A necessary and sufficient condition for the existence of a spanning tree with specified vertices having large degrees

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1 A necessary and sufficient condition for the existence of a spanning tree with specified vertices having large degrees Yoshimi Egawa Department of Mathematical Information Science, Tokyo University of Science, Shinjuku-ku, Tokyo , Japan Kenta Ozeki National Institute of Informatics, Hitotsubashi, Chiyoda-ku, Tokyo , Japan and JST, ERATO, Kawarabayashi Large Graph Project ozeki@nii.ac.jp Abstract Let G be a connected simple graph, let X V G) and let f be a mapping from X to the set of integers. When X is an independent set, Frank and Gyárfás, and independently, Kaneko and Yoshimoto gave a necessary and sufficient condition for the existence of a spanning tree T in G such that d T x) fx) for all x X, where d T x) is the degree of x in T. In this paper, we extend this result to the case where the subgraph induced by X has no induced path of order four, and prove that there exists a spanning tree T in G such that d T x) fx) for all x X if and only if for any nonempty subset S X, N G S) S fs) + 2 S ω G S) 1, where ω G S) is the number of components of the subgraph induced by S. Keywords: Degree constrained spanning tree, Submodular function 1 Introduction Throughout this paper, we consider only simple graphs. Let G be a connected graph. For x V G), a vertex v V G) {x} is called a neighbor of x in G if v is adjacent to x in G. We let N G x) denote the set of 1

2 neighbors of x in G. The degree of x is denoted by d G x); thus d G x) = N G x). For S V G), let N G S) := x S N Gx) and Γ G S) := N G S) S. Let G[S] be the subgraph of G induced by S. We denote the number of components of G[S] by ω G S). For S, R V G), the set of edges in G connecting S and R is denoted by E G S, R). Let X be a subset of V G), and let f be a mapping from X to the set of integers. In this paper, we concentrate on the existence of a spanning tree T in G such that d T x) fx) for all x X. For simplicity, we call such a spanning tree an X, f)- tree. In this context, the mapping f is referred to as a lower capacity vector or a demand vector in some papers.) More specifically, we consider a necessary and sufficient condition for the existence of an X, f)-tree. In general, it seems difficult to give such a condition. This is because the problem of determining whether a graph G has an X, f)-tree or not is at least as hard as the problem of the existence of a Hamiltonian path, which is well-known as an NP-complete problem. Formally, when we want to determine whether a given graph G has a Hamiltonian path, we construct a graph G by joining two new vertices to G, and let X := V G) and f 2. Then G has a Hamiltonian path if and only if G has an X, f)-tree. However, when we confine ourselves to the case where X is an independent set, we can obtain a necessary and sufficient condition for the existence of an X, f)-tree. Frank and Gyárfás, and independently, Kaneko and Yoshimoto gave the following result. For S X, let fs) := x S fx). Theorem 1.1 Frank and Gyárfás [1], Kaneko and Yoshimoto [2]) Let G be a connected graph, let X V G) and let f be a mapping from X to the set of integers. Then there exists an X, f)-tree in G if and only if for any nonempty subset S X, N G S) fs) + S 1. In the case where X is a non-independent set, we similarly obtain the following necessary condition for the existence of an X, f)-tree. We prove Proposition 1.2 in Section 2. Let gs; G, f) := Γ G S) fs) + 2 S ω G S). Proposition 1.2 Let G be a connected graph, let X V G) and let f be a mapping from X to the set of integers. If there exists an X, f)-tree in G, then for any nonempty subset S X, gs; G, f) 1. 1) Note that for an independent set X, condition 1) is equivalent to the necessary condition in Theorem 1.1, because Γ G S) = N G S) S = N G S) and ω G S) = S. 2

3 One might expect that condition 1) is also a sufficient condition. However, as mentioned above, it is impossible to have such a simple result unless NP = co-np). In fact, even when X induces a path consisting of four vertices, we can construct an example that satisfies condition 1) but has no X, f)-tree as follows: let G 1 be the graph in Figure 1, let X := {x 1, x 2, x 3, x 4 }, and define a mapping f by letting fx i ) = 2 for each 1 i 4. Note that for S X, we have Γ G S) 2 and fs) 2 S + ω G S) = ω G S) = 1 if S = 1 or 4, and Γ G S) 3 and fs) 2 S + ω G S) = ω G S) 2 if S = 2 or 3. Thus, for any nonempty subset S X, gs; G, f) 1, but G 1 has no X, f)-tree. More generally, when H is a graph having an induced path of order four, by considering a structure similar to G 1 in Figure 1, we can show the existence of an example of a graph G with G[X] = H that satisfies condition 1) but has no X, f)-tree. x 1 x 2 x 3 x 4 Figure 1: The graph G 1. In view of the gap between the case of an independent set and the general case, it is natural to ask what causes the gap, in other words, what properties guarantee that condition 1) is also a sufficient condition. The main purpose of this paper is to give an answer to this question by showing the following result: when G[X] has no induced path of order four, condition 1) is also a sufficient condition. As mentioned above, when X has an induced path of order four, there exists a graph satisfying condition 1) but has no X, f)-tree. Thus the condition that G[X] has no induced path of order four is best possible in this sense. Theorem 1.3 Let G be a connected graph and let X V G), and suppose that G[X] has no induced path of order four. Let f be a mapping from X to the set of integers. Then there exists an X, f)-tree in G if and only if for any nonempty subset S X, gs; G, f) = ΓG S) fs) + 2 S ωg S) 1. In Section 3, we show the submodularity of the function g, which plays a crucial role in our proof. After showing that, we prove Theorem 1.3 in Section 4. 2 A necessary condition for an X, f)-tree In this section, we prove Proposition 1.2. For reference in the proof of Theorem 1.3, we show a slightly stronger result, which includes a description of the case where 3

4 equality holds. Before stating the result, we remark that under the assumption that G is connected, the existence of an X, f)-tree is equivalent to the existence of a not necessarily spanning) forest F with X V F ) such that d F x) fx) for all x X. Proposition 2.1 Let G be a graph, let X V G) and let f be a mapping from X to the set of integers. Suppose that there exists a forest F in G with X V F ) such that d F x) fx) for all x X. Then for any nonempty subset S X, gs; G, f) = ΓG S) fs) + 2 S ωg S) 1. Moreover, if equality holds for S X, then F [S Γ G S)] is connected and Γ G S) = Γ F S). Proof of Proposition 2.1. Fix a subset S X and let l be the number of components of F [S]. Note that l ω G S) and EF [S]) = S l, because F [S] has no cycle. Hence x S N F x) S = 2 EF [S]) = 2 S l ). Let H be the graph obtained from F [S Γ F S)] by contracting each component of F [S] to one vertex. Then EH) = EF S, ΓF S) ) = NF x) NF x) S x S x S fs) 2 S + 2l. On the other hand, we obtain V H) = ΓF S) +l. Since H is a forest, it follows from these inequalities that ΓF S) + l = V H) EH) + 1 fs) 2 S + 2l + 1, or gs; G, f) = ΓG S) fs) + 2 S ωg S) ΓF S) fs) + 2 S l 1. Moreover when equality holds, H is connected and Γ G S) = Γ F S), and hence F [S Γ G S)] is connected. 3 Submodularity of the function g In this section, we show the submodularity of the function g under the assumption that G[X] has no induced path of order four. We actually prove the following slightly 4

5 stronger result, which we need in the proof of Theorem 1.3. Throughout this section, we fix a graph G and a function f; so we simply write ΓS) := Γ G S), ωs) := ω G S) and gs) := gs; G, f) = ΓS) fs) + 2 S ωs). Lemma 3.1 Let G be a connected graph and let X V G), and suppose that G[X] has no induced path of order four. Let f be a mapping from X to the set of integers. Then for any S, R X, where and ε SR := gs R) + gs R) gs) + gr) A SR ε SR, A SR := ΓS) ΓR) ΓS R), { 1 if S R = and there exists an edge connecting S and R, 0 otherwise. Before proving Lemma 3.1, we show that the assumption that G[X] has no induced path of order four is needed for the submodularity of the function g. Let G 2 be the graph in Figure 2 and let X := {x 1, x 2, x 3, x 4 }. Let fx i ) := 2 for each 1 i 4. Note that when S := {x 1, x 2, x 4 } and R := {x 1, x 3, x 4 }, we have gs) = gr) = gs R) = 1 and gs R) = 2, so g does not satisfy submodularity. x 1 x 2 x 3 x 4 Figure 2: The graph G 2. Proof of Lemma 3.1. Let B S := ΓS) R ΓS R), and B R := ΓR) S ΓS R). Since ΓS R) + ΓS R) = ΓS) + ΓR) A SR B S B R, fs R) + fs R) = fs) + fr), and S R + S R = S + R, we obtain gs) + gr) A SR ε SR gs R) gs R) = ΓS) fs) + 2 S ωs) ) + ΓR) fr) + 2 R ωr) ) A SR ε SR ) ΓS R) fs R) + 2 S R ωs R) ) ΓS R) fs R) + 2 S R ωs R) = ωs R) + ωs R) + B S + B R ωs) ωr) ε SR. 5

6 Thus it suffices to show that ωs R) + ωs R) + B S + B R ωs) + ωr) + ε SR. Now we construct two bipartite graphs H and H as follows. One partite set of H is CS) and the other is CR), where CS) is the set of components of G[S] and CR) is that of G[R]. For S i CS) and for R j CR), we let S i R j E H) if and only if I) S i R j or II) there exists an edge in G connecting S i and R j such that at least one of the end vertices is not contained in ΓS R). Edges of H satisfying I) are said to be of Type I, and other edges, that is, edges not satisfying I) and satisfying II) are said to be of Type II. Let H be a spanning forest of H such that the number of components of H is equal to that of H. Let EI := {e EH) : e is of Type I} and E II := {e EH) : e is of Type II}. Note that V H) = ωs) + ωr). Claim 1 The number of components of H is at most ωs R). Proof. Since the number of components of H is equal to that of H, it suffices to show that the number of components of H is at most ωs R). By the definition of edges of H, each component of H corresponds to some component of G[S R]. Formally, if we let S 1,..., S p CS) and R 1,..., R q CR) be such that they constitute a component of H, p i=1 S i q j=1 R j is contained in a single component of G[S R]; we consider the correspondence defined in this way. Now it suffices to show that no two components of H are mapped to the same component of G[S R] by the correspondence. Suppose that there exist two components of H which correspond to the same component of G[S R]. Again by the definition of edges of H, there exists an edge e in G connecting a component of G[S], say S i, and a component of G[R], say R j, such that S i and R j belong to distinct components of H and both end vertices of e are contained in ΓS R). Let x 1 be the end vertex of e contained in S i and let x 2 be the other end vertex of e. Since x 1, x 2 ΓS R), there exist two vertices y 1, y 2 S R such that x i N G y i ) for i = 1, 2. Note that y 1 y 2 and y 1 x 2, y 1 y 2, x 1 y 2 EG), because y 1, x 1 S i, x 2, y 2 R j, and S i and R j belong to distinct components of H. Hence {y 1, x 1, x 2, y 2 } induces a path of order four, a contradiction. Claim 2 E I ωs R). Proof. Let e = S i R j be an edge of Type I. Then S i R j. We associate with e a component of G[S R] contained in S i R j. Note that S i R j intersects with no component in CS) CR) {S i, R j }. Thus, this correspondence is injective. Hence E I ωs R). Claim 3 E II B S + B R ε SR. 6

7 Proof. We show that there exists an injective mapping h from E II to B S B R ; moreover, we show that when S R = and there exists an edge in G connecting S and R, there exists v B S B R such that v is not contained in the image of h. Let e = S i R j E II. Then S i R j =, and there exists an edge in G connecting S i and R j such that at least one of its end vertices is not contained in ΓS R). Let he) be such an edge, and let he) be an end vertex of he) not contained in ΓS R). Note that h is a mapping from E II to B S B R. Note also that even if we fix he) for every e, h is not uniquely determined, because if neither end vertex of he) is contained in ΓS R), we may choose either of the two end vertices of he) as he). Choose h so that the image of h is as large as possible. Suppose that h is not injective; that is, there exist two edges e = S i R j and e = S i R j of Type II such that he) = he ). Let u = he) = he ), and let v and v be the other end vertices of he) and he ), respectively. By symmetry, we may assume that u S, so S i = S i, v R j and v R j. Take a longest path P = v 0 v 1 v 2... v p of G with v 0 = u and v 1 = v such that for any 0 l p 1, he l ) = v l v l+1 and he l ) = v l for some e l E II. For convenience, let v 1 = v. Note that p 1, and v i R S for any odd integer i with 1 i p and v i S R for any even integer i with 1 i p. Assume for the moment that p is even. Suppose that v p ΓS R). Then there exists a vertex w S R such that v p N G w). Note that v p and v p 2 are contained in distinct components of G[S]. Hence w and v p 2 are contained in distinct components of G[S]. Since e p 1 E I, we also see that w and v p 1 are contained in distinct components of G[R]. These imply that {w, v p, v p 1, v p 2 } induces a path of order four, a contradiction. Thus v p ΓS R). Suppose that v p = he ) for some e E II, and let v p+1 be the other end vertex of he ). Then v p+1 v l for any 1 l p because H has no cycle. Hence v 0 v 1 v 2... v p v p+1 is a path longer than P, a contradiction. Thus, v p he ) for any e E II, and hence by redefining he l ) = v l+1 for 0 l p 1, we get a contradiction to the maximality of the image of h. When p is odd, we can similarly get a contradiction. Consequently, h is an injective mapping from E II to B S B R. Now we consider the case where S R = and there exists an edge in G connecting S and R. Note that ΓS R) = and E I =, and hence E II by the assumption that E G S, R). By the above argument, we can find an injective mapping h from E II to B S B R. Take e = S i R j E II. Since ΓS R) =, neither end vertex of he) is contained in ΓS R). Let u = he) and let v be the other end vertex of he). Again take a longest path P = v 0 v 1 v 2... v p of G with v 0 = u and v 1 = v such that for any 0 l p 1, he l ) = v l v l+1 and he l ) = v l for some e l E II. Since ΓS R) =, we clearly have v p B S B R. By the same argument as above, we can also show that v p is not contained in the image of h. This implies that E II B S + B R ε SR. 7

8 Note that V H) = ωs)+ωr). Thus by Claim 1, EH) V H) ωs R) = ωs) + ωr) ωs R). Therefore it follows from Claims 2 and 3 that ωs) + ωr) ωs R) EH) ωs R) + B S + B R ε SR, or ωs R) + ωs R) + B S + B R ωs) + ωr) + ε SR. This completes the proof of Lemma Proof of Theorem 1.3 In this section, we will prove Theorem 1.3 but, before proving the theorem, we mention an algorithmic aspect. As mentioned before, the submodularity of the function g plays a key role in our proof, and it is well-known that the problem of minimizing a submodular function can be solved in polynomial time. For example, see Theorem 45.1 on Page 791 of [3].) Based on this fact and the following nature of our proof, we believe that there is a polynomial time algorithm to find an X, f)-tree in a graph satisfying condition 1). Note that at each step in the proof, for a given graph G, a given set X V G) such that G[X] has no induced path of order four, and a given mapping f, we delete a vertex from X or an edge from G. To find an appropriate vertex or edge, we make use of a tight set see the definition made immediately before Claim 4), which can be found by using the problem of minimizing the submodular function g. In order to prove Theorem 1.3, it suffices to show the following theorem see the remark made immediately before the statement of Proposition 2.1). Theorem 4.1 Let G be a graph and let X V G), and suppose that G[X] has no induced path of order four. Let f be a mapping from X to the set of integers, and suppose that for any nonempty subset S X, gs, G, f) 1. Then there exists a forest F in G with X V F ) X Γ G X) and EF ) EG[X]) E G X, ΓG X) ) such that d F x) fx) for all x X. Proof of Theorem 4.1. Since removing an edge connecting two vertices of V G) X never destroys the assumption of Theorem 4.1, we may assume that V G) X is an independent set. We prove Theorem 4.1 by simultaneous induction on X and EG X, ΓG X) ). If X = 0, there is nothing to prove; so we may assume that X. 8

9 Suppose that G has no forest satisfying the properties in Theorem 4.1. We first show that EG X, ΓG X) ) 0. Suppose that EG X, ΓG X) ) = 0. This implies that Γ G X) =, and hence gx; G, f) = fx) + 2 X ω G X) 1, or fx) 2 X ω G X) 1 2 X 2. Consequently, there exists a vertex x X with fx) 1. By the induction hypothesis for X := X {x}, G has a forest F with EF ) EG[X ]) E G X, Γ G X ) ) such that d F x ) fx ) for all x X. When fx) 0 or d F x) 1, F is a forest as desired. On the other hand, when fx) = 1 and x V F ), F := F {xu} is a forest as desired, where u N G x) note that there exists such a vertex u because d G x) = Γ G x) 1 + fx) 2 + ω G {x}) = 1). This contradicts the assumption that G has no forest satisfying the properties in Theorem 4.1. Thus, EG X, ΓG X) ) 0. For a nonempty subset S X, we say that S is a tight set if gs; G, f) = 1. Let S := {S X : S is a tight set with S X}. Claim 4 For any x X S S S, N Gx) X =. Suppose that there exists a vertex x X S S S such that N Gx) X, say u N G x) X. Since x S for any S S, there exists no tight set containing x except for X. Hence for any nonempty subset R X with x R and R X, gr; G, f) 2. If u is a neighbor of some vertex in X {x}, then let f = f; otherwise let { fy) 1 if y = x, f y) := fy) otherwise. Let G be the graph obtained from G by deleting the edge xu. Then E G X, ΓG X) ) < EG X, ΓG X) ). For any nonempty subset R X with x R and R X, Γ G R) Γ G R) {u}, and hence gr; G, f ) gr; G, f) gr; G, f) 1 1. For any nonempty subset R X with x R, f R) = fr) and Γ G R) = Γ G R), and hence gr; G, f ) = gr; G, f) 1. When u is a neighbor of some vertex in X {x}, Γ G X) = Γ G X), and hence gx; G, f ) = gx; G, f) 1; otherwise Γ G X) = Γ G X) 1 and f X) = fx) 1, and hence gx; G, f ) = gx; G, f) 1. Thus, in each case, we obtain gx; G, f ) 1. Therefore G, X and f satisfy the assumption of Theorem 4.1. By the induction hypothesis, G has a forest F with EF ) EG [X]) E G X, ΓG X) ) such that d F x ) f x ) for all x X. When u is a neighbor of some vertex in X {x}, F is also a forest as desired for G, X and f; otherwise the graph F obtained from F by adding the edge xu is a forest because u is no neighbor of X {x}), and hence F is a forest as desired. This contradiction completes the proof of the claim. Claim 5 Let S S. Then the following hold. 9

10 i) There exists a tree F S with V F S ) = S Γ G S) and EF S ) EG[S]) E G S, Γ G S)) such that d FS x) fx) for all x S. ii) Let F S be as in i), and suppose that F S is chosen so that EFS ) E G S, ΓG X) ) is as small as possible. Let D be a component of G[X] with S V D). Then [ F S S ΓG S) ) V D) ] is connected. iii) Let F S be as in ii). Then E G S, ΓG X)) EF S ). Proof. i) Since S < X, we can apply the induction hypothesis to G, S and f, to obtain a forest F S with V F S ) S Γ G S) and EF S ) EG[S]) E G S, ΓG S) ) such that d FS x) fx) for all x S. By Proposition 2.1, F S is a tree and Γ G S) V F S ). Thus, V F S ) = S Γ G S). ii) We take a tree F S as in i) so that EFS ) E G S, ΓG X) ) is as small as possible. Suppose that F S [ S Γ G S) ) V D)] has at least two components, say C 1 and C 2. Since F S is connected, there exists a path P connecting C 1 and C 2 in F S with V P ) V C 1 ) = 1 and V P ) V C 2 ) = 1. For each i = 1, 2, write V P ) V C i ) = {x i } and N P x i ) = {u i }. Note that u 1, u 2 X. Since G[X] has no induced path of order four, we have x 1 x 2 EG) or there exists a vertex y X S such that y N G x 1 ) N G x 2 ). In the second case, y cannot reach x 1 in F S without passing through u 1 or cannot reach x 2 without passing through u 2, since otherwise, these two paths and P together contain at least one cycle, a contradiction. Thus, by symmetry, we may assume that y cannot reach x 1 in F S without passing through u 1. In either case, let F S be the graph obtained from F S by deleting x 1 u 1, and adding x 1 x 2 or x 1 y. Then d F S x) fx) for all x X and EF S ) E G S, ΓG X) ) is smaller than EF S ) E G S, ΓG X) ), contradicting the choice of F S. iii) Let G S be the graph obtained from G by deleting all edges in E G S, ΓG X) ) not contained in EF S ). We will show that for G S, X and f, the assumption of Theorem 4.1 holds. Take an arbitrary nonempty subset R X. If S R =, then Γ GS R) = Γ G R), and hence gr; G S, f) = gr; G, f) 1. Thus, we may assume that S R. Since the tree F S satisfies d FS x) fx) for all x S, it follows from Proposition 2.1 that gs R; G S, f) 1. On the other hand, Γ GS S) = Γ FS S) = Γ G S), and hence we have gs R; G S, f) = gs R; G, f) 1 and gs; G S, f) = gs; G, f) = 1. These inequalities and Lemma 3.1 imply that gr; G S, f) gs R; G S, f) + gs R; G S, f) gs; G S, f) 1. Therefore G S, X and f satisfy the assumption of Theorem 4.1. Hence if there exists at least one edge in E G S, ΓG X) ) not contained in EF S ), we can use the induction hypothesis, and obtain a forest as desired, a contradiction. Thus, all edges in E G S, ΓG X)) are used in F S. For each S S, we let F S be a tree as in Claim 5 i) and we assume that F S is chosen so that EFS ) E G S, ΓG X) ) is as small as possible. By Claim 5, we 10

11 obtain the following claim. Claim 6 Let S S. Then there exists no cycle in G[S Γ G S)] E G X S, ΓG X) ) containing at least one vertex in Γ G X). Proof. Suppose that there exists a cycle C in G[S Γ G S)] E G X S, ΓG X) ) containing at least one vertex in Γ G X). Let E 0 := EC) E G ΓG X), V G) ). Since Γ G X) is independent and C does not use an edge in E G ΓG X), X S ), we obtain E 0 E G S, ΓG X) ). Hence E 0 EF S ) by Claim 5 iii). On the other hand, for a component of D of G[X], let E D := EF S [V D)]) if V C) V D) ; otherwise let E D =. By the definition of E 0 and E D and by Claim 5 ii), E 0 D CX) E D contains at least one cycle, where CX) is the set of components of G[X]. This contradicts the fact that F S has no cycle. Claim 7 Let S S and let D be a component of G[X] with S V D). Then S V D) S. Proof. If S V D), then there is nothing to prove. Thus we may assume that S intersects with at least two components of G[X]. Let D 1, D 2,..., D l be the components of G[X] such that S V D i ), and let S i = S V D i ). If we have ΓG S i ) j i Γ GS j ) 2 for every 1 i l, then l i=1 Si Γ G S i ) ) contains a cycle, contradicting Claim 6. Hence there exists an index i, say i = 1, such that ΓG S 1 ) l j=2 Γ GS j ) 1. Let S := l j=2 S j. Then gs; G, f) = ΓG S) fs) + 2 S ωg S) = Γ G S 1 ) + Γ G S) Γ G S 1 ) Γ G S) fs 1 ) f S) + 2 S S ω G S 1 ) ω G S) gs 1 ; G, f) + g S; G, f) 1. Since gs; G, f) = 1, gs 1 ; G, f) 1 and g S; G, f) 1, we obtain gs 1 ; G, f) = 1 and g S; G, f) = 1, and hence S 1, S S. By applying the above argument recursively, we can prove that S V D i ) S for all 1 i l. Claim 8 There exists no cycle containing at least one vertex in Γ G X). Proof. Suppose that there exists a cycle C containing at least one vertex in Γ G X). We choose C so that V C) Γ G X) is as small as possible. Let {S 1, S 2,..., S k } be a family of sets in S such that V C) S i for each 1 i k and V C) S S S k i=1 S i. We call such a family {S 1, S 2,..., S k } a covering family of C. By Claim 7, we may assume that each S i is contained in one component of G[X]. By Claim 4, we have k 1. Let u V C) Γ G X). If N G u) S i 2 for some i, then we can find a cycle in G[S i Γ G S i )] E G X Si, Γ G X) ) containing u, contradicting Claim 6. 11

12 Thus, N G u) S i 1 for each 1 i k. In particular, we have k 2, because N G u) N C u) = 2 and N G u) k i=1 S i by Claim 4. Suppose that G[X] consists of only one component. Since N G u) 2 and N G u) S i 1 for each 1 i k, there exist two sets S i and S j, say S 1 and S 2, such that N G u) S 1, N G u) S 2 and N G u) S 1 S 2 =. Thus A S1 S 2 1, where A S1 S 2 := Γ G S 1 ) Γ G S 2 ) Γ G S 1 S 2 ). On the other hand, since S 1 and S 2 are contained in the same component of G[X] and G[X] has no induced path of order four, one of the following holds: i) S 1 S 2 ; ii) Γ G S 1 ) Γ G S 2 ) X ; or iii) there exists an edge connecting S 1 and S 2. If i) holds, gs 1 S 2 ; G, f) 1. On the other hand, when i) does not hold, we have A S1 S 2 2 if ii) holds, and ε S1 S 2 = 1 if iii) holds. In any case, it follows from Lemma 3.1 that gs 1 S 2 ; G, f) gs 1 ; G, f) + gs 2 ; G, f) gs 1 S 2 ; G, f) A S1 S 2 ε S1 S 2 0, a contradiction. Therefore G[X] has at least two components. Suppose that there exists a component of G[X] containing two sets S i and S j, say S 1 and S 2. Since G[X] has no induced path of order four, S 1 S 2, or Γ G S 1 ) Γ G S 2 ), or there exists an edge connecting S 1 and S 2. This means that gs 1 S 2 ; G, f) 1, or A S1 S 2 1, or ε S1 S 2 = 1. Consequently we obtain gs 1 S 2 ; G, f) gs 1 ; G, f) + gs 2 ; G, f) gs 1 S 2 ; G, f) A S1 S 2 ε S1 S 2 1, and hence S 1 S 2 is also a tight set. Since G[X] has at least two components and S 1 and S 2 are contained in the same component of G[X], we have S 1 S 2 X, and hence S 1 S 2 S. This implies that {S 1 S 2, S 3,..., S k } is also a covering family of C. By applying the above argument recursively, we may assume that any component of G[X] contains at most one set in {S 1, S 2,..., S k }. This choice implies that S i S j = for any 1 i < j k. Since {S 1, S 2,..., S k } is a covering family of C, changing the order of S 1, S 2,..., S k if necessary, we may also assume that there exist k distinct vertices u 1, u 2,..., u k Γ G X) such that u i Γ G S i ) Γ G S i+1 ) for each 1 i k, where S k+1 = S 1 recall that we have chosen C so that V C) Γ G X) is minimum). For 1 i k 1, let R i := i j=1 S j. By inductive argument, we will show that R i S for each 1 i k 1. When i = 1, definitely R 1 = S 1 S. Thus let i 2 and suppose that R i 1 S. By the properties of {S 1, S 2,... S k } mentioned at the 12

13 end of the preceding paragraph, we have R i 1 S i = and Γ G R i 1 ) Γ G S i ). Thus, A Ri 1 S i 1, where A Ri 1 S i := Γ G R i 1 ) Γ G S i ) Γ G R i 1 S i ), and hence it follows from Lemma 3.1 that gr i ; G, f) gr i 1 ; G, f)+gs i ; G, f) A Ri 1 S i 1. Since gr i ; G, f) 1, we obtain gr i ; G, f) = 1. Hence R i S. In particular, R k 1 S. Moreover, R k 1 S k = and A Rk 1 S k 2, where A Rk 1 S k := Γ G R k 1 ) Γ G S k ) Γ G R k 1 S k ). Again by Lemma 3.1, this implies that gr k 1 S k ; G, f) gr k 1 ; G, f) + gs k ; G, f) A Rk 1 S k 0, contradicting the assumption. We are now in a position to complete the proof of Theorem 4.1. Since EG X, Γ G X)) 0, there exists an edge xu EG) with x X and u Γ G X). Let f y) := { fy) 1 if y = x, fy) otherwise. Let G be the graph obtained from G by deleting the edge xu. Then EG X, ΓG X) ) < ) E G X, ΓX). For any nonempty subset R X with x R, ΓG R) Γ G R) {u} and f R) = fr) 1, and hence gr; G, f ) = gr; G, f) + 1 gr; G, f) 1. For any nonempty subset R X with x R, Γ G R) = Γ G R) and f R) = fr), and hence gr; G, f ) = gr; G, f) = gr; G, f) 1. Therefore G, X and f satisfy the assumption of Theorem 4.1, and by the induction hypothesis, G has a forest F with EF ) EG [X]) E G X, ΓG X) ) such that d F x ) f x ) for all x X. By Claim 8, the graph obtained from F by adding the edge xu is a forest satisfying the condition required in Theorem 4.1, which contradicts the assumption that there is no such forest. This completes the proof of Theorem 4.1. Acknowledgments The authors are grateful to the referee for useful comments. References [1] A. Frank and A. Gyárfás, How to orient the edges of a graph? Colloq. Math. Soc. Jànos Bolyai ) [2] A. Kaneko and K. Yoshimoto, On spanning trees with restricted degrees, Inform. Process. Lett ) [3] A. Schrijver, Combinatorial Optimization, Springer