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m -dominating k -ended trees of l -connected graphs
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m-dominating k-ended trees of l-connected graphs Masao Tsugaki a , ∗, Guiying Yan b a Tokyo University of Science, Kagurazaka, Shinjuku-ku, Tokyo, Japan b Institute of mathematical and system sciences, Chinese Academy of Science, Beijing, PR China
Received 29 August 2015; accepted 19 March 2017 Available online xxxxx
Abstract Let k ≥ 2, l ≥ 1 and m ≥ 0 be integers, and let G be an l-connected graph. If there exists a subgraph X of G such that the distance between v and X is at most m for any v ∈ V (G), then we say that X m-dominates G. A subset S of V (G) is said to be 2(m + 1)-stable if the distance between each pair of distinct vertices in S is at least 2(m + 1). In this paper, we prove that if G does not have a 2(m + 1)-stable set of order at least k + l, then G has an m-dominating tree which has at most k leaves. c 2017 Kalasalingam University. Publishing Services by Elsevier B.V. This is an open access article under the CC BY-NC-ND ⃝ license (http://creativecommons.org/licenses/by-nc-nd/4.0/). Keywords: Vertex dominating; m-dominating; k-ended tree
1. Introduction In this paper, we consider finite simple graphs, which have neither loops nor multiple edges. Let G be a graph, and X be a subgraph or a vertex set of G. We write |X | for the order of X . For two vertices u and v of G, let dG (u, v) denote the distance between u and v. For a vertex v of G, the distance between v and X is defined to be the minimum value of dG (v, x) for all x ∈ V (X ) or x ∈ X , and denoted by dG (v, X ). For an integer m ≥ 0, let Domi m (X ) := {v ∈ V (G) : dG (v, X ) ≤ m}. If V (G) = Domi m (X ), then we say that X m-dominates G. For an integer l ≥ 2, a subset S of V (G) is said to be l-stable if the distance between each pair of distinct vertices in S is at least l. The l-stable number of G is the cardinality of a maximum l-stable set of G, and is denoted by αl (G), that is αl (G) := max{|S| : S ⊆ V (G), S is l-stable}. Note that α 2 (G) is the independence number of G. A tree is called a k-ended tree if the number of leaves is at most k. In this paper, we investigate a stable number condition for l-connected graphs to have m-dominating k-ended trees. Concerning on this, the following three theorems are known. Theorem 1 (Kano, Tsugaki and Yan [1]). Let k ≥ 2 and m ≥ 0 be integers, and let G be a connected graph. If α 2(m+1) (G) ≤ k, then G has an m-dominating k-ended tree. Peer review under responsibility of Kalasalingam University. ∗ Corresponding author.
E-mail addresses: [email protected] (M. Tsugaki), [email protected] (G. Yan). http://dx.doi.org/10.1016/j.akcej.2017.04.004 c 2017 Kalasalingam University. Publishing Services by Elsevier B.V. This is an open access article under the CC BY-NC-ND license 0972-8600/⃝ (http://creativecommons.org/licenses/by-nc-nd/4.0/).
Please cite this article in press as: M. Tsugaki, G. Yan, m-dominating k-ended trees of l-connected graphs, AKCE International Journal of Graphs and Combinatorics (2017), http://dx.doi.org/10.1016/j.akcej.2017.04.004.
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Fig. 1. An l-connected graph which has no m-dominating k-ended tree.
Theorem 2 (Broersma [2]). Let l ≥ 1 and m ≥ 0 be integers, and let G be an l-connected graph. If α 2(m+1) (G) ≤ l + 1, then G has an m-dominating path, that is an m-dominating 2-ended tree. Theorem 3 (Win [3]). Let k ≥ 2 and l ≥ 1 be integers, and let G be an l-connected graph. If α 2 (G) ≤ k + l − 1, then G has a spanning k-ended tree, that is a 0-dominating k-ended tree. In this paper, we prove the following theorem which is a generalization of above three theorems. Theorem 4. Let k ≥ 2, l ≥ 1 and m ≥ 0 be integers, and let G be an l-connected graph. If α 2(m+1) (G) ≤ k + l − 1, then G has an m-dominating k-ended tree. The upper bound of the stable number condition of Theorem 4 is sharp. Let k ≥ 2, l ≥ 1 and m ≥ 1 be integers. For 1 ≤ i ≤ k + l, let Di,1 , Di,2 , . . . , Di,m be disjoint copies of the complete graph K l . For 1 ≤ j ≤ m − 1, join all the vertices of Di, j and all the vertices of Di, j+1 by edges. Let vi be a vertex not contained in Di,1 ∪ Di,2 ∪ · · · ∪ Di,m , and join vi and all the vertices of Di,m by edges. Let H be a complete graph of order l. For 1 ≤ i ≤ k + l, join all the vertices of H and all the vertices of Di,1 by edges. Let G denote the resulting graph (see Fig. 1). Then G is an l-connected graph, and has no m-dominating k-ended tree. On the other hand, α 2(m+1) (G) = k + l. Concerning on a degree sum condition, the following theorem is known. Theorem 5 (Broersma [2]). Let l ≥ 1 and m ≥ 0 be integers, and let G be an l-connected graph. If ∑ m x∈S |Domi (x)| ≥ |G| − l + 1 for any 2(m + 1)-stable set S of order l + 2, then G has an m-dominating path, that is an m-dominating 2-ended tree. By Theorem 4, we can obtain the following theorem which is a generalization of Theorem 5. ∑ m Theorem 6. Let k ≥ 2, l ≥ 1 and m ≥ 0 be integers, and let G be an l-connected graph. If x∈S |Domi (x)| ≥ |G| − l + 1 for any 2(m + 1)-stable set S of order k + l, then G has an m-dominating k-ended tree. Proof of Theorem 6. Suppose that G does not have an m-dominating k-ended tree. Then, by Theorem 4, α 2(m+1) (G) ≥ k + l, and hence there exists a 2(m + 1)-stable set S of order |S| = k + l. Note that for any x, y ∈ S m with x ̸= y, Domi∑ (x) and Domi m (y) are disjoint and there are no edge between them. Since G is l-connected, these imply that |G| − x∈S |Domi m (x)| ≥ l, which contradicts to the condition in Theorem 6. □ 2. Proof of Theorem 4 A system of a graph G is defined to be a set of vertex-disjoint paths and cycles of G. In this paper, we regard a vertex and an edge as a cycle, and thus a path means a path of order at least 3. Let S be a system of a graph G. For S ∈ S, we put f (S) := 2 if S is a path, and f (S) := 1 otherwise. Let PS := and CS := ⋃{S ∈ S : f (S) = 2} ⋃ {S ∈ S : fm(S) = 1}. m For X ⊆∑S and an integer m ≥ 0, we define V (X ) := V (X ), Domi (X ) := X ∈X X ∈X Domi (X ) and f (X ) := X ∈X f (X ). For an integer k ≥ 2, we call X a k-ended system if f (X ) ≤ k. Note that f (S) = 2|PS | + |CS |. Please cite this article in press as: M. Tsugaki, G. Yan, m-dominating k-ended trees of l-connected graphs, AKCE International Journal of Graphs and Combinatorics (2017), http://dx.doi.org/10.1016/j.akcej.2017.04.004.
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For an oriented path P, we write ∥P∥ for the length of P, and let a P and b P be the initial and terminal vertices of P, respectively. Let S be an oriented path (or cycle) and x ∈ V (S). We denote the successor and the predecessor of x on S by x + and x − , respectively (if exists). For convenience, let {x + } := V (S) − {x} and {x − } := V (S) − {x} if S is an edge, and x + := x and x − := x if S is a vertex. For x, y ∈ V (S), we denote by S[x, y] a path from x to y along the ← − orientation of S. The reverse sequence of S[x, y] is denoted by S [y, x]. We denote S[x, y] − {x, y}, S[x, y] − {x}, S[x, y] − {y} by S(x, y), S(x, y] and S[x, y), respectively. For convenience, let S[x, x) := ∅. Let S be a system of a graph G, and let X and Y be subsets of V (G). For a (path P of G, if )P connects a vertex of X and a vertex of Y , and all of inner vertices of P are contained in V (G) − V (S) ∪ X ∪ Y , then we call P an (X, Y( ; S C )-path of G, )and(we orient P from ) a vertex of X to a vertex of Y , that is P is a path such that a P ∈ X , b P ∈ Y and V (P) − {a P , b P } ∩ V (S) ∪ X ∪ Y = ∅. For x, y ∈ V (G), we abbreviate ({x}, Y ; S C )-path, (X, {y}; S C )-path and ({x}, {y}; S C )-path as (x, Y ; S C )-path, (X, y; S C )-path and (x, y; S C )-path, respectively. Since we can obtain a k-ended tree including V (S) from a k-ended system S, Theorem 4 is an immediate consequence of the following theorem. Theorem 7. Let k ≥ 2, l ≥ 1 and m ≥ 0 be integers, and let G be an l-connected graph. If α 2(m+1) (G) ≤ k + l − 1, then G has an m-dominating k-ended system. We now prove Theorem 7. Proof of Theorem 7. Let k ≥ 2, l ≥ 1 and m ≥ 0 be integers, and let G be a graph which satisfies the condition in Theorem 7. Suppose that G has no m-dominating k-ended system. Let S be a k-ended system in G. We give an orientation to S for each S ∈ S. Since G has no m-dominating k-ended system, there exists v ∈ V (G) − Domi m (S). Since G is C l-connected, ( exist (V (S),)v; S )-paths Q 1 , Q 2 , . . . , Q l such that they are disjoint with each other without v. Let ⋃ there D0 := 1≤i≤l V (Q i ) ∩ V (S) . For d ∈ D0 , there exists 1 ≤ i ≤ l such that Q i connects v and d, and we denote Q d := Q i . Choose such a k-ended system S, a vertex v, and paths Q 1 , Q 2 , . . . , Q l so that (S1) (S2) (S3) (S4) (S5)
|Domi m (S)| is as large as possible, |PS | is as large as possible subject to (S1), |V (PS )| is as small as possible subject to (S2), V (CS ) ∩ D0 ̸= ∅ if possible subject to (S3), and ∑ d∈V (CS )∩D0 |Q d | is as small as possible subject to (S4).
By the choice (S1), we can see that 2|PS | + |CS | = f (S) = k and b P ̸∈ D0 for any P ∈ PS , and can easily obtain the following claim. Claim 1. There exists no k-ended system S ′ which satisfies the following. Domi m (S) ∪ {v} ⊆ Domi m (S ′ ).
(1)
Let d ∈ D0 . We denote Sd be an element of S such that d ∈ V (Sd ). If there exists a successor of d on Sd which belongs to D0 , then we denote it d ′ , that is, d ′ is a vertex of V (Sd ) ∩ D0 such that V (Sd (d, d ′ )) ∩ D0 = ∅. For convenience, if there exists no successor of d on Sd which belongs to D0 , then let d ′ := b Sd in case Sd ∈ PS , d ′ := d in case Sd ∈ CS . For each P ∈ PS with V (P) ∩ D0 ̸= ∅, choose d P ∈ V (P) ∩ D0 so that |P[d P , b P ]| is as small as possible. For each d ∈ D0 , choose d ∗ ∈ V (Sd ) so that (see Fig. 2) ( ) (D1) Domi m (Sd [d + , d ∗ ]) ⊆ Domi m (V (S) − V (Sd [d + , d ∗ ))) ∪ V (Q d ) , and (D2) |V (Sd [d + , d ∗ ])| is as large as possible subject to (D1) ∼ K 1 , then let d ∗ := d). Since Domi m (d + ) ⊆ Domi m (S), there exists d ∗ for any d ∈ D0 . For (for convenience, if Sd = each d ∈ D0 , let X d := V (Sd [d + , d ∗ ]). If d ∈ D0 satisfies either “Sd ∈ PS and d ∗ = b Sd ” or “Sd ∈ CS and d ∗ = d”, then we call d is an exceptional vertex of D0 . Let D := {d ∈ D0 : d is not an exceptional vertex of D0 }. Note that by the choice (S1) and the definition of the exceptional vertex, d ∗ ∈ V (Sd [d + , d ′ )) for any d ∈ D. Note that; if d ∈ D0 is exceptional and Sd ∈ PS , then d = d Sd , and if d ∈ D0 is exceptional and Sd ∈ CS , then |V (Sd ) ∩ D0 | = 1. Please cite this article in press as: M. Tsugaki, G. Yan, m-dominating k-ended trees of l-connected graphs, AKCE International Journal of Graphs and Combinatorics (2017), http://dx.doi.org/10.1016/j.akcej.2017.04.004.
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Fig. 2. The definition of d ∗ .
( ) Claim 2. For any d ∈ D, there exists xd ∈ Domi m (X d ) such that xd ̸∈ Domi m (V (S) − X d ) ∪ V (Q d ) . Proof. Let d ∈ D. Then d is not exceptional, and( hence, d ∗ ̸= b Sd if Sd ∈) PS ; d ∗ ̸= d if Sd ∈ CS . Hence, there exists d ∗ + ∈ V (Sd ) − X d . If Domi m (X d ) ⊆ Domi m (V (S) − X d ) ∪ V (Q d ) , then d ∗ + satisfies the choice (D1), which contradicts to the choice (D2). □ ( ) Claim 3. For any C ∈ CS , there exists γC ∈ Domi m (C) such that γC ̸∈ Domi m V (S) − V (C) . ( ) Proof. If there exists C ∈ CS such that Domi m (C) ⊆ Domi m V (S) − V (C) , then S ′ := S − {C} ∪ {v} is a k-ended system which satisfies (1), a contradiction. □ ( ) Claim 4. For any (P ∈ PS , there)exist α P ∈ Domi m (a P ) and β P ∈ Domi m (b P ) such that α P ̸∈ Domi m V (S)−{a P } and β P ̸∈ Domi m V (S) − {b P } . ( ) Proof. Suppose that there exists P ∈ PS such that Domi m (a P ) ⊆ Domi m V (S)−{a P } . Let S1 := S−{P}∪{P−a P }. If |P| ≥ 4, then S1 is a k-ended system such that |Domi m (S1 )| = |Domi m (S)|, |PS1 | = |PS | and |V (PS1 )| < |V (PS )|, which contradicts to the choice (S3). If |P| = 3, then S ′ := S1 ∪ {v} is a k-ended system which satisfies (1), a contradiction. □ Let APS := {α P , β P : P ∈ PS }, ΓCS := {γC : C ∈ CS and V (C) ∩ D0 = ∅} and D ∗ := {xd : d ∈ D}. In the rest of the proof, we will investigate the distance of two distinct vertices in APS ∪ ΓCS ∪ D ∗ ∪ {v}. Now, we define a function R on APS ∪ ΓCS ∪ D ∗ ∪ {v} as follows: for P ∈ PS , C ∈ CS and d ∈ D, we put R(α P ) := {a P }, R(β P ) := {b P }, R(γC ) := V (C), R(xd ) := X d and R(v) := {v}. By Claims 2–4, and since v ̸∈ Domi m (S), we can see that this definition is well defined, and hence we can see that APS , ΓCS , D ∗ and {v} are disjoint with each other, and |APS | = 2|PS |, |ΓCS | = |{C ∈ CS : V (C) ∩ D0 = ∅}| and |D ∗ | = |D|. At first, we prepare some basic claims (Claims 5–8). Claim 5. For any x ∈ APS ∪ ΓCS ∪ D ∗ ∪ {v}, x ∈ Domi m (R(x)) and x ̸∈ Domi m (V (S) − R(x)). Proof. By Claims 2–4, and since v ̸∈ Domi m (S), we can obtain this claim. □ Claim 6. Let x, y ∈ APS ∪ ΓCS ∪ D ∗ ∪ {v} with x ̸= y. Let Q be a shortest path which connects x and y. If (V (Q) − {x, y}) ∩ V (S) ̸⊆ R(x) ∪ R(y), then dG (x, y) ≥ 2(m + 1). Proof. If there exists z ∈ (V (Q) − {x, y}) ∩ V (S) such that z ̸∈ R(x) ∪ R(y), then, by Claim 5, ∥Q∥ = ∥Q[x, z]∥ + ∥Q[z, y]∥ ≥ dG (x, z) + dG (y, z) ≥ dG (x, V (S) − R(x)) + dG (y, V (S) − R(y)) ≥ 2(m + 1). □ Please cite this article in press as: M. Tsugaki, G. Yan, m-dominating k-ended trees of l-connected graphs, AKCE International Journal of Graphs and Combinatorics (2017), http://dx.doi.org/10.1016/j.akcej.2017.04.004.
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Fig. 3. Graphs in Claim 9.
Claim 7. For any x ∈ APS ∪ ΓCS ∪ D ∗ ∪ {v}, there exists an (x, R(x); S C )-path of length at most m. Proof. By Claim 5, a shortest path which connects x and a vertex of R(x) is an (x, R(x); S C )-path of length at most m. □ Let x, y ∈ APS ∪ ΓCS ∪ D ∗ ∪ {v} with x ̸= y. If there exists no (R(x), R(y); S C )-path, then we call a pair x and y is good. Note that x ̸∈ V (S) − R(x) and y ̸∈ V (S) − R(y) by Claim 5. By Claim 7, we can see that if a pair x and y is good, then there exist no (x, R(y); S C )-path, (R(x), y; S C )-path and (x, y; S C )-path. Claim 8. Let x, y ∈ APS ∪ ΓCS ∪ D ∗ ∪ {v} with x ̸= y. If a pair x and y is good, then dG (x, y) ≥ 2(m + 1). Proof. Let Q be a shortest path which connects x and y. Since a pair x and y is good, there exists no (x, y; S C )-path, that is (V (Q) − {x, y}) ∩ V (S) ̸= ∅. By Claim 6, we may assume that (V (Q) − {x, y}) ∩ V (S) ⊆ R(x) ∪ R(y). Since a pair x and y is good, there exists no (R(x), R(y); S C )-path, and hence, (V (Q) − {x, y}) ∩ V (S) ⊆ R(x) or (V (Q) − {x, y}) ∩ V (S) ⊆ R(y) holds. By the symmetry of x and y, we may assume that (V (Q) − {x, y}) ∩ V (S) ⊆ R(x). But, then there exists a (R(x), y; S C )-path. Since a pair x and y is good, this is a contradiction. □ Some of pairs of two distinct vertices in APS ∪ ΓCS ∪ D ∗ ∪ {v} can be easily seen to be good. We enumerate these pairs in the next claim. Claim 9. A pair x and y, which satisfies one of the following, is good. (i) x ∈ APS ∪ ΓCS and y ∈ APS with x ̸= y. (ii) x ∈ APS and y = v. (iii) x = xd and y ∈ APS for some C ∈ CS and d ∈ V (C) ∩ D. (iv) x, y ∈ ΓCS with x ̸= y. (v) x = xd and y ∈ ΓCS for some C ∈ CS and d ∈ V (C) ∩ D. (vi) x = xd and y ∈ APS − {β P } for some P ∈ PS and d ∈ V (P) ∩ D. (vii) x = xd and y = β P for some P ∈ PS and d ∈ V (P) ∩ D − {d P }. Proof. See Fig. 3. By the choice (S1), we can obtain the statements (i), (ii) and (iii). By the choice (S2), we can obtain the statements (iv) and (v). Now in (vi), if there exists an (R(x), R(y); S C )-path T , then by (ii), V (T ) ∩ V (Q d ) = ∅, and hence we can construct a k-ended system S ′ with V (S ′ ) ⊆ V (S) ∪ V (Q d ) ∪ V (T ) satisfying (1), a contradiction. Similarly, in (vii), if there exists an (R(x), R(y); S C )-path T , then by (ii), V (T )∩(V (Q d )∪V (Q d P )) = ∅, and hence we can construct a k-ended system S ′ with V (S ′ ) ⊆ V (S) ∪ V (Q d ) ∪ V (Q d P ) ∪ V (T ) satisfying (1), a contradiction. □ Next, to investigate other cases, we prepare some claims (10–13). Claim 10. For any d ∈ D0 , there exists no (d, X d ; S C )-path Q such that V (Q d ) ⊆ V (Q). Please cite this article in press as: M. Tsugaki, G. Yan, m-dominating k-ended trees of l-connected graphs, AKCE International Journal of Graphs and Combinatorics (2017), http://dx.doi.org/10.1016/j.akcej.2017.04.004.
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Proof. Suppose that for some d ∈ D0 , there exists a (d, X d ; S C )-path Q such that V (Q d ) ⊆ V (Q). Let S ′ := (Sd − Sd [d, b Q ]) ∪ Q. Then f (Sd ) = f (S ′ ), and hence S ′ := S − {Sd } ∪ {S ′ } is a k-ended system which satisfies (1), a contradiction. □ Claim 11. For any d ∈ D, there exists an (xd , X d ; S C )-path Id of length at most m such that V (Q e ) ∩ V (Id ) = ∅ for any e ∈ D0 . Proof. Let d ∈ D. By Claim 7, there exists an (xd , X d ; S C )-path Id of length at most m. Since the length of Id is at most m, it follows from Claim 2 that V (Q d ) ∩ V (Id ) = ∅. Suppose that V (Q e ) ∩ V (Id ) ̸= ∅ for some e ∈ D0 with e ̸= d. Since V (Q d ) ∩ V (Id ) = ∅, there exists a (d, X d ; S C )-path Q such that V (Q d ) ⊆ V (Q), which contradicts to Claim 10. □ We omit the proof of next claim because it is obvious. Claim 12. Let x ∈ V (G) and Z 1 , Z 2 ⊆ V (G). Suppose that there exist paths R1 and R2 such that Ri connects x and a vertex of Z i for i = 1, 2, and V (R1 ) ∪ V (R2 ) ⊆ (V (G) − V (S)) ∪ Z 1 ∪ Z 2 . Then there exists a (Z 1 , Z 2 ; S C )-path R such that V (R) ⊆ V (R1 ) ∪ V (R2 ), especially, if the length of Ri is at most m for i = 1, 2, then x ∈ Domi m (a R ) ∩ Domi m (b R ). ( Claim 13. Let d, e ∈ D0 with d ̸= e. Then Domi m (X d ) ∪ Domi m (X e ) ⊆ Domi m V (S) ∪ V (Q d ) ∪ V (Q e ) − ) V (Sd [d + , d ∗ )) − V (Se [e+ , e∗ )) . ( Proof. Let F) := {z ∈ Domi m (X d ) ∪ Domi m (X e ) : z ̸∈ Domi m V (S) ∪ V (Q d ) ∪ V (Q e ) − V (Sd [d + , d ∗ )) − V (Se [e+ , e∗ )) }. Suppose that F ̸= ∅, and let x ∈ F. By the symmetry of d and e, we may assume that x ∈ Domi m (X d ). By the choice (D1) and the definition of F, x ∈ Domi m (V (Se [e+ , e∗ ))). Hence there exists a path R1 of length at most m which connects x and a vertex of V (Se [e+ , e∗ )). On the other hand, since x ∈ Domi m (X d ), there exists a path R2 of length at most m which connects x and (a vertex of V (Sd)[d +(, d ∗ )). Since the length of + ∗ R1 and R2 are ) at most (m, it follow from ) the ( definition of )F that V (R1 ) ∪ V (R2 ) ∩ V (S) − V (Sd [d , d )) − + ∗ V (Se [e , e )) = ∅ and V (R1 ) ∪ V (R2 ) ∩ V (Q d ) ∪ V (Q e ) = ∅. Hence, by Claim 12, for any x ∈ F, there exists a (V (Sd [d + , d ∗ )), V (Se [e+ , e∗ )); S C )-path Px such that x ∈ Domi m (a Px ) and V (Px ) ∩ (V (Q d ) ∪ V (Q e )) (= ∅. Choose x ∈ F so that |Sd [d, a P)x ]| is as small as possible. If Sd = Se or {Sd , Se } ∩ CS ̸= ∅, then let S ′ := (Sd ∪ Se ) − (Sd [d, a Px ] ∪ Sd [e, b Px ]) ∪ (Q d ∪ Q e ∪ Px ) and S ′ := S − {Sd , Se } ∪ {S ′ } (note that S ′ is a path if {Sd , Se } ∩ PS ̸= ∅; ← − otherwise S ′ is a cycle). If Sd ̸= Se and Sd , Se ∈ PS , then let S ′ := Sd [b Sd , a Px ]Px [a Px , b Px ]Se [b Px , b Se ], ← − ← − T ′ := Sd [a Sd , d]Q d [d, v] Q e [v, e] Se [e, a Se ] and S ′ := S − {Sd , Se } ∪ {S ′ , T ′ } (note that S ′ and T ′ are paths, and V (S ′ ) and V (T ′ ) are disjoint). In any cases, S ′ is a k-ended system, and by the choice of x, we can see F ⊆ Domi m (S ′ ), and hence S ′ satisfies (1), a contradiction. □ In the next two claims, we investigate the distance of distinct two vertices in D ∗ ∪ {v}. Claim 14. dG (x, v) ≥ 2(m + 1) for any x ∈ D ∗ . Proof. Let x ∈ D ∗ and take d ∈ D so that xd = x. Suppose that dG (x, v) < 2(m + 1). Let Q be a shortest path which connects x and v. By Claim 6, (V (Q) − {x}) ∩ V (S) ⊆ X d . Therefore, it follows from 11 and 12 (apply Claim 12 as Z 1 = V (Q d ) − {d}, Z 2 = X d , R1 = Q and R2 = Id ), that there exists a (V (Q d ) − {d}, X d ; S C )path R. Since a R ∈ V (Q d ), it follows from Claim 11 that a R ̸∈ V (Id ), and hence by Claim 12, we can see that a R ∈ V (Q). Note that ∥Q[x, a R ]∥ ≥ dG (x, a R ) ≥ dG (x, V (Q d )) ≥ m + 1 by Claim 2. If v ̸∈ Domi m (a R ), then dG (x, v) = ∥Q∥ ≥ ∥Q[x, a R ]∥ + ∥Q[a R , v]∥ ≥ m( + 1 + dG (a R , v) ≥ 2(m + 1), a contradiction. Hence ) v ∈ Domi m (a R ). Let F := {z ∈ Domi m (X d ) : z ̸∈ Domi m (V (S) − V (Sd [d + , b R ))) ∪ V (Q d [d, a R ]) }. Suppose that F = ∅. Let S ′ := (Sd − Sd [d, b R ])∪ Q d [d, a R ]R[a R , b R ] and S ′ := S −{Sd }∪{S ′ } (note that f (Sd ) = f (S ′ )). Then S ′ is a k-ended system. Since v ∈ Domi m (a R ) and F = ∅, we can see that S ′ satisfies (1), a contradiction. Hence F ̸= ∅. Let f ∈ F. By the choice (D1), f ∈ Domi m (V (Q d (a R , v])). Let R1 be a shortest path which connects f and a vertex of V (Q d (a R , v]). On the other hand, since f ∈ Domi m (X d ), there exists a path R2 of length at most m which connects f and a vertex of V (Sd [d + , b R )). Since the length of R1 and R2 are at most m, it follows from the definition of F that Please cite this article in press as: M. Tsugaki, G. Yan, m-dominating k-ended trees of l-connected graphs, AKCE International Journal of Graphs and Combinatorics (2017), http://dx.doi.org/10.1016/j.akcej.2017.04.004.
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V (R1 ) ∪ V (R2 ) ⊆ V (G) − V (S) − V (Q d ) ∪ V (Q d (a R , v]) ∪ V (Sd [d + , b R )). Therefore, by Claim 12, for any f ∈ F, there exists a (V (Q d (a R , v]), V (Sd [d + , b R )); S C )-path R f such that V (Q d ) ∩ V (R f ) = {a R f } and f ∈ Domi m (a R f ). Choose f ∈ F so that |Q d [a R f , v]| is as small as possible. Let S ′ := (Sd − Sd [d, b R f ]) ∪ Q d [d, a R f ]R f [a R f , b R f ] and S ′ := S − {Sd } ∪ {S ′ } (note that f (Sd ) = f (S ′ )). Then S ′ is a k-ended system. Since a R ∈ V (S ′ ), we can see v ∈ Domi m (S ′ ). By the choice of f , we can see F ⊆ Domi m (S ′ ), and hence S ′ satisfies (1), a contradiction. □ (
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Claim 15. dG (x, y) ≥ 2(m + 1) for any x, y ∈ D ∗ with x ̸= y. Proof. Let x, y ∈ D ∗ with x ̸= y. Take d, e ∈ D so that x = xd and y = xe . Suppose that dG (x, y) < 2(m + 1). Let Q be a shortest path which connects x and y. By Claim 6, (V (Q) − {x, y}) ∩ V (S) ⊆ X d ∪ X e . Suppose that C C V (Q) ( ∩ (V (Q d ) ∪ )V (Q( e )) = ∅. By Claim ) 11, we can see that Id is an (x, X d ; S )-path, Ie is an (y, X eC; S )-path′ and V (Id ) ∪ V (Ie ) ∩ V (Q d ) ∪ V (Q e ) = ∅. Hence, by using Q, Id and Ie , we can obtain ( an (X d , X e ; S )-path Q such that V)(Q ′ ) ∩ (V (Q d ) ∪ V (Q e )) = ∅. If Sd = Se or {Sd , Se } ∩ CS ̸= ∅, then let S ′ := (Sd ∪ Se ) − (Sd [d, a Q ′ ] ∪ Sd [e, b Q ′ ]) ∪ (Q d ∪ Q e ∪ Q ′ ) and S ′ := S − {Sd , Se } ∪ {S ′ } (note that S ′ is a path or a cycle). If Sd ̸= Se and ← − ← − ← − Sd , Se ∈ PS , then let S ′ := Sd [a Sd , d]Q d [d, v] Q e [v, e] Se [e, a Se ], T ′ := Sd [b Sd , a Q ′ ]Q ′ [a Q ′ , b Q ′ ]Se [b Q ′ , b Se ] and ′ ′ ′ ′ ′ ′ S := S − {Sd , Se } ∪ {S , T } (note that S and T are paths, and V (S ) and V (T ′ ) are disjoint). In any cases, S ′ is a k-ended system, and by Claim 13, we can see S ′ satisfies (1), a contradiction. Hence V (Q) ∩ (V (Q d ) ∪ V (Q e )) ̸= ∅. Choose qx , q y ∈ V (Q) ∩ (V (Q d ) ∪ V (Q e )) so that |Q[x, qx ]| and |Q[q y , y]| are as small as possible. Suppose that qx ∈ V (Q e ). If V (Q[x, qx ]) ∩ X e ̸= ∅, then, by using Id , we can obtain an (X d , X e ; S C )-path Q ′ such that V (Q ′ ) ∩ (V (Q d ) ∪ V (Q e )) = ∅, and can obtain a contradiction as in the case V (Q) ∩ (V (Q d ) ∪ V (Q e )) = ∅. Hence V (Q[x, qx ]) ∩ X e = ∅. Then we can apply Claim 12 as Z 1 = V (Q e ) − {e}, Z 2 = X d , R1 = Q[x, qx ] and R2 = Id , and obtain (V (Q( e ) − {e}, X d ; S C )-path ) R such that V (R) ⊆ V (Q[x, qx ]) ∪ V (Id ). By Claim 11 and the minimality of |Q[x, qx ]|, V (Q[x, qx ]) ∪ V (Id ) ∩ (V (Q d ) − {v}) = ∅, that is V (R) ∩ (V (Q d ) − {v}) = ∅. Then, by using Q d , Q e [a R , v] and R, we can obtain a (d, X d ; S C )-path Q ′ such that V (Q d ) ⊆ V (Q ′ ), which contradicts to Claim 10. Hence, qx ∈ V (Q d ). By the symmetry of x and y, we can also obtain q y ∈ V (Q e ). Then, by Claim 2, dG (x, y) = ∥Q∥ ≥ dG (x, qx ) + dG (q y , y) ≥ 2(m + 1), a contradiction. □ Now, we extend the definition of exceptional. Recall that we regard a vertex as a cycle. For P ∈ PS such that V (P) ∩ D ̸= ∅ and d P ∈ D, if a pair xd P and β P is not good, then we also call d P is exceptional. Let PS′ := {P ∈ PS : V (P) ∩ D0 ̸= ∅ and d P is exceptional} and CS′ := {C ∈ CS : V (C) ∩ D0 ̸= ∅}. For any ← − P ∈ PS′ , there exists an (X d P , b P ; S C )-path Q P , and let C P := Q P [a Q P , b P ] P [b P , a Q P ] (note that if d P ∗ = b P , then Q P = C P = b P ). By Claim 9 (i), we can obtain V (C P ) ∩ V (C Q ) = ∅ for any P, Q ∈ PS′ with P ̸= Q, and also obtain V (C P ) ∩ V (C) = ∅ for any P ∈ PS′ and C ∈ CS . By Claim 9 (ii), we can obtain V (C P ) ∩ V (Q i ) = ∅ for any P ∈ PS′ and for any 1 ≤ i ≤ l. Claim 16. |PS′ ∪ CS′ | ≤ 1. Proof. If there exist C1 , C2 ∈ CS′ with C1 ̸= C2 , then we may assume that Q 1 and Q 2 are (V (C1 ), v; S C )← − − ′ path and (V (C2 ), v; S C )-path, respectively, and let S ′ := C[a − Q 1 , a Q 1 ]Q 1 [a Q 1 , v] Q 2 [v, a Q 2 ]C[a Q 2 , a Q 2 ] and S := ′ ′ ′ ′ S − {C1 , C2 } ∪ {S } (note that f ({C1 , C2 }) = f ({S })). If there exist P1 , P2 ∈ PS with P1 ̸= P2 , then let S := ←−− ← − P1 [a P1 , d P1 ]Q d P1 [d P1 , v] Q d P2 [v, d P2 ] P2 [d P2 , a P2 ] and S ′ := S − {P1 , P2 } ∪ {S ′ , C P1 , C P2 } (note that V (S ′ ), V (C P1 ) and V (C P2 ) are disjoint with each other, and f ({P1 , P2 }) = f ({S ′ , C P1 , C P2 }) = 4). If there exist P ∈ PS′ and C ∈ CS′ , ← − then we may assume that Q 1 is a (V (C), v; S C )-path, and let S ′ := P[a P , d P ]Q d P [d P , v] Q 1 [v, a Q 1 ]C[a Q 1 , a − Q 1 ] and S ′ := S − {P, C} ∪ {S ′ , C P } (note that V (S ′ ) and V (C P ) are disjoint, and f ({P, C}) = f ({S ′ , C P }) = 3). In any cases, S ′ is a k-ended system, and by Claim 13, S ′ satisfies (1), a contradiction. □ Note that either CS′ = ∅ or PS′ = ∅ holds by Claim 16. Claim 17. If CS′ = ∅, then a pair x and v is good for any x ∈ ΓCS . Proof. Suppose that CS′ = ∅, and a pair x and v is not good for some x ∈ ΓCS . Then there exists a (R(x), v; S C )-path Q. Since CS′ = ∅, it follows from the choice (S4), that there exist 1 ≤ i ≤ l and z ∈ (V (Q i ) ∩ V (Q)) − {v}. Choose i and z so that |Q[a Q , z]| is as small as possible. Then Q ′ := Q[a Q , z]Q i [z, v] is a (V (CS ), v; S C )-path such that V (Q ′ ) ∩ V (Q j ) = {v} for any 1 ≤ j ≤ l with j ̸= i. Since CS′ = ∅, this contradicts to the choice (S4). □ Please cite this article in press as: M. Tsugaki, G. Yan, m-dominating k-ended trees of l-connected graphs, AKCE International Journal of Graphs and Combinatorics (2017), http://dx.doi.org/10.1016/j.akcej.2017.04.004.
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Table 1 ′ = ∅. Where we have already investigated or not in the case PS′ ∪ CS A PS ΓCS D∗
A PS
ΓCS
D∗
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Claims 8, 9(i) / /
Claims 8, 9(i) Claims 8, 9(iv) /
Claims 8, 9(vi), (vii), 18 Not yet Claim 15
Claims 8, 9(ii) Claims 8, 17 Claim 14
By the definition of PS′ , we can obtain the following claim. Claim 18. (i) If PS′ ∪ CS′ = ∅, then D = D0 . (ii) If PS′ = ∅, then a pair xd P and β P is good for any P ∈ PS with V (P) ∩ D0 ̸= ∅. By Claim 16, either “Case 1. PS′ ∪ CS′ = ∅”, or “Case 2. |PS′ | = 1 and CS′ = ∅”, or “Case 3. |CS′ | = 1 and PS′ = ∅” holds. We divide the proof into these three cases, and continue to investigate the distance of distinct two vertices in APS ∪ ΓCS ∪ D ∗ ∪ {v}. For the convenience of the reader, at the top of each cases, we put a table. These tables show which cases we have already finished to investigate by which claims, and we write “not yet” where we have not finish to investigate yet. Case 1. PS′ ∪ CS′ = ∅. As mentioned above, Table 1 shows where we have already investigated or not. For example, by Claims 8 and 9(i), we can see that dG (x, y) ≥ 2(m + 1) for any x, y ∈ APS with x ̸= y. In this case, at first, we investigate the relation between ΓCS and D ∗ . Claim 19. There exists z 0 ∈ ΓCS ∪ D ∗ such that a pair x and y is good for any x ∈ ΓCS − {z 0 } and y ∈ D ∗ − {z 0 }. Proof. Let D1 := {d ∈ D0 : a pair xd and x is not good for some x ∈ ΓCS } and C1 := {C ∈ CS : a pair γC and x is not good for some x ∈ D ∗ }. To prove this claim, we have only to prove that either |D1 | ≤ 1 or |C1 | ≤ 1 holds. Suppose that |D1 | ≥ 2 and |C1 | ≥ 2. Then, by the definitions of D1 and C1 , there exist d1 , d2 ∈ D1 and C1 , C2 ∈ C1 with d1 ̸= d2 and C1 ̸= C2 such that a pair xd1 and γC1 is not good, and a pair xd2 and γC2 is not good. Hence, there exist (X d1 , V (C1 ); S C )-path P1 and (X d2 , V((C2 ); S C )-path P ) 2 . By ( Claim 9 (iv), we) can see V (P1 ) ∩ V (P2 ) = ∅. Since CS′ = ∅, it follows from Claim 17 that V (P1 ) ∪ V (P2 ) ∩ V (Q d1 ) ∪ V (Q d2 ) = ∅. If ←− ← − ← − Sd1 ̸= Sd2 , then let S ′ := Sd1 [a Sd1 , d1 ]Q d1 [d1 , v] Q d2 [v, d2 ] Sd2 [d2 , a Sd2 ], T ′ := Sd1 [b Sd1 , a P1 ]P1 [a P1 , b P1 ]C1 [b P1 , b− P1 ], ← − − ′ ′ ′ ′ ′ ′ U := Sd2 [b Sd2 , a P2 ]P2 [a P2 , b P2 ]C2 [b P2 , b P2 ] and S := S − {Sd1 , Sd2 , C1 , C2 } ∪ {S , T , U } (note that V (S ), V (T ′ ) and V (U ′ ) are disjoint with each other, and f ({Sd1 , Sd2 , C1 , C2 }) = f ({S ′ , T ′ , U ′ }) = 6). If Sd1 = Sd2 (by the symmetry of d1 and d2 , we may assume that d1 and d2 are arranged in this order along Sd1 ), then let S ′ := ←− ← − ← − − ′ Sd1 [a Sd1 , d1 ]Q d1 [d1 , v] Q d2 [v, d2 ] Sd1 [d2 , a P1 ]P1 [a P1 , b P1 ]C1 [b P1 , b− P1 ], T := Sd1 [b Sd1 , a P2 ]P2 [a P2 , b P2 ]C 2 [b P2 , b P2 ] and S ′ := S −{Sd1 , C1 , C2 }∪{S ′ , T ′ } (note that V (S ′ ) and V (T ′ ) are disjoint, and f ({Sd1 , C1 , C2 }) = f ({S ′ , T ′ }) = 4). In any cases, S ′ is a k-ended system, and by Claim 13, S ′ satisfies (1), a contradiction. □ By Claims 8, 9, 14, 15 and 17–19 (see Table 1), we can see that APS ∪ (ΓCS ∪ D ∗ − {z 0 }) ∪ {v} is a 2(m + 1)-stable set, and hence, k + l − 1 ≥ α 2(m+1) (G) ≥ |APS | + (|ΓCS | + |D ∗ | − 1) + |{v}| = 2|PS | + |CS | + |D0 | = k + l, a contradiction. Case 2. |PS′ | = 1 and CS′ = ∅. Let {P0 } := PS′ and D ∗∗ := {xd : d ∈ D0 − {d P0 }}. In this case, at first, we investigate the relation between ΓCS and D ∗∗ . Claim 20. A pair x and y is good for any x ∈ ΓCS and y ∈ D ∗∗ . Proof. Suppose not. Then there exist d ∈ D0 − {d P0 }, C ∈ CS and (X d , V (C); S C )-path Q. By Claim 9(i) and Claim 17, we can see V (Q)∩V (Q P0 ) = ∅ and V (Q)∩(V (Q d )∪V (Q d P0 )) = ∅, respectively. If Sd ̸= P0 , then let S ′ := ← − ←−− ← − ′ ′ Sd [b Sd , a Q ]Q[a Q , b Q ]C[b Q , b− Q ], T := Sd [a Sd , d]Q d [d, v] Q d P0 [v, d P0 ] P0 [d P0 , a P0 ] and S := S − {Sd , P0 , C} ∪ Please cite this article in press as: M. Tsugaki, G. Yan, m-dominating k-ended trees of l-connected graphs, AKCE International Journal of Graphs and Combinatorics (2017), http://dx.doi.org/10.1016/j.akcej.2017.04.004.
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Table 2 ′ = ∅. Where we have already investigated or not in the case |PS′ | = 1 and CS APS ΓCS D ∗∗
A PS
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Claims 8, 9(i) / /
Claims 8, 9(i) Claims 8, 9(iv) /
Claims 8, 9(vi), (vii) Not yet Claim 15
Claims 8, 9(ii) Claims 8, 17 Claim 14
Table 3 ′ | = 1 and P ′ = ∅. Where we have already investigated or not in the case |CS S APS ΓCS D∗
A PS
ΓCS
D∗
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Claims 8, 9(i) / /
Claims 8, 9(i) Claims 8, 9(iv) /
Claims 8, 9(iii), (vi), (vii), 18 Claims 8, 9(v), not yet Claim 15
Claims 8, 9(ii) Not yet Claim 14
{C P0 , S ′ , T ′ } (note that V (C P0 ), V (S ′ ) and V (T ′ ) are disjoint with each other, and f ({Sd , P0 , C}) = f ({C P0 , S ′ , T ′ }) = ← − ← − ′ 5). If Sd = P0 , then let S ′ := Sd [a Sd , d]Q d [d, v] Q d P [v, d P0 ] Sd [d P0 , a Q ] Q[a Q , b Q ]C[b Q , b− Q ] and S := S −{Sd , C} 0 ′ ′ ′ ∪{C P0 , S } (note that V (C P0 ) and V (S ) are disjoint, and f ({Sd , C}) = f ({C P0 , S }) = 3). In any cases, S ′ is a k-ended system, and by Claim 13, S ′ satisfies (1), a contradiction. □ By Claims 8, 9, 14, 15, 17 and 20 (see Table 2), we can see that APS ∪ ΓCS ∪ D ∗∗ ∪ {v} is a 2(m + 1)-stable set, and hence, k + l − 1 ≥ α 2(m+1) (G) ≥ |APS | + |ΓCS | + |D ∗∗ | + |{v}| = 2|PS | + |CS | + (|D0 | − 1) + 1 = k + l, a contradiction. Case 3. |CS′ | = 1 and PS′ = ∅. Let {C0 } := CS′ . We may assume that Q 1 is a (V (C0 ), v; S C )-path. In this case, at first, we investigate the relation between ΓCS and D ∗ ∪ {v}. Claim 21. A pair xd and y is good for any d ∈ V (PS ) ∩ D0 and y ∈ ΓCS . Proof. Suppose not. Then there exist d ∈ V (PS ) ∩ D0 , C1 ∈ CS − {C0 } and (X d , V (C1 ); S C )-path Q. ⋃ If V (Q) ∩ 1≤i≤l V (Q i ) ̸= ∅, then there exists a (V (C0 ), V (C1 ); S C )-path, which contradicts to the choice ⋃ ← − (S2). Hence V (Q) ∩ 1≤i≤l V (Q i ) = ∅. Let S ′ := Sd [a Sd , d]Q d [d, v] Q 1 [v, a Q 1 ]C0 [a Q 1 , a Q 1 − ], T ′ := ← − − ′ ′ ′ ′ ′ Sd [b Sd , a⋃ Q ]Q[a Q , b Q ]C 1 [b Q , b Q ] and S := S − {Sd , C 0 , C 1 } ∪ {S , T }. Note that V (S ) ∩ V (T ) = ∅ because V (Q) ∩ 1≤i≤l V (Q i ) = ∅, and note that f ({Sd , C0 , C1 }) = f ({S ′ , T ′ }) = 4. Hence S ′ is a k-ended system which satisfies (1), a contradiction. □ Claim 22. A pair v and y is good for any y ∈ ΓCS . Proof. Suppose not. Then there exist C1 ∈ CS − {C0 } and a (v, V (C1 ); S C )-path Q. Since V (Q 1 ) ∪ V (Q) ⊆ (V (G) − V (S)) ∪ V (C0 ) ∪ V (C1 ), it follows from Claim 12 that there exists a (V (C0 ), V (C1 ); S C )-path, which contradicts to the choice (S2). □ By Claims 8, 9, 14, 15, 18, 21 and 22 (see Table 3), we can see that APS ∪ ΓCS ∪ D ∗ ∪ {v} is a 2(m + 1)-stable set, and hence, k + l − 1 ≥ α 2(m+1) (G) ≥ |APS | + |ΓCS | + |D ∗ | + |{v}| = 2|PS | + (|CS | − 1) + |D ∗ | + 1 = k + |D ∗ |, that is |D ∗ | ≤ l − 1. Since PS(′ = ∅, this implies that V (C ) 0 ) ∩ D0 has an exceptional vertex. Then, by the ( choice (D1), Domi m (C)0 ) ⊆ Domi m (V (S) − V (C0 )) ∪ V (Q 1 ) . Let F := {z ∈ Domi m (C0 ) : z ̸∈ Domi m (V (S) − V (C0 )) ∪ {a Q 1 } }. Suppose that F = ∅. Let S ′ := S − {C0 } ∪ {Q 1 [a Q 1 , a Q 1 + ]} and Q 1 ′ := Q 1 − a Q 1 . Since F = ∅, S ′ is a k-ended system such that |Domi m (S ′ )| ≥ |Domi m (S)|, |PS ′ | = |PS |, |V (PS ′ )| = |V (PS )|. Moreover, Q 1 ′ , Q 2 , . . . , Q l are (V (S ′ ), v; S ′ C )-paths such that they are disjoint with each other without v, and Q 1 ′ is a (V (CS ′ ), v; S ′ C )-path. But, |Q 1 ′ | < |Q 1 |. These contradict to the choices (S1) or (S5). Hence F ̸= ∅. Let f ∈ F. Since f ∈ Domi m (C0 ) and a Q 1 ̸∈ Domi m ( f ), there exists a path R1 of length at most m which connects f and a vertex of V (C0 ) − {a Q 1 }. On the other hand, by the definition of F, there exists a path R2 of length at most m which connects f and a vertex of V (Q 1 ) − {a Q 1 }. Since the length of R1 and R2 are at most m, it follows from the definition Please cite this article in press as: M. Tsugaki, G. Yan, m-dominating k-ended trees of l-connected graphs, AKCE International Journal of Graphs and Combinatorics (2017), http://dx.doi.org/10.1016/j.akcej.2017.04.004.
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of F that V (R1 ) ∪ V (R2 ) ⊆ (V (G) − V (S)) ∪ (V (C0 ) − {a Q 1 }) ∪ (V (Q 1 ) − {a Q 1 }). Hence, by Claim 12, there exists a m (V (C0 ) − {a Q 1 }, V (Q 1 ) − {a Q 1 }; S C )-path ⋃ R f such that f ∈ Domi (b R f ). Choose f ∈ F so that |QC1 [b R f , v]| is as small as possible. Suppose that V (R f ) ∩ 2≤i≤l V (Q i ) ̸= ∅. Then there exists a (V (C0 ) − {a Q 1 }, v; S )-path R such ← − that V (Q 1 ) ∩ V (R) = {v}. Let S ′ := Q 1 [a Q 1 , v] R [v, a R ]C0 [a R , a Q 1 ]( and S ′ := S − {C0 } ∪ {S ′ }.) Since S ′ is a cycle, we can see that S ′ is a k-ended system. Since Domi m (C0 ) ⊆ Domi m (V (S) − V (C0 )) ∪ V (Q 1 ) , we can see that S ′ ⋃ ← − satisfies (1), a contradiction. Hence V (R f ) ∩ 2≤i≤l V (Q i ) = ∅. Let S ′ := Q 1 [a Q 1 , b R f ] R [b R f , a R f ]C0 [a R f , a Q 1 ], ′ ′ ′ ′ S := S − {C0 } ∪ {S } and Q 1 := Q 1 [b R f , v]. Note that S is a cycle. By the choice of f , S ′ is a k-ended system such that |Domi m (S ′ )| ≥ |Domi m (S)|, |PS ′ | = |PS |, |V (PS ′ )| = |V (PS )|. Note that Q 1 ′ , Q 2 , . . . , Q l are (V (S ′ ), v; S ′ C )paths such that they are disjoint with each other without v, and Q 1 ′ is a (V (CS ′ ), v; S ′ C )-path such that |Q 1 ′ | < |Q 1 |. These contradict to the choices (S1) or (S5), and this completes the proof of Theorem 7. Acknowledgments The authors would like to thank the referees for their valuable comments. This work was done while the author was a foreign researcher of Chinese Academy of Science, Supported by Chinese Academy of Science Fellowship for Young International Scientists. Grant No. 2012Y1JA0004. References [1] M. Kano, M. Tsugaki, G. Yan, m-dominating k-ended trees of graphs, Discrete Math. 333 (2014) 1–5. [2] H.J. Broersma, Existence of ∆k -cycles and ∆k -paths, J. of Graph Theory 12 (4) (1988) 499–507. [3] S. Win, On a conjecture of Las Vergnas concerning certain spanning trees in graphs, Resultate Math. 2 (1979) 215–224.
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m-dominating k-ended trees of l-connected graphs Masao Tsugaki a , ∗, Guiying Yan b a Tokyo University of Science, Kagurazaka, Shinjuku-ku, Tokyo, Japan b Institute of mathematical and system sciences, Chinese Academy of Science, Beijing, PR China
Received 29 August 2015; accepted 19 March 2017 Available online xxxxx
Abstract Let k ≥ 2, l ≥ 1 and m ≥ 0 be integers, and let G be an l-connected graph. If there exists a subgraph X of G such that the distance between v and X is at most m for any v ∈ V (G), then we say that X m-dominates G. A subset S of V (G) is said to be 2(m + 1)-stable if the distance between each pair of distinct vertices in S is at least 2(m + 1). In this paper, we prove that if G does not have a 2(m + 1)-stable set of order at least k + l, then G has an m-dominating tree which has at most k leaves. c 2017 Kalasalingam University. Publishing Services by Elsevier B.V. This is an open access article under the CC BY-NC-ND ⃝ license (http://creativecommons.org/licenses/by-nc-nd/4.0/). Keywords: Vertex dominating; m-dominating; k-ended tree
1. Introduction In this paper, we consider finite simple graphs, which have neither loops nor multiple edges. Let G be a graph, and X be a subgraph or a vertex set of G. We write |X | for the order of X . For two vertices u and v of G, let dG (u, v) denote the distance between u and v. For a vertex v of G, the distance between v and X is defined to be the minimum value of dG (v, x) for all x ∈ V (X ) or x ∈ X , and denoted by dG (v, X ). For an integer m ≥ 0, let Domi m (X ) := {v ∈ V (G) : dG (v, X ) ≤ m}. If V (G) = Domi m (X ), then we say that X m-dominates G. For an integer l ≥ 2, a subset S of V (G) is said to be l-stable if the distance between each pair of distinct vertices in S is at least l. The l-stable number of G is the cardinality of a maximum l-stable set of G, and is denoted by αl (G), that is αl (G) := max{|S| : S ⊆ V (G), S is l-stable}. Note that α 2 (G) is the independence number of G. A tree is called a k-ended tree if the number of leaves is at most k. In this paper, we investigate a stable number condition for l-connected graphs to have m-dominating k-ended trees. Concerning on this, the following three theorems are known. Theorem 1 (Kano, Tsugaki and Yan [1]). Let k ≥ 2 and m ≥ 0 be integers, and let G be a connected graph. If α 2(m+1) (G) ≤ k, then G has an m-dominating k-ended tree. Peer review under responsibility of Kalasalingam University. ∗ Corresponding author.
E-mail addresses: [email protected] (M. Tsugaki), [email protected] (G. Yan). http://dx.doi.org/10.1016/j.akcej.2017.04.004 c 2017 Kalasalingam University. Publishing Services by Elsevier B.V. This is an open access article under the CC BY-NC-ND license 0972-8600/⃝ (http://creativecommons.org/licenses/by-nc-nd/4.0/).
Please cite this article in press as: M. Tsugaki, G. Yan, m-dominating k-ended trees of l-connected graphs, AKCE International Journal of Graphs and Combinatorics (2017), http://dx.doi.org/10.1016/j.akcej.2017.04.004.
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Fig. 1. An l-connected graph which has no m-dominating k-ended tree.
Theorem 2 (Broersma [2]). Let l ≥ 1 and m ≥ 0 be integers, and let G be an l-connected graph. If α 2(m+1) (G) ≤ l + 1, then G has an m-dominating path, that is an m-dominating 2-ended tree. Theorem 3 (Win [3]). Let k ≥ 2 and l ≥ 1 be integers, and let G be an l-connected graph. If α 2 (G) ≤ k + l − 1, then G has a spanning k-ended tree, that is a 0-dominating k-ended tree. In this paper, we prove the following theorem which is a generalization of above three theorems. Theorem 4. Let k ≥ 2, l ≥ 1 and m ≥ 0 be integers, and let G be an l-connected graph. If α 2(m+1) (G) ≤ k + l − 1, then G has an m-dominating k-ended tree. The upper bound of the stable number condition of Theorem 4 is sharp. Let k ≥ 2, l ≥ 1 and m ≥ 1 be integers. For 1 ≤ i ≤ k + l, let Di,1 , Di,2 , . . . , Di,m be disjoint copies of the complete graph K l . For 1 ≤ j ≤ m − 1, join all the vertices of Di, j and all the vertices of Di, j+1 by edges. Let vi be a vertex not contained in Di,1 ∪ Di,2 ∪ · · · ∪ Di,m , and join vi and all the vertices of Di,m by edges. Let H be a complete graph of order l. For 1 ≤ i ≤ k + l, join all the vertices of H and all the vertices of Di,1 by edges. Let G denote the resulting graph (see Fig. 1). Then G is an l-connected graph, and has no m-dominating k-ended tree. On the other hand, α 2(m+1) (G) = k + l. Concerning on a degree sum condition, the following theorem is known. Theorem 5 (Broersma [2]). Let l ≥ 1 and m ≥ 0 be integers, and let G be an l-connected graph. If ∑ m x∈S |Domi (x)| ≥ |G| − l + 1 for any 2(m + 1)-stable set S of order l + 2, then G has an m-dominating path, that is an m-dominating 2-ended tree. By Theorem 4, we can obtain the following theorem which is a generalization of Theorem 5. ∑ m Theorem 6. Let k ≥ 2, l ≥ 1 and m ≥ 0 be integers, and let G be an l-connected graph. If x∈S |Domi (x)| ≥ |G| − l + 1 for any 2(m + 1)-stable set S of order k + l, then G has an m-dominating k-ended tree. Proof of Theorem 6. Suppose that G does not have an m-dominating k-ended tree. Then, by Theorem 4, α 2(m+1) (G) ≥ k + l, and hence there exists a 2(m + 1)-stable set S of order |S| = k + l. Note that for any x, y ∈ S m with x ̸= y, Domi∑ (x) and Domi m (y) are disjoint and there are no edge between them. Since G is l-connected, these imply that |G| − x∈S |Domi m (x)| ≥ l, which contradicts to the condition in Theorem 6. □ 2. Proof of Theorem 4 A system of a graph G is defined to be a set of vertex-disjoint paths and cycles of G. In this paper, we regard a vertex and an edge as a cycle, and thus a path means a path of order at least 3. Let S be a system of a graph G. For S ∈ S, we put f (S) := 2 if S is a path, and f (S) := 1 otherwise. Let PS := and CS := ⋃{S ∈ S : f (S) = 2} ⋃ {S ∈ S : fm(S) = 1}. m For X ⊆∑S and an integer m ≥ 0, we define V (X ) := V (X ), Domi (X ) := X ∈X X ∈X Domi (X ) and f (X ) := X ∈X f (X ). For an integer k ≥ 2, we call X a k-ended system if f (X ) ≤ k. Note that f (S) = 2|PS | + |CS |. Please cite this article in press as: M. Tsugaki, G. Yan, m-dominating k-ended trees of l-connected graphs, AKCE International Journal of Graphs and Combinatorics (2017), http://dx.doi.org/10.1016/j.akcej.2017.04.004.
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For an oriented path P, we write ∥P∥ for the length of P, and let a P and b P be the initial and terminal vertices of P, respectively. Let S be an oriented path (or cycle) and x ∈ V (S). We denote the successor and the predecessor of x on S by x + and x − , respectively (if exists). For convenience, let {x + } := V (S) − {x} and {x − } := V (S) − {x} if S is an edge, and x + := x and x − := x if S is a vertex. For x, y ∈ V (S), we denote by S[x, y] a path from x to y along the ← − orientation of S. The reverse sequence of S[x, y] is denoted by S [y, x]. We denote S[x, y] − {x, y}, S[x, y] − {x}, S[x, y] − {y} by S(x, y), S(x, y] and S[x, y), respectively. For convenience, let S[x, x) := ∅. Let S be a system of a graph G, and let X and Y be subsets of V (G). For a (path P of G, if )P connects a vertex of X and a vertex of Y , and all of inner vertices of P are contained in V (G) − V (S) ∪ X ∪ Y , then we call P an (X, Y( ; S C )-path of G, )and(we orient P from ) a vertex of X to a vertex of Y , that is P is a path such that a P ∈ X , b P ∈ Y and V (P) − {a P , b P } ∩ V (S) ∪ X ∪ Y = ∅. For x, y ∈ V (G), we abbreviate ({x}, Y ; S C )-path, (X, {y}; S C )-path and ({x}, {y}; S C )-path as (x, Y ; S C )-path, (X, y; S C )-path and (x, y; S C )-path, respectively. Since we can obtain a k-ended tree including V (S) from a k-ended system S, Theorem 4 is an immediate consequence of the following theorem. Theorem 7. Let k ≥ 2, l ≥ 1 and m ≥ 0 be integers, and let G be an l-connected graph. If α 2(m+1) (G) ≤ k + l − 1, then G has an m-dominating k-ended system. We now prove Theorem 7. Proof of Theorem 7. Let k ≥ 2, l ≥ 1 and m ≥ 0 be integers, and let G be a graph which satisfies the condition in Theorem 7. Suppose that G has no m-dominating k-ended system. Let S be a k-ended system in G. We give an orientation to S for each S ∈ S. Since G has no m-dominating k-ended system, there exists v ∈ V (G) − Domi m (S). Since G is C l-connected, ( exist (V (S),)v; S )-paths Q 1 , Q 2 , . . . , Q l such that they are disjoint with each other without v. Let ⋃ there D0 := 1≤i≤l V (Q i ) ∩ V (S) . For d ∈ D0 , there exists 1 ≤ i ≤ l such that Q i connects v and d, and we denote Q d := Q i . Choose such a k-ended system S, a vertex v, and paths Q 1 , Q 2 , . . . , Q l so that (S1) (S2) (S3) (S4) (S5)
|Domi m (S)| is as large as possible, |PS | is as large as possible subject to (S1), |V (PS )| is as small as possible subject to (S2), V (CS ) ∩ D0 ̸= ∅ if possible subject to (S3), and ∑ d∈V (CS )∩D0 |Q d | is as small as possible subject to (S4).
By the choice (S1), we can see that 2|PS | + |CS | = f (S) = k and b P ̸∈ D0 for any P ∈ PS , and can easily obtain the following claim. Claim 1. There exists no k-ended system S ′ which satisfies the following. Domi m (S) ∪ {v} ⊆ Domi m (S ′ ).
(1)
Let d ∈ D0 . We denote Sd be an element of S such that d ∈ V (Sd ). If there exists a successor of d on Sd which belongs to D0 , then we denote it d ′ , that is, d ′ is a vertex of V (Sd ) ∩ D0 such that V (Sd (d, d ′ )) ∩ D0 = ∅. For convenience, if there exists no successor of d on Sd which belongs to D0 , then let d ′ := b Sd in case Sd ∈ PS , d ′ := d in case Sd ∈ CS . For each P ∈ PS with V (P) ∩ D0 ̸= ∅, choose d P ∈ V (P) ∩ D0 so that |P[d P , b P ]| is as small as possible. For each d ∈ D0 , choose d ∗ ∈ V (Sd ) so that (see Fig. 2) ( ) (D1) Domi m (Sd [d + , d ∗ ]) ⊆ Domi m (V (S) − V (Sd [d + , d ∗ ))) ∪ V (Q d ) , and (D2) |V (Sd [d + , d ∗ ])| is as large as possible subject to (D1) ∼ K 1 , then let d ∗ := d). Since Domi m (d + ) ⊆ Domi m (S), there exists d ∗ for any d ∈ D0 . For (for convenience, if Sd = each d ∈ D0 , let X d := V (Sd [d + , d ∗ ]). If d ∈ D0 satisfies either “Sd ∈ PS and d ∗ = b Sd ” or “Sd ∈ CS and d ∗ = d”, then we call d is an exceptional vertex of D0 . Let D := {d ∈ D0 : d is not an exceptional vertex of D0 }. Note that by the choice (S1) and the definition of the exceptional vertex, d ∗ ∈ V (Sd [d + , d ′ )) for any d ∈ D. Note that; if d ∈ D0 is exceptional and Sd ∈ PS , then d = d Sd , and if d ∈ D0 is exceptional and Sd ∈ CS , then |V (Sd ) ∩ D0 | = 1. Please cite this article in press as: M. Tsugaki, G. Yan, m-dominating k-ended trees of l-connected graphs, AKCE International Journal of Graphs and Combinatorics (2017), http://dx.doi.org/10.1016/j.akcej.2017.04.004.
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Fig. 2. The definition of d ∗ .
( ) Claim 2. For any d ∈ D, there exists xd ∈ Domi m (X d ) such that xd ̸∈ Domi m (V (S) − X d ) ∪ V (Q d ) . Proof. Let d ∈ D. Then d is not exceptional, and( hence, d ∗ ̸= b Sd if Sd ∈) PS ; d ∗ ̸= d if Sd ∈ CS . Hence, there exists d ∗ + ∈ V (Sd ) − X d . If Domi m (X d ) ⊆ Domi m (V (S) − X d ) ∪ V (Q d ) , then d ∗ + satisfies the choice (D1), which contradicts to the choice (D2). □ ( ) Claim 3. For any C ∈ CS , there exists γC ∈ Domi m (C) such that γC ̸∈ Domi m V (S) − V (C) . ( ) Proof. If there exists C ∈ CS such that Domi m (C) ⊆ Domi m V (S) − V (C) , then S ′ := S − {C} ∪ {v} is a k-ended system which satisfies (1), a contradiction. □ ( ) Claim 4. For any (P ∈ PS , there)exist α P ∈ Domi m (a P ) and β P ∈ Domi m (b P ) such that α P ̸∈ Domi m V (S)−{a P } and β P ̸∈ Domi m V (S) − {b P } . ( ) Proof. Suppose that there exists P ∈ PS such that Domi m (a P ) ⊆ Domi m V (S)−{a P } . Let S1 := S−{P}∪{P−a P }. If |P| ≥ 4, then S1 is a k-ended system such that |Domi m (S1 )| = |Domi m (S)|, |PS1 | = |PS | and |V (PS1 )| < |V (PS )|, which contradicts to the choice (S3). If |P| = 3, then S ′ := S1 ∪ {v} is a k-ended system which satisfies (1), a contradiction. □ Let APS := {α P , β P : P ∈ PS }, ΓCS := {γC : C ∈ CS and V (C) ∩ D0 = ∅} and D ∗ := {xd : d ∈ D}. In the rest of the proof, we will investigate the distance of two distinct vertices in APS ∪ ΓCS ∪ D ∗ ∪ {v}. Now, we define a function R on APS ∪ ΓCS ∪ D ∗ ∪ {v} as follows: for P ∈ PS , C ∈ CS and d ∈ D, we put R(α P ) := {a P }, R(β P ) := {b P }, R(γC ) := V (C), R(xd ) := X d and R(v) := {v}. By Claims 2–4, and since v ̸∈ Domi m (S), we can see that this definition is well defined, and hence we can see that APS , ΓCS , D ∗ and {v} are disjoint with each other, and |APS | = 2|PS |, |ΓCS | = |{C ∈ CS : V (C) ∩ D0 = ∅}| and |D ∗ | = |D|. At first, we prepare some basic claims (Claims 5–8). Claim 5. For any x ∈ APS ∪ ΓCS ∪ D ∗ ∪ {v}, x ∈ Domi m (R(x)) and x ̸∈ Domi m (V (S) − R(x)). Proof. By Claims 2–4, and since v ̸∈ Domi m (S), we can obtain this claim. □ Claim 6. Let x, y ∈ APS ∪ ΓCS ∪ D ∗ ∪ {v} with x ̸= y. Let Q be a shortest path which connects x and y. If (V (Q) − {x, y}) ∩ V (S) ̸⊆ R(x) ∪ R(y), then dG (x, y) ≥ 2(m + 1). Proof. If there exists z ∈ (V (Q) − {x, y}) ∩ V (S) such that z ̸∈ R(x) ∪ R(y), then, by Claim 5, ∥Q∥ = ∥Q[x, z]∥ + ∥Q[z, y]∥ ≥ dG (x, z) + dG (y, z) ≥ dG (x, V (S) − R(x)) + dG (y, V (S) − R(y)) ≥ 2(m + 1). □ Please cite this article in press as: M. Tsugaki, G. Yan, m-dominating k-ended trees of l-connected graphs, AKCE International Journal of Graphs and Combinatorics (2017), http://dx.doi.org/10.1016/j.akcej.2017.04.004.
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Fig. 3. Graphs in Claim 9.
Claim 7. For any x ∈ APS ∪ ΓCS ∪ D ∗ ∪ {v}, there exists an (x, R(x); S C )-path of length at most m. Proof. By Claim 5, a shortest path which connects x and a vertex of R(x) is an (x, R(x); S C )-path of length at most m. □ Let x, y ∈ APS ∪ ΓCS ∪ D ∗ ∪ {v} with x ̸= y. If there exists no (R(x), R(y); S C )-path, then we call a pair x and y is good. Note that x ̸∈ V (S) − R(x) and y ̸∈ V (S) − R(y) by Claim 5. By Claim 7, we can see that if a pair x and y is good, then there exist no (x, R(y); S C )-path, (R(x), y; S C )-path and (x, y; S C )-path. Claim 8. Let x, y ∈ APS ∪ ΓCS ∪ D ∗ ∪ {v} with x ̸= y. If a pair x and y is good, then dG (x, y) ≥ 2(m + 1). Proof. Let Q be a shortest path which connects x and y. Since a pair x and y is good, there exists no (x, y; S C )-path, that is (V (Q) − {x, y}) ∩ V (S) ̸= ∅. By Claim 6, we may assume that (V (Q) − {x, y}) ∩ V (S) ⊆ R(x) ∪ R(y). Since a pair x and y is good, there exists no (R(x), R(y); S C )-path, and hence, (V (Q) − {x, y}) ∩ V (S) ⊆ R(x) or (V (Q) − {x, y}) ∩ V (S) ⊆ R(y) holds. By the symmetry of x and y, we may assume that (V (Q) − {x, y}) ∩ V (S) ⊆ R(x). But, then there exists a (R(x), y; S C )-path. Since a pair x and y is good, this is a contradiction. □ Some of pairs of two distinct vertices in APS ∪ ΓCS ∪ D ∗ ∪ {v} can be easily seen to be good. We enumerate these pairs in the next claim. Claim 9. A pair x and y, which satisfies one of the following, is good. (i) x ∈ APS ∪ ΓCS and y ∈ APS with x ̸= y. (ii) x ∈ APS and y = v. (iii) x = xd and y ∈ APS for some C ∈ CS and d ∈ V (C) ∩ D. (iv) x, y ∈ ΓCS with x ̸= y. (v) x = xd and y ∈ ΓCS for some C ∈ CS and d ∈ V (C) ∩ D. (vi) x = xd and y ∈ APS − {β P } for some P ∈ PS and d ∈ V (P) ∩ D. (vii) x = xd and y = β P for some P ∈ PS and d ∈ V (P) ∩ D − {d P }. Proof. See Fig. 3. By the choice (S1), we can obtain the statements (i), (ii) and (iii). By the choice (S2), we can obtain the statements (iv) and (v). Now in (vi), if there exists an (R(x), R(y); S C )-path T , then by (ii), V (T ) ∩ V (Q d ) = ∅, and hence we can construct a k-ended system S ′ with V (S ′ ) ⊆ V (S) ∪ V (Q d ) ∪ V (T ) satisfying (1), a contradiction. Similarly, in (vii), if there exists an (R(x), R(y); S C )-path T , then by (ii), V (T )∩(V (Q d )∪V (Q d P )) = ∅, and hence we can construct a k-ended system S ′ with V (S ′ ) ⊆ V (S) ∪ V (Q d ) ∪ V (Q d P ) ∪ V (T ) satisfying (1), a contradiction. □ Next, to investigate other cases, we prepare some claims (10–13). Claim 10. For any d ∈ D0 , there exists no (d, X d ; S C )-path Q such that V (Q d ) ⊆ V (Q). Please cite this article in press as: M. Tsugaki, G. Yan, m-dominating k-ended trees of l-connected graphs, AKCE International Journal of Graphs and Combinatorics (2017), http://dx.doi.org/10.1016/j.akcej.2017.04.004.
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M. Tsugaki, G. Yan / AKCE International Journal of Graphs and Combinatorics (
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Proof. Suppose that for some d ∈ D0 , there exists a (d, X d ; S C )-path Q such that V (Q d ) ⊆ V (Q). Let S ′ := (Sd − Sd [d, b Q ]) ∪ Q. Then f (Sd ) = f (S ′ ), and hence S ′ := S − {Sd } ∪ {S ′ } is a k-ended system which satisfies (1), a contradiction. □ Claim 11. For any d ∈ D, there exists an (xd , X d ; S C )-path Id of length at most m such that V (Q e ) ∩ V (Id ) = ∅ for any e ∈ D0 . Proof. Let d ∈ D. By Claim 7, there exists an (xd , X d ; S C )-path Id of length at most m. Since the length of Id is at most m, it follows from Claim 2 that V (Q d ) ∩ V (Id ) = ∅. Suppose that V (Q e ) ∩ V (Id ) ̸= ∅ for some e ∈ D0 with e ̸= d. Since V (Q d ) ∩ V (Id ) = ∅, there exists a (d, X d ; S C )-path Q such that V (Q d ) ⊆ V (Q), which contradicts to Claim 10. □ We omit the proof of next claim because it is obvious. Claim 12. Let x ∈ V (G) and Z 1 , Z 2 ⊆ V (G). Suppose that there exist paths R1 and R2 such that Ri connects x and a vertex of Z i for i = 1, 2, and V (R1 ) ∪ V (R2 ) ⊆ (V (G) − V (S)) ∪ Z 1 ∪ Z 2 . Then there exists a (Z 1 , Z 2 ; S C )-path R such that V (R) ⊆ V (R1 ) ∪ V (R2 ), especially, if the length of Ri is at most m for i = 1, 2, then x ∈ Domi m (a R ) ∩ Domi m (b R ). ( Claim 13. Let d, e ∈ D0 with d ̸= e. Then Domi m (X d ) ∪ Domi m (X e ) ⊆ Domi m V (S) ∪ V (Q d ) ∪ V (Q e ) − ) V (Sd [d + , d ∗ )) − V (Se [e+ , e∗ )) . ( Proof. Let F) := {z ∈ Domi m (X d ) ∪ Domi m (X e ) : z ̸∈ Domi m V (S) ∪ V (Q d ) ∪ V (Q e ) − V (Sd [d + , d ∗ )) − V (Se [e+ , e∗ )) }. Suppose that F ̸= ∅, and let x ∈ F. By the symmetry of d and e, we may assume that x ∈ Domi m (X d ). By the choice (D1) and the definition of F, x ∈ Domi m (V (Se [e+ , e∗ ))). Hence there exists a path R1 of length at most m which connects x and a vertex of V (Se [e+ , e∗ )). On the other hand, since x ∈ Domi m (X d ), there exists a path R2 of length at most m which connects x and (a vertex of V (Sd)[d +(, d ∗ )). Since the length of + ∗ R1 and R2 are ) at most (m, it follow from ) the ( definition of )F that V (R1 ) ∪ V (R2 ) ∩ V (S) − V (Sd [d , d )) − + ∗ V (Se [e , e )) = ∅ and V (R1 ) ∪ V (R2 ) ∩ V (Q d ) ∪ V (Q e ) = ∅. Hence, by Claim 12, for any x ∈ F, there exists a (V (Sd [d + , d ∗ )), V (Se [e+ , e∗ )); S C )-path Px such that x ∈ Domi m (a Px ) and V (Px ) ∩ (V (Q d ) ∪ V (Q e )) (= ∅. Choose x ∈ F so that |Sd [d, a P)x ]| is as small as possible. If Sd = Se or {Sd , Se } ∩ CS ̸= ∅, then let S ′ := (Sd ∪ Se ) − (Sd [d, a Px ] ∪ Sd [e, b Px ]) ∪ (Q d ∪ Q e ∪ Px ) and S ′ := S − {Sd , Se } ∪ {S ′ } (note that S ′ is a path if {Sd , Se } ∩ PS ̸= ∅; ← − otherwise S ′ is a cycle). If Sd ̸= Se and Sd , Se ∈ PS , then let S ′ := Sd [b Sd , a Px ]Px [a Px , b Px ]Se [b Px , b Se ], ← − ← − T ′ := Sd [a Sd , d]Q d [d, v] Q e [v, e] Se [e, a Se ] and S ′ := S − {Sd , Se } ∪ {S ′ , T ′ } (note that S ′ and T ′ are paths, and V (S ′ ) and V (T ′ ) are disjoint). In any cases, S ′ is a k-ended system, and by the choice of x, we can see F ⊆ Domi m (S ′ ), and hence S ′ satisfies (1), a contradiction. □ In the next two claims, we investigate the distance of distinct two vertices in D ∗ ∪ {v}. Claim 14. dG (x, v) ≥ 2(m + 1) for any x ∈ D ∗ . Proof. Let x ∈ D ∗ and take d ∈ D so that xd = x. Suppose that dG (x, v) < 2(m + 1). Let Q be a shortest path which connects x and v. By Claim 6, (V (Q) − {x}) ∩ V (S) ⊆ X d . Therefore, it follows from 11 and 12 (apply Claim 12 as Z 1 = V (Q d ) − {d}, Z 2 = X d , R1 = Q and R2 = Id ), that there exists a (V (Q d ) − {d}, X d ; S C )path R. Since a R ∈ V (Q d ), it follows from Claim 11 that a R ̸∈ V (Id ), and hence by Claim 12, we can see that a R ∈ V (Q). Note that ∥Q[x, a R ]∥ ≥ dG (x, a R ) ≥ dG (x, V (Q d )) ≥ m + 1 by Claim 2. If v ̸∈ Domi m (a R ), then dG (x, v) = ∥Q∥ ≥ ∥Q[x, a R ]∥ + ∥Q[a R , v]∥ ≥ m( + 1 + dG (a R , v) ≥ 2(m + 1), a contradiction. Hence ) v ∈ Domi m (a R ). Let F := {z ∈ Domi m (X d ) : z ̸∈ Domi m (V (S) − V (Sd [d + , b R ))) ∪ V (Q d [d, a R ]) }. Suppose that F = ∅. Let S ′ := (Sd − Sd [d, b R ])∪ Q d [d, a R ]R[a R , b R ] and S ′ := S −{Sd }∪{S ′ } (note that f (Sd ) = f (S ′ )). Then S ′ is a k-ended system. Since v ∈ Domi m (a R ) and F = ∅, we can see that S ′ satisfies (1), a contradiction. Hence F ̸= ∅. Let f ∈ F. By the choice (D1), f ∈ Domi m (V (Q d (a R , v])). Let R1 be a shortest path which connects f and a vertex of V (Q d (a R , v]). On the other hand, since f ∈ Domi m (X d ), there exists a path R2 of length at most m which connects f and a vertex of V (Sd [d + , b R )). Since the length of R1 and R2 are at most m, it follows from the definition of F that Please cite this article in press as: M. Tsugaki, G. Yan, m-dominating k-ended trees of l-connected graphs, AKCE International Journal of Graphs and Combinatorics (2017), http://dx.doi.org/10.1016/j.akcej.2017.04.004.
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V (R1 ) ∪ V (R2 ) ⊆ V (G) − V (S) − V (Q d ) ∪ V (Q d (a R , v]) ∪ V (Sd [d + , b R )). Therefore, by Claim 12, for any f ∈ F, there exists a (V (Q d (a R , v]), V (Sd [d + , b R )); S C )-path R f such that V (Q d ) ∩ V (R f ) = {a R f } and f ∈ Domi m (a R f ). Choose f ∈ F so that |Q d [a R f , v]| is as small as possible. Let S ′ := (Sd − Sd [d, b R f ]) ∪ Q d [d, a R f ]R f [a R f , b R f ] and S ′ := S − {Sd } ∪ {S ′ } (note that f (Sd ) = f (S ′ )). Then S ′ is a k-ended system. Since a R ∈ V (S ′ ), we can see v ∈ Domi m (S ′ ). By the choice of f , we can see F ⊆ Domi m (S ′ ), and hence S ′ satisfies (1), a contradiction. □ (
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Claim 15. dG (x, y) ≥ 2(m + 1) for any x, y ∈ D ∗ with x ̸= y. Proof. Let x, y ∈ D ∗ with x ̸= y. Take d, e ∈ D so that x = xd and y = xe . Suppose that dG (x, y) < 2(m + 1). Let Q be a shortest path which connects x and y. By Claim 6, (V (Q) − {x, y}) ∩ V (S) ⊆ X d ∪ X e . Suppose that C C V (Q) ( ∩ (V (Q d ) ∪ )V (Q( e )) = ∅. By Claim ) 11, we can see that Id is an (x, X d ; S )-path, Ie is an (y, X eC; S )-path′ and V (Id ) ∪ V (Ie ) ∩ V (Q d ) ∪ V (Q e ) = ∅. Hence, by using Q, Id and Ie , we can obtain ( an (X d , X e ; S )-path Q such that V)(Q ′ ) ∩ (V (Q d ) ∪ V (Q e )) = ∅. If Sd = Se or {Sd , Se } ∩ CS ̸= ∅, then let S ′ := (Sd ∪ Se ) − (Sd [d, a Q ′ ] ∪ Sd [e, b Q ′ ]) ∪ (Q d ∪ Q e ∪ Q ′ ) and S ′ := S − {Sd , Se } ∪ {S ′ } (note that S ′ is a path or a cycle). If Sd ̸= Se and ← − ← − ← − Sd , Se ∈ PS , then let S ′ := Sd [a Sd , d]Q d [d, v] Q e [v, e] Se [e, a Se ], T ′ := Sd [b Sd , a Q ′ ]Q ′ [a Q ′ , b Q ′ ]Se [b Q ′ , b Se ] and ′ ′ ′ ′ ′ ′ S := S − {Sd , Se } ∪ {S , T } (note that S and T are paths, and V (S ) and V (T ′ ) are disjoint). In any cases, S ′ is a k-ended system, and by Claim 13, we can see S ′ satisfies (1), a contradiction. Hence V (Q) ∩ (V (Q d ) ∪ V (Q e )) ̸= ∅. Choose qx , q y ∈ V (Q) ∩ (V (Q d ) ∪ V (Q e )) so that |Q[x, qx ]| and |Q[q y , y]| are as small as possible. Suppose that qx ∈ V (Q e ). If V (Q[x, qx ]) ∩ X e ̸= ∅, then, by using Id , we can obtain an (X d , X e ; S C )-path Q ′ such that V (Q ′ ) ∩ (V (Q d ) ∪ V (Q e )) = ∅, and can obtain a contradiction as in the case V (Q) ∩ (V (Q d ) ∪ V (Q e )) = ∅. Hence V (Q[x, qx ]) ∩ X e = ∅. Then we can apply Claim 12 as Z 1 = V (Q e ) − {e}, Z 2 = X d , R1 = Q[x, qx ] and R2 = Id , and obtain (V (Q( e ) − {e}, X d ; S C )-path ) R such that V (R) ⊆ V (Q[x, qx ]) ∪ V (Id ). By Claim 11 and the minimality of |Q[x, qx ]|, V (Q[x, qx ]) ∪ V (Id ) ∩ (V (Q d ) − {v}) = ∅, that is V (R) ∩ (V (Q d ) − {v}) = ∅. Then, by using Q d , Q e [a R , v] and R, we can obtain a (d, X d ; S C )-path Q ′ such that V (Q d ) ⊆ V (Q ′ ), which contradicts to Claim 10. Hence, qx ∈ V (Q d ). By the symmetry of x and y, we can also obtain q y ∈ V (Q e ). Then, by Claim 2, dG (x, y) = ∥Q∥ ≥ dG (x, qx ) + dG (q y , y) ≥ 2(m + 1), a contradiction. □ Now, we extend the definition of exceptional. Recall that we regard a vertex as a cycle. For P ∈ PS such that V (P) ∩ D ̸= ∅ and d P ∈ D, if a pair xd P and β P is not good, then we also call d P is exceptional. Let PS′ := {P ∈ PS : V (P) ∩ D0 ̸= ∅ and d P is exceptional} and CS′ := {C ∈ CS : V (C) ∩ D0 ̸= ∅}. For any ← − P ∈ PS′ , there exists an (X d P , b P ; S C )-path Q P , and let C P := Q P [a Q P , b P ] P [b P , a Q P ] (note that if d P ∗ = b P , then Q P = C P = b P ). By Claim 9 (i), we can obtain V (C P ) ∩ V (C Q ) = ∅ for any P, Q ∈ PS′ with P ̸= Q, and also obtain V (C P ) ∩ V (C) = ∅ for any P ∈ PS′ and C ∈ CS . By Claim 9 (ii), we can obtain V (C P ) ∩ V (Q i ) = ∅ for any P ∈ PS′ and for any 1 ≤ i ≤ l. Claim 16. |PS′ ∪ CS′ | ≤ 1. Proof. If there exist C1 , C2 ∈ CS′ with C1 ̸= C2 , then we may assume that Q 1 and Q 2 are (V (C1 ), v; S C )← − − ′ path and (V (C2 ), v; S C )-path, respectively, and let S ′ := C[a − Q 1 , a Q 1 ]Q 1 [a Q 1 , v] Q 2 [v, a Q 2 ]C[a Q 2 , a Q 2 ] and S := ′ ′ ′ ′ S − {C1 , C2 } ∪ {S } (note that f ({C1 , C2 }) = f ({S })). If there exist P1 , P2 ∈ PS with P1 ̸= P2 , then let S := ←−− ← − P1 [a P1 , d P1 ]Q d P1 [d P1 , v] Q d P2 [v, d P2 ] P2 [d P2 , a P2 ] and S ′ := S − {P1 , P2 } ∪ {S ′ , C P1 , C P2 } (note that V (S ′ ), V (C P1 ) and V (C P2 ) are disjoint with each other, and f ({P1 , P2 }) = f ({S ′ , C P1 , C P2 }) = 4). If there exist P ∈ PS′ and C ∈ CS′ , ← − then we may assume that Q 1 is a (V (C), v; S C )-path, and let S ′ := P[a P , d P ]Q d P [d P , v] Q 1 [v, a Q 1 ]C[a Q 1 , a − Q 1 ] and S ′ := S − {P, C} ∪ {S ′ , C P } (note that V (S ′ ) and V (C P ) are disjoint, and f ({P, C}) = f ({S ′ , C P }) = 3). In any cases, S ′ is a k-ended system, and by Claim 13, S ′ satisfies (1), a contradiction. □ Note that either CS′ = ∅ or PS′ = ∅ holds by Claim 16. Claim 17. If CS′ = ∅, then a pair x and v is good for any x ∈ ΓCS . Proof. Suppose that CS′ = ∅, and a pair x and v is not good for some x ∈ ΓCS . Then there exists a (R(x), v; S C )-path Q. Since CS′ = ∅, it follows from the choice (S4), that there exist 1 ≤ i ≤ l and z ∈ (V (Q i ) ∩ V (Q)) − {v}. Choose i and z so that |Q[a Q , z]| is as small as possible. Then Q ′ := Q[a Q , z]Q i [z, v] is a (V (CS ), v; S C )-path such that V (Q ′ ) ∩ V (Q j ) = {v} for any 1 ≤ j ≤ l with j ̸= i. Since CS′ = ∅, this contradicts to the choice (S4). □ Please cite this article in press as: M. Tsugaki, G. Yan, m-dominating k-ended trees of l-connected graphs, AKCE International Journal of Graphs and Combinatorics (2017), http://dx.doi.org/10.1016/j.akcej.2017.04.004.
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Table 1 ′ = ∅. Where we have already investigated or not in the case PS′ ∪ CS A PS ΓCS D∗
A PS
ΓCS
D∗
v
Claims 8, 9(i) / /
Claims 8, 9(i) Claims 8, 9(iv) /
Claims 8, 9(vi), (vii), 18 Not yet Claim 15
Claims 8, 9(ii) Claims 8, 17 Claim 14
By the definition of PS′ , we can obtain the following claim. Claim 18. (i) If PS′ ∪ CS′ = ∅, then D = D0 . (ii) If PS′ = ∅, then a pair xd P and β P is good for any P ∈ PS with V (P) ∩ D0 ̸= ∅. By Claim 16, either “Case 1. PS′ ∪ CS′ = ∅”, or “Case 2. |PS′ | = 1 and CS′ = ∅”, or “Case 3. |CS′ | = 1 and PS′ = ∅” holds. We divide the proof into these three cases, and continue to investigate the distance of distinct two vertices in APS ∪ ΓCS ∪ D ∗ ∪ {v}. For the convenience of the reader, at the top of each cases, we put a table. These tables show which cases we have already finished to investigate by which claims, and we write “not yet” where we have not finish to investigate yet. Case 1. PS′ ∪ CS′ = ∅. As mentioned above, Table 1 shows where we have already investigated or not. For example, by Claims 8 and 9(i), we can see that dG (x, y) ≥ 2(m + 1) for any x, y ∈ APS with x ̸= y. In this case, at first, we investigate the relation between ΓCS and D ∗ . Claim 19. There exists z 0 ∈ ΓCS ∪ D ∗ such that a pair x and y is good for any x ∈ ΓCS − {z 0 } and y ∈ D ∗ − {z 0 }. Proof. Let D1 := {d ∈ D0 : a pair xd and x is not good for some x ∈ ΓCS } and C1 := {C ∈ CS : a pair γC and x is not good for some x ∈ D ∗ }. To prove this claim, we have only to prove that either |D1 | ≤ 1 or |C1 | ≤ 1 holds. Suppose that |D1 | ≥ 2 and |C1 | ≥ 2. Then, by the definitions of D1 and C1 , there exist d1 , d2 ∈ D1 and C1 , C2 ∈ C1 with d1 ̸= d2 and C1 ̸= C2 such that a pair xd1 and γC1 is not good, and a pair xd2 and γC2 is not good. Hence, there exist (X d1 , V (C1 ); S C )-path P1 and (X d2 , V((C2 ); S C )-path P ) 2 . By ( Claim 9 (iv), we) can see V (P1 ) ∩ V (P2 ) = ∅. Since CS′ = ∅, it follows from Claim 17 that V (P1 ) ∪ V (P2 ) ∩ V (Q d1 ) ∪ V (Q d2 ) = ∅. If ←− ← − ← − Sd1 ̸= Sd2 , then let S ′ := Sd1 [a Sd1 , d1 ]Q d1 [d1 , v] Q d2 [v, d2 ] Sd2 [d2 , a Sd2 ], T ′ := Sd1 [b Sd1 , a P1 ]P1 [a P1 , b P1 ]C1 [b P1 , b− P1 ], ← − − ′ ′ ′ ′ ′ ′ U := Sd2 [b Sd2 , a P2 ]P2 [a P2 , b P2 ]C2 [b P2 , b P2 ] and S := S − {Sd1 , Sd2 , C1 , C2 } ∪ {S , T , U } (note that V (S ), V (T ′ ) and V (U ′ ) are disjoint with each other, and f ({Sd1 , Sd2 , C1 , C2 }) = f ({S ′ , T ′ , U ′ }) = 6). If Sd1 = Sd2 (by the symmetry of d1 and d2 , we may assume that d1 and d2 are arranged in this order along Sd1 ), then let S ′ := ←− ← − ← − − ′ Sd1 [a Sd1 , d1 ]Q d1 [d1 , v] Q d2 [v, d2 ] Sd1 [d2 , a P1 ]P1 [a P1 , b P1 ]C1 [b P1 , b− P1 ], T := Sd1 [b Sd1 , a P2 ]P2 [a P2 , b P2 ]C 2 [b P2 , b P2 ] and S ′ := S −{Sd1 , C1 , C2 }∪{S ′ , T ′ } (note that V (S ′ ) and V (T ′ ) are disjoint, and f ({Sd1 , C1 , C2 }) = f ({S ′ , T ′ }) = 4). In any cases, S ′ is a k-ended system, and by Claim 13, S ′ satisfies (1), a contradiction. □ By Claims 8, 9, 14, 15 and 17–19 (see Table 1), we can see that APS ∪ (ΓCS ∪ D ∗ − {z 0 }) ∪ {v} is a 2(m + 1)-stable set, and hence, k + l − 1 ≥ α 2(m+1) (G) ≥ |APS | + (|ΓCS | + |D ∗ | − 1) + |{v}| = 2|PS | + |CS | + |D0 | = k + l, a contradiction. Case 2. |PS′ | = 1 and CS′ = ∅. Let {P0 } := PS′ and D ∗∗ := {xd : d ∈ D0 − {d P0 }}. In this case, at first, we investigate the relation between ΓCS and D ∗∗ . Claim 20. A pair x and y is good for any x ∈ ΓCS and y ∈ D ∗∗ . Proof. Suppose not. Then there exist d ∈ D0 − {d P0 }, C ∈ CS and (X d , V (C); S C )-path Q. By Claim 9(i) and Claim 17, we can see V (Q)∩V (Q P0 ) = ∅ and V (Q)∩(V (Q d )∪V (Q d P0 )) = ∅, respectively. If Sd ̸= P0 , then let S ′ := ← − ←−− ← − ′ ′ Sd [b Sd , a Q ]Q[a Q , b Q ]C[b Q , b− Q ], T := Sd [a Sd , d]Q d [d, v] Q d P0 [v, d P0 ] P0 [d P0 , a P0 ] and S := S − {Sd , P0 , C} ∪ Please cite this article in press as: M. Tsugaki, G. Yan, m-dominating k-ended trees of l-connected graphs, AKCE International Journal of Graphs and Combinatorics (2017), http://dx.doi.org/10.1016/j.akcej.2017.04.004.
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Table 2 ′ = ∅. Where we have already investigated or not in the case |PS′ | = 1 and CS APS ΓCS D ∗∗
A PS
ΓCS
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v
Claims 8, 9(i) / /
Claims 8, 9(i) Claims 8, 9(iv) /
Claims 8, 9(vi), (vii) Not yet Claim 15
Claims 8, 9(ii) Claims 8, 17 Claim 14
Table 3 ′ | = 1 and P ′ = ∅. Where we have already investigated or not in the case |CS S APS ΓCS D∗
A PS
ΓCS
D∗
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Claims 8, 9(i) / /
Claims 8, 9(i) Claims 8, 9(iv) /
Claims 8, 9(iii), (vi), (vii), 18 Claims 8, 9(v), not yet Claim 15
Claims 8, 9(ii) Not yet Claim 14
{C P0 , S ′ , T ′ } (note that V (C P0 ), V (S ′ ) and V (T ′ ) are disjoint with each other, and f ({Sd , P0 , C}) = f ({C P0 , S ′ , T ′ }) = ← − ← − ′ 5). If Sd = P0 , then let S ′ := Sd [a Sd , d]Q d [d, v] Q d P [v, d P0 ] Sd [d P0 , a Q ] Q[a Q , b Q ]C[b Q , b− Q ] and S := S −{Sd , C} 0 ′ ′ ′ ∪{C P0 , S } (note that V (C P0 ) and V (S ) are disjoint, and f ({Sd , C}) = f ({C P0 , S }) = 3). In any cases, S ′ is a k-ended system, and by Claim 13, S ′ satisfies (1), a contradiction. □ By Claims 8, 9, 14, 15, 17 and 20 (see Table 2), we can see that APS ∪ ΓCS ∪ D ∗∗ ∪ {v} is a 2(m + 1)-stable set, and hence, k + l − 1 ≥ α 2(m+1) (G) ≥ |APS | + |ΓCS | + |D ∗∗ | + |{v}| = 2|PS | + |CS | + (|D0 | − 1) + 1 = k + l, a contradiction. Case 3. |CS′ | = 1 and PS′ = ∅. Let {C0 } := CS′ . We may assume that Q 1 is a (V (C0 ), v; S C )-path. In this case, at first, we investigate the relation between ΓCS and D ∗ ∪ {v}. Claim 21. A pair xd and y is good for any d ∈ V (PS ) ∩ D0 and y ∈ ΓCS . Proof. Suppose not. Then there exist d ∈ V (PS ) ∩ D0 , C1 ∈ CS − {C0 } and (X d , V (C1 ); S C )-path Q. ⋃ If V (Q) ∩ 1≤i≤l V (Q i ) ̸= ∅, then there exists a (V (C0 ), V (C1 ); S C )-path, which contradicts to the choice ⋃ ← − (S2). Hence V (Q) ∩ 1≤i≤l V (Q i ) = ∅. Let S ′ := Sd [a Sd , d]Q d [d, v] Q 1 [v, a Q 1 ]C0 [a Q 1 , a Q 1 − ], T ′ := ← − − ′ ′ ′ ′ ′ Sd [b Sd , a⋃ Q ]Q[a Q , b Q ]C 1 [b Q , b Q ] and S := S − {Sd , C 0 , C 1 } ∪ {S , T }. Note that V (S ) ∩ V (T ) = ∅ because V (Q) ∩ 1≤i≤l V (Q i ) = ∅, and note that f ({Sd , C0 , C1 }) = f ({S ′ , T ′ }) = 4. Hence S ′ is a k-ended system which satisfies (1), a contradiction. □ Claim 22. A pair v and y is good for any y ∈ ΓCS . Proof. Suppose not. Then there exist C1 ∈ CS − {C0 } and a (v, V (C1 ); S C )-path Q. Since V (Q 1 ) ∪ V (Q) ⊆ (V (G) − V (S)) ∪ V (C0 ) ∪ V (C1 ), it follows from Claim 12 that there exists a (V (C0 ), V (C1 ); S C )-path, which contradicts to the choice (S2). □ By Claims 8, 9, 14, 15, 18, 21 and 22 (see Table 3), we can see that APS ∪ ΓCS ∪ D ∗ ∪ {v} is a 2(m + 1)-stable set, and hence, k + l − 1 ≥ α 2(m+1) (G) ≥ |APS | + |ΓCS | + |D ∗ | + |{v}| = 2|PS | + (|CS | − 1) + |D ∗ | + 1 = k + |D ∗ |, that is |D ∗ | ≤ l − 1. Since PS(′ = ∅, this implies that V (C ) 0 ) ∩ D0 has an exceptional vertex. Then, by the ( choice (D1), Domi m (C)0 ) ⊆ Domi m (V (S) − V (C0 )) ∪ V (Q 1 ) . Let F := {z ∈ Domi m (C0 ) : z ̸∈ Domi m (V (S) − V (C0 )) ∪ {a Q 1 } }. Suppose that F = ∅. Let S ′ := S − {C0 } ∪ {Q 1 [a Q 1 , a Q 1 + ]} and Q 1 ′ := Q 1 − a Q 1 . Since F = ∅, S ′ is a k-ended system such that |Domi m (S ′ )| ≥ |Domi m (S)|, |PS ′ | = |PS |, |V (PS ′ )| = |V (PS )|. Moreover, Q 1 ′ , Q 2 , . . . , Q l are (V (S ′ ), v; S ′ C )-paths such that they are disjoint with each other without v, and Q 1 ′ is a (V (CS ′ ), v; S ′ C )-path. But, |Q 1 ′ | < |Q 1 |. These contradict to the choices (S1) or (S5). Hence F ̸= ∅. Let f ∈ F. Since f ∈ Domi m (C0 ) and a Q 1 ̸∈ Domi m ( f ), there exists a path R1 of length at most m which connects f and a vertex of V (C0 ) − {a Q 1 }. On the other hand, by the definition of F, there exists a path R2 of length at most m which connects f and a vertex of V (Q 1 ) − {a Q 1 }. Since the length of R1 and R2 are at most m, it follows from the definition Please cite this article in press as: M. Tsugaki, G. Yan, m-dominating k-ended trees of l-connected graphs, AKCE International Journal of Graphs and Combinatorics (2017), http://dx.doi.org/10.1016/j.akcej.2017.04.004.
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of F that V (R1 ) ∪ V (R2 ) ⊆ (V (G) − V (S)) ∪ (V (C0 ) − {a Q 1 }) ∪ (V (Q 1 ) − {a Q 1 }). Hence, by Claim 12, there exists a m (V (C0 ) − {a Q 1 }, V (Q 1 ) − {a Q 1 }; S C )-path ⋃ R f such that f ∈ Domi (b R f ). Choose f ∈ F so that |QC1 [b R f , v]| is as small as possible. Suppose that V (R f ) ∩ 2≤i≤l V (Q i ) ̸= ∅. Then there exists a (V (C0 ) − {a Q 1 }, v; S )-path R such ← − that V (Q 1 ) ∩ V (R) = {v}. Let S ′ := Q 1 [a Q 1 , v] R [v, a R ]C0 [a R , a Q 1 ]( and S ′ := S − {C0 } ∪ {S ′ }.) Since S ′ is a cycle, we can see that S ′ is a k-ended system. Since Domi m (C0 ) ⊆ Domi m (V (S) − V (C0 )) ∪ V (Q 1 ) , we can see that S ′ ⋃ ← − satisfies (1), a contradiction. Hence V (R f ) ∩ 2≤i≤l V (Q i ) = ∅. Let S ′ := Q 1 [a Q 1 , b R f ] R [b R f , a R f ]C0 [a R f , a Q 1 ], ′ ′ ′ ′ S := S − {C0 } ∪ {S } and Q 1 := Q 1 [b R f , v]. Note that S is a cycle. By the choice of f , S ′ is a k-ended system such that |Domi m (S ′ )| ≥ |Domi m (S)|, |PS ′ | = |PS |, |V (PS ′ )| = |V (PS )|. Note that Q 1 ′ , Q 2 , . . . , Q l are (V (S ′ ), v; S ′ C )paths such that they are disjoint with each other without v, and Q 1 ′ is a (V (CS ′ ), v; S ′ C )-path such that |Q 1 ′ | < |Q 1 |. These contradict to the choices (S1) or (S5), and this completes the proof of Theorem 7. Acknowledgments The authors would like to thank the referees for their valuable comments. This work was done while the author was a foreign researcher of Chinese Academy of Science, Supported by Chinese Academy of Science Fellowship for Young International Scientists. Grant No. 2012Y1JA0004. References [1] M. Kano, M. Tsugaki, G. Yan, m-dominating k-ended trees of graphs, Discrete Math. 333 (2014) 1–5. [2] H.J. Broersma, Existence of ∆k -cycles and ∆k -paths, J. of Graph Theory 12 (4) (1988) 499–507. [3] S. Win, On a conjecture of Las Vergnas concerning certain spanning trees in graphs, Resultate Math. 2 (1979) 215–224.
Please cite this article in press as: M. Tsugaki, G. Yan, m-dominating k-ended trees of l-connected graphs, AKCE International Journal of Graphs and Combinatorics (2017), http://dx.doi.org/10.1016/j.akcej.2017.04.004.