ISSN 1990-4789, Journal of Applied and Industrial Mathematics, 2016, Vol. 10, No. 1, pp. 1–6. c  Pleiades Publishing, Ltd., 2016. Original Russian Text c  V.E. Alekseev, D.V. Zakharova, 2016, published in Diskretnyi Analiz i Issledovanie Operatsii, 2016, Vol. 23, No. 1, pp. 5–14. Independent Sets in Graphs without Subtrees with Many Leaves V. E. Alekseev * and D. V. Zakharova ** Lobachevsky State University of Nizhny Novgorod, pr. Gagarina 23, korp. 2, Nizhny Novgorod, 603950 Russia Received June 11, 2015; in final form, June 25, 2015 Abstract—A subtree of a graph is called inscribed if no three vertices of the subtree generate a triangle in the graph. We prove that, for fixed k, the independent set problem is solvable in polynomial time for each of the following classes of graphs: (1) graphs without subtrees with k leaves, (2) subcubic graphs without inscribed subtrees with k leaves, and (3) graphs with degree not exceeding k and lacking induced subtrees with four leaves. DOI: 10.1134/S1990478916010014 Keywords: graph, independent set, forbidden subtree, polynomial algorithm INTRODUCTION By a graph class we mean a set of graphs closed under isomorphisms. Many classes of graphs are known for which the independent set problem (in general, NP-hard) is solved in polynomial time; we call these classes IS-simple. Many IS-simple classes are described by forbidden subgraphs; i.e., by subgraphs that must not exist in the graphs of the class under consideration. We use the common notation Free(X ) for the class of all graphs not containing induced subgraphs from X (these classes are closed under vertex removal and are called hereditary). If such a set X consists of a single graph G then we simply write Free(G). One of the early results is the proof of the IS-simplicity of the class Free(K 1,3 ) [9, 10]. But already for the class Free(K 1,4 ), the independent set problem remains NP-hard (this follows, for example, from the results of [2]). Therefore, a possible direction of the extension of the above-mentioned result about graphs without K 1,3 can be the consideration of other trees with three leaves as forbidden subgraphs. Let S i,j,k denote a tree with three leaves situated at distances i, j , and k from the vertex of degree 3. Thus, K 1,3 is S 1,1,1 . In [1], the first author proved the IS-simplicity of Free(S 1,1,2 ). Moving further in this direction is still possible only under additional constraints on the class of graphs. For example, in [3], the IS-simplicity of the class of planar graphs in Free(S 1,2,k ) is proved; and, in [8], the same is done for the class of planar subcubic graphs in Free(S 2,2,k ) for every fixed k. In this article, we proceed in another direction: We consider the classes defined by forbidden trees with boundedly many leaves. This direction is suggested by the results of [5, 12] which imply that the absence of subtrees with many leaves in a graph is a necessary condition for the boundedness of the minors of the extended incidence matrix of the graph. Therefore, the polynomial solvability of the independent set problem for these graphs implies the validity of one of the partial variants of Shevchenko’s Conjecture that a problem of integer linear programming is solvable in polynomial time if the moduli of the minors of the matrix are bounded [11]. Let FT (k) denote the class of all graphs lacking subtrees with k leaves, and let FT ∗ (k) designate the class of all graphs lacking induced subtrees with k leaves. In Section 2, we show that FT (k) is IS-simple for every fixed k. We do not know whether this holds in general case for FT ∗ (k). The case of k =2 is trivial, and for k =3 the class FT ∗ (3) coincides with Free(K 1,3 ), which is IS-simple as was * E-mail: aleve@rambler.ru ** E-mail: dvzakh@rambler.ru 1