Capacitated Domination and Covering: A Parameterized Perspective

Transcription

1 Capactated Domnaton and Coverng: A Parameterzed Perspectve Mchael Dom Danel Lokshtanov Saket Saurabh Yngve Vllanger Abstract Capactated versons of Domnatng Set and Vertex Cover have been studed ntensvely n terms of polynomal tme approxmaton algorthms. Although the problems Domnatng Set and Vertex Cover have been subjected to consderable scrutny n the parameterzed complexty world, ths s not true for the capactated versons. Here we make an attempt to understand the behavor of the problems Capactated Domnatng Set and Capactated Vertex Cover from the perspectve of parameterzed complexty. The orgnal versons of these problems, Vertex Cover and Domnatng Set, are known to be fxed parameter tractable when parameterzed by a structure of the graph called the treewdth (tw). In ths paper we show that the capactated versons of these problems behave dfferently. Our results are: Capactated Domnatng Set s W[1]-hard when parameterzed by treewdth. In fact, Capactated Domnatng Set s W[1]-hard when parameterzed by both treewdth and soluton sze k of the capactated domnatng set. Capactated Vertex Cover s W[1]-hard when parameterzed by treewdth. Capactated Vertex Cover can be solved n tme 2 O(tw log k) n O(1) where tw s the treewdth of the nput graph and k s the soluton sze. As a corollary, we show that the weghted verson of Capactated Vertex Cover n general graphs can be solved n tme 2 O(k log k) n O(1). Ths mproves the earler algorthm of Guo et al. [15] runnng n tme O(1.2 k2 + n 2 ). We would also lke to pont out that our W[1]-hardness result for Capactated Vertex Cover, when parameterzed by treewdth, makes t (to the best of our knowledge) the frst known subset problem whch has turned out to be fxed parameter tractable when parameterzed by soluton sze but W[1]-hard when parameterzed by treewdth. 1 Introducton Domnatng Set (or more generally Set Cover) and Vertex Cover are problems representatve for domnaton and coverng, respectvely. Gven a graph G and an nteger k, Vertex Cover asks for a sze-k set of vertces that cover all edges of the graph, whle Domnatng Set asks for a sze-k set of vertces such that every vertex n the graph ether belongs to ths set or has a neghbor whch does. These fundamental problems n algorthms and complexty have been studed extensvely and fnd applcatons n varous domans [3, 4, 5, 8, 9, 12, 13, 15, 16, 18, 22]. Insttut für Informatk, Fredrch-Schller-Unverstät Jena, Ernst-Abbe-Platz 2, D Jena, Germany. Emal: dom@mnet.un-jena.de Department of Informatcs, Unversty of Bergen, POB 7803, 5020 Bergen, Norway. Emal: {danello,saket,yngvev}@.ub.no 1

2 Vertex Cover and Domnatng Set have a specal place n parameterzed complexty [7, 10, 21]. Vertex Cover was one of the earlest problems that was shown to be fxed parameter tractable (FPT) [7]. On the other hand, Domnatng Set turned out to be ntractable n the realm of parameterzed complexty specfcally, t was shown to be W[2]-complete [7]. Vertex Cover has been put to ntense scrutny, and many papers have been wrtten on the problem. After a long race, the currently best algorthm for Vertex Cover runs n tme O( k + kn)) [4]. Vertex Cover has also been used as a testbed for developng new technques for showng that a problem s FPT [7, 10, 21]. Though Domnatng Set s a fundamentally hard problem n the parameterzed W-herarchy, t has been used as a benchmark problem for developng sub-exponental tme parameterzed algorthms [1, 6, 11] and also for obtanng a lnear kernels n planar graphs [2, 14, 10, 21], and more generally, n graphs that exclude a fxed graph H as a mnor. Dfferent applcatons of Vertex Cover and Domnatng Set (or Set Cover) have ntated studes of dfferent generalzatons and varatons of these problems. These nclude Connected Vertex Cover, Connected Domnatng Set, Partal Vertex Cover, Partal Set Cover, Capactated Vertex Cover and Capactated Domnatng Set, to name a few. All these problems have been nvestgated extensvely and are well understood n the context of polynomal tme approxmaton [5, 12, 13, 16]. However, these problems hold a lot of promse and reman htherto unexplored n the lght of parameterzed complexty; wth exceptons that are few and far between [3, 15, 19, 22, 23]. Problems Consdered: Here we consder two problems, Capactated Vertex Cover (CVC) and Capactated Domnatng Set (CDS). To defne these problems, we need to ntroduce the notons of capactated graphs, vertex covers, and domnatng sets. A capactated graph s a graph G = (V,E) together wth a capacty functon c : V N such that 1 c(v) d(v), where d(v) s the degree of the vertex v. Now let G = (V,E) be a capactated graph, C be a vertex cover of G and D be a domnatng set of G. Defnton 1 We call C V a capactated vertex cover f there exsts a mappng f : E C whch maps every edge n E to one of ts two endponts such that the total number of edges mapped by f to any vertex v C does not exceed c(v). Defnton 2 We call D V a capactated domnatng set f there exsts a mappng f : (V \ D) D whch maps every vertex n (V \ D) to one of ts neghbors such that the total number of vertces mapped by f to any vertex v D does not exceed c(v). Now we are ready to defne Capactated Vertex Cover and Capactated Domnatng Set. Capactated Vertex Cover (CVC): Gven a capactated graph G = (V, E) and a postve nteger k, determne whether there exsts a capactated vertex cover C for G contanng at most k vertces. Capactated Domnatng Set (CDS): Gven a capactated graph G = (V, E) and a postve nteger k, determne whether there exsts a capactated domnatng set D for G contanng at most k vertces. Our Results: To descrbe our results we frst need to defne the treewdth (tw) of a graph. Let V (U) be the set of vertces of a graph U. A tree decomposton of an (undrected) graph G = (V,E) s a par (X,U) where U s a tree whose vertces we wll call nodes and X = {X V (U)} s a collecton of subsets of V such that 2

3 1. V (U) X = V, 2. for each edge {v,w} E, there s an V (U) such that v,w X, and 3. for each v V the set of nodes { v X } forms a subtree of U. The wdth of a tree decomposton ({X V (U)},U) equals max V (U) { X 1}. The treewdth of a graph G s the mnmum wdth over all tree decompostons of G. There s a tendency to thnk that most combnatoral problems, especally subset problems, are tractable for graphs of bounded treewdth (tw) when parameterzed by tw. In fact, the non-capactated versons of the problems consdered here, namely Vertex Cover and Domnatng Set, are known to be fxed parameter tractable when parameterzed by the treewdth of the nput graph. The algorthms for Vertex Cover and Domnatng Set run n tme O(2 tw n) [21] and tme O(4 tw n) [1], respectvely. In contrast, the capactated versons of these problems behave dfferently. More precsely, we show the followng: Capactated Domnatng Set s W[1]-hard when parameterzed by treewdth. In fact, CDS s W[1]-hard when parameterzed by both treewdth and soluton sze k of the capactated domnatng set. Capactated Vertex Cover s W[1]-hard when parameterzed by treewdth. Capactated Vertex Cover can be solved n tme 2 O(tw log k) n O(1) where tw s the treewdth of the nput graph and k s the soluton sze. As a corollary of the last result we obtan an mproved algorthm for the weghted verson of Capactated Vertex Cover n general graphs. Here, every vertex of the nput graph has, n addton to the capacty, a weght, and the queston s f there s a capactated vertex cover whose weght s at most k. Our algorthm runnng n tme O(2 O(k log k) n O(1) ) mproves the earler algorthm of Guo et al. [15] runnng n tme O(1.2 k2 + n 2 ). The so-called subset problems are known to go ether way, that s, FPT or W[]-hard ( 1) when parameterzed by soluton sze. However, when parameterzed by treewdth they have nvarably been FPT. Examples favorng ths clam nclude, but are not lmted to, Independent Set, Domnatng Set, Partal Vertex Cover. Contrary to these observed patterns, our hardness result for CVC when parameterzed by treewdth makes t possbly the frst known subset problem whch has turned out to be FPT when parameterzed by soluton sze, but W[1]-hard when parameterzed by treewdth. 2 Prelmnares We assume that all our graphs are smple and undrected. Gven a graph G = (V,E), the number of ts vertces s represented by n and the number of ts edges by m. For a subset V V, by G[V ] we mean the subgraph of G nduced by V. Wth N(u) we denote all vertces that are adjacent to u, and wth N[u], we refer to N(u) {u}. Smlarly, for a subset D V, we defne N[D] = v D N[v] and N(D) = N[D] \ D. Let f be the functon assocated wth a capactated domnatng set D. Gven u D and v V \ D, we say that u domnates v f f(v) = u; moreover, every vertex u D domnates tself. Note that the capacty of a vertex v only lmts the number of neghbors that v can domnate, that s, a vertex v D can domnate c(v) of ts neghbors plus v tself. Parameterzed complexty s a two-dmensonal framework for studyng the computatonal complexty of problems [7, 10, 21]. One dmenson s the nput sze n and the other one 3

4 the parameter. A problem s called fxed-parameter tractable (FPT) f t can be solved n tme f(k) n O(1), where f s a computable functon only dependng on k. Now we defne the noton of parameterzed reducton. Defnton 3 Let A, B be parameterzed problems. We say that A s (unformly many:1) reducble to B f there s an algorthm Φ whch transforms (x,k) nto (x,g(k)) n tme f(k) x α, where f,g : N N are arbtrary functons and α s a constant ndependent of x and k, so that (x,k) A f and only f (x,g(k)) B. 3 Parameterzed Intractablty Hardness Results 3.1 CDS s W[1]-hard parameterzed by treewdth and soluton sze In ths secton we show that Capactated Domnatng Set s W[1]-hard when parameterzed by treewdth and soluton sze. We reduce from k-multcolor Clque, a restrcton of the k-clque problem. Multcolor Clque: Gven an nteger k and a connected undrected graph G = (V [1] V [2] C[k],E) such that for every the vertces of V [] nduce an ndependent set, s there a k-clque C n G? In fact, we wll reduce to a slghtly modfed verson of Capactated Domnatng Set, Marked Capactated Domnatng Set where we mark some vertces and demand that all marked vertces must be n the domnatng set. We can then reduce from Marked Capactated Domnatng Set to Capactated Domnatng Set by attachng k + 1 leaves to each marked vertex and ncreasng the capacty of each marked vertex by k + 1. It s easy to see that the new nstance has a k-capactated domnatng set f and only f the orgnal one had a k-capactated domnatng set that contaned all marked vertces, and that ths operaton does not ncrease the treewdth of the graph. Thus, to prove that Capactated Domnatng Set s W[1]-hard when parameterzed by treewdth and soluton sze, t s suffcent to prove that Marked Capactated Domnatng Set s. We wll show how gven an nstance (G,k) of Multcolor Clque, we can buld an nstance (H,c,k ) of Marked Capactated Domnatng Set such that k = 7k(k 1) + 2k, G has a clque of sze k f and only f H has a capactated domnatng set of sze k, and the treewdth of H s O(k 4 ). For a par of dstnct ntegers,j, let E[,j] be the set of edges wth one endpont n V [] and the other n V [j]. Wthout loss of generalty, we wll assume that V [] = N and E[,j] = M for all, j, j. To each vertex v we assgn a unque dentfcaton number v up between N + 1 and 2N, and we set v down = 2N v up. For two vertces u and v, by addng an (A,B)-arrow from u to v we wll mean addng A subdvded edges between u and v and attachng B leaves to v (see Fg. 1). Now we descrbe how to buld the graph H for a gven nstance (G = (V [1] V [2] V [k],e),k) of Multcolor Clque. For every nteger between 1 and k we add a marked vertex ˆx that has a neghbor v for every vertex v n V []. For every j, we add a marked vertex ŷ j and a marked vertex ẑ j. Now, for every vertex v V [] and every nteger j we add a (v up,v down )-arrow from v to ŷ j and a (v down,v up )-arrow from v to ẑ j. Fnally we add a set S of k + 1 vertces and make every vertex n S adjacent to every vertex v wth v V []. See Fg. 2 for an llustraton. 4

5 u (A,B) v = u 1 2 A v 1 2 B Fgure 1: Addng an (A,B)-arrow from u to v. ˆx 2 S 2 v (v up, v down ) (v down, v up ) (v up, v down ) (v down, v up ) (v up, v down ) ŷ 21 ẑ 21 ŷ 23 ẑ 23 (v down, v up ) ŷ 2k ẑ 2k Fgure 2: Vertex selecton for color class 2. Smlarly, for every par of ntegers, j wth < j, we add a marked vertex ˆx j wth a neghbor e for every edge e n E[,j]. Moreover, we add four new marked vertces ˆp j, ˆp j, ˆq j, and ˆq j. For every edge e = {u,v} n E[,j] wth u V [] and v V [j], we add a (u down,u up )- arrow from e to ˆp j, a (u up,u down )-arrow from e to ˆq j, a (v down,v up )-arrow from e to ˆp j and a (v up,v down )-arrow from e to ˆp j. We also add a set S j of k + 1 vertces and make every vertex n S j adjacent to every vertex e wth e E[,j]. See Fg. 3 for an llustraton. Fnally, we add a marked vertex ˆr j and a marked vertex ŝ j for every j. For every j, we add (2N,0)-arrows from ŷ j to rˆ j, from ˆp j to rˆ j, from ẑ j to sˆ j, and from ˆq j to sˆ j (see Fg. 4). Ths concludes the descrpton of the graph H. We now descrbe the capactes of the vertces. For every j, the vertex ˆx has capacty N 1, the vertex ˆx j has capacty M 1, the vertces ŷ j and ẑ j both have capacty 2N 2, the vertces ˆp j and ˆq j have capacty 2NM, and both ˆr j and ŝ j have capacty 2N. For all other vertces, ther capacty s equal to ther degree n H. Observaton 1 The treewdth of H s O(k 4 ). Proof: If we remove all marked vertces ( k =1 S and j S j), a total of O(k 4 ) vertces, from H, we obtan a forest. As deletng a vertex reduces the treewdth by at most one, ths 5

6 (e = u, v) ˆx j S j e (u down, u up ) (u up, u down ) (v down, v up ) (v up, v down ) ˆp j ˆq j ˆp j ˆq j Fgure 3: Edge selecton ŷ j ˆr j ˆp j (2N, 0) (2N, 0) ẑ j ŝ j ˆq j (2N, 0) (2N, 0) Fgure 4: Vertex-Edge ncdence gadget concludes the proof. Lemma 1 If G has a multcolor clque C = {v 1,v 2,,v k } then H has a capactated domnatng set D of sze k contanng all marked vertces. Proof: For every < j let e j be the edge from v to v j n G. In addton to all the marked vertces, let D contan v and e j for every < j. Clearly D contans exactly k vertces, so t remans to prove that D s ndeed a capactated domnatng set. For every < j, let ˆx and ˆx j domnate all ther neghbors except for v and e j respectvely. The vertces v and e j can domnate all ther neghbors, snce ther capacty s equal to ther degree. Let ˆr j domnate v down of the vertces n the (2N,0)-arrow from ŷ j, and v up of the vertces of the (2N,0)-arrow from ˆp j. Smlarly let ŝ j domnate v up of the vertces n the (2N,0)-arrow from ẑ j, and v down of the vertces of the (2N,0)-arrow from ˆq j. Fnally, for every j we let ŷ j, ẑ j, ˆp j and ˆq j domnate all ther neghbors that have not been domnated yet. One can easly check that every vertex of H wll ether be a domnator or domnated n ths manner, and that no domnator domnates more vertces than ts capacty. Lemma 2 If H has a capactated domnatng set D of sze k contanng all marked vertces, then G has a multcolor clque of sze k. 6

7 Proof: Observe that for every nteger 1 k, there must be a v V [] such that v D. Otherwse we have that S D and, snce S > k, we obtan a contradcton. Smlarly, for every par of ntegers,j wth < j there must be an edge e j E[,j] such that e j D. We let e j = e j. Snce D k t follows that these are the only unmarked vertces n D. Snce all the unmarked vertces n D have capacty equal to ther degree, we can assume that each such vertex domnates all ts neghbors. We now proceed wth provng that for every par of ntegers,j wth j, e j = uv s ncdent to v. We prove ths by showng that f u V [] then v up + u down = 2N. Suppose for a contradcton that v up + u down < 2N. Observe that each vertex of T = (N(ŷ j ) N(ˆr j ) N(ˆp j )) \ (N(v ) N(e j )) must be domnated by ether ŷ j, ˆr j, or ˆp j. However, by our assumpton that v up +u down < 2N, t follows that T = 2N 2 +4N +2MN (v up + u down ) > 2N 2 + 2N + 2MN. The sum of the capactes of ŷ j, ˆr j, and ˆp j s exactly 2N 2 + 2N + 2MN. Thus t s mpossble that every vertex of T s domnated by one of ŷ j, ˆr j, and ˆp j, a contradcton. If v up + u down > 2N then v down + u up < 2N, and we can apply an dentcal argument for ẑ j, ŝ j, and ˆq j. Thus, t follows that for every j there s an edge e j ncdent both to v and to v j. Thus {v 1,v 2,,v k } forms a clque n G. As any k-clque n G s a multcolor clque ths completes the proof. Observaton 1, Lemma 1 and Lemma 2 mmedately mply the followng theorem. Theorem 1 CDS parameterzed by treewdth and soluton sze s W[1]-hard. 3.2 CVC parameterzed by treewdth s W[1]-hard Usually vertex cover problems can be seen as restrctons of domnaton problems, and therefore t s natural to expect Capactated Vertex Cover to be somewhat easer than Capactated Domnatng Set. In ths secton, we gve a result smlar to the hardness result for Capactated Domnatng Set, but weaker n the sense that we only show that Capactated Vertex Cover s hard when parameterzed by the treewdth, whle we have seen n the prevous secton that Capactated Domnatng Set s hard when parameterzed by the treewdth and the soluton sze. To obtan our result we reduce from Multcolor Clque, as n the prevous secton. Agan, we reduce to a marked verson of Capactated Vertex Cover, where we search for a sze k capactated vertex cover that contans all the marked vertces. The reducton from Marked Capactated Vertex Cover to Capactated Vertex Cover s dentcal to the reducton from Marked Capactated Domnatng Set to Capactated Domnatng Set descrbed n the prevous secton. Notce also that n Marked Capactated Vertex Cover t makes sense to have marked vertces wth capacty zero, as they wll get non-zero capacty after the reducton to Capactated Vertex Cover. We reduce by buldng for an nstance (G,k) of Multcolor Clque an nstance (H,c,k ) of Marked Capactated Vertex Cover. In fact, we construct the graph H from G n almost the same manner as n the reducton to Marked Capactated Domnatng Set. The only dfferences are: We do not add the vertex sets S and S j for every,j. When we add an (A,B)-arrow from u to v, the A vertces on the subdvded edges are marked and have capacty 1, whle the B leaves attached to v are also marked but have capacty 0. 7

8 k = 7k(k 1) + 2k + (2k 2 N + (2M + 4) k (k 1)) 2N. The new term n the value of k s smply a correcton for all the extra marked vertces n the (A,B)-arrows. Notce that n ths case the value of k s not a functon of k alone, and that therefore ths reducton does not mply that Capactated Vertex Cover s W[1]- hard parameterzed by treewdth and soluton sze. However, by applyng arguments almost dentcal to the ones n the prevous secton, we can prove the followng clams. The verfcaton of these clams s smlar to the ones we made n the last secton and hence the detals are omtted. Clam 1 The treewdth of H s O(k 3 ). Clam 2 If G has a multcolor clque C = {v 1,v 2,,v k } then H has a capactated vertex cover S of sze k contanng all marked vertces. Clam 3 If H has a capactated vertex cover S of sze k contanng all marked vertces, then G has a multcolor clque of sze k. Together, the clams mply that Capactated Vertex Cover parameterzed by treewdth s W[1]-hard. Theorem 2 CVC parameterzed by treewdth s W[1]-hard. 4 FPT Algorthm for CVC on Graphs of Bounded Treewdth In the last secton we showed that Capactated Vertex Cover, when parameterzed only by treewdth (tw) of the nput graph, s W[1]-hard. However, for Capactated Domnatng Set we were able to show that the problem remans W[1]-hard even when we parameterze by both tw and k (the soluton sze). We complement the hardness result about Capactated Vertex Cover of the last secton by gvng a tme 2 O(tw log k) n O(1) algorthm on graphs of bounded treewdth wth respect to the combned parameters tw and k, a result whch was sketched ndependently by Hannes Moser [20]. Furthermore, usng ths 2 O(tw log k) n O(1) tme algorthm for CVC on graphs of bounded treewdth, we gve an mproved algorthm for the weghted verson of Capactated Vertex Cover n general graphs. Our algorthm, runnng n tme O(2 O(k log k) n O(1) ), mproves the earler algorthm of Guo et al. [15], that runs n tme O(1.2 k2 + n 2 ). To ths end, we gve a dynamc programmng algorthm workng on a so-called nce tree decomposton of the nput graph G: A tree decomposton (X,U) s a nce tree decomposton f one can root U n such a way that the root and every nner node of U s ether an nsert node, a forget node, or a jon node. Thereby, a node of U s an nsert node f has exactly one chld j, and X conssts of all vertces of X j plus one addtonal vertex; t s a forget node f has exactly one chld j, and X conssts of all but one vertces of X j ; and t s a jon node f has exactly two chldren j 1,j 2, and X = X j1 = X j2. Gven a tree decomposton of wdth tw, a nce tree decomposton of the same wdth can be found n lnear tme [17]. In what follows, we assume that the nce tree decomposton (X,U) that we are usng has the addtonal property that the bag assocated wth the root of U s empty (such a decomposton can easly be constructed by takng an arbtrary nce tree decomposton and addng some forget nodes above the orgnal root). Smlarly, we assume that every bag assocated wth 8

9 a leaf node dfferent from the root of U contans exactly one vertex. For a node n the tree U of a tree decomposton (X,U), let Y := {v X j j s a node n the subtree of U whose root s }, Z := Y \ X, and E := {{v,w} E v Z w Z }. Startng at the leaf nodes of U that are dfferent from the root, our dynamc programmng algorthm assgns to every node of U a table A that has a column l wth l k, for every vertex v X a column vc(v) wth vc(v) {true,false}, and for every vertex v X a column s(v) wth s(v) {null,0,1,,k 1}. Every row of such a table A corresponds to a soluton (f,c) for CVC on the subgraph of G that conssts of all vertces n Y and all edges n E havng at least one endpont n Z. More exactly, for every row of a table A there s a vertex set C Y and mappng f : E C wth the followng propertes: C s a capactated vertex cover for G = (Y,E ). C l. C contans all vertces v X wth vc(v) = true and no vertex v X wth vc(v) = false. For every vertex v X C, we have {{v,w} E f({v,w} = w} = s(v), and for every vertex v X \ C, we have s(v) = null. Intutvely speakng, for a vertex v C, the varable s(v) contans the number of edges ncdent to v that are covered by vertces n Z and, therefore, do not have to be covered by v. The smple observaton that s(v) can be at most k 1 (because C can contan at most k 1 neghbors of v) s crucal for the runnng tme of the algorthm. Clearly, f the table assocated wth the root of U s nonempty, the gven nstance of CVC s a yes-nstance. We wll now descrbe the computaton of the table A for a node n U, dependng on f s a leaf node dfferent from the root, an nsert node, a forget node, or a jon node. If necessary, we wrte l, vc (v), and s (v) n order to make clear that a value l, vc(v), and s(v), respectvely, stems from a row of a table A. The node s a leaf node dfferent from the root. Let X = {v}. Then we add one row to the table A for the case that v s not part of C and one row for the case that v s part of C, provded that k > 0. Because has no chld and, hence, no neghbor of v belongs to Z, the varable s(v) s set to 0 n the case that v s part of C: 1 f k > 0: { 2 add a new row to A ; 3 update the new row n A and set vc(v) := true; s(v) := 0; l := 1; } 4 add a new row to A ; 5 update the new row n A and set vc(v) := false; s(v) := null; l := 0; The node s an nsert node. Let j be the chld of n U, and let X = X j {v}. Here we extend the table A j by addng the varables vc(v) and s(v). For every row of A j, we add one row to the table A for the case that v s not part of C and one row for the case that v s part of C, provded that l j < k. Because no neghbor of v can belong to Z, the varable s(v) s set to 0 n the case that v s part of C: 9

10 1 for every row r of A j: { 2 f l j < k: { 3 copy the row r from A j nto A ; 4 update the new row n A and set vc(v) := true; s(v) := 0; l := l + 1; } 5 copy the row r from A j nto A ; 6 update the new row n A and set vc(v) := false; s(v) := null; } The node s a forget node. Let j be the chld of n U, and let X = X j \ {v}. Clearly, all neghbors of v belong to Y j due to the defnton of a tree decomposton. What has to be done s to consder the edges {v,w} wth w X, to decde whch of them shall be covered by v, and to set the value of s j (v) accordngly. Note that ths approach ensures that for all edges {v,w} wth w Z j we have already decded n a prevous step whch of these edges are covered by v. More exactly, for every row of A j, we perform the followng steps. If vc j (v) = true, then we try all possbltes for whch edges between v and vertces w X can be covered by v and add rows to A accordngly. If vc j (v) = false, then, of course, no edge between v and vertces w X j can be covered by v, and we add one row to A. In both cases, we have to check that for every edge {v,w} wth w X that s not covered by v t holds that vc j (w) = true and the remanng capacty of w, whch can be computed from s(w) and the number of w s neghbors n Z j, s bg enough to cover {v,w}: 1 N := N(v) X ; 2 for every row r of A j: { 3 f vc j(v) = true: { 4 for every subset N of N wth N = mn{ N, cap(v) ( N(v) Z j s j(v))}: { 5 f w N \ N : vc j(w) cap(w) > N(w) Z j s j(w): { 6 copy the row r from A j nto A ; 7 for every vertex w N wth vc(w) = true: { 8 update the new row n A and set s(w) := s(w) + 1; }}} 9 else: { // vc j(v) = false 10 f w N : vc(w) = true cap(w) > N(w) Z j s j(w): { 11 copy the row r from A j nto A ; }}} The node s a jon node. Let j 1 and j 2 be the chldren of n U. Here we consder every par r 1,r 2 of rows where r 1 s from A j1 and r 2 s from A j2. We say that two rows r 1 and r 2 are compatble f for every vertex v n X t holds that vc j1 (v) = vc j2 (v). If they are compatble, then we check whether for every vertex v X wth vc j1 (v) = vc j2 (v) = true the number of edges {v,w} covered by v wth w Z j1 plus the number of edges {v,w} covered by v wth w Z j2 s at most cap(v). If ths s the case, a new row s added to A : 1 for every compatble par r 1, r 2 of rows where r 1 s from A j1 and r 2 s from A j2 : { 2 f v X : vc j1 (v) = false cap(v) N(v) Z j1 s j1 (w) + N(v) Z j2 s j2 (w): { 3 add a new row to A ; 4 update the new row n A and set l := l j1 + l j2 {v X vc j1 (v) = true} ; 5 for every vertex v X : { 6 update the new row n A and set vc(v) = vc j1 (v); s(v) = s j1 (v) + s j2 (v); }}} In all four cases ( s a leaf node dfferent from the root, an nsert node, a forget node, or a jon node), after nsertng a row to A, we delete domnated rows from A. A row r 1 s domnated by a row r 2 f r 1 and r 2 are compatble, the value of l n r 1 s equal or greater than the value of l n r 2, and for every vertex v X wth vc(v) = true the value of s(v) n r 1 s equal or less than the value of s(v) n r 2. The correctness of ths data reducton s obvous: If the soluton correspondng to r 1 can be extended to a soluton for the whole graph, then ths s also possble wth the soluton correspondng to r 2 nstead. Clearly, due to ths data reducton, the table can never contan more than k tw rows, whch leads to the followng theorem. 10

11 Theorem 3 CVC on graphs of treewdth tw can be solved n k 3 tw n O(1) tme. Proof: The correctness of the algorthm follows from the above descrpton. The runnng tme for computng one table A assocated wth a tree node s bounded from above by k 3 tw n O(1), due to the fact that every table contans at most k tw rows and that the tree decomposton has O(n) tree nodes [17]. We menton n passng that wth usual backtrackng technques t s possble to construct the mappng f and the set C after runnng the dynamc programmng algorthm. CVC n General Undrected Graphs: The algorthm descrbed above can also be used for solvng CVC on general graphs wth the followng two observatons. Frstly, the treewdth of graphs that have a vertex cover of sze k s bounded above by k, and a correspondng tree decomposton of wdth k can be found n O( k + kn) tme [4]. (For a graph G = (V,E) that has a vertex cover C wth C = k, a tree decomposton of wdth k can be constructed as follows: Let U be a path of length V \ C, and assgn to every node of U a bag X that contans C and one vertex from V \ C. The vertex cover of sze k can be found n tme O( k + kn) [4].) Secondly, Theorem 3 can easly be adapted to the weghted verson of CVC, where every vertex of the nput graph has, n addton to the capacty, a weght, and the queston s f there s a capactated vertex cover whose weght s at most k. Wth these observatons, we get the followng corollary. Corollary 1 The weghted verson of CVC on general graphs can be solved n k 3k n O(1) = 2 O(k log k) n O(1) tme. 5 Conclusons and Dscussons In ths paper we studed Capactated Vertex Cover and Capactated Domnatng Set, generalzatons of Vertex Cover and Domnatng Set, respectvely, from the parameterzed perspectve. In partcular, we focused on the parameterzed complexty when parameterzed by (a) the treewdth of the nput graph and (b) the treewdth of the nput graph and the soluton sze k. Whle the orgnal verson of these problems were known to be FPT when parameterzed by treewdth we found ther behavor n the capactated context to be surprsng. In partcular, CDS turned out to be W[1]-hard and CVC to be FPT when parameterzed by treewdth and k. An mproved algorthm for CVC n general undrected graphs was obtaned by usng the FPT algorthm for CVC (when parameterzed by treewdth and k) as a subroutne. We also observed that CVC s possbly the frst known subset problem whch has been shown to be FPT when parameterzed by soluton sze but W[1]-hard when parameterzed by treewdth. It s easy to observe that f a planar graph has a CDS of sze at most k then the treewdth of the nput graph s at most O( k) [1, 6, 11]. Hence, n order to show that CDS s FPT for planar graphs, t s suffcent to obtan a dynamc programmng algorthm for t on planar graphs of bounded treewdth. The followng queston n ths drecton remans unanswered: Is CDS n planar graphs parameterzed by soluton sze fxed parameter tractable? References [1] J. Alber, H. L. Bodlaender, H. Fernau, T. Kloks, and R. Nedermeer. Fxed parameter algorthms for DOMINATING SET and related problems on planar graphs. Algorthmca, 33(4): ,

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