1. Graph Searching Games and Probabilistic Methods
PIMS-University of Manitoba Distinguished Lecture
Graph Searching Games and Probabilistic Methods
Anthony Bonato
Ryerson University

2. Probabilistic graph searching - Anthony Bonato

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Cops and Robbers C C R
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Cops and Robbers C C R C
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Cops and Robbers C C R C
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Cops and Robbers C R C C
cop number c(G) = 3
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9. Probabilistic graph searching - Anthony Bonato
Cops and Robbers
played on an undirected graph G
two players Cops C and robber R play at alternate time-steps (cops first) with perfect information
players move to vertices along edges; may move to neighbors or pass
cops try to capture (i.e. land on) the robber, while robber tries to evade capture
minimum number of cops needed to capture the robber is the cop number c(G)
well-defined as c(G) ≤ |V(G)|
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10. Applications of Cops and Robbers
robotics
AI & mobile computing
gaming
Network interdiction
intercepting messages
or agents
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11. Probabilistic graph searching - Anthony Bonato
power of cops:
traps, photo radar, walls, capture from distance, teleportation, tandem-win, lazy cops
power of robber:
speed: fast or infinite, invisible, decoys, barricades, damage, capture cops
vertex pursuit games/processes:
Firefighting, watchman’s problem, eternal domination, seepage, robot vacuum, robot crawler, acquaintance time, Angel and Devil, burning, graph cleaning, Revolutionaries and Spies, …
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12. Randomness in vertex pursuit
DETERMINISTIC GAME/PROCESS
RANDOM
GAME/PROCESS
DETERMINISTIC BOARD
Classical model
Cops and drunk Robber, burning, Zombies and Survivors, …
RANDOM BOARD
Cops and Robbers,
Firefighter, Seepage
Robot vacuum, acquisition number, graph cleaning, acquaintance time, toppling number… ?
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13. Zombies

14. Miniconference on the Mathematics of Computation
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15. Miniconference on the Mathematics of Computation
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16. Miniconference on the Mathematics of Computation
Zombie horde
up to n/2 - 2 zombies on an induced path will never capture the survivor on a cycle
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17. The game

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Zombies and Survivors
set of zombies, one survivor
players move at alternate ticks of the clock, from vertex to vertex along edges
zombies choose their initial locations u.a.r.
at each step the zombies move along a shortest path connected to the survivor
if more than one such path, then they choose one u.a.r.
zombies win if one or more can eat the survivor
land on the survivor’s vertex
otherwise, survivor wins
NB: zombies have no strategy!
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19. (B,Mitsche,Perez-Gimenez,Pralat,16)
sk(G): probability survivor wins if k zombies play, assuming optimal play
sk+1 (G) ≤ sk (G) for all k, and sk(G) → 0 as k → ∞
zombie number of G is
z(G) = min{k ≥ c(G): sk(G) ≤ ½}
well-defined
z(G) represents the minimum number of zombies such that the probability that they eat the survivor is > ½
note that c(G) ≤ z(G)
Z(G) = z(G) / c(G): cost of being undead
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with probability
1− 5 n k = exp − 5k n +𝑂 k n 2
all zombies begin outside the cycle
implies:
𝑧 G ~ log 2 5 n
and so
𝑍 G ~ log n
n-5 leaves G
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21. Zombie number of cycles
Theorem (BMPGP,16) If n ≥ 27, then z(Cn) = 4, so Z(Cn) = 2.
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Cartesian grids
(Tosic, 87) c(G H) ≤ c(G) + c(H)
Theorem (BMPGP,16) For n ≥ 2, z(Pn Pn) = 2.
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Toroidal grids
Tn = Cn Cn
(Neufeld, 90): c(Tn) = 3
Theorem (BMPGP,16)
Let ω = ω(n) be a function tending to infinity with n.
Then a.a.s.
𝑧 𝑇 𝑛 ≥ 𝑛 / (ω log n).
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24. Toroidal grids, continued
despite the lower bound, no known subquadratic upper bound is known for the zombie number of toroidal graphs!
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25. In the beginning…

27. G(n,p) random graph model (Erdős, Rényi, 63)
Miniconference on the Mathematics of Computation
G(n,p) random graph model (Erdős, Rényi, 63)
p = p(n) a real number in (0,1), n a positive integer
G(n,p): probability space on graphs with nodes {1,…,n}, two nodes joined independently and with probability p 1 2 3 4 5
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Cop number of G(n,p)
in G(n,p), the cop number is a random variable
Theorem (Bonato, Hahn, Wang, 07)
For 0 < p < 1 a constant, then a.a.s.
c(G(n,p)) = Θ(log n).
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Sketch of proof
upper bound: uses (Dreyer, 00) bound for domination number of G(n,p)
lower bound:
G is (1,k)-e.c. if for all x and S with |S| ≤ k, there is a node z such
that R x S C z
G (1,k)-e.c. implies c(G) ≥ k
a.a.s. G(n,p) is (1,k)-e.c., where k = (1-ε)log1/1-pn
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Zig-zag
for random graphs G(n,p) with p = p(n), the asymptotic behaviour of the cop number is more complicated (Prałat, Łuczak,10)
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31. Cop-win

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Cop-win graphs
node u is a corner if there is a v such that N[v] contains N[u]
v is the parent; u is the child
a graph is dismantlable if we can iteratively delete corners until there is only one vertex
Theorem (Nowakowski, Winkler 83; Quilliot, 78)
A graph is cop-win if and only if it is dismantlable.
Idea: cop-win graphs always have corners; retract corner
and play shadow strategy
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Dismantlable graphs
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Dismantlable graphs
unique corner!
part of an infinite family that maximizes capture time (Bonato, Hahn, Golovach, Kratochvíl,09)
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35. Typical cop-win graphs
what is a random cop-win graph?
G(n,1/2) and condition on being cop-win
probability of choosing a cop-win graph on the uniform space of labeled graphs of ordered n
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Universal vertices
P(cop-win) ≥ P(universal)
= n2-n+1 – O(n22-2n+3)
= (1+o(1))n2-n+1
…this is in fact the correct answer!
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37. Almost all cop-win graphs
Theorem (B,Kemkes, Prałat,12) In G(n,1/2), P(cop-win) = (1+o(1))n2-n+1
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Corollary
Un = number of labeled graphs with a universal
vertex
Cn = number of labeled cop-win graphs
Corollary (BKP,12)
lim 𝑛→∞ 𝑈 𝑛 𝐶 𝑛 = 1.
That is, almost all cop-win graphs contain a
universal vertex.
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39. How big can the cop number be?

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c(n) = maximum cop number of a connected graph of order n
Meyniel Conjecture: c(n) = O(n1/2).
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State-of-the-art
(Lu, Peng, 13) proved that
independently proved by (Frieze, Krivelevich, Loh, 11) and (Scott, Sudakov,11)
all these proofs use the probabilistic method
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For random graphs
(Bollobás, Kun, Leader,13): if
p = p(n) ≥ 2.1log n/ n, then a.a.s.
c(G(n,p)) ≤ n1/2log n
(Prałat,Wormald,16): proved Meyniel’s conjecture a.a.s. for all p = p(n)
(Prałat,Wormald,17+): holds a.a.s. for random d-regular graphs, for d ≥ 3
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How close to n1/2?
consider a finite projective plane P
two lines meet in a unique point
two points determine a unique line
exist 4 points, no line contains more than two of them
q2+q+1 points; each line (point) contains (is incident with) q+1 points (lines)
incidence graph (IG) of P:
bipartite graph G(P) with red nodes the points of P and blue nodes the lines of P
a point is joined to a line if it is on that line
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Example
Fano plane
Heawood graph
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46. Meyniel extremal families
a family of connected graphs (Gn: n ≥ 1) is Meyniel extremal if there is a constant d > 0, such that for all n ≥ 1, c(Gn) ≥ dn1/2
IG of projective planes: girth 6, (q+1)-regular, so have cop number ≥ q+1
order 2(q2+q+1)
Meyniel extremal (must fill in non-prime orders)
other examples of Meyniel extremal families come from combinatorial designs and finite geometries (B,Burgess,2013)
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(BB,13) New ME families
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Polarity graphs
suppose PG(2,q) has points P and lines L. A polarity is a function π: P→ L such that for all points p,q, p ϵ π(q) iff q ϵ π(p).
eg of orthogonal polarity: point mapped to its orthogonal complement
polarity graph: vertices are points, x and y adjacent if
xϵ π(y)
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49. Properties of polarity graphs
order q2+q+1
(q,q+1)-regular
C4-free
diameter 2
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Meyniel Extremal
Theorem (Bonato,Burgess,13) Let q be a prime power. If Gq is a polarity graph of PG(2, q), then
q/2 ≤ c(Gq) ≤ q + 1.
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Lower bounds
Theorem (Bonato, Burgess,13)
If G is connected and K2,t-free, then
c(G) ≥ δ(G) / t.
applies to polarity graphs: t = 2
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52. Capture time

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Capture time of a graph
the length of Cops and Robbers was considered first as capture time
(B,Hahn,Golovach,Kratochvíl,09) capture time of G: length of game with c(G) cops assuming optimal play, written capt(G)
if G is cop-win, then capt(G) ≤ n - 3 if n ≥ 7 (see also (Gavanciak,10))
capt(G) ≤ n/2 for many families of cop-win graphs including trees, chordal graphs
examples of planar graphs with capt(G) = n - 3
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Trees are cop-win C
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Capture time of trees
Lemma (B, Perez-Gimenez,Reiniger,Prałat,17):
For a tree T, we have that
capt(T) = rad(T).
Proof sketch:
for capt(T) ≤ rad(T), place C on a central vertex and use the zombie strategy
for rad(T) ≤ capt(T), notice that any other initial placement of C results in R choosing a vertex distance
> rad(T) away
R stays put
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Hypercubes
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57. Cop number of products of trees
Theorem (Maamoun,Meyniel,87): The cop number of the Cartesian product of d trees is d
no reference to the length of the game; i.e capture time of grids or the hypercube
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58. Capture time of Cartesian grids
Theorem (Merhabian,10): The capture time of the Cartesian product of two trees T1 and T2 is diam(T1) + diam(T2)) / 2 . In particular, the capture time of the m x n Cartesian grid is (m + n)/2−1 .
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59. Capture time of hypercubes
Theorem (B,Gordinowicz,Kinnersley,Prałat,13) The capture time of Qn is Θ(nlog n).
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Lower bound
Theorem (BGKP,13) For b > 0 a constant, a robber can escape nb cops for at least
(1-o(1))1/2 n log n
rounds.
probabilistic method: play with a random/drunk robber
Coupon collector and large deviation bounds
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61. Proof of lower bound (sketch)
let T= 1/2(n-1)log n, ε = ln((4d+1) ln n) / ln n = o(1).
show that a random robber can play (1- ε)T rounds without being captured
can play initial round due to expansion
next consider a single cop C playing greedily
can show process of C capturing R is equivalent to the coupon collector problem
using a deviation bound, the probability single cop captures robber is exp(-(n/2)ε/4); via union bound for all nd cops this is o(1)
hence, there is SOME deterministic strategy for the robber
to survive (1- ε)T rounds
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Add more cops!
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k-capture time
define captk(G), where c(G) ≤ k ≤ γ(G)
k-capture time
capt(G) = captc(G)(G)
temporal speed-up:
as c(G) increases to γ(G),
captk(G) monotonically decreases to 1
if k > c(G), we call this Overprescribed Cops and Robbers
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Trees
for k > 0, metric k-center is a set S, |S| ≤ k, that minimizes
max 𝑣∈𝑉(𝐺) 𝑑(𝑣,𝑆)
radk(G) is this minimum
k = 1, then radk(G) = rad(G)
NP-complete to find metric k-centers (Vazirani,03)
radk is monotone on retracts
retractions monotone on walk length
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Example: k = 1
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Example: k = 2
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Example: k = 3
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Example: k = 4
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Example: k = 5 = 𝛾
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70. Retracts following theorem is key: Theorem (BGRP,17)
Suppose V can be decomposed into t-many vertex sets of retracts Gi of G. Then
captk(G)≤ max 1≤𝑖≤𝑡 captk(Gi)
Idea: Play shadow strategy in each retract.
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Trees
Corollary (BGRP,17)
For a tree T,
captk(G) = radk(G).
Idea: cover by balls (which are retracts) around vertices around metric k-center and use theorem
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Square grids
G(d,n) = d-dimensional Cartesian n-grid
(Maamoun,Meyniel,87): c(G(d,n)) = d+1 2
(Merhabian,10): capt(G(d,n)) = 1 2 nd log 2 𝑑
𝛾(G(d,n)) = Θ(nd)
use dominating sets of paths order Θ(n)
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73. k-capture time of grids
Theorem (BGRP,17) If k = O(nd), then captk(G(d,n)) = Θ(n/k1/d). Idea: cover by sub-grids (retracts) and use theorem
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74. Domination number of hypercubes
𝛾 𝑄 𝑛 is open for general n
𝛾 𝑄 𝑛 ≤ 2 𝑛−3 if n ≥ 7 n
𝛾 𝑄 𝑛 3 2 4 5 7 6 12
n= 2k-1, 2k
2n-k
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75. Temporal speed up in hypercubes
Theorem (BGRP,17)
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place cops randomly in the hypercube
cops are sufficiently dense, can occupy some Nd(R)
a.a.s. apply Hall’s condition to find a perfect matching between Nd(R) and Nd-1(R) d C
d-1 R C C C
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place cops randomly in the hypercube
cops are sufficiently dense, can occupy some Nd(R)
a.a.s. apply Hall’s condition to find a perfect matching between Nd(R) and Nd-1(R)
cops move along this matching and “close in” on R d C
d-1 R C C C
Probabilistic graph searching - Anthony Bonato

78. Where to next with Cops and Robbers?
Meyniel’s conjecture
Soft Meyniel’s conjecture: for some ε > 0,
c(n) = O(n1-ε).
topological graph theory
Schroeder’s conjecture
Lazy Cops and Robbers
planar graphs
invisible robber
0-visibility, limited visibility, hyperopic cops, localization game
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79. Contact Web: http://www.math.ryerson.ca/~abonato/
Blog:
@Anthony_Bonato
Zombies and Survivors

80. Thank you!