Hardness of peeling with stashes

•Removing a minimal stash from a graph to leave an empty k-core is NP-complete.•We construct a polynomial-time reduction from Vertex Cover.•We analyze the complexity of removing both a stash of vertices and edges.•This result holds for standard graphs and hypergraphs with one exception.

The analysis of several algorithms and data structures can be framed as a peeling process on a random graph: vertices with degree less than k and their adjacent edges are removed until no vertices of degree less than k are left. Often the question is whether the remaining graph, the k-core, is empty or not. In some settings, it may be possible to remove either vertices or edges from the graph before peeling, at some cost. For example, in hashing applications where keys correspond to edges and buckets to vertices, one might use an additional side data structure, commonly referred to as a stash, to separately handle some keys in order to avoid collisions. The natural question in such cases is to find the minimum number of edges (or vertices) that need to be stashed in order to realize an empty k-core.We show that both these problems are NP-complete for all k≥2k≥2, with the sole exception being that the edge variant of stashing is solvable in polynomial time for k=2k=2 on standard (2-uniform) graphs.