Applied Mathematics Letters 24 (2011) 908–910
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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).
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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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