Heavy cycles and spanning trees with few leaves in weighted graphs

Heavy cycles and spanning trees with few leaves in weighted graphs

Applied Mathematics Letters 24 (2011) 908–910 Contents lists available at ScienceDirect Applied Mathematics Letters journal homepage: www.elsevier.c...

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Applied Mathematics Letters 24 (2011) 908–910

Contents lists available at ScienceDirect

Applied Mathematics Letters journal homepage: www.elsevier.com/locate/aml

Heavy cycles and spanning trees with few leaves in weighted graphs✩ Binlong Li, Shenggui Zhang ∗ Department of Applied Mathematics, Northwestern Polytechnical University, Xi’an, Shaanxi 710072, PR China

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Article history: Received 12 June 2009 Received in revised form 31 December 2010 Accepted 31 December 2010 Keywords: Weighted graphs Heavy cycles Spanning trees

abstract Let G be a 2-connected weighted graph and k ≥ 2 an integer. In this note we prove that if the sum of the weighted degrees of every k + 1 pairwise nonadjacent vertices is at least m, then G contains either a cycle of weight at least 2m/(k + 1) or a spanning tree with no more than k leaves. © 2011 Elsevier Ltd. All rights reserved.

1. Introduction We use Bondy and Murty [1] for terminology and notation not defined here and consider finite simple graphs only. A weighted graph is a graph in which every edge e is assigned a nonnegative number w(e), called the weight of e. Let G be a weighted graph. For a subgraph H of G, V (H ) and E (H ) denote the sets of vertices and edges of H, respectively. The weight of H is defined by

w(H ) =



w(e).

e∈E (H )

For a vertex v ∈ V (G), NH (v) denotes the set, and dH (v) the number, of vertices in H that are adjacent to v . We define the weighted degree of v in H by dw H (v) =



w(v h).

h∈NH (v)

w When no confusion occurs, we will denote NG (v), dG (v) and dw G (v) by N (v), d(v) and d (v), respectively. An unweighted graph can be regarded as a weighted graph in which each edge e is assigned the weight w(e) = 1. Thus, in an unweighted graph, dw (v) = d(v) for every vertex v , and the weight of a subgraph is simply the number of its edges. Let G be a weighted graph. We use α(G) to denote the independence number of G. For a positive integer k, if k ≤ α(G), we define σk (G) to be the minimum value of the degree sum of any k pairwise nonadjacent vertices, and σkw (G) to be the minimum value of the weighted degree sum of any k pairwise nonadjacent vertices in G; if k > α(G), we define σk (G) and σkw (G) as ∞. Instead of σ1 (G) and σ1w (G), we use the notation δ(G) and δ w (G), respectively. A basic theorem of Dirac [2] shows the existence of long cycles in a graph under the minimum degree condition.

Theorem A (Dirac [2]). Let G be a 2-connected graph such that δ(G) ≥ d. Then G contains a cycle of length at least 2d or a Hamilton cycle.

✩ Supported by NSFC (No. 10871158).



Corresponding author. Tel.: +86 15929966707. E-mail address: [email protected] (S. Zhang).

0893-9659/$ – see front matter © 2011 Elsevier Ltd. All rights reserved. doi:10.1016/j.aml.2010.12.049

B. Li, S. Zhang / Applied Mathematics Letters 24 (2011) 908–910

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This result has been generalized to the following degree sum conditions. Theorem B (Pósa [3]). Let G be a 2-connected graph such that σ2 (G) ≥ m. Then G contains a cycle of length at least m or a Hamilton cycle. Theorem C (Fournier and Fraisse [4]). Let G be a k-connected graph such that σk+1 (G) ≥ m, where 2 ≤ k < α(G). Then G contains a cycle of length at least 2m/(k + 1) or a Hamilton cycle. Theorems A and B have been generalized to weighted graphs by the following two theorems, respectively. Theorem 1 (Bondy and Fan [5]). Let G be a 2-connected weighted graph such that δ w (G) ≥ d. Then G contains a cycle of weight at least 2d or every heaviest cycle in G is a Hamilton cycle. Theorem 2 (Bondy et al. [6]). Let G be a 2-connected weighted graph such that σ2w (G) ≥ m. Then G contains a cycle of weight at least m or a Hamilton cycle. A natural question is whether Theorem C also admits an analogous generalization for weighted graphs. This leads to the following problem. Problem 1 (Zhang et al. [7]). Let G be a k-connected weighted graph such that σkw+1 (G) ≥ m, where 2 ≤ k < α(G). Does G contain a cycle of weight at least 2m/(k + 1) or a Hamilton cycle? It seems very difficult to settle this problem, even for the case k = 2. Motivated by a result in [8], Zhang et al. [7] proved that the answer to Problem 1 in the case k = 2 is positive with the two same extra conditions as in [8]. Theorem 3 (Zhang et al. [7]). Let G be a 2-connected weighted graph which satisfies the following conditions: (1) σ3w (G) ≥ m; (2) w(xz ) = w(yz ) for every vertex z ∈ N (x) ∩ N (y) with d(x, y) = 2; (3) in every triangle T of G, either all edges of T have different weights or all edges of T have the same weight. Then G contains either a cycle of weight at least 2m/3 or a Hamilton cycle. After giving a characterization of the connected weighted graphs satisfying Conditions (2) and (3) of Theorem 3, Enomoto et al. [9] proved that the answer to Problem 1 is positive for any k ≥ 2 with these two extra conditions. For more results on the existence of heavy cycles under extra conditions, see [10–15]. In this note we prove a result on the existence of heavy cycles or spanning trees with few leaves without extra conditions. Theorem 4. Let G be a 2-connected weighted graph such that σkw+1 (G) ≥ m, where 2 ≤ k < α(G). Then G contains either a cycle of weight at least 2m/(k + 1) or a spanning tree with no more than k leaves. Note that if k = 2, then Theorem 4 says that G contains either a cycle of weight at least 2m/3 or a Hamilton path, instead of a Hamilton cycle as in Problem 1 for the case k = 2. This difference between the Hamilton cycle and Hamilton path may be helpful in either proving or constructing a counterexample to the conjecture. We postpone the proof of Theorem 4 to the next section. 2. Proof of Theorem 4 Our proof relies heavily on the following result due to Bondy et al. [6]. Lemma 1 (Bondy et al. [6]). Let G be a 2-connected weighted graph and P = v0 v1 v2 · · · vp a path in G. Define S1 = {vi |v0 vi ∈ E (G)} and S2 = {vi |vi−1 vp ∈ E (G)}. Assume each of the following conditions is satisfied. (a) NG (v0 ) \ V (P ) = NG (vp ) \ V (P ) = ∅; (b) S1 ∩ S2 = ∅ (in particular, v0 vp ̸∈ E (G)); (c) w(vi−1 vi ) ≥ w(v0 vi ) for all vi ∈ S1 , and w(vi−1 vi ) ≥ w(vi−1 vp ) for all vi ∈ S2 . Then there is a cycle C in G of weight w(C ) ≥ dw (v0 ) + dw (vp ). Before giving the proof of Theorem 4, we introduce some notation for trees. Let T be a tree. We use L(T ) to denote the set, and q(T ) the number, of the leaves of T . For two vertices u and v of T , we use PT [u, v] to denote the unique path connecting u and v in T . We use l(P ) to denote the length of a path P. Proof of Theorem 4. Suppose that every spanning tree of G has more than k leaves. Then we need only prove that G contains a cycle of weight at least k2m . +1 For a spanning tree T of G, in the following we use r (T ) to denote the number of leaves of T with weighted degree less m than k+ . Set q = min{q(T )|T is a spaning tree of G}. Then we have q ≥ k + 1. 1

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B. Li, S. Zhang / Applied Mathematics Letters 24 (2011) 908–910

Now, we choose a spanning tree T0 of G such that (1) q(T0 ) = q; (2) r (T0 ) is maximal, subject to (1). m m If r (T0 ) ≤ k − 1, let R0 = v ∈ L(T0 )|dw (v) < k+ ; otherwise let R0 be a subset of v ∈ L(T0 )|dw (v) < k+ with order 1 1 k − 1. Clearly we have q − |R0 | ≥ 2. 









Claim 1. Let T be a spanning tree of G with q leaves and R0 ⊂ L(T ), and x, y two vertices in L(T )\ R0 . Then dw (x)+ dw (y) ≥ k2m . +1 m Proof. Suppose that r (T0 ) ≤ k − 1. Then every vertex in L(T ) \ R0 has weighted degree at least k+ ; otherwise we have 1 r (T ) > r (T0 ), a contradiction. The assertion holds immediately. Suppose that r (T0 ) > k − 1. Then |R0 | = k − 1. Let u, v be two vertices in L(T ). Suppose that uv ∈ E (G). Let PT [u, v] = v0 v1 v2 · · · vp . Since T has q ≥ 3 leaves, there exists a vertex vi such that dT (vi ) ≥ 3, where 1 ≤ i ≤ p − 1. Then T ′ = T − vi−1 vi + v0 vp is a spanning tree of G with less than q leaves, a contradiction. So, we have uv ̸∈ E (G). Therefore,

dw (x) + dw (y) +



dw (v) ≥ m.

v∈R0 m Since every vertex in R0 has weighted degree less than k+ , we obtain that dw (x) + dw (y) ≥ k2m . 1 +1



For a spanning tree T with q leaves and R0 ⊂ L(T ), we choose two leaves u, v ∈ L(T ) \ R0 such that (3) PT [u, v] is as long as possible; (4) w(PT [u, v]) is as large as possible, subject to (3). We denote P (T ) = PT [u, v]. Now, we choose a spanning tree T ∗ of G with q leaves and R0 ⊂ L(T ∗ ) such that (5) P (T ∗ ) is as long as possible; (6) w(P (T ∗ )) is as large as possible, subject to (5). In the following we will prove that the path P (T ∗ ) satisfies the conditions of Lemma 1. Let P (T ∗ ) = P = v0 v1 v2 · · · vp , and S1 = {vi |v0 vi ∈ E (G)} and S2 = {vi |vi−1 vp ∈ E (G)}. (a) Suppose that NG (v0 ) \ V (P ) ̸= ∅. Let x0 be a vertex in V (G) \ V (P ) such that v0 x0 ∈ E (G), and x0 x1 x2 · · · xl be the path connecting x0 to P in T ∗ , where xl = vi . Set T ∗∗ = T ∗ − xl−1 xl + v0 x0 . If x0 ∈ L(T ∗ ) or dT ∗ (xl−1 ) ≥ 3, then T ∗∗ is a spanning tree with less than q leaves, a contradiction. So we suppose that x0 ̸∈ L(T ∗ ) and dT ∗ (xl−1 ) ≤ 2. Since xl−1 ̸∈ L(T ∗ ), we have dT ∗ (xl−1 ) = 2. Then T ∗∗ is a spanning tree with L(T ∗∗ ) = L(T ∗ ) \ {v0 } ∪ {xl−1 } and l(PT ∗∗ [xl−1 , vp ]) > l(PT ∗ [v0 , vp ]), again a contradiction. Similarly, we can prove that NG (vp ) \ V (P ) = Ø. (b) Suppose that S1 ∩ S2 ̸= ∅. Let vi be a vertex in S1 ∩ S2 . Since T ∗ has q ≥ 3 leaves, there exists a vertex vj such that dT ∗ (vj ) ≥ 3, where 1 ≤ j ≤ p − 1. If j ≤ i − 1, let T ∗∗ = T ∗ − {vi−1 vi , vj−1 vj } + {v1 vi , vi−1 vp }; if j ≥ i, let T ∗∗ = T ∗ − {vi−1 vi , vj vj+1 } + {v1 vi , vi−1 vp }. Then T ∗∗ is a spanning tree of G with less than q leaves, a contradiction. (c) Suppose that w(vi−1 vi ) ≤ w(v0 vi ), where vi ∈ S1 . Let T ∗∗ = T ∗ − vi−1 vi + v0 vi . Clearly vi−1 ̸∈ L(T ∗ ). If dT ∗ (vi−1 ) ≥ 3, then T ∗∗ is a spanning tree with q − 1 leaves, a contradiction. If dT ∗ (vi−1 ) = 2, then T ∗∗ is a spanning tree with L(T ∗∗ ) = L(T ∗ ) \ {v0 } ∪ {vi−1 }, l(PT ∗∗ [vi−1 , vl ]) = l(PT ∗ [v0 , vl ]) and w(PT ∗∗ [vi−1 , vl ]) > w(PT ∗ [v0 , vl ]), again a contradiction. Similarly, we can prove that w(vi−1 vi ) ≥ w(vi−1 vp ) for all vi ∈ S2 . So, by Lemma 1, G contains a cycle C such that w(C ) ≥ dw (v0 ) + dw (vp ) ≥ k2m . +1 References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15]

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