cops and robbers on geometric graphsabonato/presentations/siam_ab... · 2012. 6. 24. · cops and...

Cops and Robbers

Cops and Robbers on Geometric Graphs

Andrew Beveridgejoint work with

Andrzej Dudek, Alan Frieze, Tobias Müller

Department of Mathematics, Statistics and Computer ScienceMacalester College

SIAM Discrete Mathematics MeetingDalhousie University

20 June 2012

Cops and Robbers

Outline

1 Game of Cops and RobbersIntroductionAn Example GameThe Official Rules

2 Planar Graphs

3 Geometric GraphsRGG with c(G) = 1RGG with c(G) ≤ 2

4 Open Problems

Cops and Robbers

Game of Cops and Robbers

Introduction

The Game of Cops and Robbers

Introduced independently by Quilliot (1978), and Nowakowski &Winkler (1983).

Cops and Robbers

Game of Cops and Robbers

An Example Game

An Example of 2 Cop Pursuit

C1

C2

R

First the cops C1,C2 choose their starting locations.Next, the robber R chooses his starting location.

Cops and Robbers

Game of Cops and Robbers

An Example Game

2 Cop Pursuit: Turn One, Cop Move

C1

C2

R

First turn: Cops move first. Each can move to an adjacentvertex (or stay where he is).

Cops and Robbers

Game of Cops and Robbers

An Example Game

2 Cop Pursuit: Turn One, Robber Move

C1

C2

R

First turn: Robber moves second. He moves similarly.

Cops and Robbers

Game of Cops and Robbers

An Example Game

2 Cop Pursuit: Turn Two, Cop Move

C1

C2

R

Second turn: Cops move. Here C1 chooses to remainstationary.

Cops and Robbers

Game of Cops and Robbers

An Example Game

2 Cop Pursuit: Turn Two, Robber Move

C1

C2

R

Second turn: Robber move.

Cops and Robbers

Game of Cops and Robbers

An Example Game

2 Cop Pursuit: Turn Three, Cop Move

C1

C2R

Third turn: Cop move.

Cops and Robbers

Game of Cops and Robbers

An Example Game

2 Cop Pursuit: The Arrest

You havethe rightto remainsilent!

C1

C2R

Third turn: Cop move, giving a Cop Win!

Cops and Robbers

Game of Cops and Robbers

The Official Rules

How to Play Cops and Robbers

Set Up:Chose a graph G and a positive integer k .Cops C1,C2, . . . ,Ck are placed on vertices of GNext, the robber R is placed on a vertex of G.

A Game Turn:Each cop moves to an adjacent vertex, or remains in placeNext, the robber moves similarly.

Victory Conditions:Cops: one cop becomes co-located with the robberRobber: evades capture forever

Cops and Robbers

Game of Cops and Robbers

The Official Rules

How to Study Cops and Robbers

Pick a graph G and a number of cops k .Who has a winning strategy: the k cops or the robber?

DefinitionThe cop number c(G) = fewest number of cops with a winningstrategy on G.

The cop number is always defined, since |V | cops trivially win.

Objective: Given a graph G, determine its copnumber c(G).

Cops and Robbers

Planar Graphs

Cops and Robbers on Planar Graphs

Cops and Robbers

Planar Graphs

Planar G has c(G) ≤ 3

Theorem (Aigner and Fromme (1984))

If G is a planar graph then c(G) ≤ 3.

Proof Idea:Cops work together to reduce the robber’s free territory(vertices where R can move without being caught on thevery next cop move).Eventually the robber is caught.

Cops and Robbers

Planar Graphs

Path Guarding Lemma

Lemma (Path Guarding Lemma (Aigner and Fromme, 1984))

Let P be a shortest (u, v)-path on any graph G. Eventually, acop C can position himself on P so that if R steps onto P, thenhe is caught on the very next cop move.

Proof Sketch:Cop moves onto P.Cop moves along P until he reaches the vertex on Pclosest to R.The path is now guarded

If R can beat C to a vertex on P, then there must be aneven shorter (u, v)-path.

Cops and Robbers

Planar Graphs

Planar Graphs have c(G) ≤ 3

The winning 3-cop strategy for any planar graph.

u

v

First, cop C1 takes control of a shortest (u, v)-path P1.

Cops and Robbers

Planar Graphs

Planar Graphs have c(G) ≤ 3

The winning 3-cop strategy for any planar graph.

u

v

Next, cop C2 takes control of a shortest (u, v)-path P2 inG − P1.

Cops and Robbers

Planar Graphs

Planar Graphs have c(G) ≤ 3

The winning 3-cop strategy for any planar graph.

u

v

R

Robber R is trapped inside or outside the cycle P1 + P2.

Cops and Robbers

Planar Graphs

Planar Graphs have c(G) ≤ 3

The winning 3-cop strategy for any planar graph.

u

v

R

Cop C3 controls a shortest path P3 in the robber territory ofG − P1 − P2.

Cops and Robbers

Planar Graphs

Planar Graphs have c(G) ≤ 3

The winning 3-cop strategy for any planar graph.

u

v

R

C1

The choice of P3 actually frees up cop C1, who takes control ofa shortest path in the smaller robber territory inside P2 + P3.

Cops and Robbers

Geometric Graphs

Cops and Robbers on Geometric Graphs

Cops and Robbers

Geometric Graphs

Geometric Graphs

Let x1, x2, . . . , xn be points in R2.Let r ∈ R+

The geometric graph

G(x1, x2, . . . , xn; r)

is the graph with vertices

V = x1, x2, . . . , xn

with(xi , xj) ∈ E ⇐⇒ ‖xi − xj‖ ≤ r .

Cops and Robbers

Geometric Graphs

Geometric Graph Example

Cops and Robbers

Geometric Graphs

Geometric Graph Example

Cops and Robbers

Geometric Graphs

Geometric Graph Example

Cops and Robbers

Geometric Graphs

Controlling Shortest Paths on Geometric Graphs

C

R

One cop controlling a shortest path of a geometric graph doesnot necessarily trap the robber!

Cops and Robbers

Geometric Graphs

Geometric Graphs have c(G) ≤ 9

Lemma (B, Dudek, Frieze, Müller, (2012+))Three cops moving in tandem on a shortest path can preventthe robber from crossing.

u v

G

CC− C+

R

Theorem (B, Dudek, Frieze, Müller (2012+))

If G is a geometric graph then c(G) ≤ 9.

Cops and Robbers

Geometric Graphs

3 Cops Patrolling a Geometric Shortest Path

Three cops moving in tandem on a shortest path can preventthe robber from crossing.

C− CC+

vi−2vi−1 vi

vi+1 vi+2

R must cross between vi−2 and vi+2

R cannot lie in the gray regionB(vi−2, r) ∪ B(vi , r) ∪ B(vi+2, r).

Cops and Robbers

Geometric Graphs

A Geometric Graph on 1440 vertices with c(G) = 3

Aigner and Fromme (1984) : A graph with minimum degreek and girth ≥ 5 has c(G) ≥ kCreate an annular graph with repeating interior 5-cycles.Minimum degree is 3 and the girth is 5.

3 ≤ maxgeometric

c(G) ≤ 9

Cops and Robbers

Geometric Graphs

Random Geometric Graphs

Random Geometric Graph (RGG for short)

Pick n random points in[0,1]2

Consider the radius r(n)as a function of nStudy expected behavioras n→∞

Cops and Robbers

Geometric Graphs

RGG with c(G) = 1

RGGs with c(G) = 1

Theorem (BDFM (2012+))

A RGG on [0,1]2 with

r5(n) > K1log n

n

satisfies c(G) = 1 whp.

whp⇐⇒ with high probability⇐⇒ limn→∞ Pr[A] = 1

Note: This result was proven independently by Alon and Prałatusing a “lion’s move” strategy

Cops and Robbers

Geometric Graphs

RGG with c(G) = 1

Pitfall Vertex

neighborhood of v

N(v) = u ∈ V | (u, v) ∈ E)

closed neighborhood of v

N(v) = N(v) ∪ v

v ∈ V is a pitfall (or corner) if ∃w ∈ V such that

N(v) ⊆ N(w).

v

w

Cops and Robbers

Geometric Graphs

RGG with c(G) = 1

Cop Win = Dismantlable

Lemma (Quilliot, and Nowakowski & Winkler)

If v is a pitfall then c(G) = c(G − v).

G is dismantlable if G can be reduced to a single vertex bysuccessively removing pitfalls

⇒ ⇒ ⇒ ⇒

Theorem (Quilliot, and Nowakowski & Winkler)

c(G) = 1 if and only if G is dismantlable.

Cops and Robbers

Geometric Graphs

RGG with c(G) = 1

RGGs with r5(n) > K1log n

n have c(G) = 1

Claim: the RGG is dismantlable whp.

Let Nc(xi) = all neighbors of xi closer to c = (1/2,1/2).

Lemma

Suppose that r5(n) > K1log n

n . Then whp, for every xi such that‖xi − c‖ ≥ r/2, there exists xj ∈ Nc(xi) such that Nc(xi) ⊂ N(xj).

c

x1

c

xi

xj

Cops and Robbers

Geometric Graphs

RGG with c(G) = 1

RGGs with r5(n) > K1log n

n have c(G) = 1

c

x1

c

xi

xj

Using this lemma, dismantle the RGG like peeling an onion.

Order vertices by distance to cFurthest vertex is a pitfall by the lemmaRepeat until only points left are within r/2 of c.These vertices form a clique, which is dismantlable.

Cops and Robbers

Geometric Graphs

RGG with c(G) = 1

Why r5 = K1 log n/n?

cxixj c′ c xi

W

r

If xj ∈W then Nc(xi) ∈ N(xj)

Area(W ) = Ω(r5)

Cops and Robbers

Geometric Graphs

RGG with c(G) ≤ 2

RGGs with c(G) ≤ 2

Theorem (BDFM (2012+))A RGG with

r4(n) > K2log n

nsatisfies c(G) ≤ 2 whp.

Proof idea: two cops can eventually chase the robber intoa corner of [0,1]2.Why r4 > K2 log n/n?

Enables the cop to get within s ≈ r2 of a target pointEnsures that the pursuit does not go on too long (soaccumulated error is not too bad)

Cops and Robbers

Geometric Graphs

RGG with c(G) ≤ 2

RGGs with r4(n) > K2log n

n have c(G) ≤ 2 whp

C1C2

R

Cops C1,C2 start near (0,0)

The robber lines L1,L2 are the vertical and horizontallines through R

Cops and Robbers

Geometric Graphs

RGG with c(G) ≤ 2

RGGs with r4(n) > K2log n

n have c(G) ≤ 2 whp

C1C2

R

Cops C1,C2 start near (0,0)

The robber lines L1,L2 are the vertical and horizontallines through RPhase 1: The cops get within s := 5(log n/n)1/2 of L1,L2near the axes.

Cops and Robbers

Geometric Graphs

RGG with c(G) ≤ 2

RGGs with r4(n) > K2log n

n have c(G) ≤ 2 whp

C1

C2R

Phase 1: The cops get within s := 5(log n/n)1/2 of L1,L2near the axes.

Cops and Robbers

Geometric Graphs

RGG with c(G) ≤ 2

RGGs with r4(n) > K2log n

n have c(G) ≤ 2 whp

C1

C2R

P1

P2

Phase 1: The cops get within s := 5(log n/n)1/2 of L1,L2near the axes.Define P1 (P2) to be the points r/3 below (left of) R.Phase 2: The cops get within s := 5(log n/n)1/2 of P1,P2while staying near L1,L2.

Cops and Robbers

Geometric Graphs

RGG with c(G) ≤ 2

RGGs with r4(n) > K2log n

n have c(G) ≤ 2 whp

C1

C2 R

Phase 1: The cops get within s := 5(log n/n)1/2 of L1,L2near the axes.Phase 2: The cops get within s := 5(log n/n)1/2 of P1,P2while staying near L1,L2.Phase 3: R is forced into the upper right corner by C1,C2and eventually caught.

Cops and Robbers

Geometric Graphs

RGG with c(G) ≤ 2

RGGs with r4(n) > K2log n

n have c(G) ≤ 2 whp

Why can the cops gain on R in Stage 2 and Stage 3?

C1,C2 close in C1 closes in C2 closes in toward corner

This bounds the number of moves required in Stage 2 andStage 3.

Cops and Robbers

Open Problems

Open Problems

Cops and Robbers

Open Problems

Summary of Results

Arbitrary geometric graphs in [0,1]2

3 ≤ maxgeometric

c(G) ≤ 9

Random geometric graphs in [0,1]2

r5 ≥ K1logn

n→ c(G) = 1

r4 ≥ K2logn

n→ c(G) ≤ 2

We have also shown:

r2 ≤ K3log2 n

n→ c(G) ≥ 2

Cops and Robbers

Open Problems

Summary and Open Problems

Arbitrary geometric graphs in [0,1]2

What is the true upper bound for c(G)? 9 feels too big; 6seems optimistic.

Random geometric graphs in [0,1]2

Additional upper and lower bounds on the cop numberIs our 2-cop result tight?