The Game of Cops and Robbers on Graphs - Inria · 2017-11-06 · 1/19 The Game of Cops and Robbers...

The Game of Cops and Robbers on Graphs

Fionn Mc Inerney

Universite Cote d’Azur, Inria, CNRS, I3S, France

PhD Seminars of Inria Sophia-Antipolis - Mediterranee

November 6, 2017

(Modified Slides of Nicolas Nisse)

F. Mc Inerney The Game of Cops and Robbers on Graphs

What is a Graph?

A graph G

A set of vertices (dots) V (G ) connected by a set of edges(lines) E (G ).

Graphs model real-life networks: social (Facebook, Twitter),biological (food chains, protein-interaction networks),transportation (metro maps, bus routes), etc.

What is a Graph?

A graph G

A set of vertices (dots) V (G ) connected by a set of edges(lines) E (G ).

Graphs model real-life networks: social (Facebook, Twitter),biological (food chains, protein-interaction networks),transportation (metro maps, bus routes), etc.

Basic Graph Theory

Adjacency

Two vertices u and v are adjacent if they are connected by anedge.

N(v): Open-neighbourhood of v

The set of vertices adjacent to v .

N[v ]: Closed-neighbourhood of v

The set of vertices adjacent to v and v itself.

deg(v): Degree of v

|N(v)| or the number of distinct vertices adjacent to v .

Basic Graph Theory Continued

d(u, v): Distance between u and v

The length of the shortest path between u and v .

γ(G ): Domination number of a graph G

The minimum size of a set S of vertices such that all verticesin G are either in S or adjacent to a vertex in S .

v2 v3 v4 v5

v6 v7 v8

v3 is adjacent tov4, v6, v7, v8.

N(v3) = v4, v6, v7, v8deg(v3) = 4

N[v3] = v3, v4, v6, v7, v8

v2 v3 v4 v5

v6 v7 v8

N(v3) = v4, v6, v7, v8deg(v3) = 4

N[v3] = v3, v4, v6, v7, v8

v2 v3 v4 v5

v6 v7 v8

N(v3) = v4, v6, v7, v8deg(v3) = 4

N[v3] = v3, v4, v6, v7, v8

v2 v3 v4 v5

v6 v7 v8

N(v3) = v4, v6, v7, v8deg(v3) = 4

N[v3] = v3, v4, v6, v7, v8

v2 v3 v4 v5

v6 v7 v8

d(v2, v5) = 2

γ(G ) = 2

v2 v3 v4 v5

v6 v7 v8

d(v2, v5) = 2

γ(G ) = 2

Pursuit-Evasion Games

2-Player games

A team of mobile entities (Cops) track down another mobile entity (Robber)

Always one winner

Combinatorial Problem:Minimizing some resource for some Player to wine.g., minimize number of Cops to capture the Robber.

Algorithmic Problem:Computing winning strategy (sequence of moves) for some Playere.g., compute strategy for Cops to capture Robber/Robber to avoid capture.

natural applications: coordination of mobile autonomous agents(Robotics, Network Security, Information Seeking...)

but also: Graph Theory, Models of Computation, Logic, Routing...

Cops & Robbers Game [Nowakowski and Winkler; Quilliot, 1983]

Rules of the C&R game

1 Place k ≥ 1 Cops C on nodes

2 Visible Robber R at one node

3 Turn by turn(1) each C slides along ≤ 1 edge(2) R slides along ≤ 1 edge

Goal of the C&R game

Robber must avoid the Cops

Cops must capture Robber (i.e.,occupy the same node)

Cop Number of a graph G

cn(G): min # Cops to win in G

3 Turn by turn(1) each C slides along ≤ 1 edge

(2) R slides along ≤ 1 edge

Let’s play a bit

Easy remark: For any graph G , cn(G) ≤ γ(G) the size of a min dominating set of G .

Let’s play a bit

cn(tree)=1

Let’s play a bit

cn(tree)=1

cn(clique)=?

cn(cycle)=?

cn(Petersen)=?

Let’s play a bit

cn(tree)=1

cn(clique)=?

cn(cycle)=?

cn(Petersen)=?

Let’s play a bit

cn(tree)=1

cn(clique)=?

cn(cycle)=?

cn(Petersen)=?

Let’s play a bit

cn(tree)=1

cn(clique)=1

cn(cycle)=?

cn(Petersen)=?

Let’s play a bit

cn(tree)=1

cn(clique)=1

cn(cycle)=?

cn(Petersen)=?

Let’s play a bit

cn(tree)=1

cn(clique)=1

cn(cycle)=?

cn(Petersen)=?

Let’s play a bit

cn(tree)=1

cn(clique)=1

cn(cycle)=?

cn(Petersen)=?

Let’s play a bit

cn(tree)=1

cn(clique)=1

cn(cycle)=?

cn(Petersen)=?

Let’s play a bit

cn(tree)=1

cn(clique)=1

cn(cycle)=?

cn(Petersen)=?

Let’s play a bit

cn(tree)=1

cn(clique)=1

cn(cycle)=?

cn(Petersen)=?

Let’s play a bit

cn(tree)=1

cn(clique)=1

cn(cycle)=2

cn(Petersen)=?

Let’s play a bit

cn(tree)=1

cn(clique)=1

cn(cycle)=2

cn(Petersen)=?

Let’s play a bit

cn(tree)=1

cn(clique)=1

cn(cycle)=2

cn(Petersen)=?

Let’s play a bit

cn(tree)=1

cn(clique)=1

cn(cycle)=2

cn(Petersen)=?

Let’s play a bit

cn(tree)=1

cn(clique)=1

cn(cycle)=2

cn(Petersen)=?

Let’s play a bit

cn(tree)=1

cn(clique)=1

cn(cycle)=2

cn(Petersen)=?

Let’s play a bit

cn(tree)=1

cn(clique)=1

cn(cycle)=2

cn(Petersen)=3

Complexity: a graph G , cn(G ) ≤ k?

Seminal paper: k = 1 [Nowakowski and Winkler; Quilliot, 1983]

cn(G) = 1 iff V = v1, · · · , vn and, ∀i < n, ∃j > i s.t., N[vi ] ⊆ N[vj ].

(dismantable graphs) can be checked in time O(n3)

Generalization to any k [Berarducci, Intrigila’93] [Hahn, MacGillivray’06] [Clarke, MacGillivray’12]

cn(G) ≤ k? can be checked in time nO(k) ∈ EXPTIME

EXPTIME-complete in directed graphs [Goldstein and Reingold, 1995]

NP-hard and W[2]-hard [Fomin,Golovach,Kratochvil,Nisse,Suchan, 2010]

(i.e., no algorithm in time f (k)nO(1) expected)

PSPACE-hard [Mamino 2013]

EXPTIME-complete [Kinnersley 2014]

Graphs with high cop-number

Large girth (smallest cycle) AND large min degree ⇒ large cop-number

G with min-degree d and girth > 4 ⇒ cn(G) ≥ d . [Aigner and Fromme 84]

for any d , there are d-regular graphs G with cn(G) ≥ d [Aigner and Fromme 84]

cn(G) ≥ d t in any graph with min-degree d and girth > 8t − 3 [Frankl 87]

for any k, there is G with diameter 2 and cn(G) ≥ k (e.g., Kneser graph KG3k,k )

Graphs with high cop-number

Large girth (smallest cycle) AND large min degree ⇒ large cop-number

G with min-degree d and girth > 4 ⇒ cn(G) ≥ d . [Aigner and Fromme 84]

for any d , there are d-regular graphs G with cn(G) ≥ d [Aigner and Fromme 84]

cn(G) ≥ d t in any graph with min-degree d and girth > 8t − 3 [Frankl 87]

for any k, there is G with diameter 2 and cn(G) ≥ k (e.g., Kneser graph KG3k,k )

Meyniel’s Conjecture

∃ n-node graphs with min degree Θ(√n) and girth > 4

⇒ ∃ n-node graphs G with cn(G) = Ω(√n)

(e.g., projective plane, random√n-regular graphs)

Meyniel’s Conjecture

Conjecture: For any n-node connected graph G , cn(G) = O(√n) [Meyniel 85]

Link with Graph Structural Properties

Reminder: For any graph G , cn(G) ≤ γ(G) the dominating number of G .

Lemma [Aigner, Fromme 1984]

1 Cop is sufficient to “protect” a shortest path P in any graph.(after a finite number of steps, Robber cannot reach P)

⇒ cn(grid) = 2 (while γ(grid) ≈ n/5)

⇒ Cop-number related to structural properties

1 Cop can protect 1 shortest path: applications (1)

For any planar graph G (there is a drawing of G on the plane without crossing edges),there exists separators consisting of ≤ 3 shortest paths

Cop-number vs. graph structure a surprising (?) example

cn(G) ≤ 3 for any planar graph G [Aigner and Fromme 84]

G with genus ≤ g : can be drawn on a surface with ≤ g “handles”.

Cop-number vs. graph structure let’s go further

cn(G) ≤ b 3g2c+ 3 for any graph G with genus ≤ g [Schroder, 01]

Conjectures: cn(G) ≤ g + 3? cn(G) ≤ 3 if G has genus 1?

G with genus ≤ g : can be drawn on a surface with ≤ g “handles”.

Cop-number vs. graph structure let’s go further

cn(G) ≤ b 3g2c+ 3 for any graph G with genus ≤ g [Schroder, 01]

Conjectures: cn(G) ≤ g + 3? cn(G) ≤ 3 if G has genus 1?

Lemma shortest-path-caterpillar = closed neighborhood of a shortest path [Chiniforooshan 2008]

5 Cops are sufficient to “protect” 1 shortest-path-caterpillar in any graph.

Any graph can be partitioned in n/ log n shortest-path-caterpillars (consider a BFS)

For any graph G , cn(G) = O(n/ log n) [Chiniforooshan 2008]

5 Cops are sufficient to “protect” 1 shortest-path-caterpillar in any graph.Any graph can be partitioned in n/ log n shortest-path-caterpillars (consider a BFS)

Progress on Meyniel’s ConjectureMeyniel’s Conjecture [85]: For any n-node connected graph G , cn(G) = O(

cndominating set ≤ k ≤ k [folklore]

treewidth ≤ t ≤ t/2 + 1 [Joret, Kaminski,Theis 09]

chordality ≤ k < k [Kosowski,Li,Nisse,Suchan 12]

genus ≤ g ≤ b 3g2c+ 3 (conjecture ≤ g + 3) [Schroder, 01]

H-minor free ≤ |E(H)| [Andreae, 86]

degeneracy ≤ d ≤ d [Lu,Peng 12]

diameter 2 O(√n) −

bipartite diameter 3 O(√n) −

Erdos-Renyi graphs O(√n) [Bollobas et al. 08] [Luczak, Pralat 10]

Power law O(√n) [Bonato,Pralat,Wang 07]

A long story not finished yet...

cn(G) = O( nlog log n

) [Frankl 1987]

cn(G) = O( nlog n

) [Chiniforooshan 2008]

cn(G) = O( n

2(1−o(1))√

log n) [Scott, Sudakov 11, Lu,Peng 12]

note that n

2(1−o(1))√

log n≥ n1−ε for any ε > 0

Conclusion / Open problems

Meyniel’s Conjecture [1985]: For any n-node connected graph G , cn(G) = O(√n)

Conjecture [?]: For any n-node connected graph G with genus g , cn(G) ≤ g + 3

simpler(?) questions

cn(G) ≤ 3 if G has genus ≤ 1?

Directed graphs ?

Many other variants and questions... (e.g. [Clarke’09] [Bonato, et a.’13]...)

B. Alspach. Searching and sweeping graphs: a brief survey. In Le Matematiche, pages 5-37, 2004.W. Baird and A. Bonato. Meyniel’s conjecture on the cop number: a survey. http://arxiv.org/abs/1308.3385. 2013A. Bonato and R. J. Nowakowski. The game of Cops and Robber on Graphs. American Math. Soc., 2011.

Many Other Variants

Other VariantsVariant DifferencesFast Cops and Robbers Robber moves at speed sCops and Robbers at a Distance Robber can be captured at distance dLazy Cops and Robbers Only one cop may move each turnCops and Invisible Robbers Robber is invisibleCops and Robbers (capture time) Minimize capture time

Other Games on Graphs

Eternal Domination Game

Spy Game

Firefighting

Angel Problem

The Game of Cops and Robbers on Graphs - Inria · 2017-11-06 · 1/19 The Game of Cops and Robbers...

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