A note on max-leaves spanning tree problem in Halin graphs

A Halin graph H is a planar graph obtained by drawing a tree T in the plane, where T has no vertex of degree 2, then drawing a cycle C through all leaves in the plane. We write H = T ∪ C, where T is called the characteristic tree and C is called the accompanying cycle. The problem is to find a spanning tree with the maximum number of leaves in a Halin graph. In this paper, we prove that the characteristic tree in a Halin graph H is one of the spanning trees with maximum number of leaves in H, and we design a linear time algorithm to find such a tree in the Halin graph. All graphs in this paper are finite, undirected and simple. We follow the terminology and notation of [1] and [6]. A Halin graph H is a planar graph obtained by drawing a tree T in the plane, where T has no vertex of degree 2, then drawing a cycle through all leaves of T in the plane. We write H = T ∪ C, where T is called the characteristic tree and C is called the accompanying cycle. In this paper, we deal with the problem of finding a spanning tree with maximum number of leaves in a Halin graph. The problem of finding a spanning tree with maximum number of leaves for arbitrary graphs is NP-complete (see [3], Problem [ND 2]). When we confine an NP-complete problem to Halin graphs, we often get a polynomial time algorithm to solve it. (For example, see [2]). However, not every NP-complete problem confined to Halin graphs can be solved in polynomial time (see [5]). In this paper, we prove that the characteristic tree in a Halin graph H is one of the spanning trees with maximum number of leaves in H, and we design a linear time algorithm to find such a tree in the Halin graph. Theorem 1. The characteristic tree of a Halin graph H is a spanning tree with maximum number of leaves. 96 DINGJUN LOU AND HUIQUAN ZHU Proof. Let T ′ be a spanning tree with maximum number of leaves in H, V ′ 1 be the set of internal vertices of T ′ and V ′ 2 be the set of leaves of T ′. Assume that |V ′ 1 | = t′ and |V ′ 2 | = s′. Then we have s′ = ∑ v∈V ′ 1 (d(v)− 2) + 2. (1) Let T be the characteristic tree of H, V1 be the set of internal vertices of T and V2 be the set of leaves of T . Assume that |V1| = t and |V2| = s. By the same reason as above, we have s = ∑ v∈V1 (d(v)− 2) + 2. (2) Let V11 = V1 ∩ V ′ 1 , V12 = V1\V ′ 1 and V ′ 12 = V ′ 1\V1 ⊆ V2. Since T ′ is one of the spanning trees with maximum number of leaves of H, so s′ ≥ s. However, s+ t = s′ + t′ = ν. So t ≥ t′. This implies |V ′ 12| ≤ |V12|. Now, ∀v ∈ V12 and ∀x ∈ V ′ 12, dH(v) ≥ 3 = dH(x). Therefore, s = ∑ v∈V1 (d(v)− 2) + 2 = ∑ v∈V11 (d(v)− 2) + ∑ v∈V12 (d(v)− 2) + 2 ≥ ∑ v∈V11 (d(v)− 2) + ∑ v∈V ′ 12 (d(v)− 2) + 2 = ∑ v∈V ′ 1 (d(v)− 2) + 2 = s′. Since s′ ≥ s, we have s′ = s. Hence the characteristic tree T is one of the spanning trees with maximum number of leaves. By the result of Theorem 1, we can find a spanning tree with maximum number of leaves using the following algorithm: 1. Use the algorithm of Hopcroft and Tarjan [4] to find a planar embedding H̃ of the Halin graph H. 2. Traverse each face C in H̃, if T = H−E(C) is a tree, then T is the characteristic tree and C is the accompanying cycle. Note that in Step 2, we decide whether T = H − E(C) is a tree in the following way. For each face C, if ε(H) − ε(C) = ν(H) − 1, then T = H − E(C) is a tree as H − E(C) is connected for each face in a Halin graph. The correctness of the algorithm is shown in [2]. The time complexity of the above algorithm is bounded by O(ν), where ν = |V (H)|. By [4], Step 1 takes O(ν) time. Step 2 takes O(ε) time. However, in a Halin graph H, ε(H) = ε(T ) + ε(C) ≤ 2(ν − 1) = O(ν). Hence Step 2 also takes O(ν) time. MAX-LEAVES SPANNING TREE PROBLEM IN HALIN GRAPHS 97