Bounds on the Game Transversal Number in Hypergraphs (1601.04856v1)

Abstract: Let $H = (V,E)$ be a hypergraph with vertex set $V$ and edge set $E$ of order $\nH = |V|$ and size $\mH = |E|$. A transversal in $H$ is a subset of vertices in $H$ that has a nonempty intersection with every edge of $H$. A vertex hits an edge if it belongs to that edge. The transversal game played on $H$ involves of two players, \emph{Edge-hitter} and \emph{Staller}, who take turns choosing a vertex from $H$. Each vertex chosen must hit at least one edge not hit by the vertices previously chosen. The game ends when the set of vertices chosen becomes a transversal in $H$. Edge-hitter wishes to minimize the number of vertices chosen in the game, while Staller wishes to maximize it. The \emph{game transversal number}, $\tau_g(H)$, of $H$ is the number of vertices chosen when Edge-hitter starts the game and both players play optimally. We compare the game transversal number of a hypergraph with its transversal number, and also present an important fact concerning the monotonicity of $\tau_g$, that we call the Transversal Continuation Principle. It is known that if $H$ is a hypergraph with all edges of size at least~$2$, and $H$ is not a $4$-cycle, then $\tau_g(H) \le \frac{4}{11}(\nH+\mH)$; and if $H$ is a (loopless) graph, then $\tau_g(H) \le \frac{1}{3}(\nH + \mH + 1)$. We prove that if $H$ is a $3$-uniform hypergraph, then $\tau_g(H) \le \frac{5}{16}(\nH + \mH)$, and if $H$ is $4$-uniform, then $\tau_g(H) \le \frac{71}{252}(\nH + \mH)$.