We generalise the popular cops and robbers game to multi-layer graphs, where each cop and the robber are restricted to a single layer (or set of edges). We show that initial intuition about the best way to allocate cops to layers is not always correct, and prove that the multi-layer cop number is neither bounded from above nor below by any function of the cop numbers of the individual layers. We determine that it is NP-hard to decide if k cops are sufficient to catch the robber, even if all cop layers are trees. However, we give a polynomial time algorithm to determine if k cops can win when the robber layer is a tree. Additionally, we investigate a question of worst-case division of a simple graph into layers: given a simple graph G, what is the maximum number of cops required to catch a robber over all multi-layer graphs where each edge of G is in at least one layer and all layers are connected? For cliques, suitably dense random graphs, and graphs of bounded treewidth, we determine this parameter up to multiplicative constants. Lastly we consider a multi-layer variant of Meyniel's Conjecture, and show the existence of an infinite family of graphs whose multi-layer cop number is bounded from below by a constant times n / log n, where n is the number of vertices in the graph.
翻译:我们把流行的警察和强盗游戏概括为多层图,让每个警察和强盗限于一个层(或一组边缘)。我们显示,最初对警察分配到层的最佳方式的直觉并不总是正确的,并且证明多层警察数字既不受各个层警察数字的任何功能的约束,也不受各个层警察数字的任何功能的约束。我们确定很难决定K警察是否足以抓住强盗,即使所有的警察层都是树。然而,我们给出一个多元时间算法来确定当强盗层是树的时候, k警察能否赢得强盗。此外,我们调查了一个简单图表中最坏的分层问题:给一个简单的图形G,多少是警察在所有多层图中抓强盗的最大人数,而G的边缘至少是一个层,所有层都连接在一起。对于坚固的树枝根层图,我们决定了这个参数,直到多层不变的常数。最后,我们考虑一个多层图的多层图数,从一个多层图层图中显示一个多层图层图。</s>