Given a connected graph $G$ and a terminal set $R \subseteq V(G)$, {\em Steiner tree} asks for a tree that includes all of $R$ with at most $r$ edges for some integer $r \geq 0$. It is known from [ND12,Garey et. al \cite{steinernpc}] that Steiner tree is NP-complete in general graphs. {\em Split graph} is a graph which can be partitioned into a clique and an independent set. K. White et. al \cite{white} has established that Steiner tree in split graphs is NP-complete. In this paper, we present an interesting dichotomy: we show that Steiner tree on $K_{1,4}$-free split graphs is polynomial-time solvable, whereas, Steiner tree on $K_{1,5}$-free split graphs is NP-complete. We investigate $K_{1,4}$-free and $K_{1,3}$-free (also known as claw-free) split graphs from a structural perspective. Further, using our structural study, we present polynomial-time algorithms for Steiner tree in $K_{1,4}$-free and $K_{1,3}$-free split graphs. Although, polynomial-time solvability of $K_{1,3}$-free split graphs is implied from $K_{1,4}$-free split graphs, we wish to highlight our structural observations on $K_{1,3}$-free split graphs which may be used in other combinatorial problems.Comment: 12 pages, 2 figures, CALDAM 201

In this paper, we investigate the well-studied Hamiltonian cycle problem, and present an interesting dichotomy result on split graphs. T. Akiyama, T. Nishizeki, and N. Saito [22] have shown that the Hamiltonian cycle problem is NP-complete in planar bipartite graph with maximum degree 3. Using this reduction, we show that the Hamiltonian cycle problem is NP-complete in split graphs. In particular, we show that the problem is NP-complete in K1,5-free split graphs. Further, we present polynomial-time algorithms for Hamiltonian cycle in K1,3-free and K1,4-free split graphs. We believe that the structural results presented in this paper can be used to show similar dichotomy result for Hamiltonian path and other variants of Hamiltonian cycle problem. Lemma 1.[13] For a claw-free split graph G, if ∆ I = 2, then |I| ≤ 3.Observation 1 2-connected split graph G with ∆ I = 1 has a Hamiltonian cycle.Lemma 2. Let G be a K 1,3 -free split graph. G contains a Hamiltonian cycle if and only if G is 2-connected.Proof. Necessity is trivial. For the sufficiency, we consider the following cases. Case 1: |I| ≥ 4. As per Lemma 1, ∆ I = 1 and by Observation 1, G has a Hamiltonian cycle. Case 2: |I| ≤ 3. If ∆ I = 1, then by Observation 1, G has a Hamiltonian cycle. When ∆ I > 1, note that either |I| = 2 or |I| = 3. (a) |I| = 2, i.e., I = {s, t}. Since ∆ I = 2, there exists a vertex v ∈ K such that d I (v) = 2 and N I (v) = {s, t}. Let S = N (s) and T = N (t). Clearly, K = S ∪ T . Suppose the set S \ T is empty, then the vertices T ∪ {t} induces a clique, larger in size than K, contradicting the maximality of K. Therefore, the set S \

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In this paper we investigate Hamiltonian path problem in the context of split graphs, and produce a dichotomy result on the complexity of the problem. Our main result is a deep investigation of the structure of K1,4-free split graphs in the context of Hamiltonian path problem, and as a consequence, we obtain a polynomial-time algorithm to the Hamiltonian path problem in K1,4-free split graphs. We close this paper with the hardness result: we show that, unless P=NP, Hamiltonian path problem is NP-complete in K1,5-free split graphs by reducing from Hamiltonian cycle problem in K1,5-free split graphs. Thus this paper establishes a "thin complexity line" separating NP-complete instances and polynomial-time solvable instances.

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