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ON THE COP NUMBER OF GRAPHS OF HIGH GIRTH
PETER BRADSHAW, SEYYED ALIASGHAR HOSSEINI, BOJAN MOHAR∗,
AND LADISLAV STACHO†
Abstract. We establish a lower bound for the cop number of graphs of high girth in terms
of the minimum degree, and more generally, in terms of a certain growth condition. We show,
in particular, that the cop number of any graph with girth gand minimum degree δis at
least 1
g(δ−1)bg−1
4c. We establish similar results for directed graphs. While exposing several
reasons for conjecturing that the exponent 1
4gin this lower bound cannot be improved to
(1
4+ε)g, we are also able to prove that it cannot be increased beyond 3
8g. This is established
by considering a certain family of Ramanujan graphs. In our proof of this bound, we also
show that the “weak” Meyniel’s conjecture holds for expander graph families of bounded
degree.
1. Introduction
We consider the game of cops and robbers, a two-player game played with perfect informa-
tion on a finite connected graph G. In the game, the first player controls a team of cops, and
the second player controls a robber. At the beginning, the first player places each cop on her
team at a vertex of G, and then the second player places the robber at a vertex of G. The
two players take turns. On the first player’s turn, she may move each cop Cto a vertex in
the closed neighborhood of C, and on the second player’s turn, he may move the robber to
any vertex in the closed neighborhood of the robber’s current vertex. The first player wins
the game if a cop moves to the same vertex as the robber; in this case, the cop captures the
robber. The second player wins by letting the robber avoid capture indefinitely. In order
to establish a finite-time win condition, we may also say that the second player wins the
game if the same game position occurs twice; this change makes the game finite and does
not affect the strategy of either player.
The game of cops and robbers was introduced by Quilliot [19], and independently by
Nowakowski and Winkler [16]. Aigner and Fromme [1] introduced the concept of the cop
number c(G) of a graph G, which denotes the number of cops that the first player needs
to capture the robber on Gwith optimal play. Upper bounds on cop number are well-
understood for many classes of graphs. For instance, planar graphs and toroidal graphs have
cop number at most 3 [1, 11]. Recently, Bowler et al. [6] proved that the cop number of a
graph of genus gis at most 4
3g+10
3, thus improving an earlier bound of 3
2g+ 3 by Schr¨oder
[20]. In a similar flavor, Andreae [2] proved that excluding a graph Hwith tedges as a
minor in a graph Gimplies that c(G)≤t. There are also cop number bounds for highly
symmetrical graphs. For instance, Frankl [8] showed that normal Cayley graphs of degree
∗Supported in part by the NSERC Discovery Grant R611450 (Canada), and by the Research Project
J1-8130 of ARRS (Slovenia).
∗On leave from IMFM, Department of Mathematics, University of Ljubljana.
†Supported in part by the NSERC Discovery Grant R611368 (Canada).
1
arXiv:2005.10849v1 [math.CO] 21 May 2020
dhave cop number at most d. More recently, it was shown that abelian Cayley graphs
on nvertices have cop number at most 0.95√n+ 2 [7]. Lower bounds for cop number are
also known for several graph classes. For instance, projective plane incidence graphs with
2q2+ 2q+ 2 vertices have cop number at least q+ 1 [3], and certain abelian Cayley graphs
on nvertices have cop number Ω(√n), with some families achieving their cop number as
high as 1
2√n[9, 7]. Furthermore, Bollob´as et al. [4] and Pra lat and Wormald [17] showed
that random graphs in G(n, p) have cop number of order Θ(√n) a.a.s. Finally, for sparse
random graphs (random d-regular graphs with dconstant), an upper bound of O(√n) also
holds a.a.s. [18].
In this paper, we focus on lower bounds for the cop number of graphs of high girth. Aigner
and Fromme [1] proved that graphs with girth at least 5 and minimum degree δhave cop
number at least δ. Their ideas were extended later by Frankl [8], who showed more generally
that graphs with girth at least 8t−3 have cop number at least (δ−1)t+ 1. We seek to
improve these lowers bounds. Our main result, proved in Section 2, is the following lower
bound for cop number in terms of a graph’s girth gand minimum degree δ.
Theorem 1.1. Let t≥1be an integer, and let Gbe a graph of girth g≥4t+ 1 and with
minimum degree δ. Then c(G)>1
e t (δ−1)t.
Our bound is of the form Ω(g−1(δ−1)g/4), which is asymptotically superior to the afore-
mentioned bound of Frankl [8], which is of the form Ω((δ−1)g/8). One may naturally ask if
the coefficient 1
4of gis best possible. While we do not have an answer to this question, we
note that by assuming two well-known conjectures, we would be able to argue that 1
4is best
possible.
The following fundamental conjecture about cop number appeared in a paper by Frankl
[8] where it was referred to as a personal communication to the author by Henri Meyniel in
1985 (see also [3]).
Conjecture 1.2 (Meyniel’s Conjecture).For every connected graph Gof order n,c(G)≤
O(√n).
In [5], Bollob´as and Szemer´edi conjectured that for every sufficiently large g, there exists
a cubic graph of girth gwith Θ(2(1
2+o(1))g) vertices. The corresponding speculation for every
fixed degree dis also a folklore conjecture that is related to the Moore bound.
Conjecture 1.3 (Folklore).For any positive integers gand d, there exists a d-regular graph
Gof girth at least g, and order Θ((d−1)(1
2+o(1))g).
If we assume that a family of graphs with Θ((d−1)(1
2+o(1))g) vertices exists, and if we
assume that Meyniel’s conjecture holds for this family of graphs, then this family of graphs
must have cop number O((d−1)(1
4+o(1))g). It would then follow that for any lower bound
of the form Ω((d−1)cg) for the cop number of all graphs of girth gand minimum degree d,
c≤1
4, making our lower bound of 1
4best possible.
We use the same techniques as in the proof of Theorem 1.1 to show that a similar lower
bound applies to graphs of high girth with a certain growth condition, even when the mini-
mum degree may be small. We state this lower bound in Theorem 2.1. In Section 3, we apply
the techniques of Section 2 to directed graphs. The corresponding statements are given as
Theorems 3.5 and 3.6. An important observation from our proof of Theorem 3.5 is that it
2
is not girth that is important for the cop number of a digraph, but rather the novel notion
of the “trap distance” ρ∗(u, v) between the vertices of a digraph.
Finally, we compute an upper bound on the cop number of a family of Ramanujan graphs
in terms of their degree dand girth g. We state this bound in Theorem 4.6. This upper
bound, which is of order O((d−1)(3
8+o(1))g), compares quite favorably with our lower bound in
Theorem 1.1. The proof method used in establishing this upper bound works more generally
for arbitrary expander graphs.
2. Lower bounds for undirected graphs
In this section, we derive general lower bounds on the cop number of graphs of high girth.
This bound significantly improves the previous best lower bound of Frankl [8], which was
obtained in 1987.
Proof of Theorem 1.1. If t= 1, then the theorem follows from a result of Aigner and Fromme
which states that for graphs of girth at least 5, the cop number of a graph is at least its
minimum degree [1]. Otherwise, we assume that t≥2.
We will show that under the stated girth and degree conditions, the robber has a winning
strategy, provided that the number of cops is at most 1
e t (δ−1)t. We observe that our girth
condition implies that for any two vertices u, v at distance at most 2t, there is a unique
geodesic joining uto v.
We begin the game on the robber’s move with all cops on a single vertex v0∈V(G). (In
fact, we view the first move of the robber as part of his choice of the initial position.) We
let the robber begin at a vertex v1adjacent to v0. To show that the cops have no winning
strategy on G, it suffices to show that the cops cannot win from this position. At any given
state sof the game with robber to move, we assume that the robber is at a vertex vsand
that he will move to a neighboring vertex vs+1 different from vs−1(and different from vs).
After the cops move, we will reach the next state s+ 1. We will show that the robber can
move to a vertex vs+1 for which no cop is positioned at a vertex in the closed neighborhood
N[vs+1]. Thus the robber will be able to avoid capture during state s, and hence forever.
We define the following values. Let q=δ−1, r= (1 −1
t)q, and let K≤1
e t qtdenote the
number of cops. The lower bound of the theorem is obviously true if δ≤2. Thus we may
assume that δ≥3.
Suppose that the game is at a state s≥1. Let u1, . . . , uqbe distinct neighbors of vsthat
are different from vs−1. (There may be additional neighbors of vsif deg(vs)> δ, but we do
not need them for our strategy.) For each i= 1, . . . , q, let Cibe the set of cops Csuch that
dist(C, vs)≤2tand the geodesic from Cto vspasses through ui. (Note that since Cis not at
ui, such a geodesic cannot pass via vs−1.) For such a cop C∈Sq
i=1 Ci, we let ρ= dist(C, vs),
and we define the weight of Cas w(C) = rt−dρ/2e. We let kbe the number of cops that are
in none of the sets C1,...,Cq, and we define the weight of each such cop to be equal to 1. Let
Wi=k
q+PC∈Ciw(C), and let Wbe the total weight of all cops. Note that W=Pq
i=1 Wi.
At the beginning of each state s, the robber selects a neighbor ujfor which Wjis minimum
and moves to uj. If Wj< rt−1, then neither ujnor any of its neighbors contains a cop, so uj
is a safe vertex for the robber. In order to prove that the robber is never captured, it suffices
to prove that W < qrt−1, since in that case, Wj< rt−1for some 1 ≤j≤q.
Initially, C1,...,Cqare all empty (since all cops are at y1and the girth of Gis greater than
2t+ 1), and hence all cops have weight 1. Thus W < qrt−1as long as we have fewer than
3
qrt−1cops. It is easy to see by applying the inequality 1
e<(1 −1
t)t−1that this is the case:
K≤1
e tqt<(1 −1
t)t−1
tqt=qrt−1
t< qrt−1.(1)
The rest of the proof is by induction. We assume that at the current state swe have
W < qrt−1, and we let W0denote the total weight of the cops after the robber has moved to
vs+1 =ujand state shas ended. For every cop C∈ Cj, the robber moved closer to C, and
Cmay have moved closer to the robber; thus the distance ρbetween Cand the robber may
have decreased by at most 2. Consequently, the new weight w0(C) is at most rw(C). By the
girth condition of G, at the end of state s, for any other cop C0that was not in Cj, either
the geodesic from C0to vs+1 passes through vs, or dist(C0, vs+1)≥2t−1. In both cases, C’
has a weight equal to 1 at the end of state s. Then, using (1), we see that
W0≤r·Wj+K
< r ·W
q+qrt−1
t
< rt+qrt−1
t(2)
=qrt−11−1
t+1
t=qrt−1.
This completes induction and the proof.
Theorem 1.1 gives a lower bound for the cop number of a graph of girth gand minimum
degree δof the form Θ(g−1(δ−1)g/4), which is a significant improvement over Frankl’s lower
bound from [8], which is of the form Θ((δ−1)g/8). Furthermore, the method of Theorem
1.1 extends to graphs which have a fast growth property, which we define next.
Given positive integers h, q, we say that a graph Ghas (h, q)-growth if for any vertex
v∈V(G) and any neighbor uof v, the number of vertices wsatisfying dist(w, v) = hand
dist(w, u)≥his at least q. We observe that a graph of minimum degree δhas (1, δ −1)-
growth. The following theorem, stated in terms of growth, extends Theorem 1.1. The proof
of the theorem uses ideas similar to Frankl’s method in [8].
Theorem 2.1. Let t≥1,h≥1, and q≥1be integers, and let Gbe a graph of girth
g≥4h(t+ 1) −3and with (h, q)-growth. Then c(G)>1
e t qt.
Proof. If t= 1, then a method of Frankl shows that a graph with girth 8h−3 and (h, q)-
growth has cop number greater than q[8]. While Frankl only shows that the cop number is
greater than (δ−1)h, where δis the graph’s minimum degree, this more general bound in
terms of growth follows immediately from his method. Hence, we assume that t≥2.
We show that under these girth and growth conditions, the robber has a winning strategy,
provided that the number of cops is at most 1
e t qt. We observe that our girth condition implies
that for any two vertices u, v at distance at most 2h(t+ 1) −2, there is a unique geodesic
joining uto v.
We begin the game on the robber’s move with all cops on a single vertex y1∈V(G). We
let the robber begin at a vertex v1adjacent to y1. To show that the cops have no winning
strategy on G, it suffices to show that the cops cannot win from this position. At any given
state sof the game with robber to move, we assume that the robber is at a vertex vs, and
4
that on the previous move, the robber occupied a vertex ys(except when s= 1, in which
case y1is defined separately). At this point, we let the robber select a geodesic path Psof
length hfrom vsto a new vertex vs+1 that does not pass through ys. The growth condition
implies that there are at least qcandidates for the vertex vs+1. The robber and cops will
then take turns moving, making a total of hmoves each, and on each move, the robber will
move along the path Pstoward vs+1. After the robber and cops have each made hmoves,
we reach the state s+ 1. We will show that the robber is able to avoid capture during state
s, and hence forever.
We will assume a stronger condition at the beginning of each state of the game. We will
assume that at the beginning of each state s, no cop is positioned at a vertex cfor which
dist(c, vs)≤2h−2, except when the geodesic from cto vspasses through ys. The intuition
behind this condition is that we will always let our robber move away from ys, so a cop
whose geodesic to vspasses through ysposes no immediate threat to the robber due to the
large girth of G. It follows from this assumption that no cop at a vertex cwith distance less
than 2h−1 to the robber at the beginning of state scan decrease its distance to the robber
while the robber is moving along Pstoward a vertex vs+1. All other cops will have a distance
of at least 2h−1 from vs. This ensures that the robber will always be able to travel along
Psand reach vs+1 before being captured. (Here we do not exclude the possibility that the
robber is captured when he arrives to vs+1.) We have chosen our game’s initial configuration
to satisfy this condition, and we will show by induction that this condition may be satisfied
at the end of each state s.
We let r= (1 −1
t)q, and we let K≤1
e t qtdenote the number of cops.
Suppose that the game is at state s≥1. Let u1, . . . , uqbe distinct vertices at distance
exactly hfrom vsand at distance at least hfrom ys. (There may be additional such vertices,
but we do not need them for our strategy.) For each i= 1, . . . , q, let Cibe the set of cops
Csuch that dist(C, vs)≤2h(t+ 1) −2 and dist(C, ui) = dist(C, vs)−h. For such a cop
C∈Sq
i=1 Ci, we define its weight
w(C) = rt+1−dρ+2
2he
where ρ= dist(C, vs). Note that each cop in some set Cihas a unique geodesic path to
vs, and that geodesic passes through ui, but not through ys. Thus, C1,...,Cqare pairwise
disjoint. We let kbe the number of cops that are in none of the sets C1,...,Cq, and we define
the weight of each such cop to be equal to 1. Let Wi=k
q+PC∈Ciw(C), and let Wbe the
total weight of all cops. Then W=Pq
i=1 Wi.
The robber selects the vertex ujfor which Wjis minimum and moves to ujin hmoves. As
each cop Cwhose geodesic to vsdoes not pass through ysis at a distance of at least 2h−1
from vs, the robber will not be captured before reaching uj. On the other hand, for any cop
C0whose geodesic to vspasses through ys,C0will not be able to capture the robber within h
moves due to the girth condition of G. Furthermore, if Wj< rt−1when the robber is at vs,
then no cop C∈ Cjis within distance 4h−2 of vs. Therefore, the robber may safely reach
vs+1 =uj, and after reaching vs+1, no cop Cwill satisfy dist(C, vs+1 )≤2h−2, except when
the geodesic from Cto vs+1 passes through ys+1 (the vertex adjacent to vs+1 on Ps). Hence,
in order to prove that no cop Cever comes within distance 2h−2 of a vertex vs, except
when the geodesic from Cto vspasses through ys, it suffices to prove that W < qrt−1, since
in that case Wj< rt−1for some 1 ≤j≤q.
5
Initially, C1,...,Cqare all empty (since all cops are at y1and the girth of Gis greater than
2h(t+ 1) + h−1), and hence all cops have weight 1. Therefore, as in the method of (1),
W < qrt−1.
The rest of the proof is by induction. Let us assume that at the current state swe have
W < qrt−1, and let W0denote the total weight after the robber has moved to vs+1 =ujand
state shas ended. For every cop C∈ Cj, the robber moved closer to Cexactly htimes,
and Cmay have moved closer to the robber htimes; thus the distance ρbetween Cand the
robber may have decreased by at most 2h. Consequently, the new weight w0(C) is at most
rw(C). For each cop C6∈ Sq
i=1 Ci, the shortest path from the new position of Cto vs+1,
not passing through ys+1, is of length at least 2ht −1. Hence, the new weight of Cis still
w0(C) = 1. Let us finally consider a cop C∈ Ci, where i6=j. By the girth condition of G,
at the beginning of state s+ 1, a shortest path from Cto vs+1 not passing through ys+1 is
of length ρ≥4h(t+ 1) −3−(2h(t+ 1) −2+2h) = 2ht −1, since during state sthe cop C
could move at most hsteps towards vs+1. Therefore the cop Chas a new weight w0(C) = 1.
Therefore, we obtain that W0≤r·Wj+K < qrt−1by using the method of (2). This
completes the proof.
Theorem 2.1 may be applied in certain situations in which Theorem 1.1 is not useful,
namely when graphs of high girth have fast growth but low minimum degree. An example
of such a graph may be obtained from a graph of high girth and high minimum degree by
subdividing each edge approximately the same number of times.
3. Lower bounds for digraphs
The aim of this section is to show that the techniques of the previous section for undirected
graphs may also be applied to digraphs. In order to do so, we will define the dispersion of a
digraph, which is, in some sense, a directed counterpart of the girth of an undirected graph.
Roughly speaking, we do not want short cycles in the underlying undirected graph that
are composed of at most four directed geodesic paths, each of which is short. The precise
definition of dispersion requires some preparation.
Let Gbe a digraph. By vu ∈E(G) we denote the directed edge from vto u. A subgraph
of Gconsisting of two oppositely directed edges uv, vu ∈E(G) is called a digon. For vertices
v, u ∈V(G), we define the distance from vto u, denoted by dist(v, u), as the length of a
shortest directed path from vto u. A directed path from vto uis geodesic if its length is
equal to dist(v, u). A subdigraph of Gconsisting of two internally disjoint geodesic paths P
and Qis a (v, u)-trap if Pis a directed (v, x)-path with at least one edge, Qis a directed
(u, x)-path and kQk≤kPk(by k · k we denote the length of the path). The intercept x
is called the tip of the trap. The length of Pis called the length of the trap. Note that
kPk= dist(v, x) and kQk= dist(u, x). We allow for u=v, but in this case we require that
Pand Qeach have at least two edges. Note that in this case, Pand Qhave the same length.
We also allow that u=x.
Let t≥1 be an integer. We say that a digraph Gis t-dispersed if the following conditions
hold:
(1) For every v, u ∈V(G) (v6=u), Ghas no two internally disjoint (v, u)-traps, each of
length at most t.
(2) If uv ∈E(G), then Ghas no (v, u)-traps of length at most t.
6
v
u
x
y
t1
t2
t0
1
t0
2
t0
1≤t1≤t,
t0
2≤t2≤t
(a)
v
x
u=y
t1
t2
t0
1
t0
1≤t1≤t,
t2≤t
(b)
v
u
t1
t0
1
1≤t0
1≤t1≤t
(c)
v
x=u=y
t1
t1
x
Figure 1. Excluded traps in t-dispersed graphs. All thick paths shown are
geodesics. (a) Internally disjoint (v, u)-traps of lengths t1, t2≤t. (b) The
degenerate versions when ucoincides with one or both tips of the trap. (c)
(v, u)-trap of length t1≤twhen uv ∈E(G).
Forbidden subdigraphs used in the definition of t-dispersed property are illustrated in
Figure 1. Note that this property forbids certain cycles in Gthat are composed of 2, 3, or 4
geodesic directed paths of restricted length.
Lemma 3.1. Let Gbe a digraph that is t-dispersed, and let v, x ∈V(G)be vertices with
dist(v, x)≤t. Then there is a unique (v, x)-geodesic, and if uv ∈E(G), then dist(u, x)>
dist(v, x).
Proof. Suppose that there are two such geodesics Pand Q. Then it is easy to see that P∩Q
contains two vertices a, b such that the segments on the paths from ato bare internally
disjoint (and of length at least 2 since we do not have double edges). These segments would
form two internally disjoint (a, b)-traps of length at most t, contradicting property (1) from
the definition of t-dispersed. Note that in the two traps, the vertex bis the tip of both traps.
Suppose now that uv ∈E(G) and that dist(u, x)≤dist(v, x). Consider a (u, x)-geodesic
Qand let ybe the first vertex on it that intersects the (v, x)-geodesic P. Note that, since
we have geodesics, Qdoes not intersect Pat any other vertex between vand y. Thus, the
union of both geodesics from vto yand from uto ywould be a trap of length at most t,
contrary to property (2) from the definition of t-dispersed.
Lemma 3.2. Let Gbe a digraph that is t-dispersed, and let u, v be distinct vertices in G.
Then all (v, u)-traps of length at most tcontain the same outneighbor of v.
Proof. Suppose that there are two (v, u)-traps with geodesics P∪Qand P0∪Q0, respectively,
where the first edge of Pand the first edge on P0are different. We may assume that these
traps are selected to be of minimum possible lengths and that uis selected so that the lengths
of Qand Q0are also minimum. Let xand x0be the tips of the two traps, respectively. Since
every subpath of Pand P0is a geodesic, Lemma 3.1 implies that P∩P0={v}. Similarly,
we see that Q∩Q0={u}by using the fact that uis selected so that the lengths of Qand
Q0are smallest possible.
Suppose that Qintersects P0in a vertex u0. Since P∪Qis a trap, we have kPk≥kQk=
dist(u, u0) + dist(u0, x). On the other hand, dist(v, u0) + dist(u0, x)>kPkby Lemma 3.1.
7
This implies that dist(u, u0)<dist(v, u0). Our minimality choice of traps now implies that
u0=x0and also that u=x0. By repeating this argument, we end up with the conclusion
that Qis internally disjoint from P0. Similarly we see that Pand Q0are disjoint. Then the
two traps are internally disjoint, and their existence contradicts the assumption that Gis
t-dispersed.
We also define the trap distance from vto uas
ρ∗(v, u) = min{`| ∃ a (v, u)-trap of length `}.
Observe that ρ∗(v, u)≤dist(v, u) since any geodesic from vto uis a trap (with the second
path of length 0). There is also a way to define the trap distance using distances. Define
Sρ(v) (the sphere of radius ρaround v) to be the set of vertices at distance exactly ρfrom v.
Similarly, we define Bρ(v) (the ball of radius ρaround v) to be the set of vertices at distance
at most ρfrom v.
Lemma 3.3. The trap distance ρ∗(v, u)is equal to the minimum value ρfor which Sρ(v)∩
Bρ(u)6=∅.
Proof. If there is a (v, u)-trap of length ρ, then Sρ(v)∩Bρ(u)6=∅. Thus, it suffices to prove
that whenever Sρ(v)∩Bρ(u)6=∅, there is a trap of length at most ρ. Take a vertex xin
the intersection and let Pand Qbe geodesic paths from vand from uto x, respectively. If
these paths are disjoint, then we have a trap as claimed. If not, then we replace xby the
vertex yin P∩Qthat is as close as possible to v. Then we also replace the paths Pand Q
by the subpaths ending at y. Since both paths are geodesic, the number of edges that were
removed on Pand Qis the same. Thus, the new paths (which are now internally disjoint)
form a (v, u)-trap of length at most ρ.
Lemma 3.4. Let Gbe a digraph that is t-dispersed, let u, v be distinct vertices in Gand let
u0and v0be their outneighbors, respectively. If ρ∗(v0, u0)≤t, then ρ∗(v, u)≤ρ∗(v0, u0)+1.
Proof. Let t1=ρ∗(v, u) and t0
1=ρ∗(v0, u0)≤t. Suppose, for a contradiction, that t1≥t0
1+2.
Let P0∪Q0be a trap certifying that t0
1=ρ∗(v0, u0). Let P=vv0P0and Q=uu0Q0. If both
of these paths are geodesic paths, then by Lemma 3.3, ρ∗(v, u)≤t0
1+ 1, which contradicts
the assumption that t1≥t0
1+ 2. Consequently, one of the paths is not geodesic. Suppose
this is P. In that case, dist(v0, x) = t0
1≤t, which is not possible by second part of Lemma
3.1. This contradiction completes the proof. If Qis not geodesic, the proof is the same.
If the girth of the underlying undirected graph of a digraph Gis at least 4t+ 1, then Gis
t-dispersed. Therefore, the following theorems, which give lower bounds on the cop number
of a digraph in terms of its dispersion, may be seen as generalizations of the bounds from
the previous section.
Theorem 3.5. Let t≥1be an integer, and let Gbe a digraph that is t-dispersed. For each
vertex v, let qvbe equal to the outdegree d+(v)of vif vis not contained in any digons, and
be equal to d+(v)−1otherwise. If q= min{qv|v∈V(G)}, then c(G)>1
e t qt.
Proof. We first consider the case that t= 1. Suppose the robber occupies a vertex v. As G
is 1-dispersed, a single cop cannot guard more than one out-neighbor of v. Therefore, the
cop number of Gmust be at least the minimum out-degree of G; otherwise, the robber will
8
always have a safe out-neighbor to visit. The minimum out-degree of Gis greater than 1
eq,
so the theorem holds for t= 1.
Suppose that t≥2. We show that under the stated dispersion and degree conditions, the
robber has a winning strategy, provided that the number of cops is at most 1
e t qt.
We begin the game on the robber’s move with all cops on a single vertex y1∈V(G).
We let the robber begin at a vertex v1which is an out-neighbor of y1. To show that the
cops have no winning strategy on G, it suffices to show that the cops cannot win from this
position. At any given state sof the game with robber to move, we assume that the robber
is at a vertex vsand that he will move to a neighboring vertex vs+1 different from vs−1and
different from vs. After the cops move, we will reach the next state s+1. We will show that
the robber can move to a vertex vs+1 for which no cop is positioned at a vertex of the closed
in-neighborhood N−[vs+1 ]. Thus the robber will be able to avoid capture during state s, and
hence forever.
As in the proof of Theorem 2.1, we set r= (1 −1
t)qand let K≤1
e t qtdenote the number
of cops.
Suppose that the game is at a state s≥1. Let u1, . . . , uqbe distinct out-neighbors of vs
that are different from vs−1. (There may be additional out-neighbors of vs, but we do not
need these vertices for our strategy.) For each i= 1, . . . , q, let Cibe the set of cops Csuch
that
Sρ(vs)∩Sρ−1(ui)∩Bρ(C)6=∅for some ρ≤t. (3)
Note that each cop in Cihas the trap distance from vsbounded by t,ρ∗(vs, C)≤t, by
Lemma 3.3. By Lemma 3.2, each cop belongs to at most one family Ci, i.e. Ci∩ Cj=∅if
i6=j.
For a cop C∈Sq
i=1 Ci, we let ρ(C) = ρ∗(vs, C), and we define the weight w(C) = rt−ρ(C).
We let kbe the number of cops that are in none of the sets C1,...,Cq, and we define the
weight of each such cop to be equal to 1. Let Wi=k
q+PC∈Ciw(C), and let Wbe the total
weight of all cops. Then W=Pq
i=1 Wi. Now, the robber selects the vertex ujfor which
Wjis minimum and moves to uj. If Wj< rt−1, then neither ujnor any of its in-neighbors
contains a cop, since such a cop Cwould be in Cjwith ρ(C) = ρ∗(vs, C ) = 1 and would have
weight rt−1. Hence, ujis a safe vertex for the robber. In order to prove that the robber is
never captured, it suffices to prove that W < qrt−1, since in that case Wj< rt−1for some
1≤j≤q.
Initially, C1,...,Cqare all empty. To see this, note that, initially, all cops are at y1, which
is an in-neighbor of v1. If a cop Cat y1belongs to some Ci(1 ≤i≤q), then there is a
(v1, y1)-trap P∪Qof length at most t, where Puses the edge v1ui. Such a trap would
contradict property (2) from the definition of t-dispersion. Thus, we have W < qrt−1at the
start of the game, as long as we have fewer than qrt−1cops, which we see from the method
of (1).
The rest of the proof is by induction. Let us assume that at the current state s, we have
W < qrt−1, and let W0denote the total weight after the robber has moved to vs+1 =ujand
state shas ended. For every cop C∈ Cj, the robber moved closer to C, and Cmay have
moved closer to the robber. Lemma 3.4 shows that the trap distance between the robber
and Ccan decrease by at most 1, and thus, the new weight w0(C) is at most rw(C).
9
Consider now a cop C0/∈ Cj. At state s+ 1, the cops are again partitioned into the
sets C0
1,...,C0
q, defined with respect to the outgoing edges vs+1u0
1, . . . , vs+1u0
q, where the out-
neighbors u0
1, . . . , u0
qare distinct from vs. If a cop C0/∈ Cjis in some C0
i, and its weight is
w0(C0)>1, then ρ∗(vs+1, C 0)≤t−1. Lemma 3.4 shows that the trap distance between the
robber and C0could have decreased by at most 1 during the previous move. Thus, ρ∗(vs, C0)
was at most tat state sand the trap certifying this included the edge vsvs+1 =vsuj. This
shows that C0was in Cjat state s, a contradiction. We conclude that any cop Cthat was
not in Cjat state shas a new weight w0(C) = 1 at the beginning of state s+ 1.
Now, we see that W0≤r·Wj+K < qrt−1in the same way as in (2). This completes the
proof.
Similarly to Theorem 2.1, we may define a notion of growth for digraphs to generalize
Theorem 3.5 to digraphs that grow quickly but do not necessarily have high minimum out-
degree. Given positive integers h, q, we say that a digraph Ghas (h, q)-growth if for any
vertex v∈V(G) and any in-neighbor yof v, the number of vertices wsatisfying dist(v, w) = h
and dist(y, w)≥his at least q. The last condition just says that the geodesic from vto w
does not pass through y. We observe that a digraph Ghas (1, q)-growth if qis defined in
the same way as in Theorem 3.5.
Theorem 3.6. Let h≥1and t≥1be integers, and let Gbe a digraph that is (h(t+ 1) −1)-
dispersed and has (h, q)-growth. Then c(G)>1
e t qt.
Proof. First, we consider the case that t≥2. The theorem is trivial for q= 1; therefore we
assume that q≥2. We will show that under the stated dispersion and growth conditions,
the robber has a winning strategy, provided that the number of cops is at most 1
e t qt.
We begin the game on the robber’s move with all cops at a single vertex y1∈V(G). We
choose an out-neighbor v1∈N+(y1) and let the robber begin the game on v1. To show that
the cops have no winning strategy on G, it suffices to show that the cops cannot win from
this position.
At the beginning of a given state s≥1 of the game with robber to move, we assume that
the robber is at position vs, and that on the previous move, the robber occupied a vertex
ys(except when s= 1, in which case we have defined y1separately). During state s, the
robber will move along some geodesic path Psof length hwhich does not pass through ys,
from vsto a new vertex vs+1. The robber and cops will take turns moving, making a total
of hmoves each. One essential difference from the previous proofs is that the path Psis not
selected in advance, but is constructed one vertex at a time, depending on the moves of the
cops. After the robber and cops have each made hmoves, we reach the state s+ 1. We will
show that the robber is able to avoid capture during state s, and hence forever.
We will use a stronger condition at the beginning of each game state. We will assume that
at the beginning of each state s, for each vertex cwith a cop positioned at it, every (vs, c)-
trap of length at most h−1 passes through ys. Again, the intuition behind this condition is
that we will always let the robber move away from ys, so any traps containing ysare useless
for the cops. Furthermore, we would like for the robber to be able to move along the geodesic
path Psand reach vs+1 before being captured, and this restriction on traps ensures that this
will be possible. Our game’s initial configuration satisfies this condition, and we will show
by induction that this condition is satisfied for the entire duration of each state s(and hence
at the beginning of state s+ 1).
10
We let r= (1 −1
t)q, and let K≤1
e t qtdenote the number of cops. We will proceed in a
similar way as in previous proofs, but there are some subtle (yet essential) differences, and
thus we include the details. Throughout this proof we use the trap distance ρ∗(v, C) from
a vertex vand a cop C, which refers to the current position of C. Since this position is
changing over time, we have to specify at which time this trap distance is measured. We
always assume that the trap distance is measured at the time when the robber is at the
vertex vand it is the robber’s move.
Suppose that the game is at the beginning of state s≥1, and it is the robber’s move.
Recall that the robber occupies the vertex vs. Let u1, . . . , uqbe distinct vertices for which
dist(vs, ui) = hand dist(ys, ui)≥h(1 ≤i≤q). These vertices exist by the growth condition
of G. Consider the out-neighbors w1, . . . , wpof vsthat are on some (vs, ui)-geodesic. For
l= 1, . . . , p, let dl≥1 be the number of vertices uifor which the (vs, ui)-geodesic includes the
edge vswl. For each l= 1, . . . , p, let Clbe the set of cops Cfor which there is a (vs, C)-trap
of length at most h(t+ 1) −1 that uses the edge vswl. Since Gis (h(t+ 1) −1)-dispersed,
Lemma 3.2 implies that each cop belongs to at most one set Cl.
For a cop C∈Sp
l=1 Cl, we let ρ(C) = ρ∗(vs, C), and we define the weight
w(C) = rt+1−dρ(C)+1
he.(4)
We let kbe the number of cops that are in none of the sets C1,...,Cp, and we define the
weight of each such cop to be equal to 1. Let Wl=PC∈Clw(C), and let Wbe the total
weight of all cops. Then W=k+Pp
l=1 Wl. State swill consist of hsteps i,i= 1, . . . , h. If,
at the beginning of a step i, the robber occupies a vertex vand there exists a (v, C)-trap of
length at most h(t+ 1) −ifor some cop C, then we say that Cis threatening. Note that at
the beginning of step 1 of the current state s, each cop in Sp
l=1 Clis threatening. We define
the “average weight” of the threatening cops at step 1:
Z:= 1
q
p
X
l=1
Wl.
Now, at the beginning of step 1, the robber selects the vertex wjfor which Wj/djis minimum,
and the robber moves to wj. Let us note that Wj/dj≤Z. If not, then Wl> Zdlfor every
l= 1, . . . , p, and hence W−k=Pp
l=1 Wl> Z Pp
l=1 dl=Zq =W−k, a contradiction. Now
we let the cops move. After the cops move, only the cops in Cjremain threatening. Indeed,
if a cop Cis threatening at the beginning of step 2, then there exists a (wj, C)-trap of length
at most h(t+ 1) −2. Then, Lemma 3.4 tells us that at the beginning of step 1, there existed
a (vs, C)-trap of length at most h(t+1) −1, and by Lemma 3.2, this trap must have included
the edge vswj. This tells us that C∈ Cj.
Now, during steps 2 to h, we repeat this process without changing u1, . . . , uqor the
weights w(C), but updating pand the robber’s out-neighbors w1, . . . , wp, the corresponding
sets Cl, and the numbers dl. Again, the robber moves to the out-neighbor wjfor which
PC∈Cjw(C)/djis minimum. By the same arguments that we have made previously, only
the cops in Cjremain threatening, and PC∈Cjw(C)/dj≤Zstill holds. In this way, the
robber moves away from vsand reaches one of the vertices uiafter making hsteps. As
PC∈Cjw(C)/dj≤Zholds at each step, and as dj= 1 at beginning of step h, when the
robber moves to a vertex ui, the total weight of the threatening cops after step his at most
Z. Then, we set vs+1 =uiand begin the new state s+ 1.
11
We observe that for a cop C, if ρ∗(vs, C) was at most h−1 at the beginning of state s, then
Cwill not be able to capture the robber within the next hmoves, since the only short traps
pass through ys, and the robber moves away from ysat each step of state s. Each cop that is
still threatening when the robber arrives at vs+1 had weight at most Zwhen the robber was
at vs. If Z < rt−1when the robber is at vs, then each such cop Chad ρ∗(vs, C)≥2h. Hence,
at the end of state s, the trap distance from the robber to any such cop will be at least hby
Lemma 3.4. It therefore suffices to prove that W < qrt−1, since in that case Z < rt−1.
Initially, C1,...,Cpare all empty by condition (2) of our definition of dispersion. Thus
W < qrt−1as long as we have fewer than qrt−1cops, which we prove as in (1).
The rest of the proof is by induction. Let us assume that at the current state swe have
W < qrt−1. Let Dbe the set of threatening cops that have remained threatening throughout
all hsteps of state s. We consider the new weight w0(C) of each cop Cat the start of state
s+ 1. We claim that at the beginning of state s+ 1, each cop Cwith weight greater than 1
belongs to D. Indeed, suppose that a cop Chas weight greater than 1 at the beginning of
state s+ 1. By our definition (4) of weight, this implies that ρ∗(vs+1, C)≤ht −1. Then, by
Lemma 3.4, we see that ρ∗(vs, C)≤h(t+ 1) −1, and hence Cbelongs to D.
Moreover, for each cop C∈ D, the trap distance between the robber and Chas decreased
by at most h. Therefore, for a cop C∈ D, at the beginning of state s+ 1, Chas a new
weight w0(C), which is at most rw(C). Note that we have shown that the total weight from
the beginning of state sof all cops in Dis at most Z, and hence the new total weight of the
cops in Dat the beginning of state s+ 1 is at most rZ. Furthermore, as shown above, any
cop that does not belong to Dhas weight 1 at the beginning of state s+ 1. Thus, at the
beginning of state s+ 1, the new total weight W0satisfies
W0≤rZ +K < r ·W
q+qrt−1
t< qrt−1,
with the last inequality proved in the same way as (2). This completes the proof for the case
that t≥2.
In the case that t= 1, a very similar argument may be used. This time, however, we let
each cop have a weight 1 throughout the entire game. We again let the robber begin at a
vertex v1with an in-neighbor y1at which all cops begin the game. At the beginning of a
given state s≥1, we again suppose that the robber occupies a vertex vsand that the robber
occupied a vertex yson the previous move, except when s= 1, in which case y1is already
defined. We again aim to show that for any vertex cwith a cop positioned at it, every
(vs, c)-trap of length h−1 passes through ys. By using the same “average weight” argument,
we may show that the average weight Zof the threatening cops may be kept below 1 and
hence that the robber will be able to evade all threatening cops at each stage. As all other
parts of the strategy are the same as the t≥2 case, we do not include the details.
4. Bounds for expanders and an upper bound in terms of girth
In Section 2, we showed that asymptotically, a graph of girth gand minimum degree dhas
cop number Ω(g−1(d−1)g
4). As discussed at the end of the introduction, we believe that the
factor 1
4in the exponent is best possible. We are not able to prove this, but in this section
we prove that this constant factor cannot be made much larger.
We will show that if Ω((δ−1)cg) is a lower bound for the cop number of all graphs of
minimum degree δand girth g, then c≤3
8. In order to do this, we will present families
12
of d-regular graphs of fixed degree dand arbitrarily large girth g, whose cop number is at
most (d−1)cg, where ccan be made arbitrarily close to 3
8. One such family are the bipartite
Ramanujan graphs introduced by Lubotsky, Phillips, and Sarnak in [13]; these graphs are
often called LPS Ramanujan graphs. In order to show that these graphs have cop number as
small as claimed, we will present some general bounds for expander graphs and then apply
these bounds to the family of LPS Ramanujan graphs.
In proving our upper bound, we will make extensive use of the following graph parameter.
Definition 4.1. Let Gbe a graph. For a real number 0 < γ < 1, we define the parameter
hγ(G) = min
0<|S|≤n1−γ|∂S|
|S|,
where ∂S is the set of vertices in V(G)\Sthat have a neighbor in S.
For a graph G, a nonempty vertex set S⊆V(G), and a nonnegative integer r, we use
Br(S) to denote the set of vertices in V(G) whose distance from Sis at most r.
Lemma 4.2. Let Gbe a graph of order n,0< γ < 1, and let hγ(G)≥ε > 0. Let r≥0be
an integer. Then for any nonempty vertex set S⊆V(G),
|Br(S)| ≥ min{n1−γ,|S|(1 + ε)r}.
Proof. The result follows easily by induction on r.
The following theorem gives the main tool that we will use to establish our upper bound
on c. The cop strategy used in this theorem is probabilistic and borrows key ideas from the
strategies from [12], [21], and [10].
Theorem 4.3. Let ∆≥3,ε(0< ε ≤∆−2), and γ(0< γ ≤1
2(1 −log∆−1(1 + ε))) be
fixed constants. Then there is an integer n0such that for every graph Gof maximum degree
at most ∆, with hγ(G)≥εand of order n≥n0,
c(G)≤n1−1
2log∆−1(1+ε)+o(1),
where the asymptotics are considered with respect to n.
Proof. We take a small value δ > 0, and define the value κ= (1
2−2δ) log∆−1(1 + ε). Note
that γ < 1
2−κ. Next, we fix a sufficiently large graph Gof maximum degree at most ∆
which satisfies hγ(G)≥ε. We will show that c(G)≤n1−κ+o(1) =n1−1
2log∆−1(1+ε)+o(1), with
the last equality holding by letting δtend to 0.
We define r=b(1
2−δ) log n
log(∆−1) c. We will prove later that with this choice of r, any vertex
v∈V(G) has |Br(v)|<√n. Furthermore, we estimate:
(1 + ε)r= (∆ −1)rlog∆−1(1+ε)> n(1
2−2δ) log∆−1(1+ε)=nκ.
For the inequality to hold, we have assumed that nis large enough so that δlog∆−1n≥1.
Lemma 4.2 tells us that |Br(S)| ≥ |S|(1 + ε)rholds for any set S⊆V(G) that is not too
large, and therefore κis defined so that
|Br(S)| ≥ |S|nκ(5)
for such a set S.
13
Let C⊆V(G) be a random subset of vertices, obtained by taking each vertex of G
independently at random with probability p=n−κlog3n. Note that in contrast to previous
sections, here the symbol Cwill denote the set of starting positions of cops, rather than an
individual cop.
We place a cop at each vertex of C. The size of Cis a binomially distributed random
variable, and the expected size of Cis n1−κlog3n. We will estimate the size of Cwith a
Chernoff bound (c.f. [15], Chapter 5), which states that for a binomally distributed variable
Xwith mean µ, and for 0 ≤t≤µ,
Pr(|X−µ|> t)<2 exp(−t2
3µ).(6)
By (6), the probability that |C|>2n1−κlog3nis less than 2 exp −1
3n1−κlog3n=o(1),
and hence we see that a.a.s.
|C| ≤ 2n1−κlog3n. (7)
We let the robber begin the game at a vertex v∈V(G). We will attempt to capture the
robber by using the first rmoves of the game to place a cop at each vertex of Br(v). This
way, the robber will surely be captured before he has a chance to escape from Br(v). First,
we will estimate the size of Br(v). As Ghas maximum degree at most ∆, we see that
|Br(v)| ≤ 1+∆
r−1
X
j=0
(∆ −1)j= 1 + ∆
∆−2((∆ −1)r−1) ≤1 + ∆
∆−2(n1
2−δ−1) <√n,
with the last inequality holding for sufficiently large n.
We now define an auxiliary bipartite graph Hwith (disjoint) partite sets Br(v) and C.
For vertices u∈Br(v) and u0∈C, we add an edge uu0to E(H) if the distance in Gfrom u
to u0is at most r. If we can find a matching in Hthat saturates Br(v), then the cops from
Ccan fill Br(v) in the first rmoves and capture the robber.
We show that a.a.s., a matching in Hsaturating Br(v) exists. Indeed, suppose that no
such matching exists. Then by Hall’s theorem, there exists a set S⊆Br(v) such that
|Br(S)∩C|<|S|. We denote by ASthe event that |Br(S)∩C|<|S|, and we calculate
Pr(AS). The quantity |Br(S)∩C|is a binomially distributed random variable with expected
value p|Br(S)|.
By Lemma 4.2 and (5),
|Br(S)|>min{|S|(1 + ε)r, n1−γ} ≥ min{|S|nκ, n1−γ}=|S|nκ,
with the last inequality coming from the facts that γ < 1
2−κ,|S|<√n, and that nis large
enough. Therefore, p|Br(S)|>|S|log3n. By applying (6) with t=|S|(log3n−1), we can
estimate as follows:
Pr(AS)<2 exp −|S|2(log3n−1)2
3|S|log3n<exp −1
3|S|log2n.
14
Then, by considering the nonempty sets S⊆Br(v) of all sizes a≤√n, the probability that
no event ASoccurs is at least
1−
√n
X
a=1 √n
ae−1
3alog2n>1−
√n
X
a=1
e1
2alog n−1
3alog2n
>1−∞
X
a=1
e−1
6alog2n
= 1 −e−1
6log2n
1−e−1
6log2n= 1 −o(1
n).
We see that a.a.s. no event ASoccurs for the robber’s choice of v, and as the robber has n
choices for his starting vertex, we see furthermore that a.a.s. no event ASoccurs regardless
of the vertex v∈V(G) at which the robber chooses to begin the game. It follows that a.a.s.
we have a team of at most 2n1−κlog3ncops who have a strategy to capture the robber in r
moves. Recalling that κ= ( 1
2−2δ) log∆−1(1 + ε), and letting δtend to 0, we see that there
is a winning strategy with n1−1
2log∆−1(1+ε)+o(1) cops.
The last ingredient we will need is a connection between vertex expansion and the eigen-
values of a regular bipartite graph, which was proven by Tanner (see Theorem 2.1 in [22]).
Lemma 4.4. Let Hbe a d-regular bipartite graph on nvertices with a balanced bipartition
X∪Y. Let B∈RX×Ybe the square matrix whose (x, y)-entry ((x, y)∈X×Y) is 1if xis
adjacent to yand is 0otherwise. If the second largest eigenvalue of BBTis at most λ, then
any set S⊆Xhas at least f(|S|)neighbors in Y, where
f(|S|) = d2|S|
λ+ 2(d2−λ)|S|/n.
Lemma 4.5. Let 0< γ < 1be fixed. Let Gbe a connected d-regular bipartite graph.
Suppose that Ghas eigenvalues d=λ1(G)≥λ2(G)≥ ··· ≥ λn(G) = −d. Then hγ(G)≥
d
λ2(G)2−1−o(1), where the asymptotics refer to the order of the graph.
Proof. Let Ghave bipartition X∪Y. We set λ=λ2(G) and α=λ2/d. If α=d, then the
lemma is trivial, so we assume that α < d. The adjacency matrix of Gcan be written in the
form
A(G) = 0B
BT0,
where Bis defined for Gas in Lemma 4.4. It follows that
A2(G) = BBT0
0BTB.
Matrices BBTand BTBhave the same eigenvalues and it is easy to see that they are equal
to the squares of the eigenvalues of G. In particular, the second largest eigenvalue of each
of BBTand BTBis equal to λ2=αd.
Finally, we consider a set S⊆V(G),|S| ≤ n1−γ. By Lemma 4.4, S∩Xhas at least
(d
α−o(1))|S∩X|neighbors in Y, and by applying Lemma 4.4 with Xand Yexchanged,
15
S∩Yhas at least ( d
α−o(1))|S∩Y|neighbors in X. This implies that
|∂(S∩X)\S| ≥ d
α−o(1)|S∩X|−|S∩Y|
and
|∂(S∩Y)\S| ≥ d
α−o(1)|S∩Y|−|S∩X|.
By summing up these two inequalities, we see that |∂S| ≥ (d
α−1−o(1))|S|, or equivalently,
|∂S|
|S|≥d
λ2(G)2−1−o(1).
We are ready to prove our upper bound on the coefficient cof g. By letting ptend to
infinity in the following theorem, we obtain families of d-regular graphs of girth gand cop
number at most (d−1)cg, with carbitrarily close to 3
8.
Theorem 4.6. Let pbe a prime number for which p≡1 mod 4, and let d=p+ 1. There
exists an infinite family of d-regular graphs Xwith increasing girth g, whose cop number is
bounded above by
c(X)≤p(1+2 logp4+o(1)) 3
8g.
Proof. We use the Ramanujan graphs Xp,q discovered by Lubotsky, Phillips, and Sarnak in
[13]. These graphs are d-regular, where d=p+ 1 and pis any prime that p≡1 mod 4.
The second parameter qis also a prime that is larger than √p,q≡1 mod 4, and such
that the Legendre symbol of qand psatisfies q
p=−1. By the definition of Ramanujan
graphs, their second eigenvalue λ2satisfies λ2
2≤4(d−1). The graph X=Xp,q is a d-regular
bipartite Cayley graph on n=q(q2−1) vertices, and the girth gof Xp,q satisfies g≥4 log q
log p−1
[13].
We now compute an upper bound on the cop number c(X) of X. By Lemma 4.5,
hγ(Xp,q)≥d
4−1−o(1) >p+1
4−1−o(1), for any 0 < γ < 1. Then, by applying The-
orem 4.3,
c(X)≤n1−1
2logp
p+1
4+o(1) ≤n1
2+logp4+o(1).
Finally, we express this bound in terms of the girth and minimum degree of X. As
Xis a (p+ 1)-regular Cayley graph, δ(X)−1 = p. Furthermore, as qtends to infinity,
g(X)≥(4 + o(1))log q
log p. As n= (1 −o(1))q3, it follows that n≤p(3
4+o(1))g, and hence
c(X)≤p(1+2 logp4+o(1)) 3
8g. This completes the proof.
We may also use Theorem 4.3 to gain upper bounds on cop number in terms of a better
known graph parameter and give a sufficient condition for the cop number of a graph on n
vertices to be of the form O(n1−α) for some constant α > 0. The isoperimetric number h(G)
of Gis defined as follows:
h(G) = min
0<|S|≤n/2|δS|
|S|,
where δS is the set of edges with exactly one endpoint in S. (See [14] for more details.)
From Theorem 4.3 and Definition 4.1, we obtain the following corollary.
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Corollary 4.7. Let Gbe a family of graphs such that for all G∈ G, the maximum degree of
Gis at most ∆≥3. Let 0< ε ≤∆−2, and suppose that h(G)≥εfor all G∈ G sufficiently
large. Then for all G∈ G,c(G)≤n1−1
2log∆−1(1+ε/∆)+o(1).
Corollary 4.7 gives an implication about the so-called Weak Meyniel’s Conjecture. While
Meyniel’s Conjecture 1.2 asserts that the cop number of a graph on nvertices is bounded
by O(√n), its weakened version asserts that there exists a value α > 0 such that every
graph on nvertices has cop number O(n1−α) (see, e.g. [3]). Even this weakened conjecture
is still widely open. Corollary 4.7 implies that the weakened Meyniel’s conjecture holds for
expanders of bounded degree.
5. Conclusion
We have shown that in the optimal lower bound for the cop number of a graph with girth
gand minimum degree δof the form Ω((δ−1)cg), the constant csatisfies 1
4≤c≤3
8. We
suspect that 1
4is the correct answer, and it would be interesting to reduce or close this
gap between the lower and upper bounds of c. While Meyniel’s conjecture and the folklore
conjecture about degree and diameter imply that 1
4is indeed correct, it may be possible to
show c=1
4without using these conjectures.
References
[1] M. Aigner and M. Fromme. A game of cops and robbers. Discrete Applied Mathematics, 8(1):1–12,
1984.
[2] T. Andreae. On a pursuit game played on graphs for which a minor is excluded. J. Combin. Theory
Ser. B, 41(1):37–47, 1986.
[3] W. Baird and A. Bonato. Meyniel’s conjecture on the cop number: A survey. J. Comb., 3(2):225–238,
2012.
[4] B. Bollob´as, G. Kun, and I. Leader. Cops and robbers in a random graph. J. Combin. Theory Ser. B,
103(2):226–236, 2013.
[5] B. Bollob´as and E. Szemer´edi. Girth of sparse graphs. Journal of Graph Theory, 39(3):194–200, 2002.
[6] N. Bowler, J. Erde, F. Lehner, and M. Pitz. Bounding the cop number of a graph by its genus. Acta
Math. Univ. Comenian. (N.S.), 88(3):507–510, 2019.
[7] P. Bradshaw, S. A. Hosseini, and J. Turcotte. Cops and robbers on directed and undirected abelian
Cayley graphs, 2019. arXiv:1909.05342.
[8] P. Frankl. Cops and robbers in graphs with large girth and Cayley graphs. Discrete Appl. Math.,
17(3):301–305, 1987.
[9] F. Hasiri and I. Shinkar. Meyniel extremal families of Abelian Cayley graphs. arXiv e-prints, page
arXiv:1909.03027, 2019.
[10] S. A. Hosseini. Game of cops and robbers on Eulerian digraphs. PhD thesis, Simon Fraser University,
2018.
[11] F. Lehner. On the cop number of toroidal graphs, 2019. arXiv:1904.07946.
[12] L. Lu and X. Peng. On Meyniel’s conjecture of the cop number. J. Graph Theory, 71(2):192–205, 2012.
[13] A. Lubotzky, R. Phillips, and P. Sarnak. Ramanujan graphs. Combinatorica, 8(3):261–277, 1988.
[14] B. Mohar. Isoperimetric numbers of graphs. J. Combin. Theory Ser. B, 47(3):274–291, 1989.
[15] M. Molloy and B. Reed. Graph colouring and the probabilistic method, volume 23 of Algorithms and
Combinatorics. Springer-Verlag, Berlin, 2002.
[16] R. Nowakowski and P. Winkler. Vertex-to-vertex pursuit in a graph. Discrete Math., 43(2-3):235–239,
1983.
[17] P. Pra lat and N. Wormald. Meyniel’s conjecture holds for random graphs. Random Structures
Algorithms, 48(2):396–421, 2016.
17
[18] P. Pra lat and N. Wormald. Meyniel’s conjecture holds for random d-regular graphs. Random
Structures Algorithms, 55(3):719–741, 2019.
[19] A. Quilliot. Probl`emes de jeux, de point fixe, de connectivit´e et de repr´esentation sur des graphes, des
ensembles ordonn´es et des hypergraphes. PhD thesis, Universit´e de Paris VI, 1978.
[20] B. S. W. Schroeder. The copnumber of a graph is bounded by b3
2genus (G)c+ 3. In Categorical
perspectives (Kent, OH, 1998), Trends Math., pages 243–263. Birkh¨auser Boston, Boston, MA, 2001.
[21] A. Scott and B. Sudakov. A bound for the cops and robbers problem. SIAM J. Discrete Math.,
25(3):1438–1442, 2011.
[22] R. M. Tanner. Explicit concentrators from generalized N-gons. SIAM J. Algebraic Discrete Methods,
5(3):287–293, 1984.
Department of Mathematics, Simon Fraser University, Burnaby, Canada
E-mail address:pabradsh@sfu.ca
Department of Mathematics, Simon Fraser University, Burnaby, Canada
E-mail address:sahossei@sfu.ca
Department of Mathematics, Simon Fraser University, Burnaby, Canada
E-mail address:mohar@sfu.ca
Department of Mathematics, Simon Fraser University, Burnaby, Canada
E-mail address:lstacho@sfu.ca
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