Random cop-win graphs Anthony Bonato - PowerPoint Presentation

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Random cop-win graphs Anthony Bonato

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Almost all cop-win graphs contain a universal vertex

Anthony BonatoRyerson University

RSA 2011

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Cops and Robbers

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Cops and Robbers

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Cops and Robbers

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cop

number c(G)

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Cops and Robbers

played on reflexive graphs Gtwo players Cops C and robber R play at alternate time-steps (cops first) with perfect informationplayers 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 captureminimum number of cops needed to capture the robber is the cop number c(G)well-defined as c(G) ≤ δ(G)

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Fast facts about cop number

(Aigner, Fromme, 84) introduced parameter G planar, then c(G) ≤ 3(Berrarducci, Intrigila, 93), (B, Chiniforooshan,10): “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(Shroeder,01) G genus g, then c(G) ≤ ⌊ 3g/2 ⌋+3(Joret, Kamiński, Theis, 09) c(G) ≤ tw(G)/2

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Meyniel’s Conjecture

c(n) = maximum cop number of a connected graph of order nMeyniel Conjecture: c(n) = O(n1/2).

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State-of-the-art

(Lu, Peng, 09+) proved thatindependently proved by (Scott, Sudakov,10+), and (Frieze, Krivelevich, Loh, 10+)(Bollobás, Kun, Leader, 08+): if p = p(n) ≥ 2.1log n/ n, thenc(G(n,p)) ≤ 160000n1/2log n(Prałat,Wormald,11+): removed log factor

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Cop-win case

consider the case when one cop has a winning strategycop-win graphsintroduced by (Nowakowski, Winkler, 83), (Quilliot, 78) cliques, universal verticestreeschordal 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 childa graph is dismantlable if we can iteratively delete corners until there is only one vertexTheorem (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

)

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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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Cop-win Strategy (Clarke, Nowakowski, 2001)

V(G) = [n] a cop-win orderingG1 = G, i > 1, Gi: subgraph induced by deleting 1, …, i-1fi: Gi → Gi+1 retraction mapping i to a fixed one of its parentsFi = fi-1 ○… ○ f2 ○ f1 a homomorphismidea: robber on u, think of Fi(u) shadow of robbercop moves to capture shadow works as the Fi are homomorphismsresults in a capture in at most n moves of cop

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Typical cop-win graphs

what is a random cop-win graph?G(n,1/2) and condition on being cop-winprobability 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.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 vertexCn = number of labeled cop-win graphsCorollary (BKP,11+)That is, almost all cop-win graphs contain a universal vertex.

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Strategy of proof

probability of being cop-win and not having a universal vertex is very smallP(cop-win + ∆ ≤ n – 3) ≤ 2-(1+ε)nP(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:there is a cop-win ordering whose vertices in their initial segments of length 0.05n have more than 17 parents.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.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.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 thatP(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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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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Problems

higher cop number?almost all k-cop-win graphs 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,09+)characterizing cop-win planar or outer-planar graphs

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preprints, reprints, contact:Google: “Anthony Bonato”

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Geometric model for OSNs

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journal relaunch new editors accepting theoretical and empirical papers on complex networks, OSNs, biological networks