{"paper":{"title":"Revolutionaries and spies: Spy-good and spy-bad graphs","license":"http://arxiv.org/licenses/nonexclusive-distrib/1.0/","headline":"","cross_cats":["math.CO"],"primary_cat":"cs.DM","authors_text":"Daniel W. Cranston, Douglas B. West, Gregory J. Puleo, Jane V. Butterfield, Reza Zamani","submitted_at":"2012-02-14T02:02:35Z","abstract_excerpt":"We study a game on a graph $G$ played by $r$ {\\it revolutionaries} and $s$ {\\it spies}. Initially, revolutionaries and then spies occupy vertices. In each subsequent round, each revolutionary may move to a neighboring vertex or not move, and then each spy has the same option. The revolutionaries win if $m$ of them meet at some vertex having no spy (at the end of a round); the spies win if they can avoid this forever.\n  Let $\\sigma(G,m,r)$ denote the minimum number of spies needed to win. To avoid degenerate cases, assume $|V(G)|\\ge r-m+1\\ge\\floor{r/m}\\ge 1$. The easy bounds are then $\\floor{r/"},"claims":{"count":0,"items":[],"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"source":{"id":"1202.2910","kind":"arxiv","version":2},"verdict":{"id":null,"model_set":{},"created_at":null,"strongest_claim":"","one_line_summary":"","pipeline_version":null,"weakest_assumption":"","pith_extraction_headline":""},"references":{"count":0,"sample":[],"resolved_work":0,"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57","internal_anchors":0},"formal_canon":{"evidence_count":0,"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"author_claims":{"count":0,"strong_count":0,"snapshot_sha256":"258153158e38e3291e3d48162225fcdb2d5a3ed65a07baac614ab91432fd4f57"},"builder_version":"pith-number-builder-2026-05-17-v1"}