v x,v y ] v x,v y N,v x v y } of requred pars. Output: an st-path π that covers a set R ={[v x,v y ] v x,v y π} R of maxmum cardnalty. Secton A: Computer Scence Theory, Methods and Tools Downloaded from

4 1676 N. Beerenwnkel et al. (a) (b) FIGURE 1. Examples of two overlappng requred pars [u,v ]and [u,v ]. In (a) the requred pars are alternated, whle n (b)[u,v ] s nested n [u,v ]. For smplcty, n the rest of the paper, we assume that the source s and the snk t of the DAG are gven (otherwse, t s easy to fnd them). Moreover, we also assume that all the paths (unless otherwse specfed) start from s and end at t. Two requred pars [u,v ] and [u,v ]nr overlap f there exsts a path π n D such that the four vertces appear n π n one of the followng orders (assumng that the vertex u appears before u n π), where v and u are two dstnct vertces of G (see Fg. 1): () u, u,v,v (the two requred pars are sad to be alternated); () u, u,v,v (the requred par [u,v ]ssadtobe nested n [u,v ]). Note that, accordng to ths defnton, the requred pars [x, y] and [y, z] do not overlap. Moreover, we defne the overlappng degree of a requred par [u,v ] R as the number of requred pars n R that overlap wth [u,v ]. Gven a DAG D and a set R of requred pars, the compatblty relaton C R 2 on the set R s defned as follows. A par of requred pars ([u,v ], [u,v ]) belongs to C f there exsts a path π that covers both [u,v ] and [u,v ] and v appears strctly before v n π or [u,v ] s nested n [u,v ]. Note that f there exsts a path that covers two requred pars r, r, then ether (r, r ) or (r, r ) s n C. Ths defnton of compatblty among requred pars and the one proposed by [12] are closely related. However, note that, accordng to our defnton, the compatblty relaton s rreflexve (a requred par s not compatble wth tself), whle accordng to thers s symmetrc. Gven a subset R of R, we denote the restrcton of C to the elements n R as C(R ). We say that a subset C R of requred pars s a chan f C(C) s a strct total order (.e. for each r, r C wth r r, at least one of (r, r ) and (r, r ) belongs to C(C)). Fnally, we recall that a bnary relaton s a strct partal order f t s rreflexve and transtve. Please note that, n general, the compatblty relaton C s not transtve (hence, t s not a strct partal order). The compatblty relaton can be also consdered as the arc set of a dgraph (that we call the compatblty dgraph) havng as vertex set R. Such a dgraph s clearly acyclc. We call compatblty graph the undrected graph obtaned by dscardng the edge orentaton of the compatblty dgraph. In the followng, for smplcty, we wll nterchangeably consder the compatblty relaton as a bnary relaton or as the assocated dgraph. Hardness of MnRPC and MnPCRP. As mentoned n Secton 1, one of the problems we are nterested n, namely MnRPC, was ntally defned n the context of program testng [7] and ts NP-hardness was proved. From ths result, we can mmedately derve the NP-hardness of MnPCRP. Indeed, MnRPC can be easly reduced to MnPCRP by ensurng that each vertex of the graph D (nput of MnRPC) belongs to at least one requred par. Otherwse, f ths condton does not hold for some vertex v, we can modfy the graph D by contractng v (that s, removng v and addng an edge (u, z) to A, for each u, z N such that (u,v), (v,z) A). Ths mples that, snce n the resultng nstance of MnRPC all the vertces belong to some requred par, a feasble soluton of that problem covers every vertex of the graph. Then, a soluton of MnRPC s also a soluton of MnPCRP, whch mples that MnPCRP s NP-hard. Both MnRPC and MnPCRP on drected graphs (not necessarly acyclc) are as hard as MnRPC and MnPCRP, respectvely, on DAGs. In fact, snce each strongly connected component can be covered wth a sngle path, we can replace each of them wth a sngle vertex, obtanng a DAG and wthout changng the sze of the soluton. Fnally, MnRPC and MnPCRP on general graphs (wth the addtonal requrement that the coverng paths are smple) are as hard as the Hamltonan path problem, whch s NP-complete [15, probl. GT39]. 3. A SHARP TRACTABILITY BORDERLINE FOR MINRPC In ths secton, we consder the tractablty of MnRPC when the maxmum overlappng degree of the nstance s bounded. We show n Secton 3.1 that MnRPC s APX-hard (hence also NP-hard) when the maxmum overlappng degree s bounded by 1, whle n Secton 3.2 we show that MnRPC admts a polynomal tme algorthm when the maxmum overlappng degree s APX-hardness of MnRPC when the maxmum overlappng degree s bounded by 1 We wll show the APX-hardness of MnRPC when the maxmum overlappng degree s bounded by 1 by gvng an L-reducton (for detals on L-reductons we refer the reader to [16]) from the Mnmum Vertex Cover problem on Cubc Secton A: Computer Scence Theory, Methods and Tools Downloaded from

5 Coverng Pars n Drected Acyclc Graphs 1677 graphs (MnVCC). We recall that a graph s cubc when each vertex s adacent to exactly three other vertces. Gven an undrected cubc graph G = (V, E), the MnVCC problem asks for a mnmum cardnalty set V V such that for each edge {v,v } E, v V or v V. We start by showng how to transform (n polynomal tme) an nstance G = (V, E) of MnVCC nto an nstance D = (N, A), R of MnRPC such that ts maxmum overlappng degree s bounded by 1. Frst, we defne the vertex set N: N ={v,,q,v,,q {v,v } E, 1 q 4} {v,,q {v,v } E, 1 q 2} {s, t} We defne the arc set A by means of a set of paths (see Fg.. Note that the paths may share some arcs. For each edge {v,v } E, we defne two (dsont) paths π, and π, connectng the sequence of vertces s,v,,1,...,v,,4, t and s,v,,1,...,v,,4, t, respectvely. For each vertex v V, we defne four paths n D. Let {v,v }, {v,v h }, {v,v k } E, wth < h < k, bethethree edges of G ncdent to v. We defne a path: π = s,v,,1,v,,2,v,h,1,v,h,2,v,k,1,v,k,2, t Moreover, we defne three paths, called addtonal paths: () π, = s,v,,1,v,,1,v,,2,v,,4, t ; () π,h = s,v,h,1,v,h,1,v,h,2,v,h,4, t ; () π,k = s,v,k,1,v,k,1,v,k,2,v,k,4, t. The paths defned above wll be used later (see Lemmas 3.2 and 3.3) to construct a soluton of MnRPC over nstance D = (N, A), R. Fnally, we defne the set R of requred pars as follows: R ={[v,,1,v,,4], [v,,2,v,,3 ], [v,,2,v,,3 ] {v,v } E} {[v,,1,v,,2 ] v V, {v,v } E} Fgure 2 represents an extract of a drected subgraph of D assocated wth an undrected subgraph (constructed from a vertex v V )ofg. It s easy to see that, gven a graph G, the correspondng nstance D, R can be constructed n polynomal tme. Next, we show that D, R has maxmum overlappng degree 1. Lemma 3.1. degree 1. Instance D, R has a maxmum overlappng Proof. Note that the only overlappng requred pars n R are, fxed an edge {v,v } E, [v,,1,v,,2 ] and [v,,1,v,,4]. Hence, the maxmum overlappng degree n D s 1. Now, we are able to prove the two man results of the reducton. Lemma 3.2. Gven an undrected cubc graph G = (V, E) and a vertex cover V V of G, we can compute n polynomal tme a feasble soluton of the assocated nstance D = (N, A), R of MnRPC such that = 3 E + V. Proof. Consder a vertex cover V for G = (V, E). Inthe followng, we compute (n polynomal tme) a set of 3 E + V paths on D that covers all the requred pars n R. Set s constructed as follows. (1) For each vertex v / V, add to the three addtonal paths π,, π,h and π,k, where v, v h, v k are the three vertces adacent to v. ( For each vertex v V, add to the path π. (3) For each edge {v,v } E, add to the two paths π, and π,. (4) For each edge {v,v } Esuch that v,v V and <, add to the path π,. It s easy to see that covers each requred par n R. Indeed, each requred par [v,,2,v,,3 ], [v,,2,v,,3 ] s covered n step (3). Each par [v,,1,v,,2 ] s covered n step ( f v V or n step (1) f v / V. Each requred par [v,,1,v,,4] R s covered by a path added n step (1) f v / V or v / V, whle t s covered by a path added n step (4) f v V and v V. Note that, snce V s a vertex cover, at least one of v,v belongs to V. For each edge {v,v } E, steps (1) and (4) add exactly one path contanng the vertces v,,1 and v,,4. Hence, they add E paths to. Step ( adds V paths, whle step (3) adds 2 E paths. As a consequence, we have that s a feasble soluton of MnRPC and that =3 E + V. Lemma 3.3. Let G = (V, E) be an undrected cubc graph and let D = (N, A), R be the assocated nstance of MnRPC. Then, gven a set of paths of D = (N, A) whch s a soluton of MnRPC where =3 E +d, we can compute n polynomal tme a vertex cover V V for G such that V d. Proof. Note that by constructon some requred pars n R, namely [v,,2,v,,3 ], [v,,2,v,,3 ] can only be covered by the paths π, and π,, respectvely. A fortor, all these paths must belong to. Moreover, note that by constructon each requred par [v,,1,v,,4] can be covered by at most two paths, namely π, and π,. Hence, at least one of π, and π, must be n. Now, we show that we can restrct ourselves to solutons where exactly one of π, and π, s n. If both π, and π, are n, then we can compute n polynomal tme a Secton A: Computer Scence Theory, Methods and Tools Downloaded from

6 1678 N. Beerenwnkel et al. FIGURE 2. Example of the drected (acyclc) subgraph of D = (N, A) assocated wth a subgraph of a cubc graph G = (V, E) w.r.t. vertex v V. Gray boxes hghlght the pars of paths representng the arcs ncdent to v and the paths representng the two vertces v (bottom) and v (top). The requred pars (not shown) are [v,,1,v,,4], [v,,2,v,,3 ], [v,,2,v,,3 ], [v,h,1,v,h,4], [v,h,2,v,h,3 ], [v,h,2,v,h,3 ], [v,k,1,v,k,4], [v,k,2,v,k,3 ], [v,k,2,v,k,3 ], [v,,1,v,,2 ]and[v,,1,v,,2 ]. soluton such that at most one of π,, π, belongs to and such that as follows. Set s computed by replacng one of π,, π,, respectvely, wth the path π, π, respectvely. We assume w.l.o.g. that π, s replaced wth π. Clearly,. Moreover, covers each requred par n R, hence the same property holds for, snce the requred par [v,,1,v,,4] s covered by the path π, and the requred par [v,,1,v,,2 ] s covered by π. Hence, n what follows, we assume that, for each {v,v } E, contans exactly one of π,, π,. Note that the set contans the 2 E paths π,, π,, E paths π,, and d paths π. Defne the vertex cover V V as {v π }. By constructon V d. We clam that V s a vertex cover of G. Suppose, on the contrary, that V s not a vertex cover of G. Then, there exsts an edge {v,v } E such that nether v nor v are n V. It follows that nether π nor π are n. By hypothess, the set covers the requred par [v, ], hence,1,v,,2 must contan both the paths π, and π, whch are the only paths dfferent from π and π that cover the par [v,,1,v,,2 ]. Snce we assumed that does not contan both π, and π,, t follows that one of the requred par [v,,1,v,,2 ], [v,,1,v,,2 ]s not covered by. Ths fact contradcts the hypothess that s a soluton of MnRPC, hence V s a vertex cover of G and the lemma holds. Theorem 3.1. MnRPC s APX-hard even when the nput nstance has maxmum overlappng degree bounded by 1. Secton A: Computer Scence Theory, Methods and Tools Downloaded from

7 Coverng Pars n Drected Acyclc Graphs 1679 Proof. Frst, note that by Lemma 3.1 the maxmum overlappng degree of D, R s 1. Snce n a cubc graph E = 3 2 V and the cardnalty of a vertex cover V s at least 1 2 V, t follows by Lemmas 3.2 and 3.3 that we have desgned an L- reducton. Snce MnVCC s APX-hard [17], t follows that MnRPC s APX-hard even when the nput nstance has maxmum overlappng degree bounded by A polynomal tme algorthm for MnRPC wthout overlappng pars We wll show that MnRPC can be solved n polynomal tme when the nstance does not contan overlappng requred pars (.e. when the maxmum overlappng degree s 0). We obtan ths result by frst provng that, whenever the compatblty relaton of the requred pars s a strct partal order, we can compute n polynomal tme a mnmum-cardnalty set of paths whch covers all the requred pars. Then, the result follows from the fact that the compatblty relaton of a set of non-overlappng requred pars s always a strct partal order. Let D, R be an nstance of MnRPC and C be the compatblty relaton on R. The basc dea on whch the polynomal-tme algorthm s bult s that a chan C of C corresponds to a path π C n D that covers all the requred pars n C, as proved n the followng lemma. Lemma 3.4. Let D, R be an nstance of MnRPC. Then, there exts a path π C n D coverng a subset C R of requred pars f and only f C s a chan of the compatblty relaton C on R. Proof. The exstence of π C mples the exstence n the compatblty dgraph of an arc between each (unordered) par of requred pars of C. The orentaton of each arc s gven by the order of the vertces on π C (accordng to the defnton of the compatblty relaton), but, snce the compatblty dgraph s acyclc, we have that C(C) s a total order (and C s a chan). Let C be a chan of C. Snce C s a chan, gven two vertces v,v whch belong to two dfferent requred pars of C, there exsts a path between v and v (ether from v to v or from v to v ). Consder the nodes v 1,v 2,...,v l that appear n some requred par of C, sorted accordng to a topologcal order of D. Connect them to buld a path π C from s (the source of D) to t (the snk of D). Such a path exsts snce, by the prevous observaton, there exsts a path between each par of vertces v,v +1. By constructon, π C covers all the pars n C. In partcular, note that the prevous proof gves a polynomal-tme algorthm for computng a path π C whch covers a subset C of requred pars formng a chan of the relaton C. Moreover, the prevous result shows that MnRPC can be optmally solved n polynomal tme whenever C s a total order (snce n that case there exsts a unque chan contanng all the requred pars). Ths result can be generalzed for computng an optmal soluton n polynomal tme whenever C s a strct partal order, as shown n the followng theorem. Theorem 3.2. Let D, R be an nstance of MnRPC and C be the compatblty relaton of R. If C s a strct partal order, then a mnmum-cardnalty set of paths of D coverng all the requred pars n R can be computed n polynomal tme. Proof. Snce C s a strct partal order, we can compute n polynomal tme a mnmum-cardnalty set C of k chans {C 1,...,C k } coverng the partally ordered set (poset) R, C usng the classcal MnPC algorthm on DAGs [2]. By Lemma 3.4, we can then compute n polynomal tme a set of k paths of D assocated wth the chans C 1,...,C k. By constructon, these paths cover all the requred pars n R. Theset has mnmum-cardnalty because, otherwse, by Lemma 3.4 there would exst another set of k < k chans coverng the poset R, C, contradctng the mnmumcardnalty of C. Snce the compatblty relaton of a set of parwse nonoverlappng requred pars s a strct partal order, then we have the followng corollary. Corollary 3.1. An nstance D, R of MnRPC wth maxmum overlappng degree equal to 0 can be solved n polynomal tme. Proof. We clam that the compatblty relaton C of R s a strct partal order when the maxmum overlappng degree s 0. The result then follows from Theorem 3.2. By defnton, the compatblty relaton s rreflexve, thus we only have to show that C s transtve. Let r = [v 1,v2 ], r = [v 1,v2 ], r k = [vk 1,v2 k ] be three requred pars such that {(r, r ), (r, r k )} C. We have to prove that (r, r k ) belongs to C. Snce (r, r ) C and snce r and r do not overlap, there exsts a path π, connectng v 1 to v 2 and coverng both r and r. Smlarly, there exsts a path π,k connectng v 2 to vk 2 and coverng r k. The concatenaton of π, and π,k s a path whch covers r and r k. Moreover, v 2 s strctly before vk 2 on ths path, thus (r, r k ) C. 4. A SHARP TRACTABILITY BORDERLINE FOR MINPCRP In ths secton, we nvestgate the computatonal complexty of MnPCRP and we gve a sharp tractablty borderlne for k-pcrp, the restrcton of MnPCRP where we ask whether there exst k paths that cover all the vertces of the graph and all the set of requred pars. Frst, we show that 3-PCRP s NP-complete (Secton 4.1). Ths result mples that k-pcrp Secton A: Computer Scence Theory, Methods and Tools Downloaded from

8 1680 N. Beerenwnkel et al. FIGURE 3. Example of graph D = (N, A) assocated wth graph G = (V, E), n whch grey boxes represent subgraphs D,,...,D,.The central grey box shows the subgraph D, = (N,, A, ) assocated wth the edge {v,v } E. does not belong to the class XP, 1 so t s probably hopeless to look for an algorthm havng complexty O(n k ), and hence for a fxed-parameter algorthm n k. The same results hold also for k-rpc, the restrcton of MnRPC where we ask whether there exst k paths that cover all the requred pars. We complement ths result by gvng a polynomal tme algorthm for 2-PCRP (Secton 4., thus defnng a sharp borderlne between tractable and ntractable nstances of MnPCRP Hardness of 3-PCRP In ths secton, we show that 3-PCRP s NP-complete. We prove ths result va a reducton from the wellknown 3-Colorng (3C) problem whch, gven an undrected (connected) graph G = (V, E), asks for a colorng c : V {c 1, c 2, c 3 } of the vertces of G wth exactly three colors, such that, for every {v,v } E, we have c(v ) c(v ). Startng from an undrected graph G = (V, E) (nstance of 3C), we construct a correspondng nstance D = (N, A), R of 3-PCRP as follows. For each edge {v,v } E wth <, we defne a graph D, = (N,, A, ). The vertex set N, s {s,, n,, n,, f,, t, }.ThesetA, of arcs connectng the vertces of N, s (see central grey box of Fg. 3): A, ={(s,, n, ), (s,, n, ), (s,, f, ), (n,, t, ), (n,, t, ), ( f,, t, )} The whole graph D = (N, A) s constructed by concatenatng the graphs D, (for all 1 < n) accordng to the lexcographc order of ther ndces,. The snk t, of each graph D, s connected to the source of the graph whch mmedately follows D,. A dstngushed vertex s s connected to the source of the frst subgraph D,, whle the snk of the last subgraph D, s connected to a second dstngushed vertex t. Fgure 3 depcts such a constructon. The set R of requred pars s defned as 1 n R, where R ={[n,, n,h ] {v,v }, {v,v h } E}. 1 We recall that the class XP contans those problems that, gven a parameter k, can be solved n tme O(n f (k) ). The followng lemmas prove the correctness of the reducton. Lemma 4.1. Let G = (V, E) be an undrected (connected) graph and let D = (N, A), R be the correspondng nstance of 3-PCRP. Then, gven a 3-colorng of G we can compute n polynomal tme three paths of D that cover all ts vertces and every requred par n R. Proof. Consder a 3-colorng of G and let {V 1, V 2, V 3 } be the tr-partton of V nduced by the 3-colorng. We show how to compute n polynomal tme three paths π 1, π 2, π 3 that cover all the vertces of D and every requred par n R. For each v V c, path π c passes through vertces n, of subgraphs D, for every v V such that {v,v } E, whle for each subgraph D p,q such that v p,v q / V c, π c passes through vertces f p,q. Note that each π c s well-defned, snce there does not exst a par of vertces n,, n, assocated wth the same color c (otherwse {V 1, V 2, V 3 } s not a 3-colorng of G). We show that π 1, π 2, π 3 cover every vertex of N. Note that for each {v,v } E, snce v and v have dfferent colors, by constructon one of the paths π 1, π 2, π 3 passes through n, (say π c1 ), whle another one passes through n, (say π c2 ). As a consequence, by constructon we have that π c3 passes through f,. The only vertces that mght be not covered are s, and t,,for{v,v } E. However, these vertces are artculaton ponts, hence all the three paths necessarly pass through them. Now, we show that every requred par n R s covered. By constructon, the vertces n, of D assocated wth the same vertex v of G belong to the same path π c, where c s the color of v. Therefore, all the requred pars n each R are covered by one of the three paths. Lemma 4.2. Let G = (V, E) be an undrected graph and let D = (N, A), R be the correspondng nstance of 3-PCRP. Then, gven three paths n D that cover all ts vertces and every requred par n R, we can compute n polynomal tme a 3-colorng of G. Proof. Consder three paths π 1, π 2, π 3 of D that cover all the vertces of D and every requred par n R, and we show how Secton A: Computer Scence Theory, Methods and Tools Downloaded from

9 Coverng Pars n Drected Acyclc Graphs 1681 to compute n polynomal tme a correspondng 3-colorng of the graph G. Frst, we prove a property of the three paths π 1, π 2, π 3. We show that, gven a vertex v V, there exsts at least one path among π 1, π 2, π 3 that covers all the requred pars n R. Consder a vertex v V. Snce G s connected, t follows that there exsts at least one vertex adacent to v, w.l.o.g. v, such that {v,v } E. Now, consder the subgraph D,. By constructon, a soluton of MnPCRP must contan three dfferent paths, each one passng through one of the vertces n, through n,, n,, f,. Now, assume that path π 1 passes. Obvously, π 2 and π 3 cannot pass through n,. As a consequence, snce π 1 s the only path that covers n,, n,h ] for each h such and snce R contans a par [n, that {v,v h } E, t follows that all the vertces n,h wth {v,v h } E must belong to π 1. It follows that, gven a vertex v V, there exsts one path n {π 1,π 2,π 3 } that covers all the requred pars n R. Now, we defne the 3-colorng of G, where C ={c 1, c 2, c 3 } s the set of colors. If a vertex n, s covered by a path π x, 1 x 3, then we assgn the color c x to vertex v.the colorng s well-defned snce, as noted above, a sngle path covers all the vertces n, of D assocated wth the same vertex v of G. The colorng s also feasble, that s c(v ) c(v ) when {v,v } E, snce, by constructon, vertces n, and n, are covered by dfferent paths (hence c(v ) c(v )). Snce 3-PCRP s clearly n NP, the followng result s a consequence of the prevous lemmas and of the NP-hardness of 3C [15]. Theorem PCRP s NP-complete. Proof. The NP-hardness of 3-PCRP follows drectly from Lemmas 4.1 and 4.2 and from the NP-completeness of 3C [15]. 3-PCRP s n NP, snce, gven three paths π 1, π 2, π 3, we can verfy n polynomal tme that π 1, π 2, π 3 cover all the vertces of D and that every requred par n R s covered by some path n {π 1,π 2,π 3 }. The reducton can be easly modfed n order to show that also 3-RPC, that s the restrcton of MnRPC where we ask whether there exst k paths that cover all the requred pars, s NP-complete for any k > 2. Corollary RPC s NP-complete. Proof. We obtan ths result by modfyng the reducton from 3C to 3-PCRP presented above. Let G = (V, E) be the undrected graph gven as nput to 3C and let D = (N, A), R be the correspondng nstance of 3-PCRP. Frst, we can assume that G does not contan vertces wth degree 1 (.e. vertces wth only one edge ncdent to them). Otherwse, these vertces can be removed from G snce they can be always easly colored wth a color dfferent to that of ther sngle adacent vertex. As a consequence, we can assume that all the vertces and n, construct the nstance of 3-RPC n, of N belong to some requred par of R. Now, D = (N, A), ˆR, where ˆR := R {[s,, f, ] {v,v } E}. We clam that there exst three paths coverng all the vertces n N and all the requred pars n R f and only f there exst three paths coverng all the requred pars n ˆR. Note that a set of paths coverng all the vertces of D also covers the requred pars n ˆR \ R. Hence, f there exst 3 paths coverng all the vertces n N and all the requred pars n R, then the same paths cover all the requred pars n ˆR. Conversely, f there exst three paths coverng all the requred pars of ˆR, then the same paths cover all the requred pars of R. Moreover, snce all the vertces n,, n, and f, of N belong to some requred par of ˆR and snce vertces s, and t, are artculaton ponts, we have that these paths cover also all the vertces of N. Fnally, we also have that 3-RPC s clearly n NP. As a consequence, by Lemmas 4.1 and 4.2 and by the NPcompleteness of 3C [15], we have that 3-RPC s NP-hard A polynomal tme algorthm for 2-PCRP In ths secton, we gve a polynomal tme algorthm for computng a soluton of 2-PCRP. Note that 1-PCRP can be easly solved n polynomal tme, as there exsts a soluton of 1-PCRP f and only f the reachablty relaton of the vertces of the nput graph s a total order. The algorthm for solvng 2-PCRP s based on a polynomaltme reducton to the 2-Clque Partton problem, whch, gven an undrected graph G = (V, E), asks whether there exsts a partton of V n two sets V 1, V 2 both nducng a clque n G. Computng the exstence of a 2-Clque Partton over a graph G s equvalent to computng f there exsts a 2-Colorng of the complement graph G (hence decdng f G s bpartte), whch s well known to be solvable n polynomal tme [15, probl. GT15]. To perform ths reducton we assume that gven D = (N, A), R, nstance of 2-PCRP, every vertex of the graph D belongs to at least one requred par n R. Otherwse, we add to R the requred pars [s,v ] for all v N that do not belong to any requred par. Therefore, a soluton that covers all the requred pars n R covers also all the vertces, hence t s a feasble soluton of 2-PCRP. Moreover, note that ths transformaton does not affect the soluton of 2-PCRP, snce all the paths start from s and cover all the nodes of the graph, ncludng the addtonal requred pars. The algorthm computes, f exsts, a soluton for an nstance D = (N, A), R of 2-PCRP by computng a 2- Clque Partton of the compatblty graph of R. We recall that the compatblty graph s the graph obtaned from the compatblty dgraph dscardng the edge orentaton. Gven the compatblty relaton C, we denote as Cˆ the set of edges of Secton A: Computer Scence Theory, Methods and Tools Downloaded from

10 1682 N. Beerenwnkel et al. the compatblty graph (.e. the set {{r, r } (r, r ) C}). Snce the computaton of the compatblty graph and of a 2-Clque Partton can be performed n polynomal tme, the algorthm solves 2-PCRP n polynomal tme. The algorthm s based on the followng property. Lemma 4.3. Gven an nstance D = (N, A), R of 2-PCRP and the compatblty graph G = (R, C) ˆ of R, then there exsts a path π that covers a set R of requred pars f and only f R s a clque of G. Proof. If there exsts a path π whch covers all the requred pars n R, then, by defnton of the compatblty relaton, R s clearly a clque of G. We clam that, f R s a clque of G, then C(R ) s a total order. Frst, note that, for each par r, r R, we have that ether (r, r ) or (r, r ) s n C(R ). Moreover, snce the compatblty dgraph s acyclc, then also (R, C(R )) s acyclc. As a consequence, C(R ) s transtve and, by the rreflexvty of C, we conclude that C(R ) s a total order. By Lemma 3.4, snce R s a chan of C, there exsts a path π coverng the requred pars n R. From Lemma 4.3, t follows that, n order to compute the exstence of a soluton of 2-PCRP over the nstance D = (N, A), R (n whch every vertex of D belongs to at least one requred par n R), we have to compute f there exsts a 2-Clque Partton of the correspondng graph G. Snce the 2- Clque Partton problem can be solved n polynomal-tme [15, probl. GT15], we can conclude that 2-PCRP can be decded n polynomal tme. 5. PARAMETERIZED COMPLEXITY OF MAXRPSP In ths secton, we consder the parameterzed complexty of MaxRPSP. We show that, although MaxRPSP s W[1]-hard (hence unlkely fxed-parameter tractable) when parameterzed by the number of requred pars covered by a sngle path (Secton 5.1), the problem becomes fxed-parameter tractable f the maxmum overlappng degree s the parameter (Secton W[1]-hardness of MaxRPSP parameterzed by the optmum In ths secton, we nvestgate the parameterzed complexty of MaxRPSP when parameterzed by the sze of the soluton, that s the maxmum number of requred pars covered by a sngle path, and we prove that the problem s W[1]-hard (note that ths result mples the NP-hardness of MaxRPSP). Ths result shows that t s unlkely that the problem s fxed-parameter tractable, when parameterzed by the number of requred pars covered by a sngle path. For detals on the theory of fxedparameter complexty, we refer the reader to [13, 14]. We prove ths result va a parameterzed reducton from the h-clque problem to the decson verson of MaxRPSP (k-rpsp), parameterzed by the szes of the respectve solutons. Gven an undrected graph G = (V, E) and an nteger h, h-clque asks to decde f there exsts a clque C V of sze h. On the other hand, gven a DAG D, asetr of requred pars, and an nteger k, the k-rpsp problem conssts of decdng f there exsts a path n D that covers k requred pars. We recall that h-clque s known to be W[1]-hard [18]. Frst, we start by showng how to construct an nstance of k-rpsp startng from an nstance of h-clque. Gven an (undrected) graph G = (V, E) wth n vertces v 1,...,v n,we construct the assocated DAG D = (N, A) as follows. The set N of vertces s defned as: N ={v z v V, 1 z h} {s, t} Informally, N conssts of two dstngushed vertces s, t and of h copes v 1,...,vh of every vertex v of G. The set of arcs A s defned as: A ={(v z,vz+1 ) {v,v } E, 1 z h 1} {(s,v 1 ), (vh, t) v V } Informally, we connect every two consecutve copes assocated wth vertces that are adacent n G, the source vertex s to all the vertces v 1, wth 1 n, and all the vertces v h, wth 1 n, to the snk vertex t. The set R of requred pars s defned as: R ={[v x,vy ] {v,v } E, 1 x < y h} Informally, for each edge {v,v } of G there s a requred par [v x,vy ], 1 x < y h, between every two dfferent copes assocated wth v, v. By constructon, the vertces n N (except for s and t) are parttoned nto h ndependent sets I z ={v z 1 n}, wth 1 z h, each one contanng a copy of every vertex of V. Moreover, the arcs of A only connect two vertces of consecutve subsets I z and I z+1, wth 1 z h 1. Fgure 4 presents an example of drected graph D assocated wth an undrected graph G. Now, we are able to prove the man propertes of the reducton. Lemma 5.1. Let G = (V, E) be an undrected graph and D = (N, A), R be the assocated nstance of k-rpsp. Then: (1) startng from an h-clque n G we can compute n polynomal tme a path π n D that covers ( h requred pars of R; ( startng from a path π n D that covers ( h requred pars we can compute n polynomal tme an h-clque n G. Proof. (1) Startng from an h-clque C n G we show how to compute a path π n D that covers ( h requred pars of R. Secton A: Computer Scence Theory, Methods and Tools Downloaded from

11 Coverng Pars n Drected Acyclc Graphs 1683 FIGURE 4. Example of DAG D = (N, A) assocated wth an nstance G = (V, E) of the h-clque problem. Each gray box hghlghts an ndependent set I z composed of one copy of the vertces n V. Edges (v1 z,vz+1 ), (v z,vz+1 ) and (vn,v z z+1 ) are some of the drected edges n A assocated wth edges {v 1,v }, {v,v }, {v,v n } E. Let C ={v 1,...,v h } be a clque of G and let v 1,...,v h be an arbtrary orderng of C. Letπ C = s,v 1 1,...,v h h, t be a sequence of vertces obtaned by selectng the vertex v z z for each ndependent set I z, wth 1 z h (n addton to vertces s and t). Snce C s a clque of G, by constructon of D, every par of vertces (v z z,v z+1 z+1 ) s connected by an arc, hence π C s a path of D. Moreover, the path π C covers exactly ( h requred pars of R because, by constructon of R, there exsts a par between every two copes of vertces whch are adacent n G. More precsely, snce the clque C has all the possble edges among ts h vertces, the number of requred pars covered by the path π C s ( h. ( Let π be a path n D that covers a set R R of ( h) 2 requred pars, then we show how to compute n polynomal tme an h-clque C n G. Note that, by constructon of D, the path π must contan exactly one vertex v z,1 n and 1 z h, for each ndependent set I z of D. By constructon of set R, each vertex v z of π appears n at most h 1 requred pars of R. Hence, the total number of requred pars covered by the path π, whch contans exactly h nner vertces v z,sat most h(h 1)/2 = ( h.letc be the set {v v z π \{s, t}}. We clam that C s an h-clque. Frst, we prove that C contans h vertces. Suppose to the contrary that C has less than h vertces. Then, there exst two vertces v x and v y of π that correspond to the same vertex v of C, that s = =. Snce [v x,vy ] / R, t follows that each vx, vy appears n at most h 2 requred pars of R. As a consequence, the total number of requred pars covered by the path π s strctly less than ( h, volatng the ntal hypothess that π covers ( h requred pars of R. Hence C contans h vertces. As all the nternal vertces of π (.e. all ts vertces but s and t) represent dstnct vertces of G, then all the requred pars covered by π represent dstnct edges of G. The only undrected graph wth h vertces and ( h) 2 edges s the complete graph, hence C s an h-clque of G. The W[1]-hardness of k-rpsp easly follows from Lemma 5.1 and from the W[1]-hardness of h-clque when parameterzed by h [18]. Theorem 5.1. k-rpsp s W [1]-hard when parameterzed by the number of requred pars covered by a path. Proof. The result follows from Lemma 5.1 and from the W[1]- hardness of h-clque when parameterzed by h [18] An FPT algorthm for MaxRPSP parameterzed by the maxmum overlappng degree In ths secton, we propose a fxed-parameter algorthm (FPT) for the MaxRPSP problem, where the parameter s the maxmum overlappng degree. For the rest of the secton, let D = (N, A), R be an nstance of the MaxRPSP problem. For ensurng ts correctness, the algorthm wll consder the requred pars n R n an order resultng from the topologcal orderng of the compatblty dgraph of R. For ease of exposton, gven such a fxed order, we represent the th requred par of the orderng as [v 1,v2 ] and, whenever no confuson arses, we wll refer to that requred par as -par. The parameterzed algorthm s based on dynamc programmng. In fact, we can decompose a path π, startng at s, endng at a vertex v and coverng k requred pars, nto two subpaths: the frst one π 1 starts at s, ends at a vertex v, and covers k 1 requred pars, whle the other one π 2 starts at v, ends at v and covers the remanng k 2 = k k 1 requred pars (usng vertces of π 1, possbly). The key pont to defne the recurrence s that, for each requred par r, t suffces to keep track of the Secton A: Computer Scence Theory, Methods and Tools Downloaded from

12 1684 N. Beerenwnkel et al. set of requred pars overlappng r and covered by the path. To ths am, for each requred par [v 1,v2 ], we defne the set OP([v 1,v2 ]) as the set of vertces v such that v belongs to a requred par that overlaps [v 1,v2 ] and such that v2 s reachable from v. By a slghtly abuse of the notaton, we consder that OP([v 1,v2 ]) always contans vertex v1. The recurrence reles on the followng observaton. Let π be a path coverng a set R R of requred pars and let N(R ) be the set of vertces belongng to the requred pars n R. Consder two overlappng requred pars [v 1,v2 ] and [v 1,v2 ]nr, wth <. Then, ether [v 1,v2 ] s nested n [v 1,v2 ] (hence the fact that π covers the par [v1,v2 ] can be checked by the recurrence lookng only at the requred pars that overlap wth [v 1,v2 ]) or pars [v1,v2 ] and [v1,v2 ]are alternated. In the latter case, snce [v 1,v2 ]snr, we only have to consder the vertces n the set N(R ) OP([v 1,v2 ]) OP([v 1,v2 ]). Moreover, let p be the number of requred pars that overlap the requred par [v 1,v2 ]. Then OP([v1,v2 ]) s at most 2p. Hence, the cardnalty of set N(R ) OP([v 1,v2 ]) OP([v 1,v2 ]) s bounded by 2 max(p, p ). Furthermore, gven two sets S and S of vertces such that S OP([v 1,v2 ]) and S OP([v 1,v2 ]), we say that S s n agreement wth S f S (OP([v 1,v2 ]) OP([v1,v2 ])) = S (OP([v 1,v2 ]) OP([v 1,v2 ])). Informally, when S and S are n agreement, they must contan the same subset of vertces of OP([v 1,v2 ]) OP([v 1,v2 ]). Let P([v 1,v2 ], S) denote the maxmum number of requred pars covered by a path π endng at vertex v 2 and such that the set S OP([v 1,v2 ]) s covered by π. In the followng, we present the recurrence to compute P([v 1,v2 ], S). For ease of exposton, we only focus on vertces that appear as second vertces of the requred pars. In fact, paths that do not end at such vertces are not able to cover new requred pars. Furthermore, for smplcty, we consder the source s as the second vertex of a fcttous requred par (wth ndex 0) [, s] whch does not overlap any other requred par. Such a fcttous requred par does not contrbute to the total number of requred pars covered by the path. The recurrence s: P([v 1,v2 ], S) = max{p([v1,v2 ], S ) + Ov([v 1,v2 ], S, S ) } (1) for each [v 1,v2 ] and S such that: () [v 1,v2 ] not nested n [v1,v2 ] and < ; () S n agreement wth S; () there exsts a path from v 2 to v2 coverng all vertces n S \ S and where Ov([v 1,v2 ], S, S ) ={[vh 1,v2 h ] [v1 h,v2 h ] s nested n [v 1,v2 ] v1 h S v2 h S \ S }. Note that each requred par s assumed to be nested n tself. The base case of the recurrence s P([, s], ) = 0. The correctness of the recurrence derves from the followng two lemmas. Lemma 5.2. If P([v 1,v2 ], S) = k, then there exsts a path π n D endng at v 2, such that every vertex n S belongs to π and the number of requred pars covered by π s k. Proof. We prove the lemma by nducton on the ndex. It s easy to see that the base case holds. Assume that the lemma holds for ndex values less than, we prove that the lemma holds for. Let P([v 1,v2 ], S) = k. By Equaton (1), there exsts a vertex v 2 wth <, such that P([v 1,v2 ], S ) = k 1 for some set S n agreement wth S. Assume that Ov([v 1,v2 ], S, S ) =k 2, wth k 1 + k 2 = k. By nducton hypothess, snce P([v 1,v2 ], S ) = k 1, there exsts a path π endng at v 2, coverng every vertex n S, and such that π covers k 1 requred pars. Furthermore, the k 2 covered requred pars have at least one vertex n S \ S, hence the vertces of such requred pars belong to a path π whch starts at v 2 and ends at v 2 (path π exsts by hypothess). But then, the path obtaned by the concatenaton of π and π covers k 1 + k 2 requred pars. Lemma 5.3. Let π be a path n D endng at v 2 and coverng k requred pars. Let S be the set of all the vertces belongng to requred pars covered by π and overlappng [v 1,v2 ]. Then P([v 1,v2 ], S) k. Proof. We prove the lemma by nducton on the ndex. Its easy to see that the base case holds. Assume that the lemma holds for ndex values less than, we prove that the lemma holds for. Letπ be a path, endng at v 2, that covers k requred pars and let S be the set of vertces that belong to the requred pars covered by π and overlappng [v 1,v2 ]. We clam that P([v 1,v2 ], S) k. Consder the rghtmost vertex v 2 of π such that v 2 belongs to a requred par covered by π and not nested n the -par. Decompose path π nto two parts: one π from s to v 2, and the other one π from v 2 to v 2.LetS be the set of vertces that belong to the requred pars covered by π and overlappng [v 1,v2 ]. Let k be the number of requred pars covered by π and k be the number of the remanng requred pars covered by π (that s, k = k + k ). Frst, note that k = Ov([v 1,v2 ], S, S ). By nducton hypothess P([v 1,v2 ], S ) = k 1 for some k 1 k. Moreover, by constructon, S s n agreement wth S and the subpath of π from v 2 to v 2 covers all the vertces n S \ S. As a consequence, by Equaton (1), P([v 1,v2 ], S) s at least k 1 + k k + k = k, whch concludes the proof. Let p be the maxmum number of overlappng requred pars n D, R (that s, p = max {p }). Then, the cardnalty of Secton A: Computer Scence Theory, Methods and Tools Downloaded from