1. Cops and Robbers from a Distance
Miniconference on the Mathematics of Computation
EURO’09
July 2009
Cops and Robbers from a Distance
Anthony Bonato
Ryerson University
Toronto, Canada

2. Game of Cops and Robbers
R loses
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Anthony Bonato

3. cop (or search) number of G, c(G) (Aigner, Fromme,84)
some graphs need more than one cop
cop (or search) number of G, c(G) (Aigner, Fromme,84)
girth(G) ≥ 5, then c(G) ≥ δ(G)
if G is planar, then c(G) ≤ 3
structure of cop-win (c(G) = 1) finite graphs is well-known and beautiful (Nowakowski, Winkler, 83), (Quilliot, 78) C
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Anthony Bonato

4. Cops and Robber - Anthony Bonato
The infinite case
dismantling characterization, however, does not apply in the infinite case
eg: ray is dismantlable, but has infinite cop number
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Cops and Robber - Anthony Bonato

5. Cops and Robbers is a simplified model for network security:
cops (authorities/network administrator/anti-virus software) guard network from the robber (intruder/hacker/virus)
the larger the cop number, the less secure the network
large literature on the game and its variants:
surveys: Alspach, 2006; Hahn, 2007; Fomin, 2008
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Anthony Bonato

6. Application: multiple-agent moving-target search
example: in video games, player controls robber, while cops are computer generated agents
octile connected maps
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7. agents must be smart and perform calculations quickly
(Greiner et al, 08):
problem in AI
agents must be smart and perform calculations quickly
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8. Complexity
(Berrarducci, Intrigila, 93), (Goldstein, Reingold,95), (Hahn, MacGillivray, 03)
1. “c(G) ≤ s?” s fixed: running time O(n2s+3), n = |V(G)|
(Fomin, Golovach, Kratochvíl, Nisse, Suchan, 08): computing the cop number of a given graph is:
NP-hard; W[2]-hard
subexponential time: 2o(n)nO(1)
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Anthony Bonato

9. How big can c(G) be?
Meyniel Conjecture: For a connected graph G of order n,
c(G) = O(n1/2).
(Chiniforooshan, 08): c(G) = O(n / log n)
(Lu, Peng, 09?): c(G) =O(n / 2(1-o(1))(log n)1/2)
Soft Meyniel Conjecture: c(G) = O(n1-c), for some
c > 0
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Anthony Bonato

10. Miniconference on the Mathematics of Computation
G(n,p) (Erdős, Rényi, 63)
n a positive integer, p = p(n) a real number in (0,1)
G(n,p): probability space on graphs with nodes {1,…,n}, two nodes joined independently and with probability p 1 2 3 4 5
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Anthony Bonato

11. Cop number of random graphs
(Bonato, Hahn, Wang, 07): for p constant, a.a.s. (with probability tending to 1 as n → ∞)
c(G(n,p)) = Θ(log n)
(Bollobás, Kun, Leader, 08): for p = p(n),
c(G(n,p)) ≤ n1/2 + o(1)
- bound attained by sparse random graphs
(Bonato, Prałat, Wang, 09): dense random graphs, and random power law graphs
(Prałat, Łuczak, 09): zig-zag theorem
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Anthony Bonato

12. Graphs realizing the lower bound
consider a finite projective plane P with
n = q2+q+1 points, where q is a prime power
define a 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
observation: c(G(P)) = Θ(n1/2)
folklore; see also (Prałat, 09)
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Anthony Bonato

13. Variations on the game players may have variable speed
cops have partial information
players can move on edges (edge searching), or must clear edges (brush number)
players must move according to certain rules
for eg, cops must be distance at most 2 apart (tandem win graphs)
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Anthony Bonato

14. New variation: 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
k = 1 R
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Anthony Bonato

15. 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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16. Example: k = 1 C
c1(G) > 1 R
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17. Example
c1(G) = 2 R C C
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18. Polytime algorithm
Theorem (Bonato,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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Anthony Bonato

19. 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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Anthony Bonato

20. Example: s = 2, G = P3 12 13 11 C C 1 C 22 C 2 C 21 23 3 C 31 32 33
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21. Characterization
Theorem. Suppose that k, s ≥ 0. Then ck(G) > s iff there is a function
such that
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Anthony Bonato

22. 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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23. 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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24. Upper bound Theorem (BC,09) For n > 0 and k ≥ 0,
recovers upper bound from k=0:
c0(n) = O(n / log n)
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Anthony Bonato

25. Lower bound, continued Theorem (BC,Prałat,09) For k ≥ 0,
sketch of proof:
form Gk by subdividing k times each edge of G
a lemma: ck(G2k+1) ≥ c(G)
(Bollobás, Kun, Leader, 08) if p = 3log(n) / n, then a.a.s. c(G(n,p))=n1/2+o(1); now use lemma
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Anthony Bonato

26. Random graphs
for random graphs G(n,p) with p = p(n), the behaviour of distance k cop number is complicated
Theorem (BCP,09)
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27. fk(x) = log E(ck(G(n,nx-1))) / log n
Zigzag functions
for x in (0,1), define
fk(x) = log E(ck(G(n,nx-1))) / log n
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Anthony Bonato

28. Open problems distance k Meyniel conjecture: ck(n) = Θ((n/k)1/2)?
structure of ck(G) = 1 graphs, k ≥ 1:
dismantling/local-type characterization?
Goldstein, Reingold conjecture: computing the cop number of a given graph is EXPTIME-complete
is computing ck(G) EXPTIME-complete?
is computing ck(G) in FPT? Likely not, as computing c0(G) is W[2]-hard.
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30. Google: “Anthony Bonato”
preprints, reprints, contact:
Google: “Anthony Bonato”
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