conjectures on cops and robbers

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Joint Mathematics Meetings AMS Special Session. Conjectures on Cops and Robbers. Anthony Bonato Ryerson University. Cops and Robbers. C. C. R. C. Cops and Robbers. C. C. R. C. Cops and Robbers. C. R. C. C. cop number c(G) ≤ 3. Cops and Robbers. - PowerPoint PPT Presentation

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Cops and Robbers 1

Conjectures on Cops and Robbers

Anthony BonatoRyerson University

Joint Mathematics Meetings AMS Special Session

Cops and Robbers

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cop number c(G) ≤ 3

Cops and Robbers• played on a reflexive undirected graph G• two players Cops C and robber R play at alternate

time-steps (cops first) with perfect information• players move to vertices along edges; may move to

neighbors or pass • cops try to capture (i.e. land on) the robber, while

robber tries to evade capture• minimum number of cops needed to capture the

robber is the cop number c(G)– well-defined as c(G) ≤ |V(G)|

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Conjectures

• conjectures and problems on Cops and Robbers coming from 5 different directions, touch on various aspects of graph theory:

– structural, algorithmic, probabilistic, topological…

1. How big can the cop number be?

• c(n) = maximum cop number of a connected graph of order n

Meyniel Conjecture: c(n) = O(n1/2).

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Henri Meyniel, courtesy Geňa Hahn

State-of-the-art• (Lu, Peng, 13) proved that

– independently proved by (Frieze, Krivelevich, Loh, 11) and (Scott, Sudakov,11)

• (Bollobás, Kun, Leader,13): if p = p(n) ≥ 2.1log n/ n, then

c(G(n,p)) ≤ 160000n1/2log n

• (Prałat,Wormald,14+): proved Meyniel’s conjecture for all p = p(n)

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)1(1log))1(1( 22

)( ono

nnOnc

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Graph classes• (Andreae,86): H-minor free graphs have cop

number bounded by a constant.

• (Joret et al,10): H-free class graphs have bounded cop number iff each component of H is a tree with at most 3 leaves.

• (Lu,Peng,13): Meyniel’s conjecture holds for diameter 2 graphs, bipartite diameter 3 graphs.

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Questions

Soft Meyniel’s conjecture: for some ε > 0,c(n) = O(n1-ε).

• Meyniel’s conjecture in other graphs classes?– bipartite graphs– diameter 3– claw-free

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How close to n1/2?

• consider a finite projective plane P– two lines meet in a unique point– two points determine a unique line– exist 4 points, no line contains more than two of them

• q2+q+1 points; each line (point) contains (is incident with) q+1 points (lines)

• incidence graph (IG) of P:– bipartite graph G(P) with red nodes the points of P

and blue nodes the lines of P– a point is joined to a line if it is on that line

Example

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Fano plane Heawood graph

Meyniel extremal families • a family of connected graphs (Gn: n ≥ 1) is Meyniel

extremal if there is a constant d > 0, such that for all n ≥ 1, c(Gn) ≥ dn1/2

• IG of projective planes: girth 6, (q+1)-regular, so have cop number ≥ q+1– order 2(q2+q+1)– Meyniel extremal (must fill in non-prime orders)

• all other examples of Meyniel extremal families come from combinatorial designs (B,Burgess,2013)

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Minimum orders

• Mk = minimum order of a k-cop-win graph

• M1 = 1, M2 = 4• M3 = 10 (Baird, B,12)

– see also (Beveridge et al, 14+)• M4 = ?

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Conjectures on mk, Mk

Conjecture: Mk monotone increasing.

• mk = minimum order of a connected G such that c(G) ≥ k

• (Baird, B, 12) mk = Ω(k2) is equivalent to Meyniel’s conjecture.

Conjecture: mk = Mk for all k ≥ 4.

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2. 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

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Questions

Goldstein, Reingold Conjecture: if s is not fixed, then computing the cop number is EXPTIME-complete.

– same complexity as say, generalized chess

• settled by (Kinnersley,14+)

Conjecture: if s is not fixed, then computing the cop number is not in NP.

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3. Genus• (Aigner, Fromme, 84) planar graphs (genus 0)

have cop number ≤ 3.

• (Clarke, 02) outerplanar graphs have cop number ≤ 2.

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Questions• characterize planar (outer-planar) graphs with

cop number 1,2, and 3 (1 and 2)

• is the dodecahedron the unique smallest order planar 3-cop-win graph?

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Higher genus

Schroeder’s Conjecture: If G has genus k, then c(G) ≤ k +3.• true for k = 0• (Schroeder, 01): true for k = 1 (toroidal

graphs) • (Quilliot,85): c(G) ≤ 2k +3.• (Schroeder,01): c(G) ≤ floor(3k/2) +3.

5. VariantsGood guys vs bad guys 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 Robber

eternal security

fast cleaning distance k Cops and Robbers

Cops and Robbers on disjoint edge sets

The Angel and Devil

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

Hex

badgood

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Distance k Cops and Robber (B,Chiniforooshan,09)

• cops can “shoot” robber at some specified distance k

• play as in classical game, but capture includes case when robber is distance k from the cops– k = 0 is the classical game

C

R

k = 1

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Distance k cop number: ck(G)

• ck(G) = minimum number of cops needed to capture robber at distance at most k

• G connected implies ck(G) ≤ diam(G) – 1

• for all k ≥ 1, ck(G) ≤ ck-1(G)

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When does one cop suffice?

• (RJN, Winkler, 83), (Quilliot, 78)cop-win graphs ↔ cop-win orderings

• provide a structural/ordering characterization of cop-win graphs for:– directed graphs– distance k Cops and Robbers– invisible robber; cops can use traps or alarms/photo

radar (Clarke et al,00,01,06…)– infinite graphs (Bonato, Hahn, Tardif, 10)

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Lazy Cops and Robbers• (Offner, Ojakian,14+) only one can move in each

round– lazy cop number, cL(G)

• (Offner, Ojakian, 14+)

• (Bal,B,Kinnsersley,Pralat,14+) For all ε > 0,.

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Questions on lazy cops• Question: Find the asymptotic order of .

• (Bal,B,Kinnsersley,Pralat,14+) If G has genus g, then cL(G) = – proved by using the Gilbert, Hutchinson,Tarjan

separator theorem

• Question: Is cL(G) bounded for planar graphs?

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Firefighting

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A strategy• (MacGillivray, Wang, 03): If fire breaks out at (r,c),

1≤r≤c≤n/2, save vertices in following order:

(r + 1, c), (r + 1, c + 1), (r + 2, c - 1), (r + 2, c + 2), (r + 3, c -2),(r + 3, c - 3), ..., (r + c, 1), (r + c, 2c), (r + c, 2c + 1), ..., (r + c, n)

– strategy saves n(n-r)-(c-1)(n-c) vertices– strategy is optimal assuming fire breaks out in

columns (rows) 1,2, n-1, n

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¼ -grid conjecture

nPPsn nnnn 4

1)),(, (

n largefor ,0every For

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Infinite hexagonal grid

Conjecture: one firefighter cannot contain a fire in an infinite hexagonal grid.

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A. Bonato, R.J. Nowakowski, Sketchy Tweets: Ten Minute Conjectures in Graph Theory, The Mathematical Intelligencer 34 (2012) 8-15.

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