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The Robber Strikes Back

2014 SIAM Conference on Discrete Mathematics. The Robber Strikes Back. Anthony Bonato Ryerson University. C. C. R. C. C. R. R loses. R wins. Cops and Attacking Robbers. robber can attack neighbouring cop “cop number” well-defined for this game called cc(G)

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The Robber Strikes Back

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  1. 2014 SIAM Conference on DiscreteMathematics The Robber Strikes Back Anthony Bonato Ryerson University Robber Strikes Back, Anthony Bonato

  2. Robber Strikes Back Anthony Bonato

  3. C C R Robber Strikes Back, Anthony Bonato

  4. C C R R loses Robber Strikes Back, Anthony Bonato

  5. R wins Robber Strikes Back, Anthony Bonato

  6. Cops and Attacking Robbers • robber can attackneighbouringcop • “cop number” well-defined for this game • called cc(G) • a cop will not normally move to a neighbour of R unless she has “back-up” C C R Robber Strikes Back Anthony Bonato

  7. Robber Strikes Back Anthony Bonato

  8. Examples • cc(G) = 1 iffG has a universal vertex Robber Strikes Back Anthony Bonato

  9. Elementary bounds Lemma (Haidar,13) c(G) ≤ cc(G) ≤ min{2c(G), γ(G)}. Theorem (Haidar,13) • If G has girth at least 5, then cc(G) ≥ δ(G) + 1. • Isometric paths are 2-guardable. • If G is outerplanar, then cc(G) ≤ 3. Robber Strikes Back Anthony Bonato

  10. Cops and Robbers • case with one cop studied first by (Nowakowski,Winkler,83) and (Quilliot,78) • case with one cop fully characterized as dismantlable graphs • cop number introduced in (Aigner,Fromme, 84) Robber Strikes Back, Anthony Bonato

  11. Meyniel’s conjecture • c(n) = maximum cop number of a connected graph of order n Meyniel’s Conjecture: c(n) = O(n1/2). • best known upper bound: • independently proved by (Lu, Peng, 12), (Frieze, Krivelevich, Loh, 11), (Scott, Sudakov,11) Robber Strikes Back Anthony Bonato

  12. Complexity • (Berrarducci, Intrigila, 93), (Hahn,MacGillivray, 06), (B,Chiniforooshan, 09): “c(G) ≤ s?” sfixed: in P; running time O(n2s+3), n = |V(G)| • (Fomin, Golovach, Kratochvíl, Nisse, Suchan, 08): if s not fixed, then computing the cop number is NP-hard • Goldstein, Reingold Conjecture: if s is not fixed, then computing the cop number is EXPTIME-complete. • settled by (Kinnersley,14+) • not known to be in NP Robber Strikes Back Anthony Bonato

  13. Genus • (Aigner, Fromme, 84) planar graphs (genus 0) have cop number ≤ 3. • (Clarke, 02) outerplanar graphs have cop number ≤ 2. • Schroeder’s Conjecture: If G has genus k, then c(G) ≤ k + 3. • (Schroeder,01): c(G) ≤ floor(3k/2) +3. Robber Strikes Back Anthony Bonato

  14. Variety is the spice of life • traps, alarms, technology (Clarke,99,02), (Clarke,Nowakowski,01), (Musson,Tang,11) • tandem-win Cops and Robbers (Clarke,Nowakowski,05) • Complementary Cops and Robbers (Hill,08) • distance k Cops and Robbers (B,Chiniforooshan, Pralat,09), (Chalopin,Chepoi,Nisse,Vaxes,10) • Cops and Fast Robber (Alon,Mehrabian,11), (Frieze, Krivelevich,Loh,12) • play on edges (Dudek, Gordinowicz, Pralat,14) • Lazy Cops and Robbers (Offner, Ojakian,14+), (Bal,B,Kinnersley, Pralat,14+) • Cops and Invisible Robbers (Dierenowski,Dyer,Tifenbach,Yang,14+) • Hunter and Rabbit, (Babichenko,Peres,Peretz,Sousi,Winkler,14+) • Cops vs Gambler (Komarov,Winkler,14+) • Containment (Komarov,Mackey,14+) … Robber Strikes Back Anthony Bonato

  15. Graph searching games in graphs bad good

  16. Conjecture? • Question: For a graph G, cc(G) ≤ c(G) + 1?(#) • holds for cop-win graphs, outerplanar graphs,… • (wrong) idea: • play as in usual game, assuming the cops can stay distance 2 from R • use one extra cop at end for capture Robber Strikes Back Anthony Bonato

  17. Lower bounds • a hypergraph is linear if any two hyperedges intersect in at most one vertex Lemma (B,Finbow,Gordinowicz,Haidar,Kinnersley, Mitsche,Prałat,Stacho,13) Let Hbe a linear k-uniform hypergraph with minimum degree at least 3 and girth at least 5. If L(H)has domination number at least 2k, then cc(L(H)) ≥ 2k. Robber Strikes Back Anthony Bonato

  18. Sketch of proof • suppose only 2k-1 cops • R safe opening round by domination hypothesis • suppose cops win and consider the 2nd to last move of cops, and R on E • Siare disjoint cliques, • |Si| ≥ 2 • no vertex u outside Si dominates more than one vertex of Si; u cannot dominate vertices in two different Si E R C C S1 C C C Sk S2 Robber Strikes Back Anthony Bonato

  19. Counterexample to (#) P L(P) G = L(P), c(G) = 2, cc(G) = 4 Robber Strikes Back Anthony Bonato

  20. Bipartite graphs Theorem (BFGKMPS,13) For every connected bipartite graph G, we have that cc(G) ≤ c(G) + 2. Robber Strikes Back Anthony Bonato

  21. Sketch of proof • let c(G) = k, and play Cops and Attacking Robbers on G with k+2 cops: C1, C2, … , Ck, C’, C’’ • C1, C2, … , Ckwill play as in the usual game, making sure to stay “far enough away” from R • C’ and C’’ will stay on one vertex and move towards the R • WLOG, we can assume that R never passes Robber Strikes Back Anthony Bonato

  22. Sketch of proof • to each cop Ci, we associate a shadowSi • throughout the game the shadows follow a winning strategy for the ordinary game on G • shadows may be adjacent to R • let Ci(t),Si(t), and R(t) denote the positions of Ci, Si, and the robber, respectively, at the end of round t Robber Strikes Back Anthony Bonato

  23. Sketch of proof • we maintain the following invariants for all i and t: • each cop is adjacent to or equal to their shadow • if Ci(t+1) ≠Si(t+1), then Si(t+1) and R(t) belong to different partite sets of G • Ci(t+1) is not adjacent to R(t) (that is, the robber never has the opportunity to attack any cop) Robber Strikes Back Anthony Bonato

  24. Sketch of proof • in round t+1, each cop Cimoves as follows: (a) If Ci(t) ≠ Si(t), then Cimoves to Si(t) (b) if Ci(t) = Si(t), and Si(t+1) is not adjacent to R(t), then Cimoves to Si(t+1) (c) otherwise, Ciremains at her current vertex. By invariant (1), this is a legal strategy. Robber Strikes Back Anthony Bonato

  25. Sketch of proof • can argue that with the moves (a), (b), and (c), we can maintain the invariants (1)-(3). • eventually some Si captures R • either Si(t) = R(t) or Si(t+1) = R(t) • in the case Si(t) = R(t), (3) implies that Ci(t)≠Si(t) and (1) implies that Cicaptures the robber in round t + 1. • in the case when Si(t+1) = R(t), by (2) since Si(t+1) is not adjacent to R(t) we in fact have that Ci(t+1) = Si(t+1) = R(t) so the cops win. Robber Strikes Back Anthony Bonato

  26. K1,m-free, diameter 2graphs Theorem (BFGKMPS,13) Let G be a K1,m-free, diameter 2 graph, where m ≥ 3. Then cc(G) ≤ c(G) + 2m – 2. Robber Strikes Back Anthony Bonato

  27. Problems • bounds on cc(G)? • cc(G) ≤ c(G) + O(1)? • find G with c(G) > 2, cc(G) = 2c(G) • bounds for diameter 2 or claw-free graphs? • (#) hold for planar graphs? • cc(G) ≤ 4? Robber Strikes Back Anthony Bonato

  28. Problems • characterize G such that cc(G) = 2. • includes: • cop-win graphs, no universal vertex • non-cop-win graphs, domination number 2 A graph G with c(G) = cc(G) = 2 and γ(G) = 3. Robber Strikes Back Anthony Bonato

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