A Linear-Time Algorithm for the Terminal Path Cover Problem in Cographs

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1 A Linear-Time Algorithm for the Terminal Path Cover Problem in Cographs Ruo-Wei Hung Department of Information Management Nan-Kai Institute of Technology, Tsao-Tun, Nantou 54, Taiwan Abstract Let G = (V, E) be a graph with vertex set V and edge set E and let T be a subset of V. A terminal path cover PC of G with respect to T is a set of pairwise vertex-disjoint paths of G, such that all vertices of G are visited by exactly one path of PC and all vertices in T are end vertices of paths in PC. The terminal path cover problem is to find a terminal path cover of G of minimum cardinality with respect to T. The path cover problem is a special case of the terminal path cover problem with T =. A graph is a cograph if it can be described by expressions involving operations of merging disjoint graphs and taking complements of graphs. Cographs have arisen in many different areas of applied mathematics and computer science and have been independently rediscovered by various researchers. In this paper, we show that the terminal path cover problem on cographs can be solved in linear time. Introduction All graphs considered in this paper are finite and undirected, without loops or multiple edges. Let G = (V, E) denote a graph with vertex set V and edge set E. For a graph G, V (G) and E(G) denote the vertex set and edge set of G, respectively. A path cover of a graph G is a collection of vertex-disjoint paths P = (V, E ),, P k = (V k, E k ) in G whose union is V (G), where V i and E i, for i k, are the vertex and edge sets of path P i, respectively, i.e., V i V j = for i j and k i= V i = V (G). The path cover problem is to find a path cover of a graph of minimum cardinality. It is evident that the path cover problem for general graphs is NP-complete since finding a path cover, consisting of a single path, corresponds directly to the Hamiltonian path problem [5]. The Hamiltonian path problem on some special classes of graphs, including bipartite graphs [0], split graphs [6], chordal bipartite graphs [6], strong chordal split graphs [6], undirected path graphs [3], and directed path graphs [8], has been shown to be NP-complete. Hence the path cover problem on these above graphs and their superclasses of graphs is also NP-complete. However, it admits polynomial time algorithms when the input is restricted to be in some special classes of graphs, including trees [4], block graphs [5], interval graphs [, 8], circular-arc graphs [7], cographs [9, 0, 8, 3], bipartite distance-hereditary graphs [36], bipartite permutation graphs [34], and cocomparability graphs []. The path cover problem has found some applications in database designing, networking, establishing ring protocol, VLSI designing, code optimization [6], and mapping parallel programs into parallel architectures [4, 3]. Let P be a path of a graph G and let T be a subset of V (G). The first and last vertices visited by P are called the path-start and path-end of P, denoted by start(p ) and end(p ), respectively. Both of them are end vertices of P. A terminal path cover of a graph G with respect to T is a path cover of G such that all vertices in T are end vertices of paths in the path cover. A minimum terminal path cover of G with respect to T is a terminal path cover of G of minimum cardinality. Denote by π(g, T ) the cardinality of a minimum terminal path cover of G with respect to T. For simplicity, we will use π(g) to denote π(g, T ) if T =. Given a graph G and a subset T of V (G), the terminal path cover problem is to find a minimum terminal path cover of G of size π(g, T ). We call T the terminal set of G, the vertices in T the terminals, and the other vertices free vertices. The path cover problem is a special case of the terminal path cover problem with T =. The terminal path cover problem can be applied to the

2 applications of the path cover problem when some vertices corresponding to the applications are restricted to be terminals. Cographs (also called complement-reducible graphs) are defined as the class of graphs formed from a single vertex under the closure of the operations of union and complement. Cographs were introduced by Lerchs [], who studied their structural and algorithmic properties and enumerated the class. Names synonymous with cographs include D -graphs, P 4 restricted graphs, and Hereditary Dacey graphs. Several characterizations of cographs are known. For example, it is shown that G is a cograph if and only if G contains no P 4 (a path consisting of four vertices) as an induced subgraph [0]. Cographs have arisen in many disparate areas of mathematics and have been independently rediscovered by various researchers. Cographs can be recognized in linear time []. The class of cographs forms a subclass of distance-hereditary graphs [0, ] and permutation graphs, and is a superclass of threshold graphs and complete-bipartite graphs. In addition, further properties and optimization problems in these graphs have been studied [, 4, 5, 3, 4, 9,, 7, 30, 3, 33, 35, 37, 38]. Jung and Nicolai proposed linear time algorithms to solve the special cases of the terminal path cover problem with T = and T =, respectively, on cographs [8, 9]. In this paper, we show that the terminal path cover problem whose T is not restricted to on cographs can be solved in linear time. Preliminaries In this section, we define some notations and operations. Definition. [0, ] The class of cographs can be defined by the following recursive definition: () A graph consisting of a single vertex and no edges is a cograph. () If G L = (V L, E L ) and G R = (V R, E R ) are cographs, then the union G of G L and G R, denoted by G = G L G R = (V L V R, E L E R ), is a cograph. In this case, we say that G is formed from G L and G R by a union operation. (3) If G L = (V L, E L ) and G R = (V R, E R ) are cographs, then the joint G of G L and G R, denoted by G = G L G R = (V L V R, E L E R Ê), is a cograph, where Ê = {(u, v) u V L and v V R }. In this case, we say that G is formed from G L and G R by a joint operation. c a e f b (a) d i h a DT( G L ) c b d DT( G) f e (b) h DT( G R ) Figure : (a) A cograph G and a terminal set T = {a, f, i} of G, and (b) a decomposition tree DT (G) of G, where full vertices indicate the terminals in G. A cograph G can be represented by a rooted binary tree DT (G), called a decomposition tree [7, 0]. The leaf nodes of DT (G) represent the vertices of G. Each internal node of DT (G) is labeled by either or. The cograph corresponding to a -labeled (resp. -labeled) node v in DT (G) is obtained from the cographs corresponding to the children of v in DT (G) by means of a union (resp. joint) operation. A decomposition tree of a cograph can be constructed as follows: Definition. [7] The decomposition tree DT (G) of a cograph G consisting of a single vertex v is a tree of one node labeled by v. If G is formed from G L and G R by a union (resp. joint) operation, then the root of the decomposition DT (G) is a node labeled by (resp. ) with the roots of DT (G L ) and DT (G R ) being the children of the root of DT (G), respectively. Notice that the decomposition tree DT (G) of a cograph G is a rooted and unordered binary tree. Exchanging the left and right children of an internal node in DT (G) will be also a decomposition tree of G. For instance, given a cograph G shown in Figure (a), the decomposition tree DT (G) of G is shown in Figure (b). Theorem.. [7, 0] A decomposition tree DT (G) of a cograph G = (V, E) can be constructed in O( V + E ) time. A cograph is not connected if the root of its corresponding decomposition tree is a -labeled node. Hence, we assume that the root of the corresponding decomposition tree is a -labeled node. In the rest of the paper, assume that G = (V, E) is a cograph and is formed from G L and G R by either a union operation or a joint operation. We use V L and V R to denote the vertex sets of G L and i

3 G R, respectively. In other words, V = V L V R and V L V R =. Notice that every vertex in V L is adjacent to all vertices in V R if G is formed from G L and G R by a joint operation. Definition 3. A path of a cograph G is called terminal path if it visits at least two vertices and both of its end vertices are terminals. A path of G is called semi-terminal path if exactly one end vertex of it is a terminal. A path of G is called non-terminal path if neither of its end vertices is a terminal. By the above definition, a semi-terminal path may consist of only one vertex which is a terminal, a non-terminal path may visit only one free vertex, and a terminal path must contain at least two vertices which are distinct terminals. Definition 4. A terminal path cover PC of a cograph G with respect to T is called a (n, s, t)- constrained path cover if () PC = n + s + t; () there are exactly n non-terminal paths in PC; (3) there are exactly s semi-terminal paths in PC, and (4) all other paths in PC are terminal paths. A path in a terminal path cover of a cograph G is clearly either a non-terminal path, a semiterminal path or a terminal path of G. Hence, the following proposition immediately holds: Proposition.. Let G be a cograph and let T be a terminal set of G. Then, the following statements hold: ()PC is a terminal path cover of G with respect to T if and only if it is a constrained path cover of G. ()If G has a (n, s, t)-constrained path cover, then T = s + t. Definition 5. Let PC be a (n, s, t)-constrained path cover of G. We say that PC is a minimum constrained path cover of G if π(g, T ) = n + s + t. Denote by N G (PC), S G (PC), and T G (PC), the subsets of PC containing all non-terminal paths, all semi-terminal paths, and all terminal paths in PC, respectively. Define f G (PC) = n + s. Definition 6. Let P be a set of vertex-disjoint paths in G. Denote by F (P) the set of free vertices in P. Note that P might not be a terminal path cover of G. Let P be a set of vertex-disjoint paths with n non-terminal paths and s semi-terminal paths. By definition, F (P) n. Given P and a number n, where n n F (P), we can easily obtain a set P of vertex-disjoint paths with n non-terminal paths and s semi-terminal paths by splitting paths in P. Hence, the following proposition is clearly true: Proposition.3. For a set P of vertex-disjoint paths with n non-terminal paths and s semiterminal paths, there exists a set P of vertexdisjoint paths with n non-terminal paths and s semi-terminal paths such that n n F (P). Definition 7. Let PC be a constrained path cover of a cograph G, P be a path in PC, and let U be either V L or V R. A subpath P of path P is called a U-subpath of P if P visits vertices in U only. A U-subpath of P is U-maximal if it is not a proper subpath of any U-subpath of P. Denote by U(PC), the set of all U-maximal subpaths of all paths in PC. Let G be a cograph formed from G L = (V L, E L ) and G R = (V R, E R ) by either a union or a joint operation and let PC be a constrained path cover of G. By definition, V L (PC) and V R (PC) are constrained path covers of G L and G R, respectively. It is easy to see that V L (PC) V R (PC) is a constrained path cover of G and PC = V L (PC) V R (PC) if G = G L G R. Assume that G = G L G R below. Consider the process to generate PC by adding edges between V L (PC) and V R (PC), one edge at a time, to the current constrained path cover starting from V L (PC) V R (PC). The process is repeated until PC is obtained. Observe that, after adding one edge, the number of paths in the resulting constrained path cover is decreased by one. The number of edges adding to V L (PC) V R (PC) is clearly at most min{f GL (V L (PC)), f GR (V R (PC))}. Therefore, we have the following proposition: Proposition.4. Assume that G is a cograph formed from G L = (V L, E L ) and G R = (V R, E R ) by either a union or a joint operation and that PC is a constrained path cover of G. Then, V L (PC) + V R (PC) PC V L (PC) + V R (PC) min{f GL (V L (PC)), f GR (V R (PC))}. For instance, let G = G L G R be a cograph shown in Figure, T = {a, f, i} be a terminal set of G, and let PC = {P = a h c e b d f, P = i} be a constrained path cover of G, where P and P are terminal path and semiterminal path of G, respectively. Then, PC is a (0,, )-constrained path cover of G, V L (PC) = {a, c e b d f} is a (0,, 0)-constrained path cover of G L, and V R (PC) = {h, i} is a (,, 0)-constrained path cover of G R. By definition, f GL (V L (PC)) = and f GR (V R (PC)) =

4 3. Hence, min{f GL (V L (PC)), f GR (V R (PC))} =, V L (PC) + V R (PC) = 4, and V L (PC) + V R (PC) min{f GL (V L (PC)), f GR (V R (PC))} =. By Proposition.4, 4 PC =. Let Pv u denote a path of a graph with start(pv u ) = u and end(pv u ) = v. We allow that u = v only in the case that Pv u contains exactly one vertex. For two vertex-disjoint paths Pv u and Py x such that v is adjacent to x, let Pv u Py x denote the concatenation of Pv u with Py x ; that is, Py u = Pv u Py x and Py u visits every vertex of Pv u and Py x exactly once. We then define some operations on two distinct sets of vertex-disjoint paths in the following. Definition 8. ( operation) Let P = {Pv u,pv u,, Pv up p } and Q={Py x,py x,, Py xq q } be two sets of vertex-disjoint paths such that p q, v i is adjacent to x i, and both v i and x i are not terminals for i q. For an integer k q, define k to be the operation on P and Q, denoted by P k Q, that constructs a set of paths {Pv u Py x, Pv u Py x,, P u k v k P x k y k, P u k+ v k+,, Pv up p, P x k+ y k+,, Py xq q }. Figure (a) depicts the operation. Definition 9. ( operation) Let P ={Pv u,pv u,, Pv up p } and Q ={Py x,py x,, Py xq q } be two sets of vertex-disjoint paths such that p > q, both end vertices of each path in P are adjacent to all end vertices of paths in Q, and all end vertices of paths in P and Q are not terminals. For an integer k q, define k to be the operation on P and Q, denoted by P k Q, that generates a set of paths {Pv u Py x Pv u Py x P u k v k P x k y k P u k+ v k+, P u k+ v k+,, Pv up p, P x k+ y k+,, Py xq q }. The operation is revealed in Figure (b). Let P and Q be defined as the above definition except that p q k +. For simplicity, we will use P( k, )Q to denote the set of paths {Pv u Py x Pv u Py x P u k v k P x k y k P u k+ v k+ P x k+ y k+, P u k+ v k+,, Pv up p, P x k+ y k+,, P xq y q } Definition 0. ( operation) Let P ={Pv u,pv u,, Pv up p } and Q ={Py x,py x,, Py xq q } be two sets of vertex-disjoint paths such that both end vertices of each path in P are adjacent to all end vertices of paths in Q, all end vertices of paths in P are not terminals, and x i is a terminal and y i is a free vertex for i q. For an integer k min{p, q }, define k to be the operation on P and Q, denoted by P k Q, that generates a set of paths {Py x Pv u Px y, Py x3 3 Pv u Px y4 4,, P x k y k P y k x k, P u k+ v k+, P u k+ v k+,, Pv up p, P x k+ y k+,, P u k v k P xq y q }. Figure (c) shows the operation. u v u v u k v k u k+ v k+ u p v p P.. (a) Q x y x y x k y k x k+ y k+ x q y q u v u v u k v k u k+ v k+ u p v p P.. (b) Q x y x y x k y k x k+ y k+ x q y q u v u k v k u k+ v k+ u p v p.. (c) x y x y x k- y k- x k y k x k+ y k+ Figure : Operations defined on two sets P and Q of vertex-disjoint paths, where (a) P k Q, (b) P k Q, and (c) P k Q, where full vertices indicate terminals. 3 The Terminal Path Cover Problem in Cographs In [9, 0, 8, 3], they showed that the path cover problem on cographs can be solved in linear time. Note that the path cover problem is a special case of the terminal path cover problem where the terminal set of the input graph is empty. Lemma 3.. [9, 0, 8, 3] Let G L = (V L, E L ) and G R = (V R, E R ) be two cographs. Then, the following two statements hold: () If G = G L G R, then π(g) = π(g L )+π(g R ). () If G = G L G R, then π(g) = max{, π(g L ) V R, π(g R ) V L }. In addition, we can easily see that the following lemma holds: Lemma 3.. Let G L and G R be two cographs with terminal sets T L and T R, respectively. If G = G L G R, then π(g, T L T R ) = π(g L, T L ) + π(g R, T R ). By the above two lemmas, we can easily compute π(g, T ) if G = G L G R or T = T L T R =. In the rest of the paper, we focus on computing π(g, T ) under that G = G L G R and T. We first find the relation between f G (PC) and f G ( PC), where PC and PC are constrained path covers of a cograph G, PC is a minimum constrained path cover of G but PC is not mini- P Q x q y q

5 mum. Recall that f G (PC) = n + s for a (n, s, t)- constrained path cover PC of G. Lemma 3.3. Let G be a cograph with terminal set T. Suppose that PC and PC are minimum constrained path covers of G and that PC is a constrained path cover of G which is not minimum. Then, f G (PC ) = f G (PC ) and f G (PC ) < f G ( PC). Proof. Assume that PC, PC, and PC are (n, s, t )-constrained, (n, s, t )-constrained, and ( n, ŝ, t)-constrained path covers of G, respectively. By definition, f G (PC ) = n + s, f G (PC ) = n + s, and f G ( PC) = n + ŝ. We will prove the lemma by showing that n + s = n + s and n + s < n + ŝ. By Proposition., we obtain that T = s + t = s + t = ŝ + t. () Since both PC and PC are minimum constrained path covers of G and PC is not a minimum constrained path cover of G, we get that n + s + t = n + s + t, () n + s + t < n + ŝ + t. (3) By Eq. (), n = n + s + t s t. Hence, n +s = (n +s +t s t )+s = n +s + (s +t s t ). By Eq. (), s +t s t = 0. Hence, n + s = n + s. By Eq. (3), n + s < n + ŝ + t t. Hence, n + s < ( n + ŝ + t t ) s = n + ŝ + (ŝ + t t s ). By Eq. (), we have that ŝ + t t s = 0. Hence, n + s < n + ŝ. Corollary 3.4. Let G = (V, E) be a cograph with terminal set T. Suppose that PC is a minimum constrained path cover of G and that PC is a constrained path cover of G. Then, f G (PC) f G ( PC) V \ T + T. Proof. Clearly, f G ( PC) V \ T + T. By Lemma 3.3, f G (PC) f G ( PC). Hence the corollary holds. Without of loss generality, we assume the following statements hold for all lemmas in the rest of the paper. Assumption. () G L = (V L, E L ) and G R = (V R, E R ) are two cographs with terminal sets T L and T R, respectively; () G = G L G R is a cograph with terminal set T = T L T R ; (3) T and T L T R, and (4) PC L (resp. PC R ) is a minimum and (n L, s L, t L )-constrained (resp. (n R, s R, t R )-constrained) path cover of G L (resp. G R ) satisfying that there exists no minimum constrained path cover of G L (resp. G R ) with semiterminal path if s L = 0 (resp. s R = 0). In the following, we will show that given π(g L, T L ) and π(g R, T R ), π(g, T ) can be computed in constant time. We first obtain the lower bound of π(g, T ) in the following lemma: Lemma 3.5. π(g, T ) max{ T, π(g L, T L ) V R \ T R, π(g R, T R ) V L \ T L }. Proof. Clearly, π(g, T ) T. Now, we will prove that π(g, T ) π(g L, T L ) V R \ T R. Assume by contradiction that π(g, T ) < π(g L, T L ) V R \ T R. Then, G has a minimum constrained path cover PC of size p with p < π(g L, T L ) V R \ T R. Let G = G \ (V R \ T R ). Clearly, π(g, T ) π(g L, T L ). Consider removing from PC all vertices in V R \ T R. What results is a set PC of vertex-disjoint paths which is clearly a constrained path cover of G with respect to T. Since the removal of a vertex in V R \T R will increase the number of paths by at most one, we can obtain a constrained path cover of G of size at most p + V R \ T R and hence, PC p + V R \ T R. Therefore, p + V R \ T R PC π(g, T ). Since p + V R \ T R π(g, T ) and π(g, T ) π(g L, T L ), we get that p π(g L, T L ) V R \ T R. It contradicts that p < π(g L, T L ) V R \ T R. Thus, π(g, T ) π(g L, T L ) V R \ T R. Using similar arguments as those for proving that π(g, T ) π(g L, T L ) V R \ T R, we can prove that π(g, T ) π(g R, T R ) V L \ T L. Hence, π(g, T ) max{ T, π(g L, T L ) V R \ T R, π(g R, T R ) V L \ T L }. Next, we would like to find the upper bound of π(g, T ). In other words, we show that π(g, T ) max{ T, π(g L, T L ) V R \T R, π(g R, T R ) V L \ T L } except the special case of s L = T R = 0 and n L = V R. The following lemma shows that π(g, T ) = T + if s L = T R = 0 and n L = V R : Lemma 3.6. Assume that s L = T R = 0 and n L = V R. Then, π(g, T ) = T +. Proof. By Assumption, T = T L. By Proposition., T L = s L + t L. Since s L = 0, we have that t L = T > 0. We first prove that π(g, T ) T +. Assume by contradiction that π(g, T ) = T. Suppose that PC is a minimum constrained

6 path cover of G of size T. Then, V L( PC) and V R ( PC) are ( n L, ŝ L, t L )-constrained and ( n R, ŝ R, t R )-constrained path covers of G L and G R, respectively. Since T R = 0, ŝ R = t R = 0. By definition, we have that n R n R V R = n L. (4) Since PC = T, we have that PC = T G ( PC). There are two types of paths in PC: type- paths are those paths visiting vertices in V L only; and type- paths are those paths visiting vertices in both V L and V R with both end vertices in T L. The number of paths of type- is equal to t L. Suppose that f GR (V R ( PC)) < f GL (V L ( PC)). Then, n R < n L + ŝ L. By Proposition.4, PC = T L = ŝl+ t L = ŝl + t L V L ( PC) + V R ( PC) f GR (V R ( PC)) = n L + ŝ L + t L n R n R. Hence, n R n L + ŝ L and a contradiction occurs. On the other hand, suppose that f GL (V L ( PC)) < f GR (V R ( PC)). Then, n L + ŝ L < n R. By Proposition.4, PC = ŝl + t L V L ( PC) + V R ( PC) f GL (V L ( PC)) = n L + ŝ L + t L + n R n L ŝ L. Hence, n L +ŝ L n R and a contradiction occurs too. Therefore, f GL (V L ( PC)) = f GR (V R ( PC)) and we get that n L + ŝ L = n R. (5) A type- path P is clearly a terminal path of G with both end vertices in T L since PC = TL. Clearly, P contains exactly two semi-terminal paths of G L. Since V R, there exists at least one type- path in PC. Thus, ŝ L 0. By Assumption, V L ( PC) is not a minimum constrained path cover of G L. By Lemma 3.3, we obtain that n L < n L + ŝ L. (6) Combining Eqs. (4) (6), we get that n L < n L + ŝ L = n R V R = n L and hence, a contradiction occurs. Hence, π(g, T ) T, i.e., π(g, T ) T +. Next we construct a constrained path cover of G of size T + from PC L and V R. Let P = N GL (PC L )( VR, )V R. Then, P contains only a non-terminal path P n of G with one end vertex in V L and the other in V R. Let P t = v t P s be a path in T GL (PC L ), where v t T L. Let PC t = T GL (PC L ) \ {P t }, PC s = {P s, v t P n }, and let PC = PC s PC t. Then, PC is a (0,, T )- constrained path cover of G. Thus, π(g, T ) T +. It follows immediately from the above arguments that π(g, T ) = T +. For the above case of s L = T R = 0 and n L = V R, we construct a minimum constrained path cover of G with two semi-terminal paths. We then make the following assumption: Assumption. The equations of s L = T R = 0 and n L = V R do not hold at the same time. The above assumption implies that n L V R if s L = T R = 0. Let PC L and PC R be constrained path covers of G L and G R, respectively. By Corollary 3.4, n L +s L = f GL (PC L ) f GL ( PC L ) V L \T L + T L and n R + s R = f GR (PC R ) f GR ( PC R ) V R \T R + T R. Considering the relation between f GL ( PC L ) and f GR ( PC R ), we have the following four cases: Case I: V L \ T L + T L < n R + s R ; Case II: V R \ T R + T R < n L + s L ; Case III: n L + s L n R + s R V L \ T L + T L ; Case IV: n R +s R n L +s L V R \T R + T R. For Cases I and II, we will construct constrained path covers of G of sizes π(g R, T R ) V L \ T L and π(g L, T L ) V R \ T R, respectively. For the other cases, we will construct a constrained path cover of G of size T. Note that the conditions appeared in Lemma 3.6 hold only for Case III or Case IV. Following Lemma 3.5 and Lemma 3.7, we obtain that π(g, T ) = max{ T, π(g L, T L ) V R \ T R, π(g R, T R ) V L \ T L } if Assumption holds. Lemma 3.7. The following statements hold: ()If V L \ T L + T L < n R + s R, then max{ T, π(g L, T L ) V R \T R, π(g R, T R ) V L \ T L } = π(g R, T R ) V L \ T L. ()If V R \ T R + T R < n L + s L, then max{ T, π(g L, T L ) V R \T R, π(g R, T R ) V L \ T L } = π(g L, T L ) V R \ T R. Proof. We prove Statement () by showing that (π(g R, T R ) V L \ T L ) > T and π(g R, T R ) V L \ T L > π(g L, T L ) V R \ T R. We first prove that (π(g R, T R ) V L \ T L ) > T. By definition, we have that π(g R, T R ) = n R + s R + t R. (7) By assumption of the lemma, we obtain that V L \ T L + T L < n R + s R. (8) By Proposition., T L = s L + t L and T R = s R +t R. Combining Eqs. (7) and (8), we get that

7 (π(g R, T R ) V L \ T L ) = (n R + s R + t R V L \ T L ) = n R +s R + T R V L \T L > ( V L \T L + T L ) + T R V L \ T L = T L + T R = T. Next, we prove that π(g R, T R ) V L \ T L > π(g L, T L ) V R \ T R. By definition, we have that π(g L, T L ) = n L +s L +t L V L \T L + T L. (9) Combining Eqs. (8) and (9), we obtain that π(g L, T L ) + V L \ T L < n R + s R. (0) By Corollary 3.4, n R + s R V R \ T R + T R. By Eq. (8), we obtain that V L \ T L + T L < V R \ T R + T R. () Combining Eqs. (0) and (), we get that (π(g R, T R ) V L \ T L π(g L, T L ) + V R \ T R ) = (n R +s R +t R V L \T L π(g L, T L )+ V R \T R ) = n R + s R + T R V L \ T L π(g L, T L ) + V R \ T R > (π(g L, T L ) + V L \ T L ) + T R V L \ T L π(g L, T L )+ V R \T R = π(g L, T L ) V L \T L + V R \T R + T R > π(g L, T L ) V L \T L +( V L \ T L + T L ) = V L \ T L + T L π(g L, T L ) 0. Hence, π(g R, T R ) V L \ T L > π(g L, T L ) V R \ T R. By symmetry, Statement () can be proved by using arguments similar to those used in proving Statement (). Lemma 3.8. If V L \T L + T L < n R +s R, then π(g, T ) = π(g R, T R ) V L \ T L. Proof. We will prove this lemma by showing how to construct a constrained path cover of G of size π(g R, T R ) V L \ T L. By Assumption, T L T R = s R + t R. By assumption of the lemma, n R > V L \ T L + T L s R and hence, n R V L \ T L + TL sr +. We then construct from V L and PC R a constrained path cover of G of size π(g R, T R ) V L \ T L through the following procedure:. partition T L into two sets TL α and T β L such that TL α = s R and T β L = T L s R ;. PC t = TL α s R S GR (PC R ); 3. P N = N GR (PC R ) VL\T L (V L \ T L ), where P N = n R V L \ T L TL sr + and every path in P N is a non-terminal path of G with both end vertices in V R ; 4. partition P N into two sets P β N and Pγ N such that P β N = TL sr TL sr, where P γ N 5. PC t = P β N T L s R T β L and P γ N = n R V L \ T L ;, where there are exactly TL sr terminal paths in PC t ; 6. if T L s R is odd, i.e., there exists an isolated terminal v t in PC t, then 7. let P be a non-terminal path in P γ N ; 8. PC t = PC t \ {v t }; PC n = P γ N \ { P }; PC s = {v t P }; 9. PC t = PC t PC t T GR (PC R ); 0.else /* T L s R is even*/. let P t be a terminal path in PC t PC t such that P t = v t P s and v t T L ;. let P n be a non-terminal path in P γ N ; 3. PC n = P γ N \{P n}; PC s = {P s, v t P n }, and PC t = PC t PC t T GR (PC R ) \ {P t }; 4.PC = PC n PC s PC t. Following the above procedure, we construct a (n R V L \ T L TL sr,, s R + t R + TL sr )-constrained path cover PC of G if T L s R is odd; and a (n R V L \ T L TL sr,, s R + t R + TL sr )-constrained path cover of G otherwise. Hence, we can construct from V L and PC R a constrained path cover PC of G of size π(g R, T R ) V L \T L with at least a semi-terminal path. That is, π(g, T ) π(g R, T R ) V L \T L. By Lemma 3.5 and Statement () of Lemma 3.7, we get that π(g, T ) π(g R, T R ) V L \ T L. Hence, π(g, T ) = π(g R, T R ) V L \ T L. Lemma 3.9. If V R \T R + T R < n L +s L, then π(g, T ) = π(g L, T L ) V R \ T R. Proof. By Assumption, T L T R. By assumption of the lemma, n L + s L > V R \ T R + T R and hence, n L V R \ T R + TR sl +. Suppose that T R > s L. By using arguments similar to those used in proving Lemma 3.8, we can construct from V R and PC L a constrained path cover PC of G of size π(g L, T L ) V R \ T R such that PC contains at least one semi-terminal path. In the following, suppose that T R s L. Then, n L V R \ T R sl TR +. For the case of V R \ T R < sl TR, n L = 0 and we can easily obtain a minimum constrained path cover of G with semi-terminal paths as follows: () partition S GL (PC L ) into two sets S α L and Sβ L such that S α L = T R and S β L = s L T R ; () P T = S α L T R T R ; (3) P S = (V R \T R ) VR\T R S β L, and (4) PC = P T P S N GL (PC L ). Note that there exists at least one semi-terminal path in P S since V R \ T R < sl TR. In the following, we only consider the case that V R \ T R sl TR. We will construct from V R and PC L a (n, s, t)- constrained path cover of G of size π(g L, T L ) V R \ T R such that s = 0 if s L = T R = 0; and

8 s > 0 otherwise. The following procedure will construct such a constrained path cover:. partition S GL (PC L ) into two sets SL α and Sβ L such that SL α = T R and S β L = s L T R ;. PC t = SL α T R T R ; 3. partition V R \ T R into two sets V β R and V γ R such that V β R = sl TR and V γ R = V R \ T R sl TR ; 4. PC t = V β R V β R Sβ L ; 5. let P s be the semi-terminal path in PC t if s L T R is odd, and let P s = otherwise; let PC t = PC t \ {P s }; 6. P N = N GL (PC L ) V γ R V γ R, where P N = n L V R \ T R + sl TR and every path in P N is a non-terminal path of G with both end vertices in V L ; 7. PC n = P N ; PC s = {P s }; PC t = PC t PC t T GL (PC L ); 8. if s L T R is even and T R 0, then 9. let P t = v t Pŝ be a path in PC t, where v t T R and Pŝ S α L ; 0. let P n be a path in PC n ;. PC n = PC n \ {P n }; PC s = {Pŝ, v t P n }; PC t = PC t \ {P t }, and goto line 6;.if s L T R is even and s L 0, then /* T R = 0 */ 3. let P t = P s v f P s be a path in PC t, where P s, P s S β L, and v f V R ; 4. let P n be a path in PC n ; 5. PC n = PC n \{P n }; PC s = {P s v f P n, P s }; PC t = PC t \ {P t }; 6.PC = PC n PC s PC t. Following the above procedure, we obtain a constrained path cover PC of G of size π(g L, T L ) V R \ T R satisfying that it contains at least one semi-terminal path except the condition of s L = T R = 0. Hence, π(g, T ) π(g L, T L ) V R \ T R. By Lemma 3.5 and Statement () of Lemma 3.7, π(g, T ) π(g L, T L ) V R \ T R. Hence, π(g, T ) = π(g L, T L ) V R \ T R. By the above lemma, we can construct a minimum constrained path cover of G with at least one semi-terminal path except that V R \T R + T R < n L + s L and s L = T R = 0 hold together. Next, we will show that there exists no minimum constrained path cover of G with semi-terminal path if V R \ T R + T R < n L + s L and s L = T R = 0. Lemma 3.0. Assume that s L = T R = 0 and V R < n L. Then, there exists no minimum constrained path cover of G with semi-terminal path. Proof. By Assumption, T = T L > 0. Assume by contradiction that there exists a minimum and ( n, ŝ, t)-constrained path cover PC of G with ŝ 0. Then, V L ( PC) and V R ( PC) are ( n L, ŝ L, t L )-constrained and ( n R, 0, 0)-constrained path covers of G L and G R, respectively. Clearly, n R V R. Since ŝ 0, we can see that ŝ L 0. By Assumption, V L ( PC) is not a minimum constrained path cover of G L. Hence, we have that n L + ŝ L + t L > n L + t L. () By Lemma 3.9, PC = π(g L, T L ) V R = n L + s L V R. By Proposition.4, PC V L ( PC) + V R ( PC) f GR (V R ( PC)) = V L ( PC) + V R ( PC) n R = n L + ŝ L + t L n R. By Eq. (), we get that n L + s L V R = PC n L + ŝ L + t L n R > n L + t L n R. Thus, V R < n R and a contradiction occurs. Therefore, there exists no minimum constrained path cover of G with semiterminal path. Next, we will prove that under the condition of Assumption, π(g, T ) = T if n L + s L n R + s R V L \ T L + T L or n R + s R n L + s L V R \ T R + T R. Lemma 3.. Suppose that Assumption holds. If n L + s L = n R + s R, then π(g, T ) = T. Proof. By Proposition., T L = s L + t L and T R = s R + t R. By assumption of the lemma, s L + s R = (n R n L + s R ). Thus, T = T L + T R = s L + s R + (t L + t R ) = (n R n L + s R + t L + t R ) and hence, T is even. By Corollary 3.4, n R + s R V R \ T R + T R. Clearly, if n R = 0 and s R = 0, then n L = s L = 0 and hence, T GL (PC L ) T GR (PC R ) forms a (0, 0, T )- constrained path cover of G. In the following, assume that n R 0 or s R 0. We will construct a constrained path cover of G of size T from PC L and PC R in the following. Consider the following cases: Case : s L = s R. In this case, n L = n R. There are two subcases: Case.: s R = 0. By assumption, n R 0. Suppose that t R = 0; i.e., T R =. By Assumption, n L V R and hence n R V R. Hence, we can easily obtain a (n R +, 0, 0)-constrained path cover PC R of G R consisting of n R + nonterminal paths. Let P N = PC R nl N GL (PC L ). Then, P N contains only a non-terminal path P of G with both end vertices in V R. Let P t = v t P t be a terminal path in T GL (PC L ), where v t T L.

9 By Assumption, T = T L > 0 and hence P t exists. Let PC = T GL (PC L ) \ {P t } {v t P P t}. Then, PC is a constrained path cover of G of size T. On the other hand, suppose that t R 0. Let P N = N GR (PC R )( nl, )N GL (PC L ). Then, P N contains only a non-terminal path P of G with one end vertex in V L and the other in V R. By Assumption, T L T R and hence t L t R. Hence, there exist two paths P l = v l P l and P r = v r P r in T GL (PC L ) and T GR (PC R ), respectively, where v l T L and v r T R. Let PC = (T GL (PC L ) \ {P l }) (T GR (PC R ) \ {P r }) {P l P r, v l P v r }. Then, PC is a constrained path cover of G of size T. Case.: s R 0. Let P T = S GL (PC L ) sr S GR (PC R ). Let P l and P r be two semi-terminal paths in P T, where P l S GL (PC L ) and P r S GR (PC R ), and let P T = P T \ {P l, P r }. Let P N = N GR (PC R )( nl, )N GL (PC L ). Then, P N contains only a non-terminal path P of G with one end vertex in V L and the other in V R. Let PC = P T {P l P P r } T GL (PC L ) T GR (PC R ). Then, PC is a constrained path cover of G of size T. Case : s L > s R. In this case, s L s R = (n R n L ) > 0 and hence, n R n L > 0. We first partition S GL (PC L ) into two sets SL α and Sβ L such that SL α = s R and S β L = s L s R. Then we perform the following procedure: () PC t = SL α s R S GR (PC R ); () P N = N GR (PC R ) nl N GL (PC L ), where P N = n R n L ; (3) PC t = P N nr n L S β L, and (4) PC = PC t PC t T GL (PC L ) T GR (PC R ). Then, PC is a constrained path cover of G of size T. Case 3: s L < s R. By symmetry, we can obtain a constrained path cover of G of size T by arguments similar to those used in proving Case. Lemma 3.. Suppose that Assumption holds. Assume that n R = n L + sl sr if s L s R ; and n L = n R + sr sl otherwise. Then, π(g, T ) = T. Proof. First consider the case of s L s R. Then, n R = n L + sl sr. If s L s R is even, then n L + s L = n R + s R and hence, we can construct a constrained path cover of G of size T from PC L and PC R by Lemma 3.. Suppose that s L s R is odd. Let P s be a path in S GL (PC L ), PCL = PC L \{P s }, and let G = G\{P s }. By using similar arguments for proving Lemma 3., we can obtain a constrained path cover PC of G of size T from PC L and PC R. Let PC = PC {P s }. Then, PC forms a constrained path cover of G of size T. We next consider the case of s L < s R. Then, n L = n R + sr sl. If s R s L is even, then n L + s L = n R + s R and hence, we can construct a constrained path cover of G of size T from PC L and PC R by Lemma 3.. Suppose that s R s L is odd. Let s R = s R. Then, s R s L is even. The case of s R > s L can be proved by using similar arguments for proving Lemma 3.. Assume that s R = s L below. Then, s L = s R and n L = n R +. We then construct a constrained path cover of G of size T from PC L and PC R as follows: () let P s be a path in S GR (PC R ); () S R = S GR (PC R ) \ {P s }; (3) PC t = S R sl S GL (PC L ); (4) P N = N GL (PC L ) nr N GR (PC R ), where P N contains only a non-terminal path P of G with both end vertices in V L ; (5) P s = P s P ; (6) PC = PC t T GL (PC L ) T GR (PC R ) {P s }. Then, PC is a constrained path cover of G of size T. Lemma 3.3. Suppose that Assumption holds. If n L + s L < n R + s R V L \ T L + T L, then π(g, T ) = T. Proof. By Proposition., T L = s L +t L. Hence, n L + s L < n R + s R V L \ T L + s L + t L. Then, we have that n L + sl sr n R V L \ T L + sl sr + t L, n L < n R + sr sl V L \ T L + t L, ifs L s R ; (3) ifs L < s R. (4) We prove the lemma by showing how to construct a constrained path cover of G of size T from PC L and PC R, where PC L may not be a minimum constrained path cover of G L. Consider the following cases: Case : s L s R. Suppose that n R = n L + sl sr. By Lemma 3., we can obtain a constrained path cover of G of size T. In the following, assume that n R > n L + sl sr. By Eq. (3), n L < n R sl sr V L \ T L + t L. Let N S = N GL (PC L ) S GL (PC L ). Recall that F (N S) is the set of free vertices in N S. There are two subcases: Case.: F (N S) n R sl sr. In this subcase, F (N S) n R sl sr > n L. By Proposition.3, we can obtain from N S a set

10 N S of vertex-disjoint paths with n L non-terminal paths and s L semi-terminal paths such that n L = n R sl sr if s L 0; and n L = n R otherwise. Note that if s L = 0, then s R = 0 since s L s R. Let PC L = N S T GL (PC L ). Then, PC L is a ( n L, s L, t L )-constrained path cover of G L. Let N L = N GL ( PC L ) and ŜL = S GL ( PC L ). We can obtain a constrained path cover PC of G of size T from PC L and PC R through the following procedure:. if s L = 0, then /* n L = n R */. P N = N GR (PC R ) nr N L, where N L = n R and P N contains only a non-terminal path P of G with both end vertices in V R ; 3. let P t = v t P t be a terminal path in T GL ( PC L ), where v t T L and P t exists since s L = 0, T L T R, and T L + T R by Assumption ; 4. PC t = T GL ( PC L ) and PC t = PC t \ {P t } {v t P P t}; 5. PC t = PC t T GR (PC R ) and PC = PC t ; 6. else /* n L = n R sl sr */ 7. partition ŜL into two sets S α L and Sβ L such that S α L = s R and S β L = s L s R ; 8. PC t = S α L s R S GR (PC R ); 9. P N = N GR (PC R ) s L s R Sβ L ; 0. if s L s R is odd, then let P s and P n be the semi-terminal path and non-terminal path in P N, respectively; let P = P s P n, PC s = {P }, and let P N = P N \ {P s, P n };. partition P N into two sets PC t and N R such that PC t contains all terminal paths in P N ;. P = N R ( NL, ) N L, where N R = N L = n R sl sr and P contains only a non-terminal path P of G with one end vertex in V L and the other in V R ; 3. let P = P s P be a path in PC t PC t PC s such that P s is a semi-terminal path in ŜL, where P exists since s L > 0; 4. let Y be the set containing P and let Y = Y \ {P } {P s P P }; 5. PC t = PC t PC t T GL ( PC L ) T GR (PC R ) and PC = PC t PC s. Following the above procedure, we can construct a constrained path cover PC of G of size T. Case.: F (N S) < n R sl sr. By definition, V L \ T L = F (N S) F (T GL (PC L )). Let N L v and Ŝv L denote the sets of free vertices and terminals in N S, respectively. Then, N L v = F (N S) and Ŝv L = s L. Let T GL (PC L ) = {P, P,, P tl }, where P i = v i P i and v i T L for t L i. By definition, every path P i in T GL (PC L ) can be split into two semi-terminal paths v i and P i of G L. There are two subcases: Case..: n R sl sr F (N S) + t L. In this subcase, F (N S) + t L n R sl sr > F (N S). Thus, there exists a number k such that n R sl sr = F (N S) + k and t L k. Let P T = {P, P,, P k } be a subset of T GL (PC L ). For every path P i = v i P i in P T, we first split P i into two semi-terminal paths v i and P i. Then, we get a set P S of semi-terminal paths {v, v,, v k, P, P,, P k }. Let P T = T GL (PC L ) \ P T, PS = Ŝv L P S, PN = N L v, and let PC L = P N P S P T. Let ñ L = F (N S), s L = s L + k, and let t L = t L k. Then, PCL is a (ñ L, s L, t L )-constrained path cover of G L satisfying that n R sl sr = ñ L + k and hence n R = ñ L + sl sr. Since s L s R and k, we get that s L = s L + k > s R. By arguments similar to those used in proving Lemma 3., we can construct a constrained path cover of G of size T from PC L and PC R. Case..: n R sl sr > F (N S) + t L. In this subcase, we first find a minimum k, where k t L, such that F ( k i= P i) + F (N S) n R (sl+k) sr. Note that such a number k exists since F (N S) + t L < n R sl sr V L \ T L + t L = F (N S) + F (T GL (PC L )) + t L. Let P T = {P, P,, P k } be a subset of T GL (PC L ). For every path P i = v i P i in P T, we first split P i into two semi-terminal paths v i and P i. Then, we get a set P S of semi-terminal paths {v, v,, v k, P, P,, P k }. Let P T = T GL (PC L ) \ P T, PS = Ŝv L P S, PN = N L v, and let PC L = P N P S P T. Let P Ñ S = N P S, ñ L = F (N S), s L = s L + k, and let t L = t L k. Then, F (Ñ S) = F ( k i= P i) + F (N S) n R sl sr. Since t L k and n R sl sr > F (N S) + t L = ñ L + t L, we have that n R sl+k sr > ñ L. Hence, PCL is a (ñ L, s L, t L )-constrained path cover of G L satisfying that F (Ñ S) n R sl sr > ñ L and s L > s R. Then, we can construct a constrained path cover of G of size T from PC L and PC R by using arguments similar to those used in proving Case.. Case : s L < s R. By Eq. (4), n L < n R + sr sl V L \ T L + t L. Let N S = N GL (PC L ) S GL (PC L ). There are two subcases: Case.: F (N S) n R + sr sl. In this subcase, n L < n R + sr sl F (N S). By Proposition.3, we can obtain from N S a set

11 N S of vertex-disjoint paths with n L non-terminal paths and s L semi-terminal paths such that n L = n R + sr sl. Let NL and ŜL be the sets in N S containing all non-terminal and semi-terminal paths, respectively. We then construct a constrained path cover PC of G of size T through the following procedure:. partition S GR (PC R ) into two sets SR α and Sβ R such that SR α = s L and S β R = s R s L ;. partition N L into two sets N β L and N γ L such that N β L = sr sl and N γ L = n R; 3. PC t = ŜL sl SR α; 4. PC t = N β L s R s L Sβ R ; 5. if s R s L is odd, then let P s and P n be the semi-terminal path and non-terminal path in PC t, respectively; let P = P s P n, PC s = {P }, and let PC t = PC t \ {P s, P n }; 6. P N = N γ L ( n R, )N GR (PC R ), where P N contains only a non-terminal path P of G with one end vertex in V L and the other in V R if n R 0, and P N = otherwise; 7. if P N, then 8. let P t = P s P t be a path in PC s PC t such that P s S β R, where P t exists since S β R = s R s L > 0; 9. let Y be the set containing P t and let Y = Y \ {P t } {P s P P t}; 0.PC t = PC t PC t T GL (PC L ) T GR (PC R ) and PC = PC t PC s. Following the above procedure, we can obtain a constrained path cover PC of G of size T. Case.: F (N S) < n R + sr sl. By definition, V L \ T L = F (N S) F (T GL (PC L )). Let denote the sets of free vertices and N v L and Ŝv L terminals in N S, respectively. Then, N v L = F (N S) and Ŝv L = s L. Let T GL (PC L ) = {P, P,, P tl }. By definition, every path in T GL (PC L ) can be split into two semi-terminal paths. There are two subcases: Case..: n R + sr sl F (N S) + t L. By similar arguments for proving Case.., we can obtain a (ñ L, s L, t L )-constrained path cover PC L of G L satisfying that n R + sr sl = ñ L. If s R > s L, then we can construct from PC L and PC R a constrained path cover of G of size T by using arguments similar to those used in proving Lemma 3.. If s R s L and s L s R is even, then we can construct from PC L and PC R a constrained path cover of G of size T by using arguments similar to those used in proving Lemma 3.. Suppose that s R s L and s L s R is odd. Let s L = s L. Then, s L s R is even and n R + sr (s L +) = n R + sr s L = ñ L. Thus, n R = ñ L + s L sr. By similar arguments for proving Lemma 3., we can construct from PC L and PC R a constrained path cover of G of size T. Case..: n R + sr sl > F (N S) + t L. By similar arguments for proving Case.., we can obtain a (ñ L, s L, t L )-constrained path cover PC L of G L satisfying that F (N GL ( PC L ) S GL ( PC L )) n R + sr sl > ñ L, where t L < t L. Depending on whether s L s R or not, we can construct a constrained path cover of G of size T from PC L and PC R by using arguments similar to those used in proving Case. or Case.. Lemma 3.4. Suppose that Assumption holds. If n R + s R < n L + s L V R \ T R + T R, then π(g, T ) = T. Proof. This lemma can be proved by similar arguments for proving Lemma 3.3 except one case. By Proposition., T R = s R + t R. Hence, n R + s R < n L + s L V R \ T R + s R + t R. Then, we have that n R + sr sl n L V R \ T R + sr sl + t R, n R < n L + sl sr V R \ T R + t R, ifs R s L ; (5) ifs R < s L. (6) By similar arguments for proving Lemma 3.3, we show that a constrained path cover of G of size T can be constructed from PC R and PC L, where PC R may not be a minimum constrained path cover of G R. Consider the following cases: Case : s R s L. Suppose that n L = n R + sr sl. By Lemma 3., we can obtain a constrained path cover of G of size T. In the following, assume that n L > n R + sr sl. By Eq. (5), n R < n L sr sl V R \ T R + t R. Let N S = N GR (PC R ) S GR (PC R ). There are two subcases: Case.: F (N S) n L sr sl. In this subcase, F (N S) n L sr sl > n R. By Proposition.3, we can obtain from N S a set N S of vertex-disjoint paths with n R non-terminal paths and s R semi-terminal paths such that n R = n L sr sl if T R > 0; and n R = n L + otherwise. Consider the case of T R = 0. Then, s R = s L = 0. By Assumption, n L V R and hence, V R > n L > n R. Thus, there exists a number n R such that n R = n L +.

12 Let PC R = N S T GR (PC R ). Then, PCR is a ( n R, s R, t R )-constrained path cover of G R. Let N R = N GR ( PC R ) and ŜR = S GR ( PC R ). We can obtain a constrained path cover PC of G of size T from PC R and PC L through the following procedure:. if T R = 0, then /* n R = n L + */. P N = N R nl N GL (PC L ), where N R = n L + and P N contains only a non-terminal path P of G with both end vertices in V R ; 3. let P t = v t P t be a terminal path in T GL (PC L ), where v t T L and P t exists since T L by Assumption ; 4. PC t = T GL (PC L ) and PC t = PC t \ {P t } {v t P P t}; 5. PC t = PC t T GR ( PC R ) and PC = PC t ; 6. else /* n R = n L sr sl */ 7. partition ŜR into two sets SR α and Sβ R such that SR α = s L and S β R = s R s L ; 8. PC t = SR α s L S GL (PC L ); 9. P N = N GL (PC L ) s R s L Sβ R ; 0. if s R s L is odd, then let P s and P n be the semi-terminal path and non-terminal path in P N, respectively; let P = P s P n, PC s = {P }, and let P N = P N \ {P s, P n };. partition P N into two sets PC t and N L such that PC t contains all terminal paths in P N ;. P = N L ( NR, ) N R, where N L = N R = n L sr sl and P contains only a non-terminal path P of G with one end vertex in V L and the other in V R ; 3. if t R > 0, then 4. let P l = v l P l and P r = v r P r be the terminal paths in T GL (PC L ) and T GR ( PC R ), respectively, where v l T L, v r T R, and P l exists since T L T R by Assumption ; 5. PC t = PC t PC t (T GL (PC L ) \ {P l }) (T GR ( PC R ) \ {P r }) {P l r, v l P v r }; 6. else /* t R = 0 and s R > 0 */ 7. let P = P s P be a path in PC t PC t PC s such that P s is a semi-terminal path in ŜR, where P exists since s R > 0; 8. let Y be the set containing P and let Y = Y \ {P } {P s P P }; 9. PC t = PC t PC t T GL (PC L ) T GR ( PC R ); 0. PC = PC t PC s. Following the above procedure, we can construct a constrained path cover PC of G of size T. Case.: F (N S) < n L sr sl. By symmetry, this subcase can be proved by using arguments similar to those used in proving Case. of Lemma 3.3. Case : s R < s L. By similar arguments for proving Case of Lemma 3.3, this case can be proved. Following Lemma 3.6, Lemmas , and Lemmas , we conclude the following theorem: Theorem 3.5. Let G L = (V L, E L ) and G R = (V R, E R ) be two cographs with terminal sets T L and T R, respectively, and let G = G L G R. Assume that T L T R, T = T L + T R, and that PC L (resp. PC R ) is a minimum and (n L, s L, t L )-constrained (resp. (n R, s R, t R )- constrained) path cover of G L (resp. G R ) satisfying that there exists no minimum constrained path cover of G L (resp. G R ) with semi-terminal path if s L = 0 (resp. s R = 0). Then, π(g, T ) = T +,if s L = T R = 0 and n L = V R ; max{ T, π(g L, T L ) V R \ T R, π(g R, T R ) V L \ T L },otherwise. Based upon Lemma 3. and Theorem 3.5, we can construct a minimum constrained path cover of a cograph G in linear time. In the following, we will show how to find such a minimum constrained path cover of G. For a cograph G, a decomposition tree DT (G) of G can be constructed in linear time [7, 0]. Note that the leaves of DT (G) are the vertices of G and every internal node in DT (G) is labeled by either or. In the following, assume that a decomposition tree DT (G) of a cograph G and a terminal set T of G are given. For a node v in DT (G), let DT v (G) denote the subtree of DT (G) rooted at v, G v denote the subgraph of G induced by the leaves of DT v (G), and let T v denote the set of terminals in G v. Our algorithm finds a minimum constrained path cover PC v of G v for each v DT (G) such that there exists no minimum constrained path cover of G v with semiterminal path if S Gv (PC v ) =, and is sketched as follows: The algorithm visits nodes of DT (G) in a postorder sequence (i.e., bottom-up). Thus, when visiting an internal node of DT (G), both its children were visited. Assume that it is about to process internal node v. Let v L and v R be the children of v in DT (G), and let V L and V R be the vertex sets of G vl and G vr, respectively. Without loss of generality, assume that T vl T vr.

13 Our algorithm maintains the invariant that a minimum constrained path cover PC vl (resp. PC vr ) of G vl (resp. G vr ) is constructed and satisfies that there exists no minimum constrained path cover of G vl (resp. G vr ) with semi-terminal path if S GvL (PC vl ) = (resp. S GvR (PC vr ) = ). Let PC vl and PC vr be (n L, s L, t L )-constrained and (n R, s R, t R )-constrained path covers of G vl and G vr, respectively. Suppose that v is labeled by. Then, PC vl PC vr forms a minimum and (n L + n R, s L + s R, t L + t R )-constrained path cover of G v satisfying that there exists no minimum constrained path cover of G v with semi-terminal path if s L + s R = 0. In the following, assume that v is labeled by, i.e., G v = G vl G vr. In fact, the proofs of Lemma 3.6, Lemmas , and Lemmas are constructive proofs. By using these constructive proofs, our algorithm constructs a minimum and (n, s, t)-constrained path cover PC v of G v satisfying that PC v contains at least one semi-terminal path except π(g v, T v ) = Tv or (s L = T vr = 0 and V R < n L ). Lemma 3.0 shows that there exists no minimum constrained path cover of G v with semi-terminal path if s L = T vr = 0 and V R < n L. Hence, PC v is a minimum and (n, s, t)-constrained path cover of G v satisfying that there exists no minimum constrained path cover of G v with semi-terminal path if s = 0. Let Êv = {(u L, u R ) u L V L and u R V R }. In the constructive proofs of Lemma 3.6, Lemmas , and Lemmas , we can easily see that PC v can be constructed in O( Êv ) time. Hence, we conclude the following theorem: Theorem 3.6. Given a cograph G = (V, E) and a terminal set T of G, the terminal path cover problem can be solved in O( V + E )-linear time. References [] G.S. Adhar and S. Peng, Parallel algorithms for cographs and parity graphs with applications, J. Algorithms, Vol., pp. 5 84, 990. [] S.R. Arikati and C.P. Rangan, Linear algorithm for optimal path cover problem on interval graphs, Inform. Process. Lett., Vol. 35, pp , 990. [3] A.A. Bertossi and M.A. Bonuccelli, Hamiltonian circuits in interval graph generalizations, Inform. Process. Lett., Vol. 3, pp , 986. [4] B.L. Bodlaender and R.H. Mohring, The pathwidth and treewidth of cographs, SIAM J. Discrete Math., Vol. 6, pp. 8 88, 993. [5] H.L. Bodlaender, Achromatic number is NPcomplete for cographs and interval graphs, Inform. Process. Lett., Vol. 3, pp , 989. [6] F.T. Boesch and J.F. Gimpel, Covering the points a digraph with point-disjoint paths and its application to code optimization, J. ACM, Vol. 4, pp. 9 98, 977. [7] M.S. Chang, S.Y. Hsieh, and G.H. Chen, Dynamic programming on distancehereditary graphs, Lecture Notes in Comput. Sci., Vol. 350, pp , Springer, Berlin, 997. [8] M.S. Chang, S.L. Peng, and J.L. Liaw, Deferred-query: an efficient approach for some problems on interval graphs, Networks, Vol. 34, pp. 0, 999. [9] G.J. Chang and D. Kuo, The L(, )- labeling problem on graphs, SIAM J. Discrete Math., Vol. 9, pp , 996. [0] D.G. Corneil, H. Lerchs, and L.K. Stewart, Complement reducible graphs, Discrete Appl. Math., Vol. 3, pp , 98. [] D.G. Corneil, Y. Perl, and L.K. Stewart, A linear recognition algorithm for cographs, SIAM J. Comput., Vol. 4, pp , 985. [] P. Damaschke, J.S. Deogun, D. Kratsch, and G. Steiner, Finding Hamiltonian paths in cocomparability graphs using the bump number algorithm, Order, Vol. 8, pp , 99. [3] P. Damaschke, Induced subgraph isomorphism for cographs is NP-complete, Lecture Notes in Comput. Sci., Vol. 484, pp. 7 78, Springer-Verlag, Berlin, 99. [4] G. Damiand, M. Habib, and C. Paul, A simple paradigm for graph recognition: application to cographs and distance hereditary graphs, Theoret. Comput. Sci., Vol. 63, pp. 99, 00. [5] M.R. Garey and D.S. Johnson, Computers and Intractability: A Guide to the Theory of NP-Completeness, Freeman, San Francisco, CA, 979.