Cops and Robbers Cops and Robbers on Geometric Graphs Andrew Beveridge joint work with Andrzej Dudek, Alan Frieze, Tobias Müller Department of Mathematics, Statistics and Computer Science Macalester College SIAM Discrete Mathematics Meeting Dalhousie University 20 June 2012
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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?