The bondage number of random graphs

0.25 MB·English

by  Dieter Mitsche

The bondage number of random graphs
Dieter Mitsche Xavier P´erez-Gim´enez Pawe l Pra lat
∗ † ‡
6
1
0
2 Abstract

r A dominating set of a graph is a subset D of its vertices such that every vertex not in D is
a
M adjacent to at least one member of D. The domination number of a graph G is the number of
vertices in a smallest dominating set of G. The bondage number of a nonempty graph G is the

0 size ofasmallestsetofedgeswhoseremovalfromGresultsinagraphwithdominationnumber
3 greater than the domination number of G. In this note, we study the bondage number of the
binomial randomgraphG(n,p). We obtain a lower bound that matches the order of the trivial
]
O upper bound. As a side product, we give a one-point concentration result for the domination
number of G(n,p) under certain restrictions.
C
.
h
t 1 Introduction
a
m
Inthispaper,weconsidertheErd˝os-R´enyi randomgraph process,whichisastochasticprocess
[
that starts with n vertices and no edges, and at each step adds one new edge chosen uniformly at
2
random from the set of missing edges. Formally, let e ,e ,...,e be a random permutation of
v 1 2 (n)
2
9 the edges of the complete graph K . The graph process consists of the sequence of random graphs
n
6 (n)
1 ( (n,m)) 2 , where (n,m) = (V,E ), V = [n] := 1,2,...,n , and E = e ,e ,...,e . It is
0 cGlear thatm=(0n,m) is Ga graph taken umniformly at ran{dom from}the set omf all{gr1aph2s on nmv}ertices
0 G
and m edges (see, for example, [2, 6] for more details.)
.
2
Our results refer to the random graph process. However, it will be sometimes easier to work
0
5 with the G(n,p) model instead of (n,m). The (binomial) random graph G(n,p) consists of the
G
1 probability space (Ω, ,Pr), where Ω is the set of all graphs with vertex set [n], is the family of
: F F
v all subsets of Ω, and for every G Ω,
i ∈
X
r Pr(G) = p|E(G)|(1 p)(n2)−|E(G)|.
a −
This space may be viewed as the set of outcomes of n independent coin flips, one for each pair
2
u,v of vertices, where the probability of success (that is, adding edge uv) is p. Note that p = p
n
{ } (cid:0) (cid:1)
may (and usually does) tend to zero as n tends to infinity.
All asymptotics throughout are as n (we emphasize that the notations o() and O() refer
→ ∞ · ·
to functions of n, not necessarily positive unless otherwise stated, whose growth is bounded; on the
other hand, functions hidden in Θ() and Ω() notations are positive). We use the notation a b
n n
· · ∼
to denote a = (1+o(1))b . A sequence a satisfies a certain property eventually if the property
n n n
∗e-mail: [email protected], Universit´e de Nice Sophia-Antipolis, Laboratoire J.A. Dieudonn´e, Parc Valrose,
06108 Nicecedex 02, France
†e-mail: [email protected], Department of Mathematics, Ryerson University,Toronto, ON, Canada
‡e-mail: [email protected], Department of Mathematics, Ryerson University, Toronto, ON, Canada; research
partially supported byNSERCand Ryerson University
1
holds for all but finitely many terms of the sequence. We say that an event in a probability space
holds asymptotically almost surely (or a.a.s.) if the probability that it holds tends to 1 as
n goes to infinity. We often write (n,m) and G(n,p) when we mean a graph drawn from the
G
distribution (n,m) and G(n,p), respectively. All logarithms in this paper are natural logarithms.
G
A dominating set for a graph G = (V,E) is a subset D of V such that every vertex not in
D is adjacent to at least one member of D. The domination number, γ(G), is the number of
vertices in a smallest dominating set for G. The bondage number, b(G), of a non-empty graph
G is the smallest number of edges that need to be removed in order to increase the domination
number; that is,
b(G) = min B :B E,γ(G B)> γ(G) .
{| | ⊆ − }
(If G has no edges, then we define b(G) = .) This graph parameter was formally introduced
∞
in 1990 by Fink et al. [3] as a parameter for measuring the vulnerability of the interconnection
network under link failure. However, it was considered already in 1983 by Bauer at al. [1] as
“domination line-stability”. Moreover, graphs for which the domination number changes upon the
removal of a single edge were investigated by Walikar and Acharya [7] in 1979. One of the very
first observations [1, 3] is the following upper bound:
b(G) min deg(x)+deg(y) 1 ∆(G)+δ(G) 1,
≤ xy E{ − } ≤ −
∈
where ∆(G) and δ(G) are the maximum and, respectively, the minimum degree of G. Since a.a.s.
∆(G(n,p)) δ(G(n,p)) pnprovidedpn logn(thisfollowsimmediatelyfromChernoff’sbound
∼ ∼ ≫
stated below, and the union bound), we get that a.a.s.
b(G(n,p)) 2pn(1+o(1)) (1)
≤
for pn logn. For denser graphs, one can improve the leading constant of this upper bound by
≫
using the following observation of Hartnell and Rall [5]:
b(G) min deg(x)+deg(y) 1 N(x) N(y) .
≤ xy E{ − −| ∩ |}
∈
It follows that if p = Ω(1), then a.a.s.
b(G(n,p)) (2p p2)n(1+o(1)).
≤ −
Today, many properties of the bondage number are studied. For more details the reader is directed
to the survey [9] which cites almost 150 papers on the topic.
2 Results
Our goal is to investigate the bondage number of the binomial random graph on n vertices and of
the random graph process. Throughout the whole paper we will exclude the case p = p 1 and
n
→
also assume that p does not tend to zero too fast. More precisely, our main results require that
p = p eventually satisfies
n
n 1/3+ε p 1 ε, (2)
−
≤ ≤ −
for some constant ε> 0, but most arguments only require the following, milder, constraint:
log2n/√n p 1 ε.
≪ ≤ −
2
Sinceourresults areasymptotic in n,we willassumethat nis large enough sothatall requirements
intheargumentaremet. (Inparticular,thenotation “eventually” isoftenimplicitly assumedinthe
proofs and omitted.) Let be the set of dominating sets of size k of G(n,p), and let X = .
k k k
D |D |
Clearly,
n n k
f(n,k,p) := EX = 1 (1 p)k − . (3)
k
k − −
(cid:18) (cid:19)
(cid:16) (cid:17)
For a given p = p , let
n
r = r = min k N : f(n,k,p)> 1/(pn) . (4)
n
{ ∈ }
Since pn √nlog2n > 1 (eventually) and f(n,n,p) = 1, the function r is well defined for n
≫
sufficiently large.
2.1 Random Graph Process
Consider the random graph process ( (n,m)) . Clearly, the random variable γ( (n,m)) is
G 0≤m≤(n2) G
a non-increasing function of m, γ( (n,0)) = γ(K ) = n, and γ( (n, n )) = γ(K ) = 1. Suppose
G n G 2 n
that at some point the domination number drops down, that is, there exists a value of m such that
(cid:0) (cid:1)
γ( (n,m)) = k+1 but γ( (n,m+1)) = k. The random graph process continues and, as long as
G G
the domination number remains to be equal to k, the bondage number, b( (n,m+ℓ)), is a non-
G
decreasingfunctionofℓ. Moreover, wegetthatb( (n,m+ℓ)) ℓ,asonecanremovethelastℓedges
G ≤
that were added in the process (namely, e ,e ,...,e ) in order to increase the domination
m+1 m+2 m+ℓ
number. A natural and interesting question is then to ask how large the bondage number is right
before the domination number drops again; that is, what can be said about b( (n,m+ℓ)) when
G
γ( (n,m+ℓ)) = k but γ( (n,m+ℓ+1)) = k 1? It turns out that, for the range of k we are
G G −
interested in, it is of the order of the maximum degree of (n,m+ℓ), and hence it matches the
G
trivial, deterministic, upper bound mentioned in the introduction (up to a constant multiplicative
factor). Here is the precise statement.
Theorem 1. Given any constant ε > 0, let k = k be such that eventually εlogn k n1/3 ε.
n −
≤ ≤
Then, there exists m = m such that a.a.s.
n
γ( (n,m)) = k and b( (n,m)) = Θ(∆( (n,m))) = Θ(m/n).
G G G
2.2 Binomial Random Graph
Consider now the binomial random graph G(n,p). Before we state the main result for this prob-
ability space, let us mention some technical difficulties one needs to deal with. Our one-point
concentration result (below) on the domination number of G(n,p) amounts to showing that a.a.s.
X 1 (since, trivially, a.a.s. X = 0 for all i r 1). Moreover, our claim about the bondage
r i
≥ ≤ −
number requires that a.a.s. X = Ω(pn) . This follows from the fact that the number of domi-
r
→ ∞
nating sets of minimum cardinality is an upper bound on the bondage number (since each such set
D must have a vertex v / D adjacent to only one vertex in D, and thus D can be neutralized by
∈
removing a single edge). Therefore, we will restrict ourselves to situations in which EX is large
r
enough and prove concentration of X around its mean. For technical reasons of the argument, we
r
will require the aforementioned condition to hold for two consecutive values n 1 and n (see (6)
−
and (7) in Theorem 3). This motivates the assumptions on the ratio p /p in the statement of
n+1 n
Theorem 2.
We point out that, even for “natural” functions satisfying our assumptions, such as p = 1/2,
n
it is not clear whether there are always many dominating sets of minimum cardinality, or rather
3
X oscillates reaching both small and large values as n grows. All we managed to show is that,
rn
for such p , almost all values of n satisfy (6) and (7) and thus yield the bondage number as large
n
as possible. To make this precise, a set I N is said to be dense if
⊆
I [n]
lim | ∩ | = 1. (5)
n n
→∞
In view of this definition, and recalling the definition of r = r in (4), our result for the binomial
n
random graph can be stated as follows.
Theorem 2. Given any constant ε > 0, let p = p be such that eventually n 1/3+ε p 1 ε.
n −
≤ ≤ −
Moreover, suppose there exists a non-increasing non-negative sequence h = h such that p /p =
n n+1 n
1 Θ(h /n). Then, there exists a dense set I N such that, with asymptotics restricted to n I,
n
− ⊆ ∈
a.a.s.
γ(G(n,p)) = r and b(G(n,p)) = Θ(∆( (n,p))) = Θ(pn).
G
Although the conditions on p in Theorem 2 seem restrictive, many common and natural prob-
n
ability functions p satisfy it. For example, p = n 1/4, p = 1/loglogn and p = 1/2 meet the
n n − n n
requirements (by picking h = 1, h = 1/logn and h = 0, respectively). Other, seemingly more
n n n
complicated, choices such as p = (n+1) 1/4log3n+n 1/3 also satisfy our conditions. On the
n − −
other hand, mixed behaviours such as
n 1/4 n even
−
p =
n
(1/loglogn n odd
are not considered here. One can easily relax the conditions on p a bit further, but we do not aim
n
for it, as it does not appear to be possible to express that in terms of any “natural” assumptions
such as “p being non-decreasing”.
n
2.3 General Result
In fact, both Theorem 1 and Theorem 2 are implied by the following, slightly more general, result.
It is known that even for sparser graphs (namely, for p = p log2n/√n, but bounded away
n
≫
from 1) a.a.s. the domination number of G(n,p) takes one out of two consecutive integer values,
r or r +1, where r = r is defined in (4) (see [4] and also [8] for an earlier paper where denser
n
graphs were considered). The next result shows that if f(n,r,p) (that is, the expected number of
dominating sets of cardinality r) is large, then we actually have one-point concentration and the
bondage number is of order pn. Note that we may have to restrict asymptotics to an infinite subset
of N that guarantees our assumptions on f.
Theorem3. Givenanyconstantε > 0, suppose thatp = p eventuallysatisfiesn 1/3+ε p 1 ε,
n −
≤ ≤ −
and let f and r be defined as in (3) and (4). Suppose that there exists an infinite set I N and
′
⊆
ω = ω such that
n
→ ∞
EX = f(n,r,p) exp(ωlogn) (for n I ). (6)
r ′
≥ ∈
Then, a.a.s.
γ(G(n,p)) = r (for n I ).
′
∈
Moreover, suppose that
I = n N :n I , n 1 I has infinite cardinality. (7)
′ ′
{ ∈ ∈ − ∈ }
Then a.a.s.
b(G(n,p)) = Θ(∆(G(n,p))) = Θ(pn) (for n I).
∈
4
Remark 4.
(i) In many applications of Theorem 3 (for instance, in the proofs of Theorems 1 and 2), I is
′
a dense subset of N. Then, automatically I is also dense, and thus has infinite cardinality as
required.
(ii) The first part of the theorem, which characterizes the domination number of G(n,p), holds in
fact for any p = p satisfying log2n/√n p 1 ε (see Corollary 9 below).
n
≪ ≤ −
The paper is structured as follows. In Section 3, we show that the results for (n,m) and
G
G(n,p) can be obtained from Theorem 3. Section 4 develops some tools required to estimate the
second moment of X and some other random variables. Finally, Section 5 is devoted to prove
r
Theorem 3.
3 Preliminaries
In this section we are going to introduce a few inequalities used in the paper, and we show some
properties of the functions r = r and f(n,r,p) defined in (3) and (4). The function p will be
n
assumedtosatisfy(2). WewillalsoshowthatTheorem1andTheorem2areimpliedbyTheorem3.
We will use the following version of Chernoff bound (see e.g. [6]):
Lemma 5 (Chernoff Bound). If W is a binomial random variable with expectation µ, and
0 < δ < 1, then, setting ϕ(x) = (1+x)log(1+x) x for x 1 (and ϕ(x) = for x < 1),
− ≥ − ∞ −
δ2µ
Pr[W < (1 δ)µ] exp( µϕ( δ)) exp ; (8)
− ≤ − − ≤ − 2
(cid:18) (cid:19)
and if δ > 0,
δ2µ
Pr[W > (1+δ)µ] exp . (9)
≤ −2+δ
(cid:18) (cid:19)
Given p = p [0,1), definep = log 1 . Note that p p (with equality only holdingat p = 0),
n ∈ 1 p ≥
and −
p p if p = o(1),
b b
∼
p = Θ(1) if p = Θ(1) and 1 p = Θ(1), (10)

−
pb if p 1.
→ ∞ →
b
We start with a few simple observations. Let us mention that some of the properties we show

b
below are known and can be found in, for example, [4, Observation 2.1] (but mainly for p = o(1)).
We present the proof here for completeness and to prepare the reader for similar calculations later
on.
Lemma 6. Assume log2n/√n p 1 ε for some constant ε > 0, and let r be defined as in (4).
≪ ≤ −
Then, the following holds:
5
(i)
1 pn
log (1+o(1)) if p = o(1)
1 pn p log2(pn)
r = log (1+o(1)) =  (cid:18) (cid:19)
p log2(pn) 1 bpn
(cid:24) (cid:18) b (cid:19)(cid:25) pblog αlog2(pn) if p = Ω(1),
(cid:18) (cid:19)
for some 1+bo(1) α 1+o(1) = Θ(1).  b
≤ ≤ 1 p b
−
(ii)
logn log2n
r = Θ and (1 p)r = Θ .
p − pn
(cid:18) (cid:19) (cid:18) (cid:19)
In particular, r = Ω(logn) and r = o(√n/logn).
(iii) Moreover, if k = r+O(1), then
f(n,k+1,p)
= exp(Θ(log2n)).
f(n,k,p)
Proof. For a given function g = g = o(1), we define
n
1 pn
s = s (n,p) = log .
g g p log2(pn)(1+g )
(cid:24) (cid:18) n (cid:19)(cid:25)
b
First, observe that for p in the range of discussion, log(pn) = Θ(logn). Then, it follows from (10)
b
that
1 logn
s = Θ(log(pn)) O(loglog(pn)) = Θ . (11)
g
Θ(p) − p
(cid:18) (cid:19)
(cid:16) (cid:17)
Also, from the definition of p and (10), we obtain (1 p)sg = Θ log2n . Hence, part (ii) will
− pn
follow, once we show that r = sg for some function gn = o(1). In p(cid:16)articul(cid:17)ar, proving part (i) will
automatically yield part (ii). b
Given any g = o(1), define g and g+ by
n n− n
1 pn 1 pn
s = log and s 1 = log .
g p log2(pn)(1+g ) g − p log2(pn)(1+g+)
(cid:18) n− (cid:19) (cid:18) n (cid:19)
b b
Since z z < z+1 for any z R, we obtain that g g g+.
≤ ⌈ ⌉ b ∈ n− ≤ n ≤b n
Now we proceed to estimate f(n,s ,p) and f(n,s 1,p) for any g = o(1). Denoting by
g g n
−
[n] = n(n 1)...(n k+1), and using Stirling’s formula (k! √2πk(k/e)k), we observe that
k
− − ∼
f(n,s ,p) = [n]sg 1 log2(pn)(1+g ) n(1−sg/n)
g s ! − pn n−
g (cid:18) (cid:19)
nsg(1+O(s /n))sg log2(pn) log2n s
g g
= exp 1+g +O +O
(1+o(1)) 2πbsg(sg/e)sg (cid:18)− p (cid:18) n− (cid:18) pn (cid:19) n (cid:19)(cid:19)
(cid:16) (cid:17)
1+o(1) ne log2(pn) log2n
p
= exp s log 1+g +O ,
2πs g s − p b n− pn
g (cid:18) (cid:18) g(cid:19) (cid:18) (cid:18) (cid:19)(cid:19)(cid:19)
since s = Θ(logn/pp) (by (11)) and p log2n/√n, which implies that
g ≫ b
(1+O(sg/n))sg = eO(s2g/n) = eO(log2n/(p2n)) 1
∼
6
and s /n = O(logn/(pn))= o(log2n/(pn)). Hence,
g
1 log(pn)+O(loglogn) pn
f(n,s ,p) exp log +O(1)
g
∼ 2πsg p (cid:18) (cid:18)logn(cid:19) (cid:19)
p log2(pn) log2n
− b p 1+gn−+O pn
!
(cid:18) (cid:18) (cid:19)(cid:19)
p log2(pn) loglogn
= Θ logn exp − pb gn− +O logn . (12)
!
(cid:18)r (cid:19) (cid:18) (cid:18) (cid:19)(cid:19)
Moreover, the same calculations leading to (12) are valid if we replace s by s 1 and g by g+,
b g g − n− n
so we also get
p log2(pn) loglogn
f(n,s 1,p) = Θ exp g+ +O . (13)
g − logn − p n logn
!
(cid:18)r (cid:19) (cid:18) (cid:18) (cid:19)(cid:19)
In order to prove part (ii), we first take g = (loglogn)2/logn. From (13) and since g+ g , we
n b n ≥ n
obtain
f(n,s 1,p) exp Ω((log2logn)log(pn)) = o(1/(pn)),
g
− ≤ −
and thus (cid:0) (cid:1)
f(n,j,p) < 1/(pn) for all 1 j s 1,
g
≤ ≤ −
since f(n,j,p) is increasing with respect to j in that range (this can be easily checked by looking
at the ratio f(n,j + 1,p)/f(n,j,p) for j = O(logn/p) = o(√n)). Therefore, r > s 1 and,
g
−
since both r and s are natural numbers, r s = s . On the other hand, if we set
g g (loglogn)2/logn
≥
g = (loglogn)2/logn then, by (12) and since g g ,
n − n− ≤ n
f(n,s ,p) n 1/4 logn exp Ω((log2logn)log(pn)) > 1/(pn),
g −
≫ ·
and hence r s . Copmbining the(cid:0)two bounds, we conclu(cid:1)de that
(loglogn)2/logn
≤ −
r = s for some (loglogn)2/logn g (loglogn)2/logn,
g n
− ≤ ≤
which implies the first equality in part (i). The second equality follows immediately from setting
α = 1/(1+g ) and the fact that 1 1 < 1 .
n− 1+gn ≤ 1+gn− (1+gn)(1−p)
Finally, let us move to part (iii). Using part (ii), it is easy to see that for k = r+O(1) we get
f(n,k+1,p) = [n]k+1/(k+1)! 1−(1−p)k+1 n−k 1 (1 p)k+1 −1
f(n,k,p) [n] /k! 1 (1 p)k − −
k (cid:18) − − (cid:19) (cid:16) (cid:17)
n 1 (1 p)k +p(1 p)k n−k
− − −
∼ k 1 (1 p)k
(cid:18) − − (cid:19)
pn n k
= Θ 1+Θ(p(1 p)k) −
logn −
(cid:18) (cid:19)
(cid:16) (cid:17)
pn
= Θ exp Θ(pn(1 p)k)
logn −
(cid:18) (cid:19)
(cid:16) (cid:17)
pn
= Θ exp Θ(log2n) = exp Θ(log2n) .
logn
(cid:18) (cid:19)
(cid:0) (cid:1) (cid:0) (cid:1)
This finishes the proof of the lemma.
7
Now, we will show that Theorem 1 can be obtained from Theorem 3.
Proof of Theorem 1. Let k = k be such that εlogn k n1/3 ε for some ε > 0. Our goal
n −
≤ ≤
is to show that there exists m = m N such that a.a.s. γ( (n,m)) = k and b( (n,m)) =
n
∈ G G
Θ(∆( (n,m))) = Θ(m/n). We assume that Theorem 3 holds and we will use the probability space
G
G(n,p) to get the result.
It follows immediately from definition (3) that, for 1 j < n, f(n,j,p) is both a continuous
≤
and increasing function of p, taking all values between 0 and n . Then, given n N (sufficiently
j ∈
large), we can define p to be such that
+
(cid:0) (cid:1)
f(n,k 1,p )= 1/(p n). (14)
+ +
−
Moreover, straightforward computations show that, for 0< p < 1 and 0 j n/4,
≤ ≤
n
f(n,j,p) j +1
j
= < 1/2, (15)
f(n,j+1,p) ≤ n n j
(cid:0)j+(cid:1)1 −
(cid:0) (cid:1)
so in particular f(n,j,p) is increasing in j, for j in that range. Let r be defined as r in (4) for
+
p = p . From (14) and (15), we deduce that f(n,k,p ) > 1/(p n) and f(n,j,p ) 1/(p n) for
+ + + + +
≤
all j k 1, so we must have r = k. Also observe that r in (4) is a non-increasing function of p.
+
Comb≤inin−g this fact and Lemma 6(i), we conclude that n 1/3+ε′ p 1 ε, for some constant
− + ′
≤ ≤ −
ε = ε(ε), since otherwise r < εlogn or r > n1/3 ε contradicting our assumptions on k and
′ ′ + + −
the fact that k = r . Hence, in particular, 1/(p n) = o(1). It follows immediately from the first
+ +
moment method that a.a.s. G(n,p ) has no dominating set of size k 1, and then
+
−
for all 0 p p , γ(G(n,p)) k a.a.s. (16)
+
≤ ≤ ≥
since this is a non-increasing property with respect to the addition of edges. In fact, a.a.s.
γ(G(n,p )) = k but we do not prove it now, since we will need a stronger statement to hold.
+
Now, let ω = ω be a function tending to infinity sufficiently slowly in order to meet all
n
requirements in the argument. Define
2 ω√p+n 2 ωn
p := p = p 1 = p (1+o(1)),
− +− n2 + − √p+ n2 ! +
(cid:0) (cid:1) (cid:0) (cid:1)
where the last step follows from the fact that p n 1/3+ε′. Since p p , then p n 1/3+ε′/2
+ − + −
≥ − ∼ − ≥
and p is bounded away from 1. Clearly,
−
1 1
f(n,k 1,p ) f(n,k 1,p ) = < . (17)
+
− − ≤ − p+n p n
−
Let r be defined as r in (4) for p = p . Next we want to show that r = k and then that
− − −
f(n,k,p ) exp(ωlogn). First, using Lemma 6(ii) and the fact that k = r = Θ(logn/p ), we
+ +
− ≥
8
get
1 (1 p )k−1 = 1 1 p++Θ ω√p+ k−1
− − − − − n
(cid:18) (cid:19)
(cid:16) (cid:17)
k 1
= 1 (1 p )k 1 1+Θ ω√p+ −
− − + − n
(cid:16) (cid:16) (cid:17)(cid:17)
= 1 (1 p )k 1 1+Θ ωlogn
− − + − n√p+
(cid:16) (cid:16) (cid:17)(cid:17)
= 1 (1 p )k 1 Θ ωlog3n
− − + − − n2p3/2
(cid:18) + (cid:19)
= 1 (1 p )k 1 1 Θ ωlog3n .
− − + − − n2p3/2
(cid:18) (cid:18) + (cid:19)(cid:19)
(cid:16) (cid:17)
Hence,
n k+1
f(n,k 1,p ) = f(n,k 1,p ) 1 Θ ωlog3n −
− − − + (cid:18) − (cid:18)n2p3+/2 (cid:19)(cid:19)
= f(n,k 1,p ) 1 Θ ωlog3n
− + − np3/2
(cid:18) (cid:18) + (cid:19)(cid:19)
f(n,k 1,p ) = (p n) 1 (p n) 1,
+ + − −
∼ − ∼ −
as p n 1/3+ε′. Combining this with (15) and (17), we obtain that f(n,k,p ) > 1/(p n) and
+ −
≥ − −
f(n,j,p ) 1/(p n) for all j < k, so k = r . Now, using Lemma 6 again (this time part (iii)), we
− ≤ − −
get
f(n,k,p ) = f(n,k 1,p )exp(Θ(log2n)) (p n) 1exp(Θ(log2n)) exp(ωlogn),
−
− − − ∼ − ≥
as desired. The same argument holds clearly with n 1 playing the role of n. Therefore, it follows
−
from Theorem 3 that a.a.s. γ( (n,p )) = k and b(G(n,p )) = Θ(∆(G(n,p ))) cp n, for some
G − − − ≥ −
constant c = c(ε) > 0. Let Q be the graph property that we cannot destroy all dominating sets of
size k by removing any set of at most cp n edges. Clearly, this is a non-decreasing property with
−
respect to adding edges in the graph, so
for all p p 1, G(n,p) satisfies property Q a.a.s. (18)
− ≤ ≤
Finally, define
n p +p n n
+
mˆ = − = p+ ωn√p+ p+,
2 2 2 − ∼ 2
(cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19)
where at the last step we use the fact that p > n 1/3+ε. Easy manipulations yield
+ −
mˆ +ωn√p+ mˆ +(√2+o(1))ω√mˆ mˆ +ω mˆ n2 −mˆ / n2
p = = , (19)
+ n n ≥ q (cid:0)(cid:0)n (cid:1) (cid:1) (cid:0) (cid:1)
2 2 2
and similarly (cid:0) (cid:1) (cid:0) (cid:1) (cid:0) (cid:1)
p = mˆ −ωn√p+ mˆ −ω mˆ n2 −mˆ / n2 . (20)
− n ≤ q (cid:0)(cid:0)n (cid:1) (cid:1) (cid:0) (cid:1)
2 2
In view of (16), (18), (19) and (20)(cid:0), (cid:1)we can apply Propo(cid:0)sit(cid:1)ion 1.13 in [6] separately to both the
property Q and the property that γ( (n,p)) k, and we conclude that a.a.s. γ( (n,mˆ)) k
G ≥ G ≥
and (n,mˆ) satisfies property Q. These two events together imply that γ( (n,mˆ)) = k and
G G
b( (n,mˆ)) = Θ(p n)= Θ(m/n). The proof is finished.
G −
9
Now, we are going to show that Theorem 2 can be obtained from Theorem 3.
Proof of Theorem 2. Let p = p be such that n 1/3+ε p 1 ε for some ε > 0, and let r = r
n − n
≤ ≤ −
be defined as in (4). Moreover, supposethere exists a non-increasing non-negative sequence h = h
n
such that p /p = 1 Θ(h /n). Our goal is to show that there exists a positive sequence
n+1 n n
−
ω = ω and a dense set I N such that
n ′
→ ∞ ⊆
f(n,r,p) exp(ωlogn), for n I .
′
≥ ∈
The result will follow immediately from Theorem 3, and will hold for I defined as in (7). (Note
that, since I is dense, it is straightforward to verify that I must be dense too.)
′
Throughout the proof, we set ω = ω = loglogn. Note h = O(1) and so our assumptions on
n 1
p and h imply that h = O(1) and so there exists a universal constant 0 < A < 1 such that, for
n 1
every n n 3n,
′
≤ ≤ p
n′
A 1. (21)
1
≤ p ≤
n
Given any fixed j 0,1,2 , in view of our assumptions on p and h and by Lemma 6(ii), we have
∈{ }
1 (1 pn)rn+1−j = 1 1 pn+1 1+Θ(hn/n) rn+1−j
− − − −
= 1 ((cid:16)1 p )(cid:0)rn+1 j 1 Θ(cid:1)(cid:17)pn+1hn rn+1−j
− − n+1 − − n
(cid:16) (cid:16) (cid:17)(cid:17)
= 1 (1 p )rn+1 j 1 Θ hnlogn
− − n+1 − − n
(cid:16) (cid:16) h lo(cid:17)g(cid:17)3n
= 1 (1 p )rn+1 j 1+Θ n .
− − n+1 − n2p
(cid:18) (cid:18) n+1 (cid:19)(cid:19)
(cid:0) (cid:1)
Therefore,
f(n+1,r j,p ) n+1 1 (1 pn+1)rn+1−j n−rn+1+j+1
n+1 n+1 − −
− =
f(n,rn+1−j,pn) n+1−rn+1+j (cid:16) 1 (1 pn)rn+1−j(cid:17)n−rn+1+j
− −
= 1+Θ lnogpn (cid:16)1−(1−pn+1)rn+1−(cid:17)j 1−Θ hnn2lpog3n n−rn+1+j
(cid:18) (cid:18) n (cid:19)(cid:19) (cid:18) (cid:18) n+1 (cid:19)(cid:19)
(cid:16) (cid:17)
logn log2n h log3n
n
= 1+Θ 1 Θ 1 Θ
np − np − np
(cid:18) (cid:18) n (cid:19)(cid:19)(cid:18) (cid:18) n+1(cid:19)(cid:19)(cid:18) (cid:18) n+1 (cid:19)(cid:19)
g
n
= 1 Θ , (22)
− np
(cid:18) n(cid:19)
where g := log2n+h log3n. By our assumptions on h , we have log2n g = O(log3n). In
n n n n
≤
particular, for every j 0,1,2 and every n,
∈ { }
g f(n+1,r j,p ) g
n n+1 n+1 n
exp C − exp C , (23)
2 1
− np ≤ f(n,r j,p ) ≤ − np
(cid:18) n(cid:19) n+1− n (cid:18) n(cid:19)
for some universal constants C > C > 0. From (22) (with j = 0) and our assumptions on p, we
2 1
obtain
(n+1)p f(n+1,r ,p ) g 1
n+1 n+1 n+1 n
= 1 Θ 1+O < 1,
np f(n,r ,p ) − np n
n n+1 n (cid:18) (cid:18) n(cid:19)(cid:19)(cid:18) (cid:18) (cid:19)(cid:19)
10