The game of cops and robbers

Cops and Robbers:On Oriented Planar GraphsSi Young OhAdvisor: Professor. Po-Shen Loh

1

Overview

• Background:

• What is the game of cops and robbers?

• Issue:

• What is the problem? / What is known? / What do we want to

know?• Four results• Conclusion

2

Background

• Played by two distinct players.

• c cops and one robber

• Game is played on a graph G = (V, E).

• Perfect information game

3

Background

1. Cops choose where to put c cops.

Multiple cops may be on the same vertex.

2. The robber chooses a vertex.

3. Subset of cops move along edges.

4. Robber can either stay or move along a edge.

5. Repeat 3 and 4.

4

Background

• Cops win if one of cops is at the same vertex with the robber

at any moment.

• Robber wins if he can escape indefinitely.• It’s like a simple version of Pac-Man!

5

Background

• Graphs can be either undirected or oriented.

• Directed graphs are in between.

• Only connected undirected graphs are interesting.

• Only strongly connected oriented graphs are interesting.

• Multiple edges or self-loops are ignored.

• Non-empty subset of cops are moving on cops’ turn.

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Issue

• What is the smallest c(G), cop number, such that c cops are

enough to win on G?

7

Issue

• What is the smallest c(G), cop number, such that c cops are

enough to win on G?

2 2

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Issue

2 2

• What is the smallest c(G), cop number, such that c cops are

enough to win on G?

When there’s only one cop, the robber can indefinitely escape. 9

Issue

• How about this?

10

Issue

• Broader question:

• How does behave?

• It is conjectured by Meyniel that on undirected graphs.

• The best upper bound known so far is worse than for any

constant on both undirected and oriented graphs.

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Issue

• It is not trivial to find a cop number on a graph.

• It is not trivial to verify if c is a cop number on a graph.

• With c cops, is it possible for the cops to catch the robber for

every robber’s strategy?

• With (c - 1) cops, is it possible for the robber to escape

indefinitely for every cops’ strategy.

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Issue

• With n vertices and c cops, there are almost possible cop

positions.

• From each position, if each cop has k choices on a stage, there

are next stages to check.

• So, any naïve algorithms won’t work to verify if a strategy is

the best for cops.

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Issue

• The most notable strategy was suggested by Aigner and

Fromme (will be abbreviated as AF strategy):

• On an undirected graph G, if the minimum degree and there is

no 3- or 4-cycles on G, then .

• With small restrictions, the strategy guarantees high cop number.

• We can get the lower bound .

• None of the other strategies could make a better cop number.

14

Issue

• Known results on an undirected graph G with n vertices:

• It’s possible to generate a graphs so that .

• The best upper bound is .

• If G is planar, (this case is complete).

• Known results on an oriented graph G with n vertices:

• The best upper bound is .

• Little is known about games on oriented graphs.

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Issue

• Four goals of the talk:

1. Study known strategies to see advantages and disadvantages

focusing on the AF strategy.

2. Establish lower bound on general oriented graphs by modifying

the AF strategy.

3. Establish upper bounds on planar oriented graphs using Lipton

and Tarjan’s separator theorem.

4. Establish lower bounds on planar oriented graphs that does

better than the modified AF strategy. 16

Result (1) – AF strategy

• Why does it work?

• What happens when there are k – 1 cops?

d

1

2

𝑑≥𝑘

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Result (1) – AF strategy

• The robber simply stays when none of cops can catch the

robber immediately.

• The neighbor v is not available when

• A cop is at v

• A cop is at a neighbor of v

• One cop can block at most one

neighbor.d

1

2

𝑑≥𝑘

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Result (1) – AF strategy

• So, wherever the robber is, he can escape without worrying

about how cops are distributed.

• How can we use the strategy?

• It is possible for every prime p, that there exists a projective plane

of order .

• Set P = set of points, L = set of lines. ( each)

• G = (V, E) and V = P L and E = {(u,v) | u P, v L such that v contains

u}.

• Then, no 3- or 4-cycle with every vertex has degree q + 1.19

Result (1) – AF strategy

• Can we make it better?

• Without 3- or 4-cycles, with the smallest degree k, we need at

least vertices.

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Result (1) – AF strategy

• Is there some nice properties about 3- or 4-cycles?

• We first suspected that 3-cycles are good for cops.

• Both cops and a robber has same advantage when moving.

• Cops can guard vertices more efficiently.

• For example, a 4-cycle has cop number 2 but adding a diagonal

edge reduce the cop number to 1.

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Result (1) – AF strategy

• How does 3-cycles affect the cop number?

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Result (1) – AF strategy

• Conclusions about the lower bound:

• Even studying how 3-cycle affects the cop number is non-trivial.

• Without the AF strategy, we should worry about how cops are

distributed.

• It is hard to find a rule to represent the distribution on general

graphs.

• The cop number has the lower bound .

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Result (1) – AF strategy

• Cop number is at most 3 on planar undirected graphs proven

by Aigner and Fromme.

• The proof uses an algorithm to block a shortest path.

• Two disjoint paths separate a graph into two parts.

• Three cops can cooperate efficiently.

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Result (1) – AF strategy

• Using the AF strategy, we can find a planar undirected graph

such that the cop number is at least 3.

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Result (1) – AF strategy

• Conclusions about AF strategy on undirected graphs:

• Strong strategy to prove that cop number is higher than certain

numbers.

• It finds the best bound (known so far) on both general and planar

undirected graphs.

• Interesting to note that AF strategy on planar graphs cannot show

that cop number is higher than 3 because of Euler formula.

• Does it always find the best bound?

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Result (2) – Modified AF

• Let G be a strongly connected oriented graph with minimum

out-degree .

• Also, G does not contain cycles below.

• Then,

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Result (2) – Modified AF

• Why does it work?

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Result (2) – Modified AF

• Using the Euler cycle on a projective plane, we can generate

graphs with .

• Due to the same issues with undirected graphs, it is non-trivial

to find a better bound.

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Result (2) – Modified AF

• On oriented graphs, the shortest path argument does not hold

anymore.

• So, the theorem that the cop number is at most 3 on planar

undirected graphs does not hold on planar oriented graphs.

• We need to establish new bounds.

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Result (3) – Upper bound

• Lipton and Tarjan’s separator theorem:

• On planar graphs with n vertices, it is possible to partition

vertices in three sets A, B, and C such that

and

• There is no edge between A and B.

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Result (3) – Upper bound

• By continuously separating a planar oriented graph, we can

catch the robber and it needs cops.

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Result (4) – Lower bound

• Because of Euler formula, the modified AF strategy cannot

show that cop number is higher than 3.

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Result (4) – Lower bound

• Icosahedron

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Result (4) – Lower bound

35

Result (4) – Lower bound

• Icosahedron → Truncated Icosahedron

36

Result (4) – Lower bound

37

Result (4) – Lower bound

• Icosahedron → Truncated Icosahedron → Great

Rhombicosidodecahedron

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Result (4) – Lower bound

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Result (4) – Lower bound

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Result (4) – Lower bound

• Icosahedron → Truncated Icosahedron → Great

Rhombicosidodecahedron → Put shelter

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Result (4) – Lower bound

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Result (4) – Lower bound

• Icosahedron → Truncated Icosahedron → Great

Rhombicosidodecahedron → Put shelter → Give directions

(part 1)

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Result (4) – Lower bound

44

Result (4) – Lower bound

• Icosahedron → Truncated Icosahedron → Great

Rhombicosidodecahedron → Put shelter → Give directions

(part 1) → Give directions (part 2)

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Result (4) – Lower bound

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Result (4) – Lower bound

• Icosahedron → Truncated Icosahedron → Great

Rhombicosidodecahedron → Put shelter → Give directions

(part 1) → Give directions (part 2) → Final Graph.

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Result (4) – Lower bound

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Result (4) – Lower bound

• What happens when there are three cops?

• Initial:

• There are 20 shelters and one cop can block at most one shelter.

• The robber can choose a safe shelter.

• The robber waits until one cop is threatening.

• Three possible scenarios.

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Result (4) – Lower bound

• Scenario 1. All three cops are on the same unit.

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Result (4) – Lower bound

• Scenario 2. Two cops are on the same unit.

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Result (4) – Lower bound

• Scenario 3. Only one cop is on the same unit.

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Result (4) – Lower bound

• Conclusions on planar oriented graphs:

• It is possible that the cop number is at least 4.

• The upper bound is .

• We hit the same challenging issue about bounds.

53

Conclusions

• It is challenging to reduce the upper bound.

• Since independent cops are not strong enough, we should find

comprehensive rules of cooperation on all the graphs.

• It is challenging to increase the lower bound.

• We should find a rule to generate robber-favorable graphs as the

number of vertices increase indefinitely.

• We should show all the cops’ strategies do not work.

• Still, the result shows that it might be possible to make a

better robber’s strategy on non-planar graphs.54

Conclusions

• Questions?

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