TY - JOUR

T1 - Spanning trees with bounded total excess

AU - Enomoto, Hikoe

AU - Ohnishi, Yukichika

AU - Ota, Katsuhiro

PY - 2011/10/1

Y1 - 2011/10/1

N2 - Let c(H) denote the number of components of a graph H. Win proved in 1989 that if a connected graph G satisfies c(G\S)≤(k- 2)|S| + 2, for every subset S of V(G), then G has a spanning tree with maximum degree at most k. For a spanning tree T of a connected graph, the k-excess of a vertex v is defined to be max{0, degT(v) -k}. The total k-excess te(T, k) is the summation of the k-excesses of all vertices, namely, te(T, k) =Σ vεV(T)max{0. degT(v) - k}. This paper gives a sufficient condition for a graph to have a spanning tree with bounded total k-excess. Our main result is as follows. Suppose k ≥ 2, 6 ≥ 0, and G is a connected graph satisfying the following condition: For every subset S of V(G), c(G\S) ≤ (k-2)\S\ + 2 + b. Then, G has a spanning tree with total k-excess at most 6.

AB - Let c(H) denote the number of components of a graph H. Win proved in 1989 that if a connected graph G satisfies c(G\S)≤(k- 2)|S| + 2, for every subset S of V(G), then G has a spanning tree with maximum degree at most k. For a spanning tree T of a connected graph, the k-excess of a vertex v is defined to be max{0, degT(v) -k}. The total k-excess te(T, k) is the summation of the k-excesses of all vertices, namely, te(T, k) =Σ vεV(T)max{0. degT(v) - k}. This paper gives a sufficient condition for a graph to have a spanning tree with bounded total k-excess. Our main result is as follows. Suppose k ≥ 2, 6 ≥ 0, and G is a connected graph satisfying the following condition: For every subset S of V(G), c(G\S) ≤ (k-2)\S\ + 2 + b. Then, G has a spanning tree with total k-excess at most 6.

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M3 - Article

AN - SCOPUS:80053957609

VL - 102

SP - 289

EP - 295

JO - Ars Combinatoria

JF - Ars Combinatoria

SN - 0381-7032

ER -