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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
Cops and Robbers 2
C
C
C
R
Cops and Robbers
Cops and Robbers 3
C
C
C
R
Cops and Robbers
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C
C
C
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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Cops and Robbers 7
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
Cops and Robbers 8
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
Cops and Robbers 10
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
Cops and Robbers 11
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
Cops and Robbers 17
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
Cops and Robbers 18
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!
Cops and Robbers 19
Main result
Theorem (B,Kemkes, Prałat,11+)
In G(n,1/2),
P(cop-win) = (1+o(1))n2-n+1
Cops and Robbers 20
Corollaries
Corollary (BKP,11+)
The number of labeled cop-win graphs is
Cops and Robbers 21
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)
Cops and Robbers 23
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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Cops and Robbers 26
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)
Cops and Robbers 27
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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Cops and Robbers 30
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)(
Cops and Robbers 32
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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Cops and Robbers 38
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
Cops and Robbers 39
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
Cops and Robbers 43
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
Cops and Robbers 45
CharacterizationTheorem (BC,09) Suppose that k, s ≥ 0. Then
ck(G) > s iff there is a function
such that
Cops and Robbers 46
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).
Cops and Robbers 48
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
Cops and Robbers 49
Random graphs
• for random graphs G(n,p) with p = p(n), the behaviour of distance k cop number is complicated
Theorem (BCP,10)
Cops and Robbers 50
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?
Cops and Robbers 53
Number of graphs with small cop number
Cops and Robbers 54
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?
Cops and Robbers 55
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+)
Cops and Robbers 56
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
C
R
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• preprints, reprints, contact:
search: “Anthony Bonato”
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நன்றி�!