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：我们将经典的警察与小偷博弈推广到多层图，其中每个警察和小偷均被限制在单一层（即边集）内。研究表明，关于警察在层间最优分配方式的初始直觉往往不正确，并证明多层警察数既不受各层警察数的上界约束，也不受其下界约束。我们确定，即使所有警察层均为树结构，判定k个警察是否足以抓获小偷仍是NP-难问题。然而，当小偷层为树时，我们给出了判定k个警察是否获胜的多项式时间算法。此外，我们探讨了简单图最坏情况下的分层划分问题：给定简单图G，考虑所有满足G中每条边至少属于一个层且所有层均连通的多层图，抓获小偷所需的最多警察数是多少？对于团图、适当稠密的随机图以及有界树宽图，我们在乘法常数因子内确定了该参数。最后，我们研究了Meyniel猜想的变体，并证明存在无穷图族，其多层警察数下界为常数乘以n / log n，其中n为图的顶点数。