Degree Bounded Spanning Trees - CiteSeerX
Degree Bounded Spanning Trees Jun Fujisawa1
Hajime Matsumura2
Tomoki Yamashita3
Abstract In this paper, we give a sufficient condition for a graph to have a degree bounded spanning tree. Let n ≥ 1, k ≥ 3, c ≥ 0 and G be an n-connected graph. Suppose that for every independent set S ⊆ V(G) of cardinality n(k − 1) + c + 2, there exists a vertex set X ⊆ S of cardinality k such that the degree sum of vertices in X is at least |V(G)| − c − 1. Then G has a spanning tree T with maximum degree at most k + ⌈c/n⌉ and P v∈V(T ) max{dT (v) − k, 0} ≤ c.
Keywords: spanning tree, degree bounded tree, degree sum condition, independence number, total excess 1 Introduction We will generally follow notation and terminology of [3]. For a vertex x of a graph G, we denote the degree of x in G by dG (x) and the set of vertices adjacent to x in G by NG (x). Moreover, NG (x) ∪ {x} is denoted by NG [x]. For a subset S of V(G), let G − S denotes the subgraph induced by V(G) \ S . For a subgraph H of G, G − H denotes the subgraph induced by V(G) \ V(H). Here we define several invariants of a graph G. Let α(G) be the independence number of G. For a graph G with α(G) ≥ r, we define X dG (v) S is an independent set of G with |S | = r σr (G) := min v∈S
and σr (G) := ∞ if α(G) < r.
Next, for S ⊆ V(G) with |S | ≥ k, let X . dG (v) X ⊆ S , |X| = k ∆k (S ) = max v∈X
And we define
σrk (G) := min{∆k (S ) | S is an independent set of G with |S | = r} for k ≤ r and a graph G with α(G) ≥ r. If α(G) < r, we define σrk (G) := ∞. The following proposition shows the relation between σrk (G) and σr (G). Proposition 1 For any graph G and two integers r and k with k ≤ r, σrk (G) ≥ kσr (G)/r. Proof. If α(G) < r, then σrk (G) = σr (G) = ∞. So assume that α(G) ≥ r. For every independent set S P of cardinality r, ∆k (S ) ≥ k v∈S dG (v)/r ≥ kσr (G)/r. Hence by the definition of σrk (G), we obtain σrk (G) ≥
kσr (G)/r.
1 [email protected].
Department of Computer Science, Nihon University, Sakurajosui 3-25-40, Setagaya-Ku, Tokyo,
156-8550, Japan. 2 h [email protected]. Kyoto Computer Gakuin, 10-5, Teranomae-cho, Nishikujyo, Minami-ku, Kyoto, 601-8407 Japan. 3 [email protected]. Department of Mathematics, Asahi University, Gifu, 501-0296 Japan.
1
Now we introduce some theorems concerning the existence of Hamilton cycles. The following two theorems are well-known. Theorem 1 (Ore [9]) Let G be a graph with |V(G)| ≥ 3 and σ2 (G) ≥ |V(G)|. Then G has a Hamilton cycle. Theorem 2 (Chv´atal and Erd˝os [5]) Let G be an n-connected graph with |V(G)| ≥ 3 and α(G) ≤ n. Then G has a Hamilton cycle. It follows from Theorem 2 that, about the existence of Hamilton cycle, we may consider only graphs in which independence number is greater than its connectivity. In this point of view, the following theorem was obtained, which is a generalization of Theorems 1 and 2. Theorem 3 (Bondy [1]) Let G be an n-connected graph with |V(G)| ≥ 3 and n ≥ 1. If σn+1 (G) ≥ (n+1)(|V(G)|− 1)/2, then G has a Hamilton cycle. Therefore, if G is an n-connected graph, the condition on n + 1 independent vertices is essential for the existence of a Hamilton cycle. But in fact, after taking n + 1 independent vertices, there is no need to consider the degree sum of all those vertices. The following theorem suggests this assertion. Theorem 4 (Yamashita [13]) Let G be an n-connected graph with |V(G)| ≥ 3, where n ≥ 1. If σ2n+1 (G) ≥ |V(G)|, then G has a Hamilton cycle. Therefore, we have: To obtain a Hamilton cycle in an n-connected graph, degree condition on n + 1 independent vertices is essential. And among them, two vertices of high degree plays an important role. Next we turn our view to the existence of a degree bounded spanning tree. On this subject, several results are already known. For example, see [4, 6, 7, 8, 11, 12]. Here we introduce the following two results. Theorem 5 (Win [11]) Let G be a connected graph with σk (G) ≥ |V(G)| − 1, where k ≥ 2. Then G has a spanning tree with maximum degree at most k. Theorem 6 (Neumann-Lara and Rivera-Campo [8]) Let n ≥ 1, k ≥ 3, 0 ≤ c ≤ n and G be an n-connected graph with α(G) ≤ n(k − 1) + c + 1. Then G has a spanning tree with maximum degree at most k + 1 and with at most c vertices of degree k + 1. Now the situation is similar to the case of Hamilton cycle. That is, for a spanning tree with maximum degree at most k, degree sum of k independent vertices is considered in Theorem 5. However, it follows from Theorem 6 (in case of c = 0) that we may consider only graphs G which satisfy α(G) > n(k − 1) + 1, where n is the connectivity of G, and so it is natural to take degree sum of n(k − 1) + 2 vertices as a condition. Moreover, relaxing the condition on independence number to n(k − 1) + c + 1, it is considered in Theorem 6 that the case where there are at most c vertices of degree k + 1 in the resulting spanning tree. Therefore, as an extension of Theorem 6, a theorem which consider the degree sum of n(k − 1) + c + 2 independent vertices may exist: In fact, the following theorem is one of that.
2
Theorem 7 (Rivera-Campo [10]) Let n ≥ 2, k ≥ 3, 0 ≤ c ≤ n and G be an n-connected graph with σn(k−1)+c+2 (G) ≥ {n(k − 1) + c}(|V(G)| − n − c − 2)/(k − 1) + 2n + 1. Then G has a spanning tree with maximum degree at most k + 1 and with at most c vertices of degree k + 1. In case of n ≥ 2, this theorem is a common generalization of Theorems 5 and 6. However, the degree condition of this theorem is rather complicated, and there is a fraction in the statement though σn(k−1)+c+2 (G) is always integer. Therefore, it is empirically guessed that this theorem can be appropriately generalized (See [2]). 2 New Result In this paper, we give another common generalization of Theorems 5 and 6. The following is our new result. Theorem 8 Let n ≥ 1, k ≥ 3, 0 ≤ c ≤ n and G be an n-connected graph with σkn(k−1)+c+2 (G) ≥ |V(G)| − c − 1. Then G has a spanning tree with maximum degree at most k + 1 and with at most c vertices of degree k + 1. Since σkn(k−1)+c+2 (G) ≥ σk (G) (= σkk (G)) for every graph G, Theorem 8 is a generalization of Theorem 5. And if α(G) ≤ n(k − 1) + c + 1, then σkn(k−1)+c+2 (G) = ∞. Hence Theorem 8 is also a generalization of Theorem 6. Considering that σkn(k−1)+c+2 (G) is always integer, by Theorems 5, 6 and 8, we are led to the following principle: To obtain a spanning tree with maximum degree at most k in an n-connected graph, degree condition on n(k − 1) + 2 independent vertices is essential. And among them, degree sum of k vertices plays an important role. In fact, we can eliminate the condition “0 ≤ c ≤ n” in Theorem 8 and prove more general result. To describe this, we define total excess from k of a graph G as TE(G, k) :=
X
max{dG (v) − k, 0}.
v∈V(G)
Our main result in this paper is the following. Theorem 9 Let n ≥ 1, k ≥ 3, c ≥ 0 and G be an n-connected graph with σkn(k−1)+c+2 (G) ≥ |V(G)| − c − 1. Then G has a spanning tree T with maximum degree at most k + ⌈c/n⌉ and TE(T, k) ≤ c. In case of 0 ≤ c ≤ n, Theorems 8 and 9 are equivalent. The degree condition in Theorem 9 is best possible in the following sense: Let G be a complete bipartite graph with partite sets X and Y such that |X| = n and |Y| = n(k − 1) + c + 2, where n ≥ 1 and k ≥ 3. Then G is n-connected and σkn(k−1)+c+2 (G) = kn = |V(G)| − c − 2. Let T be a tree of G with maximum degree at most k + ⌈c/n⌉ and TE(T, k) ≤ c. Then the number of edges in T is at most k|X| + c = kn + c. On the other hand, T cannot span V(G) since |V(G)| = kn + c + 2. Moreover, Theorem 9 is stronger than Theorem 7. This will be proved in Section 5. 3 Preliminaries Before proving Theorem 9, we prepare some notation and facts.
3
For a tree T and u, v ∈ V(T ), we denote the unique path in T connecting u and v by T [u, v]. If u , v, then NT (u) ∩ T [u, v] consists of a single vertex and we denote this vertex by uvT . When there is no fear of confusion, we simply denote uv instead of uvT . A forest is a graph whose components are trees. For H ⊆ G and x ∈ V(G − H), a subgraph F of G is called an S (x, H)-fan of width k if F has the decomposition F = ki=1 Pi , where each Pi is a path connecting x and vi ∈ V(H), V(Pi ) ∩ V(H) = {vi } and V(Pi ) ∩ V(P j ) = {x} if i , j. We call F an (x, H)-fan of maximum width if G does not
contain an (x, H)-fan of width greater than that of F. In our proof of Theorem 9, we often consider a tree T as an outdirected tree, that is, all the edges in T are directed away from the root r ∈ V(T ). In such a case, for u ∈ V(T ) \ {r}, we denote NT+ (u) = NT (u) \ {ur } and NT− (u) = {ur }. ur is sometimes denoted by u− . And in case of u = r, we denote NT+ (u) = NT (u) and NT− (u) = ∅. Moreover, for u ∈ V(T ), we denote dT+ (u) = |NT+ (u)|. The following two facts are easy observations on trees. Fact 1 Let T be a tree and let u, v1 , v2 be distinct vertices in T . Then (NT (v1 ) \ {v1 u }) ∩ (NT (v2 ) \ {v2 u }) = ∅. Furthermore, if uv < E(T ), then (NT [v1 ] \ {v1 u }) ∩ (NT [v2 ] \ {v2 u }) = ∅. Fact 2 Let T be a tree and {v1 , v2 , . . . , vt } be an independent set of T . Then T − {v1 , v2 , . . . , vt } has exactly Pt i=1 (dT (vi ) − 1) + 1 components. 4 Proof of Theorem 9 In our proof of Theorem 9, a tree T of G is called a (k, c, n)-tree if maximum degree of T is at most k + ⌈c/n⌉ and TE(T, k) ≤ c. Take a (k, c, n)-tree T of G so that |V(T )| is as large as possible. We may assume that V(T ) , V(G) since otherwise we have nothing to prove. Claim 1 |V(T )| ≥ n. Proof. Suppose that |V(T )| < n and take u ∈ V(T ) such that dT (u) = 1. Since G is n-connected, there exists x ∈ NG (u) \ V(T ). Then T + xu is a (k, c, n)-tree larger than T , a contradiction.
Take any u0 ∈ V(G) \ V(T ) and let T 0 be the component of G − T containing u0 . Let F be a (u0 , T )-fan of maximum width and let S 1 = V(F) ∩ V(T ). Since G is n-connected and by Claim 1, |S 1 | ≥ n. We denote VT≥t := {v ∈ V(T ) | dT (v) ≥ t} and let S 2 = VT≥k+1 \ S 1 . Claim 2 S 1 ⊆ VT≥k . Proof. Suppose that dT (v) < k for some v ∈ S 1 . Then T˜ = T ∪ F[u0 , v] is a (k, c, n)-tree with |V(T˜ )| > |V(T )|, a contradiction.
Choose r ∈ S 1 so that dT (r) = minv∈S 1 dT (v). Claim 3 TE(T, k) =
P
v∈S 1 ∪S 2 (dT (v)
− k) = c.
4
Proof. Note that only the vertices of VT≥k+1 contribute to TE(T, k). Since VT≥k+1 ⊆ S 1 ∪ S 2 and dT (v) − k ≥ 0 for P every v ∈ S 1 ∪ S 2 , we obtain TE(T, k) = v∈S 1 ∪S 2 (dT (v) − k). Next, suppose that TE(T, k) < c. It follows from the fact |S 1 | ≥ n and the choice of r that dT (r) ≤ k + TE(T, k)/n < k + c/n ≤ k + ⌈c/n⌉. Thus T˜ = T ∪ F[u0 , r] is a (k, c, n)-tree with |V(T˜ )| > |V(T )|, a contradiction. Claim 4 If S 2 , ∅, then dT (r) ≤ k + ⌈c/n⌉ − 1. Proof. Assume the contrary. Since dT (v) ≥ dT (r) for any v ∈ S 1 , X
dT (v) ≥
v∈S 1
And, since S 2 , ∅,
P
v∈S 2 (dT (v)
X
dT (r) ≥ |S 1 |(k + ⌈c/n⌉) ≥ |S 1 |k + c.
v∈S 1
− k) ≥ 1. Hence it follows from Claim 3 that TE(T, k) =
X
(dT (v) − k) +
v∈S 1
=
X
(dT (v) − k)
v∈S 2
|S 1 |k + c − |S 1 |k +
X
(dT (v) − k)
v∈S 2
≥
c + 1,
which contradicts that T is a (k, c, n)-tree.
Claim 5 NT (s) ∩ S 1 = ∅ for any s ∈ S 1 . Proof. Suppose that there exist two vertices v, v′ ∈ S 1 with v′ ∈ NT (v). Then (T − vv′ ) ∪ F[u0 , v] ∪ F[u0 , v′ ] is a (k, c, n)-tree, which contradicts the maximality of T .
In the following, we consider T as an outdirected tree with the root r. Claim 6 NT+ (s) ∩ S 1 = ∅ for any s ∈ S 2 . Proof. Suppose that NT+ (s) ∩ S 1 , ∅ for some s ∈ S 2 and let s′ ∈ NT+ (s) ∩ S 1 . Let T˜ = (T − ss′ ) ∪ F[u0 , s′ ] ∪ F[u0 , r]. Then T˜ is a tree with |V(T˜ )| > |V(T )|, dT˜ (v) = dT (v) for every v ∈ V(T ) \ {r, s}, dT˜ (r) = dT (r) + 1 and dT˜ (s) = dT (s) − 1. Moreover, dT˜ (v) = 2 for every v ∈ V(T˜ ) \ V(T ). Hence Claim 3 implies that TE(T˜ , k) = TE(T, k) + 1 − 1 = c, and Claim 4 implies that dT˜ (v) ≤ k + ⌈c/n⌉ for every v ∈ V(T˜ ). Therefore, T˜ is a (k, c, n)-tree larger than T , a contradiction.
Now we make a forest T ′ from T by the following operations: 1. Delete all the vertices of S 1 . 2. For each s ∈ S 2 , choose NT+ (s)∗ ⊆ NT+ (s) so that |NT+ (s)∗ | = dT (s) − k and delete the edge set {ss′ | s′ ∈ NT+ (s)∗ }. By Claims 5 and 6, there is no duplication in edges deleted by these operations. By Fact 2 and Claim 3, T ′ has a
=
X
(dT (s) − 1) +
s∈S 1
X
s∈S 2
5
(dT (s) − k) + 1
=
X
(dT (s) − k) + (k − 1)|S 1 | + 1
s∈S 1 ∪S 2
= (k − 1)|S 1 | + c + 1 ≥ (k − 1)n + c + 1
(1)
components. Let T 1 , . . . , T a be components of T ′ , A = {0, 1, · · · , a}, and A′ = A \ {0}. Note that we can consider T i as an outdirected tree by the same orientation of T . Under this consideration, for i ∈ A′ , let ri be the root of T i and ui be a vertex of T i such that dT+i (ui ) = 0. Moreover let U = {u0 , u1 , . . . , ua }. Note that, for every i with i ∈ A′ , it follows that NT+ (ui ) ⊆ S 1 and ui < S 1 ∪ S 2 , and so we have dT (ui ) ≤ k. Suppose that i, j ∈ A′ with i , j and u′j ∈ V(T j ) is a vertex such that u′j = u j or NT+ (u′j ) ∩ S 1 , ∅. Now, for the technical reason, we define two forests F1 (ui , u′j ) and F2 (ui , u′j ) in G. To show the structure of these forests, we need some definitions. It follows from i , j that V(T [ui , u′j ]) ∩ (S 1 ∪ S 2 ) , ∅. If V(T [ui , u′j ]) ∩ S 1 , ∅, let w be the vertex of V(T [ui , u′j ]) ∩ S 1 and let w′ ∈ V(T [ui , u′j ]) ∩ NT (w). Otherwise, by the construction of T ′ , we can take w ∈ V(T [ui , u′j ]) ∩ S 2 so that V(T [ui , u′j ]) ∩ NT+ (w)∗ , ∅, and let w′ ∈ V(T [ui , u′j ]) ∩ NT+ (w)∗ . Note that ww′ < E(T ′ ) in both cases. Moreover, for v ∈ S 1 ∪ S 2 , we define 1 v if v ∈ S , f (v) = r if v ∈ S 2 .
First we describe the structure of F1 (ui , u′j ). For v ∈ V(T ), we write u ≺ v if u ∈ V(T [r, v]), and otherwise we
write u ⊀ v. Case 1. ui ⊀ u′j and u′j ⊀ ui . If dT (ui ) = dT (u′j ) = 1, then let F1 (ui , u′j ) = (T − ww′ ) ∪ F[u0 , f (w)] (See Figures 1 and 2). If dT (ui ) ≥ 2 and dT (u′j ) = 1, take any u+i ∈ NT+ (ui ). Note that u+i ∈ S 1 since NT+ (ui ) ⊆ S 1 . Let F1 (ui , u′j ) = (T − ww′ − ui u+i ) ∪ F[u0 , f (w)] ∪ F[u0 , u+i ] (See Figures 3 and 4). If dT (ui ) = 1 and dT (u′j ) ≥ 2, take any u′j + ∈ NT+ (u′j ) ∩ S 1 and let F1 (ui , u′j ) = (T − ww′ − u′j u′j + ) ∪ F[u0 , f (w)] ∪ F[u0 , u′j + ] (See Figures 5 and 6). If dT (ui ) ≥ 2 and dT (u′j ) ≥ 2, take any u+i ∈ NT+ (ui ) and u′j + ∈ NT+ (u′j ) ∩ S 1 , and let F1 (ui , u′j ) = (T − ww′ − ui u+i − u′j u′j + ) ∪ F[u0 , f (w)] ∪ F[u0, u+i ] ∪ F[u0 , u′j + ] (See Figures 7 and 8). Case 2. ui ≺ u′j . u′
In this case, we have dT (ui ) ≥ 2. Then NT+ (ui ) is not empty and contained in S 1 , and hence u+i = ui j is in S 1 . In case of dT (u′j ) = 1, let F1 (ui , u′j ) = (T − ui u+i ) ∪ F[u0 , u+i ] (See Figure 9). And in case of dT (u′j ) ≥ 2, take any u′j + ∈ NT+ (u′j ) ∩ S 1 and let F1 (ui , u′j ) = (T − ui u+i − u′j u′j + ) ∪ F[u0 , u+i ] ∪ F[u0 , u′j + ] (See Figure 10). Case 3. u′j ≺ ui . First consider the case where u′j ui is in S 1 . Let u′j + = u′j ui . In case of dT (ui ) = 1, let F1 (ui , u′j ) = (T − u′j u′j + ) ∪ F[u0 , u′j + ] (See Figure 11). And in case of dT (ui ) ≥ 2, take any u+i ∈ NT+ (ui ) and let F1 (ui , u′j ) = 6
r u0
u0 ×
× ′
w w′
w w
u′j
ui
u′j
ui
Figure 1: In case of w ∈ S 1 .
Figure 2: In case of w ∈ S 2 . r
u0
u0 × w w′
× w w′ u′j
× ui
u′j
× ui
u+i
u+i
Figure 3: In case of w ∈ S 1 .
Figure 4: In case of w ∈ S 2 . r u0
u0
× w w′
× w w′ ′ × uj
ui
′ × uj
ui
u′j +
u′j + Figure 5: In case of w ∈ S 1 .
Figure 6: In case of w ∈ S 2 . r
u0
u0 × w w′
× w w′
× ui
×
u′j
× ui
u′j +
u+i
×
u′j +
u+i
Figure 7: In case of w ∈ S 1 .
Figure 8: In case of w ∈ S 2 .
ui ui u0
×
u0
×
u+i
u+i u′j
× u′j
u′j +
Figure 9: In case of dT (u′j ) = 1.
Figure 10: In case of dT (u′j ) ≥ 2.
7
u′j
u′j u′j u0
×
u0
×
u′j +
u′j + ui
× u+i
ui
Figure 12: In case of dT (u′j ) ≥ 2. r
Figure 11: In case of dT (ui ) = 1. u′j u0
u′j
× + w u′j × ′ w
u0
× + w u′j × ′ w
ui
ui
Figure 13: In case of w ∈ S 1 .
Figure 14: In case of w ∈ S 2 . r
u′j u0
u′j
× + w u′j × ′ w ×
u0
ui
× + w u′j × ′ w ×
u+i
Figure 15: In case of w ∈ S 1 .
ui u+i
Figure 16: In case of w ∈ S 2 .
(T − u′j u′j + − ui u+i ) ∪ F[u0 , u′j + ] ∪ F[u0 , u+i ] (See Figure 12). Next consider the case where u′j ui is not in S 1 . Let u′j + be a vertex in NT+ (u′j ) ∩ S 1 . In case of dT (ui ) = 1, let F1 (ui , u′j ) = (T − ww′ − u′j u′j + ) ∪ F[u0 , u′j + ] ∪ F[u0 , f (w)] (See Figures 13 and 14). And in case of dT (ui ) ≥ 2, take any u+i ∈ NT+ (ui ) and let F1 (ui , u′j ) = (T − ww′ − u′j u′j + − ui u+i ) ∪ F[u0 , u′j + ] ∪ F[u0 , f (w)] ∪ F[u0 , u+i ] (See Figures 15 and 16). Above Case 1 to 3 cover all the cases, and so we can take F1 (ui , u′j ) in any situations. Next, we show the structure of F2 (ui , u′j ). In case of dT (ui ) ≤ 2, we take F1 (ui , u′j ) as F2 (ui , u′j ). In case of dT (ui ) ≥ 3, take any v1 , v2 ∈ NT+ (ui ). Again, note that v1 , v2 ∈ S 1 . If ui ⊀ u′j , then let F2 (ui , u′j ) = (T −ww′ −ui v1 −ui v2 )∪F[u0 , f (w)]∪F[u0 , v1 ]∪F[u0 , v2 ] (See Figures 17, 18, 19 u′
and 20). And if ui ≺ u′j , take v′1 , v′2 ∈ NT+ (ui ) so that v′1 = ui j . Let F2 (ui , u′j ) = (T −u′i v1 −ui v′2 )∪F[u0 , v′1 ]∪F[u0 , v′2 ] (See Figure 21). By the construction of F1 (ui , u′j ) and F2 (ui , u′j ), we have the following fact, which we often use in our proof. Fact 3 F1 = F1 (ui , u′j ) and F2 = F2 (ui , u′j ) satisfy all of the followings for l = 1 and 2. (i) Fl has exactly two components. (ii) V(T ) ( V(Fl ) and TE(Fl , k) ≤ TE(T, k). (iii) dFl (v) ≤ dT (v) for all v ∈ V(T ) \ {r}. Moreover, dFl (r) ≤ dT (r) if S 2 = ∅. (iv) The maximum degree of Fl is at most k + ⌈c/n⌉. 8
r u0
u0 × w w′
× w w′
ui × × v1
ui × ×
u′j v2
v1
Figure 17: In case of u′j ⊀ ui and w ∈ S 1 .
u′j v2
Figure 18: In case of u′j ⊀ ui and w ∈ S 2 . r
u′j
u′j
w × w′
u0
u0
ui × × v1 v2
w × w′ ui × × v1 v2
Figure 19: In case of u′j ≺ ui and w ∈ S 1 .
u0
Figure 20: In case of u′j ≺ ui and w ∈ S 2 .
v′2 ×
ui ×′ v1
u′j Figure 21: In case of w ∈ S 2 .
(v) ui and u′j belong to different components of Fl . (vi) dF1 (ui ) ≤ k − 1, dF1 (u′j ) ≤ k − 1, and dF2 (ui ) ≤ k − 2. (vii) If vv+ < E(T [ui , u′j ]) and v , ui , u′j for v ∈ V(T ) and v+ ∈ NT+ (v), then vv+ ∈ E(Fl ) . Moreover, E(T i ) ∪ E(T j ) ⊆ E(Fl ). The next lemma is also used several times in our proof. Lemma 1 Let i ∈ A′ and suppose that s ∈ S 1 ∪ S 2 . Then NG (ui ) ∩ (NT (s) \ {sui }) = ∅. Proof. Suppose that there exists s′ ∈ NG (ui )∩(NT (s)\{sui }). If dT (ui ) ≤ k−1, then (T − ss′ +ui s′ )+F[u0 , f (s)] is a (k, c, n)-tree, which contradicts the maximality of T . So we assume that dT (ui ) = k. Then |NT+ (ui )| ≥ k − 1 ≥ 2, and hence there exists u+i ∈ NT+ (ui ) \ {ui s }. Since u+i ∈ S 1 , (T − ss′ − ui u+i + ui s′ ) ∪ F[u0 , f (s)] ∪ F[u0 , u+i ] is a (k, c, n)-tree, which contradicts the maximality of T .
Now we continue the proof of Theorem 9. For i ∈ A and j ∈ A′ , let Wi, j = NG (ui ) ∩ V(T j ). Claim 7 W0, j = ∅ for all j ∈ A′ . Proof. Since F is a (u0 , T )-fan of maximum width, NG (u0 ) ∩ V(T ) ⊆ S 1 = V(T ) \ V(T ′ ). Hence W0, j = ∅. 9
Claim 8 Let i, j ∈ A′ with i , j. Then for any v ∈ Wi, j , NT (v) ∩ Wi, j = ∅. Proof. Suppose that v j v′j ∈ E(T ) for v j , v′j ∈ Wi, j . By (vii) of Fact 3 and the fact v j v′j ∈ E(T j ), we have the forest F2 (ui , u j ) with v j v′ ∈ E(F2 (ui , u j )). With (v) of Fact 3, we obtain that T˜ = F2 (ui , u j ) − v j v′ + ui v j + ui v′ is j
j
j
connected. It is easy to see that T˜ is a (k, c, n)-tree, which contradicts the maximality of T .
Claim 9 Let i, j ∈ A′ with i , j. For any v ∈ Wi, j \ {r j }, dT j (v) = k. Moreover, if r j ∈ Wi, j , then dT j (r j ) = k − 1. Proof. It suffices to show that dT+j (v) = k − 1 for any v ∈ Wi, j . If v ∈ S 2 , then dT+j (v) = k − 1 holds by the construction of T ′ . Hence we may assume that v < S 2 . If NT+ (v) ∩ S 1 , ∅, then we have a forest F1 (ui , v). However, F1 (ui , v) + ui v contradicts the maximality of T . Hence NT+ (v) ∩ S 1 = ∅, and so dT+ (v) = dT+j (v). If dT (v) < k, then by (iii) of Fact 3, dF1 (ui ,u j ) (v) < k, and then F1 (ui , u j ) + ui v contradicts the maximality of T . Hence dT (v) ≥ k. Since we already have v < S 1 ∪ S 2 , dT (v) ≤ k holds. This implies dT (v) = k. Therefore, since v , r, we have dT+j (v) = dT+ (v) = dT (v) − 1 = k − 1.
Claim 10 U is independent in G. Proof. By Claim 7, u0 ul < E(G) for any l ∈ A′ . Suppose that ui u j ∈ E(G) for some i, j ∈ A′ . Then F1 (ui , u j ) + ui u j contradicts the maximality of T . Therefore ui u j < E(G) for all i, j ∈ A′ , and so this claim holds. Since |U| = a + 1 ≥ n(k − 1) + c + 2 and U is independent, there exists U ∗ ⊂ U such that |U ∗ | = k and X dG (u) ≥ |V(G)| − c − 1.
(2)
u∈U ∗
Take A∗ ⊂ A so that
S
i∈A∗ {ui }
= U ∗.
For j ∈ A′ , we define ξ+j
1 = 0
if r j ∈ NG (ui ) for all i ∈ A∗ , otherwise.
Then since |A∗ | = k, it follows immediately that 1 X |Wi, j ∩ {r j }| = ξ+j . k i∈A∗
(3)
For j ∈ A′ , let
X j = {s ∈ V(T j ) ∩ S 2 | s = r−j′ for some j′ ∈ A′ with ξ+j′ = 1}.
(4)
Claim 11 |X j | ≤ 1 for each j ∈ A′ . Proof. Suppose that s′ , s′′ ∈ X j . Then there exist j′ , j′′ ∈ A′ such that s′ = r−j′ , s′′ = r−j′′ and ui r j′ , ui r j′′ ∈ E(G) for i ∈ A∗ \ {0}. By Lemma 1, we have r j′ = s′ ui and r j′′ = s′′ ui . Then T j ∪ T [s′ , ui ] ∪ T [s′′ , ui ] contains a cycle, which contradicts that T is a tree.
For j ∈ A′ , we define ξ−j = |X j |. In case of X j , ∅, i.e. ξ−j = 1, we denote the unique vertex in X j by x j . 10
Claim 12 Suppose that ξ−j = ξ+j′ = 1 and x j = r−j′ for some j, j′ ∈ A′ . Then ξ+j′′ = 0 for every j′′ ∈ A′ \ { j′ } which satisfies x j = r−j′′ . Proof. Note that x j ∈ S 2 . Let i ∈ A∗ . Then by Lemma 1, |NT (x j ) ∩ NG (ui )| ≤ 1. Since ξ+j′ = 1, r j′ ∈ NT (x j ) ∩ NG (ui ). Hence r j′′ < NT (x j ) ∩ NG (ui ). Now r j′′ ∈ NT (x j ), and so r j′′ < NG (ui ), which implies ξ+j′′ = 0. Let j ∈ A′ with ξ+j = 1. Then r−j ∈ V(T j′ ) for some j′ ∈ A′ . We denote this j′ by j− . Note that, by (4), it follows that ξ−j− = 1. And if ξ−j = 1 for some j, it follows from (4) that there exists j′ ∈ A′ such that j′ − = j. Now Claim 12 implies that such j′ is uniquely determined, and we denote it by j+ . Claim 13 Suppose that ξ−j = 1 for some j ∈ A′ . Then NT j [x j ] ∩ Wi, j = ∅ for any i ∈ A∗ \ { j+ }. Proof. Let i ∈ A∗ \ { j+ }. Since ξ−j = 1, r j+ ∈ NG (ui ). It follows from Lemma 1 and the fact x j ∈ S 2 that xuj i = r j+ . Again using Lemma 1, we obtain that v < NG (ui ) for all v ∈ NT j (x j ), and hence NT j (x j ) ∩ Wi, j = ∅. Therefore, it suffices to prove that x j < NG (ui ). Assume the contrary. It follows from x j ≺ r j+ and xuj i = r j+ that r j+ ≺ ui . With the fact r j+ ≺ u j+ , we obtain x j r j+ < E(T [ui , u j+ ]). Hence by (vii) of Fact 3 and the fact that x j , u j , we have a forest F2 (ui , u j+ ) with x j r j+ ∈ E(F2 (ui , u j+ )). Now F2 (ui , u j+ ) − x j r j+ + ui x j + ui r j+ contradicts the maximality of T , and so the claim holds.
Claim 14 For every i ∈ A∗ and j ∈ A′ with i , j, 1 k (|V(T j )| − 1 + |(Wi, j ∪ X j ) ∩ {r j }|) |Wi, j | ≤ 1 (|V(T j )| − 1 + |(Wi, j ∪ X j ) ∩ {r j }|) − ξ− j
k
if ξ−j = 1 and i = j+ , otherwise.
Proof. First, we show that right sides of the above inequalities are not less than 0. Since |V(T j )| ≥ 1 for any j ∈ A′ , 1k (|V(T j )| − 1 + |(Wi, j ∪ X j ) ∩ {r j }|) ≥ 0 holds. Suppose that ξ−j = 1, then x j ∈ S 2 . By the construction of T ′ , we have |V(T j )| ≥ k + 1 |V(T j )| ≥ k and |(Wi, j ∪ X j ) ∩ {r j }| = 1
if x j , r j , if x j = r j .
Hence 1k (|V(T j )| − 1 + |(Wi, j ∪ X j ) ∩ {r j }|) − ξ−j ≥ 0.
If i = 0, then Claim 7 implies that |Wi, j | = 0, and hence the claim holds. So we may assume that i , 0. Case 1. ξ−j = 1 and i = j+ . By Claim 8, vv′ < E(T j ) for any v, v′ ∈ Wi, j with v , v′ . Hence by Fact 1, we have (NT j [v] \ {vu j }) ∩ (NT j [v′ ] \ {v′ u j }) = ∅. It follows from Claim 10 that u j < Wi, j , thus Claim 9 implies X |V(T j )| ≥ |NT j [v] \ {vu j }| + |{u j }| = k|Wi, j | − |Wi, j ∩ {r j }| + 1. v∈Wi, j
Therefore, |Wi, j |
≤
1 1 (|V(T j)| − 1 + |Wi, j ∩ {r j }|) ≤ (|V(T j)| − 1 + |(Wi, j ∪ X j ) ∩ {r j }|). k k
Case 2. Otherwise. By Claims 8 and 13, we have X j ∩ Wi, j = ∅ and vv′ < E(T j ) for any v, v′ ∈ X j ∪ Wi, j with v , v′ . Hence by Fact 1, we have (NT j [v] \ {vu j }) ∩ (NT j [v′ ] \ {v′ u j }) = ∅. Note that, if |X j | , ∅, then x j , u j since x j ∈ S 2 . With 11
Claim 10 we have u j < X j ∪ Wi, j . Moreover, if |X j | , ∅, it follows from x j ∈ S 2 that dT j (x j ) = k in case of x j , r j , and dT j (x j ) = k − 1 in case of x j = r j . Therefore, by Claim 9, it follows that X |V(T j)| ≥ |NT j [v] \ {vu j }| + |{u j }| v∈Wi, j ∪X j
=
k|Wi, j ∪ X j | − |(Wi, j ∪ X j ) ∩ {r j }| + 1
=
k|Wi, j | + k|X j | − |(Wi, j ∪ X j ) ∩ {r j }| + 1
=
k|Wi, j | + kξ−j − |(Wi, j ∪ X j ) ∩ {r j }| + 1,
which implies |Wi, j | ≤ 1k (|V(T j )| − 1 + |(Wi, j ∪ X j ) ∩ {r j }|) − ξ−j . For j ∈ A′ , we define ξ j = ξ+j − ξ−j . Claim 15
P
i∈A∗
|Wi, j | ≤ |V(T j )| − 1 + ξ j for any j ∈ A′ \ A∗ .
Proof. If ξ−j = 1 and j+ ∈ A∗ , then by Claim 14, X X |Wi, j | = |Wi, j | + |W j+ , j | i∈A∗ i∈A∗ \{ j+ } ! X 1 1 ≤ (|V(T j)| − 1 + |(Wi, j ∪ X j ) ∩ {r j }|) − ξ−j + (|V(T j)| − 1 + |(W j+ , j ∪ X j ) ∩ {r j }|) k k i∈A∗ \{ j+ } 1X = |V(T j )| − 1 − (k − 1)ξ −j + |(Wi, j ∪ X j ) ∩ {r j }| k ∗ i∈A 1 X − = |V(T j )| − 1 − (k − 1)ξ j + |(Wi, j ∪ X j ) ∩ {r j }| . k i∈A∗
And otherwise,
X
|Wi, j |
≤
i∈A∗
= = ≤ Therefore, it suffices to show that
! X 1 (|V(T j)| − 1 + |(Wi, j ∪ X j ) ∩ {r j }|) − ξ−j k i∈A∗ 1X − |V(T j )| − 1 − kξ j + (|(Wi, j ∪ X j ) ∩ {r j }|) k ∗ i∈A 1 X − |V(T j )| − 1 − kξ j + |(Wi, j ∪ X j ) ∩ {r j }| k i∈A∗ 1 X |(Wi, j ∪ X j ) ∩ {r j }| . |V(T j )| − 1 − (k − 1)ξ −j + k i∈A∗
−(k − in both cases.
1)ξ −j
1 X + |(Wi, j ∪ X j ) ∩ {r j }| ≤ ξ j k i∈A∗
If ξ−j = 0, then by (3), 1 X 1 X −(k − 1)ξ −j + |(Wi, j ∪ X j ) ∩ {r j }| = |Wi, j ∩ {r j }| = ξ+j = ξ+j − ξ−j = ξ j . k i∈A∗ k i∈A∗ 12
And if ξ−j = 1, then −(k −
1)ξ −j
$ % 1 X 1 + |(Wi, j ∪ X j ) ∩ {r j }| ≤ −(k − 1) + · k = −(k − 2) ≤ −1 ≤ −ξ−j + ξ+j = ξ j . k i∈A∗ k
Hence the claim holds.
Claim 16 ξ−j = 0 for any j ∈ A∗ \ {0}. u
Proof. Suppose that ξ−j = 1, i.e. X j , ∅, for some j ∈ A∗ \ {0}, then r j+ u j ∈ E(G). However r j+ , x j j , which contradicts Lemma 1.
For each j ∈ A∗ \ {0}, we define W∗, j =
[
Wi, j .
(5)
i∈A∗ \{ j}
Claim 17 Let j ∈ A∗ \ {0} such that |V(T j )| ≥ 2. Then, |W j, j | ≤ |V(T j )| − (k − 1)|W∗, j | + |W∗, j ∩ {r j }| − 1. Furthermore, if r j < NG (u j ) and r j
Hajime Matsumura2
Tomoki Yamashita3
Abstract In this paper, we give a sufficient condition for a graph to have a degree bounded spanning tree. Let n ≥ 1, k ≥ 3, c ≥ 0 and G be an n-connected graph. Suppose that for every independent set S ⊆ V(G) of cardinality n(k − 1) + c + 2, there exists a vertex set X ⊆ S of cardinality k such that the degree sum of vertices in X is at least |V(G)| − c − 1. Then G has a spanning tree T with maximum degree at most k + ⌈c/n⌉ and P v∈V(T ) max{dT (v) − k, 0} ≤ c.
Keywords: spanning tree, degree bounded tree, degree sum condition, independence number, total excess 1 Introduction We will generally follow notation and terminology of [3]. For a vertex x of a graph G, we denote the degree of x in G by dG (x) and the set of vertices adjacent to x in G by NG (x). Moreover, NG (x) ∪ {x} is denoted by NG [x]. For a subset S of V(G), let G − S denotes the subgraph induced by V(G) \ S . For a subgraph H of G, G − H denotes the subgraph induced by V(G) \ V(H). Here we define several invariants of a graph G. Let α(G) be the independence number of G. For a graph G with α(G) ≥ r, we define X dG (v) S is an independent set of G with |S | = r σr (G) := min v∈S
and σr (G) := ∞ if α(G) < r.
Next, for S ⊆ V(G) with |S | ≥ k, let X . dG (v) X ⊆ S , |X| = k ∆k (S ) = max v∈X
And we define
σrk (G) := min{∆k (S ) | S is an independent set of G with |S | = r} for k ≤ r and a graph G with α(G) ≥ r. If α(G) < r, we define σrk (G) := ∞. The following proposition shows the relation between σrk (G) and σr (G). Proposition 1 For any graph G and two integers r and k with k ≤ r, σrk (G) ≥ kσr (G)/r. Proof. If α(G) < r, then σrk (G) = σr (G) = ∞. So assume that α(G) ≥ r. For every independent set S P of cardinality r, ∆k (S ) ≥ k v∈S dG (v)/r ≥ kσr (G)/r. Hence by the definition of σrk (G), we obtain σrk (G) ≥
kσr (G)/r.
1 [email protected].
Department of Computer Science, Nihon University, Sakurajosui 3-25-40, Setagaya-Ku, Tokyo,
156-8550, Japan. 2 h [email protected]. Kyoto Computer Gakuin, 10-5, Teranomae-cho, Nishikujyo, Minami-ku, Kyoto, 601-8407 Japan. 3 [email protected]. Department of Mathematics, Asahi University, Gifu, 501-0296 Japan.
1
Now we introduce some theorems concerning the existence of Hamilton cycles. The following two theorems are well-known. Theorem 1 (Ore [9]) Let G be a graph with |V(G)| ≥ 3 and σ2 (G) ≥ |V(G)|. Then G has a Hamilton cycle. Theorem 2 (Chv´atal and Erd˝os [5]) Let G be an n-connected graph with |V(G)| ≥ 3 and α(G) ≤ n. Then G has a Hamilton cycle. It follows from Theorem 2 that, about the existence of Hamilton cycle, we may consider only graphs in which independence number is greater than its connectivity. In this point of view, the following theorem was obtained, which is a generalization of Theorems 1 and 2. Theorem 3 (Bondy [1]) Let G be an n-connected graph with |V(G)| ≥ 3 and n ≥ 1. If σn+1 (G) ≥ (n+1)(|V(G)|− 1)/2, then G has a Hamilton cycle. Therefore, if G is an n-connected graph, the condition on n + 1 independent vertices is essential for the existence of a Hamilton cycle. But in fact, after taking n + 1 independent vertices, there is no need to consider the degree sum of all those vertices. The following theorem suggests this assertion. Theorem 4 (Yamashita [13]) Let G be an n-connected graph with |V(G)| ≥ 3, where n ≥ 1. If σ2n+1 (G) ≥ |V(G)|, then G has a Hamilton cycle. Therefore, we have: To obtain a Hamilton cycle in an n-connected graph, degree condition on n + 1 independent vertices is essential. And among them, two vertices of high degree plays an important role. Next we turn our view to the existence of a degree bounded spanning tree. On this subject, several results are already known. For example, see [4, 6, 7, 8, 11, 12]. Here we introduce the following two results. Theorem 5 (Win [11]) Let G be a connected graph with σk (G) ≥ |V(G)| − 1, where k ≥ 2. Then G has a spanning tree with maximum degree at most k. Theorem 6 (Neumann-Lara and Rivera-Campo [8]) Let n ≥ 1, k ≥ 3, 0 ≤ c ≤ n and G be an n-connected graph with α(G) ≤ n(k − 1) + c + 1. Then G has a spanning tree with maximum degree at most k + 1 and with at most c vertices of degree k + 1. Now the situation is similar to the case of Hamilton cycle. That is, for a spanning tree with maximum degree at most k, degree sum of k independent vertices is considered in Theorem 5. However, it follows from Theorem 6 (in case of c = 0) that we may consider only graphs G which satisfy α(G) > n(k − 1) + 1, where n is the connectivity of G, and so it is natural to take degree sum of n(k − 1) + 2 vertices as a condition. Moreover, relaxing the condition on independence number to n(k − 1) + c + 1, it is considered in Theorem 6 that the case where there are at most c vertices of degree k + 1 in the resulting spanning tree. Therefore, as an extension of Theorem 6, a theorem which consider the degree sum of n(k − 1) + c + 2 independent vertices may exist: In fact, the following theorem is one of that.
2
Theorem 7 (Rivera-Campo [10]) Let n ≥ 2, k ≥ 3, 0 ≤ c ≤ n and G be an n-connected graph with σn(k−1)+c+2 (G) ≥ {n(k − 1) + c}(|V(G)| − n − c − 2)/(k − 1) + 2n + 1. Then G has a spanning tree with maximum degree at most k + 1 and with at most c vertices of degree k + 1. In case of n ≥ 2, this theorem is a common generalization of Theorems 5 and 6. However, the degree condition of this theorem is rather complicated, and there is a fraction in the statement though σn(k−1)+c+2 (G) is always integer. Therefore, it is empirically guessed that this theorem can be appropriately generalized (See [2]). 2 New Result In this paper, we give another common generalization of Theorems 5 and 6. The following is our new result. Theorem 8 Let n ≥ 1, k ≥ 3, 0 ≤ c ≤ n and G be an n-connected graph with σkn(k−1)+c+2 (G) ≥ |V(G)| − c − 1. Then G has a spanning tree with maximum degree at most k + 1 and with at most c vertices of degree k + 1. Since σkn(k−1)+c+2 (G) ≥ σk (G) (= σkk (G)) for every graph G, Theorem 8 is a generalization of Theorem 5. And if α(G) ≤ n(k − 1) + c + 1, then σkn(k−1)+c+2 (G) = ∞. Hence Theorem 8 is also a generalization of Theorem 6. Considering that σkn(k−1)+c+2 (G) is always integer, by Theorems 5, 6 and 8, we are led to the following principle: To obtain a spanning tree with maximum degree at most k in an n-connected graph, degree condition on n(k − 1) + 2 independent vertices is essential. And among them, degree sum of k vertices plays an important role. In fact, we can eliminate the condition “0 ≤ c ≤ n” in Theorem 8 and prove more general result. To describe this, we define total excess from k of a graph G as TE(G, k) :=
X
max{dG (v) − k, 0}.
v∈V(G)
Our main result in this paper is the following. Theorem 9 Let n ≥ 1, k ≥ 3, c ≥ 0 and G be an n-connected graph with σkn(k−1)+c+2 (G) ≥ |V(G)| − c − 1. Then G has a spanning tree T with maximum degree at most k + ⌈c/n⌉ and TE(T, k) ≤ c. In case of 0 ≤ c ≤ n, Theorems 8 and 9 are equivalent. The degree condition in Theorem 9 is best possible in the following sense: Let G be a complete bipartite graph with partite sets X and Y such that |X| = n and |Y| = n(k − 1) + c + 2, where n ≥ 1 and k ≥ 3. Then G is n-connected and σkn(k−1)+c+2 (G) = kn = |V(G)| − c − 2. Let T be a tree of G with maximum degree at most k + ⌈c/n⌉ and TE(T, k) ≤ c. Then the number of edges in T is at most k|X| + c = kn + c. On the other hand, T cannot span V(G) since |V(G)| = kn + c + 2. Moreover, Theorem 9 is stronger than Theorem 7. This will be proved in Section 5. 3 Preliminaries Before proving Theorem 9, we prepare some notation and facts.
3
For a tree T and u, v ∈ V(T ), we denote the unique path in T connecting u and v by T [u, v]. If u , v, then NT (u) ∩ T [u, v] consists of a single vertex and we denote this vertex by uvT . When there is no fear of confusion, we simply denote uv instead of uvT . A forest is a graph whose components are trees. For H ⊆ G and x ∈ V(G − H), a subgraph F of G is called an S (x, H)-fan of width k if F has the decomposition F = ki=1 Pi , where each Pi is a path connecting x and vi ∈ V(H), V(Pi ) ∩ V(H) = {vi } and V(Pi ) ∩ V(P j ) = {x} if i , j. We call F an (x, H)-fan of maximum width if G does not
contain an (x, H)-fan of width greater than that of F. In our proof of Theorem 9, we often consider a tree T as an outdirected tree, that is, all the edges in T are directed away from the root r ∈ V(T ). In such a case, for u ∈ V(T ) \ {r}, we denote NT+ (u) = NT (u) \ {ur } and NT− (u) = {ur }. ur is sometimes denoted by u− . And in case of u = r, we denote NT+ (u) = NT (u) and NT− (u) = ∅. Moreover, for u ∈ V(T ), we denote dT+ (u) = |NT+ (u)|. The following two facts are easy observations on trees. Fact 1 Let T be a tree and let u, v1 , v2 be distinct vertices in T . Then (NT (v1 ) \ {v1 u }) ∩ (NT (v2 ) \ {v2 u }) = ∅. Furthermore, if uv < E(T ), then (NT [v1 ] \ {v1 u }) ∩ (NT [v2 ] \ {v2 u }) = ∅. Fact 2 Let T be a tree and {v1 , v2 , . . . , vt } be an independent set of T . Then T − {v1 , v2 , . . . , vt } has exactly Pt i=1 (dT (vi ) − 1) + 1 components. 4 Proof of Theorem 9 In our proof of Theorem 9, a tree T of G is called a (k, c, n)-tree if maximum degree of T is at most k + ⌈c/n⌉ and TE(T, k) ≤ c. Take a (k, c, n)-tree T of G so that |V(T )| is as large as possible. We may assume that V(T ) , V(G) since otherwise we have nothing to prove. Claim 1 |V(T )| ≥ n. Proof. Suppose that |V(T )| < n and take u ∈ V(T ) such that dT (u) = 1. Since G is n-connected, there exists x ∈ NG (u) \ V(T ). Then T + xu is a (k, c, n)-tree larger than T , a contradiction.
Take any u0 ∈ V(G) \ V(T ) and let T 0 be the component of G − T containing u0 . Let F be a (u0 , T )-fan of maximum width and let S 1 = V(F) ∩ V(T ). Since G is n-connected and by Claim 1, |S 1 | ≥ n. We denote VT≥t := {v ∈ V(T ) | dT (v) ≥ t} and let S 2 = VT≥k+1 \ S 1 . Claim 2 S 1 ⊆ VT≥k . Proof. Suppose that dT (v) < k for some v ∈ S 1 . Then T˜ = T ∪ F[u0 , v] is a (k, c, n)-tree with |V(T˜ )| > |V(T )|, a contradiction.
Choose r ∈ S 1 so that dT (r) = minv∈S 1 dT (v). Claim 3 TE(T, k) =
P
v∈S 1 ∪S 2 (dT (v)
− k) = c.
4
Proof. Note that only the vertices of VT≥k+1 contribute to TE(T, k). Since VT≥k+1 ⊆ S 1 ∪ S 2 and dT (v) − k ≥ 0 for P every v ∈ S 1 ∪ S 2 , we obtain TE(T, k) = v∈S 1 ∪S 2 (dT (v) − k). Next, suppose that TE(T, k) < c. It follows from the fact |S 1 | ≥ n and the choice of r that dT (r) ≤ k + TE(T, k)/n < k + c/n ≤ k + ⌈c/n⌉. Thus T˜ = T ∪ F[u0 , r] is a (k, c, n)-tree with |V(T˜ )| > |V(T )|, a contradiction. Claim 4 If S 2 , ∅, then dT (r) ≤ k + ⌈c/n⌉ − 1. Proof. Assume the contrary. Since dT (v) ≥ dT (r) for any v ∈ S 1 , X
dT (v) ≥
v∈S 1
And, since S 2 , ∅,
P
v∈S 2 (dT (v)
X
dT (r) ≥ |S 1 |(k + ⌈c/n⌉) ≥ |S 1 |k + c.
v∈S 1
− k) ≥ 1. Hence it follows from Claim 3 that TE(T, k) =
X
(dT (v) − k) +
v∈S 1
=
X
(dT (v) − k)
v∈S 2
|S 1 |k + c − |S 1 |k +
X
(dT (v) − k)
v∈S 2
≥
c + 1,
which contradicts that T is a (k, c, n)-tree.
Claim 5 NT (s) ∩ S 1 = ∅ for any s ∈ S 1 . Proof. Suppose that there exist two vertices v, v′ ∈ S 1 with v′ ∈ NT (v). Then (T − vv′ ) ∪ F[u0 , v] ∪ F[u0 , v′ ] is a (k, c, n)-tree, which contradicts the maximality of T .
In the following, we consider T as an outdirected tree with the root r. Claim 6 NT+ (s) ∩ S 1 = ∅ for any s ∈ S 2 . Proof. Suppose that NT+ (s) ∩ S 1 , ∅ for some s ∈ S 2 and let s′ ∈ NT+ (s) ∩ S 1 . Let T˜ = (T − ss′ ) ∪ F[u0 , s′ ] ∪ F[u0 , r]. Then T˜ is a tree with |V(T˜ )| > |V(T )|, dT˜ (v) = dT (v) for every v ∈ V(T ) \ {r, s}, dT˜ (r) = dT (r) + 1 and dT˜ (s) = dT (s) − 1. Moreover, dT˜ (v) = 2 for every v ∈ V(T˜ ) \ V(T ). Hence Claim 3 implies that TE(T˜ , k) = TE(T, k) + 1 − 1 = c, and Claim 4 implies that dT˜ (v) ≤ k + ⌈c/n⌉ for every v ∈ V(T˜ ). Therefore, T˜ is a (k, c, n)-tree larger than T , a contradiction.
Now we make a forest T ′ from T by the following operations: 1. Delete all the vertices of S 1 . 2. For each s ∈ S 2 , choose NT+ (s)∗ ⊆ NT+ (s) so that |NT+ (s)∗ | = dT (s) − k and delete the edge set {ss′ | s′ ∈ NT+ (s)∗ }. By Claims 5 and 6, there is no duplication in edges deleted by these operations. By Fact 2 and Claim 3, T ′ has a
=
X
(dT (s) − 1) +
s∈S 1
X
s∈S 2
5
(dT (s) − k) + 1
=
X
(dT (s) − k) + (k − 1)|S 1 | + 1
s∈S 1 ∪S 2
= (k − 1)|S 1 | + c + 1 ≥ (k − 1)n + c + 1
(1)
components. Let T 1 , . . . , T a be components of T ′ , A = {0, 1, · · · , a}, and A′ = A \ {0}. Note that we can consider T i as an outdirected tree by the same orientation of T . Under this consideration, for i ∈ A′ , let ri be the root of T i and ui be a vertex of T i such that dT+i (ui ) = 0. Moreover let U = {u0 , u1 , . . . , ua }. Note that, for every i with i ∈ A′ , it follows that NT+ (ui ) ⊆ S 1 and ui < S 1 ∪ S 2 , and so we have dT (ui ) ≤ k. Suppose that i, j ∈ A′ with i , j and u′j ∈ V(T j ) is a vertex such that u′j = u j or NT+ (u′j ) ∩ S 1 , ∅. Now, for the technical reason, we define two forests F1 (ui , u′j ) and F2 (ui , u′j ) in G. To show the structure of these forests, we need some definitions. It follows from i , j that V(T [ui , u′j ]) ∩ (S 1 ∪ S 2 ) , ∅. If V(T [ui , u′j ]) ∩ S 1 , ∅, let w be the vertex of V(T [ui , u′j ]) ∩ S 1 and let w′ ∈ V(T [ui , u′j ]) ∩ NT (w). Otherwise, by the construction of T ′ , we can take w ∈ V(T [ui , u′j ]) ∩ S 2 so that V(T [ui , u′j ]) ∩ NT+ (w)∗ , ∅, and let w′ ∈ V(T [ui , u′j ]) ∩ NT+ (w)∗ . Note that ww′ < E(T ′ ) in both cases. Moreover, for v ∈ S 1 ∪ S 2 , we define 1 v if v ∈ S , f (v) = r if v ∈ S 2 .
First we describe the structure of F1 (ui , u′j ). For v ∈ V(T ), we write u ≺ v if u ∈ V(T [r, v]), and otherwise we
write u ⊀ v. Case 1. ui ⊀ u′j and u′j ⊀ ui . If dT (ui ) = dT (u′j ) = 1, then let F1 (ui , u′j ) = (T − ww′ ) ∪ F[u0 , f (w)] (See Figures 1 and 2). If dT (ui ) ≥ 2 and dT (u′j ) = 1, take any u+i ∈ NT+ (ui ). Note that u+i ∈ S 1 since NT+ (ui ) ⊆ S 1 . Let F1 (ui , u′j ) = (T − ww′ − ui u+i ) ∪ F[u0 , f (w)] ∪ F[u0 , u+i ] (See Figures 3 and 4). If dT (ui ) = 1 and dT (u′j ) ≥ 2, take any u′j + ∈ NT+ (u′j ) ∩ S 1 and let F1 (ui , u′j ) = (T − ww′ − u′j u′j + ) ∪ F[u0 , f (w)] ∪ F[u0 , u′j + ] (See Figures 5 and 6). If dT (ui ) ≥ 2 and dT (u′j ) ≥ 2, take any u+i ∈ NT+ (ui ) and u′j + ∈ NT+ (u′j ) ∩ S 1 , and let F1 (ui , u′j ) = (T − ww′ − ui u+i − u′j u′j + ) ∪ F[u0 , f (w)] ∪ F[u0, u+i ] ∪ F[u0 , u′j + ] (See Figures 7 and 8). Case 2. ui ≺ u′j . u′
In this case, we have dT (ui ) ≥ 2. Then NT+ (ui ) is not empty and contained in S 1 , and hence u+i = ui j is in S 1 . In case of dT (u′j ) = 1, let F1 (ui , u′j ) = (T − ui u+i ) ∪ F[u0 , u+i ] (See Figure 9). And in case of dT (u′j ) ≥ 2, take any u′j + ∈ NT+ (u′j ) ∩ S 1 and let F1 (ui , u′j ) = (T − ui u+i − u′j u′j + ) ∪ F[u0 , u+i ] ∪ F[u0 , u′j + ] (See Figure 10). Case 3. u′j ≺ ui . First consider the case where u′j ui is in S 1 . Let u′j + = u′j ui . In case of dT (ui ) = 1, let F1 (ui , u′j ) = (T − u′j u′j + ) ∪ F[u0 , u′j + ] (See Figure 11). And in case of dT (ui ) ≥ 2, take any u+i ∈ NT+ (ui ) and let F1 (ui , u′j ) = 6
r u0
u0 ×
× ′
w w′
w w
u′j
ui
u′j
ui
Figure 1: In case of w ∈ S 1 .
Figure 2: In case of w ∈ S 2 . r
u0
u0 × w w′
× w w′ u′j
× ui
u′j
× ui
u+i
u+i
Figure 3: In case of w ∈ S 1 .
Figure 4: In case of w ∈ S 2 . r u0
u0
× w w′
× w w′ ′ × uj
ui
′ × uj
ui
u′j +
u′j + Figure 5: In case of w ∈ S 1 .
Figure 6: In case of w ∈ S 2 . r
u0
u0 × w w′
× w w′
× ui
×
u′j
× ui
u′j +
u+i
×
u′j +
u+i
Figure 7: In case of w ∈ S 1 .
Figure 8: In case of w ∈ S 2 .
ui ui u0
×
u0
×
u+i
u+i u′j
× u′j
u′j +
Figure 9: In case of dT (u′j ) = 1.
Figure 10: In case of dT (u′j ) ≥ 2.
7
u′j
u′j u′j u0
×
u0
×
u′j +
u′j + ui
× u+i
ui
Figure 12: In case of dT (u′j ) ≥ 2. r
Figure 11: In case of dT (ui ) = 1. u′j u0
u′j
× + w u′j × ′ w
u0
× + w u′j × ′ w
ui
ui
Figure 13: In case of w ∈ S 1 .
Figure 14: In case of w ∈ S 2 . r
u′j u0
u′j
× + w u′j × ′ w ×
u0
ui
× + w u′j × ′ w ×
u+i
Figure 15: In case of w ∈ S 1 .
ui u+i
Figure 16: In case of w ∈ S 2 .
(T − u′j u′j + − ui u+i ) ∪ F[u0 , u′j + ] ∪ F[u0 , u+i ] (See Figure 12). Next consider the case where u′j ui is not in S 1 . Let u′j + be a vertex in NT+ (u′j ) ∩ S 1 . In case of dT (ui ) = 1, let F1 (ui , u′j ) = (T − ww′ − u′j u′j + ) ∪ F[u0 , u′j + ] ∪ F[u0 , f (w)] (See Figures 13 and 14). And in case of dT (ui ) ≥ 2, take any u+i ∈ NT+ (ui ) and let F1 (ui , u′j ) = (T − ww′ − u′j u′j + − ui u+i ) ∪ F[u0 , u′j + ] ∪ F[u0 , f (w)] ∪ F[u0 , u+i ] (See Figures 15 and 16). Above Case 1 to 3 cover all the cases, and so we can take F1 (ui , u′j ) in any situations. Next, we show the structure of F2 (ui , u′j ). In case of dT (ui ) ≤ 2, we take F1 (ui , u′j ) as F2 (ui , u′j ). In case of dT (ui ) ≥ 3, take any v1 , v2 ∈ NT+ (ui ). Again, note that v1 , v2 ∈ S 1 . If ui ⊀ u′j , then let F2 (ui , u′j ) = (T −ww′ −ui v1 −ui v2 )∪F[u0 , f (w)]∪F[u0 , v1 ]∪F[u0 , v2 ] (See Figures 17, 18, 19 u′
and 20). And if ui ≺ u′j , take v′1 , v′2 ∈ NT+ (ui ) so that v′1 = ui j . Let F2 (ui , u′j ) = (T −u′i v1 −ui v′2 )∪F[u0 , v′1 ]∪F[u0 , v′2 ] (See Figure 21). By the construction of F1 (ui , u′j ) and F2 (ui , u′j ), we have the following fact, which we often use in our proof. Fact 3 F1 = F1 (ui , u′j ) and F2 = F2 (ui , u′j ) satisfy all of the followings for l = 1 and 2. (i) Fl has exactly two components. (ii) V(T ) ( V(Fl ) and TE(Fl , k) ≤ TE(T, k). (iii) dFl (v) ≤ dT (v) for all v ∈ V(T ) \ {r}. Moreover, dFl (r) ≤ dT (r) if S 2 = ∅. (iv) The maximum degree of Fl is at most k + ⌈c/n⌉. 8
r u0
u0 × w w′
× w w′
ui × × v1
ui × ×
u′j v2
v1
Figure 17: In case of u′j ⊀ ui and w ∈ S 1 .
u′j v2
Figure 18: In case of u′j ⊀ ui and w ∈ S 2 . r
u′j
u′j
w × w′
u0
u0
ui × × v1 v2
w × w′ ui × × v1 v2
Figure 19: In case of u′j ≺ ui and w ∈ S 1 .
u0
Figure 20: In case of u′j ≺ ui and w ∈ S 2 .
v′2 ×
ui ×′ v1
u′j Figure 21: In case of w ∈ S 2 .
(v) ui and u′j belong to different components of Fl . (vi) dF1 (ui ) ≤ k − 1, dF1 (u′j ) ≤ k − 1, and dF2 (ui ) ≤ k − 2. (vii) If vv+ < E(T [ui , u′j ]) and v , ui , u′j for v ∈ V(T ) and v+ ∈ NT+ (v), then vv+ ∈ E(Fl ) . Moreover, E(T i ) ∪ E(T j ) ⊆ E(Fl ). The next lemma is also used several times in our proof. Lemma 1 Let i ∈ A′ and suppose that s ∈ S 1 ∪ S 2 . Then NG (ui ) ∩ (NT (s) \ {sui }) = ∅. Proof. Suppose that there exists s′ ∈ NG (ui )∩(NT (s)\{sui }). If dT (ui ) ≤ k−1, then (T − ss′ +ui s′ )+F[u0 , f (s)] is a (k, c, n)-tree, which contradicts the maximality of T . So we assume that dT (ui ) = k. Then |NT+ (ui )| ≥ k − 1 ≥ 2, and hence there exists u+i ∈ NT+ (ui ) \ {ui s }. Since u+i ∈ S 1 , (T − ss′ − ui u+i + ui s′ ) ∪ F[u0 , f (s)] ∪ F[u0 , u+i ] is a (k, c, n)-tree, which contradicts the maximality of T .
Now we continue the proof of Theorem 9. For i ∈ A and j ∈ A′ , let Wi, j = NG (ui ) ∩ V(T j ). Claim 7 W0, j = ∅ for all j ∈ A′ . Proof. Since F is a (u0 , T )-fan of maximum width, NG (u0 ) ∩ V(T ) ⊆ S 1 = V(T ) \ V(T ′ ). Hence W0, j = ∅. 9
Claim 8 Let i, j ∈ A′ with i , j. Then for any v ∈ Wi, j , NT (v) ∩ Wi, j = ∅. Proof. Suppose that v j v′j ∈ E(T ) for v j , v′j ∈ Wi, j . By (vii) of Fact 3 and the fact v j v′j ∈ E(T j ), we have the forest F2 (ui , u j ) with v j v′ ∈ E(F2 (ui , u j )). With (v) of Fact 3, we obtain that T˜ = F2 (ui , u j ) − v j v′ + ui v j + ui v′ is j
j
j
connected. It is easy to see that T˜ is a (k, c, n)-tree, which contradicts the maximality of T .
Claim 9 Let i, j ∈ A′ with i , j. For any v ∈ Wi, j \ {r j }, dT j (v) = k. Moreover, if r j ∈ Wi, j , then dT j (r j ) = k − 1. Proof. It suffices to show that dT+j (v) = k − 1 for any v ∈ Wi, j . If v ∈ S 2 , then dT+j (v) = k − 1 holds by the construction of T ′ . Hence we may assume that v < S 2 . If NT+ (v) ∩ S 1 , ∅, then we have a forest F1 (ui , v). However, F1 (ui , v) + ui v contradicts the maximality of T . Hence NT+ (v) ∩ S 1 = ∅, and so dT+ (v) = dT+j (v). If dT (v) < k, then by (iii) of Fact 3, dF1 (ui ,u j ) (v) < k, and then F1 (ui , u j ) + ui v contradicts the maximality of T . Hence dT (v) ≥ k. Since we already have v < S 1 ∪ S 2 , dT (v) ≤ k holds. This implies dT (v) = k. Therefore, since v , r, we have dT+j (v) = dT+ (v) = dT (v) − 1 = k − 1.
Claim 10 U is independent in G. Proof. By Claim 7, u0 ul < E(G) for any l ∈ A′ . Suppose that ui u j ∈ E(G) for some i, j ∈ A′ . Then F1 (ui , u j ) + ui u j contradicts the maximality of T . Therefore ui u j < E(G) for all i, j ∈ A′ , and so this claim holds. Since |U| = a + 1 ≥ n(k − 1) + c + 2 and U is independent, there exists U ∗ ⊂ U such that |U ∗ | = k and X dG (u) ≥ |V(G)| − c − 1.
(2)
u∈U ∗
Take A∗ ⊂ A so that
S
i∈A∗ {ui }
= U ∗.
For j ∈ A′ , we define ξ+j
1 = 0
if r j ∈ NG (ui ) for all i ∈ A∗ , otherwise.
Then since |A∗ | = k, it follows immediately that 1 X |Wi, j ∩ {r j }| = ξ+j . k i∈A∗
(3)
For j ∈ A′ , let
X j = {s ∈ V(T j ) ∩ S 2 | s = r−j′ for some j′ ∈ A′ with ξ+j′ = 1}.
(4)
Claim 11 |X j | ≤ 1 for each j ∈ A′ . Proof. Suppose that s′ , s′′ ∈ X j . Then there exist j′ , j′′ ∈ A′ such that s′ = r−j′ , s′′ = r−j′′ and ui r j′ , ui r j′′ ∈ E(G) for i ∈ A∗ \ {0}. By Lemma 1, we have r j′ = s′ ui and r j′′ = s′′ ui . Then T j ∪ T [s′ , ui ] ∪ T [s′′ , ui ] contains a cycle, which contradicts that T is a tree.
For j ∈ A′ , we define ξ−j = |X j |. In case of X j , ∅, i.e. ξ−j = 1, we denote the unique vertex in X j by x j . 10
Claim 12 Suppose that ξ−j = ξ+j′ = 1 and x j = r−j′ for some j, j′ ∈ A′ . Then ξ+j′′ = 0 for every j′′ ∈ A′ \ { j′ } which satisfies x j = r−j′′ . Proof. Note that x j ∈ S 2 . Let i ∈ A∗ . Then by Lemma 1, |NT (x j ) ∩ NG (ui )| ≤ 1. Since ξ+j′ = 1, r j′ ∈ NT (x j ) ∩ NG (ui ). Hence r j′′ < NT (x j ) ∩ NG (ui ). Now r j′′ ∈ NT (x j ), and so r j′′ < NG (ui ), which implies ξ+j′′ = 0. Let j ∈ A′ with ξ+j = 1. Then r−j ∈ V(T j′ ) for some j′ ∈ A′ . We denote this j′ by j− . Note that, by (4), it follows that ξ−j− = 1. And if ξ−j = 1 for some j, it follows from (4) that there exists j′ ∈ A′ such that j′ − = j. Now Claim 12 implies that such j′ is uniquely determined, and we denote it by j+ . Claim 13 Suppose that ξ−j = 1 for some j ∈ A′ . Then NT j [x j ] ∩ Wi, j = ∅ for any i ∈ A∗ \ { j+ }. Proof. Let i ∈ A∗ \ { j+ }. Since ξ−j = 1, r j+ ∈ NG (ui ). It follows from Lemma 1 and the fact x j ∈ S 2 that xuj i = r j+ . Again using Lemma 1, we obtain that v < NG (ui ) for all v ∈ NT j (x j ), and hence NT j (x j ) ∩ Wi, j = ∅. Therefore, it suffices to prove that x j < NG (ui ). Assume the contrary. It follows from x j ≺ r j+ and xuj i = r j+ that r j+ ≺ ui . With the fact r j+ ≺ u j+ , we obtain x j r j+ < E(T [ui , u j+ ]). Hence by (vii) of Fact 3 and the fact that x j , u j , we have a forest F2 (ui , u j+ ) with x j r j+ ∈ E(F2 (ui , u j+ )). Now F2 (ui , u j+ ) − x j r j+ + ui x j + ui r j+ contradicts the maximality of T , and so the claim holds.
Claim 14 For every i ∈ A∗ and j ∈ A′ with i , j, 1 k (|V(T j )| − 1 + |(Wi, j ∪ X j ) ∩ {r j }|) |Wi, j | ≤ 1 (|V(T j )| − 1 + |(Wi, j ∪ X j ) ∩ {r j }|) − ξ− j
k
if ξ−j = 1 and i = j+ , otherwise.
Proof. First, we show that right sides of the above inequalities are not less than 0. Since |V(T j )| ≥ 1 for any j ∈ A′ , 1k (|V(T j )| − 1 + |(Wi, j ∪ X j ) ∩ {r j }|) ≥ 0 holds. Suppose that ξ−j = 1, then x j ∈ S 2 . By the construction of T ′ , we have |V(T j )| ≥ k + 1 |V(T j )| ≥ k and |(Wi, j ∪ X j ) ∩ {r j }| = 1
if x j , r j , if x j = r j .
Hence 1k (|V(T j )| − 1 + |(Wi, j ∪ X j ) ∩ {r j }|) − ξ−j ≥ 0.
If i = 0, then Claim 7 implies that |Wi, j | = 0, and hence the claim holds. So we may assume that i , 0. Case 1. ξ−j = 1 and i = j+ . By Claim 8, vv′ < E(T j ) for any v, v′ ∈ Wi, j with v , v′ . Hence by Fact 1, we have (NT j [v] \ {vu j }) ∩ (NT j [v′ ] \ {v′ u j }) = ∅. It follows from Claim 10 that u j < Wi, j , thus Claim 9 implies X |V(T j )| ≥ |NT j [v] \ {vu j }| + |{u j }| = k|Wi, j | − |Wi, j ∩ {r j }| + 1. v∈Wi, j
Therefore, |Wi, j |
≤
1 1 (|V(T j)| − 1 + |Wi, j ∩ {r j }|) ≤ (|V(T j)| − 1 + |(Wi, j ∪ X j ) ∩ {r j }|). k k
Case 2. Otherwise. By Claims 8 and 13, we have X j ∩ Wi, j = ∅ and vv′ < E(T j ) for any v, v′ ∈ X j ∪ Wi, j with v , v′ . Hence by Fact 1, we have (NT j [v] \ {vu j }) ∩ (NT j [v′ ] \ {v′ u j }) = ∅. Note that, if |X j | , ∅, then x j , u j since x j ∈ S 2 . With 11
Claim 10 we have u j < X j ∪ Wi, j . Moreover, if |X j | , ∅, it follows from x j ∈ S 2 that dT j (x j ) = k in case of x j , r j , and dT j (x j ) = k − 1 in case of x j = r j . Therefore, by Claim 9, it follows that X |V(T j)| ≥ |NT j [v] \ {vu j }| + |{u j }| v∈Wi, j ∪X j
=
k|Wi, j ∪ X j | − |(Wi, j ∪ X j ) ∩ {r j }| + 1
=
k|Wi, j | + k|X j | − |(Wi, j ∪ X j ) ∩ {r j }| + 1
=
k|Wi, j | + kξ−j − |(Wi, j ∪ X j ) ∩ {r j }| + 1,
which implies |Wi, j | ≤ 1k (|V(T j )| − 1 + |(Wi, j ∪ X j ) ∩ {r j }|) − ξ−j . For j ∈ A′ , we define ξ j = ξ+j − ξ−j . Claim 15
P
i∈A∗
|Wi, j | ≤ |V(T j )| − 1 + ξ j for any j ∈ A′ \ A∗ .
Proof. If ξ−j = 1 and j+ ∈ A∗ , then by Claim 14, X X |Wi, j | = |Wi, j | + |W j+ , j | i∈A∗ i∈A∗ \{ j+ } ! X 1 1 ≤ (|V(T j)| − 1 + |(Wi, j ∪ X j ) ∩ {r j }|) − ξ−j + (|V(T j)| − 1 + |(W j+ , j ∪ X j ) ∩ {r j }|) k k i∈A∗ \{ j+ } 1X = |V(T j )| − 1 − (k − 1)ξ −j + |(Wi, j ∪ X j ) ∩ {r j }| k ∗ i∈A 1 X − = |V(T j )| − 1 − (k − 1)ξ j + |(Wi, j ∪ X j ) ∩ {r j }| . k i∈A∗
And otherwise,
X
|Wi, j |
≤
i∈A∗
= = ≤ Therefore, it suffices to show that
! X 1 (|V(T j)| − 1 + |(Wi, j ∪ X j ) ∩ {r j }|) − ξ−j k i∈A∗ 1X − |V(T j )| − 1 − kξ j + (|(Wi, j ∪ X j ) ∩ {r j }|) k ∗ i∈A 1 X − |V(T j )| − 1 − kξ j + |(Wi, j ∪ X j ) ∩ {r j }| k i∈A∗ 1 X |(Wi, j ∪ X j ) ∩ {r j }| . |V(T j )| − 1 − (k − 1)ξ −j + k i∈A∗
−(k − in both cases.
1)ξ −j
1 X + |(Wi, j ∪ X j ) ∩ {r j }| ≤ ξ j k i∈A∗
If ξ−j = 0, then by (3), 1 X 1 X −(k − 1)ξ −j + |(Wi, j ∪ X j ) ∩ {r j }| = |Wi, j ∩ {r j }| = ξ+j = ξ+j − ξ−j = ξ j . k i∈A∗ k i∈A∗ 12
And if ξ−j = 1, then −(k −
1)ξ −j
$ % 1 X 1 + |(Wi, j ∪ X j ) ∩ {r j }| ≤ −(k − 1) + · k = −(k − 2) ≤ −1 ≤ −ξ−j + ξ+j = ξ j . k i∈A∗ k
Hence the claim holds.
Claim 16 ξ−j = 0 for any j ∈ A∗ \ {0}. u
Proof. Suppose that ξ−j = 1, i.e. X j , ∅, for some j ∈ A∗ \ {0}, then r j+ u j ∈ E(G). However r j+ , x j j , which contradicts Lemma 1.
For each j ∈ A∗ \ {0}, we define W∗, j =
[
Wi, j .
(5)
i∈A∗ \{ j}
Claim 17 Let j ∈ A∗ \ {0} such that |V(T j )| ≥ 2. Then, |W j, j | ≤ |V(T j )| − (k − 1)|W∗, j | + |W∗, j ∩ {r j }| − 1. Furthermore, if r j < NG (u j ) and r j
