The computational complexity of the parallel knock-out problem

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1 The computatonal complexty of the parallel knock-out problem Hajo Broersma 1 Matthew Johnson 1 Danël Paulusma 1 Ian A. Stewart 1 1 Department of Computer Scence Durham Unversty South Road, Durham, DH1 3LE, U.K. {hajo.broersma,matthew.johnson2,danel.paulusma,.a.stewart}@durham.ac.uk July 24, 2007 Abstract We consder computatonal complexty questons related to parallel knock-out schemes for graphs. In such schemes, n each round, each remanng vertex of a gven graph elmnates exactly one of ts neghbours. We show that the problem of whether, for a gven bpartte graph, such a scheme can be found that elmnates every vertex s NP-complete. Moreover, we show that, for all fxed postve ntegers k 2, the problem of whether a gven bpartte graph admts a scheme n whch all vertces are elmnated n at most (exactly) k rounds s NP-complete. For graphs wth bounded tree-wdth, however, both of these problems are shown to be solvable n polynomal tme. We also show that r-regular graphs wth r 1, factor-crtcal graphs and 1-tough graphs admt a scheme n whch all vertces are elmnated n one round. Keywords: parallel knock-out; graphs; computatonal complexty AMS Subject Classfcatons: 05C85, 03D15 1 Introducton In ths paper, we consder parallel knock-out schemes for fnte undrected smple graphs. These were ntroduced by Lampert and Slater [9]. Such a scheme proceeds n rounds: n the frst round each vertex n the graph selects exactly one of ts neghbours, and then all the selected vertces are elmnated smultaneously. In subsequent rounds ths procedure s repeated n the subgraph nduced by those vertces not yet elmnated. The scheme contnues untl there are no vertces left, or untl an solated vertex s obtaned (snce an solated vertex wll never be elmnated). A graph s called KO-reducble or smply reducble f there exsts a parallel knock-out scheme that elmnates the whole graph. The parallel knock-out number of a graph G, denoted by pko(g), A prelmnary verson of ths paper was presented at the 7th Latn Amercan Theoretcal Informatcs Symposum 2006 and an extended abstract appeared n Lecture Notes n Computer Scence 3887 (2006), pp

2 s the mnmum number of rounds n a parallel knock-out scheme that elmnates every vertex of G. If G s not reducble, then pko(g) =. Our man motvaton for studyng the concept of reducblty s ts ntmate relaton to wellstuded concepts n structural and algorthmc graph theory, lke matchngs and cycles. To llustrate ths, we note that a graph G wth a perfect matchng has pko(g) = 1, as each vertex can select the vertex t s matched wth n a perfect matchng of G. Smlarly, a graph G wth a hamltonan cycle has pko(g) = 1, as each vertex can select ts successor on a hamltonan cycle of G wth some fxed orentaton. Whereas t s easy to check (.e., by a polynomal algorthm) whether a graph admts a perfect matchng, t s NP-complete to decde whether a graph has a hamltonan cycle. What can be sad about the complexty of decdng whether a graph G has a fnte parallel knock-out number? Or about determnng (an upper bound on) the value of pko(g)? These complexty questons are our man concern n ths paper and wll be answered n Secton 4. We wll also consder several structural propertes related to reducblty, but only wth some relaton to complexty questons. Other structural propertes related to reducblty can be found n [3] and [5]. 1.1 Complexty questons related to reducblty Consder the followng decson problem. Parallel Knock-Out (PKO) Instance: A graph G. Queston: Is G reducble? In [9], whch appeared n 1998, t was clamed that PKO s NP-complete even when restrcted to the class of bpartte graphs. No proof was gven; the reader was referred to a paper that was n preparaton. Our attempts to obtan and verfy ths proof have been unsuccessful. We shall obtan the result as a corollary to a stronger theorem (Theorem 1 below) by consderng a related problem, whch s defned for each postve nteger k. Parallel Knock-Out (k) (PKO(k)) Instance: A graph G. Queston: Is pko(g) k? Our frst result classfes the complexty of PKO(k), k 2. Theorem 1 For k 2, PKO(k) s NP-complete even f nstances are restrcted to the class of bpartte graphs. The proof s postponed to Secton 4. By usng almost the same arguments, we wll also show that decdng whether pko(g) = k s polynomally solvable for k = 1 and NP-complete for any fxed k 2 that s not part of the nput. As a matter of fact t s not dffcult to show that a graph G has pko(g) = 1 f and only f G contans a [1,2]-factor,.e., a spannng subgraph n whch every component s ether a cycle or an edge: smply note that a vertex u that selects a vertex v s ether selected by v or by a vertex w {u, v}, and combne ths observaton wth the fact that every vertex selects exactly one other vertex and that all the graphs we consder are fnte. The problem of decdng whether G contans a [1,2]-factor s a folklore problem appearng n many standard books on combnatoral optmzaton. For convenence we shortly dscuss t below. 2

3 Let V (G) = {v 1,v 2,...,v n }. Defne a bpartte graph G wth vertex set V (G ) = {u 1,u 2,...,u n, w 1,w 2,...,w n } n whch u w j E(G ) and u j w E(G ) f and only f v v j E(G). A [1,2]-factor n G corresponds to a perfect matchng n G. Hence the related decson problem and consequently PKO(1), can be decded n polynomal tme. Ths s also clear from the followng polynomally solvable decson problem (see [7], problem GT13, page 193): gven a drected graph D, decde whether V (D) can be parttoned nto dsjont sets of cardnalty at least 2 such that each of the sets nduces a subgraph wth a drected hamltonan cycle. To show the ntmate relaton wth knock-out schemes, replace each edge of G by two oppostely drected arcs. Clearly G has a [1,2]- factor f and only f the related drected graph has such a partton nto hamltonan cycles. In [3], t was shown, usng a dynamc programmng approach, that the parallel knock-out number for trees can be computed n polynomal tme. The authors presented an O(n 3.5 log 2 n) algorthm for computng the parallel knock-out number of an n-vertex tree, and asked whether there exsts a substantally faster algorthm for ths problem, wth a tme complexty of, say, O(n log n) or O(n 2 )? Our next result mples that there exsts a lnear tme algorthm for ths problem. A key ngredent of the dynamc program for trees n [3] s the reducton to a number of polynomally solvable bpartte matchng problems. For hgher tree-wdths, these bpartte matchng problems have no natural polynomally solvable analogues. Therefore the dynamc program for trees does not carry over to the bounded tree-wdth classes. In [3] t was asked whether one can avod the computaton of perfect matchngs n auxlary bpartte graphs whle computng pko(t) for a tree T. And can one then generalze such a method to graphs of bounded tree-wdth? In our second result, we gve an affrmatve answer, although we do not provde an explct algorthm. Theorem 2 The problem PKO(k) can be solved n lnear tme on graphs wth bounded tree-wdth. We wll also show that PKO can be solved n polynomal tme on graphs wth bounded tree-wdth. 1.2 Structural propertes related to reducblty As noted above, there s an ntmate relatonshp between reducblty and other structural propertes, lke the exstence of a [1,2]-factor, a noton that s a common generalzaton of a perfect matchng and a hamltonan cycle. Apart from hamltonan graphs and graphs that have a perfect matchng, for example also all k-traversable graphs have been shown to have a [1,2]-factor [4]. A graph s k-traversable f t admts a closed walk n whch every vertex occurs exactly k tmes. These graphs were also studed n [8]. We establsh several other results n Secton 6 that explore the relatonshp between reducble graphs and propertes related to the exstence of near perfect matchngs or hamltonan cycles. For example we show that all factor-crtcal and all 1-tough graphs have a [1,2]-factor,.e., have parallel knock-out number 1. We refer to Secton 6 for defntons. We also show there that r-regular graphs wth r 1 have a [1,2]-factor, and we have a closer look at almost regular bpartte graphs,.e., bpartte graphs n whch all vertces n the same bpartton class have the same degree. 1.3 Organzaton of the paper In the next two sectons we ntroduce a number of defntons and prelmnary observatons. In Secton 4 and Secton 5 are the proofs and corollares of Theorems 1 and 2, respectvely. Secton 6 deals wth factor-crtcal graphs, 1-tough graphs, r-regular graphs and almost regular bpartte graphs. 3

4 2 Prelmnares Graphs n ths paper are denoted by G = (V,E). An edge jonng vertces u and v s denoted uv. For graph termnology not defned below, we refer to [2]. For convenence we allow graphs to have an empty vertex set. We say that G = (V,E) s the null graph f V = E =. For a vertex u V we denote ts neghbourhood, that s, the set of adjacent vertces, by N(u) = {v uv E}. The degree d G (v) of a vertex v n G s the number of edges ncdent wth t, or, equvalently, the sze of ts neghbourhood. A maxmal connected subgraph of a graph G s called a component of G. Adoptng the termnology and notaton from [3], for a graph G, a KO-selecton s a functon f : V V wth f(v) N(v) for all v V. If f(v) = u, we say that vertex v fres at vertex u, or that vertex u s knocked out by vertex v. For a KO-selecton f, we defne the correspondng KO-successor of G as the subgraph of G that s nduced by the vertces n V \ f(v ); f H s the KO-successor of G we wrte G H. Note that every graph wthout solated vertces except for the null graph has at least one KO-successor. A graph G s called KO-reducble, f there exsts a fnte sequence G G 1 G 2 G r, where G r s the null graph. If no such sequence exsts, then pko(g) =. Otherwse, the parallel knock-out number pko(g) of G s the smallest number r for whch such a sequence exsts. A sequence of KO-selectons that transform G nto the null graph s called a KO-reducton scheme. A sngle step n ths sequence s called a round of the KO-reducton scheme. A subset of V s knocked out n a certan round f every vertex n the subset s knocked out n that round. We make some smple observatons that we wll use later on. Observaton 3 Let G = (V,E) be a KO-reducble graph, and let V 1 = {v V d(v) = 1}. Then n the frst round of any KO-reducton scheme each vertex of V 1 s knocked out by ts unque neghbour n G. Proof. Ths s clear, snce otherwse some vertex v V 1 wll be an solated vertex after the frst round, as the neghbour of v s knocked out by v n the frst round. Observaton 4 Let G be a graph on at least three vertces. If G contans two vertces of degree 1 that share the same neghbour, then G s not KO-reducble. Proof. Suppose G s KO-reducble. Then by Observaton 3, the shared neghbour knocks out both vertces of degree 1, a contradcton. Observaton 5 Let u 1,u 2,u 3,u 4 be four vertces of a KO-reducble graph G such that N(u 2 ) = {u 1,u 3 }, N(u 3 ) = {u 2,u 4 } and N(u 4 ) = {u 3 }. If u 1 s knocked out n the frst round of a KOreducton scheme, then u 1 fres at u 2 n the frst round. Proof. By Observaton 3, u 3 and u 4 knock each other out n the frst round, so u 3 does not knock out u 2. If u 1 s knocked out n the frst round of a KO-reducton scheme, then u 1 fres at u 2 n the frst round; otherwse u 2 wll be an solated vertex after the frst round. An odd path u 1 u 2...u 2k+1 s called a centred path of G wth centrevertex u k+1 f G {u k+1 } contans as components the path u 1 u 2... u k and the path u k+2 u k+3...u 2k+1. 4

5 Observaton 6 Let P = u 1 u 2...u 7 be a centred path of a KO-reducble graph G. In the frst round of any KO-reducton scheme u 1 and u 2 fre at each other, u 3 fres at u 2, u 6 and u 7 fre at each other, u 5 fres at u 6, u 4 fres at u 3 or u 5, and u 4 wll not be knocked out. In the second round of any KO-reducton scheme u 4 and ts remanng neghbour n P fre at each other. Proof. By Observaton 3, u 1 and u 2, and u 6 and u 7 knock each other out n the frst round. Suppose u 3 fres at u 4 n the frst round. Then u 4 has to fre at u 3 ; otherwse u 3 wll be an solated vertex after the frst round. But now u 5 wll be an solated vertex after the frst round. Hence u 3 fres at u 2, and smlarly u 5 fres at u 6. So at least one of u 3 and u 5 survves the frst round. Ths mples that u 4 has to survve the frst round as well. The result now follows by applyng Observatons 3 and 4 to the KO-successor of G. 3 NP-complete problems In ths secton, we consder two NP-complete problems that wll play a key role n our proof of Theorem 1. We refer to [7] and [10] for further detals. The frst problem concerns domnatng sets. A set S V s a domnatng set of a graph G = (V,E) f every vertex of G s n S or adjacent to a vertex n S. We wll make use of the followng NP-complete decson problem. Domnatng Set (DS) Instance: A graph G = (V,E) and a postve nteger p. Queston: Does G have a domnatng set of cardnalty at most p? The second problem concerns hypergraph 2-colourngs. A hypergraph J = (Q, S) s a par of sets where Q = {q 1,...,q m } s the vertex set and S = {S 1,...,S n } s the set of hyperedges. Each member S j of S s a subset of Q. A 2-colourng of J = (Q, S) s a partton of Q nto sets B and W such that, for each S S, B S and W S. We wll also make use of the followng NP-complete decson problem. Hypergraph 2-Colourablty (H2C) Instance: A hypergraph J = (Q, S). Queston: Is there a 2-colourng of J = (Q, S). Before we turn to our proofs of the complexty results n the next secton, we need a few more defntons. The ncdence graph I of a hypergraph J = (Q, S) s a bpartte graph wth vertex set Q S where (q,s) forms an edge f and only f q S. Wth a hypergraph J = (Q, S) we can assocate another hypergraph J = (X, Z) called the trple of J; trples of hypergraphs wll play a crucal role n our NP-completeness proofs n the next secton. It requres a lttle effort to defne the vertex set X and hyperedge set Z of the trple of J. Recall that Q = {q 1,...,q m } and S = {S 1,...,S n }. For 1 m, let l() be the number of hyperedges n S that contan q, let Q = {q 1 all such sets s the vertex set of J, that s Now the hyperedges. X =,...,ql() } and let U = {u 1 m (Q U ). =1 5,...,ul() }. The unon of

6 T 1 1 T 2 1 T 3 1 S R 1 1 R 2 1 P1 1 S 4 P S 7 P1 3 R 3 1 q 1 1 u 1 1 q 2 1 u 2 1 q 3 1 u 3 1 Fgure 1: Part of the ncdence graph of the trple of a hypergraph. Let us frst defne the followng sets: for 1 m, for 1 k l(), let P k = {q k,uk }, for 1 m, for 1 k l() 1, let R k = T k = {u k,qk+1 }, and for 1 m, let R l() Let P = {P 1 = T l() = {u l(),q 1 }. l(),...,p }, R = {R 1,...,Rl() }, and T = {T 1 l(),...,t }, and let m m m P = P, R = R, T = T. =1 =1 For 1 j n, let us also defne a set S j. If n J, S j contans q, then n J, S j contans a vertex of Q. In partcular, f S j s the kth hyperedge that contans q n J, then S j contans qk. For example, f q 1 s n S 1, S 4 and S 7 (only) n J, then l(1) = 3 and n J there are vertces q1 1,q2 1,q3 1 wth q1 1 S 1, q2 1 S 4, and q3 1 S 7. Let S = {S 1,...,S n}. The set of hyperedges for J s Z = S P R T. We denote the ncdence graph of the trple J of J by I. See Fgure 1 for an example that llustrates the case where q 1 belongs to S 1, S 4 and S 7. Proposton 7 The hypergraph J = (Q, S) has a 2-colourng B W f and only f ts trple J = (X, Z) has a 2-colourng B W such that for each 1 m ether Q B and U W, or Q W and U B. Proof. Suppose B W s a 2-colourng of J. Defne a partton B W of X as follows. If q s n B, then each q k s n B and each u k s n W. If q s n W, then each q k s n W and each u k s n B. Obvously, B W s a 2-colourng of J wth the desred property. =1 6

7 Suppose we have a 2-colourng B W of J such that for each 1 m ether Q B and U W, or Q W and U B. Then let q B f and only f Q B, and let W = Q \ B. Clearly, f S j contans only elements from B (respectvely W), then S j would contan only elements from B (respectvely W ). Hence B W s a 2-colourng of J. 4 Complexty classfcaton We now have all the ngredents to prove our man complexty result. We repeat t here for convenence. Theorem 8 For k 2, PKO(k) s NP-complete even f nstances are restrcted to the class of bpartte graphs. Proof. It s clear that PKO(k) s n NP. The rest of the proof s n two cases. We gve separate proofs for the cases k = 2 and k 3. Case 1. k = 2. We use reducton from DS. Gven G = (V,E) and a postve nteger p V, we shall complete the proof by constructng a bpartte graph B such that pko(b) = 2 f and only f G has a domnatng set D wth D p. Let the vertex set of B be the dsjont unon of V = {v 1,...,v n }, V = {v 1,...,v n } and W = {w 1,...,w n p }. Let the edge set of B consst of v v, 1 n, v v j and v v j, for each edge v v j E, and v w h, 1 n, 1 h n p. Suppose that G has a domnatng set D = {v 1,...,v d } where d p. Note that every vertex n V s adjacent to a vertex of D n B. We shall descrbe a 2-round KO-reducton scheme for B. In the frst round for 1 n, v fres at v, for 1 j p, v j fres at v j, for p + 1 j n, v j fres at a vertex n D, and for 1 h n p, w h fres at a vertex n D. Thus each vertex n {v 1,...,v p } and V s elmnated n the frst round, and each vertex n V \ {v 1,...,v p } and W survves to round 2. As the survvng vertces nduce the balanced complete bpartte graph K n p,n p n B, t s clear that every survvng vertex can be elmnated n one further round. Now suppose that B has a 2-round KO-reducton scheme. Let D be the subset of V contanng vertces that are fred at n round 1. As every vertex n V fres at and so s adjacent to a vertex n D, D s a domnatng set n G (snce each vertex n V s joned only to copes of tself and ts neghbours). We complete the proof of Case 1 by showng that D p. Let V S = V \ D and V S V W be the sets of vertces that survve round 1. As round 2 s the fnal round, V S = V S. (1) 7

8 H S H P T H T S P R H R I q y q 1 H q u H u Fgure 2: The graph G n Case 2. As V W = 2n p and at most n vertces n V W are fred at n round 1, V S n p. Thus, by (1), V S n p. Therefore D = V V S n (n p) = p. Case 2. k 3. We use reducton from H2C. Let J = (Q, S) be an nstance of H2C. Let I be the ncdence graph of ts trple J = (X, Z). Recall that Z = S P R T. From I, we obtan another bpartte graph G by addng X + Z mutually vertex-dsjont paths and connectng each vertex of I wth one of these added paths as follows: For each vertex x n X, add a path H x = y x 1 yx 2 yx 3 and jon x to yx 1. For each vertex R n R, add a path H R = y R 1... yr 4 and jon R to yr 1. For each vertex T n T, add a path H T = y T 1...yT 4 and jon T to yt 1. For each vertex P n P, add a path H P = y P 1...yP 7 and jon P to the centrevertex yp 4. For each vertex S n S, add a path H S = y S 1...yS 7 and jon S to the centrevertex y S 4. Fgure 2 llustrates G. We complete the proof by showng that J s 2-colourable f and only f pko(g) k. Throughout the proof, G 1 and G 2 denote the graphs nduced by the survvng vertces of G after, respectvely, one and two rounds of a KO-reducton scheme. Suppose B W s a 2-colourng of J. By Proposton 7, J has a 2-colourng B W. We defne a three-round KO-reducton scheme for G, so we show that n ths case pko(g) 3 k. 8

9 Round 1. Vertces of degree 1 and ther neghbours fre at each other. Each H P wth P P and each H S wth S S s a centred path of G, and the vertces fre as n Observaton 6. For each z R T, vertex y1 z fres at yz 2 and yz 2 fres at yz 3. Each vertex n Z fres at one of ts neghbours n B. Each vertex x n X fres at ts neghbour y1 x n Hx. Each y1 x wth x B fres at x. Each y1 x wth x W fres at y2 x. Thus every vertex n W and no vertex n B survves the frst round. Also every vertex n Z survves the frst round. After the frst round, each vertex z R T s adjacent to a vertex y1 z of degree 1, and each vertex z S P s adjacent to a vertex y4 z whose only other neghbour s a vertex y3 z (or yz 5 ) of degree 1. Round 2. Because B W s a 2-colourng of J = (X, Z), every vertex n Z has a neghbour n W n G 1. For each S j S we choose one neghbour n W and let W be the set of selected vertces. Snce no two vertces n S have a common neghbour n X, W = n. The vertces n G 1 fre as follows. Vertces of degree 1 and ther neghbours fre at each other. Each vertex P P wth a neghbour n W \W fres at ths neghbour. Otherwse P fres at y4 P. Each x X fres at ts neghbour n P. Each S S fres at y4 S. Thus the vertex set of G 2 s W S. Round 3. Each S S and ts unque neghbour n W fre at each other, whch leaves us wth the null graph. Now we suppose that pko(g) k. We assume that a partcular KO-reducton scheme for G s gven and prove that J has a 2-colourng. We start wth the followng useful property. Clam 1. If a vertex of a set Q s knocked out n the frst round, then all the vertces of Q are knocked out n the frst round. Proof of Clam 1. Suppose that a vertex q k Q s knocked out n the frst round. We prove the clam by showng that q k+1 (wth q l()+1 = q 1 ) s also knocked out n the frst round. If q k Q s knocked out n the frst round, then, by Observaton 5, q k fres at y qk 1. Suppose q k+1 s not knocked out n the frst round. Observaton 6 mples that P k+1 must fre at u k+1, and P k must fre at ether q k or u k. If P k fres at u k, then, by Observaton 5, uk fres at y qk 1. Snce vertces n H P k must fre as n Observaton 6, ths means that G 1 contans a component somorphc to a path on three vertces. By Observaton 4, G 1 s not KO-reducble. Hence, P k fres at q k. y T k+1 For the same reason R k+1 or T k+1 cannot fre at u k 1, respectvely. Due to Observaton 5 ths mples that y Rk+1 y T k+1 2., and consequently they fre at yrk+1 1 and 1 fres at y Rk+1 2, and y T k+1 1 fres at In G 1, T k and R k have exactly the same neghbours, namely uk and qk+1. If T k and R k fre at a dfferent neghbour n the second round, then due to Observaton 5 both wll be solated vertces n G 2. Suppose T k and R k fre at the same neghbour. Then n all possble schemes G 2 wll contan two vertces of degree 1 havng the same neghbour. Observaton 4 mples that G 2 s not KO-reducble. We conclude that q k+1 must be knocked out n the frst round as well. Usng the same arguments, we get the followng clam. Clam 2. If a vertex n a set U s knocked out n the frst round, then all vertces n U are knocked out n the frst round. 9

10 By Clam 1 and Clam 2 we may defne a set B X as follows. All vertces of a set Q or U are n B f and only f the set s knocked out n the frst round. Let W = X\B. We need one more clam. Clam 3. For all 1 m, ether Q B and U W, or Q W and U B. Proof of Clam 3. Let 1 m. By Observaton 6, each vertex P k P must fre at ether q k or u k n the frst round. The prevous two clams mply that Q or U s knocked out n the frst round. Suppose both sets are knocked out n the frst round. Then, by Observaton 5, u 1 fres at y u1 1, and q1 fres at y q1 1. Then, by Observaton 6, P 1 wll not be knocked out n any round. The clam s proved. By Clam 3, all vertces n Z\S have one neghbour n B and one neghbour n W. Let S j be a vertex n S. By Observaton 6, S j fres at a neghbour n m =1 Q. By defnton, ths neghbour s n B. By Observatons 5 and 6, S j s knocked out by a neghbour n m =1 Q that s not knocked out n the frst round. By defnton, ths neghbour s n W. It s now clear that B W s a 2-colourng of J such that for each 1 m ether Q B and U W, or Q W and U B. Hence, by Proposton 7, the hypergraph J also has a 2-colourng. Ths completes the proof of Theorem 1. Theorem 1 has the followng two easy consequences. Corollary 9 The problem PKO s NP-complete, even f nstances are restrcted to the class of bpartte graphs. Proof. The problem PKO s clearly n NP. We use reducton from H2C. From an nstance J = (Q, S) we construct the graph G as n the proof of Theorem 1. We clam that J s 2-colourable f and only f G s KO-reducble. Suppose that J s 2-colourable. As we have seen n the proof of Theorem 1 ths mples that pko(g) 3. Hence G s KO-reducble. Suppose that G s KO-reducble. We copy the proof of Case 2 of Theorem 1. The second corollary of Theorem 1 nvolves the followng decson problem. Exact Parallel Knock-Out (k) (EPKO(k)) Instance: A graph G. Queston: Is pko(g) = k? Corollary 10 The problem EPKO(k) s polynomally solvable for k = 1 and s NP-complete for k 2, even f nstances are restrcted to the class of bpartte graphs. Proof. We already observed n Secton 1 that EPKO(1) s polynomally solvable. Ths mples that EPKO(2) s NP-complete snce PKO(2) s NP-complete. For the case k 3 we make use of a famly of trees Y l wth pko(y l ) = l that have been constructed n [3]. We add a dsjont copy of the tree Y k to the graph G constructed n the proof of Case 2 n Theorem 1. The new graph G has pko(g ) = k f and only f pko(g) k. Note that the sze of a tree Y k only depends on k and not on the sze of our nput graph G (so we do not need the exact descrpton of ths famly). We can even make the nstance graph connected by addng an edge between the neghbor of a leaf n Y k and the neghbor of a degree-one vertex n G. Note that H2C remans NP-complete for connected hypergraphs. Also note that by 10

11 Observaton 3, n any KO-reducton scheme of the new graph a degree 1 vertex and ts neghbour knock each other out n the frst round, so the added edges do not change the KO-reducblty propertes of the graph. 5 Bounded tree-wdth In ths secton we use monadc second-order logc; that s, that fragment of second-order logc where quantfed relaton symbols must have arty 1. For example, the followng sentence, whch expresses that a graph (whose edges are gven by the bnary relaton E) can be 3-coloured, s a sentence of monadc second-order logc: { ( R W B x (R(x) W(x) B(x)) (R(x) W(x)) ) ( (R(x) B(x)) (W(x) B(x)) x y E(x,y) )} ( (R(x) R(y)) (W(x) W(y)) (B(x) B(y))) (the quantfed unary relaton symbols are R, W and B, and should be read as sets of red, whte and blue vertces, respectvely). Thus, n partcular, there exst NP-complete problems that can be defned n monadc second-order logc. A semnal result of Courcelle [6] s that on any class of graphs of bounded tree-wdth, every problem defnable n monadc second-order logc can be solved n tme lnear n the number of vertces of the graph. Moreover, Courcelle s result holds not just when graphs are gven n terms of ther edge relaton, as n the example above, but also when the doman of a structure encodng a graph G conssts of the dsjont unon of the set of vertces and the set of edges, as well as unary relatons V and E to dstngush the vertces and the edges, respectvely, and also a bnary ncdence relaton I whch denotes when a partcular vertex s ncdent wth a partcular edge (thus, I V E). The reader s referred to [6] for more detals and also for the defnton of tree-wdth whch s not requred here. To prove Theorem 2, we need only prove the followng proposton. Proposton 11 For k 1, PKO(k) can be defned n monadc second order logc. Proof. Recall that a parallel knock-out scheme for a graph G = (V,E) s a sequence of graphs G G 1 G 2 G r, where G r s the null graph. Let W 0 = V and, for 1 r, let W be the vertex set of G. If we can wrte a formula Φ(W,W +1 ) of monadc second-order logc that says there exsts a KO-selecton f on W such that the vertex set of the KO-successor s W +1, then we could prove the proposton wth the followng sentence Ω k whch s satsfed f and only f G s n PKO(k): W 0 W 1 W k { v(w 0 (v) V (v)) Φ(W 0,W 1 ) Φ(W 1,W 2 ) Φ(W k 1,W k ) ( v( W k (v) V (v))) } 11

12 (Here and elsewhere we have presupposed that each W s a set of vertces; we could easly nclude addtonal clauses to check ths explctly.) The followng clam wll help us wrte Φ(W,W +1 ). Clam 4. There s a KO-selecton f on W such that W +1 s the vertex set of the KO-successor f and only f there s a partton V 1,V 2,V 3 of W and subsets E 1,E 2,E 3 of E such that (a) for j = 1,2,3, each vertex n V j s ncdent wth exactly one edge of E j, ths edge jons t to a vertex n W \ V j, and ths accounts for every edge n E j (so V j = E j ). (b) W +1 W and, for j = 1,2,3, W +1 V j s the set of vertces n V j not ncdent wth edges n E j for any j j. We wll prove the clam later. Frst we use t to wrte Φ(W,W +1 ). The followng formula ψ(v 1,E 1,V 2,E 2,V 3,E 3,W ) checks that the sets V 1,V 2 and V 3 partton W, that the sets E 1,E 2,E 3 are edges n the graph, and that (a) s satsfed. v((v 1 (v) V 2 (v) V 3 (v)) W (v)) v( (V 1 (v) V 2 (v)) (V 1 (v) V 3 (v)) (V 2 (v) V 3 (v))) x((e 1 (x) E 2 (x) E 3 (x)) E(x)) x(e 1 (x) u v(v 1 (u) (V 2 (v) V 3 (v)) I(u,x) I(v,x))) x(e 2 (x) u v(v 2 (u) (V 1 (v) V 3 (v)) I(u,x) I(v,x))) x(e 3 (x) u v(v 3 (u) (V 1 (v) V 2 (v)) I(u,x) I(v,x))) v(v 1 (v)!x(i(v,x) E 1 (x))) v(v 2 (v)!x(i(v,x) E 2 (x))) v(v 3 (v)!x(i(v,x) E 3 (x))) (The semantcs of! s there exsts exactly one ; clearly, ths abbrevates a more complex though routne frst-order formula.) The followng formula checks that (b) s satsfed and s denoted χ(v 1,E 1,V 2,E 2,V 3,E 3,W,W +1 ). And now we can wrte Φ(W,W +1 ): v(w +1 (v) (W (v) ((V 1 (v) x((e 2 (x) E 3 (x)) I(v,x))) (V 2 (v) x((e 1 (x) E 3 (x)) I(v,x))) (V 3 (v) x((e 1 (x) E 2 (x)) I(v,x)))))). V 1 E 1 V 2 E 2 V 3 E 3 (ψ(v 1,E 1,V 2,E 2,V 3,E 3,W ) χ(v 1,E 1,V 2,E 2,V 3,E 3,W,W +1 )). It only remans to prove Clam 4. Suppose that we have sets V 1,V 2,V 3,E 1,E 2 and E 3 that satsfy the condtons of the clam. Then to defne the KO-selecton f, for j = 1,2,3, for each vertex v V j, let v fre at the unque neghbour joned to v by an edge n E j. It s easy to check that W +1 s the vertex set of the KO-successor. Now suppose that we have a KO-selecton f. Let H be the spannng subgraph of G wth edge set {vf (v) v W }. The frng can be represented as an orentaton of H: orent each edge from 12

13 Fgure 3: A representaton of vertces frng v to f (v) (some edges may be orented n both drectons). As each vertex has exactly one edge orented away from t, each component of the orented graph contans one drected cycle, of length at least 2, wth a pendant n-tree attached to each vertex of the cycle; see Fgure 3. We fnd the sets V 1,V 2,V 3,E 1,E 2,E 3 ; the edge sets contan only edges of H. We may assume that H s connected (else we can fnd the sets componentwse). Let the vertces of the unque cycle n the orentaton be v 1,...,v c where the edges are v l v l+1, 1 l c 1, and v c v 1. So H contans vertces v 1,...,v c wth a pendant tree (possbly trval) attached to each. For 1 l c, let U l e be the set of vertces n the pendant tree attached to v l whose dstance from v l s even (but not zero), and let U l o be the vertces n the tree at odd dstance from v l. Let V 1 = Uo l l odd l even V 2 = Ue l l odd l even V 3 = {v c }, U l e U l o {v l : l s even,l c}, {v l : l s odd,l c}, and and, for = 1,2,3, let E contan vf (v) for each v V. It s clear that the sets we have chosen satsfy the condtons of the clam. Ths completes the proof of the clam and of the proposton. Theorem 2 follows from the proposton. And, notng that EPKO(k) s defned by the monadc second-order sentence Ω k Ω k 1, we have the followng result. Corollary 12 For k 1, EPKO(k) s solvable n lnear tme on any class of graphs wth bounded tree-wdth. In partcular, we obtan the followng result for trees, answerng an open queston n [3]. Corollary 13 For k 1, EPKO(k) s solvable n lnear tme for trees. 13

14 Fnally, we note that to check whether a graph G s reducble t s suffcent to check whether pko(g) = k, for 1 k, where s the maxmum degree of G. Thus G s reducble f and only f the sentence Ω Ω 1 Ω 1 s satsfed. Ths gves us our last result of ths secton. Corollary 14 On any class of graphs wth bounded tree-wdth, PKO can be solved n polynomal tme. 6 Graphs wth a small parallel knock-out number As we noted n the ntroducton, graphs wth a [1,2]-factor, or more partcularly wth a perfect matchng or wth a hamltonan cycle have parallel knock-out number 1. We start ths secton by studyng the related classes of factor-crtcal graphs and of 1-tough graphs. A graph G s sad to be factor-crtcal f G v has a perfect matchng for every vertex v of G. A graph G = (V,E) s called 1-tough f ω(g S) S for every nonempty subset S of V, where ω(g S) denotes the number of components of the graph G S. Clearly, every hamltonan graph s 1-tough and every factor-crtcal graph has a matchng leavng only one vertex unmatched. A natural queston s whether factor-crtcal graphs and 1-tough graphs have a small parallel knock-out number. The next results show that these graphs n fact have a [1,2]-factor,.e., have parallel knock-out number 1 (unless they are trval,.e., contan only one vertex). We start wth factor-crtcal graphs. Theorem 15 Let G be a nontrval factor-crtcal graph and v V (G). Then G has a [1,2]-factor consstng of an odd cycle C contanng v and a perfect matchng n G V (C). Proof. Let M be a perfect matchng n G v. If some neghbours of v are matched by M, we mmedately fnd the desred odd cycle (trangle) C and perfect matchng n G V (C), and we are done. Suppose ths s not the case. Then we take an arbtrary neghbour x of v, and note that there exsts a perfect matchng M n G x. Clearly x s matched to a vertex y v under M and v s matched to a vertex p under M. By our assumpton we may assume p y; otherwse we fnd a trangle C wth the desred propertes. Snce both M and M saturate all vertces except for v and x, respectvely, there exsts an (M,M)-alternatng path P between v and y begnnng and endng wth an edge of M. Now P together wth the edges between x and v and x and y forms an odd cycle C, and the remanng edges of M form a perfect matchng n G V (C). The above result also mples that nontrval 1-tough graphs on an odd number of vertces have a [1,2]-factor. In order to prove ths, we need the followng well-known result of Tutte [11]. Let ω o (G) denote the number of odd components of a graph G,.e., the number of components contanng an odd number of vertces. Theorem 16 ([11]) A graph G has a perfect matchng f and only f ω o (G S) S for all S V (G). Ths theorem has the followng consequence. Corollary 17 If G s a 1-tough graph on an odd number of vertces, then G s factor-crtcal. 14

15 Proof. Suppose G s 1-tough on an odd number of vertces, but not factor-crtcal. Then there exsts a vertex v V (G) such that G = G v has no perfect matchng. Thus by Theorem 16 there exsts a set X V (G ) wth ω o (G X ) = X k, for some nteger k 0. Settng X = X {v}, and lettng ω e denote the number of even components, we have ω(g X) = ω o (G X) + ω e (G X) = ω o (G X ) + ω e (G X ) = X k + ω e (G X ) = X + k + ω e (G X ) X 1. Snce G s 1-tough, k = 0 and ω e (G X ) = 0; otherwse ω(g X) > X 1, a contradcton. Let H 1,...,H X +1 denote the odd components of G X. Then V (G) = 1+ X + X +1 =1 V (H ) = X +1 =1 ( V (H ) + 1). Snce all V (H ) are odd, V (G) s even, a contradcton. Corollary 18 Every nontrval 1-tough graph has a [1,2]-factor. Proof. Consder a nontrval 1-tough graph G on n vertces. If n s odd the result follows by combnng Theorem 15 and Corollary 17. If n s even G has a perfect matchng by Theorem 16. We now turn to regular graphs and almost regular bpartte graphs. Frst we note that the trck ntroduced n Secton 1 mmedately mples that every r-regular graph G wth r 1 has a [1,2]-factor,.e., pko(g) = 1. Proposton 19 Every r-regular graph G wth r 1 has a [1,2]-factor. Proof. let G be an r-regular graph wth r 1 and wth V (G) = {v 1,v 2,...,v n }. Defne a bpartte graph G wth vertex set {u 1,u 2,...,u n, w 1,w 2,...,w n } n whch u w j E(G ) and u j w E(G ) f and only f v v j E(G). Then G s an r-regular bpartte graph and has a perfect matchng (See, e.g., [2] Exercse 5.2.3(a)). Ths matchng corresponds to a [1,2]-factor n G. The above result also mmedately mples the followng statement for graphs that contan a k- factor,.e., a spannng k-regular subgraph. Corollary 20 Every nontrval graph wth a k-factor has a [1,2]-factor. Our complexty results for bpartte graphs motvated us to consder bpartte graphs that are almost regular n the followng sense. A bpartte graph G s called (r,s)-regular f all vertces n one class of the bpartton have degree r and all other vertces have degree s. By Proposton 19 any (r,r)-regular bpartte graph G wth r 1 has pko(g) = 1, and one easly checks that any (1,s)-regular bpartte graph G wth s 2 has pko(g) =. Wth the next result we characterze all reducble (2, s)-regular bpartte graphs, notng that (2, s)-regular graphs wth s 2, 3 are not reducble. Let G be a (2,3)-regular bpartte graph, and let L denote the vertces wth degree 2 (left vertces) and R the vertces wth degree 3 (rght vertces). Then E(G) = 2 L = 3 R, so R = 2k and L = 3k for some postve nteger k. We call a subset A of R wth k vertces that has the whole set L as ts neghbourhood a k-star cover of G. Clearly ths mples that all vertces of A 15

16 have mutually dsjont neghbours n L. We wll also need the noton of an f-factor and a result due to Ore. If f s an nteger valued functon on the set V (G) such that 0 f(v) d G (v) for each vertex v V (G), then a spannng subgraph F of G s called an f-factor of G f d F (v) = f(v) for each vertex v V (G). The followng theorem of Ore (see, e.g., [1], Theorem 7.2.2) characterzes bpartte graphs wth an f-factor. In ths theorem E(U,y) denotes the set of edges between a vertex set U and a vertex y, and f(u) = v U f(v). Theorem 21 Let G be a bpartte graph wth bpartton (L,R). Then G has an f-factor f and only f f(l) = f(r) and for any set U of R: f(u) y L mn(f(y), E(U,y) ). Wth the adopton of the above conventons and result we can prove the followng result for (2, 3)-regular bpartte graphs. Theorem 22 Let G be a (2,3)-regular bpartte graph. Then G s reducble f and only f pko(g) = 2 f and only f G has a k-star cover. Moreover, we can determne pko(g) n polynomal tme. Proof. Let G be a (2,3)-regular bpartte graph, and let L denote the vertces wth degree 2 and R the vertces wth degree 3. If G s reducble, then pko(g) = 2: t cannot be 1 snce L > R, and t cannot be larger than 2 snce n every round the degree of the vertces decreases by at least 1 and the vertces n R cannot elmnate each other snce R s an ndependent set. Ths clearly proves the frst equvalence of the statement. For the same reasons, f pko(g) = 2, then n the second round the remanng vertces of L have degree 1, and those vertces and the remanng vertces of R elmnate each other along a perfect matchng M. We now show that M = k. Frst of all, M k snce R = 2k and hence the vertces of R can elmnate at most 2k vertces of L n the frst round; secondly, M k snce otherwse all 3k vertces n L elmnate fewer than k from R n round 1, contradctng that every vertex of R has degree 3 n G. Snce the remanng k vertces of R are saved n round 1, all 3k left vertces elmnate together only R k = k rght vertces. Snce all vertces n R have degree 3, ths scheme corresponds to a k-star cover of G. Conversely, suppose G has a k-star cover A n R. Then B = R\A s also a k-star cover of G. Now we set f(x) = 1 for all x A L, f(x) = 2 for all x B. Then f(l) = 3k and f(r) = k +2k = 3k, so f(l) = f(r). For a set U R, f(u) = U A + 2 U B 3max( U A, U B ). Snce A s a k-star cover, all neghbours of the vertces of U A n L are dstnct, and the same holds for B and U B. Any neghbour y L of each of the vertces from U A or U B clearly has mn(f(y), E(U,y) ) 1 snce f(y) = 1 and y s a neghbour of a vertex of U. So y L mn(f(y), E(U,y) ) 3max( U A, U B ). Usng Theorem 21, we conclude that G has an f-factor. Consder the followng KO-scheme for G: n round 1, all vertces n L fre at A, whle vertces n A fre va matchng edges of the f-factor and vertces n B fre along one of the edges of the f-factor. After round 1, all remanng vertces at the rght sde are precsely the set B. Because of the f-factor n G and the frng n the frst round, the vertces of B form a perfect matchng M wth the remanng vertces n L. We use M to elmnate all remanng vertces n the second round. Ths completes the proof of the second equvalence of the statement. Determnng whether G has a k-star cover s a problem that can be solved n polynomal tme. Snce both A and B must be k-star covers, one can start by puttng one arbtrary vertex v of R n A, and then puttng all vertces of R that have a common neghbour wth v n B, and so on, untl all vertces have been allocated or a conflct occurs. 16

17 The cases of reducble (r,s)-regular bpartte graphs for other values of r and s do not seem to admt a smlar characterzaton. We leave them as nterestng open problems. 7 Conclusons In ths paper we have studed the computatonal complexty of problems related to the parallel knock-out number pko(g) of a graph G. We have shown that determnng whether pko(g) = 1 s polynomally solvable, whereas determnng whether pko(g) k (or pko(g) = k) s NP-complete for any fxed k 2 that s not part of the nput, even when restrcted to the class of bpartte graphs. We also showed that the latter problems restrcted to graphs wth bounded tree-wdth are solvable n lnear tme, by formulatng them n monadc second-order logc. Moreover, we studed some specal graph classes wth small parallel knock-out numbers. An nterestng open problem s the computatonal complexty of both decson problems when restrcted to planar graphs. Snce outer-planar graphs have bounded tree-wdth, both problems can be solved n lnear tme when restrcted to outer-planar graphs. Snce 4-connected planar graphs are hamltonan, pko(g) = 1 for a 4-connected planar graph G. From a result n [3] we can easly deduce that pko(g) 20log n for any reducble planar graph G on n vertces. References [1] A.S. Asratan, T.M.J. Denley, and R. Häggkvst (1998). Bpartte Graphs and ther Applcatons. Cambrdge Unversty Press, Cambrdge. [2] J.A. Bondy and U.S.R.Murty (1976). Graph Theory wth Applcatons. Macmllan, London and Elsever, New York. [3] H.J. Broersma, F.V. Fomn, R. Královč, and G.J. Woegnger (2007). Elmnatng graphs by means of parallel knock-out schemes, Dscrete Appled Mathematcs 155 (2), [4] H.J. Broersma and F. Göbel (1990). k-traversable graphs, Ars Combnatora 29A, [5] H.J. Broersma, M. Johnson, and D. Paulusma (2007). Upper bounds and algorthms for parallel knock-out numbers, Proceedngs of SIROCCO 2007: 14th Internatonal Colloquum on Structural Informaton and Communcaton Complexty, Lecture Notes n Computer Scence 4474, [6] B. Courcelle (1990). The monadc second-order logc of graphs. I. Recognzable sets of fnte graphs, Informaton and Computaton 85, [7] M.R. Garey and D.S. Johnson (1979). Computers and Intractablty: A Gude to the Theory of NP-Completeness. Freeman, San Francsco. [8] B. Jackson and N.C. Wormald (1990). k-walks, Australasan J. of Combnatorcs 2, [9] D.E. Lampert and P.J. Slater (1998). Parallel knockouts n the complete graph, Amercan Mathematcal Monthly 105,

18 [10] L. Lovász (1973). Coverng and colorng of hypergraphs, Proceedngs of the 4th Southeastern Conference on Combnatorcs, Graph Theory, and Computng, Utltas Mathematca, [11] W.T. Tutte (1947). The factorzaton of lnear graphs, J. London Math. Soc. 22,