Cops and Robbers 1

Catch me if you can!The Game of Cops and Robbers on Graphs

Anthony BonatoRyerson University

ICMCM’11 December 2011

Cops and Robbers

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C

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R

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C

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R

cop number c(G) ≤ 3

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Cop number > 2

• no dominating set (i.e. every vertex joined to some vertex in the set) of order 2, so R is safe on first move with only 2 cops

• no 3- or 4-cycles and 3-regular, so robber can escape each round:– one cop can cover at most

one of neighbour of R– always a node for R to move

to

CC

R

Cops and Robbers

• played on reflexive graphs G• two players Cops C and robber R play at alternate

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

moved 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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Basic facts on the cop number

• c(G) ≤ γ(G) (the domination number of G)– far from sharp: paths

• trees have cop number 1– one cop chases the robber to an end-vertex

• cop number can vary drastically with subgraphs– add a universal vertex

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Applications: multiple-agent moving-target search

• octile connected maps

• example: in video games, player controls robber, while cops are computer generated agents

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(Greiner et al, 08), (Moldenhauer et al, 09):

• problem in AI

• agents must be smart and perform calculations quickly

• other applications:

−missile defense

−counter-terrorism

−robotics

More facts about cop number

• (Aigner, Fromme, 84) introduced parameter

– G planar, then c(G) ≤ 3

– no 3- or 4-cycles, then c(G) ≥ minimum degree

• (Berrarducci, Intrigila, 93), (B, Chiniforooshan,09):

“c(G) ≤ s?” s fixed: 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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Cop-win case

• consider the case when one cop has a winning strategy– cop-win graphs

• introduced by (Nowakowski, Winkler, 83), (Quilliot, 78) – cliques, universal vertices– trees– chordal graphs

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Characterization

• node u is a corner if there is a v such that N[v] contains N[u]– v is the parent; u is the child

• a graph is dismantlable if we can iteratively delete corners until there is only one vertex

Theorem (Nowakowski, Winkler 83; Quilliot, 78)

A graph is cop-win if and only if it is dismantlable.

idea: cop-win graphs always have corners; retract corner

and play shadow strategy;

- dismantlable graphs are cop-win by induction

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Dismantlable graphs

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Dismantlable graphs

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• unique corner!• part of an infinite family that maximizes capture time

(Bonato, Hahn, Golovach, Kratochvíl,09)

Cop-win orderings

• a permutation v1, v2, … , vn of V(G) is a

cop-win ordering if there exist vertices w1, w2, …, wn such that for all i, wi is the parent of vi in the subgraph induced V(G) \ {vj : j < i}.

– a cop-win ordering dismantlability

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1

23

4

5

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G(n,p) random graphs(Erdős, Rényi, 63)

• p = p(n) a real number in (0,1), n a positive integer• G(n,p): probability space on graphs with nodes {1,

…,n}, two nodes joined independently and with probability p

Typical cop-win graphs

• what is a random cop-win graph?

• G(n,1/2) and condition on being cop-win

• probability of choosing a cop-win graph on the uniform space of labeled graphs of ordered n

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Cop number of G(n,1/2)

• (B,Hahn, Wang, 07), (B,Prałat, Wang,09)

A.a.s. (i.e. probability tending to 1 as n → ∞)

c(G(n,1/2)) = (1+o(1))log2n.

-matches the domination number

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Universal vertices

• P(cop-win) ≥ P(universal)

= n2-n+1 – O(n22-2n+3)

= (1+o(1))n2-n+1

• …this is in fact the correct answer!

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Main result

Theorem (B,Kemkes, Prałat,11+)

In G(n,1/2),

P(cop-win) = (1+o(1))n2-n+1

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Corollaries

Corollary (BKP,11+)

The number of labeled cop-win graphs is

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Corollaries

Un = number of labeled graphs with a universal

vertex

Cn = number of labeled cop-win graphs

Corollary (BKP,11+)

That is, almost all cop-win graphs contain a

universal vertex.Cops and Robbers 22

.1lim

n

n

n C

U

Strategy of proof

• probability of being cop-win and not having a universal vertex is very small

1. P(cop-win + ∆ ≤ n – 3) ≤ 2-(1+ε)n

2. P(cop-win + ∆ = n – 2) = 2-(3-log23)n+o(n)

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P(cop-win + ∆ ≤ n – 3) ≤ 2-(1+ε)n

• consider cases based on number of parents:

a. there is a cop-win ordering whose vertices in their initial segments of length 0.05n have more than 17 parents.

b. there is a cop-win ordering whose vertices in their initial segments of length 0.05n have at most 17 parents, each of which has co-degree more than n2/3.

c. there is a cop-win ordering whose initial segments of length 0.05n have between 2 and 17 parents, and at least one parent has co-degree at most n2/3.

d. there exists a vertex w with co-degree between 2 and n2/3, such that wi = w for i ≤ 0.05n.

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P(cop-win + ∆ = n – 2) ≤ 2-(3-log23)n+o(n)

Sketch of proof: Using (1), we obtain that there is an ε > 0

such that

P(cop-win) ≤ P(cop-win and ∆ ≤ n-3) + P(∆ ≥ n-2)

≤ 2-(1+ε)n + n22-n+1

≤ 2-n+o(n) (*)• if ∆ = n-2, then G has a vertex w of degree n-2, a unique

vertex v not adjacent to w.– let A be the vertices not adjacent to v (and adjacent to w)– let B be the vertices adjacent to v (and also to w)

• Claim: The subgraph induced by B is cop-win.

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A B

w

v

x

Proof continued

• n choices for w; n-1 for v

• choices for A

• if |A| = i, then using (*), probability that B is cop-win is at most 2-n+2+i+o(n)

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2

0

2n

i i

n

Meyniel’s Conjecture

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

graph of order n

• Meyniel Conjecture: c(n) = O(n1/2).

• deepest conjecture on the cop number

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Henri Meyniel, courtesy Geňa Hahn

State-of-the-art

• (Lu, Peng, 11+) proved that

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

• (Bollobás, Kun, Leader, 11+): if

p = p(n) ≥ 2.1log n/ n, then a.a.s.

c(G(n,p)) ≤ 160000n1/2log n

• (Prałat,Wormald,11+): removed log factor

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no

nOnc

2log))1(1(2)(

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Incidence graphs

• 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 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

Graphs with large cop number

• (Prałat,09) c(G(P)) = q+1– lower bound: girth = 6, δ = q+1

• P only known to exist for q prime power• using Bertrand’s postulate,

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2,8

)(nn

nc

Affine planes

• affine plane: – q2 points, each pair of points determines a unique line– each line has q points, q2 +q lines, each point on q+1 lines

• q+1 parallel classes: each contains q lines

• delete k parallel classes from affine plane A,

form incidence graph: G(A)-k

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Example: q=3, k=11 2 3

4 5 6

7 8 9

Meyniel extremal families

• a family of connected graphs (Gn: n ≥ 1) is Meyniel extremal if for large n, c(Gn) ≥ dn1/2

• (Baird, B, 11+) If k=o(q), then G(A)-k has order 2q2+(1-k)q, is (q+1-k,q)-regular and

q+1-k ≤ c(G(A)-k) ≤ q

– gives infinitely many distinct Meyniel extremal families

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Distance k Cops and Robber

• 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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A new parameter: 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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Example: k = 1

C

R

c1(G) > 1

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Example

C C

R c1(G) = 2

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Polytime algorithm

Theorem (B,Chiniforooshan,09) Given G as input with k ≥ 0 and s > 0 integers, there is a O(n2s+3) algorithm to determine if ck(G) ≤ s.

• generalizes algorithm in case k = 0

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Strong products

• sth strong power of G:

– vertices: s-tuples from V(G)– edges: two s-tuples are joined if they are

equal or adjacent in each coordinate • idea: set of s cops moving in G move as one

cop moving in the sth strong power of G

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Example: s = 2, G = P3

1

2

3

11 12 13

2122

23

31 32 33

C

C

C C

C

C

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CharacterizationTheorem (BC,09) Suppose that k, s ≥ 0. Then

ck(G) > s iff there is a function

such that

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Algorithm

• finds a function Ψ from satisfying (1), (2) from the theorem

• at each step, for any function Ψ’ satisfying (1), (2) of Theorem, Ψ’(T) is a subset of Ψ(T) for all T

• ck(G) > s iff final value of Ψ satisfies (1), (2)

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ck(n)

• ck(n) = maximum value of ck(G) over connected G of order n

• Meyniel conjecture:

c0(n) = O(n1/2).

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Upper bound

Theorem (BC,09) For n > 0 and k ≥ 0,

Theorem (BC,Prałat,10) For k ≥ 0, )1(2/1

)(o

k k

nnc

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Random graphs

• for random graphs G(n,p) with p = p(n), the behaviour of distance k cop number is complicated

Theorem (BCP,10)

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Zig-zag functions

• for x in (0,1), define

fk(x) = log E(ck(G(n,nx-1))) / log n

Five problems on cop number

1) Do almost all graphs with cop number k (k-cop-win) contain a dominating set of order k?– would imply that the number of labeled k-cop-win

graphs of order n is

– difficulty: no simple elimination ordering for k > 1 (Clarke, MacGillivray,11+)

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Minimum orders

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

M1 = 1, M2 = 4,

M3 = 10 (Baird, B,11+)

• Petersen graph unique

minimum order 3-cop-win

2) M4 = ?

• Are the Mk monotone increasing?

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Number of graphs with small cop number

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Planar graphs

• (Aigner,Fromme, 84): planar graphs have cop number ≤ 3

3) Characterize planar graphs with cop number 1,2, and 3.

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

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Distance k cop-win

• 4) Characterize graphs where ck(G) = 1– open even if k = 1

• c1(G) =1 characterized in bipartite case by

(Chalopin, Chepoi, Nisse,Vaxés,11+)

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The robber fights back!

• robber can attack neighbouring cop

• one more cop needed in this graph (check)

5) Does any graph G need c(G)+1 many cops in this game to win?

C

C

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• preprints, reprints, contact:

search: “Anthony Bonato”

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