Induced stars in trees

We show that for n ~ 2, every n-vertex tree has a star forest of order exceeding n /2. There are many theorems in graph theory in which a graph is required to have a particular subgraph. A stronger requirement, however, is that it be an induced subgraph: i.e. a subgraph in which two vertices must be adjacent if they are adjacent in the big graph. A common problem of Ramsey theory, for instance, is to show that for a particular graph H, any graph G having enough vertices must have as a subgraph a copy of H or its complement HC. Typically, the copy of H (or HC) will contain extraneous edges, and thus will not be an induced subgraph of G. It is trivial to partition the vertices of a tree into stars. (A star, KI,n is graph with a vertex of degree n ~ 1 and all its neighbors of degree one.) But this might not give an induced sub graph in which each component is a star. For instance consider the double-star Sm,n, the graph with a vertex of degree m adjacent to a vertex of degree n, and all its other vertices of degree one. The vertices of S3,3 can be partitioned into two disjoint copies of the star K I ,2. But this is not an induced subgraph. The largest induced subgraph consisting of disjoint stars is the single star K I ,3. We define a star-forest in a graph to be an induced subgraph in which each component is a star. In this note we show that within any tree we can find a large star-forest. Extending this result to graphs other than trees seems surprisingly resistant to easy solution. THEOREM. Every n-vertex tree has a star-forest of order strictly greater than n/2 (for n ~ 2). The double-stars Sn,n and Sn+l,n show this bound is best possible. We will refer to a vertex of degree one as an end-vertex, and any other vertex as an interior vertex. To prove the theorem, let T be a counter-example with as few vertices as possible, say n vertices. First we claim that every interior vertex of T must have at least one neighbor that is an end-vertex. Australasian Journal of Combinatorics 18(1998), pp.275-276 If not, let v be an interior vertex none of whose neighbors are end-vertices. Then each component of T v has at least two vertices, so by the minimality of T the theorem is true for each component. Say component Ti has ni vertices (i = 1 to p), and thus has a star forest with Si vertices where Si > nd2. Note that 2:f=l ni = n-1, and since Si is an integer we have Si ~ (ni + 1) /2. The union of the star-forests for the components gives a star-forest in T, as vertices in different components cannot be adjacent in T. But the number of vertices in the union is at least 2:f=l Si ~ 2:f=l (ni + 1)/2 ~ (2:f=l ni)/2+ 1 = (n-1)/2+ 1 > n/2. Thus the theorem is true for T, proving the claim that each interior vertex has at least one neighbor that is an end-vertex. Now, we can partition the vertices of T into disjoint stars by taking each interior vertex v to be the center of a star whose other vertices are the end-vertices adjacent to it. To get a star-forest, consider a two-coloring of the vertices of T. Let Ti be the forest induced by the stars whose centers are colored i (for i = 1 or 2). Since Tl and T2 partition the vertices of T, one of them has at least n/2 vertices. If each has exactly n/2 vertices, let v be an interior vertex whose neighbors include exactly one interior vertex. (I.e. v is an end-vertex in the subgraph of T obtained by deleting all its end-vertices.) Choose the forest Ti that does not include v, and thus does include the star centered at v', the one interior vertex adjacent to v. Then Ti + v is a star-forest with v added to the star centered at v' and thus having more than n/2 vertices, as desired. How does this problem generalize to graphs in general? Is there a positive constant c such that every n-vertex graph G (without isolated vertices) must have a star-forest with at least cn vertices? If so there would be constant c' such that G has an induced subgraph H with at least c'n vertices in which each vertex has odd degree. (This would be achieved by deleting one end vertex from each of the even stars.) But this is an old conjecture [2] which has received some attention recently with quite limited success. It has been proved for trees [3] and for graphs of maximum degree three [1]; for general graphs it is known only that the constant, if it exists, is no greater than 2/7.