Cops and Robbers is EXPTIME-complete

We investigate the computational complexity of deciding whether k cops can capture a robber on a graph G. In 1995, Goldstein and Reingold conjectured that the problem is EXPTIME-complete when both G and k are part of the input; we prove this conjecture.Comment: v2: updated figures and slightly clarified some minor point

To catch a falling robber

We consider a Cops-and-Robber game played on the subsets of an $n$-set. The robber starts at the full set; the cops start at the empty set. On each turn, the robber moves down one level by discarding an element, and each cop moves up one level by gaining an element. The question is how many cops are needed to ensure catching the robber when the robber reaches the middle level. Aaron Hill posed the problem and provided a lower bound of $2^{n/2}$ for even $n$ and $\binom{n}{\lceil n/2 \rceil}2^{-\lfloor n/2 \rfloor}$ for odd $n$. We prove an upper bound (for all $n$) that is within a factor of $O(\ln n)$ times this lower bound.Comment: Minor revision

Bounds on the length of a game of Cops and Robbers

In the game of Cops and Robbers, a team of cops attempts to capture a robber on a graph G. All players occupy vertices of G. The game operates in rounds; in each round the cops move to neighboring vertices, after which the robber does the same. The minimum number of cops needed to guarantee capture of a robber on G is the cop number of G, denoted c(G), and the minimum number of rounds needed for them to do so is the capture time. It has long been known that the capture time of an n-vertex graph with cop number k is O(nk+1). More recently, Bonato, Golovach, Hahn, and Kratochvíl ([3], 2009) and Gavenčiak ([10], 2010) showed that for k = 1, this upper bound is not asymptotically tight: for graphs with cop number 1, the cop can always win within n − 4 rounds. In this paper, we show that the upper bound is tight when k ≥ 2: for fixed k ≥ 2, we construct arbitrarily large graphs G having capture time at least (|V (G)|/40k4 )k+1. In the process of proving our main result, we establish results that may be of independent interest. In particular, we show that the problem of deciding whether k cops can capture a robber on a directed graph is polynomial-time equivalent to deciding whether k cops can capture a robber on an undirected graph. As a corollary of this fact, we obtain a relatively short proof of a major conjecture of Goldstein and Reingold ([11], 1995), which was recently proved through other means ([12], 2015). We also show that n-vertex strongly-connected directed graphs with cop number 1 can have capture time Ω(n2), thereby showing that the result of Bonato et al. [3] does not extend to the directed setting

Game Brush Number

We study a two-person game based on the well-studied brushing process on graphs. Players Min and Max alternately place brushes on the vertices of a graph. When a vertex accumulates at least as many brushes as its degree, it sends one brush to each neighbor and is removed from the graph; this may in turn induce the removal of other vertices. The game ends once all vertices have been removed. Min seeks to minimize the number of brushes played during the game, while Max seeks to maximize it. When both players play optimally, the length of the game is the game brush number of the graph $G$, denoted $b_g(G)$. By considering strategies for both players and modelling the evolution of the game with differential equations, we provide an asymptotic value for the game brush number of the complete graph; namely, we show that $b_g(K_n) = (1+o(1))n^2/e$. Using a fractional version of the game, we couple the game brush numbers of complete graphs and the binomial random graph $\mathcal{G}(n,p)$. It is shown that for $pn \gg \ln n$ asymptotically almost surely $b_g(\mathcal{G}(n,p)) = (1 + o(1))p b_g(K_n) = (1 + o(1))pn^2/e$. Finally, we study the relationship between the game brush number and the (original) brush number.Comment: 20 pages, 3 figure

Domination Game: A proof of the 3/5-Conjecture for Graphs with Minimum Degree at Least Two

In the domination game on the graph G, the players Dominator and Staller alternately select vertices of G. Each vertex chosen must strictly increase the number of vertices dominated. This process eventually produces a dominating set of G; Dominator aims to minimize the size of this set, while Staller aims to maximize it. The size of the dominating set produced under optimal play is the game domination number of G, denoted by γg(G). In this paper, we prove that γg(G) ≤ 2n/3 for every n-vertex isolate-free graph G. When G has minimum degree at least 2, we prove the stronger bound γg(G) ≤ 3n/5; this resolves a special case of a conjecture due to Kinnersley, West, and Zamani [SIAM J. Discrete Math. 27 (2013), 2090–2107]. Finally, we prove that if G is an n-vertex isolate-free graph with vertices of degree 1, then γg(G) ≤ 3n/5 + [ℓ/2] + 1; in the course of establishing this result, we answer a question of Brešar, Dorbec, Klavžar, and Košmrlj [Discrete Math.330 (2014), 1–10.]

Toppling Numbers of Complete and Random Graphs

We study a two-person game played on graphs based on the widely studied chip-firing game. Players Max and Min alternately place chips on the vertices of a graph. When a vertex accumulates as many chips as its degree, it fires, sending one chip to each neighbour; this may in turn cause other vertices to fire. The game ends when vertices continue firing forever. Min seeks to minimize the number of chips played during the game, while Max seeks to maximize it. When both players play optimally, the length of the game is the toppling number of a graph G, and is denoted by t(G). By considering strategies for both players and investigating the evolution of the game with differential equations, we provide asymptotic bounds on the toppling number of the complete graph. In particular, we prove that for sufficiently large n 0.596400n2 \u3c t(Kn) \u3c 0.637152n2. Using a fractional version of the game, we couple the toppling numbers of complete graphs and the binomial random graph G(n,p). It is shown that for pn ≥ n2 / √ log n asymptotically almost surely t(G(n,p)) = (1+o(1))pt(Kn)