the critical behavior of random digraphs

The Critical Behavior of RandomDigraphs*

Tomasz Łuczak,1 Taral Guldahl Seierstad2,3

1Department of Discrete Mathematics, Adam Mickiewicz University, 61-614Poznan, Poland; e-mail: [email protected]

2Humboldt-Universität zu Berlin, Institut für Informatik, Unter den Linden 6,10099 Berlin, Germany

3University of Oslo, Institute of Basic Medical Sciences, Department of Biostatistics,Postboks 1122, Blindern N-0317, Oslo, Norway; e-mail: [email protected]

Received 14 February 2007; accepted 1 February 2008; received in final form 23 February 2009Published online 23 July 2009 in Wiley InterScience (www.interscience.wiley.com).DOI 10.1002/rsa.20283

ABSTRACT: We study the critical behavior of the random digraph D(n, p) for np = 1 + ε, whereε = ε(n) = o(1). We show that if ε3n → −∞, then a.a.s. D(n, p) consists of components which areeither isolated vertices or directed cycles, each of size Op(|ε|−1). On the other hand, if ε3n → ∞, thena.a.s. the structure of D(n, p) is dominated by the unique complex component of size (4 + o(1))ε2n,whereas all other components are of size Op(ε

−1). © 2009 Wiley Periodicals, Inc. Random Struct. Alg.,35, 271–293, 2009

Keywords: random digraphs; phase transition; critical behavior

1. INTRODUCTION

Let D(n, p) be a random digraph with vertex set [n] = {1, 2, . . . , n}, where each of then(n−1)possible arcs is included in the digraph, independently of each other, with probabilityp = p(n). We say that a property A of D(n, p) holds a.a.s. if the probability of A tends to 1as n → ∞. For a random variable Xn and a positive function f (n) we write Xn = Op(f (n))

[Xn = �p(f (n))] if for every function ω(n) → ∞ a.a.s. Xn ≤ ω(n)f (n) [Xn ≥ f (n)/ω(n)].By a component of a digraph D we mean a maximal, strongly connected subgraph of D. Acomplex component is a component which is neither a single vertex nor a cycle.

Correspondence to: T. Łuczak*Supported by KBN Grant (1 P03A 025 27); DFG (GRK 588/2); Research Council of Norway (170620/V30);Foundation for Polish Science.© 2009 Wiley Periodicals, Inc.

271

272 ŁUCZAK AND SEIERSTAD

The phase transition phenomenon in D(n, p) was studied by Łuczak [9] and Karp [7](for an analogous result for a multiparameter generalization of D(n, p), see Cohen, Łuczak[3]). They proved that if np → c < 1, then a.a.s. D(n, p) contains no complex components,and each component of D(n, p) has Op(1) vertices. On the other hand, if np → c > 1,then a.a.s. D(n, p) contains a unique complex component on (α(c) + o(1))n vertices forsome explicitly defined positive constant α(c), whereas all other components of D(n, p) areisolated vertices or cycles of size Op(1). Karp [7] also considered the structure of D(n, p)

when np = 1 + ε and ε = ε(n) is a function which slowly tends to 0 as n → ∞. Ourmain result states that the component structure of D(n, p) changes when np = 1 + ε andε = ε(n) = �(n−1/3).

Theorem 1. Let np = 1 + ε, such that ε = ε(n) → 0.

i. If ε3n → −∞, then a.a.s. every component in D(n, p) is either a vertex or a cycle oflength Op(1/|ε|).

ii. If ε3n → ∞, then a.a.s. D(n, p) contains a unique complex component, of order(4 + o(1))ε2n, whereas every other component is either a vertex or a cycle of lengthOp(1/ε).

It should be mentioned that the fact that in the supercritical phase, when np = 1 + ε,ε3n → ∞, ε = o(1), the largest component of D(n, p) is of size (4 + o(1))ε2n is not verysurprising. Let us recall that for such a parameter p = p(n) the probability that a vertex ofthe undirected random graph G(n, p) belongs to the giant component is (2 + o(1))ε. In thefollowing section we observe that this is precisely the probability that from a given vertexv one can reach a lot of vertices of D(n, p) and, by symmetry, it is also the probability thatfrom many vertices of D(n, p) one can reach v. Thus, the probability that a vertex v hasboth a lot of descendants and ancestors should be, clearly, (4 + o(1))ε2. Moreover, one canexpect that the giant component of D(n, p) consists only of such vertices and, conversely, all(or almost all) such “large” vertices belong to the giant component. Thus, the main difficultyis to define appropriately when a vertex is large, and argue that the number of such verticesis a.a.s. (4 + o(1))ε2n, as expected.

Finally, we remark that, unlike in the undirected case, in a directed graph �D the factthat the set of vertices W is a component of �D does not only depend on the arcs incidentto W , but may also depend on the arcs with neither end on W . Hence many techniquesused to study the critical behavior of undirected random graphs (such as direct counting orgenerating function methods) seem to not be very useful in the directed case and we arebasically doomed to use the branching process approach, whose analysis near the criticalpoint is delicate and rather technical.

2. D(n, p) AND G(n, p)

To outline our argument more precisely, we need to introduce some notation. If v and w aretwo vertices, and V is a set of vertices in D(n, p), then dV (v, w) denotes the length of theshortest path from v to w in D(n, p) which visits only vertices in V ; if no such path exists,we put dV (v, w) = ∞. We also set

−→Sd (V , v) = {w : dV (v, w) = d}

Random Structures and Algorithms DOI 10.1002/rsa

CRITICAL BEHAVIOR OF RANDOM DIGRAPHS 273

and

←−Sd (V , v) = {w : dV (w, v) = d}.

Thus,−→S (V , v) = ∪d≥0

−→Sd (V , v) and

←−S (V , v) = ∪d≥0

←−Sd (V , v) stand for the sets of all

vertices that can be reached from v in V and all vertices in V which can reach v, respectively.

If V is the whole vertex set [n], we often omit writing [n]; for example we write−→S (v) =−→

S ([n], v). The vertices in−→S (v) are called the descendants of v and the vertices in

←−S (v) are

the ancestors of v. Note that v is an element of both−→S (v) and

←−S (v) and that there may be

other vertices which are both descendants and ancestors of v; these are precisely the verticescomprising the component containing v. By S(v; n, p) we denote the random variable which

counts the vertices in−→S (v) for a vertex v in D(n, p); in a similar way Sd(v; n, p) denotes the

size−→Sd (v) in D(n, p). Clearly, the choice of v has no effect on the distribution of S(v; n, p)

and Sd(v; n, p); we therefore let S(n, p) and Sd(n, p) be random variables with the samedistribution as S(v; n, p) and Sd(v; n, p), respectively.

Let us recall that the random (undirected) graph G(n, p) is a graph with vertex set [n]such that each of the

(n2

)possible edges is included with probability p, independently of

each other. Note that for every integer k, the probability that−→S (v) has k elements in D(n, p)

equals the probability that v is contained in a component of order k in G(n, p). Indeed, let us

assume that we want to find all vertices of D(n, p) contained in−→S (v). A natural way of doing

this is using a sequential search, say, the breadth-first search. That is, we determine the sets−→S1 (v),

−→S2 (v), . . ., in that order. Assume that we have found the sets

−→Si (v) for every i ≤ i0

for some i0 ≥ 0, where we put−→S0 (v) = {v}. Then, for every w ∈ Si0(v) and x /∈ ∪i≤i0 Si(v),

we ask whether the arc (w, x) is in D(n, p) (which happens with probability p independentlyfor each such pair). Clearly, if (w, x) belongs to D(n, p) for at least one w ∈ Si0(v), thenx ∈ Si0+1. We continue in the same manner until for some i we get Si(v) = ∅, in which casewe have found all descendants of v.

Exposing the component containing v in G(n, p) can be done using the same type ofbreadth-first search and because the probability of an affirmative answer is the same inG(n, p) and D(n, p) whenever we ask about the presence of an edge, respectively, an arc,the probability of finding, say, k vertices by this process is the same in D(n, p) and G(n, p).This argument holds as long as, in D(n, p), we do not check whether the arc (v, w) and thearc (w, v) are both contained in D(n, p).

We often use the above observation to deduce properties of D(n, p) from results onG(n, p), whose structure is nowadays well studied and understood. The following theoremfrom Łuczak [8] (see also [6] Chapter 5), which supplemented an earlier result of Bollobás[1] (see also [2]), can be considered an analog of Theorem 1 for the undirected case.

Theorem 2. Let np = 1 + ε, such that ε = ε(n) → 0 but n|ε|3 → ∞, and k0 =2ε−2 log n|ε|3.

i. If nε3 → −∞, then G(n, p) a.a.s. contains no component of order greater than k0.Moreover, a.a.s. each component of G(n, p) is either a tree, or contains precisely onecycle.

ii. If nε3 → ∞, then G(n, p) a.a.s. contains exactly one component of order greater thank0. This component a.a.s. has (2 + o(1))εn vertices.



Let us observe that from Theorem 2(i) it follows that if v is a given vertex in D(n, p),

then the probability that |−→S (v)| > k0 is o(1). However, this does not directly imply that

D(n, p) a.a.s. contains no vertices v such that |−→S (v)| > k0.In this article we also use some more technical results, which have been used in the

studies of the critical behavior of G(n, p). We first remark that from the calculations used inthe proof of Theorem 2(ii), contained mainly in [1] (see also [2], [6] and [12]), it follows thatin the supercritical phase — that is for np = 1+ε, where ε3n → ∞ — the expected numberof vertices in components of G(n, p) larger than k0 is (2+o(1))εn, and, moreover, for everyconstant δ > 0 the expected number of vertices which are contained in components ofG(n, p) whose sizes belong to [0.1k0, (2 − δ)εn] ∪ [(2 + δ)εn, n] is o(εn). Consequently,the following holds.

Lemma 3. If np = 1 + ε, where ε = ε(n) → 0 but ε3n → ∞, then the probability thata randomly chosen vertex v from G(n, p) is contained in a component of size larger than0.1k0 is (2 + o(1))ε. Moreover, the probability that v is contained in a component of size(2 + o(1))εn is also (2 + o(1))ε.

Our next result bounds from above the probability that a vertex v has many descendantsin D(n, p).

Lemma 4. Let np = 1 + ε, where |ε| ≤ 1/2, k ≤ |ε|n/6. Then, for some absoluteconstant c, we have

P[S(n, p) = k] ≤ ck−3/2 exp(−kε2/12) ≤ ck−3/2. (1)

Proof. As we have already noticed, the probability that S(n, p) = k is equal to the proba-bility that a vertex in G(n, p) is contained in a component of order k. Let Xk be the numberof components of order k in G(n, p). Bollobás [1] (see also [2] p.132) showed that for someabsolute constant c′

E[Xk] ≤ c′nk−5/2 exp(−kε2/2 + kε3/3 + k2ε/2n).

As for |ε| ≤ 1/2 and k ≤ |ε|n/6 we have

−kε2/2 + kε3/3 + k2ε/2n ≤ −kε2/12

and

P[S(n, p) = k] = k

nE[Xk],

so (1) follows.

From Lemmas 3 and 4 we get the following result.

Lemma 5. Let np = 1 + ε, where ε = ε(n) → 0, and k = o(n2/3). Then, for someabsolute constant c,

E[|−→S (v)|||−→S (v)| ≤ k] ≤ 3c√

k. (2)



Proof. From Lemmas 3 and 4 we get

E[|−→S (v)|||−→S (v)| ≤ k] ≤k∑

k=1

kP[|−→S (v)| = k]P[|−→S (v)| ≤ k]

≤ 1

1 − 2ε + o(ε)

k∑k=1

ck−1/2 ≤ (2 + o(1))c√

k,

so the assertion follows.

Suppose we use a breadth-first search starting from the vertex v in order to determinewhether v belongs to a large component. In view of Lemma 3, we may assume that v belongsto the large component once we have exposed k0 vertices. It will be, however, convenientto count the number of generations the breadth-first search lasts, rather than the numberof vertices that are exposed. The next lemma provides a connection between these twoviewpoints, but first we need a few inequalities.

Let g = 2 log log nε3/ε. The expected number of neighbors of a vertex is (n−1)p ≤ 1+ε,so the expected number of vertices in the ith generation of the breadth-first search startingat v is bounded from above by (1 + ε)i. Thus

E[S≤g(n, p)] ≤g∑

i=0

(1 + ε)i = (1 + ε)g+1 − 1

ε= O

(log2 nε3

ε

). (3)

By Markov’s inequality,

P

[S≤g(n, p) ≥ εn

log nε3

]≤ E[S≤g(n, p)]

εn/ log nε3= O(log3 nε3)

nε2= o(ε). (4)

Lemma 6. Let v be a vertex of D(n, p) and let C1 denote the event that−→Sg (v) �= ∅ and C2

denote the event that |−→S (v)| ≥ k0. Then,

P[(C1 ∧ C2) ∨ (C1 ∧ C2)] = o(ε).

Proof. Suppose first that−→Sg (v) �= ∅. Thus the breadth-first search starting at v lasts for at

least g generations. Because of the duality of the models D(n, p) and G(n, p), the probabilitythat this happens in D(n, p) is equal to the probability that it happens in G(n, p). In the modelG(n, p), this can only happen if v is in a component of diameter at least g. Let Zn,p(g, k0) bethe number of components in G(n, p) with fewer than k0 vertices and diameter at least g.Łuczak [11] showed that

E[Zn,p(g, k0)] = (2 + o(1))nε3(1 − ε)g.

The probability that the breadth-first search starting at v lasts at least g generations but doesnot generate k0 vertices is therefore bounded from above by

k0E[Zn,p(g, k0)]n

= O

(ε

log nε3

)= o(ε),



and hence

P[−→Sg (v) �= ∅ ∧ |−→S (v)| < k0] = o(ε),

proving one part of the lemma.

On the other hand, suppose that−→Sg (v) = ∅. We have P[−→Sg (v) = ∅] = 1 − o(1), so

by (4),

P[−→S≤g(v) ≥ εn/ log nε3 | −→Sg (v) = ∅] ≤ P[−→S≤g(v) ≥ εn/ log nε3]

P[−→Sg (v) = ∅]= o(ε),

whereas the probability that k0 ≤ |−→S (v)| ≤ εn/ log nε3 is o(ε) by Lemma 3. Thus

P[−→Sg (v) = ∅ ∧ |−→S (v)| ≥ k0] = o(ε),

which completes the proof.

As by Lemma 3 we have P[S(n, p) ≥ εn] = (2 + o(1))ε, and

P[S(n, p) ≥ εn] ≤ P[S≤g(n, p) ≥ εn] + P[Sg(n, p) > 0],it follows from (4) that

P[Sg(n, p) > 0] ≥ (2 + o(1))ε. (5)

On the other hand, by Lemmas 3 and 6, we have

P[−→Sg (v) �= ∅] = P[−→Sg (v) �= ∅ ∧ −→S (v) < k0] + P[−→Sg (v) �= ∅ ∧ −→

S (v) ≥ k0]≤ o(ε) + P[−→S (v) ≥ k0] = (2 + o(1))ε,

so together with (5), we get

P[Sg(n, p) > 0] = (2 + o(1))ε. (6)

3. THE SUBCRITICAL CASE

In this section we prove Theorem 1(i). In fact, we show the following slightly strongerresult. Let Ls, s ≥ 1, be the sth largest component of D(n, p); in case there are more thanone component of the same size, we order them lexicographically.

Theorem 7. Let np = 1 − ε, where ε = ε(n) → 0 but ε3n → ∞. Assume also thata > 0 is a constant and let Xs(n), s ≥ 1, denote the size of Ls. Then a.a.s. D(n, p) containsno complex components and

limn→∞ P(Xs < a/ε) =

s−1∑i=0

λia

i! e−λa , (7)

where λa = ∫ ∞a

e−x

x dx.



Proof. Note that each complex component contains a pair of directed cycles C1 and C2

such that the intersection of the sets of their vertices spans a directed path (which, perhaps,consists of a single vertex). It is easy to see that there are at most k2k! such pairs on agiven set of k vertices. Hence, the expected number of such pairs in D(n, p), np = 1 − ε, isbounded from above by

n∑k=3

(n

k

)k2k!pk+1 ≤ 1

n

n∑k=3

k2(1 − ε)k

≤ 1

n

n∑k=3

k2 exp(−εk) = O

(1

ε3n

).

Thus, the probability that D(n, p) contains a complex component is bounded from aboveby O(1/(ε3n)) and tends to 0 as n → ∞. This completes the proof of the first part ofTheorem 7.

Now let the random variable Ya(n) count the number of directed cycles larger thanm = �a/ε�. Then

EYa(n) =n∑

k=m

(n

k

)(k − 1)!pk = (1 + o(1))

n1/3∑k=m

1

ke−εk

= (1 + o(1))

∫ ∞

a

e−x

xdx = (1 + o(1))λa .

One can easily check that for a given r ≥ 2, the rth factorial moment of Ya(n) tends toλr

a as n → ∞. Consequently, the random variable Ya(n) tends in distribution to a Poissonvariable with expectation λa, and since clearly

P(Xs(n) < a/ε) = P(Ya(n) ≤ s − 1),

the assertion follows.

4. THE SUPERCRITICAL CASE

Throughout this section we study the structure of D(n, p) in the supercritical phase whennp = 1+ε and ε = ε(n) → 0 but ε3n → ∞ as n → ∞. Let us recall that k0 = 2ε−2 log nε3.

Let X denote the number of “large” vertices, which have both more than 1.9εn descen-dants and more than 1.9εn ancestors. The main difficulty in the proof of Theorem 1(ii) is toestimate the number of large vertices, which is done in the following lemma, whose proofwe postpone until the next section.

Lemma 8. If np = 1 + ε, where ε = ε(n) → 0 but ε3n → ∞, then the expectation of Xis given by EX = (4 + o(1))ε2n, whereas for the variance of X we have VarX = o((EX)2).

In particular, a.a.s. X = (4 + o(1))ε2n.

Once we assume that Lemma 8 holds, the rest of the proof of Theorem 1(ii) is not sodifficult. Let us start with the two following observations.



Lemma 9. If np = 1 + ε, where ε = ε(n) → 0 but ε3n → ∞, then a.a.s. at most onecomponent of D(n, p) contains cycles longer than 2 log log nε3/ε.

Proof. Let v denote a large vertex of D(n, p) (from Lemma 8 we know that a.a.s. at least

one such vertex exists). If we remove from D(n, p) all vertices of−→S (v), we get a digraph

with n′ = n − |−→S (v)| ≤ n − 1.9εn vertices. As

n′p ≤ np − |−→S (v)|p ≤ 1 − 0.8ε,

from Theorem 1(i) we deduce that a.a.s. D(n, p)\−→S (v) contains no cycles longer than, say,

2 log log nε3/ε. An analogous argument shows that a.a.s. no such cycles are contained in

D(n, p)\←−S (v). Hence, a.a.s. all cycles longer that 2 log log nε3/ε must belong to a strongly

connected component containing v.

Lemma 10. If np = 1 + ε, where ε = ε(n) → 0 but ε3n → ∞, then a.a.s. o(ε2n) largevertices of D(n, p) are not contained in cycles longer than 2 log log nε3/ε.

Proof. Let v be a vertex of D(n, p). Now generate−→S (v) in the breadth-first search process.

The probability that |−→S (v)| ≥ 1.8εn is (2+o(1))ε (see Lemma 3). For any vertex v, clearly

either←−S (v) ∩ −→

S (v) = {v} or some in-neighbors of v are in−→S (v).

If−→S (v) is large, then the probability that

←−S (v) is large and

←−S (v) ∩ −→

S (v) = {v} isbounded from above by the probability that the component of G(n′, p), n′p = 1 − 0.8ε,containing a given vertex v, is larger than 1.9εn′, which is o(ε) according to Lemma 3.Indeed, using Lemma 3 we can crudely estimate this probability by, say, ε/ log ε3n.

Now recall that g = 2 log log nε3/ε. The expected number of vertices in−→S≤g(v),

conditioned on−→S (v) being large (i.e. the process does not become extinct quickly) is

given by

E[|−→S≤g(v)||−→Sg (v) �= ∅] = E[|−→S≤g(v)|]P[−→Sg (v) �= ∅]

≤ O( g∑

i=1

(1 + ε)i

2ε

)

≤ O( (1 + ε)g

ε2

)≤ O

( log2 nε3

ε2

),

as P[−→Sg (v) �= ∅] = (2 + o(1))ε by (6). Consequently, the probability that there is an arc

starting at a vertex in−→S≤g(v) with the end at v is bounded from above by

O( log2 nε3

nε2

)= o(ε).

Thus, given that−→S (v) is large, with probability o(ε) either (i)

←−S (v) is large but

←−S (v) ∩−→

S (v) = {v} or (ii) v belongs to a directed cycle of length at most g. But if v is a largevertex which is not contained in a directed cycle longer than g, it must satisfy either (i) or

(ii). Since the probability that−→S (v) is large is (2 + o(1))ε, the expected number of such

vertices is

2εno(ε) = o(ε2n).

By Markov’s inequality, the number of such vertices is then o(ε2n).



Proof of Theorem 1(ii). From Lemmas 8 and 10 we infer that a.a.s. (4 + o(1))ε2n largevertices belong to components which contain cycles longer than 2 log log nε3/ε, whileLemma 9 ensures that such a component is unique.

We need to argue, however, that a.a.s. all remaining components are cycles of sizeOp(1/ε). Our argument is similar to the one used in the proof of Lemma 9. Let W be theset of vertices of the component of D(n, p) containing cycles longer than 2 log log nε3/ε,and let

−→W [

←−W ] denote the set of all descendants [ancestors] of vertices from W . Finally, let

W = [n] \ (W ∪ −→W ∪ ←−

W ). Note that a.a.s. W contains a large vertex, so we have both

m′ = |W ∪ ←−W | ≥ 1.9εn.

and

m′′ = |W ∪ −→W | ≥ 1.9εn.

Observe also that the digraph induced in D(n, p)by W∪−→W can be identified with D(n−m′, p),

conditioned on the event A that it contains no cycles longer than 2 log log nε3/ε. However,we have (n − m′)p ≤ 1 − 0.8ε. Hence, Theorem 1(i) implies that the probability of A tendsto 1 as n → ∞, so, using Theorem 1(i) once again, we infer that a.a.s. all components ofD(n, p) contained in W ∪−→

W are isolated vertices or cycles of length Op(1/ε). An analogous

argument shows that this is also the case for components contained in W ∪ ←−W .

Thus, to complete the proof, it is enough to show that the giant component is a.a.s.complex. To this end observe that D(n, p), np = 1 + ε, can be viewed as obtained fromthe digraph D(n, p′) where np′ = 1 + ε/2, by adding to it arcs of D(n, p′′), where 1 − p =(1 − p′)(1 − p′′). Thus, a.a.s. the largest component W of D(n, p), which has (4 + o(1))ε2nvertices, contains the largest component W ′ of D(n, p′), which has only (1 + o(1))ε2nvertices. But then, clearly, W cannot be a directed cycle, so it must be complex.

5. PROOF OF THE MAIN LEMMA

This section is devoted to the proof of Lemma 8. Let A denote the set of all large vertices ofD(n, p), that is the set of those vertices which have both more than 1.9εn descendants andmore than 1.9εn ancestors. Instead of calculating the expectation and variance of X = |A|,we estimate the number Y of elements in a certain set B closely related to A, and then showthat the symmetric difference between A and B is likely to be small.

Let g = 2ε−1 log log nε3, and let B denote the set of those vertices from which the breadth-first search lasts for at least g generations in both directions. In other words, B is the set of

vertices v such that−→Sg (v) �= ∅ and

←−Sg (v) �= ∅. We show that a.a.s. Y = |B| = (4+o(1))ε2n,

and then prove that a.a.s. |X − Y | = o(ε2n).Before calculating the expectation and variance of Y , we state a number of inequalities

which we shall use later on. Let v be a vertex in D(n, p), and let D′ be a set of vertices inD(n, p) such that |D′| = n′ ≤ n. Then |−→S (D′, v)| and |←−S (D′, v)| have the same distributionas the random variable S(n′, p), while for every d ≥ 0, |−→Sd (D′, v)| and |←−Sd (D′, v)| have thesame distribution as Sd(n′, p).

Lemma 3 implies that

P[S(n′, p) ≥ k0] ≤ (2 + o(1))ε.



If we moreover assume that n′ = n − o(εn), then

P[S(n′, p) ≥ k0] = (2 + o(1))ε.

Similarly, we see that if n′ ≤ n, then clearly (3) and (4) both hold with n exchanged withn′, while if in addition n′ = n − o(εn), then also (6) holds. Stated precisely, we have

E[Sg(n′, p)] ≤ (1 + ε)g = O(log2 nε3), (8)

E[S≤g(n′, p)] = O

(log2 nε3

ε

), (9)

P[S≤g(n′, p) ≥ εn/ log nε3] = o(ε), (10)

whenever n′ ≤ n and

P[Sg(n′, p) > 0] = (2 + o(1))ε (11)

when n′ = n − o(εn).Let us note that if Z is a non-negative random variable and A is any event with P[A] > 0,

then

E[Z | A] ≤ E[Z]P[A] . (12)

Moreover, if A is the event that Z > 0, then in (12) the equality holds. Conditioning on thebreadth-first search from v lasting at least g generations, we get from (8)

E[Sg(n′, p) | Sg(n

′, p) > 0] = E[Sg(n′, p)]P[Sg(n′, p) > 0] = O

(log2 nε3

ε

), (13)

and from (9)

E[S≤g(n′, p) | Sg(n

′, p) > 0] ≤ E[S≤g(n′, p)]P[Sg(n′, p) > 0] = O

(log2 nε3

ε2

)= o(nε). (14)

If we instead condition on the process dying out before g generations, we obtain

E[S(n′, p) | Sg(n′, p) = 0] = E[S≤g(n

′, p) | Sg(n′, p) = 0]

≤ E[S≤g(n′, p)]P[Sg(n′, p) = 0] = O

(log2 nε3

ε

). (15)

Finally, notice that Lemma 5 gives

E[S(n′, p) | S(n′, p) < k0] = O

(log1/2 nε3

ε

). (16)

To prove Lemma 8, we first estimate the expectation of the number Y of vertices in B.Let v be a vertex in D(n, p). Using the breadth-first search we first expose the set

−→V := −→

S≤g(v),



and then the set←−V := ←−

S≤g(D \ −→V , v).

Let V = −→V ∪ ←−

V . Note that in the process of exposing the sets−→V and

←−V no potential arc

was tested more than once.We moreover let

−→Vg = −→

Sg (v) and←−Vg = ←−

Sg (v). The significance of these sets is that the

vertices in−→Vg are the only vertices in

−→V from which there can be an arc to D\−→

V . Similarly,

the vertices in←−Vg are the only vertices in

←−V to which there can be an arc from D \ V .

Lemma 11. E[Y ] = (4 + o(1))ε2n.

Proof. Let v be a vertex, and let−→V and

←−V be defined as above. From (11) it follows that

the probability that−→Vg �= ∅ is (2 + o(1))ε. If, in addition, |−→V | = o(εn), the probability that←−

Vg �= ∅ is also (2 + o(1))ε. The probability that |−→V | ≥ εnlog nε3 is o(ε) by (10), so it follows

that the probability that both−→Vg �= ∅ and

←−Vg �= ∅ is (4 + o(1))ε2n.

The set←−V may be strictly smaller than

←−S≤g(v), so it is possible that

←−Sg (v) �= ∅ although←−

Vg = ∅. This may happen only if there are arcs from−→V to

←−V . As noted earlier there can

be no arc from−→V<g to D \ −→

V , so we only have to consider arcs going from−→Vg to

←−V . The

expected number of such arcs is E[|−→Vg | · |←−V |]p. The random variables |−→Vg | and |←−V | are not

independent, but we note that the bound for |←−V | which follows from (15) holds regardlessof the size of

−→Vg . Hence, the probability that there is an arc from

−→Vg to

←−V is bounded from

above by

O

(log2 nε3

ε

)O

(log2 nε3

ε

)p = O

(log4 nε3

nε2

)= o(ε)

according to (13) and (15). This completes the proof of the lemma.

The next step in the proof is to bound the variance of Y .

Lemma 12. VarY = o(E[Y ]2).

Proof. We prove the lemma by showing that E[Y(Y − 1)] = (1 + o(1))E[Y ]2. That is, we

calculate the number of pairs of distinct vertices v and w such that−→Sg (v) �= ∅,

←−Sg (v) �= ∅,−→

Sg (w) �= ∅, and←−Sg (w) �= ∅.

To this end, we assume that v is a vertex such that−→Sg (v) �= ∅ and

←−Sg (v) �= ∅, and

count the number of vertices w �= v such that−→Sg (w) �= ∅ and

←−Sg (w) �= ∅. Our goal is to

show that the expected number of such vertices w is (4 + o(1))ε2n. As previously, we firstexpose the descendants and ancestors of v, producing the sets

−→V and

←−V . We then expose

the descendants and ancestors of w in a similar way, letting

−→W = −→

S≤g(D \ V , w)

and←−W = ←−

S≤g(D \ (V ∪ −→W ), w).



We also define the sets−→Wg = −→

Sg (D \ V , w) and←−Wg = ←−

Sg (D \ (V ∪ −→W ), w). Finally let

W = −→W ∪ ←−

W and D′ = D \ (V ∪ W).By (11), we have

P[−→Wg �= ∅ | v ∈ B ∧ w /∈ V ] = (2 + o(1))ε

and

P[←−Wg �= ∅ | v ∈ B ∧ w /∈ V ∧ −→Wg �= ∅] = (2 + o(1))ε,

so clearly

P[−→Wg �= ∅ ∧ ←−Wg �= ∅ | v ∈ B ∧ w /∈ V ] = (4 + o(1))ε2.

But the breadth-first searches used to construct−→Wg and

←−Wg are restricted to vertices that have

not yet been exposed. It is therefore entirely possible that−→Sg �= ∅ even though

−→Wg = ∅,

and similarly that←−Sg �= ∅ even though

←−Wg = ∅. Our task is to show that these events are

not likely, or more precisely to show that they happen with probability o(ε).Thus, we will show the following. Let A1 be the event that v ∈ B, w /∈ V , and

−→Wg = ∅.

Then we will show that

P[−→Sg (w) �= ∅ | A1] = o(ε). (17)

Likewise, letting A2 be the event that v ∈ B, w /∈ V ,−→Sg (w) �= ∅ and

←−Wg = ∅, we will show

that

P[←−Sg (w) �= ∅ | A2] = o(ε). (18)

Finally we will have to consider those vertices which are contained in V ; thus, we will showthat

E[B ∩ V ] = o(nε2). (19)

Hence, if v ∈ B and (17), (18), and (19) hold, then the expected number of vertices wdistinct from v such that w ∈ B as well, is

E[Y − 1 | v ∈ B] ≤ nP[−→Sg (w) �= ∅ ∧ ←−Sg (w) �= ∅ | w /∈ V ] + E[B ∩ V ]

= nP[−→Sg (w) �= ∅ | w /∈ V ]P[←−Sg (w) �= ∅ | w /∈ V ∧ −→Sg (w) �= ∅] + o(nε2)

≤ n(P[−→W �= ∅ | w /∈ V ] + P[−→Sg (w) �= ∅ | A1])× (P[←−W �= ∅ | w /∈ V ∧ −→

Sg (w) �= ∅] + P[←−Sg (w) �= ∅ | A2]) + o(nε2)

= n((2 + o(1))ε + o(ε))((2 + o(1))ε + o(ε)) + o(nε2) = (4 + o(1))nε2.

Thus, to prove the lemma, it is sufficient to show (17), (18), and (19).



1. Proof of (17). If A1 holds and−→Sg (w) �= ∅, it must be because there are arcs from−→

W to V . Note that there can be no arcs from−→W to

←−V<g, as the origin of any such arc would

be in←−V . We therefore only have to consider arcs from

−→W to

←−Vg and

−→V . As we shall soon

see the probability that there are arcs to←−Vg is o(ε). The probability that there are arcs to

−→V

is significantly greater, so we will have to look at this case much more carefully.We distinguish between two types of vertices in

−→V , malevolent and benevolent ones.

Loosely speaking, the malevolent vertices are those which have many descendants, whereasthe benevolent vertices have few descendants. We will show that w has a malevolent descen-dant with probability o(ε). On the other hand, w may have several benevolent descendants,but it is unlikely that it has so many of them that the total number of its descendants becomesk0 or more. Because of Lemma 6, this will imply the claim.

Recall that the set−→V was found using a breadth-first search starting with the vertex v. In

the course of exposing the vertices in−→V we have therefore exposed several arcs between

vertices in−→V , such that we have a directed tree, rooted at v. Let us call this tree

−→T . This

tree has the same vertex set as−→V , whereas the set of arcs of

−→T is a subset of the set of arcs

of D(n, p) joining vertices in−→V .

If x is a vertex in−→V , we define

−→Tx to be the subtree of

−→T rooted at x. In precise terms,

−→Tx

consists of those vertices y ∈ −→T such that the unique path from v to y in

−→T passes through

x. Let−→Tx

′ = −→Tx ∩−→

Vg be the set of descendants of x in−→T which are in the last generation of−→

V . These are the only vertices in−→Tx from which there can be an arc to D \ −→

V . Note thatthis set may be empty. We moreover define the set

−→Ux by

−→Ux = −→

Tx ∪⋃y∈−→

Tx′

−→S≤g(D

′, y).

Thus, to obtain−→Ux , we first find the subtree of

−→T rooted at x and then continue the breadth-

first search for up to g generations in D′. We say that x ∈ −→V is malevolent if

−→Tx

′ is nonemptyand Sg(D′, y) �= ∅ for some y ∈ Tx

′. Otherwise we say that x is benevolent.

Let us first identify the vertices in−→V which are descendants of w. Let us expose all arcs

from−→W to

−→V and denote by

−→Z1 the set of vertices in

−→V to which there is an arc with origin

in−→W . Then we expose the descendants of these vertices, but without exposing any more

internal arcs in−→V , letting

−→U1 =

⋃x∈−→

Z1

−→Ux .

Having found these extra descendants of w, we expose the possible arcs going from−→U1 to−→

V , and continue as above. We define the sets−→Z2 ,

−→Z3 , . . ., and

−→U2,

−→U3, . . ., as follows: For

m ≥ 2, let−→Zm be the set of vertices in

−→V \ (

−→U1 ∪ . . . ∪ −−→

Um−1) to which there is an arc from−−→Um−1, and let

−→Um =

⋃x∈−→

Zm

−→Ux \

m−1⋃i=1

−→Ui .



Finally, let−→U = ∪m≥1

−→Um be the set of all descendants of w found by this procedure. Observe

that by this process we are guaranteed to expose the entire set of−→U , even though

−→Ux and

−→Uy

may overlap for many choices of x and y. Furthermore, observe that if none of the verticesin

−→U are malevolent, then the breadth-first searches used to expose the sets

−→U1,

−→U2, . . ., are

allowed to continue until no more descendants can be found in D′. In this case there are noarcs from

−→U to the rest of D′. Thus, if there is also no arc from

−→W ∪ −→

U to←−V or

←−W , then−→

W ∪ −→U comprises all descendants of w. We can therefore conclude that if

−→Sg (w) �= ∅ and

A1 holds, at least one of the following events must happen.

i. w has a malevolent descendant.ii. There is an arc from

−→W or

−→U to

←−V or

←−W .

iii. |−→W ∪ −→U | ≥ k0.

iv.−→Sg (w) �= ∅ and |−→S (w)| < k0.

We show that each of these events happens with probability o(ε). That (iv) holds withprobability o(ε) follows from Lemma 6, so we only have to consider the three first events.From (13) and (16) we have

E[|−→Vg ||A1] = O

(log2 nε3

ε

)(20)

and

E[|−→W ||A1] = O

(log1/2 nε3

ε

). (21)

Let us first count the expected number of malevolent vertices. If y ∈ −→Vg , the probability

that y is malevolent is simply the probability that−→Sg (D′, y) �= ∅, which is bounded from

above by (2 + o(1))ε according to (11). Every vertex in−→Vg has g ancestors in

−→T , so by

(20) the expected number of malevolent vertices in−→V is bounded from above by

O

(log2 nε3

ε

)(2 + o(1))εg ≤ O

(log3 nε3

ε

). (22)

Now let us count the expected number of vertices in−→U1,

−→U2, . . .. Suppose that x is a

randomly chosen vertex in−→V and let y be a vertex in

−→Vg . As y has g ancestors in

−→T , the

probability that x is an ancestor of y in−→T is g/|−→V |. The expected number of descendants

of x in−→Tx

′ is therefore

E[|−→Tx′||A1] = gE[|−→Vg ||A1]

|−→V |= O

(ε−2 log3 nε3

)|−→V |

. (23)

There are at most g vertices in the unique path in−→Tx from x to y for every y ∈ −→

Tx′, so

E[|−→Tx ||A1] ≤ E[|−→Tx′||A1]g = O

(ε−3 log4 nε3

)|−→V |

. (24)



If x is benevolent, then−→Sg (D′, y) = ∅ for every y ∈ −→

Tx′. In this case, by (15),

E[|−→Ux||A1] ≤ E[|−→Tx ||A1] + E[|−→Tx′||A1]O

(log2 nε3

ε

)

= O(ε−3 log4 nε3

)|−→V |

+ O(ε−2 log3 nε3

)|−→V |

O

(log2 nε3

ε

)

= O(ε−3 log5 nε3

)|−→V |

. (25)

The expected number of arcs going from−→W to

−→V is E[|−→W | · |−→V ||A1]p. Note that (21)

holds even if we condition on the size of−→V . Thus

E[|−→Z1 ||A1] ≤ O

(log1/2 nε3

ε

)|−→V |. (26)

For m ≥ 1, let Bm be the event that all the sets−→U1, . . . ,

−→Um contain only benevolent vertices.

Then, from (25) and (26),

E[|−→U1||A1 ∧ B1] ≤ E[|−→Z1 || A1] E[|−→Ux|| A1]

= O

(log11/2 nε3

nε4

). (27)

Generally, if Bm+1 holds, then

E[|−−→Zm+1||A1 ∧ Bm+1] ≤ E[|−→Um| · |−→V ||A1 ∧ Bm]p. (28)

Therefore, according to (25),

E[|−−→Um+1||A1 ∧ Bm+1] ≤ E[|−−→Zm+1||A1 ∧ Bm+1] E[|−→Ux||x is benevolent]

= E[|−→Um||A1 ∧ Bm]O(

log5 nε3

nε3

). (29)

By induction it follows that

E[|−→Um||A1 ∧ Bm] ≤ O

(ε−1Cm log5m+1/2 nε3

(nε3)m

)(30)

for m ≥ 1, where C is the constant implicit in (29).We can now calculate the probability that w has a malevolent descendant. The expected

number of arcs from−→W to a malevolent vertex in

−→V is, according to (21) and (22),

O

(log1/2 nε3

ε

)O

(log3 nε3

ε

)p = O

(log7/2 nε3

nε2

).



If we assume that Bm holds, then the expected number of arcs from−→Um to a malevolent

vertex is bounded from above by

O

(ε−1Cm log5m+1/2 nε3

(nε3)m

)O

(log3 nε3

ε

)p = O

(εCm log5m+7/2 nε3

(nε3)m+1

).

Hence, the probability that−→U contains a malevolent vertex is

O

(ε∑m≥0

Cm log5m+7/2 nε3

(nε3)m+1

)= O

(log7/2 nε3

nε2

)= o(ε).

This proves that the event (i) happens with probability o(ε).Now we consider the cases (ii) and (iii). From (30) we find that

E[|−→U ||A1] = O

(ε−1

∑m≥1

Cm log5m+1/2 nε3

(nε3)m

)= O

(log11/2 nε3

nε4

), (31)

so by (21) and (31),

E[|−→W | + |−→U ||A1] = O

(log1/2 nε3

ε

). (32)

Markov’s inequality then gives us

P[|−→W | + |−→U | ≥ k0] ≤ E[|−→W | + |−→U |]k0

= O(ε−1 log1/2 nε3

)2ε−2 log nε3

= o(ε),

thereby proving that (iii) happens with probability o(ε).What remains is to calculate the probability that there is an arc from

−→W ∪ −→

U to←−V or←−

W . We first note that there can be no arcs from−→W ∪ −→

U to←−V<g, except from vertices in

−→Vg .

The equation (20) also holds for←−Vg , so together with (32) this shows that the probability

that there is an arc from−→W ∪ −→

U to←−Vg is

O

(log1/2 nε3

ε

)O

(log2 nε3

ε

)p = O

(log5/2 nε3

nε2

)= o(ε).

We have made no assumptions regarding the order of←−W , so the expected size is given by

(9). The probability that there is an arc from−→W ∪ −→

U to←−W is therefore

O

(log1/2 nε3

ε

)O

(log2 nε3

ε

)p = O

(log5/2 nε3

nε2

)= o(ε).

The only arcs we have yet to consider are therefore the ones from−→U ∩ −→

Vg to←−V<g. Let−→

Z = ⋃m≥1

−→Zm. From (26), (28), and (30) it follows that E[−→Z ] = O(log1/2 nε3/(nε))|−→V |.

Then, by (23), we get

E[|−→U ∩ −→Vg ||A1] = O

(log7/2 nε3

nε3

).



The expected size of←−V<g is bounded by (14), so the probability that there is an arc from−→

U ∩ −→Vg to

←−V<g is bounded by

O

(log7/2 nε3

nε3

)O

(log2 nε3

ε2

)p = o(ε).

This completes the proof of (17).

Proof of (18). We now assume that w is a vertex such that−→Sg (w) �= ∅, and we want to

calculate the probability that←−Sg (w) �= ∅, given that

←−Wg = ∅.

We want to avoid exposing arcs leading to vertices which we have exposed before, toavoid dependencies. In the previous step, we may have had to expose some of the vertices in

D′ in order to determine whether−→Sg (w) �= ∅. To be precise, we may have exposed some of

the vertices in−−→S≤2g(v)\−→

V , but we do not know how many. However, to keep our calculations

as tidy as possible, we assume that we have exposed all of the vertices in−−→S≤2g(D′ ∪ −→

V , v),

and we call this set−→V ′. In addition we let

−→V2g

′ = −→S2g(D′ ∪ −→

V , v). We have

E[|−→V ′||−→Sg (v) �= ∅] ≤E

[∑2gi=0(1 + ε)i

]P[−→Sg (v) �= ∅]

= O

(log4 nε3

ε2

)(33)

and similarly

E[|−→V2g′||−→Sg (v) �= ∅] = O

(log4 nε3

ε

). (34)

We moreover let D′′ = D′ \ V ′.The calculations needed to prove (18) are very similar to those in the previous case, and

we use the same notation as earlier, with the arrows reversed: in the course of exposing←−V ,

we have exposed a directed tree, rooted at v, which we call←−T . If x is a vertex in

←−V , we

define←−Tx to be the subtree of

←−T rooted at x, and we let

←−Tx

′ = ←−Tx ∩ ←−

Vg . Furthermore←−Ux is

defined such that←−Ux = ←−

Tx ∪⋃y∈←−

Tx′

←−S≤g(D

′′, v).

A vertex y ∈ ←−Vg is malevolent if

←−Sg (D′′, y) �= ∅. A vertex x ∈ ←−

V<g is malevolent if it has a

malevolent ancestor in the tree←−T . Vertices in

←−V which are not malevolent are benevolent.

We let←−Z1 be the set of vertices x in

←−V such that there is an arc from x to some vertex in←−

W , and we let←−U1 =

⋃x∈←−

Z1

←−Ux .

For m ≥ 1,←−−Zm+1 is the set of vertices x in

←−V \ (

←−U1 ∪ . . . ∪ ←−

Um), such that there is an arcfrom x to some vertex in

←−Um, and for m ≥ 2,

←−Um =

⋃x∈←−

Zm

←−Ux \

m−1⋃i=1

←−Ui .



Finally←−Z = ⋃

m≥1

←−Zm and

←−U = ⋃

m≥1

←−Um. As in the previous case, we see that if A2 is

satisfied and←−Sg (w) �= ∅, then at least one of the following events must happen.

i. w has a malevolent ancestor.ii. There is an arc from

−→V ′ or

−→W to

←−W or

←−U .

iii. |←−W ∪ ←−U | ≥ k0.

iv.←−Sg (w) �= ∅ and |←−

S≤g(w)| < k0.

Most of the calculations which show that each of these events happens with probabilityo(ε) are basically identical to those in the previous case, so we skip them. For example, theprobability that there is an arc from

−→Vg to

←−W is virtually the same as the probability that, in

the previous case, there was an arc from−→W to

←−Vg . The only differences from the previous

case is that we have−→Wg �= ∅ by assumption, and we have to consider the possibility of arcs

from−→V2g

′ to←−U and from

−→V ′ to

←−U ∩ ←−

Vg .By similar calculations as earlier, we obtain

E[|−→Wg||A2] = O

(log2 nε3

ε

), (35)

E[|←−W ||A2] = O

(log1/2 nε3

ε

), (36)

E[|←−U ||A2] = O

(log11/2 nε3

nε4

), (37)

and

E[|←−U ∩ ←−Vg ||A2] = O

(log7/2 nε3

nε3

). (38)

Thus, by (35), (36), and (37), the probability that there is an arc from−→Wg to

←−W or

←−U is

O

(log2 nε3

ε

)O

(log1/2 nε3

ε

)p = O

(log5/2 nε3

nε2

)= o(ε).

By (34) and (37), the probability that there is an arc from−→V2g

′ to←−U is

O

(log4 nε3

ε

)O

(log11/2 nε3

nε4

)p = O

(log19/2 nε3

n2ε5

)= o(ε).

By (33), (14) and (37), the probability that there is an arc from−→V ′ or

−→W to

←−U ∩ ←−

Vg is(O

(log4 nε3

ε2

)+ O

(log1/2 nε3

ε2

))O

(log7/2 nε3

nε3

)p = o(ε).

This implies that (18) holds.



Proof of (19). It remains to calculate the expected number of vertices w ∈ V such that

w ∈ B. Let us first count the number of such vertices in−→V . As by assumption

←−Sg (v) �= ∅,

for every vertex w in−→V we have

←−Sg (w) �= ∅, so we have to bound the probability that−→

Sg (w) �= ∅. Recall that−→T is the rooted tree on the same vertex set as

−→V consisting of the

arcs of−→V which were exposed in the course of the breadth-first search beginning at v, with−→

Tw being the subtree of−→T rooted at w.

If x ∈ −→Vg , the probability that

−→Sg (D \ V , x) �= ∅ is at most (2 + o(1))ε by (11). We reuse

the term malevolent to mean vertices x such that either x ∈ −→Vg and

−→Sg (D \ V , x) �= ∅, or

x ∈ −→V<g and x has a descendant y in

−→T such that y ∈ −→

Vg and−→Sg (D \ V , y) �= ∅. A vertex

x ∈ −→V is benevolent if it is not malevolent. Every vertex y in

−→Vg has at most g ancestors in−→

Tg , so the expected number of malevolent vertices is at most

E[|−→Vg ||−→Vg �= ∅](2 + o(1))εg = O

(log3 nε3

ε

)= o(nε2).

Suppose now that w is a benevolent vertex. We let

−→U (1)

w = −→Tw ∪

⋃x∈−→

Tw∩−→Vg

S≤g(D \ V , x).

For m ≥ 1 we let−→Z (m)

w be the set of vertices x in−→V such that there is an arc from a vertex

in−→U (m)

w to x, and we let

−→U (m+1)

w =⋃

x∈−→Z (m)

x

−→U (1)

x \m⋃

i=1

−→U (i)

w .

For m ≥ 1, let B(m)w be the event that none of the sets

−→U (1)

w , . . .−→U (m)

w contains malevolentvertices. Using the same calculations as in (24) and (25), we get

E[∣∣−→U (1)

w

∣∣|v ∈ B ∧ B(1)w

] = O(ε−3 log5 nε3)

|−→V |, (39)

and we can show that

E[∣∣−→U (m)

w

∣∣|v ∈ B ∧ B(m)w

] = n

|−→V |O

(Cm log5m nε3

(nε3)m

)(40)

for a constant C. Suppose now that B(m)w holds. The probability that there is an arc from−→

U (m)w to a malevolent vertex is

1

|−→V |O

(Cm log5m nε3

(nε3)m

)O

(log3 nε3

ε

).

The probability that one of the sets−→U (1)

w ,−→U (2)

w , . . ., contains a malevolent vertex is therefore

O

(log3 nε3

ε|−→V |

)O

(∑m≥1

Cm log5m nε3

(nε3)m

)= 1

|−→V |O

(log8 nε3

nε4

).



Thus, the expected number of vertices w in−→V which have a malevolent descendant is

O

(log8 nε3

nε4

)= o(nε2). (41)

If we now let B(∞)w be the event that the sets

−→U (m)

w for m = 1, 2, . . ., do not contain any

malevolent vertices, and let−→Uw = ⋃

m≥1

−→U (m)

w , then by (39) and (40),

E[|−→Uw||v ∈ B ∧ B(∞)

w

] = 1

|−→V |O

(log5 nε3

ε3

).

If B(∞)w holds, the probability that |−→Uw| ≥ k0 is therefore

O(ε−3 log5 nε3)

|−→V |k0

= 1

|−→V |O

(log4 nε3

ε

).

The expected number of vertices w in−→V such that none of the sets

−→U (1)

w ,−→U (2)

w , . . ., contains

malevolent vertices, and |−→Uw| ≥ k0 is thus

O

(log4 nε3

ε

)= o(nε2). (42)

The number of vertices w in−→V such that w ∈ B is counted by (41) and (42), and is o(nε2).

By symmetry, the same holds for←−V . This completes the proof of the lemma.

Proof of Lemma 8. Lemmas 11 and 12 together with Chebyshev’s inequality imply that

the number Y of vertices v such that−→Sg (v) �= ∅ and

←−Sg (v) �= ∅ is a.a.s. (4+o(1))ε2n. Thus,

to complete the proof of Lemma 8, it is enough to show that a.a.s. |X − Y | < o(ε2n).

Let us calculate first the expected number of vertices v in D(n, p) such that−→Sg (v) �= ∅,←−

Sg (v) �= ∅ and |−→Sg (v)| < 1.9εn. From the remarks preceding Lemma 3 it follows that

P[k0 ≤ |−→S (v)| < 1.9εn] = o(ε). On the other hand, Lemmas 3 and 6 imply that the

probability that−→Sg (v) �= ∅ and |−→S (v)| < 1.9εn is o(ε). Furthermore, if we assume that

|−→S (v)| < 1.9εn, then

P[←−Sg (v) �= ∅||−→S (v)| < 1.9εn] ≤ P[|Sg(n, p)| > 0] + P[∃ an arc from−→S (v) to v]

≤ (2 + o(1))ε + 1.9ε = (3.9 + o(1))ε.

Thus the probability that−→Sg (v) �= ∅,

←−Sg (v) �= ∅ and |−→S (v)| < 1.9εn is o(ε2). By symmetry,

the probability that−→Sg (v) �= ∅,

←−Sg (v) �= ∅ and |←−S (v)| < 1.9εn is also o(ε2). Hence, by

Markov’s inequality, a.a.s. |B \ A| = o(ε2n).Before we show that a.a.s. |A\B| = o(ε2n), let us define C as the set of vertices v of D(n, p)

such that v belongs to a cycle �C of length at most g and either |−→S (v, D \ V(�C))| ≥ 1.9εn,

or |←−S (v, D \ V(�C))| ≥ 1.9εn. Note that by Lemma 3, the expectation of |C| is boundedfrom above by

E|C| ≤∑k≤g

k

(nk

)k!pk4ε ≤ g2ε(1 + ε)g ≤ log3 nε3

ε= o(nε2), (43)



so, by Markov’s inequality, a.a.s. |C| = o(nε2). Now let us estimate the probability that for

a vertex v /∈ C we have−→Sg (v) = ∅, |−→S (v)| ≥ 1.9εn, and |←−S (v)| ≥ 1.9εn. Lemma 6 asserts

that the probability that−→Sg (v) = ∅ and |−→S (v)| ≥ 1.9εn is o(ε). On the other hand, as there

are no arcs from−→S (v) to v, by Lemma 3, the probability that |←−S (v)| ≥ 1.9εn conditioned

on the fact that−→Sg (v) = ∅ and |−→S (v)| ≥ 1.9εn, is at most (2 + o(1))ε. In the same way,

one can argue that the probability that for v /∈ C we have←−Sg (v) = ∅, |←−S (v)| ≥ 1.9εn and

|−→S (v)| ≥ 1.9εn is o(ε2). Thus, by Markov’s inequality, a.a.s. |A \ B| = o(ε2n) + |C| =o(ε2n).

We conclude this section with the remark that simple calculations similar to that presentedin (43) show that, as in the case of random graph (see [4] and [10]), in the supercriticalcase a.a.s. long cycles of D(n, p) are contained in the giant component while short cycleslie outside it.

6. THE CRITICAL PHASE

In this final section we comment briefly on the critical behavior of D(n, p), i.e., on thecomponent structure of D(n, p) when np = 1 + O(n−1/3). Although using our method wecannot get theorems so strong as the ones obtained for G(n, p) in [5] and [13], nonethelessTheorems 13 and 14 below show that the critical behavior of both D(n, p) and G(n, p) issomewhat similar.

Let us define the excess of a digraph as the difference between the number of its edges andthe number of its vertices. An elementary walk is a walk joining two vertices of combineddegree at least three, such that each of its vertices, except of the ends, has both in- and out-degree one. Note that each elementary walk is either a directed path or a directed cycle. Adigraph is complex if its excess is positive. The spread of a complex digraph is the length ofits shortest elementary walk if the maximum degree of the digraph is three, and 0 otherwise.

The following result provides some information on the number, size, and structure ofcomplex components of D(n, p) in the critical phase.

Theorem 13. Let pa = pa(n) be such that npa = 1 + (a + o(1))n−1/3, where a is a (notnecessarily positive) constant. Then

i. the complex components in D(n, pa) (if there exist any) have Op(n1/3) verticescombined;

ii. the spread of each complex component in D(n, pa) is �p(n1/3);iii. the combined excess of complex components in D(n, pa) is Op(1); in particular, there

are Op(1) of them.

Proof. Let ω = ω(n) be any function which tends to infinity as n → ∞. Consider a graphD(n, p′), where np′ = 1 + ω1/5n−1/3. Because of Theorem 1, a.a.s. D(n, p′) contains theunique complex component of size smaller than 5ω2/5n1/3. As we can view D(n, pa) as asubgraph of D(n, p′), all complex components of D(n, pa) must be contained in D(n, p′), sothey a.a.s. must have fewer than 5ω2/5n1/3 vertices combined. Thus (i) follows.

Note that each complex component of spread smaller than contains a subgraph ofexcess one and spread smaller than , which either is a directed cycle with a path diagonal



or consists of two directed cycles joined by a directed path. Let X(k, ) denote the numberof such structures on k vertices in D(n, pa). Then,

EX(k, ) ≤(

nk

)k!6kpk+1

a ≤ 6eak/n. (44)

Since we have shown that a.a.s. each component has size smaller that k0 = 5ω2/5n1/3 it isenough to show that a.a.s. D(n, pa) contains no such structures with 0 = n1/3/ω and k ≤ k0.But from (44) we get that the probability of the existence of such structure is bounded fromabove by

k0∑k=1

6eak0/n ≤ 75eaω−1/5 → 0,

so (ii) follows.Finally, (iii) is an easy consequence of (i) and (ii) and the observation that a graph of k

vertices and excess m > 0 has spread bounded above by k/(3m).

Our last result states that the complex components can emerge at every stage of thecritical phase.

Theorem 14. Let pa = pa(n) be such that npa = 1 + (a + o(1))n−1/3, where a is a (notnecessarily positive) constant. Let P(n, a) denote the probability that D(n, pa) contains acomplex component. Then,

0 < lim infn→∞ P(n, a) ≤ lim sup

n→∞P(n, a) < 1.

Proof. From Theorems 1 and 13 it follows that there exist constants b > a, x, and s > 0,such that if npb = 1 + bn−1/3, then for large enough n with probability at least 0.9 therandom digraph D(n, pb) contains a unique complex component C of size smaller than5b2n1/3, which has excess smaller than x and spread larger than sn1/3. Note that D(n, pa) canbe viewed as obtained from D(n, pb) by deleting its edges independently with probability

1 − pa/pb = (b − a + o(1))n−1/3.

Thus, the probability that we delete none from at most 5b2n1/3 + x edges of C is boundedaway from 0 and, consequently, the probability that D(n, pa) contains a unique complexcomponent is bounded away from zero as well. Moreover, as each elementary walk inC has at least sn1/3 edges, the probability that we destroy such a walk in the process ofobtaining D(n, pa) from D(n, pb) is bounded away from 0. Thus, the probability that wedestroy all elementary walks in C is also bounded away from zero, and so is the probabilitythat D(n, pa) contains no complex components.

In a similar way, one can show, for instance, that with probability bounded away fromboth 0 and 1 D(n, pa) contains, say, precisely three complex components, two of excessthree and one of excess eleven. Nevertheless, from our argument it does not follow thatP(n, a) defined in Theorem 14 tends to a limit as n → ∞, as it is undoubtedly the case (forthe proof of the analogous result for G(n, p) see [5] and [13]).



ACKNOWLEDGMENTS

This research started during the second author’s stay at Adam Mickiewicz University andwas completed when both the authors visited the Mittag-Leffler Institute; we would like tothank both institutions for their hospitality. Finally, we wish to thank the referees for theirvaluable comments and suggestions.

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