the robber strikes back

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The Robber Strikes Back

Anthony BonatoRyerson University

2014 SIAM Conference on Discrete Mathematics

Robber Strikes Back, Anthony Bonato

Robber Strikes Back Anthony Bonato

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C

CR


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C

CR

R loses


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R wins

Cops and Attacking Robbers

• robber can attack neighbouring cop• “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”

Robber Strikes Back Anthony Bonato

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C RC

Robber Strikes BackAnthony Bonato

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Examples

• cc(G) = 1 iff G has a universal vertex


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Elementary bounds

Lemma (Haidar,13)c(G) ≤ cc(G) ≤ min{2c(G), γ(G)}.

Theorem (Haidar,13)1. If G has girth at least 5, then

cc(G) ≥ δ(G) + 1.2. Isometric paths are 2-guardable.3. If G is outerplanar, then cc(G) ≤ 3.


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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)


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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)


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Complexity• (Berrarducci, Intrigila, 93), (Hahn,MacGillivray, 06),

(B,Chiniforooshan, 09):

“c(G) ≤ s?” s fixed: 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


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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.


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+) …


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Graph searching games in graphs

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slow medium fast helicopter

slow traps, tandem-win, Lazy Cops and Robbers

medium robot vacuum Cops and Robbers edge searching, Cops and Fast Robbers

eternal security, Hunter and Rabbit

fast cleaning distance k Cops and Robbers, Wall Cops and Robbers

Cops and Robbers on disjoint edge sets

The Angel and Devil

helicopter Seepage, Wall Cops and Robbers

Helicopter Cops and Robbers, Marshals, The Angel and Devil,Firefighter

Hex

badgood

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


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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 H be 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.


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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


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E

S1S2

Sk

• Si are disjoint cliques, |Si| ≥ 2• no vertex u outside Si

dominates more than one vertex of Si; u cannot dominate vertices in two different Si

C

R

CC

CC

Counterexample to (#)


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G = L(P), c(G) = 2, cc(G) = 4

P L(P)

Bipartite graphs

Theorem (BFGKMPS,13)For every connected bipartite graph G, we have that

cc(G) ≤ c(G) + 2.


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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, … , Ck will 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


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Sketch of proof• to each cop Ci, we associate a shadow Si

• 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


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Sketch of proof• we maintain the following invariants for all i and t:

1. each cop is adjacent to or equal to their shadow

2. if Ci(t+1) ≠ Si(t+1), then Si(t+1) and R(t) belong to different partite sets of G

3. Ci(t+1) is not adjacent to R(t) (that is, the robber never has the opportunity to attack any cop)


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Sketch of proof• in round t+1, each cop Ci moves as follows:

(a) If Ci(t) ≠ Si(t), then Ci moves to Si(t)

(b) if Ci(t) = Si(t), and Si(t+1) is not adjacent to R(t), then Ci moves to Si(t+1)

(c) otherwise, Ci remains at her current vertex.

By invariant (1), this is a legal strategy.Robber Strikes Back

Anthony Bonato24

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 Ci captures 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.


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K1,m-free, diameter 2 graphs

Theorem (BFGKMPS,13)Let G be a K1,m-free, diameter 2 graph, where m ≥ 3. Then

cc(G) ≤ c(G) + 2m – 2.


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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?


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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.


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the robber strikes back

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