the game of cops and robbers
DESCRIPTION
This is the slides from my master's thesis defense. The topic is about the game of cops and robbers focusing on if we can get a better strategy than Aigner and Fromme's one. It concludes that in planar oriented setting, we can beat the strategy and opens up another question of establishing new bounds.TRANSCRIPT
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.
6
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
• 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.
11
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.
12
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.
13
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.
15
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
• 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
18
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.
20
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.
21
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 .
23
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.
24
Result (1) – AF strategy
• Using the AF strategy, we can find a planar undirected graph
such that the cop number is at least 3.
25
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?
26
Result (2) – Modified AF
• Let G be a strongly connected oriented graph with minimum
out-degree .
• Also, G does not contain cycles below.
• Then,
27
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.
29
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.
30
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.
31
Result (3) – Upper bound
• By continuously separating a planar oriented graph, we can
catch the robber and it needs cops.
32
Result (4) – Lower bound
• Because of Euler formula, the modified AF strategy cannot
show that cop number is higher than 3.
33
Result (4) – Lower bound
• Icosahedron → Truncated Icosahedron → Great
Rhombicosidodecahedron → Put shelter
41
Result (4) – Lower bound
• Icosahedron → Truncated Icosahedron → Great
Rhombicosidodecahedron → Put shelter → Give directions
(part 1)
43
Result (4) – Lower bound
• Icosahedron → Truncated Icosahedron → Great
Rhombicosidodecahedron → Put shelter → Give directions
(part 1) → Give directions (part 2)
45
Result (4) – Lower bound
• Icosahedron → Truncated Icosahedron → Great
Rhombicosidodecahedron → Put shelter → Give directions
(part 1) → Give directions (part 2) → Final Graph.
47
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.
49
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