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Cops and Robbers1 Catch me if you can! The Game of Cops and Robbers on Graphs Anthony Bonato Ryerson University ICMCM’11 December 2011.

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Presentation on theme: "Cops and Robbers1 Catch me if you can! The Game of Cops and Robbers on Graphs Anthony Bonato Ryerson University ICMCM’11 December 2011."— Presentation transcript:

1 Cops and Robbers1 Catch me if you can! The Game of Cops and Robbers on Graphs Anthony Bonato Ryerson University ICMCM’11 December 2011

2 Cops and Robbers 2 C C C R

3 3 C C C R

4 4 C C C R cop number c(G) ≤ 3

5 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 Cops and Robbers5 C C R

6 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)| Cops and Robbers6

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 Robbers7

8 8 Applications: multiple-agent moving-target search octile connected maps example: in video games, player controls robber, while cops are computer generated agents

9 Cops and Robbers9 (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

10 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(n 2s+3 ), n = |V(G)| (Fomin, Golovach, Kratochvíl, Nisse, Suchan, 08): if s not fixed, then computing the cop number is NP- hard Cops and Robbers10

11 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 Robbers11

12 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 Cops and Robbers12

13 Dismantlable graphs Cops and Robbers13

14 Dismantlable graphs Cops and Robbers14 unique corner! part of an infinite family that maximizes capture time (Bonato, Hahn, Golovach, Kratochvíl,09)

15 Cop-win orderings a permutation v 1, v 2, …, v n of V(G) is a cop-win ordering if there exist vertices w 1, w 2, …, w n such that for all i, w i is the parent of v i in the subgraph induced V(G) \ {v j : j < i}. –a cop-win ordering dismantlability Cops and Robbers15 1 2 3 4 5

16 Cops and Robbers16 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

17 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 Robbers17

18 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))log 2 n. -matches the domination number Cops and Robbers18

19 Universal vertices P(cop-win) ≥ P(universal) = n2 -n+1 – O(n 2 2 -2n+3 ) = (1+o(1))n2 -n+1 …this is in fact the correct answer! Cops and Robbers19

20 Main result Theorem (B,Kemkes, Prałat,11+) In G(n,1/2), P(cop-win) = (1+o(1))n2 -n+1 Cops and Robbers20

21 Corollaries Corollary (BKP,11+) The number of labeled cop-win graphs is Cops and Robbers21

22 Corollaries U n = number of labeled graphs with a universal vertex C n = number of labeled cop-win graphs Corollary (BKP,11+) That is, almost all cop-win graphs contain a universal vertex. Cops and Robbers22

23 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-log 2 3)n+o(n) Cops and Robbers23

24 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 n 2/3. 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 n 2/3. d.there exists a vertex w with co-degree between 2 and n 2/3, such that w i = w for i ≤ 0.05n. Cops and Robbers24

25 P(cop-win + ∆ = n – 2) ≤ 2 -(3-log 2 3)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 + n 2 2 -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. Cops and Robbers25

26 Cops and Robbers26 A B w v x

27 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 Robbers27

28 Meyniel’s Conjecture c(n) = maximum cop number of a connected graph of order n Meyniel Conjecture: c(n) = O(n 1/2 ). deepest conjecture on the cop number Cops and Robbers28

29 Cops and Robbers29

30 Cops and Robbers30 Henri Meyniel, courtesy Geňa Hahn

31 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)) ≤ 160000n 1/2 log n (Prałat,Wormald,11+): removed log factor Cops and Robbers31

32 Cops and Robbers32 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 q 2 +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

33 Example Cops and Robbers33 Fano plane Heawood graph

34 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, Cops and Robbers34

35 Affine planes affine plane: –q 2 points, each pair of points determines a unique line –each line has q points, q 2 +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 Cops and Robbers35

36 Cops and Robbers36 Example: q=3, k=1 1 2 3 4 56 7 8 9

37 Meyniel extremal families a family of connected graphs (G n : n ≥ 1) is Meyniel extremal if for large n, c(G n ) ≥ dn 1/2 (Baird, B, 11+) If k=o(q), then G(A) -k has order 2q 2 +(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 Cops and Robbers37

38 Cops and Robbers38 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

39 Cops and Robbers39 A new parameter: c k (G) c k (G) = minimum number of cops needed to capture robber at distance at most k G connected implies c k (G) ≤ diam(G) – 1 for all k ≥ 1, c k (G) ≤ c k-1 (G)

40 Cops and Robbers40 Example: k = 1 C R c 1 (G) > 1

41 Cops and Robbers41 Example C C R c 1 (G) = 2

42 Cops and Robbers42 Polytime algorithm Theorem (B,Chiniforooshan,09) Given G as input with k ≥ 0 and s > 0 integers, there is a O(n 2s+3 ) algorithm to determine if c k (G) ≤ s. generalizes algorithm in case k = 0

43 Cops and Robbers43 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

44 Cops and Robbers44 Example: s = 2, G = P 3 1 2 3 11 12 13 21 22 23 31 32 33 C C C C C C

45 Cops and Robbers45 Characterization Theorem (BC,09) Suppose that k, s ≥ 0. Then c k (G) > s iff there is a function such that

46 Cops and Robbers46 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 c k (G) > s iff final value of Ψ satisfies (1), (2)

47 Cops and Robbers47 c k (n) c k (n) = maximum value of c k (G) over connected G of order n Meyniel conjecture: c 0 (n) = O(n 1/2 ).

48 Cops and Robbers48 Upper bound Theorem (BC,09) For n > 0 and k ≥ 0, Theorem (BC,Prałat,10) For k ≥ 0,

49 Cops and Robbers49 Random graphs for random graphs G(n,p) with p = p(n), the behaviour of distance k cop number is complicated Theorem (BCP,10)

50 Cops and Robbers50 Zig-zag functions for x in (0,1), define f k (x) = log E(c k (G(n,n x-1 ))) / log n

51 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+) Cops and Robbers51

52 Minimum orders M k = minimum order of a k-cop-win graph M 1 = 1, M 2 = 4, M 3 = 10 (Baird, B,11+) Petersen graph unique minimum order 3-cop-win 2) M 4 = ? Are the M k monotone increasing? Cops and Robbers52

53 Number of graphs with small cop number Cops and Robbers53

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 Robbers54

55 Distance k cop-win 4) Characterize graphs where c k (G) = 1 –open even if k = 1 c 1 (G) =1 characterized in bipartite case by (Chalopin, Chepoi, Nisse,Vaxés,11+) Cops and Robbers55

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? Cops and Robbers56 C C C R

57 Cops and Robbers57 preprints, reprints, contact: search: “Anthony Bonato”

58 Cops and Robbers58

59 நன்றி ! Cops and Robbers59


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