Transcript
Page 1: Conjectures on  Cops and Robbers

Cops and Robbers 1

Conjectures on Cops and Robbers

Anthony BonatoRyerson University

Joint Mathematics Meetings AMS Special Session

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cop number c(G) ≤ 3

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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…

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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

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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

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Example

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Fano plane Heawood graph

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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.

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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

22

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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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