A STUDY OF RANDOM HYPERGRAPHS AND
DIRECTED GRAPHS
DISSERTATION
Presented in Partial Fulfillment of the Requirements for the Degree Doctor of
Philosophy in the Graduate School of the Ohio State University
By
Daniel J Poole, MS
Graduate Program in Mathematics
The Ohio State University
2013
Dissertation Committee:
Boris Pittel, Advisor
Neil Falkner
Matt Kahle
c Copyright by
Daniel J Poole
2013
ABSTRACT
We establish and describe the likely structure of random hypergraph and directed
graph models, expanding upon classical results that pertain to the likely structure of
the random undirected graph models.
The random d-uniform hypergraph process, H̃n , begins as an empty hypergraph
on n vertices. At each step, a new hyperedge of cardinality d is added; each new
hyperedge is chosen uniformly at random among all non-present hyperedges. We
prove a hypergraphic analogue of the Bollobás-Thomason Theorem: for each fixed
k ∈ N, with high probability, the stopping time for H̃n having minimum degree at
least k coincides with the stopping time for H̃n being k-connected. Also, the diameter
at the moment of connectivity is shown to be one of at most nine deterministic values
near ln n/ ln((d − 1) ln n). Finally, the sharp threshold of existence of a weak (Berge)
Hamilton cycle is established.
We study the random directed graph model D(n, m = cn), a random instance of a
directed graph on n vertices with m arcs. Karp and Luczak independently found that
for c > 1, with high probability, there is a unique strong component with size linear
in n. We prove that the pair of the numbers of the vertices and arcs in the largest
strong component, once properly centered and scaled, is asymptotically Gaussian
distributed.
ii
To Nikki, Loki, and Osiris
iii
ACKNOWLEDGMENTS
First, I would like to thank my advisor, Boris Pittel. My interest in Random
Graphs was bred by his infectious enthusiasm for the field. Without his guidance and
encouragement this work would not have been possible.
I would also like to thank my fellow students, Huseyin Acan, Nicholas Peterson,
and Christopher Ross, whose comments and suggestions over the past couple of years
in our research seminars were profoundly helpful. Additionally, I am grateful to
Neil Falkner and Matthew Kahle for reviewing my dissertation despite their busy
schedules.
iv
VITA
1984 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Born in Kettering, OH
2007 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
B.Sc. in Mathematics, Chemistry,
The Ohio State University
2011 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
MA in Mathematics,
The Ohio State University
2007-Present . . . . . . . . . . . . . . . . . . . . . . . . . .
Graduate Teaching Associate,
The Ohio State University
FIELDS OF STUDY
Major Field: Mathematics
Specialization: Combinatorial Probability
v
TABLE OF CONTENTS
Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
ii
Dedication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
iii
Acknowledgments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
iv
Vita . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
v
List of Figures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
ix
CHAPTER
1
2
3
PAGE
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1.1 Basic Hypergraph Definitions . . . . . . . . . . . . . . . . . . . . .
1.2 Hypergraph Results . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Digraph Results . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
5
9
Stopping Time of k-Connectivity . . . . . . . . . . . . . . . . . . . . .
13
2.1
2.2
2.3
2.4
2.5
2.6
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Diameter at the Moment of Connectedness . . . . . . . . . . . . . . . .
47
3.1
3.2
3.3
3.4
3.5
3.6
3.7
47
48
50
54
60
68
72
Overview . . . . . . . . . . . . . . . .
Minimum Degree Through the Critical
With High Probability τk = τk0 . . . .
Deleting k − 1 Vertices . . . . . . . .
With High Probability τk = Tk . . . .
Sharp Threshold of k−connectivity .
Overview . . . . . . . . . . . . . . . .
Breadth First Search in Hypergraphs .
Likely Upper Bound on |Γi | . . . . . .
Typical |Γ3 | is Zero or Logarithmically
Likely Lower Bound on |Γi | . . . . . .
Diameter of Hd (n, m) . . . . . . . . .
Diameter of H̃ (T1 ) . . . . . . . . . . .
vi
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Window
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Large
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4
Threshold of Weak Hamiltonicity . . . . . . . . . . . . . . . . . . . . .
4.1
4.2
4.3
4.4
4.5
4.6
4.7
4.8
4.9
5
Overview . . . . . . . . . . . . . . . . . . . . .
Hypergraph Analogue of De la Vega’s Theorem
Pósa’s Lemma . . . . . . . . . . . . . . . . . .
No Small Non-expanding Sets . . . . . . . . . .
Bounding P (a, b) . . . . . . . . . . . . . . . . .
No Medium Non-expanding Sets . . . . . . . .
No Large Non-expanding Sets . . . . . . . . . .
Completing the proof of the main result . . . .
Relation to Stopping Times . . . . . . . . . . .
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. 75
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Asymptotic Normality of Strong Giant . . . . . . . . . . . . . . . . . . 114
5.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . .
5.2 The Greedy Deletion Algorithm . . . . . . . . . . . .
5.3 Enumerating digraphs with constrained in/out-degrees
5.4 Asymptotic Transition Probabilities . . . . . . . . . .
5.5 The Characteristic Function Recursion . . . . . . . . .
5.6 Approximating the PDE for fj . . . . . . . . . . . . .
5.7 Solving the PDE for fj . . . . . . . . . . . . . . . . .
5.8 The PDE for ψj,k . . . . . . . . . . . . . . . . . . . .
5.9 The Special Case and c ↓ 1 . . . . . . . . . . . . . . .
5.10 Large Deviation Bounds . . . . . . . . . . . . . . . . .
5.11 Asymptotic Distribution of (ν̂, µ̂) starting at generic s
5.12 Asymptotic Distribution of s(D(n, m = cn)) . . . . . .
5.13 Asymptotic Distribution of (ν̂, µ̂) of D(n, m = cn) . .
5.14 Asymptotic Distribution of the Strong Giant . . . . .
5.15 Asymptotic Behavior of the Covariance Matrix, B(c) .
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114
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Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195
6.1 Possible Extensions to Hypergraph Results . . . . . . . . . . . . . 195
6.2 Possible Extensions to Digraph Results . . . . . . . . . . . . . . . 202
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 206
APPENDICES
A
Finding the (1, 1)−core Suffices . . . . . . . . . . . . . . . . . . . . . . 211
A.1
A.2
A.3
A.4
Overview . . . . . . . . . . . .
The Depth-First Search . . . .
Concentration of the size of L .
Number of vertices in cycles . .
vii
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211
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A.5 Descendant sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228
A.6 Can stop the deletion process early . . . . . . . . . . . . . . . . . . 232
B
Partial Derivatives from Deletion Process . . . . . . . . . . . . . . . . . 235
C
Glossary of terms and notations . . . . . . . . . . . . . . . . . . . . . . 239
viii
LIST OF FIGURES
FIGURE
PAGE
1.1
A 4-uniform hypergraph . . . . . . . . . . . . . . . . . . . . . . . . .
2
2.1
A 4-uniform hypergraph with ess-deg(v) = 3 but deg(v) = 4 . . . . .
14
3.1
A breadth-first search in a 3-uniform hypergraph . . . . . . . . . . .
48
4.1
A weak cycle in a 4-uniform hypergraph . . . . . . . . . . . . . . . .
75
5.1
Strong Connectivity Issues . . . . . . . . . . . . . . . . . . . . . . . . 115
5.2
A matching and its corresponding directed multigraph . . . . . . . . 127
6.1
A 2-overlapping cycle in a 5-uniform hypergraph . . . . . . . . . . . 199
6.2
An unlikely birth of a first cycle . . . . . . . . . . . . . . . . . . . . . 203
ix
CHAPTER 1
INTRODUCTION
In this dissertation, we generalize classical results in the theory of random (undirected) graphs to random hyper graphs and directed graphs. The field of random
graphs is commonly thought to have been founded by Paul Erdős and Alfréd Rényi
in a series of papers starting in 1959; in this field, some probability distribution is
assigned to some collection of graphs and one is interested in the “typical” structure,
usually as the number of the vertices tends to infinity.
1.1
Basic Hypergraph Definitions
A hypergraph, H = (V, E), is an ordered pair of sets consisting of a set of vertices,
V , and a set of hyperedges, E, where hyperedges are subsets of V . Unless otherwise
stated, we will be using V = [n] := {k ∈ N : k ≤ n} for some n ∈ N. A d-uniform
hypergraph is a hypergraph where each hyperedge has cardinality d. In particular,
[n]
a d-uniform hypergraph has hyperedge set E ⊂ E(n, d) :=
, the collection of
d
all subsets of [n] of cardinality d. From now on, we concern ourselves only with duniform hypergraphs for some d ≥ 2. Of particular interest, a graph is a 2-uniform
hyperedges and its hyperedges are simply called edges. See Figure 1.1 for a sketch of
a 4-uniform hypergraph; as in figures throughout this dissertation, the vertices are
denoted by dots, while the hyperedges are curves encompassing some vertices.
1
Figure 1.1: A 4-uniform hypergraph
A subhypergraph (or simply subgraph) of a hypergraph, H = (V, E), is a hy 0
V
0
0
0
0
0
pergraph, H = (V , E ), such that V ⊂ V and E ⊂ E ∩
. We say that
d
P = (v0 , e1 , v1 , . . . , el , vl ) is a path of length l from v0 to vl in H if vi ∈ V, ei ∈ E
such that the vi0 s are distinct and vi−1 , vi ∈ ei . We say that P spans the vertices
v0 , v1 , . . . , vl . A distinction between graphs and d-uniform hypergraphs with d ≥ 3 is
that the hyperedges in a path can contain vertices other than v0 , v1 , . . . , vl . A hypergraph is connected if there is a path between any pair of vertices. A component is a
maximal connected subgraph. A trivial component is a component consisting of a single vertex. For a vertex, v, we define the degree of v to be deg(v) = |{e ∈ E : v ∈ e}|.
There are three random hypergraph models that we will consider. We define
Hd (n, p) as the random hypergraph on vertex set [n], where each potential hyperedge
e ∈ E(n, d) is present with probability p independent of all other elements of E(n, d).
In particular, if H is a d-uniform hypergraph on [n], then
n
P (Hd (n, p) = H) = pe(H) (1 − p)( d )−e(H) ,
where e(H) is the number of hyperedges of H. Due to the distribution of the whether
2
a specific possible hyperedge is present, this model is sometimes called the Bernoulli
random hypergraph model. Let Hd (n, m) be the random hypergraph distributed
uniformly on the set of all hypergraphs, (V, E), with E ⊂ E(n, d) and |E| = m. The
(nd)
model Hd (n, m) can be gainfully viewed as a snapshot of the process {Hd (n, µ)}µ=0
(furthermore denoted H̃n or simply H̃), where Hd (n, µ + 1) is obtained from Hd (n, µ)
n
by inserting an extra hyperedge chosen uniformly at random among all
− µ still
d
available elements of E(n, d). When discussing a snapshot of the random process, we
will denote Hd (n, m) by H̃(m). For d = 2, these hypergraph models are the usual
(n2 )
well-studied random graph models, G(n, p), G(n, m), and G̃ = {G(n, µ)}µ=0
.
Typically, we consider the parameters, p and m, as depending on n (the number
of vertices) and wish to determine the likely structure of these random hypergraph
models with these parameters as n tends to infinity. Formally, a property, P, is a
subset of hypergraphs. For some p = p(n), we say that with high probability (denoted
w.h.p.) the model Hd (n, p) has property P if lim P (Hd (n, p) ∈ P) = 1. Similarly, we
n→∞
say w.h.p. Hd (n, m) has property P if lim P (Hd (n, m) ∈ P) = 1. By close analogy
n→∞
with definition of almost surely in probability spaces, some authors use the phrase
asymptotically almost surely instead of with high probability.
In the hypergraph process, H̃, we typically care about the stopping time, TP , for
some property, P, defined by TP = min{m : H̃(m) ∈ P}. Note that H̃(TP ) is the
random hypergraph process stopped at TP , the first moment the hypergraph develops
the property P. Viewed yet another way, the stopping time TP is the random variable
of the number of hyperedges present at the first time the hypergraph process has the
property P. We say w.h.p. H̃(TP ) has property Q if lim P (H̃(TP ) ∈ Q) = 1;
n→∞
in words, we say that w.h.p. at the first moment that the hypergraph process has
property P, the hypergraph also has property Q.
These three random hypergraph models are closely related. In fact, Hd (n, p)
3
conditioned on having exactly m present hyperedges (i.e. e(Hd (n, p)) = m) is equal
n
in distribution to Hd (n, m). The expected number of hyperedges of Hd (n, p) is
p,
d
n
which leads us to suspect that for m ≈
p, the structures of Hd (n, p) and Hd (n, m)
d
are asymptotically similar, provided that e(Hd (n, p)) is sharply concentrated around
n
p. This informal relation will be rigorously shown later to convert results between
d
these two models.
For m1 < m2 (or p1 < p2 ), we can view Hd (n, m1 ) (resp. Hd (n, p1 )) as a subgraph
of Hd (n, m2 ) (resp. Hd (n, p2 )). This is obvious for Hd (n, m), by our hypergraph
process H̃. For Hd (n, p), introduce the family, {Xe }e∈E(n,d) , of independent [0, 1]uniform random variables. Define E(p) = {e ∈ E(n, d) : Xe ≤ p}. Clearly each
[n], E(p) has the same distribution of Hd (n, p) and moreover E(p) increases with p.
An increasing property, P, is a property closed under the addition of hyperedges;
formally, P is an increasing property if for each H ∈ P and potential hyperdege
e ∈ E(n, d), we have H ∪ {e} ∈ P. Similarly, a decreasing property is a property
where for each H ∈ P and e ∈ E(n, d), we have H \ e ∈ P. For any increasing
property P and any p1 < p2 , m1 < m2 , we have that
P (Hd (n, p1 ) ∈ P) ≤ P (Hd (n, p2 ) ∈ P),
and
P (Hd (n, m1 ) ∈ P) ≤ P (Hd (n, m2 ) ∈ P),
because of the subgraph containment noted above.
In the field of random graphs, one often desires to find when a certain property
arises. The function m(n) is a threshold of property P, if for any ω(n) → ∞,
lim P (Hd (n, m(n)/ω(n)) ∈ P) = 0, lim P (Hd (n, ω(n)m(n)) ∈ P) = 1,
n→∞
n→∞
4
and m(n) is a sharp threshold if for each > 0,
lim P (Hd (n, (1 − )m(n)) ∈ P) = 0, lim P (Hd (n, (1 + )m(n)) ∈ P) = 1.
n→∞
n→∞
In other words, for m0 much smaller than threshold m, w.h.p. Hd (n, m0 ) does not
have the property, P, while if m0 is much larger than m, then w.h.p. Hd (n, m0 ) does
have this property. Analogously, the function p(n) is a threshold for property P, if
for any ω(n) → ∞,
lim P (Hd (n, p(n)/ω(n)) ∈ P) = 0, lim P (Hd (n, ω(n)p(n)) ∈ P) = 1,
n→∞
n→∞
and p(n) is a sharp threshold if for each > 0, we have
lim P (Hd (n, (1 − )p(n)) ∈ P) = 0, lim P (Hd (n, (1 + )p(n)) ∈ P) = 1.
n→∞
n→∞
Suppose f (x) > 0 and consider x → a, where a is a real number or infinite. We say
g(x) = O(f (x)) as x → a, if there is some constant C > 0 such that |g(x)| ≤ Cf (x)
for all x near a; i.e. if a is finite, for all x in some interval around a and if a is infinite,
for all x sufficiently large. We will also be using the notation |f | ≤b g for f = O(g).
In words, one says that f is at most on order of g. We say that f is on the order of
g, denoted g(x) = Θ(f (x)), as x → a, if there are some positive constants c and C
such that cf (x) ≤ g(x) ≤ Cf (x) for all x near a. Likewise, we define g(x) = o(f (x))
g(x)
= 0.
as x → a, if lim
x→a f (x)
1.2
Hypergraph Results
In one of the very first papers on random graphs, Erdős and Rényi [18] showed that
n
m = ln n is a sharp threshold for connectivity for G(n, m). Later, Stepanov [59]
2
established the sharp threshold of connectivity for G(n, p) among other results. In
n
this thesis, we prove that m = ln n is a sharp threshold for connectivity in Hd (n, m);
d
5
this result has been claimed before [12], but no proof has been given, as far as we
n
know. We establish a stronger result: for m = (ln n + cn ) , cn = O(1), w.h.p.
d
Hd (n, m) consists of only one non-trivial component and a set of isolated vertices
whose size is asymptotically Poisson distributed. This suggests that the most likely
barrier for connectedness is not the existence of two or more non-trivial components
but rather the presence of isolated vertices. In fact, Bollobás and Thomason [9] would
find that w.h.p. at the moment that the graph process G̃ loses its last isolated vertex,
this process also becomes connected. In other words, w.h.p. the stopping time for G̃
having minimum degree at least 1 coincides with the stopping time for connectedness.
We prove the analogous result for the hypergraph process, H̃.
There are various measures for the strength of connectedness of a connected graph,
but here we will focus on k-(vertex)connectivity. Suppose H = (V, E) is a hypergraph
and V0 ⊂ V . We define the hypergraph obtained by deleting the vertices V0 from
H as the hypergraph H 0 = (V \ V0 , E 0 ), where E 0 = {e ∈ E : e ∩ V0 = ∅}. In
other words, we also delete any hyperedges containing a vertex from V0 . For k ∈ N,
a hypergraph is k-connected if deleting any set of k − 1 vertices from H results in
a connected hypergraph. Note that the definition of 1-connectedness coincides with
connectedness.
n
(ln n + (k − 1) ln ln n + ω(n)), and ω(n) → ∞ however
d
slowly, then w.h.p. Hd (n, m) is k-connected; also we find the analogous statement for
We will find that, if m =
Hd (n, p) extending the known results of Erdős-Rényi [21], Ivchenko [26], and Bollobás
[4]. Let τk be the stopping time for H̃ having minimum degree at least k and let Tk
be the stopping time for H̃ being k-connected. In [9], Bollobás and Thomason show
that in the graph process G̃, for any k = k(n) ∈ [1, n], we have that w.h.p. τk = Tk .
We will show that for d ≥ 3 and fixed k, with high probability, τk = Tk . In other
6
words, at the precise moment the hypergraph process H̃ loses its last vertex of degree
less than k, the hypergraph becomes k-connected.
Among extremal characteristics of graphs, the diameter has played an especially
important role. The distance between vertices v and w, denoted d(v, w), is the length
of the shortest path connecting v and w, and is defined to be infinite if no such path
exists. The diameter of a hypergraph is the maximum distance between all pairs of
vertices. For a connected finite graph, the diameter is finite, while the diameter of a
disconnected graph is infinite by definition. The diameter of G(n, p) has been studied
in great detail; in particular, see Burtin [10] and Bollobás [5].
Notice that in the hypergraph process, H̃, the stopping time for connectedness, T1 ,
is also the first moment that the diameter is finite. We will find that the diameter of
H̃(T1 ) is approximately Dn := ln n/ ln((d − 1) ln n). Moreover, w.h.p. the diameter is
one of at most 9 values in the vicinity of Dn extending a similar result for G̃ proved by
Bollobás [7]. In addition, we will determine the concentration of the diameter of both
Hd (n, p) and Hd (n, m), for p and m sufficiently close to but above their respective
connectedness thresholds.
Existence problems of Hamilton cycles have a rich history in random graphs. First
posed by Erdős and Rényi [20], even a correct threshold of Hamiltonicity resisted the
efforts of researchers until a breakthrough by Pósa [57] and Korshunov [35], who
found that w.h.p. G(n, m = cn ln n) has a Hamilton cycle if c is sufficiently large.
Later Korshunov [36], Komlós and Szemerédi [33], and Bollobás [6] established the
sharp theshold for Hamiltonicity as well as the probability of Hamiltonicity within
the narrow window enclosing the critical p (resp. m) in G(n, p) (resp. G(n, m)). As
a culmination point, Bollobás [7] showed w.h.p. the first moment the graph process,
G̃, has minimum degree at least 2 coincides with the moment the process develops a
Hamilton cycle.
7
As for hypergraphs, the first problem that arises is how to even define a cycle, let
alone a Hamilton cycle. In this thesis we mainly use the notion of a Berge cycle, see
Berge [2]. In Chapter 6, we will comment on another commonly used cycle notion,
the so-called l-overlapping cycles.
For l ≥ 3, a (Berge) cycle, C = (v0 , e1 , v1 , . . . , el , vl = v0 ), of length l is an
alternating sequence of vertices and hyperedges such that v1 , . . . , vl−1 are distinct,
e1 , . . . , el are distinct, and vi−1 , vi ∈ ei . A weak cycle is defined similarly except we do
not require the hyperedges to be distinct. We say that the (weak) cycle C spans the
vertices v0 , v1 , . . . , vl−1 .
For d = 2 (the graph case), the edges of a weak cycle must be distinct, so the
definition of weak cycle and cycle coincide. A Hamilton cycle is a cycle on the entire
vertex set and a hypergraph is called Hamiltonian if there is a Hamilton cycle. Similarly a weak Hamilton cycle is a weak cycle on the entire vertex set and a hypergraph
is weak Hamiltonian if there is a weak Hamilton cycle. One crucial difference between
the definitions of cycle and weak cycle is that for d ≥ 3, it is possible for a vertex
with degree 1 to be in weak cycle, while necessarily this vertex cannot be in a cycle.
n
For d ≥ 3, we will prove that if m = (ln n + ω(n)) and ω(n) → ∞ however
d
slowly, then w.h.p. Hd (n, m) contains a weak Hamilton cycle. This contrasts with
n
the result for d = 2 that w.h.p. G(n, m = (ln n + ln(ln n) + ω)) is Hamiltonian, but
2
n
G(n, m = (ln n + ln(ln n) − ω)) is not Hamiltonian. Unlike the graph case, the most
2
likely barrier for weak Hamiltonicity is isolated vertices rather than vertices of degree
n
1. We will show that w.h.p. Hd (n, m = (ln n + c)), for c ∈ R, has a weak cycle
d
on the set of all non-isolated vertices. Furthermore, for the stopping time for weak
Hamiltonicity, T , we show that (T − τ1 )/n converges to 0 in probability, although we
strongly suspect that P (T − τ1 = 0) → 1.
8
1.3
Digraph Results
First, we begin with some definitions. A directed graph (digraph in short) is an ordered
pair of sets, D = (V, E), consisting of a set, V , of vertices along with a set, E, of
arcs which are ordered pairs of distinct vertices, i.e. E ⊂ V × V \ {(v, v) : v ∈ V }.
A path, P = (x = v0 , e1 , v1 , . . . , el , vl = y), from x to y of length l is an alternating
sequence of vertices and arcs so that vi ∈ V, ei ∈ E where v0 , v1 , . . . , vl are distinct and
ei = (vi−1 , vi ). As opposed to paths in graphs and hypergraphs, direction matters!
For instance, in a digraph, there can be a path from x to y, even though there is not
a path from y to x.
A cycle, C = (v0 , e1 , . . . , el , vl = v0 ), of length l is an alternating sequence of
vertices and arcs where v0 , v1 , . . . , vl−1 are distinct and ei = (vi−1 , vi ). A digraph is
strongly connected if for each v, w ∈ V , there is a path from v to w and a path from
w to v. A strong component is a maximal strongly connected subdigraph. Strong
components made up of a single vertex are called trivial.
While there is a simple way to decide whether a set of vertices form a component
of a graph, determining the strong components is a little more cumbersome. For
instance, a vertex set V0 in graph G form a component if and only if the induced
graph on V0 is connected and there are no edges from V0 to V \ V0 . In digraphs, a
vertex set V0 forms a strong component if and only if the induced digraph on V0 is
strongly connected and there are no vertices, v, in V \ V0 such that there is a path
from v to V0 and a path from V0 to v. This path condition makes proving results
about strong components much more difficult.
For a vertex, v, we define the in-degree of v, denoted in-deg(v), as the number of
arcs that lead into v. Formally, we have that
in-deg(v) := |{e = (w, v) ∈ E : w ∈ V }|.
9
Similarly the out-degree of v, denoted out-deg(v), is the number of arcs leaving v.
Necessarily, each vertex in a non-trivial strong component must have both in-degree
and out-degree at least 1. However, vertices can have positive in-degree and positive
out-degree and still be in a trivial component.
The random digraph models are defined similarly to our random hypergraph models. We define D(n, p) as the random digraph on vertex set [n], where each of the
potential arcs is present with probability p independently of one another. In particular, if D is a digraph on [n] and e(D) denotes the number of arcs in D, then
P (D(n, p) = D) = pe(D) q n(n−1)−e(D) .
Let D(n, m) be the random digraph with arc-set chosen uniformly at random among
n(n − 1)
all
m-element subsets of [n] × [n] \ {(v, v) : v ∈ [n]}. For p1 < p2 and
m
m1 < m2 , we can couple the models in such a way that D(n, p1 ) and D(n, m1 ) are
subgraphs of D(n, p2 ) and D(n, m2 ), respectively.
In one of the early works in random graphs, Erdős and Rényi [19] discovered the
double-jump phase transition for G(n, m) that occurs near m ∼ n/2. For instance,
if c < 1, then w.h.p. the largest component of G(n, m = cn/2) has size on the
order of ln n; if c = 1, the largest component has size on the order of n2/3 ; and if
c > 1, the largest component has size on the order of n, while the other components
have size at most on the order of ln n. Later, Bollobás [7] found similar results for
c sufficiently close to 1, namely n1/3 (c − 1) = Θ (ln n)1/2 . For c > 1, we call this
unique component, of order n, the giant component since it is so much larger than
the other components. There are many results concerning the structure of this giant
component. Of particular interest here is the k-core, which is the maximal subgraph
with minimum degree k, and the excess of the giant component, which is the number
of edges in the giant component subtracted by the number of vertices in the giant.
10
Although there are many more characteristics, our results on random digraphs are
more closely related to these.
Stepanov [60] and Pittel [50] showed that the size of the giant component, once
centered and scaled, has an asymptotically normal distribution. Moreover, Pittel
proved that the giant component consists of a giant 2-core with a mantle of trees
of total size on the order of n, plus possibly a few unicyclic graphs sprouting off
this 2-core. Janson, Knuth, Luczak, and Pittel [29] proved many results about the
structure of G̃ through the double jump, including the asymptotic normality of the
excess. Later Pittel and Wormald [56] established the asymptotic joint Gaussian
distribution of the size of the 2-core, size of the tree mantle, and the excess. An
immediate consequence is that the joint distribution of the number of vertices and
number of edges of the giant component is asymptotically Gaussian.
Analysis of the phase transition for random digraphs has proven to be more difficult and results are relatively scarce. Most notably, Karp [31] and Luczak [39] showed
that if p = c/n and c > 1, then w.h.p. D(n, p) has a strong component of size on
the order of n. For c < 1, the largest strong component of D(n, p = c/n) has size
bounded in probability; in sharp contrast, the largest component of the undirected
random graph in the analogous range has size with exact order ln n. More recently,
Luczak and Seierstad [43] investigated the behavior of D(n, p = c/n) when c → 1
slowly, namely n1/3 |c − 1| → ∞. Though these papers provide concentration results
for the size of the giant strong component, its limiting distribution has remained undetermined. In this thesis, we show that the number of vertices, N , and the number
of arcs, M , of the giant strong component of D(n, m = cn), c > 1, is asymptotically
2-dimensional Gaussian. More precisely, if θ = θ(c) is the unique root in (0, 1) of
1 − θ = e−cθ for c > 1, then we show that
N − θ2 n M − cθ2 n
D
,
=⇒ N (0, B),
1/2
1/2
n
n
11
where N (0, B) is a 2-dimensional Gaussian random variable with mean 0 and a
certain 2 × 2 covariance matrix B = B(c) determined later.
To prove this result we develop and analyze a deletion algorithm, of a kind that
has been found useful in studies of undirected graphs. At each step, we will delete
some vertices with in-degree or out-degree zero. Necessarily these deleted vertices
could not have been in a non-trivial strong component in the initial digraph. Using
the Markov nature of this algorithm, we will approximate the likely realization of the
deletion process by a deterministic trajectory which is a solution of a certain system
of ordinary differential equations. Associated with this system are integral formulas
for entries of a covariance matrix. To show convergence to the Gaussian distribution,
we prove convergence of the 2-dimensional characteristic function, of the number of
vertices and the number of arcs in a terminal directed graph. To turn this result into
the main claim on (N, M ), we show that w.h.p. the terminal digraph consists of a
strong giant component and some negligible “residue,” i.e. some vertices and arcs,
whose size orders are dwarfed by the random fluctuations of the number of arcs and
vertices in the terminal digraph.
12
CHAPTER 2
STOPPING TIME OF k-CONNECTIVITY
2.1
Overview
In [9], Bollobás and Thomason proved for any k = k(n) ∈ [1, n−1] w.h.p. the stopping
time of the graph process, G̃, being k−connected coincides with the stopping time
of G̃ having minimum degree at least k. In this thesis, we establish a hypergraphic
counterpart of their result for any fixed k (and hence, trivially, for any kn bounded).
Earlier, we defined the stopping times τk = min{m : H̃(m) has min deg ≥ k} and
Tk = min{m : H̃(m) is k − connected} (as a reminder, a hypergraph is k−connected
if whichever k − 1 vertices are deleted, the induced hypergraph is connected). Deterministically, τk ≤ Tk , because if a vertex, v, has degree at most k − 1, we can delete
one vertex from each of v 0 s incident hyperedges to isolate v. The goal of this chapter
is to show w.h.p. τk = Tk . To this end, we will need to define a different, albeit
stronger notion of degree. The essential degree of a vertex v, denoted by ess-deg(v),
is the maximum number of hyperedges containing v, whose pairwise intersections are
only {v}. Formally, if E is the set of hyperedges containing v in H, then
ess-deg(v) =
max
E⊂E,∀e,f ∈E,e6=f =⇒ f ∩e=v
|E|.
Trivially, ess-deg(v) ≤ deg(v). See Figure 2.1 for an example of a hypergraph where a
13
vertex has essential degree less than its degree; notice that the pairwise intersections
of e1 , e3 , e4 is exactly {v}. We define the stopping time
τk0 = min{m : H̃(m) has min ess-deg ≥ k}.
e2
e1
v
e3
e4
Figure 2.1: A 4-uniform hypergraph with ess-deg(v) = 3 but deg(v) = 4
Our argument is as follows: First, we will find the typical range of τk and show
that w.h.p. τk = τk0 . Then, we demonstrate that prior to this typical range of τk0 ,
w.h.p. the random hypergraph process is “almost” k−connected, in the sense that
upon deletion of any k − 1 vertices, there remains a component containing almost all
left-over vertices. Finally, using this structural result about the hypergraph process
before τk0 , we can show that the main obstacle to k-connectedness is the existence of
vertices with essential degree less than k in the sense that once τk0 is reached, it is
likely that the process is k-connected.
14
2.2
Minimum Degree Through the Critical Window
In this section, the minimum degree of the random hypergraph Hd (n, m) is determined
for a certain interval of m and as a consequence the typical range of τk is found.
First, a Poisson random variable with parameter λ, denoted P oi(λ), is an nonnegative-valued random variable, where for all j ∈ {0, 1, 2, . . .},
P (P oi(λ) = j) = e−λ
λj
.
j!
n
(ln n + (k − 1) ln ln n + cn ), where cn → c ∈ R, and let Z
d
be the number of vertices of degree less than or equal to k − 1 in Hd (n, M ). Then Z
Lemma 2.2.1. Let M =
converges in distribution to a Poisson random variable with parameter
e−c
.
(k − 1)!
Also, w.h.p. there are no vertices of degree less than k − 1.
Proof. Note that Z = X +Y , where X is the number of vertices of degree k−1 and
Y is the number of vertices of degree less than k − 1. By the Converging Together
Lemma, see Durrett[17], it suffices to show that X converges in distribution to a
Poisson random variable and P (Y = 0) → 1. While not true for all distributions,
convergence of the moments of X to the moments of a Poisson random variable implies
convergence in distribution of X to this Poisson random variable. In particular, it
suffices to show that for each fixed r ∈ N,
lim E[(X)r ] =
n→∞
e−c
(k − 1)!
r
,
where (ν)r = ν(ν − 1) · · · (ν − r + 1) is the r−falling factorial of ν (see Bollobás [8]).
Claim 2.2.2. For each fixed r ∈ N,
lim E[(X)r ] =
n→∞
15
e−c
(k − 1)!
r
.
Proof. By symmetry,
E[(X)r ] =
X
P (vi1 , . . . , vir have degree k − 1)
i1 <i2 <...<ir
=
X
P (v1 , . . . , vr have degree k − 1)
i1 <i2 <...<ir
= (n)r P (v1 , . . . , vr have degree k − 1).
Now we have that
P (v1 , . . . , vr have degree k − 1) = P1 + P2 ,
where P1 is the probability that each vertex of {v1 , v2 , . . . , vr } has degree k − 1 and
that there is no hyperedge containing at least two vertices of {v1 , v2 , . . . , vr }, and P2
is the probability these vertices have degree k − 1 but there is at least one hyperedge
containing at least two vertices of {v1 , . . . , vr }. That is, P1 corresponds to the event
that there are r(k − 1) hyperedges on {v1 , . . . , vr } while P2 corresponds to the case
where the number of such hyperedges is less than r(k − 1). The reason for considering
P1 and P2 is that we anticipate, and will prove, that P2 = o(P1 ). Note that if r = 1,
then the event corresponding to P2 is empty, so P2 = 0. By the construction of our
random hypergraph Hd (n, M ), we have that
n−1
P1 = |H| ·
d
M
,
(2.2.1)
where H is the collection of hypergraphs on n vertices with m hyperedges (of cardinality d) such that v1 , . . . , vr have degree k −1 and no hyperedge contains at least two
of v1 , . . . , vr . To compute this probability, we need to find the number of such hyper
n−r
graphs. There are
potential hyperedges that contain vi and d − 1 vertices
d− 1
n−r
of [n] \ [r] and
potential hyperedges contained entirely within [n] \ [r]. Each
d
hypergraph can be uniquely constructed from choosing k − 1 hyperedges from the
16
n−r
potential
hyperedges on vi , i ∈ [1, r], and M − r(k − 1) hyperedges among
d−1
potential hyperedges contained entirely within [n] \ [r]. Hence
|H| =
n−r
d−1
r n−r
d
.
M − r(k − 1)
k−1
(2.2.2)
Using (2.2.2), the probability in (2.2.1) becomes
r
(n−r
d−1 )
P1 =
(n−r
d )
M −r(k−1)
k−1
.
(nd)
(2.2.3)
M
This second binomial expression in the numerator above can be expanded as follows
n−r
d
n−r
M − r(k − 1) + 1
d
= n−r
− M + r(k − 1) M − r(k − 1) + 1
d
n−r
(M − r(k − 1) + 2)2
d
= n−r
= ...
− M + r(k − 1) 2 M − r(k − 1) + 2
d
n−r
(M )r(k−1)
d
.
= n−r
( d − M + r(k − 1))r(k−1) M
M − r(k − 1)
Since k and r are fixed, we have that as n tends to infinity,
(M )r(k−1) = (1 + o(1))
n
d
ln n
r(k−1)
,
d r(k−1)
n−r
n
− M + r(k − 1)
= (1 + o(1))
d
d! ,
r(k−1)
and
n−r
d−1
r
k−1
= (1 + o(1))
1
(k − 1)!
nd−1
(d − 1)!
k−1 !r
.
Applying these asymptotics to the probability yield
P1 = (1 + o(1))
1
(k − 1)!
= (1 + o(1))
r
1
(k − 1)!
nd−1 nd ln n
(d − 1)! nd!d
r
(ln n)r(k−1)
!r(k−1) (n−r)
d
M
(n)
d
M
(n−r
d )
M
.
(n)
d
M
17
(2.2.4)
Now to complete finding an asymptotic expression of P1 , we need to find a sharp
bound on the fraction of binomial expressions above. Note that
M
−1 n−r
M
−1 n−r
(n−r
d )
Y
Y
−i
M
(n − r)d
d
d
M
M
=
=
=
+O
n n
n
(d)
(n)d
nd
−i
d M
d
i=0
i=0
M
M 2 M
(n − r)d
1+O
=
(n)d
nd
and
d−1
d−1 (n − r)d Y n − r − i Y
1
r
=
=
1− +O
(n)d
n−i
n
n2
i=0
i=0
d
r
rd
1
1
= 1− +O
=
exp
−
+
O
.
n
n2
n
n2
Since d ≥ 3 and M = O (n ln n), we have that
r
(n−r
d )
e−cn
rd
M
= (1 + o(1)) exp − M = (1 + o(1))
.
(n)
n
n(ln n)k−1
d
M
Hence (2.2.4) becomes
r
r
−cn r
1
1
e
e−cn
r(k−1)
P1 = (1 + o(1))
(ln n)
= (1 + o(1)) r
,
k−1
(k − 1)!
n(ln n)
n (k − 1)!
r
1
−c
e
.
which implies that lim (n)r P1 =
n→∞
(k − 1)!
Note: The methods and bounds used in computing P1 and analyzing its asymptotics will be utilized many times throughout this thesis.
To complete the proof of the claim, it suffices to show that (n)r P2 → 0 as n → ∞.
To deal with the overlap of the hyperedges on [r], introduce ij , j ∈ [1, d], the number
of hyperedges with j elements of [r] and d − j elements of [n] \ [r]. On the event
corresponding to P2 , necessarily at least one of i2 , i3 , . . . , id is non-zero. We break up
the probability P2 according to the different values that i1 , i2 , . . . , id can take on this
event. Note that
i1 + 2i2 + · · · did =
d
X
k=1
18
deg(vk ).
Let
I=
d
X
ij ,
J=
j=1
d
X
jij .
j=1
Then on the event corresponding to P2 , we must have J = r(k − 1) and at least one of
i2 , i3 , . . . , id is non-zero and so J − I ≥ 1. We break up the probability P2 as follows:
X
P2 =
P (i),
i≥0:
i1 <J=r(k−1)
where P (i) is the probability that there exist exactly ij hyperedges with exactly
j elements of [r]. We compute P (i) in a similar fashion to P1 before. There are
r
n−r
potential hyperedges that contain one vertex in [r] and d − 1 vertices
1 d − 1 r
n−r
from [n] \ [r],
potential hyperedges that contain two vertices of [r] and
2 d−2
d − 2 vertices from [n] \ [r] and so on. Hence
(r)(n−r)
n−r r
n−r (r1)(n−r
d−1 )
2
d−2
· · · (d)(d−d) ( d )
i1
P (i) =
i2
id
M −I
(2.2.5)
(nd)
M
Whereas for P1 we desired sharp bounds, for our claim an upper bound on (2.2.5) is
sufficient. First, note that
r
n−r
d−j
j
ij
≤ rj nd−j
ij
=
r jij
n
ndij .
As before,
n−r
d
M −I
(M )I
= n−r
( d − M + I)I
n−r
d
M
≤
M
n−r
−M
d
Since
M
ln n
= (1 + o(1))(d − 1)! d−1
n
−M
n−r
d
and
(n−r
d )
rd
M
= (1 + o(1)) exp − M ,
(n)
n
d
M
19
!I n−r
d
M
.
uniformly over our range of i,
!I (n−r)
I
d
ln
n
rd
M
M
≤b (d − 1)! d−1 exp − M
n n−r
(
n
n
−
M
d)
d
M
I r
ln n
1
≤b
.
nd−1
n(ln n)k−1
Applying these bounds to (2.2.5) yield
!
I r
d Y
1
r jij dij
ln n
P (i) ≤b
n
n
nd−1
n(ln n)k−1
j=1
I r J−I
1
1
1
ndI ln n
=
≤b J
≤ n−r−1 .
n
nd−1
n(ln n)k−1
n ln n
nr
Hence P2 = o(1/nr ), as desired.
We have just shown that X converges in distribution to a Poisson random variable.
To complete the proof of the lemma, it suffices to show that P (Y = 0) → 1. For
k = 1, necessarily, there are no vertices of degree less than 0. Suppose k ≥ 2. Since
Y is non-negative integer-valued, in order to show that P (Y = 0) → 1, it suffices to
show that E[Y ] tends to zero because P (Y ≥ 1) ≤ E[Y ]. Note that
E[Y ] =
k−2
X
nP (v1 has degree j).
j=0
Just as in (2.2.3) for r = 1, we find that
(n)−(n−1)
(n−1
d
d−1
d−1 )
P (v1 has degree j) =
M −j
j
n
d
( )
M
(nd)−(n−1
d−1 )
(M
)
j
M
= n(d−1)j n
n n−1
(
−
−
M
+
j
d)
d
d−1
j
M
!j
!
n−1
M
M
≤ n(d−1)j
exp − d−1n
n−1
n
− d−1 − M
d
d
j
ln n
1
1
≤b n(d−1)j
e−cn ≤b
.
d−1
k−1
n
n(ln n)
n ln n
Hence E[Y ] → 0, completing the proof of the lemma.
20
n
Corollary 2.2.3. Let ω = ω(n) → ∞, M0 = (ln n + (k − 1) ln ln n − ω) and M1 =
d
n
(ln n + (k − 1) ln ln n + ω). Then w.h.p. τk ∈ [M0 , M1 ].
d
Proof. It suffices to show that w.h.p. there exists a vertex of degree at most k − 1
in Hd (n, M0 ) and w.h.p. all vertices have degree at least k in Hd (n, M1 ).
n
Let c ∈ R and M c := (ln n + (k − 1) ln ln n + c). For large enough n, we have
d
c
that M0 ≤ M ≤ M1 . Since having a vertex of degree less than k is a monotone
decreasing property,
P (Hd (n, M0 ) has min deg ≤ k − 1) ≥ P (Hd (n, M c ) has min deg ≤ k − 1)
≥ P (Hd (n, M1 ) has min deg ≤ k − 1).
However from Lemma 2.2.1, we know that
P (Hd (n, M ) has min deg ≤ k − 1) → 1 − exp −
c
e−c
(k − 1)!
.
So for any c ∈ R,
lim inf P (Hd (n, M0 ) has min deg ≤ k − 1) ≥ 1 − exp −
n→∞
e−c
(k − 1)!
lim sup P (Hd (n, M1 ) has min deg ≤ k − 1) ≤ 1 − exp −
n→∞
e−c
(k − 1)!
,
.
By letting c tend to −∞ and ∞, the desired probabilities are obtained.
By noticing that P (τk ≤ m) = P (Hd (n, m) is k − connected), the following corollary is immediate.
n
d
(ln n + (k − 1) ln ln n)
. Then Xn converges in
n
distribution to a random variable X with cumulative distribution function FX (x) =
exp −e−x /(k − 1)! .
Corollary 2.2.4. Let Xn :=
τk −
Now, suppose that p = (d − 1)!
ln n + (k − 1) ln ln n + cn
for some cn → c. For
nd−1
this parameter p,
n
n
E[e(Hd (n, p))] =
p ≈ (ln n + (k − 1) ln ln n + cn ) .
d
d
21
One expects that the likely structure of Hd (n, p) mirrors the typical structure of
Hd (n, m) for m near the expected number of hyperedges of Hd (n, p). In fact, in an
argument very similar to proof in Lemma 2.2.1, one can prove the following lemma:
Lemma 2.2.5. Let Z denote the number of vertices of degree less than or equal to
k − 1 in Hd (n, p), where p is defined above. Then Z converges in distribution to a
e−c
. Also, w.h.p. there are no vertices
Poisson random variable with parameter
(k − 1)!
of degree less than k − 1.
Sometimes an argument can not be as easily adapted from one hypergraph model
to another.
The following lemma will be useful in converting from Hd (n, p) to
Hd (n, m).
Lemma 2.2.6. Suppose Q is a property of hypergraphs. Suppose there are positive
1
constants a and b such that an ln n ≤ M ≤ bn ln n, and p = n M . If with probability
d
at least 1 − o n−1/2 (ln n)−1/2 , Hd (n, p) has property Q, then with high probability
Hd (n, M ) has property Q.
Proof. The probability that Hd (n, p) has does not have property Q can be written
in terms of the probability that Hd (n, m) does not have property Q as in
P (Hd (n, p) ∈
/ Q) =
(nd)
X
P (Hd (n, m) ∈
/ Q)P (e(Hd (n, p)) = m),
m=0
because Hd (n, p) conditioned on e(Hd (n, p)) = m is distributed as Hd (n, m). As a
trivial consequence, we have
P (Hd (n, p) ∈
/ Q) ≥ P (Hd (n, M ) ∈
/ Q)P (e(Hd (n, p) = M ).
Now we wish to determine the the asymptotic behavior of the probability that
e(Hd (n, p)) = M . A Binomial random variable on ν trials with success probability ρ,
denoted Bin(ν, ρ), is an integer-valued random variable where for each j ∈ [0, ν],
ν j
P (Bin(ν, ρ) = j) =
ρ (1 − ρ)ν−j .
j
22
Since the hyperedges of Hd (n, p) are present with probability p independent of one
another, the number of hyperedges of Hd (n, p) is a binomially distributed random
n
variable on
trials with success probability p. So we find
d
n
n
d
P (e(Hd (n, p)) = M ) =
pM q ( d )−M
M
M
1
n
n
= (1 + o(1))
p
exp −p
.
M!
d
d
Using Stirling’s Formula on M ! yields
e M n M
n
1
p
exp −p
P (e(Hd (n, p)) = M ) = (1 + o(1)) √
d
d
2πM M
!!M
n
n
p
p
1
d
exp 1 − d
= (1 + o(1)) √
.
M
M
2πM
n
However we chose p in such a way that
p = M , so
d
P (e(Hd (n, p)) = M ) = (1 + o(1)) √
1
.
2πM
Hence
√
P (Hd (n, M ) ∈
/ Q) ≤b M P (Hd (n, p) ∈
/ Q).
√
√
Notice that M = O
n ln n and from the hypothesis, P (Hd (n, p) ∈
/ Q) =
o n−1/2 (ln n)−1/2 . Thus P (Hd (n, M ) ∈
/ Q) = o(1) as desired.
We will need a stronger result about exactly how many vertices have degree k − 1
just prior to the typical window of τk .
n
(ln n + (k − 1) ln ln n − ω) such that
d
ω ≤ ln ln n. Let X be the number of vertices in Hd (n, M0 ) with degree k − 1. Then
Lemma 2.2.7. Let ω = ω(n) → ∞ and M0 =
for each > 0,
P |X − E[X]| > E[X] → 0.
Proof. Let > 0. We will show this result by Chebyshev’s inequality involving
the first and second moments of X. We start with computing E[X].
23
Claim 2.2.8.
E[X] =
1+O
ln ln n
ln n
eω
.
(k − 1)!
Proof. We have that E[X] = nP1 , where P1 is the probability that a specific vertex
has degree k − 1. We computed this probability (albeit, for a different parameter m)
in equation (2.2.3) and we will use similar asymptotics here. So
(n−1) (n)−(n−1)
P1 =
d−1
d
k−1
M0 −k+1
d−1
(nd)
.
(2.2.6)
M0
As before, we find that
n−1
d−1
k−1
1
n(d−1)(k−1)
= 1+O
,
n
(k − 1)!((d − 1)!)k−1
and
n
− n−1
d−1
=
M0 − k + 1
n
− n−1
(M0 )k−1
d
d−1
n
n−1
M0
−
−
M
+
k
−
1
0
d
d−1
k−1
k−1 n
− n−1
M0
1
d
d−1
d! d
.
= 1+O
n
n
M0
d
Applying these asymptotics to (2.2.6) gives
k−1 (nd)−(n−1
d−1 )
1
d
1
M0
M0
P1 = 1 +
.
(nd)
n
(k − 1)! n
(2.2.7)
M0
Now we wish to analyze this last fraction of binomial expressions.
M −1 n
!
(nd)−(n−1
M
n−1
n−1
0
0 −1
d−1 )
Y
Y
−
−
i
M0
d
d−1
=
=
1 − nd−1
n
(n)
−i
−i
d
M0
i=0
M
0 −1
Y
d
d
i=0
n−1
!
M0
d
ln n
1−
=
+O
= 1− +O
n
nd
i=0
ln n
d
ln n
eω
= 1+O
exp − M0 = 1 + O
.
n
n
n
n(ln n)k−1
d−1
n
d
nd−1 i
n2d
Hence (2.2.7) becomes
k−1
ln n
eω
d
1
M0
P1 = 1 + O
n
(k − 1)! n
n(ln n)k−1
ln ln n
eω 1
= 1+O
.
ln n
(k − 1)! n
24
Thus
E[X] =
1+O
ln ln n
ln n
eω
.
(k − 1)!
Now to find the variance, we first find the second factorial moment.
Claim 2.2.9.
E[X(X − 1)] =
1+O
ln ln n
ln n
(E[X])2 .
Proof. We start be noticing that
E[X(X − 1)] =
X
P (vi , vj have degree k − 1)
i6=j
= n(n − 1)P (v1 , v2 have degree k − 1).
We find this probability in a similar fashion to P (i) in (2.2.5). Consider the event
where the number of hyperedges that contain both v1 and v2 is i ≤ k − 1. Each of
these i hyperedges contain d − 2 elements of the remaining n − 2 vertices. Then v1 is
in k − 1 − i hyperedges that do not contain v2 and similarily for v2 . The number of
hyperedges that contain v1 or v2 is k − 1 − i + k − 1 − i + i = 2(k − 1) − i. Clearly,
P (v1 , v2 have degree k − 1) =
k−1
X
P (i),
i=0
where P (i) is the probability that v1 and v2 have degree k − 1 and there are exactly i hyperedges that contain both v1 and v2 . To find P (i), notice that there
n−2
n−1
n−2
are
potential hyperedges that contain both v1 and v2 ,
−
d−2
d−1
d−2
potential hyperedges that contain v1 and not v2 and vice versa. Also, there are
n
n−1
n−2
−2
+
potential hyperedges that contain neither v1 or v2 .
d
d−1
d−2
Hence the number of hypergraphs on [n] with M0 hyperedges where the degrees of v1
and v2 are k − 1 and there are i hyperedges containing both v1 and v2 is
n−2 n−1
n−1
n−2 n−2 n−1
n−2 n
−
2
+
−
−
d
d−1
d−2
d−1
d−2
d−1
d−2
d−2
.
i
k−1−i
k−1−i
M0 − 2(k − 1) + i
25
(2.2.8)
We have that
n−2
d−2
i
1
1 n(d−2)i
= 1+O
n
i! ((d − 2)!)i
and
n−2
d−1
− n−2
1
1
n(d−1)(k−i−1)
d−2
= 1+O
.
n
(k − 1 − i)! ((d − 1)!)k−i−1
k−1−i
Now we simplify the last binomial expression of (2.2.8),
2(k−1)−i
− 2 n−1
+ n−2
d!M0
1
d−1
d−2
= 1+O
n
nd
M0 − 2(k − 1) + i
n
−2
× d
n
d
n−1
d−1
+
n−2
d−2
.
n−2
d−2
,
M0
So
2(k−1)−i
n−1
n−2 −
2
+
ln n
ln ln n
d
d−1
d−2
(d − 1)! d−1
= 1+O
ln n
n
M0 − 2(k − 1) + i
n
− 2 n−1
+
d
d−1
×
M0
n
and
+ n−2 (nd)−2(n−1
d−1 ) ( d−2 )
M0
(nd)
=
ln n
n
2d
exp − M0
n
ln n
n
e2ω
.
n2 (ln n)2k−2
1+O
M0
=
1+O
After applying these asymptotics and much cancellation, we find that
n−2 n−1
n−1
n−2 n−2 n−1
n−2 n
−
2
+
−
−
1
d
d−1
d−2
d−1
d−2
d−1
d−2
P (i) = n d−2
(d)
i
k−1−i
k−1−i
M0 − 2(k − 1) + i
M0
ln ln n
e2ω
= 1+O
c(k, i) 2+i
,
ln n
n (ln n)2i
where c(k, i) is some constant that depends on k and i. We find that
ln ln n
1
1
P (0) = 1 + O
e2ω ,
2
2
ln n
n ((k − 1)!)
26
and for i ∈ [1, k − 1],
P (i) = O
1
n3
.
So we have that
ln ln n
1
E[X(X − 1)] = 1 + O
e2ω ,
ln n
((k − 1)!)2
or using the asymptotics for E[X], we have that
ln ln n
(E[X])2 .
E[X(X − 1)] = 1 + O
ln n
Using this second factorial moment, we find that
ln ln n
var[X] = E[X(X − 1)] + E[X] − (E[X]) = E[X] + O (E[X])
ln n
2
2
.
By Chebyshev’s inequality we have that
1 var[X]
1
1
P (|X − E[X]| ≥ E[X]) ≤ 2
≤ 2·
+O
2
E[X]
E[X]
ln ln n
ln n
,
and in particular P (|X − E[X]| ≥ E[X]) → 0.
Combining these previous results we get the following corollary.
Corollary 2.2.10. Let ω → ∞ such that ω ≤ ln ln ln n. If
M0 :=
n
(ln n + (k − 1) ln ln n − ω) ,
d
then with high probability, Hd (n, M0 ) has minimum degree k −1 and there are at most
2eω vertices of degree k − 1.
Proof. By Lemma 2.2.7, w.h.p. minimum degree is at most k − 1. If k = 1, necesn
sarily minimum degree is at least 0. For k ≥ 2, let M 0 = (ln n + (k − 2) ln ln n + ω) .
d
0
Since ω ln ln n, for sufficiently large n, M < M0 and by Corollary 2.2.3, minimum
degree is at least k − 1 in Hd (n, M 0 ). This completes showing that w.h.p. minimum
degree is k − 1. The bound on the number of vertices with degree k − 1 is established
by Lemma 2.2.7.
27
2.3
With High Probability τk = τk0
Let ω = ω(n) → ∞ such that ω ≤ ln ln ln n. Let
M0 =
n
(ln n + (k − 1) ln ln n − ω)
d
M1 =
n
(ln n + (k − 1) ln ln n + ω) .
d
and
First, we will show a series of lemmas about some structural elements of Hd (n, M0 )
which we will later use to show w.h.p. τk = τk0 .
Lemma 2.3.1. With high probability in Hd (n, M0 ), there are no vertices, v, with
ess-deg(v) ≤ k − 1 and deg(v) ≥ess-deg(v) + 2.
Proof. Since a vertex with non-zero degree must have essential degree at least
one, the statement is trivially true for k = 1. Suppose k ≥ 2. For 1 ≤ l ≤ k − 1, let
Yl denote the number of vertices, v, such that ess-deg(v) = l and deg(v) ≥ l + 2. We
wish to show that w.h.p. Y1 = Y2 = . . . = Yk−1 = 0, so as before it suffices to show
that E[Yl ] tends to zero for 1 ≤ l ≤ k − 1. Note that
E[Yl ] = nP (v1 has essential degree l and degree ≥ l + 2).
(2.3.1)
Whereas before we found the exact probability and then found an upper bound for
this explicit expression, here we find an upper bound on the number of hypergraphs
on [n] with M0 hyperedges where v1 has essential degree l and degree at least l + 2.
First, we choose l potential hyperedges on v1 whose pairwise intersections are only
v1 . We can do this in
n − 1 n − d n − 2d + 1
n − (l − 1)d + l − 2
···
d−1
d−1
d−1
d−1
28
ways. Next we choose two hyperedges who contain v1 and at least one other vertex
of the previously chosen l hyperedges. There are less than
(d − 1)l
1
2
n−2
d−2
n
choices in this step. Finally, we choose M0 − (l + 2) out of all remaining
− (l + 2)
d
possible hyperedges. This argument overcounts the number of desired hypergraphs
because essential degree of v1 may have increased with addition of these last hyperedges. So we find that
P (v1 has essential degree l and degree ≥ l + 2) ≤
2 (nd)−(l+2)
n−1
n − (l − 1)d + l − 2
(d − 1)l n − 2
M0 −(l+2)
. (2.3.2)
···
(nd)
d−1
d−1
1
d−2
M0
From equations (2.3.1) and (2.3.2), we find
(nd)−(l+2)
E[Yl ] ≤ n · n(d−1)l (d − 1)l n
d−2 2
·
M0 −(l+2)
(nd)
(nd) ≤b ndl+2d−l−3
M0
= ndl+2d−l−3
≤b
n
d
(M0 )l+2
≤ ndl+2d−l−3
− M0 + l + 2 l+2
n
d
M0 −l−2
(nd)
M0
!l+2
M
0
− M0
(ln n)l+2
.
n
Lemma 2.3.2. With high probability in Hd (n, M0 ), minimum essential degree is k −1
and each vertex with essential degree k − 1 also has degree k − 1.
Proof. Essential degree is zero if and only if degree is zero, so for k = 1, the
second part of the lemma is trivial. In this case, the first part of lemma is established
by Corollary 2.2.10. Now suppose k ≥ 2. By Corollary 2.2.10 in conjunction with
Lemma 2.3.1, w.h.p. minimum essential degree is at least k − 2. For the first half of
29
the lemma, we wish to show that w.h.p. there are no vertices with essential degree
k − 2 and degree k − 1 (from Lemma 2.3.1, there are not any with degree ≥ k w.h.p.).
Once we have this, to show the second part of the lemma, we wish to show w.h.p.
there are no vertices with essential degree k − 1 and degree k.
Let Zl be the number of vertices, v, with ess-deg(v) = l, and deg(v) = l + 1, for
l = k − 2, k − 1. It suffices to show that E[Zl ] = o(1). For k = 2, Z0 = 0 because
essential degree zero implies degree zero. Suppose l > 0.
E[Zl ] = nP (v1 has ess-deg = l and degree = l + 1).
We bound this probability in a similar fashion to the probability in the last lemma,
except we pick the remaining M0 − (l + 1) hyperedges among possible hyperedges
that do not contain v1 rather than any other potential hyperedges. Now
P (v1 has ess-deg = l and degree = l + 1) ≤
(nd)−(n−1
d−1 )
n−1 n−d
n − (l − 1)d + l − 2 l(d − 1) n − 2 M0 −(l+1)
,
···
(nd)
d−1
d−1
d−1
1
d−2
M0
and so
E[Zl ] ≤b n n
d−1 l
nd−2
!l+1 (n)−(n−1)
d
d−1
M0
n
d
M0
n−1
d−1
(nd)
−
− M0
M0
!
n−1
ω
M0
e
≤b (ln n)l+1 exp − d−1n
≤ (ln n)l+1−(k−1) → 0.
n
d
Now we have sufficient knowledge about the structure of H̃(M0 ) to be able to
show w.h.p. τk = τk0 .
Lemma 2.3.3. With high probability τk = τk0 .
Proof. Let An be the event that τk ∈ [M0 , M1 ] and H̃n (M0 ) has minimum degree
k − 1 and minimum essential degree k − 1, where each vertex with essential degree
30
k −1 also has degree k −1, as well as there are at most 2eω vertices of degree k −1. By
Corollary 2.2.3, Corollary 2.2.10 and Lemma 2.3.2, we have that w.h.p. An occurs.
On the event An , if τk0 > τk , then at some m ∈ (M0 , M1 ], we must have chosen a
hyperedge that contains a vertex with degree k − 1 and one of its neighors. Let Ṽ0
be the vertices of degree k − 1 in H̃(M0 ). Suppose V0 ⊂ [n] with |V0 | ≤ 2eω . If em
denotes the hyperedge added to H̃(m − 1) to obtain H̃(m), then by a union bound,
P (An ∩ {τk < τk0 } ∩ {Ṽ0 = V0 }) ≤
M1
X
P (An ∩ {em has a vertex of V0 and one of its neighbors}).
m=M0
To bound these latter probabilities, note that em must have at least one vertex from
V0 and at least one vertex from its (l − 1)(d − 1) adjacent vertices in H̃(M0 ). Hence
we have that
P (An ∩ {τk < τk0 } ∩ {Ṽ0 = V0 }) ≤
|V0 |
1
M1
X
(k−1)(d−1)
1
n
−M
d
M =M0
n−2
d−2
n−2
d−2
≤b (M1 − M0 ) eω
n
d
− M1
≤b nωeω
ln ln n · ln ln ln n
1
≤
.
2
n
n
This bound is uniform over all such sets V0 . Hence
X
P (An ∩ {τk < τk0 }) =
P (An ∩ {τk < τk0 } ∩ {Ṽ0 = V0 }) = o(1).
V0 ,|V0 |≤2eω
Since
{τk < τk0 } ⊂ Acn ∪ (An ∩ {τk < τk0 }) ,
where the probability of each of the last two events tend to zero, the probability that
τk 6= τk0 tends to zero, as well.
2.4
Deleting k − 1 Vertices
In this section, we will show that likely H̃(M0 ) is almost k−connected. Concisely, we
prove that w.h.p. whichever k − 1 vertices are deleted, there remains a component
31
that contains almost all the left over vertices. We obtain a relatively crude lower
bound on the size of the component left after deleting k − 1 vertices from H̃(m) for m
close to M0 /2. And then we use this bound to get an even better lower bound on the
size of the leftover component in H̃(M0 ) by analyzing the likely effect of throwing in
the remaining M0 /2 hyperedges. While possible to complete in one step, we employ
this two-step estimate to simplify the calculations.
To handle these estimates for H̃(m) and H̃(M0 ), we consider the Bernoulli random
−1
−1
n
n
∗
∗
hypergraph modelsHd (n, p ) and Hd (n, p) where p = m
and p = M0
,
d
d
because the Bernoulli random hypergraph models are more amenable to asymptotic estimates. Then we will use our conversion lemma to convert these results
to Hd (n, m). Let
p∗ =
1 n ln n
.
n
2d
d
Lemma 2.4.1. With probability at least 1−n−1 , Hd (n, p∗ ) has the property “whichever
8
.”
k − 1 vertices are deleted, there remains a component of size ≥ n − n
ln n
Proof. Given a set of k − 1 vertices, w = (w1 , . . . , wk−1 ), let F(w, ν) be the
event that if w1 , w2 , . . . , wk−1 are deleted along with any hyperedges containing them,
then there remains no components of size ≥ n − ν. For ease of computation, let
= 8/ ln n. In particular, for this lemma, we wish to show that with large enough
probability, Hd (n, p) is not in F (w, n) for any w. This is equivalent to showing that
the probability of the union of F(w, n) tends to zero suitably fast. Using the union
bound and symmetry, we find that
P ∪ F (w, n) ≤
w
X
P (F (w, n)) =
w
n
P (F (w∗ , n)),
k−1
for w∗ = (v1 , . . . , vk−1 ). Note that the random hypergraph Hd (n, p∗ )\{v1 , . . . , vk−1 } is
distributed as Hd (n−k +1, p). So the probability of F (w∗ , n) in Hd (n, p∗ ) is equal to
32
the probability that there are no components of size at least n−n in Hd (n−k+1, p∗ ).
We now wish to show this latter probability is sufficiently small.
Claim 2.4.2. The probability that all components of Hd (n − k + 1, p∗ ) have size less
than n − n is at most e−n .
Before we prove this claim, notice that if this claim is true, then the probability
of the union of F(w, n) is at most
nk e−n n−1 ,
as desired. So to complete the proof of the lemma, it suffices to prove this claim.
Proof. We will show that any hypergraph without such a large component will
have two relatively large vertex sets with no hyperedge between them, which will lead
to the exponentially decaying bound on this probability.
Suppose H is a d-uniform hypergraph on n − k + 1 vertices such that each component has size less than n − n, and L1 , . . . , Ll are the vertex sets of the compoJ
nents of H. Then there is some minimal J such that | ∪ Li | ≥ n. At this point,
i=1
2
J
| ∪ Li | < n + (1 − )n = 1 −
n, otherwise minimality of J is contradicted.
i=1
2
2
J
J
In particular, | ∪ Li | is suitably bounded away from 0 and n. Let S = ∪ Li and
i=1
i=1
U = V (H) − S. Necessarily there cannot be a present hyperedge, e, such that both
e ∩ S 6= ∅ and e ∩ U 6= ∅.
For ease of notation, let A denote the event that each component has size less
than n − n in Hd (n − k + 1, p∗ ). By the above comments, on the event A, there
h n
n i
is some vertex set S ⊂ [n − k + 1] such that |S| ∈
and there are no
,n −
2
2
hyperedges including a vertex from S and a vertex from S c . By using a union bound
33
(over |S|) and then again over all subsets of cardinality |S|,
n− n
2
P (A) ≤
X
P (∃S ⊂ [n − k + 1], |S| = s, no hyperedge between S & S c )
s= n
2
n− n
2
X n − k + 1
P (no hyperedge between [s] and [n − k + 1] \ [s]).
≤
s
n
s=
2
s
n−k+1−s
There are
potential hyperedges contained within [s] and
potend
d
tial hyperedges contained entirely within [n − k + 1] \ [s]. So the number of potential
hyperedges that have non-trivial intersection with both [s] and [n − k + 1] \ [s] is
n−k+1−s
n−k+1
s
−
−
.
d
d
d
Hence
n− n
2
X n − k + 1 n−k+1
s
n−k+1−s
P (A) ≤
(1 − p∗ )( d )−(d)−( d ) .
s
n
s=
(2.4.1)
2
Trivially, we see that if
Bn :=
n−k+1
d
(1 − p∗ )(
max
)−(ds)−(n−k+1−s
),
d
s∈[n/2,n−n/2]
then
n−n/2 P (A) ≤ Bn
X
s=n/2
n−k+1
s
≤ 2n Bn .
Now it remains to show that Bn decays relatively quickly. Since 1 − x ≤ e−x and
x → ex is an increasing function, we have that
n−k+1
s
n−k+1−s
∗
Bn ≤
max
exp −p
−
−
s∈[n/2,n−n/2]
d
d
d
n−k+1
s
n−k+1−s
∗
≤ exp
max
−p
−
−
s∈[n/2,n−n/2]
d
d
d
=: exp (bn ) .
First notice that by symmetry (replacing s with n − k + 1 − s) we can consider just
the maximum over just s in [n/2, (n − k + 1)/2]. In fact, if we denote the largest
34
integer less than or equal to x, i.e. the floor of x, by bxc, then we can consider only
s in I := [n/2, b(n − k + 1)/2c]. Hence
n−k+1
s
n−k+1−s
∗
−
−
.
bn = max −p
s∈I
d
d
d
(2.4.2)
We can use the following claim to simplify finding an upper bound on bn .
Claim 2.4.3. For ν ≥ d,
ν d − d2 ν d−1 ≤ (ν)d ≤ ν d .
Proof. We have that
d
d
d2
d
(ν)d = ν(ν − 1) · · · (ν − d + 1) ≥ ν 1 −
≥ν 1−
= ν d − d2 ν d−1 .
ν
ν
d
The penultimate inequality is simply Bernoulli’s inequality. Also the upper bound
for (ν)d is trivial.
By applying the claim to (2.4.2), we obtain that
∗
p
d
2
d−1
d
d
bn ≤ max −
(n − k + 1) − d (n − k + 1)
− s − (n − k + 1 − s)
.
s∈I
d!
The following claim will show that this maximum is obtained when s = n/2.
Claim 2.4.4. Let f (x) = xd + (n − k + 1 − x)d . Then f (x) is decreasing on the
n n − k + 1
,
.
interval
2
2
Proof. We have that f 0 (x) = d xd−1 − (n − k + 1 − x)d−1 and if x is in the above
range, then this derivative is negative.
Hence
n d p∗
n d
d
2
d−1
bn ≤ −
(n − k + 1) − d (n − k + 1)
−
− n−k+1−
d!
2
2
∗
d
p
n
=−
nd − n −
+ O p∗ d nd + O p∗ nd−1
d!
2
∗
d
p d
=− n 1− 1−
+ O (n/ ln n) .
d!
2
35
Since tends to zero, we have that
1−
d
d
=1−
+ O 2 .
2
2
Hence
bn ≤ −
p∗ nd d
+ O (n/ ln n) = −2n + O (n/ ln n) .
d! 2
In particular, for all sufficiently large n, bn ≤ −1.9n and Bn ≤ exp (−1.9n). Hence
P (A) ≤ 2n Bn ≤ exp ((ln 2 − 1.9)n) ≤ e−n ,
as desired.
Now we increase p to the edge of the transition and show that the large component
after deleting k − 1 vertices is even closer to n. Let σ = k ln n/ ln ln n and ω = ln ln σ.
n
1
Let M0 = (ln n + (k − 1) ln ln n − ω) and p = n M0 . Note that a later lemma will
d
d
n
require that we set
p = M0 .
d
Lemma 2.4.5. With probability at least 1 − O n−2/3 , Hd (n, p) has the property
“whichever k − 1 vertices are deleted, there remains a component of size at least
n − σ.”
Proof. The property “whichever k − 1 vertices are deleted, there remains a component of size ≥ n − n” is an increasing property and p∗ < p. So by Lemma 2.4.1,
Hd (n, p) also has this property with probability at least 1 − O (1/n). It suffices to
show that with probability at least 1 − O n−2/3 , there is not a component of size in
[n−n, n−σ] after deleting the k−1 vertices, where as before = 8/ ln n. In particular,
we wish to show that with large enough probability, there are no w = (w1 , . . . , wk−1 )
such that F(w, σ) \ F(w, n) holds. Using an union bound yields
P ∪ F(w, σ) \ F(w, n) ≤
w
n
P (F(w∗ , σ) \ F(w∗ , n)),
k−1
36
(2.4.3)
where w = (v1 , . . . , vk−1 ). Just as before, the probability of F(w∗ , σ) \ F(w∗ , n) is
the same as the probability that there is a component of size in [n − n, n − σ] in
Hd (n − k + 1, p). We wish to show that this probability is sufficiently small. For ease
of computation, let n0 = n − k + 1. We wish to show that
Claim 2.4.6. The probability that H(n0 , p) has a component of size in [n − n, n − σ]
is at most n−k+1/3 .
Proof. Let A be this event. Note that if there is a component of size in [n −
n, n − σ], then there is a vertex set S with |S| ∈ [σ, n], such that there are no
hyperedges containing an element of S and an element of S C . In particular S could
be the vertices not in the component of size in [n − n, n − σ]. In bounding P (A), we
take the union bound over all such sets S. Note that
n 0 X
n (nd0 )−(ds)−(n0d−s)
,
P (A) ≤
q
s
s=σ
and so we find that
P (A) ≤
n 0 X
n
s=σ
s
0
d
n
s
−
exp −p
1−O
nd
d
n0 −s
d
n0
d
!!
.
(2.4.4)
Uniformly over s ∈ [σ, n], we have that
d−1
d−1 n0 −s
Y n0 − s − i Y
s
si
d
=
=
1− 0 − 0 0
n0
n0 − i
n
n (n − i)
d
i=0
i=0
2
s d
s0
s
s
=1−d 0 +O
.
= 1− +O 2
n
n
n
n2
Now let’s find the asymptotics of the exponent in (2.4.4).
0 2 2 ln n
s
s
s
n
= ln n − ω + O
s+O
p
d +O
2
n
n
n
n
d
= (ln n − ω) s + O s2 /n + O(1).
So there is some γ > 0 such that for all s ∈ [σ, n],
0 2 n
s
s
p
d +O
≥ (ln n − ω)(s − γs2 /n) − γ.
d
n
n2
37
(2.4.5)
µ
ν
ν·e
Using the estimate (2.4.5) along with the binomial bound,
≤
, in
µ
µ
equation (2.4.4) gives
s
n X
s2
en0
exp −s ln n + ωs + γ + γ (ln n − ω) .
(2.4.6)
P (A) ≤
s
n
s=σ
For sufficiently large n, we have that
s ≤ 8n/ ln n ≤
9
n,
ln n − ω
which implies that
ln n − ω 2
s ≤ 9s.
n
Our bound in (2.4.6) becomes
P (A) ≤
n X
e s
s=σ
s
exp (ωs + γ + 9γs) = e
γ
n 1+9γ+ω s
X
e
s=σ
s
.
ω
Note that e = ln σ, so we have that
s
σ
1+9γ
n 1+9γ
X
e
ln σ
e
ln σ
P (A) ≤b
≤b
.
s
σ
s=σ
By choice of σ = k ln n/ ln ln n, the bound above gives
P (A) ≤b exp (−k ln n + δ ln n) ,
for any δ > 0. In particular, for n sufficiently large P (A) ≤ n−k+1/3 .
Now applying the claim to (2.4.3) gives for sufficiently large n,
1
P ∪ F(w, σ) \ F(w, n) ≤ nk−1 · n−k+1/3 = 2/3 ,
w
n
as desired.
Now we wish to convert this last result about Hd (n, p) to the model Hd (n, m).
Corollary 2.4.7. Let σ = k ln n/ ln ln n and ω = ln ln σ. Then with high probability,
n
whichever k − 1 vertices are deleted in Hd (n, M0 = (ln n + (k − 1) ln ln n − ω)),
d
there remains a component of size at least n − σ.
38
Proof. Let Q be the property that “whichever k − 1 vertices are deleted, there
remains a component of size at least n − σ”. This corollary follows from applying the
conversion Lemma 2.2.6 to Lemma 2.4.5.
2.5
With High Probability τk = Tk
Now that we have sufficient information about the likely structure of H̃(M0 ) and
know that w.h.p. τk = τk0 , we can establish that the probability that the hypergraph,
at the first moment that minimum degree is at least k, is not k−connected tends to
zero; i.e. P (τk < Tk ) tends to zero. As before, we let σ = k ln n/ ln ln n, ω = ln ln σ
and
Mi =
n
ln n + (k − 1) ln ln n + (−1)i+1 ω ,
d
for i = 0, 1. The next lemma helps us deal with the event {τk ≤ M < Tk } on a
certain likely event.
Let Am be the event that minimum essential degree of Hd (n, m) is at least k,
Hd (n, m) is not k−connected, and Hd (n, m) has property Q, where Q is the property
“whichever k −1 vertices are deleted, there remains a component of size at least n−σ.
Lemma 2.5.1. Uniformly over M ∈ [M0 , M1 ],
P (AM ) ≤b
σ dk+1 (ln n)k
.
nd−1
Proof. On the event that Hd (n, M ) has property Q and is not k−connected,
there exist some k − 1 vertices w1 , . . . , wk−1 such that upon their deletion, there is a
component of size n − s for some 1 ≤ s ≤ σ. Let S be the vertices not in this large
component. If essential degree is at least k, then there must be some hyperedge, e, on
w which does not include an element of w1 , . . . , wk−1 and this hyperedge remains in
the induced hypergraph. Hence for any w ∈ S, there is a hyperedge containing w in
39
the induced hypergraph and necessarily e is contained in S. This tells us that S must
have size at least d. By using an union bound for both w1 , . . . , wk−1 and |S| = s, we
obtain that
X
σ n
n − (k − 1)
P (s),
k − 1 s=d
s
P (AM ) ≤
where P (s) is the probability that Hd (n, M ) has minimum essential degree at least
k, and {k, k + 1, . . . , k + n − s − 1} is a component of size n − s after deleting the
first k − 1 vertices, [k − 1]. We will now bound this P (s).
Suppose H is some hypergraph in the event corresponding to P (s) and w = vn
which is in S. From before, we know that there is at least one hyperedge on w contained in S, and that any hyperedge containing w, before deletion, either contains a
deleted vertex or is contained entirely within S. Suppose there are exactly i hyperedges containing w that are completely contained in S. So i ≥ 1 and since ess-deg(w)
is at least k, if i < k, then are at least k − i hyperedges containing w (before deletion)
that contain a deleted vertex. Therefore, we can break up the probability P (s) into
P (s) ≤
k−1
X
P (s, i) + P (s, ≥ k),
i=0
where P (s, i) corresponds to when there are i hyperedges on w contained entirely
within S and k − i hyperedges on w containing a deleted vertex, and P (s, ≥ k)
corresponds to the event that there are at least k hyperedges containing w that are
contained entirely within S.
Hence we have that
s 1
P (s, i) ≤
d
i
1
k−1
1
n−2
d−2
n
d
−
s
1
n−(k−1)−s
d−1
M −k
k−i
1
,
(nd)
M
40
(2.5.1)
because there can be no hyperedges (before deletion) that contain one vertex of S
and d − 1 vertices from the large remaining component. So we find that
M
P (s, i) ≤b sdi n(d−2)(k−i)
n
d
−s
n−(k−1)−s
d−1
−M
!k (n)−s(n−(k−1)−s)
d
d−1
M
(n)
M1
≤ σ dk n(d−2)(k−1)
n
d
−s
d
M
!k
n−(k−1)−s
d−1
− M1
exp −
s
n−(k−1)−s
d−1
n
d
M0
!
.
Note that this bound no longer depends on M or i. The same upper bound works
for P (s, ≥ k) (in fact, plugging i = k in (2.5.1) is an upper bound for P (s, ≥ k)). For
sufficiently large n, uniformly for s ∈ [d, σ],
k
esω
ln n
P (s) ≤b σ n
nd−1
ns (ln n)s(k−1)
ω s
1 1
1
e
≤b σ dk (ln n)k s k−1 d−1
.
n n
n
ln n
dk (d−2)(k−1)
Since ω = ln ln σ, eω = ln σ ≤ ln n and
P (s) ≤b σ dk (ln n)k
1 1
1
.
s
k−1
d−1
n n
n
Hence for large enough n,
X
σ n
n − (k − 1) dk
1
1 1
σ (ln n)k s k−1 d−1
k − 1 s=d
n n
n
s
P (AM ) ≤b
≤
σ
X
s=d
σ dk (ln n)k
1
nd−1
≤
σ dk+1 (ln n)k
,
nd−1
as desired.
Corollary 2.5.2. The probability that H̃(M0 ) has property Q and τk0 ∈ [M0 , M1 ], but
τk0 < Tk tends to zero; in other words,
P ({H̃(M0 ) ∈ Q} ∩ {τk0 < Tk } ∩ {τk0 ∈ [M0 , M1 ]}) = o(1).
41
Proof of corollary. We will show this probability tends to zero by finding a suitable
union bound. In fact, the event in question will be contained in the union of Am from
the previous lemma. Now
{τk0 ∈ [M0 , M1 ]} ∩ {τk0 < Tk } ⊂
M1
∪ {τk0 ≤ M < Tk },
M =M0
and since property Q is an increasing property, for M ≥ M0 ,
{H̃(M0 ) ∈ Q} ⊂ {H̃(M ) ∈ Q}.
Hence
{H̃(M0 ) ∈ Q} ∩ {τk0 < Tk } ∩ {τk0 ∈ [M0 , M1 ]} ⊂
M1
∪ {H̃(M ) ∈ Q} ∩ {τk0 ≤ M < Tk }.
M =M0
However, note that these last events are precisely AM from the last lemma. Consequently,
P ({H̃(M0 ) ∈ Q} ∩
{τk0
< Tk } ∩
{τk0
∈ [M0 , M1 ]}) ≤
M1
X
P (AM ).
M =M0
Using the uniform bound from the last lemma, we find that
M1
X
P (AM ) ≤b (M1 − M0 )
M =M0
σ dk+1 (ln n)k
ωσ dk+1 (ln n)k
≤
= o(1),
nd−1
nd−2
since d ≥ 3.
At this point, we have essentially proved the ultimate result. We need only put
the pieces together.
Theorem 2.5.3. With high probability τk = Tk .
Proof. Since (deterministically) τk ≤ Tk , we wish to show that P (τk < Tk ) → 0.
Note that on the event τk < Tk , at least one of the following events must hold:
42
• τk 6= τk0 ,
• τk ∈
/ [M0 , M1 ],
• H̃(M0 ) does not have property Q, or
• τk < Tk , but τk = τk0 ∈ [M0 , M1 ] and H̃(M0 ) does have property Q.
This last event is contained in the event that τk0 < Tk , τk0 ∈ [M0 , M1 ] and H̃(M0 ) has
property Q. Formally we have that
{τk < Tk } ⊂ {τk 6= τk0 } ∪ {τk ∈
/ [M0 , M1 ]} ∪ {H̃(M0 ) ∈
/ Q}∪
{τk0 < Tk , τk0 ∈ [M0 , M1 ], H̃(M0 ) ∈ Q}.
By Lemma 2.3.3 and Corollaries 2.2.3, 2.4.7, 2.5.2, we have shown that the probabilities of these four events tend to zero. Hence w.h.p. τk = Tk , as desired.
Just as for τk , the following corollary is immediate in light of the theorem.
n
d
(ln n + (k − 1) ln ln n)
. Then Yn converges in
n
distribution to a random variable X with FX (x) = exp e−x /(k − 1)! .
Corollary 2.5.4. Let Yn :=
2.6
Tk −
Sharp Threshold of k−connectivity
As a consequence of the previous result about coincidence of the stopping times of
k−connectedness and H̃ having minimum degree at least k, we will establish the
sharp threshold of k−connectedness of Hd (n, p) and Hd (n, m), as well as find the
probability of k−connectedness for parameters, p and m, in the critical window.
n
(ln n + (k − 1) ln ln n + cn ) where cn → c ∈ R. Then
d
e−c
P (Hd (n, M ) is k-connected) → exp −
.
(k − 1)!
Lemma 2.6.1. Let M =
43
Proof. By the previous theorem,
P (Hd (n, M ) is k-connected) = P (Tk ≤ M ) = P (τk ≤ M ) + o(1)
= P (Hd (n, M ) has min degree ≥ k) + o(1).
From Lemma 2.2.1, we know that
P (Hd (n, M ) has min degree ≥ k) → exp −
e−c
(k − 1)!
,
as desired.
n
ln n + (k − 1) ln ln n + (−1)i+1 ω for
d
i = 0, 1. Then w.h.p. Hd (n, M0 ) is not k−connected and Hd (n, M1 ) is k−connected.
n
As a consequence, M = ln n is a sharp threshold for k−connectedness.
d
Corollary 2.6.2. Let ω → ∞ and Mi =
The proof of this corollary follows from the previous lemma as well as an argument
similar to the proof of Corollary 2.2.3.
Now we wish to show a similar result for Hd (n, p).
Theorem 2.6.3. Let p =
(d − 1)!
(ln n + (k − 1) ln ln n + cn ) where cn → c ∈ R.
nd−1
Then
P (Hd (n, p) is k-connected) → exp −
e−c
(k − 1)!
.
Proof. We will prove this theorem as a consequence of the previous theorem. As
noted in the conversion lemma (Lemma 2.2.6), the random number of present hyper n
edges, e(Hd (n, p)), is binomially distributed with
trials and success probability
d
p. Let
n
n
2/3
m0 =
p − n , m1 =
p + n2/3 .
d
d
We will show that the number of present hyperedges is likely between m0 and m1
and then bound the property that Hd (n, p) is k−connected, by the probabilities that
Hd (n, mj ) is k−connected. First, by Chebyshev’s Inequality,
n
Var
Bin
,p
n d
P e(Hd (n, p)) −
p ≥ n2/3 ≤
.
2/3
2
(n )
d
44
Note that the variance of the binomial distribution Bin(ν, ρ) is νρ(1 − ρ). Hence
n
p(1 − p)
ln n
2/3
d
P e(Hd (n, p)) ≥ n
≤
≤b 1/3 = o(1).
4/3
n
n
In other words, P (e(Hd (n, p)) ∈
/ [m0 , m1 ]) = o(1). For ease of notation, let Pk be the
property of k−connectedness. Now
P (Hd (n, p) ∈ Pk ) =
=
(nd)
X
P (Hd (n, m) ∈ Pk )P (e(Hd (n, p)) = m)
m=0
m1
X
P (Hd (n, m) ∈ Pk )P (e(Hd (n, p)) = m) + o(1).
(2.6.1)
m=m0
Since k-connectedness is an increasing property, for m ∈ [m0 , m1 ],
P (Hd (n, m0 ) ∈ Pk ) ≤ P (Hd (n, m) ∈ Pk ) ≤ P (Hd (n, m1 ) ∈ Pk ).
Hence
P (Hd (n, m0 ) ∈ Pk ) + o(1) =
≤
m1
X
m=m0
m1
X
P (Hd (n, m0 ) ∈ Pk )P (e(Hd (n, p)) = m)
P (Hd (n, m) ∈ Pk )P (e(Hd (n, p)) = m),
m=m0
and similarly,
P (Hd (n, m1 ) ∈ Pk ) + o(1) =
≥
m1
X
m=m0
m1
X
P (Hd (n, m1 ) ∈ Pk )P (e(Hd (n, p)) = m)
P (Hd (n, m) ∈ Pk )P (e(Hd (n, p)) = m).
m=m0
Combining these last inequalities with equation (2.6.1) yields
P (Hd (n, m0 ) ∈ Pk ) + o(1) ≤ P (Hd (n, p) ∈ Pk ) ≤ P (Hd (n, m1 ) ∈ Pk ) + o(1).
For both j = 0 and j = 1, we notice that
mj =
n
(ln n + (k − 1) ln ln n + c + o(1)) ,
d
45
so by the previous theorem, the probability that Hd (n, mj ) is k−connected tends to
e−c
e− (k−1)! .
Hence the probability that Hd (n, p) is k−connected also tends to the same limit.
The next corollary is immediate given the lemma.
ln n + (k − 1) ln ln n + (−1)i+1 ω
.
nd−1
Then w.h.p. Hd (n, p0 ) is not k−connected and Hd (n, p1 ) is k−connected.
Corollary 2.6.4. Let ω → ∞ and pi = (d − 1)!
46
CHAPTER 3
DIAMETER AT THE MOMENT OF CONNECTEDNESS
3.1
Overview
As defined earlier, T1 is the stopping time for H̃ being connected. We defined the
distance between two vertices v and w, denoted d(v, w), to be the length of the
shortest path from v to w. Also, the diameter of a connected hypergraph is the
maximum distance between any two vertices, while the diameter of a disconnected
hypergraph is defined to be infinite. In this chapter, we wish to find likely values of
the diameter of H̃(T1 ), i.e. the hypergraph at the first moment the diameter of the
hypergraph process is finite. From the last chapter, we know that we may consider
the diameter of H̃(τ1 ) instead, since w.h.p. T1 = τ1 .
The bulk of the proof finding this likely diameter is based on an analysis of the
typical structure of a breadth-first search in Hd (n, p) like studies of the graph case,
see for instance, Burtin [10], Bollobás [5], Chung and Lu [14], and Fernholz and
Ramachandran [22]. In particular, we will establish both likely upper and lower
bounds on the number of vertices of distance i from any given vertex.
In this chapter, we consider the Bernoulli random hypergraph model, Hd (n, p),
instead of Hd (n, m), because independence of potential hyperedges makes analyzing
this breadth-first search simpler. We begin with the natural definition of a breadthfirst search in a hypergraph.
47
3.2
Breadth First Search in Hypergraphs
A breadth-first search on a hypergraph, H, starting at a vertex v0 , is the collection
of sets, (Γi (v0 ))ni=0 , where Γi (v0 ) is the set of vertices of distance i from v0 . Formally,
i
Γi (v0 ) = {w ∈ V (H) : d(w, v0 ) = i}. We also define the sets Ni (v0 ) = ∪ Γj (v0 ).
j=0
More importantly, there is a process to find Γi (v0 ). See Figure 3.1 for a sketch of a
breadth-first search up to Γ2 (v0 ).
−Γ0 (v0 )
v0
e2
e1
v1
e3
v5
v2
e4
v6
v7
v3
v8
e5
v4
−Γ1 (v0 )
e6
v9
−Γ2 (v0 )
Figure 3.1: A breadth-first search in a 3-uniform hypergraph
We start with Γ0 (v0 ) = {v0 } and N0 (v0 ) = {v0 } and determine the sets Γi (v0 ) and
Ni (v0 ), recursively. Suppose that Γi (v0 ) and Ni (v0 ) are defined for some i. Necessarily,
vertices in [n] \ Ni (v0 ) are at distance at least i + 1 from v0 ; moreover, these vertices
can not be in a hyperedge with a vertex of distance less than i from v0 . Hence a
necessary and sufficient condition for a vertex, w, to be at distance i + 1 from v0 is for
w to be in [n] \ Ni (v0 ) and be in a hyperedge with a vertex of distance i, where this
48
hyperedge can only contain vertices of Γi (v0 ) and [n] \ Ni (v0 ); so Γi+1 (v0 ) is the set
of such vertices, and we can find these vertices by checking the potential hyperedges
from Γi (v0 ) to [n] \ Ni (v0 ).
Formally, let Ei denote the set of potential hyperedges contained in Γi ∪ ([n] \ Ni )
that contain at least one vertex of Γi and at least one vertex from [n] \ Ni . In other
words,
Ei = {e ∈ E(n, d) : e ⊂ Γi (v0 ) ∪ ([n] \ Ni (v0 )), e ∩ Γi (v0 ) 6= ∅, e ∩ [n] \ Ni (v0 ) 6= ∅}.
Note that
Ei =
|Γi | n − |Ni |
|Γi | n − |Ni |
|Γi |
n − |Ni |
+
+ ... +
.
1
d−1
2
d−2
d−1
1
We will find that during most of the breadth-first search in our random model,
Hd (n, p), potential hyperedges of Ei that contain exactly one vertex of Γi (v0 ) and
d − 1 vertices of [n] \ Ni (v0 ) are the dominant contribution to the growth of |Γj (v0 )|.
In other words, hyperedges like e6 in Figure 3.1 contribute little to the growth of the
size of Γi (v0 ). Let Ei denote the subset of Ei that are present in the hypergraph. In
particular, the vertices in the hyperedges of Ei are either of distance i or i + 1 from
v0 and any vertex of distance i + 1 from v0 must necessarily be in some hyperedge
in Ei . Hence Γi+1 (v0 ) is the set of vertices in the hyperedges of Ei that are not in
Γi (v0 ); in other words,
!
Γi+1 (v0 ) =
[
e
\ Γi (v0 ).
e∈Ei
Then we note that Ni+1 (v0 ) = Ni (v0 ) ∪ Γi+1 (v0 ).
From now on, unless otherwise stipulated, we consider the breadth-first search in
Hd (n, p) and when v0 is known or fixed, we denote Γi (v0 ), Ni (v0 ) by Γi , Ni , respectatively. We will consider the random variable Γi+1 conditioned on (Ni , Γi ). In fact,
the process {(Γi , Ni )} is a Markov process.
49
Although the distribution of Γi+1 conditioned on (Γi , Ni ) is slightly complicated,
we will determine stochastic bounds which are suitably simpler to analyze. For instance, conditioned on (Γi , Ni ), the potential hyperedges of Ei are present with probability p independently of one another. Hence, the number of present hyperedges from
Γi to [n] \ Ni , denoted by |Ei |, is binomially distributed with |Ei | trials and success
probability p. Unlike the distribution of Γi+1 , binomial distributions are much easier
to analyze, so our goal is to relate these two random variables. For example, each
hyperedge of Ei can add at most d − 1 vertices to Γi+1 . So |Γi+1 | ≤ (d − 1)|Ei |. This
argument allows us to consider a relatively nice binomial distribution rather than
trying to deal with the actual distribution of |Γi+1 |.
However, there is not a comparably simple argument for determining a likely lower
bound on |Γi+1 | in terms of |Ei |. In fact, a large portion of this chapter deals with
finding a lower bound on likely |Γi |. For instance, due to overlap, a vertex in [n] \ Ni
can be in many hyperedges from Γi to Ni . For example, the hyperedge e6 in Figure
3.1 adds no unique vertices to Γ2 . For the first three steps, we will show that there
is likely not much overlap to obtain our lower bound on |Γi |. In the later steps, we
define a greedy algorithm to check potential hyperedges, where we never even check
a potential hyperedge unless its presence adds d − 1 vertices to our intermediate Γi+1 .
3.3
Likely Upper Bound on |Γi |
In finding a likely upper bound on |Γi |, we will need the following lemma about
unlikely events for a binomially distributed random variable, see Chung and Lu[14].
Lemma 3.3.1. Let Bin(ν, ρ) be a binomially distributed random variable with ν trials
and success probability ρ. For a > 0,
a2
1) P (Bin(ν, ρ) < νρ − a) ≤ e− 2νρ ,
2
a
− 2νρ
+
2) P (Bin(ν, ρ) > νρ + a) ≤ e
a3
(νρ)3
.
50
Suppose that we condition on the values of Γ1 (v), . . . , Γi (v). As noted before, the
number of hyperedges between Γi (v) and [n] \ Ni (v), Ei , (conditioned on Γ1 , . . . , Γi )
is binomially distributed on |Ei | trials with success probability p; also each such
hyperedge of Ei adds at most d − 1 vertices to Γi+1 . Hence |Γi+1 | ≤ (d − 1)|Ei |. Also,
since each possible hyperedge of Ei contains a vertex of Γi , we have that the number
|Γi | n − 1
of trials, |Ei |, is at most
.
1
d−1
n−1
Note that E[# of hyperedges containing v0 ] =
p. As long as p is not too
d−1
large where we expect to have large overlap of hyperedges containing v0 , we have
n−1
E[# of children of v0 ] ≈ (d − 1)
p. We anticipate that if this expectation is
d−1
large enough, then (by close analogy with Galvin-Watson branching processes) Γi
grows geometrically and should be relatively close to (E[# of children])i for some
time.
Lemma 3.3.2. Fix a vertex v ∈ [n]. Let p be such that
n−1
5
ln n ≤
p.
6
d−1
Then with probability at least 1 − O n−2 in Hd (n, p), for all 0 ≤ j ≤ n,
j
n−1
|Γj (v)| ≤ 10 (d − 1)
p ,
d−1
j
n−1
|Nj (v)| ≤ 11 (d − 1)
p .
d−1
Proof. Fix a vertex v ∈ [n]. For simplicity, we will denote Γi (v) and Ni (v) by Γi
and Ni . Introduce {bj }j≥0 , a positive increasing sequence and define
j
n
Bj := bj (d − 1)
p .
d−1
Consider the random variable Γi+1 conditioned on (Γi , Ni ) = (A, A0 ) for some |A| ≤
Bi . Since |Γi+1 | ≤ (d − 1)|Ei |,
P (|Γi+1 | ≥ Bi+1 |Γi = A, Ni = A0 ) ≤ P ((d − 1)|Ei | ≥ Bi+1 |Γi = A, Ni = A0 ) .
51
As previously noted, |Ei | conditioned on the value of Γi and Ni is binomially distributed, where the number of trials is |Ei |. But notice that
n−1
n−1
|Ei | ≤ |Γi |
≤ Bi
,
d−1
d−1
where increasing the number of trials increases the probability this binomial random
variable is at least Bi+1 . Hence
0
P (|Γi+1 | ≥ Bi+1 |Γi = A, Ni = A ) ≤ P
n−1
1
Bin Bi
,p ≥
Bi+1 .
d−1
d−1
Note that the above no longer depends on our choice of A with |A| ≤ Bi . Now
i+1
1
n−1
n−1
bi+1 − bi
Bi+1 − Bi
p=
(d − 1)
p
.
(3.3.1)
d−1
d−1
d−1
d−1
In using the tail bound on the binomial distribution Lemma 3.3.1, we want the
quantity above to be large enough to get a sufficiently good bound on the probability
yet small enough so that the sequence bi does not grow too fast. This leads us to
define bi recursively. Let b0 = 1 and
√
bi
bi+1 = bi + λ
(d − 1)
where λ =
,
i+1
p 2
n−1
d−1
p
6(d − 1) ln n. Clearly, the sequence bi is increasing. We will show that
bi ≤ 10 by induction. Suppose bj ≤ 10 for j ≤ i. Then
p
j+1
i
i
X
X
2
√
bj
1
≤
1
+
λ
10
bi+1 = 1 + λ
j+1
n−1
(d − 1) 65 ln n
2
j=0
j=0 (d − 1) d−1 p
s
p
√
60(d − 1) ln n
60
≤1+ q
<1+
73 < 10.
5 +1 = 1+
5
6
(d − 1) 6 ln n − 1
We will prove this lemma by showing the following technical claim.
Claim 3.3.3. For each 0 ≤ i ≤ n, with probability at least 1 −
|Γj | ≤ Bj ,
for 0 ≤ j ≤ i.
52
i (8(d−1))3/2
e
,
n3
Proof. We prove this claim by induction on i. The statement of the claim is
trivially true for i = 0. Suppose the statement is true up to i. Note that
P (for some 0 ≤ j ≤ i, |Γj | > Bj ) ≤ P (for some 0 ≤ j ≤ i, |Γj | > Bj )
+ P (|Γi+1 | > Bi+1 and for all j ≤ i, |Γj | ≤ Bj ). (3.3.2)
~ = (A0 , A1 , A2 , . . . , Ai ), Aj ⊂ [n], let Γ(A)
~ denote the event that Γj (v) = Aj
For A
~ : Aj pairwise disjoint, |Aj | ≤ Bj }. By the inductive
for 0 ≤ j ≤ i. Let A = {A
hypothesis,
P (for some 0 ≤ j ≤ i, |Γj | > Bj ) =
X
~ ∈A
A
/
~ ≤ i e(8(d−1))3/2 .
P (Γ(A))
n3
(3.3.3)
It remains to show that the probability of the event where |Γj | ≤ Bj for j ≤ i, but
|Γi+1 | > Bi+1 is suitably small. For this case, we will examine Γi+1 conditioned on
~ for some A
~ ∈ A. As before
Γ(A)
n−1
1
~
Bi+1 .
P |Γi+1 | ≥ Bi+1 Γ(A) ≤ P Bin Bi
,p ≥
d−1
d−1
Notice that from our defintion of bi as well as the equation (3.3.1),
s n−1
n−1
1
1
Bi+1 = Bi
.
p + λ Bi
p
d−1
d−1
d−1 d−1
(3.3.4)
(3.3.5)
Applying Lemma 3.3.1 to (3.3.4) with (3.3.5) yields
~
P |Γi+1 | ≥ Bi+1 |Γ A
≤ exp
≤ exp
−λ2
+
2(d − 1)
Bi
λ3
!
3/2
p(d − 1)
!
λ3
1
−λ2
3/2
+
≤
exp
(8(d
−
1))
.
3/2
5
2(d − 1)
n3
ln n
n−1
d−1
6
~ ∈ A, so we can break up the probability as
This bound works for all A
X ~ P (Γ(A))
~
P |Γi+1 | ≥ Bi+1 , |Γj | ≤ Bj , for j ≤ i =
P |Γi+1 | ≥ Bi+1 |Γ(A)
~
A∈A
≤
X
−3 (8(d−1))3/2
n e
~
A∈A
53
3/2
~
P Γ(A) ≤ n−3 e(8(d−1)) . (3.3.6)
Combining (3.3.3) and (3.3.6) in (3.3.2) yields
P (|Γj | > Bj , for some 0 ≤ j ≤ i + 1) ≤
i + 1 (8(d−1))3/2
e
.
n3
This concludes the inductive step.
By taking i = n in the claim, with probability at least 1 − O
1
, for all
n2
0 ≤ j ≤ n,
j
n−1
|Γj | ≤ Bj ≤ 10 (d − 1)
p .
d−1
For 0 ≤ j ≤ n and for n sufficiently large,
j
X
j
X
l
n−1
|Nj | =
|Γi | ≤
10 (d − 1)
p
d
−
1
i=0
l=0
!i
j X
∞
n−1
1
≤ 10 (d − 1)
p
d−1
(d − 1) n−1
p
d−1
i=0
j
j 5
ln n
n−1
n−1
6
≤ 11 (d − 1)
p .
≤ 10 (d − 1)
p 5
d−1
d−1
ln n − 1
6
3.4
Typical |Γ3 | is Zero or Logarithmically Large
Finding a lower bound on typical |Γi | is much more difficult than the upper bound.
One issue is that at the lower extreme of the admissible p, there will likely be isolated
vertices. Obviously, for those vertices Γi = ∅ for each i ≥ 1. Also, there will likely be
vertices of small degree making it possible that it takes a couple steps until |Γi | starts
growing quickly. In turns out, on the other hand, that conditioned on |Γ2 | ≥ 6(!),
with high probability |Γi |, (i ≥ 3) grows geometrically fast with i. We will prove this
property via a sequence of intermediate results; in these results, we determine the
likely structure of Hd (n, p) with probability at least 1 − o(n−3/2 (ln n)−1/2 ), because
we will wish to use our conversion lemma to convert our results to Hd (n, m).
54
As noted in the previous section, during the breadth-first search, conditioned on
Γi , Ni , the number of hyperedges, Ei , from Γi to [n] \ Ni is binomially distributed.
However we wish to determine the number of vertices in these hyperedges other than
those in Γi . Each of these present hyperedges can add at most d − 1 vertices to
Γi+1 , but due to overlap with other such hyperedges, it could add zero. Early in the
depth-first search, we anticipate that there will not be much if any overlap. To deal
with this possible overlap, we use the following claim.
n−1
5
p ≤ L ln n. Then with
Claim 3.4.1. Let L > 1 be fixed, and ln n ≤
6
d−1
probability at least 1 − o n−3/2 (ln n)−1/2 , for all pairs of vertices v, w, we have that
the number of hyperedges containing both v and w is at most 3. In other words,
|{e : {v, w} ⊂ e}| ≤ 3,
for all pairs of vertices v and w.
Proof. Let D be the event that each pair of vertices is in at most 3 common
hyperedges. On the complimentary event, Dc , there is necessarily a pair of vertices
that are in at least 4 common hyperedges. Hence by the union bound,
P (Dc ) ≤
X
P (vi , vj are in ≥ 4 hyperedges together)
i6=j
4
n
n n−2
(ln n)4
≤
P (|{e : {v1 , v2 } ∈ e}| ≥ 4) ≤
p4 ≤b
.
2
2
d−2
n2
Lemma 3.4.2. Let L > 1 be fixed, and p is such that
5
ln n ≤
6
n−1
p ≤ L ln n.
d−1
For every fixed vertex v ∈ [n], with probability at least 1 − o n−3/2 (ln n)−1/2 , either
v is isolated or |Γ2 (v)| ≥ 6.
55
Proof. For brevity, we denote Γi (v) by Γi . If Γ1 is non-empty, then necessarily
Γ1 contains at least d − 1 vertices. Let D be the event that every pair of vertices
are covered by at most 3 common hyperedges. In addition, let C be the event that
|Γ1 | ≤ 10(d − 1)L ln n. By Claim 3.4.1 and Lemma 3.3.2, we have that
P (DC ) + P (C C ) = o(n−3/2 (ln n)−1/2 ).
For ease of computation, let K = 10(d − 1)L.
Now
P |Γ2 | < 6, |Γ1 | > 0 ≤ P {|Γ2 | < 6, |Γ1 | > 0} ∩ C ∩ D
+ o(n−3/2 (ln n)−1/2 ). (3.4.1)
We break up the second probability in (3.4.1) as
P {|Γ2 | < 6, |Γ1 | > 0} ∩ C ∩ D =
K
ln n
X
P {|Γ2 | < 6, |Γ1 | = l} ∩ D
l=d−1
=
K
ln n
X
l=d−1
X
P {|Γ2 | < 6, Γ1 = A} ∩ D . (3.4.2)
A⊂[n]\{v},
|A|=l
Note that
P {|Γ2 | < 6, Γ1 = A} ∩ D = P {|Γ2 | < 6} ∩ DΓ1 = A P (Γ1 = A).
The term P {|Γ2 | < 6} ∩ DΓ1 = A is more amenable for analysis. In the next
claim, we find a bound for this probability.
Claim 3.4.3. Uniformly over A with |A| = l ∈ [d − 1, K ln n],
P {|Γ2 | < 6} ∩ DΓ1 = A ≤ n−5/3 .
Proof. Since d ≥ 3, there must be at least two vertices in A. Let u, w be two
vertices in A. To find a typical lower bound on |Γ2 | (conditioned on Γ1 = A) we will
consider only potential hyperedges from u, w to [n] \ ({v0 } ∪ A).
56
Let E be the set of potential hyperedges that contain exactly one of u, w and
d − 1 vertices from [n] \ ({v} ∪ A) and let E be the set of present hyperedges of E.
Conditioned on (Γ1 , Γ2 ) = ({v}, A), the random variable |E| is binomially distributed
n − |A| − 1
on 2
trials with success probability p. Each hyperedge of E contains
d−1
at least 1 vertex from [n] \ ({v} ∪ A) which must necessarily be in Γ2 . On the event
D, each vertex from [n] \ ({v} ∪ A) can be in at most 3 hyperedges with u and at
most 3 hyperedges with w. Hence, on the event D, if |E| ≥ 62 , then |Γ2 | ≥ 6. Now
n−l−1
P {|Γ2 | < 6} ∩ DΓ1 = A ≤ P Bin 2
, p ≤ 35 .
d−1
Uniformly over l ∈ [d − 1, K ln n],
35 n−l−1
X
n−l−1
2 d−1
n−l−1
P Bin 2
, p ≤ 35 =
pk q 2( d−1 )−k
d−1
k
k=0
n−1
≤b exp −2
p ≤ n−5/3 .
d−1
Now applying this claim in (3.4.2), we find that
X
P {|Γ2 | < 6, |Γ1 | > 0} ∩ C ∩ D ≤
n
−5/3
P Γ1 = A
A⊂[n]\{v},
d−1≤|A|≤K ln n
= n−5/3 P (d − 1 ≤ |Γ1 | ≤ K ln n) ≤ n−5/3 .
Using this bound in (3.4.1) finishes the proof of the lemma.
Lemma 3.4.4. Let L > 1 and p be such that
5
n−1
p ≤ L ln n.
ln n ≤
6
d−1
Fix vertex v ∈ [n]. With probability at least 1 − o n−3/2 (ln n)−1/2 , either v is isolated
or
1
n−1
|Γ3 (v)| ≥ (d − 1)
p.
20
d−1
57
d−1 n−1
Proof. For ease of computation, let b =
p. By the previous lemma,
20 d − 1
it suffices to show that the probability that |Γ2 | ≥ 6 but |Γ3 | < b is at most
o(n−3/2 (ln n)−1/2 ). By Lemma 3.3.2, with probability at least 1 − o(n−3/2 (ln n)−1/2 ),
we have that
2
n−1
|N2 | ≤ 11 (d − 1)
p ≤ 11(d − 1)2 L2 (ln n)2 =: N 0 .
d−1
~ = (A1 , A2 ) disjoint set of vertices ⊂ [n] \ {v} such that
Let A be the collection of A
1 + |A1 | + |A2 | ≤ N 0 and |A2 | ≥ 6. Again we let D denote the event that every pair
of vertices are in at most 3 common hyperedges together. So we have that
P {|Γ3 | < b, |Γ2 | ≥ 6} ≤ P {|Γ3 | < b, (Γ1 , Γ2 ) ∈ A} ∩ D
+ o(n−3/2 (ln n)−1/2 ). (3.4.3)
Note that
P {|Γ3 | < b, (Γ1 , Γ2 ) ∈ A} ∩ D =
X
P {|Γ3 | < b} ∩ D(Γ1 , Γ2 ) = (A1 , A2 ) P Γ1 = A1 , Γ2 = A2 . (3.4.4)
(A1 ,A2 )∈A
~
~ denotes the event that (Γ1 , Γ2 ) = (A1 , A2 ). We now
Let A ∈ A and as before Γ A
will consider the random variable |Γ3 | conditioned on Γ1 = A1 and Γ2 = A2 . By
definition of A, there are distinct vertices w1 , w2 , . . . , w6 in A2 . To obtain a lower
bound on |Γ3 |, we will consider only potential hyperedges from wi0 s to [n]\({v0 }∪A1 ∪
A2 ). Let E be the set of potential hyperedges that contain exactly one of w1 , . . . , w6
and d − 1 vertices from [n] \ ({v} ∪ A1 ∪ A2 ) and let E be the set of present hyperedges
of E. Conditioned on (Γ1 , Γ2 ) = (A1 , A2 ), the random variable |E| is binomially
distributed on
n − 1 − |A1 | − |A2 |
6
d−1
58
trials with success probability p.
Each hyperedge of E contains d − 1 vertices from [n] \ ({v} ∪ A1 ∪ A2 ) which must
necessarily be in Γ3 . On the event D, each vertex from [n] \ ({v} ∪ A1 ∪ A2 ) can be
in at most 3 · 6 = 18 hyperedges with vertices of w1 , . . . , w6 . Hence, on the event D,
(d − 1)|E| ≤ 18|Γ3 |.
In particular,
P
o
n d − 1
~
~
|E| ≤ b ∩ DΓ A
|Γ3 | < b ∩ DΓ A
≤P
18
!
n − 1 − |A1 | − |A2 |
18
b .
≤ P Bin 6
,p ≤
d−1
d−1
By definition of A, we have that 1 + |A1 | + |A2 | ≤ N 0 , so
P
n − N0
18
~
b .
|Γ3 | < b ∩ DΓ A ≤ P Bin 6
,p ≤
d−1
d−1
~ in A. In determining the
Notice that this bound does not depend on the choice of A
probability that this binomial random variable is smaller than 18b/(d − 1), we notice
that the expectation of this binomial random variable,
n − N0
n−1
6
p∼6
p,
d−1
d−1
is much larger than
18
9 n−1
b=
p.
d−1
10 d − 1
So by Lemma 3.3.1,
n − N0
9 n−1
P Bin 6
,p ≤
p
10 d − 1
d−1

 1 6
≤ exp −
2
59
n−N 0
d−1
p−
6
2 
p 
.
9 n−1
10 d−1
n−N 0
d−1
p
Since −(a + b)2 /a ≤ −a + 2b (for a, b > 0), we find that
1 6
−
2
n−N 0
d−1
6
p0 −
2
p
9 n−1
10 d−1
n−N 0
d−1
p0
9 n−1
6 n − N0
p+
p
≤−
2 d−1
10 d − 1
n−1
9
5
≤
p −3 + o(1) +
≤ − ln n.
d−1
10
3
~ ∈ A, we have
Hence, uniformly over A
P
~
|Γ3 | < b ∩ DΓ A
≤ n−5/3 .
Applying this bound to (3.4.4) yields
P {|Γ3 | < b, (Γ1 , Γ2 ) ∈ A} ∩ D ≤ n−5/3 ,
which along with (3.4.3) completes the proof of this lemma.
3.5
Likely Lower Bound on |Γi |
When |Γ3 | is logarithmically large, we can show from this moment onwards, Γi likely
grows geometrically. To show this, we consider a special subset of vertices in Γi+1
that are found in a process that only checks potential hyperedges from Γi to [n] \ Ni
which give rise to d − 1 new members of Γi+1 .
First, we define this greedy algorithm. Let H be a hypergraph and v ∈ V (H) and
suppose that we have just defined Γi (v). Let E be the set of potential hyperedges
from Γi to [n] \ Ni , which have exactly one vertex of Γi and d − 1 vertices of [n] \ Ni ;
i.e.
E = {e = (w1 , . . . , wd ) : w1 ∈ Γi , w2 , . . . , wd ∈ [n] \ Ni }.
We check whether the potential hyperedges of E are present in the following way;
if e is non-present, we continue onto the next potential hyperedge to check, but if
e is present, for the remainder of the process, we no longer consider any potential
60
hyperedge that contains a vertex in e∩([n]\Ni ). In particular, each hyperedge that we
find to be present gives rise to d − 1 vertices that are in Γi+1 . In particular, if E is the
set of hyperedges that we find to be present in our process, then (d − 1)|E| ≤ |Γi+1 |.
Now let us define this greedy algorithm formally. Order the potential hyperedges
of E in any way (for example, lexicographically). During the process, we determine
the set of seen present hyperedges at step j, denoted Fj , and the set of potential
hyperedges that we have either checked or know that we will not check, denoted Lj .
Specifically, we start with F0 = L0 = ∅, and find Fj and Lj recursively.
Suppose that Fj and Lj have just been defined. If we no longer have any potential
hyperedges that we wish to check for existence (in other words, Lj = E), then we
define Fj+1 = Fj and Lj+1 = Lj . Suppose that we still have potential hyperedges
to check, and let e be the minimum element (with respect to our order) of E \ Lj .
We now check whether e is present in the hypergraph. If e is not present, we define
Fj+1 = Fj and Lj+1 = Lj ∪ {e}. If e is present, then we define Fj+1 = Fj ∪ {e} and
Lj+1 = Lj ∪ {e} ∪ A, where A is the set of potential hyperedges from Γi to [n] \ Ni
which have non-empty intersection with e in [n] \ Ni ; in other words,
A = {f ∈ E : f ∩ e ∩ ([n] \ Ni ) 6= ∅}.
So A includes the vertices that we will not check later in the process. Since e has
d − 1 vertices of [n] \ Ni , we have that
n − |Ni | − 1
|A| ≤ (d − 1)|Γi |
.
d−2
Now we define J to be the first moment that Lj = E. Note that each element of
FJ is a present hyperedge that gives rise to d − 1 vertices in Γi+1 . In particular,
(d − 1)|FJ | ≤ |Γi+1 |.
Now we consider this process on the random hypergraph Hd (n, p) conditioned on
Γi and Ni . First, we let E be the set of present hyperedges of E. Note that |E| is
61
n − |Ni |
binomially distributed with |Γi |
trials with success probability p. As we
d−1
mentioned before, each present hyperedge that we find in the process decreases the
numberof potential hyperedge checks in the process by at most
n − |Ni | − 1
(d − 1)|Γi |
.
d−2
Furthermore, J is the number of hyperedge checks in the process, so
n − |Ni | − 1
J ≥ |E| − |E| · (d − 1)|Γi |
.
d−2
Also, we see that |FJ | is binomially distributed with J trials and success probability
p. In the proof of the following lemma, we will find our likely lower bound on |FJ |
by considering both |E| and J. In this lemma, we only consider finding a likely lower
bound on |Γi | up to i = 0.51 ln n/ ln ln n. Our goal of this chapter is to establish that
the diameter is roughly ln n/ ln ln n for our given range of the parameter p. Because
of the nature of our argument, we will need to have a lower bound on |Γi | up to about
half our anticipated diameter. However, as i gets closer to the suspected diameter,
finding lower bounds on |Γi | becomes more difficult, and in fact, for large enough i,
|Γi | should be zero. For this reason, we stop at 0.51 ln n/ ln ln n.
Lemma 3.5.1. With the same conditions from Lemma 3.4.4, namely fix L > 1 and
let p be such that
5
ln n ≤
6
n−1
p ≤ L ln n.
d−1
For every fixed vertex v ∈ [n], with probability at least 1 − o(n−3/2 (ln n)−1/2 ), either v
is isolated or for 3 ≤ i ≤ 0.51 ln n/ ln ln n,
i−2
1
n−1
|Γi (v)| ≥
(d − 1)
p
.
21
d−1
Proof. As before, we denote Γi (v), Ni (v) by Γi , Ni . Let Bi be the event that for
0 ≤ j ≤ i,
i
n−1
|Γi | ≤ 10 (d − 1)
p ,
d−1
62
n
and let B = ∩ Bi . Note that B ⊂ Bn−1 ⊂ . . . ⊂ B2 ⊂ B1 and by Lemma 3.3.2,
i=0
−3/2
P (B ) = o(n
(ln n)−1/2 ). In particular, on the event B, we have that
0.51 lnlnlnnn
ln n
n−1
≤ 11 exp 0.51
ln (d − 1)L ln n
|Ni | ≤ 11 (d − 1)
p
ln ln n
d−1
c
≤ 11 exp (0.52 ln n) = 11n0.52 ≤ n3/4 .
Also, let C denote the event that
n−1
1
p.
|Γ3 | ≥ (d − 1)
20
d−1
By Lemma 3.4.4, with probability at least 1 − o(n−3/2 (ln n)−1/2 ), either v is isolated
or C occurs. To prove the lemma, we will show the following claim, which is more
receptive to an inductive argument. For ease of notation, for 3 ≤ j ≤ ln n, let
!
j
Y
300
d
Πj =
1−
.
j−2 −
1/4
n
2
(ln
n)
k=3
Note that Πj is decreasing in j, is less than 1, and
!
ln n
ln n
Y
X
300
d
Πln n =
1−
≥1−
k−2 −
n1/4
(ln n) 2
k=3
k=3
300
(ln n)
j−1
2
+
d
!
n1/4
301
≥ 0.99,
≥1− √
ln n
for sufficiently large n. Let Aj denote the event that
j−2
1
n−1
|Γj | <
(d − 1)
p
Πj .
20
d−1
Claim 3.5.2. For each 3 ≤ i ≤ 0.51 ln n/ ln ln n,
i
2i
P
∪ Aj ∩ Bi ∩ C ≤ 2 .
j=3
n
Before we prove the claim, let’s consider why the lemma follows from this claim.
i
c
Consider the event ∪ Aj . On this event, for each 3 ≤ j ≤ i, we have
j=3
j−2
j−2
Πj
n−1
0.99
n−1
|Γj | ≥
(d − 1)
p
≥
(d − 1)
p
20
d−1
20
d−1
j−2
1
n−1
≥
(d − 1)
p
21
d−1
.
63
Let D be the event that v is not isolated. Notice that
i
i
c
P
∩ Aj ∩ D = P (D) − P
∪ Aj ∩ D
j=3
j=3
i
≥ P (D) − P
∪ Aj ∩ Bi ∩ C − P (Bic ) − P (C c ) .
j=3
By taking i = 0.51 ln n/ ln ln n in the claim as well as our bounds on the probability
of B c and C c , we find that
i
c
P
∩ Aj ∩ D ≥ P (D) − o(n−3/2 (ln n)−1/2 ).
j=3
Hence
i
c
P (D ) + P D ∩ ∩ Aj
≥ 1 − o(n−3/2 (ln n)−1/2 ).
c
j=3
In other words, with probability at least 1 − o(n−3/2 (ln n)−1/2 ), either v is isolated
(i.e. the event Dc ) or for 3 ≤ j ≤ 0.51 ln n/ ln ln n,
j−2
n−1
1
(d − 1)
p
|Γj | ≥
21
d−1
.
Now we prove the claim by induction on i.
Proof. The statement of the claim is trivially true for i = 3, because A3 ∩ C = ∅.
Suppose the statement is true up to i. Notice the following containments
i+1
i+1
∪ Aj ∩ Bi+1 ∩ C ⊂ ∪ Aj ∩ Bi ∩ C
j=3
j=3
i
i
c
⊂
∪ Aj ∩ Bi ∩ C ∪ Ai+1 ∩ ∩ Aj ∩ Bi .
j=3
j=3
By the inductive hypothesis,
P
i
∪ Aj
j=3
∩ Bi ∩ C
≤
2i
.
n2
So our task is to show that
i
2
c
P Ai+1 ∩ ∩ Aj ∩ Bi ≤ 2 .
j=3
n
64
(3.5.1)
Let A be the collection of elements of the form (A1 , . . . , Ai ) where Aj are disjoint sets
of vertices such that
" j−2
j #
n−1
n−1
Πj
(d − 1)
p
, 10 (d − 1)
p
.
|Aj | ∈
20
d−1
d−1
i
c
Note that the event Bi ∩ ∩ Aj is the event that (Γj )ij=0 ∈ A. To complete the
j=3
~ ∈ A,
claim, it suffices to show that uniformly over A
i+1 !
Πi+1
n−1
2
~
P |Γi+1 | <
(d − 1)
p
≤ 2,
Γ A
20
d−1
n
(3.5.2)
because
X i
c
~
P Ai+1 ∩ Γ(A)
P Ai+1 ∩ ∩ Aj ∩ Bi =
j=3
=
X
~
A∈A
~
A∈A
P
Πi+1
|Γi+1 | ≤
20
i+1 ! n−1
~
~ .
(d − 1)
p
P Γ A
Γ A
d−1
Our task for the remainder of this proof to show show the bound in (3.5.2). For
~ ∈ A. Since we have that
this we consider our process from before. Fix some A
(d − 1)|FJ | ≤ |Γi+1 |,
i+1 !
n−1
Πi+1
~
≤
(d − 1)
p
P |Γi+1 | ≤
Γ A
20
d−1
i+1 !
Πi+1
n−1
~
P (d − 1)|FJ | ≤
(d − 1)
p
.
Γ A
20
d−1
As previously mentioned, we have that |FJ | is binomially distributed on J trials with
success probability p, but J is a random variable with
n − |Ni |
n − |Ni | − 1
J ≥ |Γi |
− |E| · (d − 1)|Γi |
,
d−1
d−2
n − |Ni |
where |E| is binomially distributed on |Γi |
trials. We find that
d−1
n − 1 ~ ≤
P |E| ≥ 2|Γi |
p Γ A
d−1
n−1
n−1
P Bin |Γi |
, p ≥ 2|Γi |
p .
d−1
d−1
65
Using the tail bounds from Lemma 3.3.1, we have that
P
n − 1 ~ |E| ≥ 2|Γi |
p Γ A
≤ exp 1 −
d−1
≤ exp 1 −
n−1
1
|Γi |
p
2
d−1
1d−1 5
2
(ln n) ≤ n−2 ,
2 4 36
~ ∈ A (and i). On the event that
for sufficiently large n, uniformly over choice of A
n−1
|E| < 2|Γi |
p,
d−1
we have that
n − |Ni |
n−1
n − |Ni | − 1
J ≥ |Γi |
− 2|Γi |
p · (d − 1)|Γi |
d−1
d−1
d−2
3/4
n
n−n
− 2(d − 1)L|Γi |2 (ln n)
=: N0 .
≥ |Γi |
d−1
d−2
Claim 3.5.3.
n−1
12
d
P Bin(N0 , p) ≤ |Γi |
p 1−
−
≤ n−2 .
d−1
|Γi |1/2 n1/4
√
Proof. Let λ = 4 ln n. Then for sufficiently large n, (does not depend on i or
~
choice of A),
3/2
2(d − 1)L|Γ|
2
(ln n)
n
n
d−1
9/10
2
p
p ≤ 2(d − 1)Ln (ln n)
n−d+2 d−1
d−2
n9/10
≤ 2(d − 1)2 L2 (ln n)3
≤ 1.
n−d+2
Hence
2(d − 1)L|Γi |2 (ln n)2
s n
n − n3/4
1/2
p.
p ≤ |Γi | ≤ λ |Γi |
d−2
d−1
By Lemma 3.3.1,
2
p
λ
P (Bin(N0 , p) ≤ N0 p − λ N0 p) ≤ exp −
.
2
66
Note that
p
p
n − n3/4
n
2
N0 p − λ N0 p = |Γi |
p − 2(d − 1)L|Γi | (ln n)
p − λ N0 p
d−1
d−2
s n − n3/4
n − n3/4
≥ |Γi |
p − 2λ |Γi |
p.
d−1
d−1
With probability at least 1 − e−λ
2 /2
= 1 − n−2 ,
s 3/4
n − n3/4
n−n
p
Bin(N0 , p) ≥ |Γi |
p − 2λ |Γi |
d−1
d−1


3/4
n−n
2λ

≥ |Γi |
p 1 − q
d−1
n−n3/4
|Γi | d−1 p
n − n3/4
12
≥ |Γi |
p 1−
.
d−1
|Γi |1/2
In addition, we have
d−1
d−2
n
n3/4
n − n3/4
n − 1 Y n − n3/4 − i
≥
1−
=
d−1
n−d
d−1
d − 1 i=0 n − 1 − i
n−1
d
≥
1 − 1/4 .
n
d−1
Hence with probability at least 1 − n−5/2 for sufficiently large n,
d
n−1
12
−
.
Bin(N0 , p) ≥ |Γi |
p 1−
|Γi |1/2 n1/4
d−1
Combining these last two results, we have that
n−1
12
2
d
~
≤ 2.
p 1 − 1/2 − 1/4 Γ A
P |Γi+1 | ≤ (d − 1)|Γi |
d−1
|Γ|
n
n
Note that
1−
12
d
300
d
− 1/4 ≤ 1 −
,
i−2 −
1/2
1/4
|Γi |
n
n
(ln n) 2
which implies that
P
Πi+1
|Γi+1 | ≤
20
i+1 !
n−1
2
~
(d − 1)
p
≤ 2.
Γ A
d−1
n
67
~ in A. This completes the
This bound works uniformly over i and the choice of A
inductive step of the claim and hence the proof of the lemma.
3.6
Diameter of Hd (n, m)
If we combine our results on likely lower and upper bounds on |Γi | (Lemma 3.3.2 and
Lemma 3.5.1) along with a union bound over all vertices, we obtain the following
corrolary.
Corollary 3.6.1. Fix L > 1. Let p be such that
n−1
5
ln n ≤
p ≤ L ln n.
d−1
6
With probability at least 1 − o n−1/2 (ln n)−1/2 in Hd (n, p), for every vertex v ∈ [n],
we have that for 0 ≤ i ≤ n,
i
n−1
|Γi (v)| ≤ 10 (d − 1)
p ,
d−1
and either v is isolated or for 3 ≤ i ≤ 0.51 ln n/ ln ln n,
i−2
n−1
1
.
(d − 1)
p
|Γi (v)| ≥
21
d−1
Now that we have sufficiently good bounds on the growth of the typical breadthfirst search, we can determine typical values of the diameter. We begin by showing
that likely each pair of non-isolated vertices is connected by a path. Before we prove
this result, we need the following definition. We call the smallest integer greater than
or equal to x the ceiling of x, which is denoted by dxe. Let
k1 =
l 1 ln(212 · 22 · n) m
+2
2 ln((d − 1) n−1
p)
d−1
and
k2 =
l ln(212 · 22 · n) m
+ 4 − k1 .
ln((d − 1) n−1
p)
d−1
Note that k1 ≤ k2 .
68
Lemma 3.6.2. Let p be such that
n−1
p ≤ L ln n.
d−1
With probability at least 1 − o n−1/2 (ln n)−1/2 , for each v, w which are not isolated,
5
ln n ≤
6
there exists a path connecting v and w of length at most
&
'
ln (212 · 22 · n)
+ 5.
k1 + k2 + 1 =
ln (d − 1) n−1
p
d−1
Proof. By Corollary 3.6.1, with sufficiently high probability, for any vertex v ∈ [n],
we have that either v is isolated or
ki −2
n−1
1
(d − 1)
p
|Γki (v)| ≥
d−1
21
!
1
n−1
1 ln(212 · 22 · n)
≥
exp ln (d − 1)
p ·
21
d−1
2 ln (d − 1) n−1
p
d−1
=
1
21 · 2 · n1/2 = 2n1/2 .
21
and |Nki (v)| ≤ n3/4 for i = 1, 2.
Fix v, w ∈ [n]. We run the breadth-first search algorithm on v up to Nk1 and
w up to Nk2 . Consider the event where Γk1 (v) = A, Γk2 (w) = B, Nk1 (v) = C and
Nk2 (w) = D, where 2n1/2 ≤ |A|, |B| and |C|, |D| ≤ n3/4 . If C ∩ D 6= ∅, then
necessarily, there must be a path between v and w of length at most k1 + k2 . Suppose
C ∩ D = ∅. Since we have only run the breadth-first search up to Nk1 (v) and
Nk2 (w), we have not considered any potential hyperedge containing one vertex of
A, one vertex of B and d − 2 vertices from [n] \ (C ∪ D) in Hd (n, p). If any such
hyperedge is present, there is a path from v to w of length k1 + k2 + 1. Conditioned
on Γ(v) = A, Γ(w) = B, Nk1 (v) = C, Nk2 (w) = D, the probability that there is no
hyperedge containing a vertex of A and a vertex of B is bounded above by (uniformly
over all such sets)
q
|A|·|B|·(n−|C|−|D|
)
d−2
n − 2n3/4
≤ exp −p · 4n
≤ exp (−4 ln n) .
d−2
69
Taking the union bound over all such pairs of vertices v, w, we obtain the desired
result.
Corollary 3.6.3. Let M be such that
5n
n
ln n ≤ M ≤ L ln n.
6d
d
In Hd (n, m), with high probability there is a path between any two non-isolated vertices
of length at most
l ln (212 · 22 · n) m
+ 5.
ln (d − 1) nd M
Proof. Let Q be the property ”any two non-isolated vertices can be connected by
a path of length at most k1 + k2 + 1” and apply Lemma 2.2.6 and Lemma 3.6.2 to
n
n n−1
M=
p=
p.
d
d d−1
Lemma 3.6.4. Let p be such that
n−1
p ≤ L ln n.
d−1
With probability at least 1 − o n−1/2 (ln n)−1/2 in Hd (n, p), the diameter is at least
5
ln n ≤
6
l :=
l
m
ln(n/22)
.
ln((d − 1) n−1
p)
d−1
Proof. For each v ∈ [n], with sufficiently large probability by Corollary 3.6.1,
l−1
n−1
|Nl−1 (v)| ≤ 11 (d − 1)
p
d−1
≤ 11 exp
!
ln(n/22)
n−1
n
ln((d − 1)
p) = .
n
d−1
2
ln((d − 1) d−1 p)
Since this is less than n, there must be vertices not reached within distance l − 1 of
v. Hence the diameter must be at least l.
70
Corollary 3.6.5. Let
5n
n
ln n ≤ M ≤ L ln n.
6d
d
W.h.p. in Hd (n, M ), the diameter is at least
l ln(n/22) m
.
ln((d − 1) nd M )
Proof. As before, use Lemma 2.2.6 to transfer the previous lemma from Hd (n, p)
to Hd (n, M ).
Theorem 3.6.6. Let ω → ∞, ω = o (ln n) and let M be such that
n
n
(ln n + ω) ≤ M ≤ (L ln n) .
d
d
Then with high probability, the diameter of Hd (n, M ) is in the interval
"
#
l
m l ln (212 · 22 · n) m
ln(n/22)
,
+5 .
ln (d − 1) nd M
ln (d − 1) nd M
Moreover, this interval has length at most 6.
Proof. The lower bound is given by Corollary 3.6.5. By Corollary 2.2.3, with
high probability, there are no isolated vertices and then the upper bound is given by
Corollary 3.6.3. Note that this is an alternative proof that Hd (n, M ) is connected for
this range of M . Since dxe − dye ≤ dx − ye, the difference between the upper and
lower bound is less than or equal to
l
m
l ln (212 · 22 · n)
ln(n/22) m
11
−
+5≤
+ 5 = 6.
ln ((d − 1)M )
ln ((d − 1)M )
ln ((d − 1)M )
One can prove the following theorem using an analogous argument.
Theorem 3.6.7. Let ω → ∞, ω = o(ln n) and let p such that
n−1
ln n + ω ≤
p ≤ L ln n.
d−1
Then with high probability, the diameter of Hd (n, p) is in the interval
"
#
l
m l ln (212 · 22 · n) m
ln(n/22)
,
+5 .
ln (d − 1) n−1
p
ln (d − 1) n−1
p
d−1
d−1
Moreover, this interval has length at most 6.
71
3.7
Diameter of H̃ (T1 )
We will now be able to find the likely diameter at the first moment the diameter is
finite in the hypergraph process.
Theorem 3.7.1. As before let τ1 = min{m : H̃n (m) has min degree ≥ 1}. Let > 0,
then with high probability, the diameter of H̃(τ1 ) is in the following interval
ln n
ln n
− ,
+8+ .
ln((d − 1) ln n)
ln((d − 1) ln n)
Proof. The lower bound will be simple to establish since the diameter is decreasing
with the addition of hyperedges, which will allow us to consider the diameter of
Hd (n, m) for m large enough to likely be connected.
n
Let ω → ∞, ω = o (ln ln ln n) and Mi =
ln n + (−1)i+1 ω for i = 0, 1. By
d
Corollary 2.2.3, with high probability τ1 ≤ M1 . Since diameter decreases as hyperedges are added, on the event that {τ1 ≤ M1 },
diameter of H̃(τ1 ) ≥ diameter of H̃(M1 ).
In addition, since H̃(M1 ) is distributed as Hd (n, M1 ), by Corollary 3.6.5 with high
probability,
diameter of H̃(τ1 ) ≥
l
m
ln(n/22)
.
ln((d − 1)(ln n + ω)
Now we need to establish the upper bound. To this end, we consider H̃(M0 ). Likely
H̃(M0 ) is not connected, but each pair of non-isolated vertices is connected by a path
of length at most
ln (212 · 22 · n)
K :=
+ 5,
ln ((d − 1) (ln n − ω))
by Corollary 3.6.3. As hyperedges are added and isolated vertices become nonisolated, the distance this newly non-isolated vertex and other non-isolated vertices
could be larger than K. However, if at the moment that each vertex is non-isolated,
72
each isolated vertex of H̃(M0 ) is in a hyperedge with a non-isolated vertex of H̃(M0 ),
the diameter is at most K + 2. We will now show that we likely have this case.
Suppose Ṽ0 are the isolated vertices of H̃(M0 ) and Ṽ1 = [n] \ Ṽ0 . With high
probability the hypergraph induced on Ṽ1 is connected and has diameter at most K,
by Corollary 3.6.3. If, in H̃(τ1 ), each of the vertices in Ṽ0 are now in a hyperedge with
at least one vertex from Ṽ1 , then necessarily H̃(τ1 ) is connected and has diameter at
most K + 2.
Claim 3.7.2. With high probability, no hyperedge added between M0 and M1 in H̃
contains at least two vertices of Ṽ0 .
Let M ∈ (M0 , M1 ]. From Corollary 2.2.10, there are at most 2eω isolated vertices
in Hd (n, M0 ). Consider the event Ṽ0 = V0 , for some V0 ⊂ [n] with |V0 | ≤ 2eω .
P (eM
|V0 | n − 2
1
contains two member of V0 , Ṽ0 = V0 ) ≤
n
2
d−2 d −M
e2ω
1
≤b 2 .
≤ 2e2ω nd−2 n
n
− M1
d
This bounding constant is uniform over M ∈ (M0 , M1 ] and sets V0 with |V0 | ≤ 2eω .
By taking a union bound,
P (∃M ∈ (M0 , M1 ], eM contains two vertices of Ṽ0 , |Ṽ0 | ≤ 2eω ) ≤b (M1 − M0 )
e2ω
n2
= o(1).
Since P (|Ṽ0 | > 2eω ) = o(1), this completes the proof of the claim. As a consequence,
this is an alternate proof that w.h.p. H̃(τ1 ) is connected. Hence with high probability,
the diameter of H̃(τ1 ) is in the following interval
l
m l
m
ln(n/22)
ln (212 · 22 · n)
,
+7 .
ln((d − 1)(ln n + ω)
ln ((d − 1) (ln n − ω))
73
Since ω = o(ln ln n),
m
ln(212 · 22 · n)
ln(212 · 22 · n)
≤
+1
ln((d − 1)(ln n − ω))
ln((d − 1)(ln n − ω))
ln n
≤
+ 1 + ,
ln((d − 1) ln n)
l
m
ln(n/22)
ln(n/22)
≥
ln((d − 1)(ln n + ω)
ln((d − 1)(ln n + ω)
ln n
≥
− .
ln((d − 1) ln n)
l
Thus with high probability, the diameter of H̃(τ1 ) is in the desired range.
We have shown that w.h.p H̃(τ1 ) is connected in two different topological ways.
In the previous chapter, we showed that w.h.p. there is exactly one component in
H̃(τ1 ), while in this chapter, we showed that w.h.p. there is a path between any two
vertices in H̃(τ1 ).
We have the following immediate consequence:
Theorem 3.7.3. Let > 0, then with high probability, the diameter of H̃(T1 ) is in
the following interval
ln n
ln n
− ,
+8+ .
ln((d − 1) ln n)
ln((d − 1) ln n)
74
CHAPTER 4
THRESHOLD OF WEAK HAMILTONICITY
4.1
Overview
A weak cycle, C = (v0 , e1 , v1 , . . . , el , vl = v0 ), of length l ≥ 3, is an alternating
sequence of vertices and hyperedges such that v0 , v1 , . . . , vl−1 are distinct and vi−1 , vi ∈
ei (the ei need not be distinct). Note that hyperedges in the weak cycle may include
vertices outside of v0 , v1 , . . . , vl−1 . We say that v0 , v1 , . . . , vl−1 is the vertex set of the
cycle or that C spans v0 , v1 , . . . , vl−1 . See Figure 4.1 for an example of a weak cycle,
which is denoted by a dotted closed curve.
v8
v1
v2
e3
e2
v7
e1
v3
v6
v5
v4
Figure 4.1: A weak cycle in a 4-uniform hypergraph
75
In this chapter, we establish the sharp threshold for presence of a weak cycle on the
entire vertex set of Hd (n, p); in other words, when a weak Hamilton cycle develops.
To find this sharp threshold, we will follow an argument similar to Bollobás’ proof
that there is a Hamilton cycle in G(n, p) for suitable p (see [6] and [8]). For the graph
case, the primary barrier for Hamiltonicity is the existence of vertices with degree
less than 2. In particular, if p = (ln n + ln ln n + c)/n, then
−c
P (G(n, p) is Hamiltonian) = P (min deg (G(n, p)) ≥ 2) + o(1) = e−e
+ o(1).
In apparent contrast, we wish to show that, for d ≥ 3:
(d − 1)!
(ln n + c), then w.h.p. there is a weak cycle in
nd−1
Hd (n, p) spanning all of the non-isolated vertices. Consequently
Theorem 4.1.1. If p =
P (Hd (n, p) is weak Hamiltonian) = P (min deg (Hd (n, p)) ≥ 1) + o(1)
−c
= e−e
+ o(1).
Whereas vertices of degree 1 are an issue for Hamiltonicity of G(n, p), we see that
their mere existence is not an issue for weak Hamiltonicity of Hd (n, p) for d ≥ 3.
Although this seems like a behaviorial difference between the graph case and the
hypergraph case, notice that a vertex in a graph has at least 2 neighbors if and only
if its degree is at least 2, but for d ≥ 3, a vertex has at least 2 neighbors if and only
if its degree is at least 1. In both cases (d = 2 and d ≥ 3), the main barrier for weak
Hamiltonicity is vertices with less than 2 neighbors.
After establishing the sharp threshold in Hd (n, p), we will convert these results
to Hd (n, m) and then to the hypergraph process H̃. We will prove this theorem by
showing the following:
1. First, via an analogue of de la Vega’s Theorem, it will be shown that there is
likely a path of length n − o(n) for p sufficiently close to but less than the sharp
threshold of connectivity.
76
2. Second, we obtain a necessary condition for an extremal path starting at a fixed
vertex, which actually says that the set of endpoints of all paths obtained via
rotations is “non-expanding.”
3. Then, we show that all non-expanding sets are sufficiently large, which means
that the set of endpoints is large as well. This will imply that for any such graph,
there are sufficiently many non-present hyperedges such that the addition of any
one of these absent hyperedges will increase the length of a longest path.
4. By increasing p in small increments, we will be assured of likely increasing the
length of some longest path at each step until we end with a weak cycle on the
set of non-isolated vertices.
We start with a few definitions. For a hypergraph, H, and a subset of its vertex
set, V 0 ⊂ V (H), we define Γ(V 0 ) as the set of vertices in a hyperedge with vertices in
V 0 and N (V 0 ) as the set of neighbors of V 0 . Formally,
Γ(V 0 ) = {w ∈ V (H) : ∃e ∈ E(H), ∃v ∈ V 0 such that w, v ∈ e}
and N (V 0 ) = Γ(V 0 ) \ V 0 . We say that a set of vertices, A, is non-expanding if the
neighbor set of A has cardinality less than twice the cardinality of A; in other words,
|N (A)| < 2|A|. Before we prove the hyper graphic version of de la Vega’s Theorem,
let’s prove a consequence of minimal non-expanding sets.
Lemma 4.1.2. Suppose H is a hypergraph with a non-empty non-expanding set, A,
that has no proper non-expanding subsets. The induced hypergraph on A ∪ N (A) is
connected.
Proof. We prove this lemma by contradiction. Suppose the induced hypergraph
is not connected. Then we can decompose the induced hypergraph into 2 disjoint
hypergraphs. For example, T = A ∪ N (A) = T1 ∪ T2 , where Ti are non-empty and
77
there are no hyperedges between the vertex sets T1 and T2 (using only hyperedges
contained within T ). Let Ai = A ∩ Ti . Note that (trivially) Γ(Ai ) ⊂ T. In fact,
Γ(Ai ) ⊂ Ti , because otherwise, there is a hyperedge in the induced hypergraph on T
from T1 to T2 . So
Ai ∪ N (Ai ) = Ai ∪ Γ(Ai ) ⊂ Ti ,
which implies that the vertex sets A1 ∪ N (A1 ) and A2 ∪ N (A2 ) are disjoint. Hence
|A1 ∪ N (A1 )| + |A2 ∪ N (A2 )| = |A1 ∪ A2 ∪ N (A1 ) ∪ N (A2 )| = |A ∪ Γ(A)|
= |A| + |N (A)| < 3|A| = 3|A1 | + 3|A2 |.
Hence either |A1 ∪ N (A1 )| < 3|A1 | or |A2 ∪ N (A2 )| < 3|A2 |. Consequently, either |N (A1 )| < 2|A1 | or |N (A2 )| < 2|A2 |, which contradicts minimality of the nonexpanding set A.
4.2
Hypergraph Analogue of De la Vega’s Theorem
Prior to the random hypergraph becoming connected, we will show that there is a
path of length n − o(n). In the proof of Hamiltonicity of G (n, p) , where
p=
ln n + ln ln n + ω
,
n
with ω → ∞, this is established by de la Vega’s theorem. Here, we prove a hypergraphic version of de la Vega’s theorem. In fact, it turns out that it is possible to use
de la Vega’s theorem to prove its hypergraph version after a reduction argument. We
start with the statement of de la Vega’s Theorem.
Theorem 4.2.1 (de la Vega’s Theorem). Let Θ = Θ(n) ∈ (4 ln 2, ln n − 3 ln ln n),
4 ln 2
then w.h.p. there is a path in G(n, p = Θ/n) of length at least n 1 −
.
Θ
See Bollobás [8] for a proof of this theorem. After our reduction step, we will be
using the following immediate corollary of de la Vega’s Theorem.
78
1
Corollary 4.2.2. Let Θ ∈ 4 ln 2, ln n , then with high probability, there is a path
2
in G (bn/2c, p = Θ/bn/2c) of length at least
n
2
3
1−
.
Θ
Proof. For sufficiently large n, we have that
1
ln n ≤ lnbn/2c − 3 ln lnbn/2c
2
and
4 ln 2
n
3
bn/2c 1 −
≥
1−
.
Θ
2
Θ
Applying the theorem yields the desired result.
In the following hypergraph analogue of de la Vega’s Theorem, we do not try
to obtain the best possible bounds on a likely long path, instead we care only to
find sufficient bounds for our final result. The key part of the following proof is the
reduction to random graphs.
Lemma 4.2.3 (de la Vega for Hypergraphs). Suppose σ(n) → ∞ as n → ∞ with
θ
σ = o(ln n). Let θ = θ(n) ∈ [σ, 2 ln n] and p = (d − 1)! d−1 . With high probability in
n
2d+2
Hd (n, p), there is a path of length at least n 1 −
.
θ
Proof. Let θ(n) ∈ [σ, 2 ln n], p = (d − 1)!
θ
nd−1
and p1 = p/3. Let H i , for i = 1, 2, 3,
be random hypergraphs on [n], which are independent and distributed as Hd (n, p1 ).
Let H be a random hypergraph on [n], where a potential hyperedge is present in H
if and only if the hyperedge is present in at least one of H i . We will show that likely
there is a long path spanning only vertices from {1, 2, . . . , n/2 } in H 1 and a long
path spanning only vertices from { n/2 + 1, . . . , n} in H 2 . Using H 3 we will likely
concatenate these long paths to obtain a long path in H of suitable length.
79
By independence of H i ’s, H is equal in distribution to Hd (n, p0 := 1 − q13 ). Moreover
p0 = 1 − (1 − p1 )3 ≤ 3p1 = p.
We couple H and Hd (n, p) such that H is a sub-hypergraph of Hd (n, p). Since containing such a long path is an increasing property, it suffices to show that there is a
long enough path in H.
Now we reduce to finding a long path in a random graph. We construct a random
graph, G1 , on bn/2c by the following: for each possible edge {i, j} ⊂ bn/2c , i 6= j,
we define {i, j} ∈ E(G1 ) if and only if there is some hyperedge in H 1 that contains
i and j but no other vertex of bn/2c , i.e. hyperedges of the form {i, j, v3 , . . . , vd }
where {v3 , . . . , vd } ⊂ n/2 +1, . . . , n . Each potential edge in G1 considers distinct
potential hyperedges in H 1 , so the potential edges in G1 are present independently
dn/2e
1
of one another. So the edge {i, j} is not in G if and only if all
potential
d−2
hyperedges of the form {i, j, w3 , . . . , wd } are not present. Hence
dn/2e
P ({i, j} ∈
/ E) = (1 − p1 )( d−2 ) =: 1 − p∗ .
Consequently, G1 is equal in distribution to G
n/2 , p∗ . Note that
dn/2e
2 d−2
p∗ = 1 − e−p1 ( d−2 )+O(p1 n )
θ
(ln n)2
nd−2
1
+O
= 1 − exp − (d − 1)! d−1 d−2
3
n
2 (d − 2)!
n2
d−1 θ
(ln n)2
+
O
.
=
3 · 2d−2 n
n2
For large enough n, we have that
p∗ ≥ 0.9
θ
d−1
.
d−1
3·2
bn/2c
In particular, we meet the conditions to apply the corollary to de la Vega’s theorem
(Corollary 4.2.2) to G1 . So w.h.p. there is a path in G1 of length at least
n
10 · 2d−1
1−
,
2
(d − 1)θ
80
which corresponds to a path in H 1 of same length spanning only vertices in n/2
(note that the hyperedges that make up this path will include larger index vertices).
Similarly, we construct another random graph, G2 , on bn/2c by the following:
for each possible edge {i, j} ⊂ bn/2c , we define {i, j} ∈ E(G2 ) if and only if there is
n
n
some hyperedge in H 2 of the form d e + i, d e + j, v3 , . . . , vd where {v3 , . . . , vd } ⊂
2
2
dn/2e . By a similar argument, there is a path in G2 of length at least
n
10 · 2d−1
1−
,
2
(d − 1)θ
which corresponds to a path in H 2 of same length spanning only vertices in dn/2e +
1, · · · , n .
Using H 3 , we will concatenate these paths together. Let E denote the event that
these long paths exists in H 1 and H 2 . On E, let A be the last n/ ln n vertices of such
a long path in H 1 and B be the first n/ ln n vertices of a such long path in H 2 (both
n
of these are defined because the length of these paths is − o(n) n/ ln n). Then
2
3
1
by independence of potential hyperedges of H from H and H 2 ,
n
n−2
n
n
ln n
P (E ∩ {no hyperedge between A, B}) ≤ (1 − p1 )( ln n )( ln n )( d−2 )
p n2
nd−2
d − 1 θn
≤ exp −
+ o(1) ≤ exp −
= o(1).
3 (ln n)2 (d − 2)!
4 (ln n)2
Hence, with high probability these two long paths on distinct vertices exists and
there is a hyperedge that contains one of the last n/ ln n vertices, a, of the first path
and one of the first n/ ln n vertices, b, of the second path. We construct our path
in H by following the path in H 1 until we reach the vertex a, then choosing this
connecting hyperedge containing a and b, then following the path in H 2 until it ends.
This path has length at least
n
10 · 2d−1
n
10 · 2d−1
n
2d+2
2
1−
−
+1≥n−
n−2
≥n−
n,
2
(d − 1)θ
ln n
θ(d − 1)
ln n
θ
as desired.
81
Corollary 4.2.4. Let ω → ∞, such that ω = o(ln n), and p = (d − 1)!
ln n − ω
. Then
nd−1
w.h.p. Hd (n, p) has a path of length at least
2d+3
.
n 1−
ln n
Proof. Let Θ = ln n − ω and notice that
2d+2
2d+3
≤
,
Θ
ln n
for sufficiently large n.
4.3
Pósa’s Lemma
Now that we have established that paths of length n − o(n) are likely to exist near the
connectedness threshold of Hd (n, p), Pósa’s Lemma will tell us that the neighbor sets
of endpoints of longest paths are relatively small, which we will later use to extend
these long paths. Note that any neighbor of an endpoint of a longest path must
necessarily be in said path or else we could extend this supposely longest parth and
reach a contradiction.
The hypergraph version of Pósa’s Lemma is essentially the same as the graph
version. Let H be a hypergraph and let P = (v0 , e1 , v1 , . . . , eh , vh ) be a longest path
starting from v0 . Suppose there is some hyperedge, e ∈ E(H), such that for some
0 ≤ i < h, the vertices vi and vh are in e. We say that the path
P 0 = (v0 , e1 , v1 , . . . , ei , vi , e, vh , eh−1 , vh−1 , . . . , ei+1 , vi+1 )
is a rotation of P by {vi , vh }. Note that the vertex set of the path is unchanged after
rotation. Each hyperedge may give rise to more than one such rotation for a given
hyperpath (which is a difference between graphs and hypergraphs). Furthermore, e
could have already been present within the path P .
82
Our proof of Pósa’s Lemma requires that we allow the possibility of repeated
hyperedges in the paths which in turn gives rise to our definition of a “weak” cycle
allowing duplicate hyperedges as well. In fact, this is the precise moment in which
we turned to “weak” cycles.
To proceed, let P(P, v0 ) be the set of paths obtained by the path P , starting at v0
via any number of rotations. Let S = S(P, v0 ) be the set of all endpoints of P(P, v0 )
excluding v0 . We call this set S a Pósa set. Let T be the neighbor set of S; in other
words
T = {t ∈ V (H) \ S : t is in an hyperedge with a vertex of S}.
Lemma 4.3.1 (Pósa’s Lemma). |T | < 2|S|.
Proof. For sake of self-containment, we include the proof of Pósa’s Lemma for
hypergraphs even though it essentially follows from the graph version; in fact, the
proof presented here follows the proof in [8] closely. It suffices to show that each
vertex of T is in the original path P and that one of its adjacent neighbors in P is
in S, because each vertex of S in the path gives rise to at most 2 vertices of T in the
path except the endpoint in P, which can give rise to only one vertex of T .
Let t ∈ T and suppose that s ∈ S is a neighbor of t. Since s ∈ S, there is some
path P 0 starting at v0 that ends in s. If t is not in this path, then we can extend the
path starting at v0 by one, which contradicts the maximality of the path P . Hence t
is in P 0 and since rotations preserve the vertex set of paths, we must have t ∈ P as
well.
Claim 4.3.2. One of the neighbors of t in P is in S.
Proof. We prove this claim by contradiction. First suppose that t 6= v0 and both
neighbors of t in P are not in S. So
P = (v0 , e1 , v1 , · · · , u, f1 , t, f2 , v, · · · , eh , vh ),
83
where u, v ∈
/ S. Notice that any rotation by {vi , vh }, where vi ∈
/ {u, t}, will keep the
three vertices u, t, v together in the path except possibly changing the order in which
they appear in the path. Rotation by {u, s} gives a path with t as its endpoint,
contradicting t ∈ T , and rotation by {t, s} gives v as its endpoint, contradicting
v∈
/ S.
We can do some number of rotations until we obtain a path P 0 with s (an
S−neighbor of t) as its endpoint. From the previous argument, we must have either u, f1 , t, f2 , v or v, f2 , t, f1 , u appearing in P 0 . Rotation by {t, s} yields a path
ending with either v or u. Hence one of either u or v must be in S.
The case t = v0 is similar (and simplier) to the above argument.
This inequality given by Pósa’s Lemma motivated our definition of non-expanding.
In order to show that the set of endpoints formed by rotations of a longest path must
likely be large, we will want to show the absence of non-expanding sets of “small”
size. For this, we need the following two definitions: For a hypergraph H, let V1 (H)
denote the set of non-isolated vertices of H and we say a hypergraph is empty if there
are no hyperedges, i.e. V1 (H) is empty.
For hypergraph, H, we define the following function which measures how expanding the hypergraph is:
u(H) := max{u : if A ⊂ V1 (H), |A| < u, then |N (A)| ≥ 2|A|}.
In other words, u(H), is the size of the smallest non-expanding set of non-isolated
vertices. As a consequence of Pósa’s Lemma, for any longest path, P , starting at nonisolated v0 , we have that |S(P, v0 )| ≥ u(H). One important property of the measure
u(H) is that if H is a hypergraph without isolated vertices, then adding hyperedges
to H can only decrease u.
The following corollary to Pósa’s Lemma for hypergraph is similar to another
corollary to Pósa’s Lemma due to Bollobás [8].
84
Corollary 4.3.3. Let P be a longest path in non-empty hypergraph, H, on [n], and
suppose the length of this path is h. If H does not have a weak cycle of length h + 1,
then there are at least
u(H)
n−1
d−1
n−1−u(H)
d−1
−
d
absent hyperedges of H, and the addition of any one of these non-present hyperedges
creates a weak cycle of length h + 1.
Proof. Let u = u(H) and v0 be one of the two endpoints of P . Let S be defined
as before. By Pósa’s Lemma, we know that |S| ≥ u. Let w1 , w2 , · · · , wu be distinct
vertices of S and P1 , P2 , · · · , Pu be paths starting at v0 that end in w1 , w2 , · · · , wu .
Let Si = S(Pi , wi ) for 1 ≤ i ≤ u. As before, by Pósa’s Lemma, we have that |Si | ≥ u,
as well. Let ti1 , ti2 , · · · tiu be u vertices of Si . Addition of any hyperedge that contains
wi and at least one element of ti1 , ti2 , · · · , tiu creates a weak cycle of length h + 1. Let
Ei be the collection of absent hyperedges e ∈ E(n, d), such that wi ∈ e and for some
j, tij ∈ e. Note that
u n−1−u
u n−1−u
u
n−1−u
|Ei | =
+
+ ··· +
1
d−2
2
d−3
d−1
0
n−1
n−1−u
=
−
.
d−1
d−1
Since each absent hyperedge of Ei contains wi , any e ∈ ∪ Ei can be in at most d
i
different Ei ’s. Hence we have that
u
[
u
E
i ≥
n−1
d−1
i=1
−
d
n−1−u
d−1
.
Necessarily none of these possible hyperedges can be present in H giving the desired
bound.
Corollary 4.3.4. Let D = D(d) > 1 be a constant. Then there is some C = C(d) > 0
such that if H is a hypergraph on [n] with at most 2 ln n isolated vertices which has
85
n
and contains a longest path of length h, but no weak cycle of length h + 1,
D
then there are at least Cnd absent hyperedges in H such that the addition of any one
u(H) ≥
of these non-present hyperedges creates a weak cycle of length h + 1.
Proof. If n0 ≥ n − 2 ln n is the number of non-isolated vertices of H, then the
previous lemma gives the number of such absent beneficial hyperedges is at least
0 0 n
n0 −1− D
n −1
n −1
n0 −1−u(H)
n
− d−1
u(H) d−1 −
D
d−1
d−1
≥
d
d
n
n
0
0
(n − 1)d−1 − n − 1 −
≥
D · d!
D d−1
d
!
n
1
ln
n
≥
nd−1 1 − 1 −
−O
D · d!
D
n
≥ Cnd ,
1
for any C <
D · d!
4.4
d !
1
1− 1−
and n sufficiently large.
D
No Small Non-expanding Sets
As noted before, by Pósa’s Lemma, if P is a longest path, then the set of endpoints
formed by rotations of P is non-expanding. We wish to show that this set of endpoints
must likely be sufficiently large (positive fraction of n) so that when we incrementally
increase our parameter p, we will be assured that we likely increase the length of
longest paths. For this, we will show that likely any set of non-isolated vertices of
size less than n/3d is expanding. For ease of notation, let ln(3) n denote ln ln ln n.
Lemma 4.4.1. Let ω(n) ∈ [− ln(3) n, ln(3) n] and p = (d − 1)!
ln n + ω
. Then w.h.p.
nd−1
u(Hd (n, p)) ≥ n/3d .
These next few sections deal with showing the absence of non-expanding sets
according to their size. Specifically, we break up this range into small [1, n1/5 ] ,
86
medium [n1/5 , 6n/ ln n] and large [6n/ ln n, n/3d ] sized sets. The method used
for medium and large sets are similar but the bounds used in each are different.
In showing the absence of small-sized sets, we use the consequence of minimal nonexpanding sets from Lemma 4.1.2. We begin by showing the likely absence of small
non-expanding sets of non-isolated vertices.
ln n + ω
. Then w.h.p.
nd−1
there are no non-expanding sets consisting of non-isolated vertices size at most n1/5 .
Lemma 4.4.2. Let ωn ∈ [− ln(3) n, ln(3) n] and p = (d − 1)!
Proof. In particular, it suffices to show w.h.p. there are no minimal non-expanding
sets of non-isolated vertices of size at most n1/5 . If A is a minimal non-expanding
set of non-isolated vertices of size at most n1/5 , then by Lemma 4.1.2, the induced
hypergraph on T := A ∪ N (A) is connected and necessarily
d ≤ |T | < n1/5 + 2n1/5 ;
we obtain the lower bound on |T | because there must be at least one hyperedge on
each vertex of A, while the upper bound is forced by definition of non-expanding.
Furthermore, each vertex of A has no neighbors outside of T . As a consequence, it
“almost” suffices to show that there are no such sets T with:
• |T | =: t ∈ l := max{4, d}, 3n1/5 ,
• the induced hypergraph on T is connected, and
• at least dt/3e vertices of T have no neighbors outside of T .
We say almost, because in addition, we need to show that for d = 3, w.h.p. there are
no non-expanding sets such that |A ∪ N (A)| = 3.
Claim 4.4.3. Let E be the expected number of such sets T. Then E → 0.
87
Proof. First, let’s compute this expectation. For ease of notation, let n1 = 3n1/5 .
Then by an union bound over all sets of cardinality t and all further subsets of
cardinality dt/3e,
n1 X
n
t
P,
E≤
t
dt/3e
t=l
(4.4.1)
where P is the probability that the induced hypergraph on v1 , v2 , . . . , vt is connected
and there are no hyperedges containing a vertex of v1 , . . . , vdt/3e and a vertex of
vt+1 , . . . , vn . Notice that the set of potential hyperedges contained entirely within
v1 , . . . , vt is disjoint from the set of potential hyperedges containing a vertex of
v1 , . . . , vdt/3e and a vertex of vt+1 , . . . , vn ; so these two events are independent! Hence
P = P (induced hypergraph on v1 , . . . , vt is connected) q N ,
(4.4.2)
where N is the number of sets of cardinality d that contain at least one vertex of
v1 , . . . , vdt/3e and at least one vertex of vt+1 , . . . , vn . Notice that the induced hypergraph on v1 , . . . , vt is distributed as Hd (t, p). Let Pt (p) be the probability that Hd (t, p)
is connected.
Now let’s find a nice closed form for N . We find this closed form by finding a
more general formula: Suppose A and B are disjoint vertex sets of [n]. Let N 0 be
the number of d element subsets of [n] that contain at least one vertex of A and at
n
least one vertex of [n] \ (A ∪ B). Note that
− N 0 is the number of potential
d
hyperedges either contained entirely in A ∪ B or [n] \ A. In addition, any potential
hyperedges contained entirely within both A ∪ B and [n] \ A must be contained in B.
Hence
n
|A| + |B|
n − |A|
|B|
0
−N =
+
−
,
d
d
d
d
or simply
n
n − |A|
|A| + |B|
|B|
N =
−
−
+
.
d
d
d
d
0
88
Hence (4.4.2) becomes
).
)−(dt )+(t−dt/3e
d
n−dt/3e
d
n
P = Pt (p)q ( d )−(
(4.4.3)
Now let’s consider Pt (p). One isolated hyperedge will make a component of size d
and any additional hyperedge adds at most d−1 new vertices to the component. So for
lt−1m
a hypergraph on t vertices to be connected, there must be at least
hyperedges
d−1
present. Moreover note that Pt (p) is increasing in p. For ease of calculation, let
pi = (d − 1)!
ln n + (−1)i ln(3) n
,
nd−1
for i = 1, 2. Hence
Pt (p) ≤ Pt (p2 ) ≤
t−1
t
e
d
d
p2 d−1
t−1
d d−1 e
t−1
!d d−1
e
d
e td!
≤
t−1
d d−1
e
p2
.
Taking the ceilings off in both terms above increases this expression. In particular,
we have that
Pt (p) ≤
etd p2
t−1
t−1
d−1
.
Notice that
n
t
nt
t!
≤
≤ nt .
t
dt/3e
t! dt/3e! (t − dt/3e)!
Furthermore, uniformly over the range of t, we have
n
n−dt/3e
t
t−dt/3e
n
−
dt/3e
t
n
−
−
+
(
)
(
)
(
)
(
)
d
d
d
q d
≤ exp −p1
−
+ p2
d
d
d
2 n
d
t
d
≤ exp −p1
dt/3e + O
+ p2 n1
d
n
n2
t
nd−1
≤b exp − p1
.
3 (d − 1)!
Using these last three inequalities along with (4.4.1), we find that
E ≤b
n1
X
t=l
n
t
etd p2
t−1
89
t−1
d−1
t
n−1
e− 3 p1 ( d−1 ) .
Using the definition of p1 , we find that
t−1
d d−1
n1
n1
X
X
t
2
et p2
t
ln(3) n
3
3
e
E ≤b
=:
ct ,
n
t−1
t=l
t=l
(4.4.4)
We will show that the dominant contribution comes from cl . We see that
2
l
3
l−1
d−1
cl ≤b n p2 e
l
3
ln(3) n
(ln n)3
≤ 1/3 .
n
We bound the sum of ct by considering the ratio of consecutive terms,
t−1
t−1
d−1
d d−1
d
1
2
(t
+
1)
t
−
1
ct+1
1
1/3
= n 3 (ep2 ) d−1
(t + 1) d−1
1 (ln ln n)
d
ct
t
t
t d−1
t−1
3
1
2
d
d−1
≤ n 3 (ep2 ) d−1 e t
(t + 1) 2 · 1 · 1 · (ln ln n)1/3
≤b n2/3
(ln n)1/(d−1) 1/5 3/2
n
ln n.
n
For large enough n, the ratio between consecutive terms is less than (ln n)2 /n1/30 . So
(4.4.4) becomes
E ≤b
X
cl
j=0
(ln n)2
n1/30
j
≤b cl ≤b
(ln n)3
.
n1/3
Since this expectation tends to zero, w.h.p. there are no minimal non-expanding
sets of non-isolated vertices, A, such that the size of A∪N (A) is between l = max{4, d}
and 3n1/5 . This is precisely the location where our right endpoint value n1/5 comes
into play. If our right endpoint is much larger, the ratio of consecutive terms would
start to be larger than 1 and we could not simply bound the sum by considering only
the first summand.
To complete the proof of this lemma, we need to show for d = 3, w.h.p. there
are no minimal non-expanding sets, A, such that |A ∪ N (A)| = 3. In this situation,
either |A| = 2 or |A| = 3. For |A| = 2, this case corresponds to a pair of vertices of
degree 1 in a hyperedge together.
Claim 4.4.4. W.h.p. there is a not a pair of vertices of degree 1 in a common
hyperedge.
90
Proof. We prove this claim by using a first moment argument. Let X be the
number of pairs of vertices of degree 1 in an hyperedge together. Then
n n−2
E[X] =
P {v1 , v2 , v3 } is present, deg(v1 ) = deg(v2 ) = 1
2
1
n−2
n
n
n
−
2
−1
−
3
3
−
−1
≤ n pq ( 3 ) ( 3 ) ≤ n p2 exp −p1
3
3
1
ln n · (ln ln n)2
3
2
≤ n p2 exp −p1 n 1 + O
≤b
= o(1).
n
n
The case |A| = 3 corresponds to 3 vertices of degree 1 in a hyperedge together.
Claim 4.4.5. W.h.p. there is not a triplet of vertices of degree 1 in an hyperedge
together.
Proof. As before, this claim will be shown by a first moment method. Let X be
the number of triplets of vertices of degree 1 in an hyperedge together. So
n
E[X] =
P {v1 , v2 , v3 } is present, deg(v1 ) = deg(v2 ) = deg(v3 ) = 1
3
n
n
n−3
−(n−3
3
3
(
)
)
3
3
≤ n pq
≤ n p2 exp −p1
−
3
3
ln n · (ln ln n)3
1
3
≤b
= o(1).
≤ n3 p2 exp −p1 n2 1 + O
2
n
n2
This concludes the leftover cases and hence the proof of the lemma.
4.5
Bounding P (a, b)
In the previous section, we found that w.h.p. there are no non-expanding sets of
non-isolated vertices of size less than n1/5 by consider the expectation of number of
such sets T = A ∪ N (A). Now let’s consider the number, X, of non-expanding sets of
91
size in [n0 , n1 ], where n0 and n1 are specified later. Notice that we are not restricting
our non-expanding sets to only non-isolated vertices here. So
n1
X
E[X] =
2|A|−1
X
P (the neighbor set of A is B)
A,|A|=n0 B,|B|=0
n1 2a−1
X
X nn − a
=
a=n0 b=0
a
b
n
+ b
)−(a+b
d ) (d) ,
n−a
d
P (a, b)q ( d )−(
where P (a, b) is the probability that a and b fixed vertices, each of the b vertices is in
a hyperedge (⊂ A ∪ B) with at least one of the a vertices, with the understanding of
defining P (a, 0) = 1. The only non-explicit term is this P (a, b) term. In this section,
we determine an useful bound on P (a, b) for showing the absense of medium sized
non-expanding sets. This analysis of P (a, b) makes showing the absense of medium
non-expanding sets the deepest part of our argument. For showing the absense of
large sized non-expanding sets, we use the trivial inequality P (a, b) ≤ 1.
First, we bound P (a, b) in relative generality. Suppose A is a set of a vertices and
B is a set of b vertices where a, b ≥ 1 and a + b ≥ d. We wish to find an upper bound
on the probability that each vertex in B is in an hyperedge with at least one vertex
a+b
of A, where each of the
potential hyperedges is present with probability
d
p ∈ (0, 1) independently of one another. Let E be this event and so P (a, b) = P (E).
Lemma 4.5.1.
a+b−1
b−1
P (a, b) ≤ 1 − q ( d−1 )−(d−1)
b
d d−1
e
.
Before we prove this lemma, we note that this inequality reminds us of Stepanov’s
Inequality in random graphs. In Chapter 6, we discuss a hypergraph version of
Stepanov’s Inequality.
Proof. This proof is algorithmic in nature. On the event in question, there must
b exist at least
hyperedges and moreover that many hyperedges must be found
d−1
in the following greedy algorithm. First, we start at a vertex of B and check potential
92
hyperedges on this vertex one at a time until we find a present hyperedge, then we
move to another unseen vertex of B (i.e. a vertex not in any found hyperedges yet)
and continue looking at possible hyperedges, and so on.
Formally, we define the following process, ~t, ~e , in this probability space. Let
w1 , w2 , · · · , wb be the vertices of B and let t1 := w1 . We check the possible hyperedges
containing t1 and at least one vertex of A one after another according to any ordering
of the potential hyperedges. If none of these potential hyperedges are present, we say
that the process fails (note that this process can not fail at this step on the event E).
Suppose e1 is the first potential hyperedge that we see is present (we don’t check any
of the later possible hyperedges containing t1 at this point). We look at up to
a
a
b−1
a
b−1
a+b−1
b−1
+
+ ···
=
−
d−1
d−2
1
1 d−2
d−1
d−1
potential hyperedges. Note that on E, each vertex of B \e1 must be in some unchecked
but present hyperedge with a vertex of A.
As long as the process doesn’t fail at step 1, let t2 be the lowest index element of
B \e1 . After defining t2 , we know some potential hyperedges (maybe none) containing
both t2 and t1 are not present from step 1. Let M2 be the number of hyperedges
that we know are not present that contain t2 from step 1. Now we check possible
hyperedges containing t2 one after another (except those that we know aren’t present
in the previous step) to see if any are present. If none of these potential hyperedges
are present, the process fails (again this process can not fail at this step on E, since
if there is any hyperedge containing t2 and a vertex of A, we will find at least one).
Suppose e2 is the first present hyperedge containing t2 that we find. Note that we
a+b−1
b−1
can look at up to
−
− M2 hyperedges in step 2. Again, on E,
d−1
d−1
each vertex of B \ (e1 ∪ e2 ) must be in some unchecked but present hyperedge with a
vertex of A.
Then we let t3 be the lowest index element of B \ (e1 ∪ e2 ) and M3 be the number
93
of hyperedges that we know are not present that contain t3 from step 1 and 2. And
so on.
On E, we can define t0 s up to td
b
e
d−1
, because each ei contains at least 1 element
of A and so ei contains at most d − 1 elements of B. We say that we succeed at step
j if ej is defined by this process and we define the event SUCCESS if we have defined
ed
b
e
d−1
. Necessarily to succeed in step j, we must succeed in each previous step. Note
that on the event E is contained in the event SUCCESS. For ease of notation, let
P1 = P (succeed at step 1) and for j ≥ 2,
Pj = P (succeed at step j|succeed at steps 1, . . . , j − 1).
Hence
P (a, b) ≤ P (SU CCESS) = P succeed at steps 1, 2, · · · ,
= Pd
b
e
d−1
· Pd
b
e−1
d−1
b d−1
· · · P2 · P1 .
(4.5.1)
(4.5.2)
Note that we succeed in step 1 if any of the hyperedges containing t1 and at least one
vertex of A is present. Each of these potential hyperedges are present with probability
p. So
a+b−1
d−1
P1 = 1 − q (
b−1
)−(d−1
).
Now consider Pj for some j ≥ 1. For ease of computation, let
a
b−1
a
b−1
0
M :=
+ ··· +
.
d−2
1
1 d−1
Note that when Mj is defined, necessarily Mj ≤ M 0 , because M 0 is the number of
hyperedges that contain a specific vertex of B along with at least one other vertex of
B and at least one vertex of A. Note that
0
Pj =
M
X
P (succeed at j, Mj = M |succeed at 1, 2, · · · , j − 1)
M =0
0
=
M
X
P (succeed at j|Mj = M )P (Mj = M |succeed at 1, 2, . . . , j − 1).
M =0
94
Conditioned on Mj = M , we succeed in step j if any of the remaining unseen
b−1
a+b−1
−
− M potential hyperedges are present. Again, these potential
d−1
d−1
hyperedges are present with probability p independent of each other. Hence
a+b−1
d−1
P (succeed at step j|Mj = M ) = 1 − q (
b−1
− b−1
)−M ≤ 1 − q (a+b−1
)−(d−1
d−1 ) (d−1) .
Hence we find that
0
Pj ≤
M X
1 − q(
a+b−1
d−1
b−1
)−(d−1
) P (M = M |succeed at steps 1, 2, · · · , j − 1)
j
M =0
a+b−1
d−1
= 1 − q(
b−1
)−(d−1
) .
Thus (4.5.1) becomes
b
d d−1
e
a+b−1
b−1
P (a, b) ≤ 1 − q ( d−1 )−(d−1)
.
For the following specific range of p, a and b, we can use the following simplier
bound.
1
Corollary 4.5.2. If a ≥ d, b ∈ [1, 2a] such that 2a · p d−1 ≤ 1, then
b
1
P (a, b) ≤ 2ap d−1 .
Proof. From the previous lemma,
a+b−1
b−1
P (a, b) ≤ 1 − q ( d−1 )−(d−1)
b
d d−1
e
.
By Bernoulli’s inequality,
a
a
b−1
1
1 − (1 − p)(d−1)+(d−2)(
b−1
)+···+(a1)(d−2
)≤
a
a
b−1
a
b−1
+
+ ··· +
p.
d−1
d−2
1
1 d−2
95
In addition, for b ≤ 2a,
a
a
b−1
a
b−1
+
+ ···
≤ ad−1 1 + 2 + 4 + · · · 2d−2
d−1
d−2
1
1 d−2
< ad−1 2d−1 .
Hence
b
d d−1
e
a
a
b−1
a
b−1
P (a, b) ≤
+
+ ···
p
d−1
d−2
1
1 d−2
d b e
≤ ad−1 2d−1 p d−1 .
By hypothesis, ad−1 2d−1 p ≤ 1, so we can take off the ceilings in the above statement
to obtain the desired inequality.
4.6
No Medium Non-expanding Sets
As before, let ω = ω(n) ∈ [− ln(3) n, ln(3) n], where ln(3) n = ln ln ln n. Also, let
ln n + ω
p = (d − 1)! d−1 and
n
pi = (d − 1)!
ln n + (−1)i ln(3) n
,
nd−1
for i = 1, 2.
Lemma 4.6.1. W.h.p. there are no non-expanding sets of size in n1/5 , 6n/ ln n .
Proof. We introduce the right endpoint for a couple of reasons. First, our bound
on P (a, b) is good when a is not too close to the order of n. Furthermore, for our
method in the next section, we will require that a is at least 6n/ ln n to show that a
certain function is increasing in a. For ease of writing, let n0 = n1/5 and n1 = 6n/ ln n.
Let X denote the number of non-expanding sets, A ⊂ [n] in Hd (n, p), such that
|A| ∈ [n0 , n1 ]. As before,
E[X] =
n1 X
2a X
n n−a
a=n0 b=0
a
b
96
n
n−a
d
P (a, b)q ( d )−(
+ b
)−(a+b
d ) (d) ,
(4.6.1)
where P (a, b) is the probability that for a and b fixed vertices, each of the b vertices is
in a hyperedge (⊂ A ∪ B) with at least one of the a vertices with the understanding
that P (a, 0) = 1. For a in this range, we have that
1
1
2ap d−1 ≤ 2n1 p2d−1 < 1,
so by applying Corollary 4.5.2 to (4.6.1), we obtain that
E[X] ≤
n1 b=2a
X
X nn − a a=n0 b=0
a
b
1
2ap d−1
b
n
q ( d )−(
+ b
)−(a+b
d ) (d) =:
n−a
d
X
γba
(4.6.2)
a,b
By considering the ratio of consecutive terms, we find
a
a+b
b
γb+1
1
1
n−a−b
n − 3a
d−1
=
2ap
2ap d−1 · 1
q −(d−1)+(d−1) ≥
a
γb
b+1
2a
1
d−1
ln n − ln(3) n
1
1
≥ (ln n) d−1 ,
≥ (n − 3n1 )
n
2
for large enough n over the entire range for a, b. So we can bound E[X] in (4.6.2) by
!j
n1
n1
X
X
X
2
a
a
E[X] ≤
≤
2
γ2a
γ2a
.
(4.6.3)
1
d−1
(ln
n)
a=n0
a=n0
j=0
Now
a
γ2a
2a n n−a 3a 2a
1
n n−a q ( d )−( d )−( d )+( d )
2ap d−1
=
a
2a
en a en 2a 2a
n
n−a
3a
2a
1
≤
2ap d−1
e−p(( d )−( d )−( d )+( d ))
a 2a
3 3
2
n
n−a
3a
2a
e n d−1
= exp a ln
p
−p
−
−
+
.
d
d
d
d
a
We will use the following proposition to obtain a good bound on the binomial expresa
sions in γ2a
.
Proposition 4.6.2. As m → ∞, we have that
d
d
(m)d = m −
md−1 + O md−2 .
2
97
Proof. Clearly (m)d is a polynomial in m of degree d. So we care only to find the
first two coefficients.
(m)d =
d−1
Y
(m − i) = md − (1 + 2 + · · · (d − 1)) md−1 + O md−2
i=0
d
=m −
md−1 + O md−2 .
2
d
Uniformly over a ∈ [n0 , n1 ], the quantities n, n − a, 2a, 3a tend to infinity. By
Proposition 4.6.2,
d d−1
d
d
(n)d − (n − a)d = n −
n
− (n − a) +
(n − a)d−1 + O nd−2
2
2
d
d−1
X d
d X d−1
i d−i
=−
(−a) n +
(−a)i nd−1−i
i
2
i
i=1
i=1
d
+ O(nd−2 ),
and similarly,
d
d
d−1
d
−(3a)d + (2a)d = −(3a) +
(3a)
+ (2a) −
(2a)d−1 + O nd−2
2
2
d
= 2d − 3d ad +
3d−1 − 2d−1 ad−1 + O nd−2 .
2
Note that O pnd−2 = o(1). Hence (4.6.3) becomes
d
E[X] ≤ 3
n1
X
exp (a · fn (a)) ,
a=n0
for sufficiently large n, where
!
3 3
d 2
p X d
e n d−1
p
−
(−a)i−1 nd−i
fn (a) := ln
a
d! i=1 i
!
X
d−1 p
d
d−1
+
(−a)i−1 nd−1−i
d!
2 i=1
i
d−2
d−1
p
d
d
d
d−1
d−1
−
2 −3 a
+
3
−2
a
.
2
d!
Now we will show that fn is convex and is small enough that E[X] → 0.
98
(4.6.4)
Claim 4.6.3. For x ∈ [n0 , n1 ], fn (x) is convex.
Proof of claim. This will be shown by computing fn00 .
!
d X
1
d
p
fn00 (a) = 2 −
(i − 1)(i − 2)(−a)i−3 nd−i
a
d! i=3 i
!
X
d−1 d
d−1
p
(i − 1)(i − 2)(−a)i−3 nd−1−i
+
d!
2 i=3
i
p
d
d
d
d−3
d−1
d−1
d−4
,
− (d − 2) 2 − 3 (d − 1)a
+
3
−2
(d − 3)a
d!
2
where the second sum and the last term are understood to be zero when d = 3. Note
that since a ≤ 6n/ ln n, we have that
1
1 (ln n)2
≥
.
a2
36 n2
Therefore
fn00 (a)
d d−1 2 d−4
d d−1 2 d−4
1 (ln n)2 p2
d 2 d−3
−
2 dn
+
2 dn
+
3 dn
≥
36 n2
d!
2
2
ln n
1 (ln n)2
−O
.
≥
36 n2
n2
For large enough n, this quantity is positive over the range of a.
As a consequence of convexity,
max fn (a) = max{fn (n0 ), fn (n1 )}.
a∈[n0 ,n1 ]
In particular, (4.6.4) becomes
E[X] ≤ 3
n1
X
exp (a max{fn (n0 ), fn (n1 )}) .
a=n0
99
Computing fn (n0 ) and fn (n1 ) will complete the proof of the lemma. For sufficiently
large n, we have that

2 
d−1
n
 p1 d−1
 − dn
d!
(3)
ln n + ln
 n3
fn (n0 ) ≤ O (1) + ln  1/5 ·
n
n2
d−1
1
+ O p2 n 5 nd−2 + O p2 nd−2 + O p2 n 5
1
1
= − ln n + O (ln ln n) ≤ − ln n,
5
6
3 3 2
e n d−1
p1 d−1
nd−1
fn (n1 ) ≤ ln
p
ln n − dn
+ O p2
6n 2
d!
ln n
d−1
n
+ O p2 nd−2 + O p2
(ln n)d−1
1
= − ln n + O (ln ln n) ≤ − ln n.
2
Hence
E[X] ≤ 3
n1 X
n
− 16
a
1
1/5
≤ b n− 6 n
= o(1).
a=n0
4.7
No Large Non-expanding Sets
As before, let ω ∈ [− ln(3) n, ln(3) n], p = (d − 1)!
pi = (d − 1)!
ln n + ω
and for i = 1, 2,
nd−1
ln n + (−1)i ln(3) n
.
nd−1
Finally let’s look at the third and final part of the range.
Lemma 4.7.1. W.h.p. there are no non-expanding sets of size in 6n/ ln n, n/3d .
Proof. For ease of notation, let n0 := 6n/ ln n, D = 3d and n1 := n/D. Let X
denote the number of non-expanding sets in Hd (n, p) of size in [n0 , n1 ]. As usual, it
suffices to show that E[X] → 0. In the previous section, we found that
n1 X
2a X
n
n−a
a+b
b
n n−a
E[X] =
P (a, b)q ( d )−( d )−( d )+(d) ,
a
b
a=n b=0
0
100
but our previous bound for P (a, b) is not useful for this range. Instead, we notice
that (trivially) P (a, b) ≤ 1, and so we have that
n1 X
2a X
X
n n (nd)−(n−a
− a+b + b
d ) ( d ) (d) =:
q
E[X] ≤
γba .
a
b
a=n b=0
a,b
(4.7.1)
0
Uniformly over a ∈ [n0 , n1 ], with n sufficiently large,
n
2
a
b
a+b
γb+1
n − b −(d−1
) ≥ n − 2 D · 1 ≥ 1 − D ≥ 3.
)+(d−1
=
q
2
γba
b+1
2 Dn + 1
+ n1
D
Hence
n1 X
2a
X
n1
3X
a
γ2a
.
(4.7.2)
3
2
a=n0 b=0
a=n0
a=n0
b=0
ν
eν µ
Combining (4.7.1) and (4.7.2) as well as the binomial inequality
≤
gives
µ
µ
γba
≤
n1
X
a
γ2a
2a 2a−b
X
1
≤
3 3 n1
3X
en
E[X] ≤
exp a ln
×
2 a=n
4a3
0
p
exp −
(n)d − (n − a)d − (3a)d + (2a)d . (4.7.3)
d!
By applying Proposition 4.6.2 on the falling factorials in (4.7.3), we obtain that
E[X] ≤ 2
n1
X
exp (a · gn (a)) ,
(4.7.4)
a=n0
where
!
d e3 n3
p X d
i−1 d−i
gn (a) := ln
−
(−a) n
4a3
d! i=1 i
!
X
d−1 d
d−1
p
(−a)i−1 nd−1−i
+
d!
2 i=1
i
d−1
d−2
p
d
d
d
d−1
d−1
−
2 −3 a
+
3
−2
a
.
d!
2
We will bound this sum by noticing that gn is increasing on this range and finding a
bound on gn . In fact, for some α ∈ (0, 1) fixed, we have that
!
d p X d
p
gn (αn) = −
(−α)i−1 nd−1 + (3d − 2d ) (αn)d + O(1).
d! i=1 i
d!
101
Hence
!
d X
d i−1
p2 nd−1 d−1 d
α
+
α 3 + O(1)
i
d!
i=2
!
d
X d
ln n
ln n 2 d
≤ − ln n +
·α
+
α 3 + o(ln n).
d
i
d
i=0
p2 nd−1
p1 nd−1
+
gn (αn) ≤ −
(d − 1)!
d!
In particular, we notice that for large enough n, we have that
1
d
2 d
gn (αn) ≤ ln n −1 +
α · 2 + α 3 + 0.1 .
d
By choosing α = 1/3d = 1/D, we have that
gn (n/D) ≤ ln n −1 +
1
2
!!
2
2
1
2
+ 2 + 0.1
≤ − ln n.
3
3
3
So as long as gn is increasing on [n0 , n1 ], using (4.7.4), we see that
E[X] ≤ 2
n1
X
a=n0
a
n1 X
1
= o(1).
exp a · max gn (a) ≤ 2
a∈[n0 ,n1 ]
n2/3
n
0
All that remains of the proof of this lemma is to show that gn is indeed increasing.
Claim 4.7.2. gn is increasing on [n0 , n1 ].
Proof. We will prove this claim by showing that gn0 (x) > 0 for x ∈ [n0 , n1 ]. For
sufficiently large n, uniformly over the range of a and p,
3
p
gn0 (a) = − −
a d!
p
−
d!
−
d X
d
i=2
i
!
(i − 1)(−a)i−2 nd−i
2d − 3d (d − 1)ad−2 + o(ln n/n).
We wish to find an lower bound on gn0 (a), so we find
ln n p1 d d−2 p2
0
gn (a) ≥ −
+
n
−
2n
d! 2
d!
+ o(ln n/n).
102
d X
d
(i − 1)ai−2 nd−i
i
i=3
!
Notice that
d X
d
i
i=3
(i − 1)a
i−2 d−i
n
d nd−2 X d
nd−2
· d2d−1 .
≤
i≤
D i=3 i
D
Hence
gn0 (a)
ln n p1 d d−2 d2d−1 p2 d−2
+
n
−
n
+ o(ln n/n)
≥−
2n
d! 2
D d!
d !
ln n
1 d−1 1 2
≥
− +
−
+ o(ln n/n).
n
2
2
2 3
Hence for sufficiently large n, uniformly over the range of a and p, we have
gn0 (a) ≥ ln n/(6n).
In summary, we have just finished the proof of Lemma 4.4.1 that says for p near
ln n
(d − 1)! d−1 w.h.p. the smallest non-expanding set of non-isolated vertices has size at
n
least n/3d (i.e. u(Hd (n, p)) ≥ n/3d ). In other words, likely Pósa sets have size linear
in n. By our deterministic result about the number of missing beneficial hyperedges,
if we increase our hyperedge probability p, we will likely create a cycle which we will
break apart to obtain a longer path.
4.8
Completing the proof of the main result
We can now prove our main result.
ln n + cn
. With high probability,
nd−1
Hd (n, p) has a weak cycle that spans the non-isolated vertices. As a consequence,
Theorem 4.8.1. Let cn → c ∈ R and p = (d − 1)!
−c
P (Hd (n, p) has a weak Hamilton cycle) → e−e .
Proof. We prove this theorem with a series of lemmas. We start with hyperedge
probability p0 relatively far below p and incrementally increase our hyperedge probability. Due to the nature of our argument, we focus on only newly formed hyperedges
103
contained entirely within the non-isolated vertices of p0 . After sufficiently many steps,
we will be assured that for some hyperedge probability p1 between p0 and p, there is
a weak cycle in Hd (n, p1 ) spanning the non-isolated vertices of Hd (n, p0 ). However,
since p0 is too far below p1 , there is a non-trivial limiting probability that an isolated
vertex of Hd (n, p0 ) is no longer isolated in Hd (n, p1 ). At this point, we can increment
the hyperedge probability even further to p from p1 , so that we obtain a weak cycle
spanning the non-isolated of Hd (n, p1 ). But now, since p1 is relatively close to p, the
probability that an isolated vertex in Hd (n, p1 ) becomes non-isolated after increment
hyperedge probability to p tends to zero, which will finish the proof of the theorem.
Truth be told, we increment our hyperedge probability in two steps, unlike Bollobás’ proof [8] in the graph case which increments in one step, because we wish to
determine the probability of weak Hamiltonicity in the critical window (where limiting
probability is strictly between 0 and 1), rather than show that w.h.p. Hd (n, (n ln n +
nω)/n) is weak Hamiltonian for ω → ∞, which can be done in one increment step as
well.
Since both increment steps are quite similar, we first start with showing that for
p1 near but below p, the isolated vertices in Hd (n, p1 ) stay isolated in Hd (n, p). Let
V0 (H) denote the isolated vertices of the hypergraph H.
(ln n)3
. With high probability, the isolated vertices of
nd
Hd (n, p1 ) are still isolated in Hd (n, p) (under the usual containment). In other words,
Lemma 4.8.2. Let p1 := p −
w.h.p. V0 (Hd (n, p1 )) = V0 (Hd (n, p)).
Proof. Note that we construct Hd (n, p) from independent Hd (n, p1 ) and Hd (n, p∗ )
with
p∗ =
p − p1
(ln n)3
≤b
,
1 − p1
nd
where a hyperedge is present in Hd (n, p) if and only if this hyperedge is present in at
least one of Hd (n, p1 ) and Hd (n, p∗ ).
104
By Lemma 2.2.5, the number of isolated vertices in Hd (n, p1 ) is asymptotically
Poisson. So we have w.h.p. V0 (Hd (n, p1 )) ≤ ln n. By independence of present hyperedges of Hd (n, p1 ) and Hd (n, p∗ ), for any set of vertices A,
P ({V0 (Hd (n, p1 )) = A} ∩ {there is a non-isolated vertex in A in Hd (n, p∗ )})
= P ({V0 (Hd (n, p1 )) = A}) P ({there is a non-isolated vertex in A in Hd (n, p∗ )}) .
Note that as long as |A| ≤ ln n, we have that
P (there is a non-isolated vertex in A in Hd (n, p∗ )) ≤b
(ln n)4
.
n
Hence after “summing” over the sets A with |A| ≤ ln n, we have that
P (|V0 (Hd (n, p1 ))| ≤ ln n, one of V0 (Hd (n, p1 )) is non-isolated in Hd (n, p∗ ))
≤b
(ln n)4
.
n
Combining this result with the fact that w.h.p. |V0 (Hd (n, p1 ))| ≤ ln n finishes the
proof of the lemma.
Now let p0 = p1 −
2d+4 ln n
, where C = C(d) is defined in Corollary 4.3.4.
C nd
Lemma 4.8.3. (i). With high probability, there is a weak cycle in Hd (n, p1 ) spanning
the non-isolated vertices of Hd (n, p0 ).
(ii). With high probability, there is a weak cycle in Hd (n, p) spanning the nonisolated vertices of Hd (n, p1 ).
Before we prove this lemma, notice that the second part of the lemma in conjugation with the previous lemma completes the proof that Hd (n, p) has a weak cycle
spanning the non-isolated vertices of Hd (n, p).
Proof. (i). In this proof, we will incrementally increase the hyperedge probability until we are likely to obtain a weak cycle spanning the non-isolated vertices of
105
Hd (n, p0 ). At each step, we will likely increase the length of the longest path until
we reach the weak cycle. We will use the following results on the likely structure of
2d+3
n
Hd (n, p0 ). From Corollary 4.2.4, w.h.p., there is a path of length at least n −
ln n
in Hd (n, p0 ). In addition, from Lemma 2.2.5, the number of vertices of degree zero
in Hd (n, p0 ) is at most ln n. By Lemma 3.6.2, w.h.p. the non-isolated vertices of
Hd (n, p0 ) form a component. Also from Lemma 4.4.1, w.h.p. u(Hd (n, p0 )) ≥ n/3d .
Let E be the intersection of these “with high probability” events about Hd (n, p0 ).
2 ln n
Now let’s incrementally increase our parameter from p0 to p1 . Let ∆p =
,
C nd
where C = C(d) is defined in Corollary 4.3.4. We will incrementally increase our
j
n k
times. Let H 0 be distributed as Hd (n, p0 ) and for each
parameter k := 2d+3
ln n
i ∈ {1, 2, . . . , k}, let H i be independent copies of Hd (n, ∆p). Furthermore, let H(j)
be the random hypergraph where a hyperedge is in H(j) if and only if this hyperedge
is present in at least one of H 0 , H 1 , . . . , H j . By construction, we have that
H(0) ⊂ H(1) ⊂ . . . ⊂ H(k).
In moving from H(j) to H(j + 1), we will show that we add a hyperedge that turns a
longest path into a cycle. Each potential hyperedge of H(j) is present independently
of each other hyperedge, and P e ∈ E(H(j)) = 1 − q0 · (1 − ∆p)j . Let
p0 = 1 − q0 · (1 − ∆p)k .
So we have that
p0 ≤ 1 − (1 − p0 )(1 − k∆p) ≤ k∆p + p0 (1 − kp1 ) ≤ p0 + k∆p ≤ p1 .
Since having weak cycle spanning the non-isolated vertices of H(0) is a monotone
increasing property, it suffices to show that w.h.p. there is a weak cycle spanning the
non-isolated vertices of H(0) in H(k).
For hypergraph, H, and a subset, W , of its vertex set, we denote the subgraph
induced on W by (H)W and as before let V1 (H) be the vertices of degree at least 1 in
106
H. We will restrict our attention to potential hyperedges contained in V1 (Hd (n, p0 )).
This is because once restricted, our measure u is (non-strictly) increasing with the
addition of hyperedges.
Consider the event {V1 (Hd (n, p0 )) = A} ∩ E for a set of vertices A with size at
least n − ln n. Define lj as the length of the longest path in (H(j))A if this induced
hypergraph is not weak Hamiltonian and if it is weak Hamiltonian, let lj = |A|.
Consider the event lj = lj+1 < |A| − 1 as well as V1 (H(0)) = A and E. First, since
(H(0))A is connected (by E), so is (H(j))A . If (H(j))A has a weak cycle of length
lj + 1 ≤ |A| − 1, then since (H(j))A is connected, we can break apart this weak cycle
and extend this long path by 1 contradicting maximum length of lj . So (H(j))A can
not have a weak cycle of length lj + 1. Likewise, there can not be a cycle of length
lj+1 + 1 = lj + 1 in (H(j + 1))A . Thus, by Corollary 4.3.4, there are at least Cnd
absent hyperedges of (H(j))A that must also remain absent in (H(j + 1))A .
Now consider the event lj = lj+1 = |A| − 1 as well as V1 (Hd (n, p0 )) = A and E.
Since lj = lj+1 = |A| − 1, we have a Hamilton path in (H(j))A and (H(j + 1))A but
no weak Hamilton cycle in either. Also by Corollary 4.3.4, there are Cnd beneficial
absent hyperedges of (H(j))A that must also remain absent in (H(j + 1))A .
Hence on the event
{V1 (H(0)) = A} ∩ E ∩ {lj = lj+1 < |A|},
there are C(d)nd hyperedges missing in H j+1 . Since the hyperedges of H j+1 are
independent from the events that depend on H(0), H(1), . . . , H(j),
P ({V1 (H(0)) = A} ∩ E ∩ {lj = lj+1 < |A|})
d
≤ (1 − ∆p)Cn P ({V1 (H(0)) = A} ∩ E) .
Note that
d
(1 − ∆p)Cn ≤ exp −∆p · Cnd = 1/n2 .
107
Hence
P ({V1 (H(0)) = A} ∩ E ∩ {lj = lj+1 < |A|}) ≤
1
P ({V1 (H(0)) = A} ∩ E) .
n2
(4.8.1)
Furthermore, the event that we do not have a weak cycle of length |A| in H(k) is
contained in the event that at some intermediate step, our longest paths are not long
enough and every longest path does not increase in the following step. In other words,
{lk < |A|} ⊂ {l0 < |A| − k} ∪ {l0 ≥ |A| − k, lk < |A|}
k−1
⊂ {l0 < |A| − k} ∪ ∪ {lj = lj+1 < |A|} ,
j=0
where the last term comes from
|A| − k ≤ l0 ≤ l1 ≤ · · · ≤ lk < |A|.
On E, we know that l0 ≥ n − k ≥ |A| − k, and so
P ({V1 (H(0)) = A} ∩ E ∩ {lk < |A|})
≤
k−1
X
P ({V1 (H(0)) = A} ∩ E ∩ {lj = lj+1 < |A|}).
j=0
Using (4.8.1), we find that
P ({V1 (H(0)) = A} ∩ E ∩ {lk < |A|}) ≤ P ({V1 (H(0)) = A} ∩ E)
k
.
n2
(4.8.2)
Hence by “summing” (4.8.2) over all such vertex sets A, with |A| ≥ n − ln n, gives
P ({|V1 (H(0))| ≥ n − ln n} ∩ E ∩ {no weak cycle in H(k) spanning V1 (H(0))})
≤ P ({|V1 (H(0))| ≥ n − ln n} ∩ E)
k
. (4.8.3)
n2
The event that |V1 (H(0))| ≥ n − ln n is one of the likely events that make up E.
Since P (E) → 1 and k = O (n/ ln n), with high probability in H(k), there exists a
108
weak cycle spanning V1 (Hd (n, p0 )). This completes the proof of the first part of the
lemma.
(ii). In proving the second part of the lemma, we essentially use the same method
as before, but now we only need to increment ln n times which changes the hyperedge
probability a relatively neglible amount rather than about roughly n/ ln n times.
Now let H 0 be distributed as Hd (n, p1 ) and k1 = dln ne. As before, let H i be
independent random hypergraphs on [n] distributed as Hd (n, ∆p), for i ≤ k1 , and
H(j) be the random hypergraph that a hyperedge e ∈ E(H(j)) if and only if e ∈
E(H i ) for at least one i ∈ {0, 1, . . . , j}. We let E be the event that H(0) has only
one non-trivial component, has a path of length of length at least n − ln n in H(0)
(w.h.p. by first part of lemma), and that u(H(0)) ≥ n/D. By first part of lemma
and the comments before, we have that P (E) → 1. Note that
p1 < p1 + ∆p < p1 + 2∆p < . . . < p1 + k1 ∆p ≤ p.
It suffices to show that w.h.p. there is a weak cycle in H(k1 ) spanning the non-isolated
vertices of H(0).
Consider the event {V1 (Hd (n, p1 )) = A} ∩ E where A is a set of vertices with
|A| ≥ n − ln n. Just as before,
P (V1 (H(0)) = A, E, lj = lj+1 < |A|) ≤
1
P (V1 (H(0)) = A, E),
n2
and
P (V1 (H(0)) = A, E, lk1 < |A|) ≤
kX
1 −1
P (V1 (H(0)) = A, E, lj = lj+1 < |A|)
j=0
≤ P (V1 (H(0)) = A, E)
k1
.
n2
Again “summing” over all such vertex sets A, with |A| ≥ n − ln n, gives
P (E, lk1 < |V1 (H(0))|) ≤ P (E)
109
k1
.
n2
Since P (E) → 1, with high probability there is a weak cycle in Hd (n, p) containing
the entire vertex set of V1 (Hd (n, p1 )) as desired.
It remains to show that the last statement of the theorem that the probability of
−c
weak Hamiltonicity tends to e−e .
Claim 4.8.4. Let p = (d − 1)!
ln n + cn
where cn → c ∈ R. Then
nd
−c
P (Hd (n, p) is weak Hamiltonian) → e−e .
Proof. By the first part of the theorem that we have shown, the probability that
Hd (n, p) is weak Hamilton is within an additive error of o(1) from the probability
that Hd (n, p) has no vertex of degree zero. By Lemma 2.2.5, this latter probability
tends to the desired limit.
Now let’s convert our result to Hd (n, m).
Corollary 4.8.5. Let M =
n
(ln n + cn ) where cn → c ∈ R. Then
d
−c
P (Hd (n, M ) is weak Hamiltonian) → e−e .
Proof. Earlier, when we converted results about Hd (n, p) to Hd (n, m), we showed
that if the property held with very high probability for Hd (n, p), then with high probability Hd (n, m) also has this property. It is easier to convert increasing properties
between the models. Note that for m1 < m2 ,
P (Hd (n, m1 ) is weak Hamilton) ≤ P (Hd (n, m2 ) is weak Hamilton).
Hence for any p, M 0 and any increasing property Q,
P (Hd (n, p) ∈ Q) =
(nd)
X
P (Hd (n, m) ∈ Q)P (e(Hd (n, p)) = m)
m=0
≤ P (Hd (n, M 0 ) ∈ Q) + P (e(Hd (n, p)) > M 0 ),
110
and
P (Hd (n, p) ∈ Q) ≥ P (Hd (n, M 0 ) ∈ Q) (1 − P (e(Hd (n, p)) > M 0 )) .
For i = 1, 2, let
ln n + cn + (−1)i n−1/4
.
pi = (d − 1)!
nd−1
The random variable e(Hd (n, p)) is binomially distributed, and so by Chebyshev’s
inequality, one can easily show that
P (e(Hd (n, p1 )) > M ) → 0
and
P (e(Hd (n, p2 )) ≤ M ) → 0.
Hence we get
P (Hd (n, p1 )is weak Hamilton) − o(1) ≤ P (Hd (n, M ) is weak Hamilton)
≤ P (Hd (n, p2 ) is weak Hamilton) + o(1).
Since cn − n−1/4 , cn + n−1/4 → c, by Claim 4.8.4, P (Hd (n, p) is weak Hamilton) tends
−c
to e−e , which forces P (Hd (n, M ) is weak Hamilton) to tend to the desired limit as
well.
4.9
Relation to Stopping Times
Let T = min{m : H̃(m) is Hamilton}. An immediate consequence of Corollary 4.8.5
is that Xn := (T − n ln n/d) /d converges in distribution to a random variable X with
−x
cumulative distribution function FX (x) = e−e .
Necessarily, we know that τ1 ≤ T , where τ1 is the stopping time for having
minimum degree at least 1. Unlike the previous results on stopping times, here we
do not prove coincidence, although we suspect that likely τ1 = T .
111
Theorem 4.9.1. There is some δ = δ(n) → 0, such that with high probability
τ1 ≤ T ≤ τ1 + δn.
In other words, w.h.p. T − τ = o(n).
Proof. By Corollary 4.8.5, for M =
n
(ln n + cn ) with cn → c ∈ R,
d
−c
P (T ≤ M ) = P (Hd (n, M ) has a weak Hamilton cycle) = e−e
+ o(1)
= P (Hd (n, M ) has min degree at least 1) + o(1)
= P (τ1 ≤ M ) + o(1).
Claim 4.9.2. If > 0, then P (T − τ1 ≥ n) → 0. In other words, (T − τ1 )/n
converges to 0 in probability.
Proof. We prove this claim by contradiction. Suppose that
lim sup P (T − τ1 ≥ n) = α > 0.
n→∞
Then there is some A ∈ N such that
−A
A
e−e + 1 − e−e
Define Mi =
≤
α
.
2
n
ln n + (−1)i A for i = 1, 2. Then by definition of A,
d
α
A
−A
lim P (τ1 ∈
/ [M1 , M2 ]) = e−e + 1 − e−e ≤ .
n→∞
2
So
lim sup P (T − τ1 ≥ n, τ ∈ [M1 , M2 ]) ≥ lim sup P (T − τ1 ≥ n)
n→∞
n→∞
− lim P (τ1 ∈
/ [M1 , M2 ]) ≥
n→∞
n
Let B = d2/e and mi =
d
i
ln n − A +
B
{τ1 ∈ [M0 , M1 ]} =
α
.
2
for 0 ≤ i ≤ 2AB. Since
2AB−1
∪ {τ1 ∈ [mi , mi+1 ]},
i=0
112
(4.9.1)
by pigeonhole principle, there is some I ∈ [0, 2AB − 1], such that
lim sup P (T − τ1 ≥ n, τ1 ∈ [mI , mI+1 ]) ≥
n→∞
α
.
4AB
In particular, {T − τ1 ≥ n, τ1 ∈ [mI , mI+1 ]} ⊂ {T − mI ≥ n, τ1 ≤ mI+1 }. Hence
lim sup P (T ≥ mI + n, τ1 ≤ mI+1 ) ≥
n→∞
α
.
4AB
However
mI+1 = mI +
1
n
≤ mI +
< mI + n,
B
2d
and so
P (T ≥ mI + n, τ1 ≤ mI+1 ) ≤ P (T > mI+1 , τ1 ≤ mI+1 )
= P (τ1 ≤ mI+1 ) − P (T ≤ mI+1 , τ1 ≤ mI+1 )
= P (τ1 ≤ mI+1 ) − P (T ≤ mI+1 ).
But these two probabilities tend to the same number. Hence we must have
P (T − τ1 ≥ n) → 0.
Now we will find an δ = δ(n) → 0 such that P (T − τ1 ≥ δn) → 0. We find
the sequence of natural numbers, Nm , recursively. Let N1 = 1. We know for a fixed
1
m ∈ N, that P (T − τ1 ≥ n) → 0. Suppose Nm is defined. Let Nm+1 > Nm so that
m
1
1
sup P T − τ1 ≥
n ≤
.
m+1
m+1
n≥Nm+1
Defining δ(j) =
1
for N (k) ≤ j < N (k + 1), yields
k
P (T − τ1 ≥ δn) → 0.
113
CHAPTER 5
ASYMPTOTIC NORMALITY OF STRONG GIANT
5.1
Overview
Karp [31] and Luczak [39] proved that if p = c/n and c > 1 (fixed), then w.h.p.
D(n, p) has a strong component with size on the order of n, while the size of the
second largest strong component is bounded in probability. Furthermore, if θ = θ(c)
is the unique root in (0, 1) of the equation
1 − θ = e−cθ ,
(5.1.1)
then the size of the strong giant component divided by n tends to θ2 in probability. In
this chapter, we show that the distribution of the size of the strong giant component,
properly centered and scaled, is asymptotically normal. In fact, we will prove the
following stronger result.
Theorem 5.1.1. Let c > 1 be fixed (does not depend on n). Let L1 = (V1 , A1 ) be the
largest strong component of D(n, m = cn), where V1 is its vertex set and A1 is its arc
set. Then
|V1 | − θ2 n |A1 | − cθ2 n
,
n1/2
n1/2
D
=⇒ N (0, B) ,
where N (0, B) is a 2−dimensional Gaussian random variable with mean 0 and covariance matrix B = B(c) given by integral expressions determined later in this chapter.
114
In other words, we will show that the number of vertices and arcs of the strong
giant component is asymptotically jointly Gaussian.
Results about the strong giant component have been relatively sparse, especially
in light of the depth of knowledge about the structure of the giant component of
G(n, p) and G(n, m). One reason for this deficiency is that determing the strong
components requires much more “global” information. As Pittel notes in [52], “the
graph component notion morphs into two, harder-to-handle, dual notions of a sinkset and a source-set, the subsets of vertices with no arc going outside, and no arcs
coming from outside, respectively.” A digraph is strongly connected precisely when
there are only trivial source-sets and sink-sets. In fact, an addition of an arc to a
digraph can connect many (more than 2) strong components; in particular, it can
connect strong components relatively “far apart”. For an example, see Figure 5.1,
where the addition of ~e (strongly) connects the digraph.
~e
Figure 5.1: Strong Connectivity Issues
This issue with determining the strong components makes the problem of analyzing the sizes of the strong components intrinsically more difficult than their graph
counterparts.
115
We need a few definitions. For a digraph, D, let Oi = Oi (D) (resp. Oo ) denote
the vertices of D with zero in-degree (resp. out-degree). A vertex with zero in-degree
or zero out-degree (i.e. in Oi ∪ Oo ) is called semi-isolated. An isolated vertex has
both zero in-degree and zero out-degree (i.e. in Oi ∩ Oo ).
The usefulness of adopting random algorithms for distributional results in random
graphs motivated us to try to find a deletion algorithm which ends with the largest
strong component. However the random algorithms that we wished to utilize can only
use “local” information. We considered a greedy deletion algorithm which sucessively
deletes semi-isolated vertices. This algorithm stops when there are not any more semiisolated vertices left to delete (i.e. both Oi and Oo are empty). We noticed that the
vertices of the strong giant component will be ever be deleted in this process. However,
any other non-trivial strong component (i.e. of size at least 2) will also not be deleted.
In fact, the terminal digraph of this deletion algorithm is the maximal sub(di)graph of
the starting digraph that has minimum in-degree and out-degree at least 1, which we
call the (1, 1)−core. The (1, 1)−core of a digraph is precisely the non-trivial strong
components along with directed paths between these components. Although this
(1, 1)−core can contain vertices outside of the largest strong component, we noticed
that the number of such vertices and arcs incident to these vertices are relatively
small compared to the vertices and arcs of the strong giant component. In fact, we
show that
Theorem 5.1.2. Fix c > 1. Let L1,1 = (V1,1 , A1,1 ) denote the (1, 1)−core of D(n, m =
cn). Then with high probability,
0 ≤ |V1,1 | − |V1 | ≤ 2(ln n)8 , 0 ≤ |A1,1 | − |A1 | ≤ 2(ln n)10 .
In other words, we will show that there are likely at most 2(ln n)8 vertices and
2(ln n)10 arcs outside of the strong giant component in the (1, 1)−core.
116
Due to this result, as well as our suspicion that the natural fluctuations of the
size of the strong giant component is on the order of n1/2 , which dwarfs (ln n)10 , we
turn our attention to finding the distribution of the (1, 1)−core, which can be found
using this deletion algorithm that considers only local information!
The observation that the (1, 1)−core has a massive strong component using almost
all vertices is very similar to the fact that there is likely a massive strong component
in a digraph chosen to have minimum in-degree and out-degree at least one, which
was recently established and used in papers on asymptotically counting the number
of strongly connected digraphs, by Pittel [52] and Pérez-Giménez & Wormald [48].
For our method, we will consider a randomized deletion algorithm where at each
step, we delete a random uniformly chosen semi-isolated vertex, along with all newly
formed isolated vertices. By construction, this process is necessarily Markov, and as
a consequence of its Markov nature, there is a natural recurrence relation for the joint
characteristic function of terminal |V1,1 | and |A1,1 |. As Pittel noticed in [51], such a
recurrence can be gainfully used to show asymptotic normality. Although Pittel uses
the moment generating function rather than our characteristic functions, he gives
the method that one can use to approximate the actual characteristic function by
a normal characteristic function which “almosts” satisfies this recurrence relation.
To establish that this Gaussian characteristic function closely fits the recurrence, we
approximate the likely realization of the deletion process by a deterministic trajectory
that is a solution of a certain system of ordinary differential equations. The solution
of such a system of differential equation will imply that such a Gaussian characteristic
function exists and is approximately the characteristic function of |V1,1 | and |A1,1 | in
the sense of pointwise limits.
Leaving technical issues aside for now, the nature of the approximation to the
random realization of the deletion process will force us to stop our deletion process
117
just shy of its terminal digraph. However, the following lemma tells us that even
when stopped early, we are not too far away from the terminal digraph.
Lemma 5.1.3. Let c > 1. With high probability, D(n, m = cn) is such that if D(t),
which is the digraph after some t deletions, has at most (ln n)2 semi-isolated vertices,
then
0 ≤ |V (D(t))| − |V1,1 | ≤ 2(ln n)4 ,
0 ≤ |A(D(t))| − |A1,1 | ≤ 4(ln n)6 .
Note that this result is does not depend on the choices of the deletion process,
just that the number of semi-isolated vertices is below some polynomial logarithm
expression.
Neither the proof of Theorem 5.1.2 or Lemma 5.1.3 analyze the deletion algorithm.
Because of that, their proofs are relegated to Appendix A.
5.2
The Greedy Deletion Algorithm
Before we describe the greedy deletion algorithm, we start with a few definitions.
A vertex w is an in-neighbor of v if the arc (v, w) is present. The descendant set
of vertex, v, is the set of vertices reachable from v; in other words, the descendant
set of v is {w ∈ [n] : ∃ path from v to w}, with the understanding that v is in its
own descendant set. Out-neighbor and ancestor set are defined analogously. For a
digraph, D, let S := (V, Oi , Oo , µ) denote the 4−vector composed of the vertices of
D, the vertices of in-degree zero, the vertices of out-degree zero, and the number of
arcs.
Our deletion step runs on a digraph without isolated vertices and returns a subgraph also without isolated vertices. We define the following step, made up of two
substeps, on a digraph, D, without isolated vertices.
118
Substep 1. Choose a semi-isolated vertex and delete this vertex along with all incident arcs obtaining an intermediate digraph D∗ with parameter S∗ = (ν ∗ , νi∗ , νo∗ , µ∗ )
along with I ∗ = Oi∗ ∩ Oo∗ , the set of newly formed isolated vertices in D∗ .
Substep 2. Delete these newly isolated vertices, I ∗ , from D∗ obtaining a digraph,
D0 , without isolated vertices, with parameter S0 = (ν 0 , νi0 , νo0 , µ0 ).
As long as this D0 still has semi-isolated vertices (i.e. Oi0 ∪ Oo0 6= ∅), we can run
this step again. If we continue running this greedy deletion algorithm until there are
no semi-isolated vertices left to delete, this terminal digraph will be the (1, 1)−core
of D (and the (1, 1)−core of the intermediate digraphs as well).
Let us have a closer look at a generic step. Suppose we delete a vertex, v, with
in-degree zero in substep 1 (the case where we delete a vertex with out-degree zero
is analogous). Any vertex other than v has the same out-degree in D∗ as in D, so
Oo = Oo∗ . However vertices from Oo whose only in-neighbor in the original digraph,
D, is the deleted vertex, v, now also have zero out-degree, and hence are in Oi∗ and I ∗ .
In fact, all isolated vertices of D∗ are born this way. Non-semi-isolated vertices of D
join Oi∗ if their only in-neighbor in D is v. In the second substep, we delete I ∗ from the
digraph to obtain D0 . In particular V 0 = V ∗ \ I ∗ , Oi0 = Oi∗ \ I ∗ , Oo0 = Oo∗ \ I ∗ , µ0 = µ∗ .
At the end, a vertex v is deleted from Oi , along with vertices, B, from Oo whose
only in-neighbor was v, so that Oo0 = Oo \ B. Also, vertices of D which have v as
its only in-neighbor, denoted R, now have in-degree zero; so Oi0 = (Oi \ {v}) ∪ R.
The number of arcs must decrease by at least max{1, |B| + |R|} (we supposed that
D had no isolated vertices, so v had at least one out-neighbor and at least one arc is
deleted). In particular, we have V 0 = V \ ({v} ∪ B), Oi0 = (Oi \ {v}) ∪ R, Oo0 = Oo \ B,
and µ − µ0 ≥ |B| + |R|. There is an similar description if we initially delete a vertex
of zero out-degree.
In summary, for S0 to be able to follow from S after one step, we must have
119
V 0 = V \ (A ∪ B) , Oi0 = (Oi \ A) ∪ R, Oo0 = (Oo \ B) ∪ T, where A ⊂ Oi and B ⊂ Oo
and that at least one of A and B have cardinality 1 and R, T ⊂ V \ (Oi ∪ Oo ) , with
at least of R and T being empty. Also µ − µ0 ≥ max{|A| + |T |, |B| + |R|}. If S0 can
follow from S, these sets A, B, R, T are uniquely determined from (S, S0 ).
We did not mention how we choose the first semi-isolated vertex to be deleted. In
our random deletion process, we choose this semi-isolated vertex uniformly at random
among all semi-isolated vertices (i.e. the probability of a specific semi-isolated vertex
being chosen is 1/(|Oi | + |Oo |)).
We define the process {D(t)} recursively.
Suppose D(0) is a digraph with-
out isolated vertices, then we obtain D(t + 1) by running one step on D(t), along
with the understanding that if D(t) does not have any semi-isolated vertices, then
D(t + 1) = D(t). Clearly, this process is Markov. There is, however, a much simpler Markov chain that rules the deletion process. Specifically, we will show that
{(|V (t)|, |Oi (t)|, |Oo (t)|, µ(t))} is a Markov process. The following reduction argument to this Markov chain is similar to the reduction argument by Aronson, Frieze
and Pittel [1] on a random sequence model.
(in)
Let DS denote the set of digraphs with parameters S. For D0 ∈ DS0 , let NS,S0 (D0 )
denote the number of digraphs D ∈ DS such that D0 can be obtained from D via
one step of this deletion algorithm by initially deleting a vertex with in-degree zero.
(out)
Similarly, NS,S0 (D0 ) is number of digraphs, D, such that D0 can be obtained after
(in)
(out)
deleting a vertex with zero out-degree. Let NS,S0 = NS,S0 + NS,S0 . We start with
showing that these numbers do not depend on the choice of D0 ∈ DS0 .
(in)
(in)
(out)
(out)
(in)
Claim 5.2.1. NS,S0 (D0 ) = NS,S0 and NS,S0 (D0 ) = NS,S0 for some numbers NS,S0 and
(out)
NS,S0 that depend only on S and S0 and not on the choice of D0 ∈ DS0 .
(in)
Proof. The two cases are very similar, so we only show that NS,S0 (D0 ) does
not depend on D0 . Unless S and S0 are such that |A| = 1, |T | = 0 and µ − µ0 ≥
120
(in)
max{1, |B| + |R|}, NS,S0 must be zero. Let’s denote the single vertex of A as v. For
any starting digraph D, necessarily v is the vertex deleted in substep 1. In D, each
vertex in B and R has v as its only in-neighbor, so v must have |B| = |Oo | − |Oo0 |
arcs ending at B and |R| = |Oi0 | − |Oi | + 1 arcs ending at R. Other arcs incident to
v must end at a vertex in
V \ ({v} ∪ Oi0 ∪ B) = V \ (Oi ∪ R ∪ B).
Hence any such D can be constructed by distributing these additional arcs among
the vertices of V \ (Oi ∪ R ∪ B). Thus

|V | − |Oi | − |B| − |R|



, if |A|=1, |T |=0, µ − µ0 ≥ max(1, |B| + |R|)
0 − |B| − |R|
(in)
µ
−
µ
NS,S0 =


0,
otherwise,
and NS,S0 does not depend on the choice of D0 ∈ DS0 , as desired. Likewise, one can
show that
(out)
NS,S0

|V | − |Oo | − |A| − |T |



, if |B|=0, |R|=0, µ − µ0 ≥ max(1, |A| + |T |)
0 − |A| − |T |
µ
−
µ
=


0,
otherwise.
For the process, {D(t)}, we define the process {S(t)} by
S(t) = (V (t), Oi (t), Oo (t), µ(t)).
Lemma 5.2.2. Suppose D(0) is uniform on DS(0) when conditioned on S(0). Then
D(t) conditioned on S(0), . . . , S(t) is distributed uniformly on DS(t) .
Proof. We prove this lemma by induction on t. The base case (t = 0) is assumed
to be true. Suppose statement is true up to t ≥ 0. Let S = S(t) = (V, Oi , Oo , µ) and
S0 = S(t + 1) = (V 0 , Oi0 , Oo0 , µ0 ). Let D0 ∈ DS0 . Then
X
P D(t + 1) = D0 |S(0), . . . , S(t) =
P D(t + 1) = D0 , D(t) = D|S(0), . . . , S(t) ,
D∈DS
121
but we can break up the probability inside the sum on the right as follows
P D(t + 1) = D0 , D(t) = D|S(0), . . . , S(t) =
P D(t + 1) = D0 |D(t) = D P D(t) = D|S(0), . . . , S(t) .
Using the inductive hypothesis, i.e. P D(t) = D|S(0), . . . , S(t) = |DS |−1 , for each
D ∈ DS , we have that
X
1
.
P D(t + 1) = D0 |S(0), . . . , S(t) =
P D(t + 1) = D0 |D(t) = D
|DS |
D∈D
S
To finish the proof of the lemma, it suffices to show that
X
P (D(t + 1) = D0 |D(t) = D) =
D∈DS
NS,S0
|Oi | + |Oo |
(in)
(out)
NS,S0
NS,S0
+
,
=
|Oi | + |Oo | |Oi | + |Oo |
(5.2.1)
because if so, the probability that D(t + 1) = D0 (conditioned on S(0), . . . , S(t))
depends only upon S and S0 and not on choice of D0 ∈ DS0 .
Now we prove this last equality. First, we break up the event {D(t+1) = D0 } into
two events depending on whether we delete a vertex with in-degree zero or out-degree
zero in the first substep; let C denote the event that we delete a vertex of in-degree
zero first. By symmetry, it suffices to show that
(in)
X
D∈DS
NS,S0
P ({D(t + 1) = D } ∩ C|D(t) = D) =
.
|Oi | + |Oo |
0
(in)
(in)
(in)
By the definition of NS,S0 (D0 ) = NS,S0 , we know that exactly NS,S0 of these summands
are non-zero. Furthermore, if D is such that the probability is non-zero, then because
we know (from S and S0 , as well as C) exactly which semi-isolated vertex is deleted
first, this probability is precisely the probability of choosing this vertex in the first
substep, which is exactly (|Oi | + |Oo |)−1 .
122
Lemma 5.2.3. {S(t)} is a Markov process.
Proof. We compute
P (S(t + 1) = S0 | S(0), . . . , S(t)) =
X
P (D(t + 1) = D0 | S(0), . . . , S(t))
D0 ∈DS0
=
X
X
P (D(t + 1) = D0 , D(t) = D | S(0), . . . , S(t))
D0 ∈DS0 D∈DS(t)
=
X
X
P (D(t + 1) = D0 | D(t) = D) · P (D(t) = D | S(0), . . . , S(t))
D0 ∈DS0 D∈DS(t)
=
X
X
P (D(t + 1) = D0 | D(t) = D) · |DS(t) |−1 ,
D0 ∈DS0 D∈DS(t)
which depends on S(t) only.
As mentioned before, we will in fact show that the cardinalities of S(t) form a
Markov chain. Before showing this result, we need a couple definitions. For S =
(V, Oi , Oo , µ), let ν = |V |, νi = |Oi |, νo = |Oo | and s = s(S) = (ν, νi , νo , µ). Let
g(ν, νi , νo , µ) denote the number of digraphs with vertex set [ν] with µ arcs such that
{1, . . . , νi } are the vertices of in-degree zero and {νi +1, . . . , νi +νo } are the vertices of
out-degree zero. Note that the number of digraphs, |DS |, with parameter S depends
only upon the cardinalities in S; moreover, this number is precisely g(s(S)).
Lemma 5.2.4. Let ν(t) = |V (t)|, νi (t) = |Oi (t)|, νo (t) = |Oo (t)|, and s(t) =
(ν(t), νi (t), νo (t), µ(t)). Then {s(t)} is a Markov process.
Proof. Using the last equality in the proof that {S(t)} is Markov as well as
equation (5.2.1), we find that
P (S(t + 1) = S0 | S(0), . . . , S(t)) =
X
P (D(t + 1) = D0 | D(t) = D) ·
D 0 ∈DS0 ,
D∈DS(t)
=
X
D0 ∈DS0
NS,S0 (D0 )
1
NS,S0
g(s(S0 ))
=
.
|Oi | + |Oo | g(s(t))
νi (t) + νo (t) g(s(t))
123
1
g(s(t))
Hence
X
P s(t + 1) = s0 |S(t) =
S0 :s(S0 )=s0
NS,S0 g(s0 )
νi + νo g(s)
(in)
X
=
S0 :s(S0 )=s0
NS,S0 g(s0 )
+
νi + νo g(s)
(out)
X
S0 :s(S0 )=s0
NS,S0 g(s0 )
.
νi + νo g(s)
(in)
Let us consider the first sum. As mentioned before, NS,S0 = 0 unless
ν − ν 0 = 1 + b, νi0 − νi = r − 1, νo0 = νo − b, µ0 = µ − k,
for some b, r ≥ 0 and k ≥ max{1, b + r} (consider |A|, |B|, |R|, |T | = a, b, r, t from
before). Note that we can solve for b and r here in terms of s and s0 . For any S and
(in)
S0 , so that NS,S0 6= 0,
(in)
NS,S0
ν − νi − b − r
=
.
k−b−r
(in)
We wish to find the number of S0 with s(S0 ) = s0 so that NS,S0 6= 0. Any such S0 can
be found by choosing exactly 1 vertex from Oi to be deleted in substep 1, choosing b
vertices to be deleted out of Oo to obtain Oo0 (i.e. choosing the set B), and choosing
r vertices out of V \ (Oi ∪ Oo ) that are added to obtain Oi0 (i.e. choosing the set R).
This yields
0
0
0
|{S : s(S ) = s
(in)
, NS,S0
νi
νo
ν − νi − νo
6= 0}| =
.
1
b
r
Hence, we find that
X
S0 :s(S0 )=s0
(in)
NS,S0 g(s0 )
1 g(s0 ) ν − νi − b − r νi
νo
ν − νi − νo
=
.
νi + νo g(s)
νi + νo g(s)
k−b−r
1
b
r
By symmetry,
X
S0 :s(S0 )=s0
(out)
NS,S0 g(s0 )
1 g(s0 ) ν − νi − a − t νi
νo
ν − νi − νo
=
.
νi + νo g(s)
νi + νo g(s)
k−a−t
1
a
t
Both of these sums depend on S(t) only through s(t). Hence conditioning on s(t) is
the same as conditioned on S(t).
124
We define the following transition probability components Pi and Po , where Pi
corresponds to the transition probability from s to s0 when deleting a vertex of indegree zero first in substep 1 and Po corresponds to the transition probability where
a zero out-degree vertices is deleted first. Specifically,
ν − νi − νo
νi g(s0 ) ν − νi − b − r νo
Pi (a, b, r, t, k|s) =
k−b−r
b
r
νi + νo g(s)
(5.2.2)
if a = 1, t = 0, k ≥ max{1, b + r}, otherwise zero, and
ν − νi − νo
νo g(s0 ) ν − νo − a − t νi
Po (a, b, r, t, k|s) =
νi + νo g(s)
k−a−t
a
t
(5.2.3)
if b = 1, r = 0, k ≥ max{1, a + t}, otherwise zero.
Hence P s(t+1) = s0 |s(t) = s = Pi (a, b, r, t, k|s)+Po (a, b, r, t, k|s). An important
remark is that this Markov chain is that it is time-homogeneous in the sense that
our transition probabilities depend only on the states s and s0 and not on t. For
ease of notation, we will denote Pi (a, b, r, t, k|s) and Po (a, b, r, t, k|s) as Pi (∆s|s) and
Po (∆s|s), where s0 = s + ∆ and a, b, r, t, k are defined by the following:
For Pi (∆s|s) to be non-zero, we must have non-negative integers b, r, k such that
∆T = (−1 − b, r − 1, −b, −k),
where k ≥ max{1, b + r}. For this case, we set Pi (∆s|s) = Pi (a = 1, b, r, t = 0, k|s).
Likewise, for Po (∆s|s) to be non-zero, we must have non-negative integers a, t, k such
that
∆T = (−1 − a, −a, t − 1, −k),
where k ≥ max{1, a + t}. We then set Po (∆s|s) = Po (a, b = 1, r = 0, t, k|s).
A couple of difficulties arise when we wish to find expectations on this probability space. The primary issue concerns the ratio g(s0 )/g(s), for which there does not
seem to be a closed-form expression. As a consequence, our next goal is to find an
125
asymptotic formula for this ratio g(s0 )/g(s), then establish that the resulting asymptotic formula for the transition probability is relatively close to the actual transition
probability, in the sense that expectations of certain desired random variables can be
approximated by sums over the asymptotic expressions for the transition probabilities.
5.3
Enumerating digraphs with constrained in/out-degrees
Now the only non-explicit term in our transition probabilities ((5.2.2) and (5.2.3))
for our Markov chain, {s(t)}, is g(s) and g(s0 ). As a reminder, g(s) is the number of
digraphs on [ν] with µ arcs where {1, 2, . . . , νi } are the vertices of zero in-degree and
{νi +1, νi +2, . . . , νi +νo } are the vertices of zero out-degree. In this section, we will find
estimates for g(s), in a near identical fashion to how Pittel [52] determines both crude
and sharp bounds on g(ν, 0, νo , µ) in his proof finding a sharp asymptotic formula for
the number of strongly connected moderately sparse digraphs. We will need the
following theorem counting the number of directed graphs with a specific in/outdegree sequence which is an important case of McKay’s (see [44], [45]) asymptotic
formula for the number of 0 − 1 matrices with given row and column sums.
Theorem 5.3.1. Let δ := (δ1 , δ2 , . . . , δν ) ≥ 0 and ∆ := (∆1 , . . . , ∆ν ) ≥ 0 be such
that
X
X
δi = µ,
i
∆i = µ,
i
where µ ≥ ν. Introduce g(δ, ∆), the number of digraphs on [ν] with µ arcs with indegree sequence δ and out-degree sequence ∆. If D := max δi + max ∆i = O(µ1/4 ),
i
then


g(δ, ∆) = µ! 
Y
i∈[ν]
1 
H(δ, ∆),
δi !∆i !
126
i
where the “fudge factor”, always 1 at most, is given by
!
X
1 X
1X
δi ∆i − 2
(δi )2
(∆j )2 + O D4 /µ .
H(δ, ∆) = exp −
µ i
2µ i
j
(5.3.1)
There is no general form for H(δ, ∆), but the above asymptotic formula works
for a wide range of degree sequences.
McKay proves this theorem using a random matching scheme. First, one starts
with two copies, [µ]1 and [µ]2 , of [µ] along with partitions [µ]1 = ∪ Ii , [µ]2 = ∪ Oi ,
i
i
where |Ii | = δi and |Oi | = ∆i . Each of the µ! matchings from [µ]1 to [µ]2 gives rise
to a directed multigraph (where multiple arcs between pairs of vertices and loops are
allowed) where we collapse each Ii and Oi to a single vertex, vi , and keep all present
arcs. See Figure 5.2, for an example of a matching and its corresponding directed
multigraph.
∆1
δ1
∆2
v1
v2
v3
v4
δ2
∆3
δ3
δ4
[6]1
∆4
[6]2
Figure 5.2: A matching and its corresponding directed multigraph
127
By construction, necessarily this directed multigraph has in-degree sequence δ and
out-degree sequence ∆. Each digraph, without loops or multiple arcs, corresponds to
Y
exactly
(δ!∆!) matchings. So H(δ, ∆) is precisely the probability that a matching
i
chosen uniformly at random among all possible matchings gives rise to a digraph
without loops and multiple arcs. As such, necessarily we have H(δ, ∆) ≤ 1, which
gives rise to the inequality:
g(δ, ∆) ≤ Q
µ!
.
i (δi !∆i !)
(5.3.2)
This last inequality gives rise to the following useful upper bound on g(ν, νi , νo , µ),
which holds for any parameters ν, νi , νo and µ.
Lemma 5.3.2. For each x, y > 0,
(ex − 1)ν−νi (ey − 1)ν−νo
.
g(ν, νi , νo , µ) ≤ µ!
xµ
yµ
(5.3.3)
Proof. First, we see that
X
g(ν, νi , νo , µ) =
g(δ, ∆),
δ,∆∈D
where D is all allowed in/out-degree sequences for our parameter (ν, νi , νo , µ). Formally, D is the set of in/out-degree sequences such that
ν
X
δj =
j=1
ν
X
∆j = µ
j=1
and
δj = 0, for j ∈ {1, . . . , νi },
δj > 0 for j > νi ,
∆j = 0 for j ∈ {νi + 1, . . . , νi + νo }, and
∆j > 0 for j ∈
/ {νi + 1, . . . , νi + νo }.
128
To simplify the proof, we will use the operator [xk ]. For k ∈ {0, 1, . . . , }, we define
X
the operator [xk ] on a (formal) power series,
ak xk , by
k
[xk ]
X
ak x k = ak .
k
Since µ = δ1 + . . . + δν = ∆1 + . . . , ∆ν , we see that
g(s) = µ![xµ y µ ]
X
H(δ, ∆)
Y xδ j y ∆ j
j
δ,∆∈D
δj !∆j !
≤ µ![xµ y µ ]
X Y x δj y ∆j
·
.
δj ! ∆j !
δ,∆∈D j
(5.3.4)
Now this operator allows us to switch the order of the sum and the product! So
νi X ∆j
νY
ν
i +νo X
Y
Y
X x δj y ∆j
y
xδ j
·
·
g(s) ≤ µ![x y ]
∆j ! j=ν +1 δ ≥1 δj ! j=ν +ν +1 δ ,∆ ≥1 δj !∆j !
j=1 ∆ ≥1
µ µ
i
j
j
i
o
j
j
= µ![xµ y µ ] (ey − 1)νi · (ex − 1)νo · ((ex − 1)(ey − 1))ν−νi −νo .
Since this bivariate power series on the RHS has only non-negative coefficients, we
obtain the following, Chernoff-type, bound for g(ν, νi , νo , µ) :
µ µ
ν−νi
x
[x y ] (e − 1)
y
ν−νo
(e − 1)
(ex − 1)ν−νi (ey − 1)ν−νo
≤
·
,
xµ
yµ
for any positive x and y.
Naturally, we wish to determine the values of x and y which minimize the RHS
in (5.3.3).
Claim 5.3.3. Suppose µ > ν − νi , ν − νo > 0. If
fi (x) :=
(ex − 1)ν−νi
,
xµ
fo (y) :=
(ey − 1)ν−νo
,
yµ
then fi (x)fo (y) is minimized at x = zi , y = zo , where zi , zo are the unique positive
roots of
zi ezi
µ
zo ezo
µ
,
=
.
=
z
z
e i −1
ν − νi e o − 1
ν − νo
129
Proof. Since fi and fo are positive functions, it suffices to minimize fi (x) and
fo (y) seperately, and by symmetry, we consider just fi (x). We find that
d
ex
µ
ν − νi
xex
µ
ln fi (x) = (ν − νi ) x
− =
−
.
dx
e −1 x
x
e x − 1 ν − νi
xex
µ
is strictly increasing on (0, ∞), where h(0+ ) = 1 <
x
e −1
ν − νi
and h(∞) = ∞. Hence, the minimum of f (x) is attained at the unique root of
µ
h(x) =
.
ν − νi
This last claim gives rise to the following two important variables. For a given
The function h(x) =
s = (ν, νi , νo , µ) that mets the conditions of previous claim, we define zi = zi (s) and
zo = zo (s) as the unique roots of
zi ezi
µ
=
,
z
i
e −1
ν − νi
zo ezo
µ
=
.
z
o
e −1
ν − νo
Even though these variables arise in obtaining asymptotic bounds for g(s), they
are also indispensable in determining the sharp asymptotics of g(s) as well as the
analysis of the greedy deletion algorithm. Such seemingly “hidden” parameters have
been found important for other problems of asymptotic graph enumeration, and their
use in various deletion algorithms (see Aronson, Frieze, and Pittel [1], and Pittel and
Wormald [56]).
There is a natural way to view the parameters zi and zo . Namely, note that the
number of vertices with positive in-degree is a digraph corresponding to parameter s
is precisely ν − νi . So the average in-degree of the ν − νi vertices with positive inµ
degree is
. If we chose a digraph uniformly at random among all digraphs with
ν − νo
parameter s, then the in-degree of a vertex, conditioned on having positive in-degree,
is asymptotic to a Poisson random variable with parameter zi , conditioned on this
Poisson variable being positive.
The following theorem gives an asymptotic formula for g(s) for a wide range of
parameters s.
130
Theorem 5.3.4. Let ν − νi − νo = Θ(ν), µ − ν → ∞, and µ = O(ν). Introduce
Zi , Zo , two independent truncated Poissons, with parameters zi and zo , i.e.
P (Zi = j) =
zij /j!
,
ezi − 1
P (Zo = j) =
zoj /j!
,
ezo − 1
(j ≥ 1).
Then
g(ν, νi , νo , µ) =
1+O
(ln ν)6
ν
(ezi − 1)ν−νi (ezo − 1)ν−νo
ziµ
zoµ
e−η
, (5.3.5)
× p
2π (ν − νi )V ar[Zi ](ν − νo )V ar[Zo ]
µ!
where
η=
µ(ν − νi − νo )
zi zo
+
.
(ν − νi )(ν − νo )
2
The statement of this theorem is a nearly trivial extension of Theorem 2.2 in
Pittel [52]. The proof of the theorem here barely deviates from Pittel’s proof, so only
an outline is presented. Note that the equality in (5.3.4) can be rewritten as
g(s) = µ!fi (zi )fo (zo )[xµ y µ ]
X
H(δ, ∆)
δ,∆∈D
Y (zi x)δj /δj ! Y (zo y)∆j /∆j !
·
,
e zi − 1
ezo − 1
j:∆ >0
j:δ >0
j
j
where
fi (x) =
(ex − 1)ν−νi
,
xµ
fo (y) =
(ey − 1)ν−νo
,
yµ
(zi )δj /δj !
is precisely the probability that Zi equals δj .
ezi − 1
In view of the product of j, it is natural to estimate this sum using a sequence
as before. We notice that
of independent Zi and Zo . Introduce Zνi i +1 , Zνi i +2 , . . . , Zνi as independent copies of
Zi and Z1o , . . . , Zνoi , Zνoi +νo +1 , . . . , Zνo as independent copies of Zo , and denote Zi =
(νi zeroes, Zνi i +1 , . . . , Zνi ), and Zo = (Z1o , . . . , Zνoi , νo zeroes, Zνoi +νo +1 , . . . , Zνo ). For the
coordinates that are zero, we define Zji and Zjo to be zero.
The sum above (along with the operator [xµ , y µ ]) collapses into the expectation
131
of H(Z i , Z o ) on the set that |Z i | :=
X
Zji = µ, |Z o | = µ. We find asymptotics for
j
this expectation by noticing that
h
i
i
o i
o
i
o
E H(Z , Z )1{|Z i |=|Z o |=µ} = E H(Z , Z )|Z | = |Z | = µ
× P (|Z i | = µ)P (|Z o | = µ).
Via a local limit theorem, one can show that
1 + O(1/ν)
P (|Z i | = µ) = p
,
2π(ν − νi )V ar[Zi ]
1 + O(1/ν)
P (|Z o | = µ) = p
.
2π(ν − νo )V ar[Zo ]
Just as in Pittel [52], one can show this conditional expectation of H(Z i , Z o ) is within
a multliplicative factor of (1 + O((ln ν)6 /ν)) from
"
"
#!
#
X
X
1
1
1
exp − E
Zji Zjo −
E
Zji 2
µ
2
µ
j
j
"
#!!
X
1
E
Zjo 2
.
µ
j
Since Zji and Zjo are independent, we have E[Zji Zjo ] = E[Zji ]E[Zjo ]. For the first νi +νo
coordinates, either E[Zji ] = 0 or E[Zjo ] = 0. For the last ν − νi − νo entries,
E[Zji ] = E[Zi ] =
µ
zi
=
−z
i
1−e
ν − νi
and
E[Zjo ] = E[Zo ] =
µ
.
ν − νo
Hence
"
#
X
1
µ(ν − νi − νo )
E
Zji Zjo =
.
µ
(ν − νi )(ν − νo )
j
Furthermore, if Zji is non-zero, then
E[(Zji )2 ] = E[(Zi )2 ] = zi2 /(1 − e−zi ) = zi µ/(ν − νi ).
Since exactly ν − νi of the Zji are non-zero, we have that
"
#
X
1
ν − νi
zi µ
E
(Zji )2 =
·
= zi .
µ
µ
ν
−
ν
i
j
132
By symmetry, we have
"
#
X
1
E
(Zjo )2 = zo
µ
j
which completes the proof of the lemma.
5.4
Asymptotic Transition Probabilities
Now that we have found an asymptotic formula for the non-explicit terms g(s) and
g(s0 ), we can find asymptotic formulas for our transition probabilities defined in (5.2.2)
and (5.2.3). We begin with finding the asymptotics of the ratio g(s0 )/g(s), which is
present in the transition probabilities, for values of s that satisify the conditions of
the Theorem 5.3.4, along with the stronger condition that µ − ν = Θ(ν) rather than
simply µ − ν → ∞. Note that µ − ν = Θ(ν) implies that µ = O(ν) (which is one of
the previous conditions).
Lemma 5.4.1. Uniformly over s, with ν − νi − νo , µ − ν = Θ(ν),
(i) we have that
zik
zok
ν
g(s0 )
≤b
.
g(s)
(µ)k (ezi − 1)b+r (ezo − 1)a+t
(ii) If in addition, s0 is such that a = 1, t = 0 and max{1, b + r} ≤ k ≤ ln ν, then
k
g(s0 )
(ln ν)6
zi zo
1
1
.
(5.4.1)
= 1+O
z
z
g(s)
ν
µ
e o − 1 (e i − 1)b+r
Likewise, if s0 is such that b = 1, r = 0 and max{1, a + t} ≤ k ≤ ln ν, then
k
g(s0 )
(ln ν)6
zi zo
1
1
.
= 1+O
g(s)
ν
µ
ezi − 1 (ezo − 1)b+r
(5.4.2)
Proof. (i) First, by Theorem 5.3.4, we have that
g(s) =
1+O
(ln ν)6
ν
µ!fi (s, zi )fo (s, zo )
×
e−η
p
, (5.4.3)
2π (ν − νi )V ar[Zi ](ν − νo )V ar[Zo ]
133
where
fi (s, zi ) =
(ezi − 1)ν−νi
,
ziµ
fo (s, zo ) =
(ezo − 1)ν−νo
.
zoµ
Note that for s0 to follow from s, we must have that ν 0 = ν − a − b, νi0 = νi − a + r,
νo0 = νo − b + t and µ0 = µ − k where either a = 1, t = 0 or b = 1, r = 0 and
k ≥ max{b + r, a + t}. From Lemma 5.3.2, we have that
g(s0 ) ≤ (µ − k)!
(ezi − 1)ν−νi −b−r (ezo − 1)ν−νo −a−t
ziµ−k
zoµ−k
,
where zi = zi (s) and zo = zo (s) are defined for s rather than s0 . It is easy to show
that uniformly over the s that meet the conditions of the lemma, the variables zi , zo
are bounded away from 0 and ∞. In particular, uniformly over all such s,
zoµ
ν
ziµ
1
≤b
.
g(s)
µ! (ezi − 1)ν−νi (ezo − 1)ν−νo
Hence
g(s0 )
ν
zik
zok
≤b
.
g(s)
(µ)k (ezi − 1)b+r (ezo − 1)a+t
(ii) By symmetry, we need to only show the first asymptotic expression (5.4.1).
Now suppose that s and s0 are such that a = 1, t = 0 and k ≤ ln ν. In particular, s0
also meets the conditions of Theorem 5.3.4. To obtain our sharp asymptotic formula
for this case, we will need to analyze each term and show that the difference between
evaluating at zi and at zi0 = zi (s0 ) is minimal. We begin with noticing that
ν 0 = ν + O(ln ν), νi0 = νi + O(ln ν), νo0 = νo + O(ln ν), µ0 = µ + O(ln ν).
Now we have that
0
µ0
µ−k
µ
zi ezi
zi0 ezi
=
=
=
+
O((ln
ν)/ν)
=
+ O((ln ν)/ν).
0
ν 0 − νi0
ν − νi − b − r
ν − νi
ezi − 1
ezi − 1
zez
(with the definition
Furthermore, since the derivative of the function, h(z) = z
e −1
h(0) = 1), is bounded between 1/2 and 1 for z ≥ 0, the previous equation implies
that zi0 = zi + O((ln ν)/ν). As a consequence,
V ar[Zi0 ] = V ar[Zi ](1 + O((ln ν)/ν)), V ar[Zo0 ] = V ar[Zo ](1 + O((ln ν)/ν)),
134
and
ν 0 = ν + O((ln ν)/ν).
Now we need to consider the fraction fi (s0 , zi0 )/fi (s, zi ) and fo (s0 , zo0 )/fo (s, zo ). We
begin by noticing that
0
0
0
0
fi (s0 , zi0 )
(ezi − 1)ν −νi /(zi0 )µ
zik
zik
fi (s0 , zi0 )
= z
=
0
0
fi (s, zi )
fi (s0 , zi ) (ezi − 1)b+r
(e i − 1)ν −νi /(zi )µ0 (ezi − 1)b+r
and
0
0
0
0
(ezo − 1)ν −νo /(zo0 )µ zok
fo (s0 , zo0 ) zok
fo (s0 , zo0 )
= zo
=
.
fo (s, zo )
(e − 1)ν 0 −νo0 /(zo )µ0 ezo − 1
fo (s0 , zo ) ezo − 1
To complete the proof of the lemma, it suffices to prove the following claim:
Claim 5.4.2. Uniformly over s and s0 in question, we have that
fo (s0 , zo0 )
fi (s0 , zi0 )
3
=
1
+
O
(ln
ν)
/ν
,
= 1 + O (ln ν)3 /ν .
0
0
fi (s , zi )
fo (s , zo )
Proof. First, we expand the exponent of fi (s0 , z) about z = zi . Now
0
fi (s0 , zi0 ) = exp (ν 0 − νi0 ) ln ezi − 1 − µ0 ln zi0
= exp ((ν 0 − νi0 ) ln (ezi − 1) − µ0 ln zi )
ezi
0
0
0
0
0
2
× exp
(ν − νi ) zi
− µ /zi (zi − zi ) + O ν · (zi − zi )
.
e −1
Since zi0 − zi = O((ln ν)/ν), this last error term is O((ln ν)2 /ν). Then we have that
(ν 0 − νi0 )
µ0
ezi
µ
e zi
−
=
(ν
−
ν
)
−
+ O(ln ν)
i
ezi − 1 zi
ezi − 1 zi
ν − νi
zi ezi
µ
=
−
+ O(ln ν),
zi
ezi − 1 ν − νi
but by the definition of zi , the expression inside the parentheses is zero. Hence
fi (s0 , zi0 ) = fi (s0 , zi ) exp O((ln ν)2 /ν) = fi (s0 , zi ) 1 + O (ln ν)2 /ν ,
as desired. The corresponding ratio for fo is done in a similar fashion.
135
Other than the ratio g(s0 )/g(s), the terms in the transition probabilities (see
(5.2.2) and (5.2.3)) are simply binomial expressions. If we look at Pi , we see that we
have the binomial expressions:
ν − νi − b − r
ν − νi − νo
νo
,
, and
.
k−b−r
r
b
For s and s0 that meet the conditions of part (ii) of the previous lemma (most notably,
we have ν − νi − νo = Θ(ν) and k − b − r, r = O(ln ν)), we have that
k−b−r−1
ν − νi − b − r
(ν − νi )k−b−r Y ν − νi − b − r − i
=
(k − b − r)! i=0
ν − νi
k−b−r
=
(ν − νi )k−b−r
1 + O (ln ν)2 /ν ,
(k − b − r)!
(5.4.4)
and similarly,
ν − νi − νo
(ν − νi − νo )r
1 + O (ln ν)2 /ν .
(5.4.5)
=
r!
r
νo
We can not approximate
in the same fashion because we know that close to
b
the end of the deletion process, νo will be on order much smaller than ν (in fact, we
stop when both νi and νo are zero). Motivated by these expressions and the sharp
asymptotics of the ratio g(s0 )/g(s), we define the following auxilliary transition “probabilities,” expecting their sharp approximation for the actual transition probability.
Let
b r
1
1 ν − νi − νo zi zo
νo
1 zi zo
νi
qi (∆s|s) =
νi + νo e z o − 1 b
ezi − 1 µ
r!
ezi − 1
µ
k−b−r
1
zi zo
·
(ν − νi )
,
(k − b − r)!
µ
(5.4.6)
if ∆s is such that a = 1, t = 0 and k ≥ max{1, b + r}, and qi = 0 for any other ∆s.
Also let
a t
νi
1 zi zo
1 ν − νi − νo zi zo
νo
1
qo (∆s|s) =
νi + νo e z i − 1 a
ezo − 1 µ
t!
e zo − 1
µ
k−a−t
1
zi zo
·
(ν − νo )
,
(k − a − t)!
µ
136
(5.4.7)
if ∆s is such that b = 1, r = 0 and k ≥ max{1, a + t}, and qo = 0 for any other ∆s.
The following lemma tells us that this approximation is sharp for the vast majority
of “probable” cases for “good” s.
Lemma 5.4.3. (i) Uniformly over s with ν − νi − νo , µ − ν = Θ(ν) and ∆s such that
k ≤ ln ν,
qi (∆s|s) = Pi (∆s|s) 1 + O (ln ν)6 /ν
qo (∆s|s) = Po (∆s|s) 1 + O (ln ν)6 /ν
,
.
(ii) Uniformly over s with ν − νi − νo , µ − ν = Θ(ν),
2
Pi (∆s|s) ≤ exp − (ln ν)(ln ln ν)
qi (∆s|s),
3
∆s:k≥ln ν
∆s:k≥ln ν
X
X
and
2
Po (∆s|s) ≤ exp − (ln ν)(ln ln ν) .
qo (∆s|s),
3
∆s:k≥ln ν
∆s:k≥ln ν
X
X
Proof. By symmetry, we need only consider qi and Pi for both parts of the lemma.
(i) The proof is immediate after part (ii) of Lemma 5.4.1 and the binomial term
approximations (5.4.4) and (5.4.5).
(ii) Now we wish to show the sums of Pi and qi over ∆s such that k ≥ ln ν are
negligible. First we have that
g(s0 ) ν − νi − b − r νo
ν − νi − νo
Pi (∆s|s) ≤
g(s)
k−b−r
b
r
0
g(s0 ) ν − νi
g(s ) ν − νi − b − r ν − νi
≤
≤
.
g(s)
k−b−r
b+r
g(s)
k
Now applying the bound from part (i) of Lemma 5.4.1, we have that
Pi (∆s|s) ≤
ν
(zi zo )k
(ν − νi )k
ν
(zi zo )k
≤
.
(µ)k (ezi − 1)b+r (ezo − 1)
k!
k! (ezi − 1)b+r (ezo − 1)
137
Trivially, since x < ex − 1, we have (ex − 1)−1 < x−1 . Also, as mentioned before,
there is some A > 1 such that uniformly over s, both zi and zo are bounded above
by A. Hence
Pi (∆s|s) ≤
ν
ν
(zi )k−b−r (zo )k−1 ≤
· A2k .
k!
k!
Note this last bound does not depend on the value of b, r (other than that b + r ≤ k,
of course). Also for sufficiently large ν, this bound is decreasing in k. In particular,
XX
X
Pi (∆s|s) ≤ ν 3
b=0 r=0 k=max{b+r,ln ν}
ν
· A2 ln ν .
(ln ν)!
Using Stirling’s formula on (ln ν)! yields
X
Pi (∆s|s) ≤ ν
4 (eA
∆s:k≥ln ν
2 ln ν
)
= exp (−(ln ν)(ln ln ν) + O(ln ν)) .
(ln ν)ln ν
Now we need bound the analogous sum of qi . Here
b r
k
1
zi zo
1
νo
ν − νi − νo
k−b−r
qi (∆s|s) ≤
((ν − νi ))
zo b!r!(k − b − r)! ezi − 1
e zi − 1
µ
k
1
1 1
zob+r (zi zo )k−b−r ≤ 3k · A2k .
≤
zo k! b, r, k − b − r
k!
We bound the sum over the RHS in the above in the same fashion as before.
Focusing on random variables that are either linear or quadratic polynomials of
∆s, we want their expected values to be sharply approximated with their corresponding sums over qi + qo . For notation sake, for any function f (a, b, r, t, k), we denote the
expectation of f over P by Es [f ]. In particular,
Es [f ] =
X
f (a, b, r, t, k) (Pi (∆s|s) + Po (∆s|s)) .
∆s
Also we define the q−expectation of f as
Esq [f ] :=
X
f (a, b, r, t, k) (qi (∆s|s) + qo (∆s|s)) .
∆s
This latter expression is not an expectation in the traditional sense because we will
notice that qi + qo is substochastic, in the sense that Esq [1] < 1.
138
Lemma 5.4.4. Suppose f (a, b, r, t, k) is a function of a, b, r, t, k such that
|f (a, b, r, t, k)| ≤ (ln ν)2 , for k ≤ ln ν,
|f (a, b, r, t, k)| ≤ ν 2 , for k ≥ ln ν.
Then uniformly over s with ν − νi − νo , µ − ν = Θ(ν), we have that
Es [f ] = Esq [f ] + O((ln ν)8 /ν).
Proof. Trivially in light of the previous lemma, the sums of f · (Pi + Po ) and
f · (qi + qo ) over ∆s with k ≥ ln ν are on order at most
(ln ν)8
2
.
ν exp − ln ν(ln ln ν) 3
ν
3
Now, for k ≤ ln ν, it suffices to show by symmetry that
X
f (a, b, r, t, k)Pi (∆s|s) =
∆s:k≤ln ν
X
f (a, b, r, t, k)qi (∆s|s) + O
∆s:k≤ln ν
(ln ν)8
ν
.
By the triangle inequality, we have that
X
f (a, b, r, t, k) (qi (∆s|s) − Pi (∆s|s)) ≤
X
(ln ν)2 qi (∆s|s) − Pi (∆s|s).
∆s:k≤ln ν
∆s:k≤ln ν
From the previous lemma, uniformly for such s and ∆s with k ≤ ln ν, we have
(ln ν)6
q i − Pi = Pi · O
. Hence
ν
X (ln ν)8
(ln ν)8
f (a, b, r, t, k) (qi (∆s|s) − Pi (∆s|s)) ≤b
· Pi (∆s|s) ≤
.
ν
ν
∆s
∆s:k≤ln ν
X
Let’s summarize what we have done so far. First, we defined our greedy deletion
algorithm which terminates at the (1, 1)−core. Then we noticed that instead of
analyzing the sequence of digraphs, we can instead consider the simpler process {s(t)},
which is the 4−vector composed of the numbers of all vertices, vertices of in-degree
139
zero, vertices of out-degree zero, and of arcs present at the beginning of the t−th
step. Then, we found the transition probability of this process, which contains the
non-explicit term g(s0 )/g(s). Although this ratio does not seem to have a closedform expression, for a wide range of parameter values, we gave both crude and sharp
bounds on this ratio. Finally, we established that for certain desired random variables,
we can convert expectations over the transition probabilities to their corresponding
sums over our auxiliary approximate “transition probabilities”.
5.5
The Characteristic Function Recursion
Let’s recall that our deletion process is supposed to run until all semi-isolated vertices
are deleted and the (1, 1)−core is obtained. After this point, each following step
simply returns the same digraph (i.e. D(t + 1) = D(t), if D(t) has no semi-isolated
vertices). Now we need to introduce the following stopping time, τ̄ , which is the first
moment that there are no more semi-isolated vertices. Formally,
τ̄ := min{t : νi (t) = νo (t) = 0}.
As just noted, we have s(t) = s(τ̄ ) for all t ≥ τ̄ . We define the random variables,
ν̄ and µ̄, as the terminal number of vertices and arcs (i.e. ν̄ = ν(τ̄ ), µ̄ = µ(τ̄ )).
Let ϕs (u = (u1 , u2 )T ) denote the joint characteristic function of (ν̄, µ̄) starting with
s(0) = s. Formally,
ϕs (u) := Es [eiu1 ν̄+iu2 µ̄ ],
where as before, Es [·] denotes the expectation according to probability distribution
of the process, {s(t)}, starting at s.
Since this Markov process is time homogeneous, we have the following recursion,
ϕs (u) =
X
ϕs0 (u)P (s(t + 1) = s0 |s(t) = s) ,
s0
140
(5.5.1)
for any time t ≥ 0. Note that P = Pi + Po , where Pi and Po are defined in (5.2.2)
and (5.2.3).
We anticipate that for our starting s(0), the random variables ν̄ and µ̄ are asymptotic jointly Gaussian with means and covariances linear in n. This leads us to believe
that there are smooth functions fj and ψj,k , for j, k = 1, 2 (ψ1,2 = ψ2,1 ), of the scaled
initial state (s/n) such that
2
X
n
ϕs (u) ≈ exp in
uj fj (s/n) −
2
j=1
2
X
!!
uj uk ψj,k (s/n)
=: Ḡn (s/n, u),
j,k=1
for all u1 , u2 = O(n−1/2 ). In vector notation, we can write Ḡn (s/n, u) as
n T
exp inu f (s/n) − u ψ(s/n)u ,
2
T
where f T = (f1 , f2 ) and ψ is the 2 × 2 matrix with entries ψj,k , j, k ∈ {1, 2}. Necessarily, such fj and ψj,k must satisfy the boundary conditions
f1 (α, 0, 0, γ) = α,
f2 (α, 0, 0, γ) = γ,
ψj,k (α, 0, 0, γ) = 0.
(5.5.2)
One way to find such functions f and ψ is to determine functions so that Ḡ would
nearly satisfy the recursion relation (5.5.1). This method of finding such functions,
f and ψ, is detailed in Pittel [51] and gainfully used in Pittel [49], and Pittel and
Weishaar [55].
However our sharp asymptotics for g(s) require that s has ν −νi −νo , µ−ν = Θ(ν).
So we can afford to run this process until the possibly earlier moment when s no
longer meets these conditions. Our hope is that at this earlier moment, out state s(t)
is sufficiently close to its terminal state of the process allowed to run until completion.
This leads us to define the set of such good s. Fix δ, δ 0 ∈ (0, 1). Then, let S be
the set of s = (ν, νi , νo , µ) such that ν, νi , νo , µ ≥ 0, and
δ 0 ν ≤ µ − ν ≤ ν/δ 0 .
ν − νi − νo ≥ δν,
141
From now on, it is silently assumed that ν ≤ n and µ = O(ν).
We define the stopping time τ ∗ as the first moment that the process leaves S.
Formally, τ ∗ := min{t : s(t) ∈
/ S}. We will consider the deletion process up to time τ ∗
and will use P ∗ for its transition probabilities and Es∗ [◦] for expectations according
to P ∗ . Specifically, we have that
P ∗ (s(t + 1) = s0 |s(t) = s) =



P (s(t + 1) = s0 |s(t) = s), if s ∈ S


1s (s0 ),
if s ∈
/ S,
where 1s (s0 ) = 1 if s0 = s and 0 if s0 6= s.
Since this stopping time only considers the current location of s(t), the process
{s(t)}0≤t≤τ ∗ is also a time-homogeneous Markov process. We define ν ∗ , µ∗ as the
terminal parameters according to this process (i.e. ν(τ ∗ ), µ(τ ∗ ) respectively). Again,
we notice that the characteristic function for the pair (ν ∗ , µ∗ ), denoted ϕ∗s (u), satisfies
the recurrence relation
ϕ∗s (u) =
X
ϕ∗s0 (u)P ∗ (s(1) = s0 |s(0) = s) .
(5.5.3)
s0
We conjecture the existence of smooth functions fj (α, βi , βo , γ) and ψj,k (α, βi , βo , γ),
for 1 ≤ j, k ≤ 2, such that for u1 , u2 = O(n−1/2 ),
ψs∗ (u)
n T
≈ Gn (s/n, u) := exp inu f − u ψ(s/n)u ,
2
T
(5.5.4)
where uT = (u1 , u2 ), f T = (f1 , f2 ) and ψ is the 2 × 2 matrix with entries {ψj,k }.
We will prove this conjecture by producing fj and ψj,k such that G(s/n, u) almost
satisfies (5.5.3), i.e.
G(s/n, u) −
X
G(s0 /n, u)P ∗ (s(1) = s0 |s(0) = s) = o(1/n),
(5.5.5)
s0
uniformly for all “good” s (s ∈ S) and ||u|| = O(n−1/2 ). We need this error to be
o(1/n) since we will use an union bound over all steps whose total number is of order
n.
142
As discussed before, s0 can follow from s only if s0 = s + ∆, where
∆ := (−a − b, r − a, t − b, −k)
is such that k ≥ max{a + t, b + r} and either a = 1, t = 0 or b = 1, r = 0. Denote the
gradient of fj by ∇fj ; as long as k ≤ ln n, we have that
nuj fj (s0 /n) = nuj fj (s/n) + nuj (∆/n)T ∇fj + O(n|uj |(ln n)n−2 )
= nuj fj (s/n) + uj ∆T ∇fj + O((ln n)n−3/2 ).
Likewise
nuj uk ψj,k (s0 /n) = nuj uk ψj,k (s/n) + uj uk ∆T ∇ψj,k + O((ln n)n−2 ).
Therefore, for k ≤ ln n,
X
1X
G(s0 /n, u)
= exp i
uj ∆T ∇fj −
uj uk ∆T ∇ψj,k + O
G(s/n, u)
2
j
j,k
=1+i
X
j
ln n
n3/2
!
1X
uj ∆ ∇fj −
uj uk ∆T ∇fj ∆T ∇fk + ∆T ∇ψj,k + O
2 j,k
T
ln n
n3/2
.
Since ||u|| = O(n−1/2 ), for G to closely fit our recurrence relation (in the sense of
(5.5.5)), we would need that
!
X
∆T P ∗ (a, b, r, t, k|s) Dfj = Es∗ [∆T ]∇fj = o(n−1/2 )
(5.5.6)
a,b,r,t,k
and
Es∗ [∆T ]∇ψj,k + Es∗ [ ∆T ∇fj
∆T ∇fk ] = o(1).
(5.5.7)
Along with the boundary conditions (5.5.2), these are the partial differential equations
for f and ψ.
Now to establish the existence of such functions fj and ψj,k , we will need to find
the expectations of linear and quadratic polynomials of the variables a, b, r, t and k.
143
Here is where our ability to approximate these linear and quadratic polynomials will
be crucial. In the next section, we will find these expectation (up to a negligible
additive error) and find a homogeneous partial differential equation that we wish the
functions f1 and f2 satisfy.
5.6
Approximating the PDE for fj
For our partial differential equations for fj , we need to compute Es∗ [∆T ], defined by
Es∗ [∆T ] = (Es∗ [−a − b], Es∗ [r − a], Es∗ [t − b], Es∗ [−k]).
So we wish to find the expectation of the random variables −a − b, r − a, t − b and
−k over the probability space P ∗ (◦|s), where s ∈ S. Trivially, if s ∈
/ S, then with
probability (according to P ∗ ) one, each of the parameters a, b, r, t, and k are zero.
However if s ∈ S, then due to Lemma 5.4.4, we can approximate these expectations
over P ∗ by their q−expected analogues. By symmetry of “in” and “out”, it suffices
to evaluate the averages of a, b, r, t, k with respect to qi . To this end, we first obtain a
closed form expression for the trivariate generating function of the parameters b, r, k
with respect to qi . Let
F (x, y, w) : =
XX
X
xb y r wk qi (∆s|s).
b=0 r=0 k=max{1,b+r}
Using the definition of qi (5.4.6), we find that
(
X X X νo 1 zi zo b
1
νi
·
xw
F (x, y, w) =
νi + νo e z o − 1
b
ezi − 1 µ
b=0 r=0 k−b−r=0
)
r
k−b−r !
1 ν − νi − νo zi zo
1
zi zo
·
yw
(ν − νi )
w
−1 ,
r!
ezi − 1
µ
(k − b − r)!
µ
where we add and subtract the expression
νi
1
,
z
νi + νo e o − 1
144
so that our triple sum is over all k ≥ b + r, rather than k ≥ max{1, b + r}. Therefore,
we find that
1
νi
F (x, y, w) =
z
o
νi + νo e − 1
(
xw zi zo
1 + zi
e −1 µ
νo
+ (ν − νi )
exp
zi zo
w
µ
ν − νi − νo zi zo
yw
ezi − 1
µ
!
)
−1 .
(5.6.1)
With this function, F (x, y, w), we can compute the qi -expected value by evaluating
F and its partial derivatives at x = y = w = 1. To approximate the resulting
expressions, we will need to impose the condition that ν = Θ(n), in addition to
needing that s ∈ S. We will use the following asymptotic expression to simplify our
asymptotic expressions: uniformly over s ∈ S with ν = Θ(n), we have that
ν
zi2 zo2
νo zi zo νo
1
1 zi zo o
−2
−
+O n
= exp zi
1 + zi
e −1 µ
e −1 µ
2 (ezi − 1)2 µ2
νo zi zo
= exp zi
+ O n−1 ,
e −1 µ
and
1 zi zo
1 + zi
e −1 µ
νo −1
= exp
νo zi zo
z
e i −1 µ
+ O n−1 .
For ease of notation, each of the following sums are over ∆s where a = 1, t = 0,
b ≥ 0, r ≥ 0 and k ≥ max{1, b + r}. Necessarily qi = 0 unless t = 0, so trivially,
X
tqi (∆s|s) = 0.
Also qi = 0 unless a = 1, so uniformly over s ∈ S with ν = Θ(n),
X
aqi (∆s|s) =
=
X
qi (∆s|s) = F (1, 1, 1)
νi
1
νi + νo ezo − 1
exp
νo zi zo ν − νi − νo zi zo
+
ezi − 1 µ
ezi − 1
µ
!
zi zo + (ν − νi )
− 1 + O n−1 .
µ
145
However, the exponent is
νo zi zo ν − νi − νo zi zo
zi zo
zi zo
1
+
+ (ν − νi )
= (ν − νi )
+1
e zi − 1 µ
e zi − 1
µ
µ
µ
e zi − 1
ν − νi zi ezi
= zo , (!)
= zo
µ ezi − 1
since
zi ezi
µ
. Hence
=
ezi − 1
ν − νi
X
aqi (∆s|s) =
νi
+ O(n−1 ).
νi + νo
We determine the last three sums by taking partial derivatives of F .
X
X
∂ X b r k
∂F
(1, 1, 1)
x y w qi (∆s|s)
=
∂x
∂x
x=y=w=1
ν −1
νi
1
1 zi zo o
1 zi zo
=
νo 1 + z i
z
z
o
i
νi + νo e − 1
e −1 µ
e −1 µ
!
ν − νi − νo zi zo
zi zo
· exp
+ (ν − νi )
e zi − 1
µ
µ
νi
ezo
νo zi zo
−1
+
O
n
=
;
νi + νo ezo − 1 ezi − 1 µ
bqi (∆s|s) =
∂ X b r k
∂F
(1, 1, 1)
rqi (∆s|s) =
x y w qi (∆s|s)
=
∂y
∂y
x=y=w=1
ν
νi
1
1 zi zo o
=
1 + zi
νi + νo ezo − 1
e −1 µ
!
zi zo
ν − νi − νo zi zo
ν − νi − νo zi zo
· exp
+ (ν − νi )
z
i
e −1
µ
µ
ezi − 1
µ
νi
ezo ν − νi − νo zi zo
−1
=
+
O
n
;
νi + νo ezo − 1 ezi − 1
µ
X
∂ X b r k
∂F
kqi (∆s|s) =
x y w qi (∆s|s)
=
(1, 1, 1)
∂w
∂w
x=y=w=1
νi
zo ezo
−1
=
+
O
n
.
νi + νo e z o − 1
146
The sum of aqo (∆s|s) (resp. bqo ) is found by switching i and o in the formula for the
sum of bqi (∆s|s) (resp. aqi ). Similarly, we determine the sums over qo of the other
variables. So we define the following asymptotic expectations.
Es [a] : =
=
Es [b] : =
=
Es [r] : =
Es [t] : =
Es [k] : =
=
νo
ezi
νi
νi zi zo
+
,
νi + νo νi + νo e z i − 1 e z o − 1 µ
νi
νi νo
µ
e−zo
+
,
νi + νo νi + νo ν − νi ν − νo
νi
e zo
νo zi zo
νo
+
,
z
z
νi + νo e o − 1 e i − 1 µ
νi + νo
νi νo
µ
e−zi
νo
+
,
νi + νo νi + νo ν − νi ν − νo
νi
µ ν − νi − νo −zi
e ,
νi + νo ν − νo ν − νi
µ ν − νi − νo −zo
νo
e ,
νi + νo ν − νo ν − νi
νi
e zo
νo
e zi
zi ,
z
+
o
νi + νo ezo − 1
νi + νo ezi − 1
µ
νo
νi
+
.
νi + νo ν − νi ν − νo
We have just shown that each of q−expected values of these parameters a, b, r, t, k
differ from their asymptotic expectations by an additive error of order at most n−1 ,
as long as s ∈ S with ν = Θ(n). Furthermore, these linear functions a, b, r, t, k meet
the conditions of Lemma 5.4.4 so that their expectations over P differ from their
q−expected values by an additive error of order at most (ln n)8 /n. Effectively, we
have shown that uniformly over s ∈ S with ν = Θ(n),
Es [∆T ] = Es [∆T ] + O((ln n)8 /n).
(5.6.2)
As a consequence, to establish the existence of such fj to satisfy the original PDE
(5.5.6), it suffices to show the existence of such fj that satisfy the homogeneous partial
differential equation:
Es [−(a + b)](fj )α + Es [r − a](fj )βi + Es [t − b](fj )βo + Es [−k](fj )γ = 0,
147
(5.6.3)
where fj has the boundary conditions, f1 (α, 0, 0, γ) = α and f2 (α, 0, 0, γ) = γ.
Note that each of the defined E[◦] are scale-free, in the sense that E[a](ν, νi , νo , µ) =
E[a](ν/n, νi /n, νo /n, µ/n).
5.7
Solving the PDE for fj
We will solve the homogeneous partial differential equation (5.6.3), by using the
method of characteristics. In the first part of this section, we will analyze the characteristics of the PDE and afterwards, in the second part, we will determine explicit
expressions for f1 and f2 .
Even though we solve the PDE for generic parameters α, βi , βo , and γ, our likely
starting parameters coming from D(n, m = cn) will be near
1 − e−2c , e−c (1 − e−c ), e−c (1 − e−c ), c .
Also, if everything works out, we expect to have f1 = θ2 at this point, where θ = θ(c)
is the unique positive root of
1 − θ = e−cθ ,
by the previously known results by Karp [31] and Luczak [39] that the size of the
strong giant component of D(n, m = cn) is concentrated around θ2 n. However, we
are not aware of any results concerning the number of arcs in the strong component,
so we do not have any strong intuition about f2 at this point.
Part 1. The Characteristics We wish to solve the system of ordinary differential equations given by
dβi
dβo
dγ
dfj
dα
= Es [−a − b],
= Es [r − a],
= Es [t − b],
= Es [−k],
= 0.
dt
dt
dt
dt
dt
148
After simpifying the asymptotic expectations in this equation, we find that
dα
dt
dβi
dt
dβo
dt
dγ
dt
βi βo
γ
(e−zi + e−zo ),
βi + βo (α − βi )(α − βo )
βi
γ α − βi − βo −zi
γ
e−zo
=
e − 1 − βo
,
βi + βo α − βo α − βi
α − βi α − βo
βo
γ α − βi − βo −zo
γ
e−zi
=
e − 1 − βi
,
βi + βo α − βi α − βo
α − βo α − βi
γ
βo
βi
=−
+
,
βi + βo α − βi α − βo
= −1 −
(5.7.1)
where zi and zo are defined by
zi ezi
γ
=
,
z
e i −1
α − βi
γ
zo ezo
=
.
z
e o −1
α − βo
Although this system of ordinary differential equations looks formidable, we can show
two relatively simple functions of these variables, along with zi and zo , must be
constant along a trajectory. As a consequence, we will be able to determine the
functions f1 and f2 , which as mentioned before meet the boundary conditions
f1 (α, 0, 0, γ) = α,
f2 (α, 0, 0, γ) = γ.
Note that we can write this system of ordinary differential equations as
dw
= Es [∆] =: H(w),
dt
where w = (α, βi , βo , γ).
Just as we defined the set of “good” s as S, we define the corresponding set of
“good” w:
n
W := w = (α, βi , βo , γ) ≥ 0 : α − βi − βo ≥ δ and
o
γ
γ
,
≥ 1 + δ0 ,
α − βi α − βo
for some positive constants δ and δ 0 , determined later. First, we show that trajectories
starting in the interior of W exists and are unique up to reaching the boundary of
W.
149
Lemma 5.7.1. Let w(0) be in the interior of W. Then there is a unique w(t) such
dw
that
= H(w) and w(t) exists up to boundary of W.
dt
Proof. We notice the interior of W, denoted int(W), is the union (over k ∈ N) of
n
Wk := w : α, βi , βo , γ ∈ [1/k, k], α − βi − βo ≥ δ and
o
γ
γ
,
≥ 1 + δ0 .
α − βi α − βo
It suffices to show that trajectories starting in Wk exist and are unique up to leaving
Wk . One can easily see that each Wk is compact and that the partial derivatives
of the components of H(w) are bounded on Wk . By Existence and Uniqueness Theorem of Systems of differential equations, see for instance Coddington[11], at each
point w(0) ∈ Wk , there exists a h0 > 0 such that there is an unique trajectory,
w(t), starting at w(0), that exists on some interval t ∈ (−h0 , h0 ). Using a standard
compactness argument, one can show that trajectories are unique up to and leaving
Wk .
For w(0) in the interior of W, let tW = tW (w(0)) be the first moment we reach
the boundary; formally
tW = sup{t : w(t0 ) ∈ int (W) for t0 ≤ t}.
We have that tW is well-defined, because trajectories in the interior of W exist and are
unique. However if the trajectory stays in the interior indefinitely, then tW = ∞. The
next two propositions tell us that along the trajectory (before tW ), certain functions
of w are constant.
Proposition 5.7.2. Let w(0) ∈ int (W). Then, for t < tW , we have that
zi (t)zo (t)
zi (0)zo (0)
=
.
γ(t)
γ(0)
Proof. We will show this by computing
150
dzi
. The straightforward expression for
dt
this derivative is quite long, but it simplifies immensely. Since the parameter zi is
zi ezi
γ
defined by zi
, we have that
=
e −1
α − βi
dzi
1
1 dγ
γ
dα dβi
= zez 0
−
−
.
dt
(α − βi )2 dt
dt
( ez −1 ) |zi α − βi dt
After plugging in our expressions for these derivatives from (5.7.1), we obtain
(ezi − 1)2
γ
βo
βi
dzi
−1
= zi zi
+
·
dt
e (e − 1 − zi )
α − βi βi + βo α − βi α − βo
γ
βi βo
γ
−zi
−zo
−
−1 −
(e + e )
(α − βi )2
βi + βo (α − βi )(α − βo )
βi
γ α − βi − βo −zi
γ
e
+
2
(α − βi ) βi + βo α − βo α − βi
βi
βi βo
γ
e−zo
−
+
.
βi + βo βi + βo α − βi α − βo
Now if we expand this last expression, we find that
(ezi − 1)2
γ
−βo
−βi
dzi
= zi zi
·
+
dt
e (e − 1 − zi ) α − βi
(βi + βo )(α − βi ) (βi + βo )(α − βo )
1
βi βo γ
+
(e−zi + e−zo )
+
α − βi (βi + βo )(α − βi )2 (α − βo )
βi
γ
e−zi
βi βo γe−zi
+
−
βi + βo α − βo α − βi (βi + βo )(α − βo )(α − βi )2
βi
βi βo γe−zo
−
−
.
(βi + βo )(α − βi ) (α − βi )2 (βi + βo )(α − βo )
In the above, the terms within the curly brackets that have (α − βi )2 in the denominator cancel. Moreover we see that
1
βi
βo
−
−
= 0.
α − βi (βi + βo )(α − βi ) (βi + βo )(α − βi )
With these simplifications, we obtain that
dzi
(ezi − 1)2
γ
−βi
βi
γ
e−zi
= zi zi
·
+
.
dt
e (e − 1 − zi ) α − βi
(βi + βo )(α − βo ) βi + βo α − βo α − βi
After pulling out common factors, as well as using the defintion of zi , we find that
dzi
(ezi − 1)2
zi ezi
βi
zi
= zi zi
−1 + zi
,
dt
e (e − 1 − zi ) ezi − 1 (βi + βo )(α − βo )
e −1
151
which implies that
dzi
e zi − 1
−ezi + 1 + zi
zi βi
= zi
,
dt
e − 1 − zi (βi + βo )(α − βo )
ezi − 1
or simply
zi βi
1 dzi
βi
dzi
=−
=⇒
=−
.
dt
(βi + βo )(α − βo )
zi dt
(βi + βo )(α − βo )
(5.7.2)
Analogously, we have that
1 dzo
βo
=−
.
zo dt
(βi + βo )(α − βi )
(5.7.3)
Adding both equations and using the fourth equation of (5.7.1), we find that
1 dzo
1
1 dzi
+
=
zi dt
zo dt
γ
−
γ
βi
γ
βo
−
βi + βo α − βo βi + βo α − βi
=
1 dγ
,
γ dt
or simply
d
[ln zi + ln zo − ln γ] = 0.
dt
Therefore
z1 z2
≡ constant.
γ
Proposition 5.7.3. Let w(0) ∈ int (W). Then, for t < tW , we have that
γ(0)(α(0) − βi (0) − βo (0))
γ(t)(α(t) − βi (t) − βo (t))
=
.
(α(t) − βi (t))(α(t) − βo (t))
(α(0) − βi (0))(α(0) − βo (0))
Proof. By adding the second and third equations from (5.7.1), we find that
d(βi + βo )
γβi βo (e−zo + e−zi )
γ(α − βi − βo ) (βi e−zi + βo e−zo )
= −1 −
+
.
dt
(βi + βo )(α − βi )(α − βo )
(βi + βo )(α − βi )(α − βo )
Now using the first equation from (5.7.1), we find that
d(βi + βo )
dα γ(α − βi − βo ) (βi e−zi + βo e−zo )
=
+
.
dt
dt
(βi + βo )(α − βi )(α − βo )
152
After rearranging, we see that
d(α − βi − βo )
γ(βi e−zi + βo e−zo )
1
=−
.
α − βi − βo
dt
(βi + βo )(α − βi )(α − βo )
(Notice that α − βi − βo is the scaled number of vertices with both positive in-degree
and positive out-degree. We see that this proportion decreases along the trajectory
as it should.)
Thus using the definition of zi , zo as well as (5.7.2), (5.7.3),
d
γe−zi
−βi
γe−zo
−βo
+
ln [α − βi − βo ] =
dt
α − βi (βi + βo )(α − βo )
α − βo (βi + βo )(α − βi )
1 dzo
1 dzi
+ zo
.
= zi
e − 1 dt
e − 1 dt
The above can be rewritten as
d ln (α − βi − βo ) − ln 1 − e−zi − ln 1 − e−zo = 0;
dt
whence
e zo
e zi
ln (α − βi − βo ) zi
≡ constant.
e − 1 ezo − 1
Using the definition of zi and zo again, we see that
(α − βi − βo )
e zo
α − βi − βo
γ γ(α − βi − βo )
e zi
γ2
=
=
.
ezi − 1 ezo − 1
zi zo
(α − βi )(α − βo )
zi zo (α − βi )(α − βo )
By the previous proposition, γ/(zi zo ) is constant, so the desired quantity is constant
as well.
Thus we have proved that the following integrals are constant along any trajectory
starting in the interior of W, up to reaching the boundary,
I1 (α, βi , βo , γ) :=
zi zo
,
γ
I2 (α, βi , βo , γ) :=
γ(α − βi − βo )
.
(α − βi )(α − βo )
Note. The constancy of Ij along the trajectory means (is equivalent to) the
following: the gradient of Ij is orthogonal to the velocity along the trajectory; i.e.
the RHS of (5.7.1). In particular,
∇Ij (w) · E[∆] = 0.
153
(5.7.4)
It’s worth commenting on the nature of this constant I2 . If the trajectory ever
reaches the terminal state, on the boundary, where βi = βo = 0, and α, γ > 0, then
γ(α − βi − βo )
γ
= ,
(α − βi )(α − βo )
α
which would correspond to the arc density of the terminal digraph. Also at this end,
we would have that
γ
zi ezi
γ
=
=
ezi − 1
α − βi
α
zo ezo
γ
γ
=
=
.
e zo − 1
α − βo
α
and
In this case, both zi and zo end at z(γ/α) = z(I2 (α, βi , βo , γ)), where z(η), for η > 1,
is defined as the unique positive root of
zez
= η.
ez − 1
This leads us to introduce
∗
z =z
γ(α − βi − βo )
(α − βi )(α − βo )
,
a constant along the trajectory. Since
γ(α − βi − βo )
≤ min
(α − βi )(α − βo )
γ
γ
,
α − βi α − βo
,
and z(η) is increasing, we have zi , zo ≥ z ∗ along the trajectory.
Now, the fact that both integrals, I1 and I2 , are constant along trajectories allows
us to define an alternative (closed) set W ⊂ W with the property that a trajectory
starting in int(W ) never leaves W , a property which W itself does not have. Furthermore, for these trajectories, we want to reach the boundary in finite time and our
position at this point to be of the form
(α, 0, 0, γ),
154
for some positive α and γ, so that we can use our boundary conditions for f1 and f2 .
Fix σ > 1, and let = (n) > 0 be decreasing and such that σ − (0) > 1. Here is
the definition of W ,
n
o
W = W(n) := w ≥ 0 : I1 (w), I2 (w) ∈ [σ − , σ + ] and α − βi − βo ≥ 0 .
We split the proof of invariance of W and further properties into several claims. Just
as for W, we define tW = tW (w(0)) as the first time that the trajectory starting at
w(0) hits the boundary of W . Formally, we have that
tW = sup{t : w(s) ∈ int (W ) for s ≤ t}.
Our first goal is to show the times tW and tW are equal. Note that these times are
necessarily positive, but could be infinite. To this end, it suffices to show that
• W ⊂ W, which implies that tW ≤ tW , and
• Suppose w(0) is in the interior of W such that tW < ∞. If w(tW −) exists and
is finite, then w(tW −) is on the boundary of W. This implies that tW ≥ tW .
First, we wish to show that W ⊂ W. To this end, by the definition of W, it
suffices to prove the following claim:
Claim 5.7.4. There are some positive numbers, constant with respect to n, δ =
δ(σ, (0)) and δ 0 = δ 0 (σ, (0)) such that for all w ∈ W ,
α − βi − βo ≥ δ and
γ
γ
,
≥ 1 + δ0.
α − βi α − βo
Proof. First, we see that
γ
γ
γ(α − βi − βo )
,
≥
≥ σ − (0).
α − βi α − βo
(α − βi )(α − βo )
155
Since σ − (0) > 1, we can choose any δ 0 so that 0 < δ 0 < σ − 1 − (0). The lower
bound on γ/(α − βi ) and γ/(α − βo ) implies the existence of some ρ > 0, such that
zi , zo ≥ ρ for all w ∈ W . We see that
I2 (w)/I1 (w) =
γ(α − βi − βo )
ezi
e zo
= (α − βi − βo ) zi
.
(α − βi )(α − βo )
e − 1 ezo − 1
By the definition of W , this ratio is in the interval
σ − (0) σ + (0)
,
,
σ + (0) σ − (0)
and since zi , zo ≥ ρ, we have that
σ − (0)
≤ I2 /I1 ≤ (α − βi − βo )
σ + (0)
eρ
eρ − 1
2
.
Hence we can choose
σ − (0)
δ=
σ + (0)
eρ − 1
eρ
2
,
so to have that α − βi − βo ≥ δ for all w ∈ W .
Claim 5.7.5. Suppose w(0) is in the interior of W such that tW < ∞. If w(tW −)
exists and is finite, then w(tW −) is on the boundary of W. Furthermore, w(tW −)
must have that α, γ > 0 and at least one of βi and βo is zero.
Proof. By the definition of W , if w = (α, βi , βo , γ) := w(tW −) is on the boundary of W , then one of the following conditions must hold
• at least one of α, βi , βo , γ is zero,
• at least one of I1 (w) and I2 (w) is either σ − or σ + , or
• α − βi − βo = 0.
However, the integrals I1 (w(t)) and I2 (w(t)) are constant along any trajectory starting from the interior, so by continuity, we have that Ij (w) ∈ (σ − , σ + ). Since
W ⊂ W, we have that, along the trajectory before tW ,
α(t) − βi (t) − βo (t) ≥ δ > 0.
156
Again by continuity, we have that
α − βi − βo ≥ δ > 0.
Therefore at this point, at least one of α, βi , βo and γ must be zero as so we are on
the boundary of W as well. Now to prove the last statement of the claim, we notice
that (trivially)
α ≥ α − βi − βo ≥ δ > 0,
and using the lower bound on zi , zo found in Claim 5.7.4 along with constancy of the
integral I2 , we see that
γ ≥ zi zo (σ − (0)) ≥ ρ2 (σ − (0)) > 0.
Therefore at least one of βi and βo are zero.
We have just shown that for trajectories starting in int(W ), the times tW and
tW are equal. For w(0) ∈ int (W ), we define
T = T (w(0)) = tW (w(0)),
as this common time. To be able to determine our functions f1 and f2 , we would like
to know that at the end of our trajectory, we arrive at the set where we define the
boundary conditions for fj . To this end, we would need to know that T is finite, and
furthermore that w(T −) exists and is of the form
w(T −) = (α, 0, 0, γ),
for some positive α and γ. We have shown that if T is finite and w(T −) exists and
is finite, then α(T −), γ(T −) > 0 and at least one of βi (T −) and βo (T −) is zero. We
will use the next proposition to show that the ratio βi /βo heads towards 1 so that if
one of βi or βo tends to zero, the other must as well. Then we will show that T is
finite and that all the components, which are non-negative, of w decrease along the
trajectory implying that w(T −) exists and is finite.
157
Proposition 5.7.6. If the initial point w(0) is in the interior of W , then the ratio
βi (t)/βo (t) varies monotonically in the direction toward 1.
Proof. It suffices to show that if βi /βo < 1, then
d βi
> 0,
dt βo
and if βi /βo > 1, then
d βi
< 0.
dt βo
The two cases are treated similarly. Here, let’s suppose that βi /βo < 1. First, from
the second formula of (5.7.1),
γ(α − βi − βo )e−zi
1
1 dzo
1 dβi
=
−
+ zo
βi dt
(βi + βo )(α − βi )(α − βo ) βi + βo e − 1 dt
1 dzo
zi
dzi
1
γ(α − βi − βo )
+ zo
−
−
−
,
=
(βi + βo )(α − βi )(α − βo ) (βi + βo )
dt
βi + βo e − 1 dt
and analogously,
1 dβo
1 dzi
γ(α − βi − βo )
zo
dzo
1
+ zi
=
−
−
−
.
βo dt
(βi + βo )(α − βi )(α − βo ) (βi + βo )
dt
βi + βo e − 1 dt
By subtracting these two equations, we see that
1 dβi
1 dβo
zo − zi
dzi
1 dzi dzo
1 dzo
−
=
−
− zi
+
+ zo
βi dt
βo dt
βi + βo
dt
e − 1 dt
dt
e − 1 dt
zi
zo
e
dzi
e
zo − zi
dzo
− zi
+ zo
.
=
βi + βo e − 1 dt
e − 1 dt
Hence
d
ln
dt
βi (ezi − 1)
βo (ezo − 1)
=
zo − zi
.
βi + βo
By definition of zi and zo , βi < βo iff zi < zo . Suppose βi /βo < 1, then we have that
d
βi (ezi − 1)
ln
> 0.
(5.7.5)
dt
βo (ezo − 1)
This previous equation is equivalent to
d βi
>0
dt βo
and
158
d e zi − 1
> 0,
dt ezo − 1
as desired.
With this proposition at hand, we are ready to prove that we do reach this terminal
point.
Lemma 5.7.7. Let w(0) be in the interior of W . Then
(i) T is finite, and
(ii) w(T −) exists and βi (T −) = βo (T −) = 0.
Proof. (i) First, it is clear, that the RHS of the first and fourth equations in
(5.7.1) as well as (5.7.2) and (5.7.3) are negative, so each of α(t), γ(t), zi (t), zo (t) are
decreasing on (0, T ). Note that
βi
γ(α − βi − βo ) −zi
γβo e−zo
dβi
=
e −1−
dt
βi + βo (α − βi )(α − βo )
(α − βi )(α − βo )
βi
γ(α − βi − βo ) −zi
e −1 .
≤
βi + βo (α − βi )(α − βo )
By the definition of z ∗ , we have that
∗
γ(α − βi − βo ) −zi
z ∗ ez −zi
z∗
e = z∗
≤ z∗
,
(α − βi )(α − βo )
e −1
e −1
where in the last inequality, we use the fact that zi ≥ z ∗ along the trajectory. Uniformly over W , there is a lower bound on z ∗ , so there is some 0 > 0 such that
z∗
≤ 1 − 0 ,
∗
z
e −1
for all w ∈ W . As a consequence, we have that
βi
dβi
≤ −0
.
dt
βi + βo
Hence βi is decreasing along the trajectory. Analogously, we find that
dβo
βo
≤ −0
,
dt
βi + βo
159
and βo is decreasing as well. Furthermore, along the trajectory,
d(βi + βo )
≤ −0 .
dt
Since we certainly stop when βi + βo is zero, this last equation implies that T is finite.
Since each component of w(t) is non-negative and decreasing, we have that w(T −)
exists. We can define w(T ) as this limit so that w(t) is continuous on [0, T ].
(ii) Necessarily, w(T ) is on the boundary of W. As previously mentioned, we
have that at least one of βi (T ) and βo (T ) is zero. However, by Proposition 5.7.6,
they must both be zero.
Part 2. Determining f1 and f2 .
Now we can find our functions f1 and f2 . Let w(0) in the interior of W . As
just shown α(T ), γ(T ) > 0 and βi (T ) = βi (T ) = 0. Since z ∗ is constant along the
trajectory, we have that
γ(T )(α(T ) − βi (T ) − βo (T ))
γ(0)(α(0) − βi (0) − βo (0))
=
(α(0) − βi (0))(α(0) − βo (0))
(α(T ) − βi (T ))(α(T ) − βo (T ))
γ(T )
=
,
α(T )
(5.7.6)
and zi (T ) = zo (T ) = z ∗ . By constancy of zi zo /γ along the trajectory, we find that
γ(T ) = γ(0)
(z ∗ )2
zi (T )zo (T )
= γ(0)
.
zi (0)zo (0)
zi (0)zo (0)
Using this equation, if we solve for α(T ) in (5.7.6), we find that
(z ∗ )2 (α(0) − βi (0))(α(0) − βo (0))
zi (0)zo (0) γ(0)(α(0) − βi (0) − βo (0))
(z ∗ )2 (α(0) − βi (0))(α(0) − βo (0))
=
.
zi (0)zo (0)
α(0) − βi (0) − βo (0)
α(T ) = γ(0)
We could also write α(T ) as
γ(0) z ∗ (ez∗ − 1)
.
α(T ) =
zi (0)zo (0)
ez ∗
160
Along the trajectory, the functions fj are constant. By the boundary conditions, we
have that for w(0) in the interior of W ,
f1 (α(0), βi (0), βo (0), γ(0)) = f1 (α(T ), 0, 0, γ(T )) = α(T )
f2 (α(0), βi (0), βo (0), γ(0)) = f2 (α(T ), 0, 0, γ(T )) = γ(T ).
Using the equations for α(T ) and γ(T ) above, we have the following explicit formulas
for f1 and f2 : for any w = (α, βi , βo , γ) in the interior of W ,
2
γ(α−βi −βo )
z (α−β
(α − βi )(α − βo )
i )(α−βo )
·
,
f1 (α, βi , βo , γ) = γ
γ
α
−
β
−
β
i
o
z α−βi z α−βo
2
γ(α−βi −βo )
z (α−β
i )(α−βo )
· γ,
f2 (α, βi , βo , γ) = γ
γ
z α−βi z α−βo
where once again, z(η) is the unique positive root of
zez
= η.
ez − 1
Note : For our special parameters
(α, βi , βo , γ) = (1 − e−2c , e−c (1 − e−c ), e−c (1 − e−c ), c),
we find that
z
f1 =
c(1−2e−c +e−2c )
(1−e−c )2
z
2
c
1−e−c
2
2
(1 − e−c )
z (c)2
·
=
2 .
1 − 2e−c + e−2c
z 1−ec −c
We see that z(c/(1 − e−c )) is the unique positive root of
zez
cec
c
=
=
,
ez − 1
1 − e−c
ec − 1
and so z(c/(1 − e−c )) = c. Now z(c) is the unique positive root of
zez
= c,
ez − 1
161
which implies that z(c) is a root of
1−
z
= e−z = e−c(z/c) .
c
This last equation is exactly how we defined θ = θ(c). Hence z(c) = cθ(c). Therefore
for these special parameters, we find that
f1 =
c2 θ 2
= θ2 ,
c2
just as we expected from the results by Karp [31] and Luczak [39]. Also at this point,
we see that
f2 = c
z(c)2
= cθ2 .
z(c/(1 − e−c ))2
This means that the average in-degree and out-degree of the terminal digraph
should be asymptotic to c, the original average in/out-degree of the initial digraph
D(n, m = cn).
5.8
The PDE for ψj,k
Now that we have established that such functions f1 and f2 exists, we turn our
attention to establishing the existence of such a ψj,k that satisfies the PDE (5.5.7)
with the boundary conditions (5.5.2); namely,
Es∗ ∆T ∇ψj,k + Es∗ ∆T ∇fj ∆T ∇fk = o(1),
where ∆T = (−a − b, r − a, t − b, −k). Just as for the PDE for f1 , f2 (5.5.6), it suffices
to find ψj,k that satisfy the non-homogeneous PDE:
Es [∆T ]∇ψj,k + Es [(∆T ∇fj )(∆T ∇fk )] = 0,
together with the boundary conditions: for α, γ > 0,
ψj,k (α, 0, 0, γ) = 0.
162
(5.8.1)
Notice that the linear operator Es [∆T ]∇ is exactly the same as the PDE for f1 and
f2 . Applying the method of characteristics to (5.8.1), we have the same trajectory,
but ψj,k is not constant along the trajectory. In fact,
dψj,k
= −E[(∆T ∇fj )(∆T ∇fk )].
dt
For starting points in int(W ), we found that at t = T , both βi (T ) and βo (T ) are
zero and α(T ) and γ(T ) are positive. So by the boundary condition, we have that
ψj,k (α(T ), βi (T ), βo (T ), γ(T )) = 0.
Therefore
dψj,k T
ψj,k (α(0), βi (0), βo (0), γ(0)) = −
dt
t=0
Z T
Z
T
T
=
E[(∆ ∇fi )(∆ ∇fj )]dt =:
0
T
Ψj,k dt.
0
Note: (i) The fact that we can write ψj,k as the above implies that the corresponding
matrix {ψj,k } is non-negative definite as expected, since it is supposed to be the
limiting covariance matrix of the scaled initial state. Specifically, we see that



!2 
2
X
Ψ
Ψ
1,2 
 1,1
uT 
uj ∆T ∇fj  ≥ 0.
u = E 
Ψ2,1 Ψ2,2
j=1
(ii) The endpoint T is a function of the initial state s(0), which we have not
determined in an explicit form. Earlier we showed that zi monotonically decreases
along the trajectory and ends at z ∗ . So we can reparameterize the trajectory in terms
of zi and find that
Z
zi (0)
ψj,k (α(0), βi (0), βo (0), γ(0)) =
Ψj,k
z∗
where we use our formula for
dzi
from (5.7.2).
dt
163
(βi + βo )(α − βo )
dzi ,
βi zi
Let’s briefly describe the derivation of formulas for Ψj,k . Since ∆ and ∇fi are
4−dimensional, there are 16 summands in E[(∆T ∇fj )(∆T ∇fk )] for each j, k. Each
summand is a quadratic polynomial in a, b, r, t, k multiplied by a partial derivative of
fj and a partial derivative of fk . Explicitly,
(∆T ∇fj )(∆T ∇fk ) = (a + b)2 (fj )α (fk )α + (r − a)(−a − b)(fj )βi (fk )α
+ (t − b)(−a − b)(fj )βo (fk )α + k(a + b)(fj )γ (fk )α + (−a − b)(r − a)(fj )α (fk )βi
+ (r − a)2 (fj )βi (fk )βi + (t − b)(r − a)(fj )βo (fk )βi − k(r − a)(fj )γ (fk )βi
+ (−a − b)(t − b)(fj )α (fk )βo + (r − a)(t − b)(fj )βi (fk )βo + (t − b)2 (fj )βo (fk )βo
− k(t − b)(fj )γ (fk )βo + (−a − b)(−k)(fj )α (fk )γ + (r − a)(−k)(fj )βi (fk )γ
+ (t − b)(−k)(fj )βo (fk )γ − k(−k)(fj )γ (fk )γ .
Clearly, this is a quadratic polynomial with respect to a, b, r, t, and k with coefficients
depending on the current state s. Arguing as in the case of linear polynomials, we
find the approximate expectations of all the quadratic polynomials above. Again,
we find their corresponding qi −expected values by taking partial derivatives of the
generating function F (x, y, w) defined in (5.6.1). For example, let’s find Es [a2 ]. Like
our calculation of Esq [a], we have that
νi
+ O n−1 ,
νi + νo
X
X
X
∂F ∂ 2 F +
b2 qi (∆s|s) =
b(b − 1)qi (∆s|s) +
bqi (∆s|s) =
∂x2 (1,1,1) ∂x (1,1,1)
!
2
νi
νo zi zo
νo zi zo
ezo
−1
=
+
+
O
n
.
νi + νo e z o − 1
e zi − 1 µ
ezi − 1 µ
X
a2 qi (∆s|s) =
X
qi (∆s|s) =
The sum of a2 qo is computed from the sum of b2 qi by switching each instance of i
with o, and o with i. Hence
Esq [a2 ] = Es [a2 ] + O(n−1 ),
164
where
νi
νo
ezi
Es [a2 ] :=
+
νi + νo νi + νo ezi − 1
νi zi zo
z
o
e −1 µ
2
νi zi zo
+ zo
e −1 µ
!
.
We compute the other approximate expectations using this same method; a list of
explicit formulas of these expectations is given in Appendix B.
One can easily compute the partial derivatives of f1 and f2 using standard differentiation techniques. Again, see Appendix B for a list of explicit formulas for these
partial derivatives.
These integrand functions Ψj,k are clearly continuous, so by the method of characteristics, such functions ψj,k exists, although unlike fj , we have only the integral
expressions.
In summary, we have determined fj and ψj,k which are solutions to the partial
differential equations (5.5.6) and (5.5.7), meeting the boundary conditions (5.5.2). In
fact, we have shown the following more explicit bounds: uniformly over s ∈ S, with
ν = Θ(n) and s/n ∈ W ,
Es∗ [∆T ]∇fj (s/n) = O (ln n)8 /n ,
(5.8.2)
and
Es∗ [∆T ]∇ψj,k + Es∗ [ ∆T ∇fj
∆T ∇fk ] = O (ln n)8 /n .
(5.8.3)
For these such s, our Gaussian characteristic function G(s/n, u) satisfies the recurrence relation (5.5.5). In fact due to the bounds in (5.8.2) and (5.8.3), uniformly over
||u|| = O(n−1/2 ) and s ∈ T = {s ∈ S : s/n ∈ W , ν = Θ(n)}, we have that
G(s/n, u) −
X
G(s0 /n, u)P (s(t + 1) = s0 |s(t) = s) = O((ln n)8 /n3/2 ).
(5.8.4)
s0
Remark 5.8.1. Clearly, we need to show then that, for s(0) from a subset of T, w.h.p.
s(t) is in T for as long as the deterministic trajectory can be a good approximation
of the deletion process.
165
Leaving the proof of this fact till Section 5.10, we turn our attention to computing
ψj,k at the likely starting point induced by D(n, m = cn), as well as determining the
asymptotic behavior of this quantity as c decreases to 1.
5.9
The Special Case and c ↓ 1
In this section, we consider ψj,k (1−e−2c , e−c −e−2c , e−c −e−2c , c) for c > 1, which is the
special case that comes from typical starting points corresponding to D(n, m = cn).
c
For ease of notation, let’s denote this quantity by ψj,k
. Previously, we found that
f1 = θ2 and f2 = cθ2 at this starting point, where θ is the unique positive root of
1 − θ = e−cθ .
c
To find ψj,k
, we need to first consider the trajectory starting at this point. Since βi
and βo start out equal, throughout the trajectory this equality holds by Proposition
5.7.6. So we let βi = βo =: β, which tells us that zi = zo =: z. From these initial
conditions, we have that
γ(α − 2β)
z2
=c=
,
γ
(α − β)2
throughout the trajectory. We can use these equalities to solve for α, β, and γ in
terms of z. In fact, we have that
γ(α − 2β)
2(α − β)
α
2 zez
α z 2 e2z
c=
=γ
−
=γ
−
(α − β)2
(α − β)2
(α − β)2
γ ez − 1 γ 2 (ez − 1)2
2zez
cα z 2 e2z
ce2z
2z z
= z
−
= z
−α + z (e − 1) .
e − 1 z 2 (ez − 1)2
(e − 1)2
ce
Solving for α in the above yields
α=
(ez − 1)2
2z z
ez − 1
(e
−
1)
−
=
(2zez − c(ez − 1)) .
cez
e2z
ce2z
In a similar way, we find that
ez − 1
β=
(zez − c(ez − 1)) ,
2z
ce
166
along the trajectory. Also γ = z 2 /c. We notice that
−βz
−zγ
−c zez
−cez
dz
=
=
=
=
.
dt
(β + β)(α − β)
2γ(α − β)
2z ez − 1
2(ez − 1)
This allows us to convert the integral finding ψj,k that is initially over t to one that
is over z that starts at c and ends at cθ, because
2(ez − 1)
dz = dt.
−cez
Hence we write each term in E[(∆T ∇fj )(∆T ∇fk )] in terms of just z. (!)
Z c
2(ez − 1)
c
E[(∆T ∇fj )(∆T ∇fk )](z) ·
ψj,k =
dz.
cez
cθ
(5.9.1)
In this last integral, the integrand is some explicit function of just z. Each term
of E[(∆T ∇fj )(∆T ∇fk )] is given in Appendix B, in terms of α, βi , βo , γ, which in turn
are explicit functions of z. The integrand can be written as a fraction N (z)/D(z),
where D(z) = z 4 (ez − 1)6 (ez − 1 − z)2 and N (z) is a polynomial (over Q) in z, ez
and e−z . Although we have an explicit function, there are many terms in N (z) and
attempts to simplify this integrand were futile. In fact, there are thousands of terms
when the polynomial N (z) is expanded out. Still, we have an explicit integral formula
for the covariance term ψj,k which allows numerical evaluation for fixed c > 1.
However, we can say substantially more about fj and ψj,k when c is close to 1
c
from above. Using Mathematica, we can determine the behavior of ψj,k
as c decreases
to 1. First, since θ is defined as the unique positive root of 1 − θ = e−cθ , we have
θ θ2
− ln(1 − θ)
=1+ +
+ ...
c=
θ
2
3
In particular, as c decreases to 1, θ ∼ 2(c − 1) or more explicitly
8
θ = 2(c − 1) − (c − 1)2 + O((c − 1)3 ).
3
We could instead view this situation as θ tends to zero and define c in terms of θ.
We know that f1 = θ2 and that f2 = cθ2 = θ2 + O(θ3 ) as θ → 0+ .
167
To analyze the integral in (5.9.1), we use Mathematica to find the power series in
z and θ of the integrand. The power series centered at z = 0 for both
1
1
−
,
ez − 1 z
and
2
1
2
−
,
+
ez − 1 − z z 2 3z
converge for |z| ≤ 1.2. We may suppose for the rest of this section that θ is close
enough to zero, so that c ≤ 1.2 and these power series converge for z in [cθ, c].
Furthermore the power series for c in terms of θ converges as long as θ < 1. The
following expression also appears in the integrand,
1 − (1 − θ)
θ
ecθ − 1
=
=
=
cθ
e −c
1 − c(1 − θ)
1 − c(1 − θ)
1
1
2
+
θ
2·3
+
θ2
3·4
+
θ3
4·5
+ ...
=
X
aj θ j ,
j≥0
where the above power series converges as long as θ < 1. As a consequence, for θ
small enough we can write
E[(∆T ∇fj )(∆T ∇fk )]
2(ez − 1)
= a2j,k (z)θ2 + a3j,k (z)θ3 + a4j,k (z)θ4 + . . . ,
z
e
where alj,k (z) is a function of the form
alj,k (z) = blj,k (−5)
X
1
1
1
l
l
+
b
(−4)
+
...
+
b
(−1)
+
blj,k (i)z i ,
j,k
j,k
5
4
1
z
z
z
i≥0
for some constants b, where the series converges for z ∈ [0, 1.2]. By counting multiplicity of the poles at z = 0, we see why we need to go up to z −5 . Furthermore, each
partial derivative of fi is multipled by θ, so that our power series in θ starts at θ2 .
Using Mathematica, we are able to determine these constants and study the behavior
as θ → 0+ . In particular, we find that
a21,1 (z) = a21,2 (z) = a22,2 (z) =
168
32
+ h(z),
z2
where h(z) has a power series representation. We find that
c
ψ1,1
c
c
= ψ2,1
ψ1,2
c
ψ2,2
Z
1
h(z)dz θ2 + O θ3 ln(1/θ) ,
= 20θ + −65 +
0
Z 1
= 20θ + −46 +
h(z)dz θ2 + O θ3 ln(1/θ) ,
0
Z 1
= 20θ + −27 +
h(z)dz θ2 + O θ3 ln(1/θ) ,
0
or rather in terms of (c − 1), up to additive error of order (c − 1)3 ln
c
ψ1,1
c
ψ1,2
c
ψ2,2
Z 1
940
= 40(c − 1) + −
h(z)dz (c − 1)2 ,
+4
3
0
Z 1
712
= 40(c − 1) + −
+4
h(z)dz (c − 1)2 ,
3
0
Z 1
484
+4
h(z)dz (c − 1)2 .
= 40(c − 1) + −
3
0
1
,
c−1
(5.9.2)
To give an idea about how complicated the original integrand is, we have that
32
− e2z + 3e3z − 3e4z + e5z + z − 4ez z + 5e2z z − 2e3z z
e2z z 3 (ez − 1)3
32
+ 2ez z 2 − 6e2z z 2 + 7e3z z 2 − 3e4z z 2 + e2z z 3 − 2e3z z 3 + 2e4z z 3 − 2 .
z
h(z) =
The sum within the braces has a Taylor series starting with z 4 and h(z) is analytic
at z = 0 with h(0) = −4/3. Numerically integrating h(z) gives
Z
1
h(z)dz = 0.5025...
0
5.10
Large Deviation Bounds
With evaluation of ψj,k out of the way, we turn now to show that for s(0) in some
subset of T, w.h.p. s(t) is in T until almost the end of the deletion process. To this
γ(α − βi − βo )
end, we will consider the integrals I1 =
and I2 = zi zo /γ, which we
(α − βi )(α − βo )
found to be constant along the deterministic trajectory. These constancies makes it
169
highly plausible that along the realization of the Markov process itself w.h.p. the
corresponding functions,
µ(ν − νi − νo )
,
(ν − νi )(ν − νo )
zi zo
F2 (ν, νi , νo , µ) =
,
µ/n
F1 (ν, νi , νo , µ) =
of the unscaled parameters, should stay relatively constant too (note that Fj (s) =
Ij (s/n)). If this is the case, then we have a solid chance to prove that, even stronger,
the scaled Markov process {s(t)/n} is w.h.p. close to the deterministic trajectory in
the sense that G(s/n, u) nearly satisfies the recurrence relation.
We will prove that F1 and F2 are essentially constant by building a suitable supermartingale from Fj and applying the Optional Sampling Theorem to this supermartingale. This kind of supermartingale and stopping time method was already
used by Pittel et al. [54] and Aronson et al. [1] for deletion processes for k−core and
matching problems.
Our definition of W leads us to consider the corresponding set of s, specifically
S , where
(
Sη :=
)
s ≥ 0 : F1 (s), F2 (s) ∈ [σ−η, σ+η], ν −νi −νo ≥ 0, νi +νo > (ln n)2 . (5.10.1)
In particular, S differs from W due to appearance of extra constraint; namely,
νi + νo > (ln n)2 . We need this constraint to show that the process starting in S
is sufficiently close to the deterministic trajectory (our approach is guided by the
intuition that when the number of semi-isolated vertices drops below (ln n)2 , the
number of remaining vertices and arcs are close to those of the (1, 1)−core). Most
importantly, note that if s ∈ S , then s/n ∈ W . Just as we did for W and W in
Claim 5.7.4, one can show that S ⊂ S. Furthermore, it is easy to show that uniformly
over s ∈ S , µ, ν and ν − νi − νo are on the order of n and ∇Fi is on the order of n−1 .
Therefore S ⊂ T.
170
We can use our approximation (5.8.4) as long as s(t) ∈ T. This leads us to define
τ̂ as the stopping time for leaving S , and consider the process up to the stopping
time τ̂ . Let P̂ be the corresponding transition probability for the process that stops
at τ̂ . Since S ⊂ S, necessarily τ̂ ≤ τ ∗ .
Now let us prove that w.h.p. Fj (s(t)) stays relatively close to Fj (s(0)) for all
t ≤ τ̂ . To this end, introduce
Qj (s) := exp L Fj (s) − Fj (s(0)) ,
where L = Ln = o((ln n)(ln ln n)) to be specified shortly; also, we consider the process
Qj (t) := Qj (s(t))1{t<τ̂ } .
Lemma 5.10.1. For j = 1, 2, the process
Rj (t) =
Qj (t)
1+
1
t
n1.99
is a non-negative supermartingale.
Proof. There is little difference between the proofs for j = 1 and j = 2, so here
we prove the lemma for j = 1. It suffices to show that
Ê[Q1 (t + 1)|{s(t0 )}t0 ≤t ] ≤ Q1 (t)(1 + n−1.99 ),
(5.10.2)
for all t, since (5.10.2) can be rewritten as
Ê[R1 (t + 1)|{s(t0 )}t0 ≤t ] ≤ R1 (t).
Trivially if s(t) ∈
/ S , then with probability 1, we have that Q1 (t + 1) = Q1 (t), so
that
Ê[Q1 (t + 1)|{s(t0 )}t0 ≤t ] = Qi (t).
So if s(t) ∈
/ S , then
Ê[Q1 (t + 1)|{s(t0 )}t0 ≤t ] ≤ Qi (t)(1 + n−1.99 ).
171
(5.10.3)
Now let s(t) ∈ S , in which case our expectation over P̂ is the same as the
expectation over P . We have that
Q1 (t + 1) = 1{t+1<τ ∗ } eL(Fi (s(t+1))−Fi (s(t))) Q1 (t).
Hence, we find that
E[Q1 (s(t + 1))|{s(t0 )}t0 ≤t ] = Q1 (t)E[1{t+1<τ ∗ } eL(F1 (s(t+1))−F1 (s(t))) |s(t)].
(5.10.4)
So this latter expectation is
E[1{t+1<τ ∗ } eL(F1 (s(t+1))−F1 (s(t))) |s(t)] =
X
0
eL(F1 (s )−F1 (s(t))) P (s(t + 1) = s0 |s(t)).
s0 ∈S
Furthermore,
E[1{t+1<τ ∗ } eL(F1 (s(t+1))−F1 (s(t))) |s(t)] ≤
0
X
eL(F1 (s )−F1 (s(t))) P (s(t + 1) = s0 |s(t))
s0 :k≤ln n
+
X
0
eL(F1 (s )−F1 (s(t))) P (s(t + 1) = s0 |s(t)). (5.10.5)
s0 ∈S
k>ln n
We begin by considering this first sum. Uniformly over s(t) ∈ S and s0 such that
k ≤ ln n,
(µ − k)(ν − νi − νo − r − t)
µ0 (ν 0 − νi0 − νo0 )
=
0
0
0
0
(ν − νi )(ν − νo )
(ν − νi − b − r)(ν − νo − a − t)
µ(ν − νi − νo )
ln n
ln n
=
+O
= F1 (s(t)) + O
.
(ν − νi )(ν − νo )
n
n
F1 (s0 ) =
For these s(t) and s0 , we have that
0
L (F1 (s ) − F1 (s(t))) = O
(ln n)2 ln ln n
n
.
Therefore,
L(F1 (s0 )−F1 (s(t)))
e
(ln n)5
= 1 + L (F1 (s ) − F1 (s(t))) + O
n2
(ln n)5
0
= 1 + L∇F1 (s(t)) (s − s(t)) + O
.
n2
0
172
So
0
X
eL(F1 (s )−F1 (s(t))) P (s(t + 1) = s0 |s(t)) = P (k ≤ ln n|s(t))
s0 :k≤ln n
X
+ L∇F1 (s(t))
(s0 − s(t))P (s(t + 1) = s0 |s(t)) + O((ln n)5 /n2 ).
s0 :k≤ln n
However as we found earlier in (5.6.2), uniformly over s(t) ∈ S,
X
s0 − s(t) P (s(t + 1) = s0 |s(t)) = E[∆] + O((ln n)8 /n).
s0 :k≤ln n
dw
= E[∆], we have
dt
that ∇I1 · E[∆] = 0, or equivalently ∇F1 · E[∆] = 0. Combining this with the fact
As noted in (5.7.4), since Ij is constant along the trajectory
that ∇F1 = O(n−1 ), we have
X
0
eL(F1 (s )−F1 (s(t))) P (s(t + 1) = s0 |s(t)) = P (k ≤ ln n|s(t)) + O((ln n)10 /n).
s0 :k≤ln n
Here, by the second part of Lemma 5.4.3, we have that
P (k > ln n|s(t)) =
2
X
P (s(t + 1) = s0 |s(t)) ≤ e− 3 (ln n)(ln ln n) ,
(5.10.6)
s0 :k>ln n
which implies that
L(F1 (s0 )−F1 (s(0))
X
e
0
P (s(t + 1) = s |s(t)) = 1 + O
s0 :k≤ln n
(ln n)10
n2
.
Now we turn our attention to the second sum of (5.10.5). Since s0 ∈ S and L =
o((ln n)(ln ln n)), we have that
0
0
eL(F1 (s )−F1 (s(0))) ≤ eLF1 (s ) ≤ eL(σ+) ≤ eo((ln n)(ln ln n)) ,
and so, by (5.10.6),
X
2
0
eL(F1 (s )−F1 (s(0))) P (s(t + 1) = s0 |s(t)) ≤b e− 3 (ln n)(ln ln n) s0 :k>ln n
173
(ln n)10
.
n2
Therefore, if s(t) ∈ S , then we have that
L(F1 (s(t+1)−F1 (s(t))))
E 1{t+1<τ ∗ } e
=1+O
(ln n)10
n2
.
(5.10.7)
Hence, by (5.10.3) and (5.10.7), for any s(t),
Ê[Q (t + 1)|s(t)] ≤ Q (t) 1 +
1
1
1
n1.99
.
Therefore {R1 (t)} is a non-negative supermartingale.
To use the full power of this lemma, we need to send L to ∞, but still have that
L = o((ln n)(ln ln n)). This leads us to define
L :=
(ln n)(ln ln n)
.
ln ln ln n
The next lemma will show that Fj is relatively constant for t ≤ τ̂ .
Lemma 5.10.2. Let j = 1 or j = 2. For any a > 0, uniformly over s(0) ∈ S , with
probability at least 1 − n−a , the process starting from s(0) satisfies
|Fj (s(t)) − Fj (s(0))| ≤
1
,
ln ln ln n
for all t ≤ τ̂ .
Proof. Let ρn = 1/ ln ln ln n and let’s consider the stopping time τ , the first
moment that Fj (s(t))−Fj (s(0)) > ρn , if such a t exists. Let τ 0 = min{τ, τ̂ }. Applying
the Optional Sampling Theorem (see Durrett [17]), we have that
τ 0
n
E[Qj (τ 0 )] = E[ 1 + n−1.99 Rj (τ 0 )] ≤ 1 + n−1.99 E[Rj (0)]
n
n
= 1 + n−1.99 E[Qj (0)] = 1 + n−1.99 ≤ e.
On the event {τ ≤ τ̂ }, we have that Qj (τ 0 ) ≥ eLρn and so
eLρn P (τ ≤ τ̂ ) ≤ E[Qj (τ 0 )] ≤ e.
174
Equivalently,
P max[Fj (s(t)) − Fj (s(0))] > ρn = P τ ≤ τ̂ ≤b e−Lρn
t≤τ̂
−(ln n)
≤b e
ln ln n
(ln ln ln n)2
n−a ,
∀ (fixed) a > 0.
The case Fj (s(t)) − Fj (s(0)) ≤ −ρn is treated similarly.
Now that we have proved Lemma 5.10.2, we can prove more information on s(τ̂ )
assuming a stronger constraint on s(0). Specifically, if 0 < 0 < and s(0) ∈ S0 (see
definition of Sη in (5.10.1)), then w.h.p. at the first moment that we leave S (i.e. τ̂ ),
the current number of semi-isolated vertices has dropped below (ln n)2 . If this is the
case, then our approximation (5.8.4) w.h.p. works until the point where the number
of semi-isolated vertices has dropped below (ln n)2 .
Lemma 5.10.3. Let 0 < 0 < with 0 possibly tending to zero (with respect to n),
but fixed (does not depend on n). For any a > 0, and n sufficiently large, uniformly
over s(0) ∈ S0 , with probability at least 1 − n−a ,
Fi (s(τ̂ )) ∈ [σ − , σ + ],
νi (τ̂ ) + νo (τ̂ ) < (ln n)2 .
Proof. Let a > 0. By Lemma 5.10.2, uniformly over s(0) ∈ S0 , with probability
at least 1 − n−a /2, we have that for all t ≤ τ̂ ,
|Fi (s(t)) − Fi (s(0))| < ρn .
For n large enough, we have that
0 + ρn ≤ ∗ :=
0 + .
2
Therefore with probability at least 1 − n−a /2, for all t ≤ τ̂ , we have that
Fi (s(t)) ∈ [σ − 0 − ρn , σ + 0 + ρn ] ⊂ [σ − , σ + ].
175
Necessarily for these such {s(t)}, we have that s(τ̂ − 1) resides in S∗ . Uniformly over
s(t) ∈ S∗ ⊂ S, the probability that we have a transition to s0 where k ≥ ln n, is at
2
most e− 3 (ln n)(ln ln n) , by the second part of Lemma 5.4.3. By a union bound over the
2
number of steps, with probability at least 1−ne− 3 (ln n)(ln ln n) ≥ 1−n−a /2, any process
that starts in S∗ leaves S∗ through a transition where k ≤ ln n. For any s ∈ S∗
and s0 where k ≤ ln n, necessarily Fi (s0 ) ∈ [σ − , σ + ]. Invoking the definition of S
(see (5.10.1)), if s ∈ S∗ and s0 ∈
/ S , yet k ≤ ln n, then necessarily νi0 + νo0 < (ln n)2 .
Therefore with probability at least 1 − n−a , νi (τ̂ ) + νo (τ̂ ) < (ln n)2 .
In the language of Remark 5.8.1, we have proved that as a smaller subset of T,
we can choose S0 , in which for s(0) ∈ S0 , w.h.p. s(t) ∈ T for t ≤ τ̂ and at t = τ̂ , the
number of semi-isolated vertices drops below (ln n)2 .
We introduce the random variables ν̂ = ν(τ̂ ) and µ̂ = µ(τ̂ ), the number of vertices
and arcs when the process stops upon leaving S . By the previous lemma, we suspect
that these parameters will likely be near the terminal parameters where we allow the
deletion process to continue until there are no more semi-isolated vertices to delete. In
the next section, we find the asymptotic distribution of (ν̂, µ̂) starting from a generic
s ∈ S0 .
5.11
Asymptotic Distribution of (ν̂, µ̂) starting at generic s
In this section, our goal is to show for a certain desired range of starting s(0) (which
in the next section, we will show that the parameters induced by D(n, m = cn) are
likely within), the terminal pair (ν̂, µ̂), once centered and scaled, is asymptotically
Gaussian. We begin by showing that according to transition probability P̂ , the
Gaussian charactertistic function nearly satisfies the recurrence relation. Namely,
176
Proposition 5.11.1. Uniformly over s ∈ S and ||u|| = O(n−1/2 ),
X
X
(n)
(i) G(s/n,
u)
−
·
·
·
G(s
/n,
u)
P̂
({s(t)}
=
{s
})
≤ n−0.49 .
s0
s(n)
Proof. For s ∈ S , we have that s/n ∈ W and ν = Θ(n). So by (5.8.4), uniformly
over s ∈ S and ||u|| = O(n−1/2 ),
G(s/n, u) −
X
G(s0 /n, u)P (s(t + 1) = s0 |s(t) = s) = O((ln n)8 /n3/2 ).
s0
Since P̂ is the transition probability for the process that stops at τ̂ , the moment s(t)
leaves S , we have: for s ∈
/ S ,
G(s/n, u) =
X
G(s0 /n, u)1s (s0 ) =
X
G(s0 /n, u)P̂ (s(t + 1) = s0 |s(t) = s).
s0
s0
Therefore, uniformly over all s and ||u|| = O(n−1/2 ), we have that
G(s/n, u) −
X
G(s0 /n, u)P̂ (s(t + 1) = s0 |s(t) = s) = O((ln n)8 /n3/2 ),
s0
and in particular, for n large enough,
X
G(s/n, u) −
G(s0 /n, u)P̂ (s(t + 1) = s0 |s(t) = s) ≤ n−1.49 .
s0
Using this “near” recursion for G(s0 /n, u) for all s0 , by the triangle inequality, we find
that
XX
G(s/n, u) −
G(s00 /n, u)P̂ (s00 |s0 )P̂ (s0 |s) ≤ 2n−1.49 .
s0
s00
If we can keep using the recurrence, then we obtain that
X
X
G(s/n, u) −
···
G(s(n) /n, u)P̂ (s(n) |s(n−1) ) · · · P̂ (s0 |s) ≤ n−.49 ,
s0
(5.11.1)
s(n)
as desired.
Note that we use this near recursion n times because deterministically the deletion
process induced by D(n, m = cn) lasts at most n steps.
177
By the (time homogeneous) Markov nature of {s(t)} over P̂ , we have the following
decomposition,
ϕ̂s (u) =
X
ϕ̂s0 (u)P̂ (s0 |s).
s0
Using this exact recurrence n times, we find that
ϕ̂s (u) −
X
···
s0
X
ϕ̂s(n) (u)P̂ (s(n) |s(n−1) ) · · · P̂ (s0 |s) = 0.
(5.11.2)
s(n)
To be able to show that G(s/n, u) is a good approximation to ϕ̂s (u), for the
starting points s that we will care about, we will show that the largest contribution
to these n-fold sums comes from terminal s(n) , where ϕ̂s(n) (u) is near G(s(n) /n, u). In
particular, we notice that for all s, by (5.11.1) and (5.11.2), we have that uniformly
over ||u|| = O(n−1/2 ),
|G(s/n, u) − ϕ̂s (u)| ≤
X
···
s0
X
G(s(n) /n, u) − ϕ̂s(n) (u)P̂ {s(t)} = {s(i) }
s(n)
+ n−.49 . (5.11.3)
Theorem 5.11.2. Let 0 < 0 < with 0 = 0 (n) possibly tending to zero, but fixed
(does not depend on n). For ||u|| = O n−1/2 , uniformly over s(0) ∈ S0 ,
|G(s(0)/n, u) − ϕ̂s(0) (u)| ≤b n−.49 .
Proof. In light of (5.11.3), we wish to show that for likely endpoints, s(n) , the
difference between the characteristic function of (ν̂, µ̂) of the process starting at s(n)
and the Gaussian approximation G(s(n) /n, u) is small enough to get the bounds in
the statement of the theorem.
By Lemma 5.10.3, with probability at least 1 − n−1 , uniformly over s(0) ∈ S0 , we
have that s(τ̂ ) satisfies
Fi (s(τ̂ )) ∈ [σ − , σ + ], and νi (τ̂ ) + νo (τ̂ ) < (ln n)2 .
178
Let S denote the set of s such that Fi (s) ∈ [σ − , σ + ] and νi + νo < (ln n)2 . So we
have that P (s(τ̂ ) ∈
/ S) ≤ n−1 uniformly over s(0) ∈ S0 . First, we break up the sum
in (5.11.3) into whether s(n) is in S or not. Second, for s(n) ∈
/ S, we use the trivial
inequality
|G(s(n) /n, u) − ϕ̂s(n) (u)| ≤ 2.
Hence
|G(s/n, u) − ϕ̂s (u)| ≤
X
···
s0
X G(s(n) /n, u) − ϕ̂s(n) (u)P̂ {s(t)} = {s(i) }
s(n) ∈S
+2
X
···
s0
X
P̂ {s(t)} = {s(i) } + n−0.49 . (5.11.4)
s(n) ∈S
/
This last sum is precisely the probability that s(τ̂ ) is not in S, which we just noted
is at most n−1 .
For the first sum, we need to bound |G(s/n, u)− ϕ̂s (u)| for s ∈ S. By the definition
of S, if s ∈ S, then s ∈
/ S , and so the process that starts at s stays at s. So for any
s ∈ S,
ϕ̂s (u) = Ês [exp (iu1 ν̂ + iu2 µ̂)] = exp (iu1 ν + iu2 µ) ,
n
Turn to G(s/n, u) = exp iuT f (s/n) − uT ψ(s/n)u . Uniformly over s ∈ S,
2
ν
ν ν ν µ
µ
(ln n)2
i
o
f1 (s/n) = f1
, , ,
= f1
, 0, 0,
+O
n n n n
n
n
n
2
ν
(ln n)
= +O
.
n
n
Similarly, we find that
µ
f2 (s/n) = + O
n
(ln n)2
n
and ψj,k (s/n) = O
(ln n)2
n
In particular, uniformly over s ∈ S and ||u|| = O(n−1/2 ),
G(s/n, u) = exp iu1 ν + iu2 µ + O((ln n)2 /n1/2 )
= exp (iu1 ν + iu2 µ) + O((ln n)2 /n1/2 ).
179
.
(This is the spot where the usage of characteristic functions rather than Laplace
transform is absolutely crucial!)
Therefore uniformly over s ∈ S and ||u|| = O(n−1/2 ), we find that
G(s/n, u) = ϕ̂s (u) + O((ln n)2 /n1/2 ).
Hence
XX
s0
···
s00
X G(s(n) /n, u) − ϕ̂s(n) (u)P̂ ({s(t)} = {s(i) })
s(n) ∈S
≤b
X
(ln n)2
(ln n)2 X
√
···
P̂ {s(t)} = {s(i) } ≤ √ . (5.11.5)
n s0
n
(n)
s
∈S
Using this bound from (5.11.5) in (5.11.4), we find that uniformly over s(0) ∈ S0 and
||u|| = O(n−1/2 ),
|G(s(0)/n, u) − ϕ̂s(0) (u)| ≤b n−0.49 ,
as desired.
A near immediate consequence of this theorem is that the pair of random variables
(ν̂, µ̂) starting from a fixed s(0) ∈ S0 is asymptotically Gaussian, which we prove in
the next corollary.
Corollary 5.11.3. Let c > 1 and s ∈ S0 , where 0 = n−1/3 . Then
ν̂ − nf1 (s/n) µ̂ − nf2 (s/n)
√
√
,
n
n
D
=⇒ N (0, ψ) ,
where N (0, ψ) is a Gaussian distribution with mean 0 and covariance matrix ψ =
(ψj,k (s/n)).
Proof. Fix t1 , t2 ∈ R and ui = ti n−1/2 , then by the previous theorem,
h
it·
Es e
(ν̂,µ̂)
√
n
i
√ T
1 T
− exp i nt f (s/n) − t ψ(s/n)t = O(n−0.49 ).
2
180
(5.11.6)
Since the functions fi are real, and |eix | = 1 for x ∈ R, equation (5.11.6) is equivalent
to
h
it·
Es e
(ν̂,µ̂)−nf (s/n)
√
n
i
1 T
− exp − t ψ(s/n)t = O(n−0.49 ).
2
(5.11.7)
By Multivariate Continuity Theorem, (see Durrett [17]), pointwise convergence of the
characteristic function implies convergence in distribution. Namely, the pair
ν̂ − nf1 (s/n) µ̂ − nf2 (s/n)
√
√
,
n
n
converges in distribution to a Gaussian random variable, N (0, ψ(s/n)).
In summary, we have proved subject to certain constraints on s(0), the pair (ν̂, µ̂)
associated with stopping time τ̂ is jointly asymptotically Gaussian with parameters
depending on the scaled state s(0)/n. However our goal is to prove a similar result for
the deletion process applied to a random instance of the digraph with m = cn arcs, in
which case s(0) is random itself. This is why in the next section, we discuss the joint
distribution of s(0) arising from D(n, m = cn). While the scaled initial state will
be concentrated around some deterministic point, the random fluctuations will be on
the order of n−1/2 . This will give rise to fluctuations of order n−1/2 in the parameters
fj (s(0)/n) and ψj,k (s(0)/n). Although the variation of ψj,k is negligible in the limit,
the fluctuations in the mean vector will affect the limiting ultimate covariance matrix.
5.12
Asymptotic Distribution of s(D(n, m = cn))
First of all, we have that
s(D(n, m = cn)) = (X, Xi , Xo , cn),
where X, Xi , Xo are the numbers of non-isolated vertices, vertices with zero in-degree
and non-zero out-degree, and vertices with zero out-degree and non-zero in-degree of
D(n, m = cn).
181
Lemma 5.12.1. Let c > 1. Then
X − (1 − e−2c )n Xi − (e−c − e−2c )n Xo − (e−c − e−2c )n
√
√
√
,
,
n
n
n
D
=⇒ N (0, K)
where N (0, K) is a 3-dimensional Gaussian distribution with 0 and a covariance
matrix K = K(c), determined later.
Proof. Note that X − Xi − Xo is the total number of vertices of positive in/outdegree. It can be easily shown that
E[Xi ] = E[Xo ] = ne−c (1 − e−c ) + O(1),
E[X] = n(1 − e−2c ) + O(1)
and that
V ar(Xi ) = V ar(Xo ) = O(n),
V ar(X) = O(n).
Given the generic values νi , νo , ν of Xi , X0 , X, we have that
P (Xi = νi , X0 = ν0 , X = ν) =
×
1
(n)2
m
X
n!
g(n, m, δ, ∆). (5.12.1)
νi !νo !(ν − νi − νo )!(n − ν)! δ,∆
Here δ, ∆ denote the generic values of in-degrees and out-degrees of a digraph
with m arcs and specified vertex subsets of the respective cardinalities νi , νo , ν, and
g(n, m, δ, ∆) is the total number of such digraphs. In particular,
δr > 0, ∆r = 0,
r ∈ {1, . . . , νi };
∆s > 0, δs = 0,
s ∈ {νi + 1, . . . , νi + νo };
∆t > 0, δt > 0,
t ∈ {νi + νo + 1, . . . , ν};
δu = ∆u = 0, u ∈ {ν + 1, . . . , n},
X
X
X
X
δr +
δt =
∆s +
∆t = m.
r
t
s
182
t
We find a sharp asymptotic formula for this sum in a similar fashion to our sharp
asymptotics for g(s) determined in Theorem 5.3.4. From Theorem 5.3.1, we know
that
g(n, m, δ, ∆) = H(δ, ∆)m!
Y 1 Y 1 Y 1
,
δ
!
∆
!
δ
!∆
!
r
s
t
t
s
t
r
(5.12.2)
where the “fudge factor” H(δ, ∆) ≤ 1. We need to obtain a sharp asymptotic formula
for this sum, assuming that
νi − ne−c (1 − e−c )
νo − ne−c (1 − e−c )
=
O(1),
x
:=
= O(1),
o
n1/2
n1/2
ν − n(1 − e−2c )
= O(1).
x :=
n1/2
xi :=
First, consider the sum without the factor H(δ, ∆). Then
m!
XY 1 Y 1 Y 1
= m! [zim zom ](ezi − 1)ν−νi (ezo − 1)ν−νo .
δ
!
∆
!
δ
!∆
!
r
s
t
t
s
t
δ,∆ r
We introduce a truncated Poisson Z, where
P (Z = j) =
e−c cj /j!
cj /j!
=
,
1 − e−c
ec − 1
so that
ezc − 1
E[z ] = c
;
e −1
Z
in particular,
c
cec
=
.
E[Z] = c
e −1
1 − e−c
Then
a
a
(ec − 1)a m
(ec − 1)a m ezc − 1
[z
]
[z ] E[z Z ]
[z ](e − 1) =
=
m
c
m
c
e −1
c


(ec − 1)a X
=
P
Zt = m ,
cm
m
z
a
t∈[a]
where Z1 , . . . , Za are independent copies of Z. We use this formula for
ai = ν − νi = n(1 − e−c ) + (x − xi )n1/2 ,
ao = ν − νo = n(1 − e−c ) + (x − xo )n1/2 ,
183
(5.12.3)
so that
m − ai E[Z] =cn − n(1 − e−c ) + (x − xi )n1/2
c
1 − e−c
c
(x − xi )n1/2 ,
1 − e−c
m − ao E[Z] =cn − n(1 − e−c ) + (x − xo )n1/2
=−
=−
c
1 − e−c
c
(x − xo )n1/2 .
1 − e−c
For notation sake, we denote “f = (1 + o(1))g” uniformly over xi , xo , x = O(1) by
simply f ∼ g. By the local limit theorem,




X
X
P
Zt = m = P
(Zt − E[Z]) = m − ai E[Z]
t∈[ai ]
t∈[ai ]
(m − ai E[Z])2
∼p
exp −
2ai V ar(Z)
2πai V ar(Z)
1
c2 (x − xi )2
∼p
exp −
,
2(1 − e−c )3 V ar(Z)
2πn(1 − e−c )V ar(Z)
1
and likewise


2
2
X
c
(x
−
x
)
1
o
.
P
exp −
Zt = m ∼ p
−c )V ar(Z)
2(1 − e−c )3 V ar(Z)
2πn(1
−
e
t∈[ao ]
By these last two equations, equation (5.12.3) becomes
m!
X Y 1 Y 1 Y 1
δr ! s ∆s ! t δt !∆t !
r
δ,∆
"
#
c2 (x − xi )2 + (x − x0 )2
(ec − 1)2ν−νi −νo
m!
exp −
. (5.12.4)
∼ 2m ·
c
2πn(1 − e−c )V ar(Z)
2(1 − e−c )3 V ar(Z)
Recall though that we need an asymptotic formula for the sum on the RHS
in (5.12.1), so we need to engage the fudge factor H(δ, ∆). To this end, we notice that the dominant contribution to the RHS of (5.12.4) comes from δ, ∆ with
max δv , max ∆v ≤ 2 ln n in a similar fashion to the second part of Lemma 5.4.3.
Indeed if d ≥ 2 ln n, then
d! ≥ (d − [ln n])!(ln n)ln n ,
184
and the argument that led to (5.12.4) shows that the total contribution of δ, ∆ with
max δv , max ∆v ≤ 2 ln n is at most of order
n · n−1 m!(ln n)− ln n n−1 m!.
For max δv , max ∆v ≤ 2 ln n, we have the sharp McKay formula
ln H(δ, ∆) = −
1
−
2m2
1 X
∆t δt
m t
!
!
X
X
(δr )2 +
(δt )2
r
X
X
(∆s )2 +
(∆t )2
t
s
+ O(m−1 (ln n)4 ).
t
Defining H(δ, ∆) = 0 for max δv > 2 ln n or max ∆v > 2 ln n, we write
X
g(n, m, δ, ∆) =
δ,∆
m!(ec − 1)ai +ao E H(Zi , Zo )1{|Zi |=|Zo |=m)} .
2m
c
(5.12.5)
Here Zi (Zo resp.) consists of ai = ν − νi (resp. ao = ν − ν0 ) independent copies of
Z. Now, with probability 1 − O(n−b ), for any b > 0,
X
∆t δt ∼ (ν − νi − νo )E 2 [Z],
t
and
X
X
(Zri )2 +
(Zti )2 ∼ ao E[(Z)2 ],
r
X
t
s
(Zso )2 +
X
(Zto )2 ∼ ai E[(Z)2 ].
t
Since
P (|Zi | = |Zo | = m) = Θ(n−1 )
is only polynomially small, we obtain that
E H(Zi , Zo )1{|Zi |=|Zo |=m)} ∼ P (|Zi | = |Zo | = m)
ai ao 2
ν − νi − νo 2
× exp −
E [Z] −
E [(Z)2 ] .
m
2m2
185
Therefore (5.12.5) becomes
X
(ν − νi )(ν − νo ) 2
ν − νi − νo 2
E [Z] −
E [(Z)2 ]
g(n, m, δ, ∆) ∼ exp −
2
m
2m
δ,∆
"
#
c2 (x − xi )2 + (x − x0 )2
m!
(ec − 1)2ν−νi −νo
× 2m ·
exp −
. (5.12.6)
c
2πn(1 − e−c )V ar(Z)
2(1 − e−c )3 V ar(Z)
Using Stirling Formula yields
m!
(n)2
m
−c−c2 /2
∼ 2πm e
c 2m
e
.
(5.12.7)
√
Let β̄i = β̄o = e−c (1 − e−c ) and ᾱ = 1 − e−2c , so νi = β̄i n + xi n. Applying
Stirling’s Formula to νi ! gives
√
ν νi
i
νi ! = (1 + O (1/n)) 2πνi
e
q
√
1
xi
∼ 2π β̄i n νi exp (β̄i n + xi n) ln n + ln β̄i + ln 1 + √
e
β̄i n
√ 1
√
√
√
∼ Ci n νi exp β̄i n ln n + β̄i n ln β̄i + xi n ln n + xi n ln β̄i + xi n
e
1
+ x2i /β̄i − x2i /β̄i ,
2
for some constant Ci > 0. There are similar asymptotic formulas for νo !, (ν − νi − νo )!
and (n − ν)!. After some simplifying, we find that for some constant C > 0,
Cn2
νi !νo !(ν − νi − νo )!(n − ν)! = 1 + O n−1/2
en
(
√
−c
−2ce−c
−c 2−2e
+ n(2x − xi − xo ) ln (ec − 1)
× exp n ln n + n ln e
1−e
+
1
2
2
2
x2i
x2o
x
(x − xi − xo )
+
+
+
1 − ᾱ
ᾱ − β̄i − β̄o
β̄i
β̄o
)
.
(5.12.8)
Note that
(ec − 1)2ν−νi −νo
= exp
e2m
√ 2ᾱ − β̄i − β̄o n + (2x − xi − xo ) n ln (ec − 1) − 2cn ,
and
(ec − 1)2ᾱ−β̄i −β̄o e−2c = (ec − 1)2−2e
−c
186
−c
e−2c = e−2ce
1 − e−c
2−2e−c
.
Using the asymptotic behavior in equations (5.12.6), (5.12.7) and (5.12.8), we find
that (5.12.1) becomes, after simplifying,
−3/2
P (X = ν, Xi = νi , Xo = νo ) ∼ A(2πn)
1 T
exp − x Rx ,
2
for some cumbersome constant A and symmetric positive-definite 3 × 3 matrix R
defined by
c2 (x − xo )2
x2
c2 (x − xi )2
+
+
2(1 − e−c )3 V ar[Z] 2(1 − e−c )3 V ar[Z] 1 − ᾱ
(x − xi − xo )2 x2i
x2
+
+
+ o . (5.12.9)
β̄i
β̄o
ᾱ − β̄i − β̄o
√
Necessarily, A = detR and the local limit convergence above implies convergence
xT Rx =
in distribution. Namely,
X − ᾱn Xi − β̄i n Xo − β̄o n
√
√
, √
,
n
n
n
D
=⇒ N (0, K),
where K = R−1 .
From the quadratic form of R in (5.12.9), we find an explicit expression for R.
Then using Mathematica, we find that

K(c) = R−1

 A B B 


e −1
 B C D ,
= 4c c

e (2e − c − 2) 


B D C
c
where
A = 2e2c − cec − 2 − 3c,
B = (2 + c)ec − 2 − 3c,
C = 2e3c + (−4 − 2c)e2c + (4 + 3c)ec − 3c − 2,
D = −2ec + 2 + 3c.
187
5.13
Asymptotic Distribution of (ν̂, µ̂) of D(n, m = cn)
In summary, we have determined the asymptotic distribution of (ν̂, µ̂) starting from a
generic s(0) as well as the asymptotic distribution of the starting initial state induced
from D(n, m = cn). Now we combine these two results to obtain that the distribution
of (ν̂, µ̂) starting from s(D(n, m = cn)) is asymptotically jointly Gaussian as well.
Our argument is similar to the proof of Lemma 2 in Pittel [50].
Lemma 5.13.1. Let c > 1. Let ν̂, µ̂ be the number of vertices and arcs left at the end
of the process according to transition probability P̂ starting from s(D(n, m = cn)).
Then we have that
ν̂ − θ2 n µ̂ − cθ2 n
√
, √
n
n
D
=⇒ N (0, B),
where N (0, B) is a Gaussian distribution with mean 0 and covariance matrix B =
B(c), determined in the proof.
Proof. Let t1 , t2 ∈ R. We wish to show that
2
2
1 T
√ n +it2 µ̂−cθ
√ n
it1 ν̂−θ
n
n
ϕ̃(t1 , t2 ) := E e
= exp − t B t + o(1),
2
where this expectation corresponds to the process starting at s(D(n, m = cn)). Since
D(n, m = cn) conditioned on {s(D(n, m = cn)) = s} is uniform among possible
digraphs with parameter s, we have the decomposition:
2
2n
X it ν̂−θ
µ̂−cθ
√ n
1 √n +it2
n
P s(D(n, m) = s .
ϕ̃(t) =
Es e
s
Let’s denote this latter expectation in the sum by ϕ̃s (t). Since this sum is over all
possible s, we begin by breaking the sum into two parts, one for likely s and the
other for unlikely s induced from D(n, m = cn). Let a > 0 and let S(a) be the set of
s = (ν, νi , νo , µ = cn) such that
q
√
(ν − ᾱn)2 + (νi − β̄i n)2 + (νo − β̄o n)2 ≤ a n,
188
where ᾱ = 1 − e−2c and β̄i = β̄o = e−c (1 − e−c ). One can easily show that S(a) ⊂ S0 ,
where σ = c and 0 = n−1/3 . Now for the “unlikely” s ∈
/ S(a), we have that
X
X
ϕ̃
(t)P
s(D(n,
m
=
cn))
=
s
P s(D(n, m = cn)) = s
≤
s
s∈S(a)
/
s∈S(a)
/
→ P |N (0, K)| ≥ a ,
(5.13.1)
which is small if a is large.
Now we consider s in S(a); we can rewrite ϕ̃s (t) as
ϕ̃s (t) = exp itT f (s/n) − (θ2 , cθ2 )T
√ n
× Es [exp itT (ν ∗ , µ∗ )T − nf (s/n) n−1/2 ]. (5.13.2)
This last expectation is the characteristic function of centered (ν̂, µ̂) starting from s.
In the proof of Corollary 5.11.3, in (5.11.7), we found that uniformly over s ∈ S0 ,
Es [exp itT ((ν̂, µ̂) − nf (s/n)) n−1/2 ]
1 T
= exp − t ψ(s/n) t + O(n−0.49 ). (5.13.3)
2
Now uniformly over s ∈ S(a), we (trivially) have that
s/n = (ᾱ, β̄i , β̄o , c) + O(n−1/2 ).
As a consequence we have that
ψ(s/n) = ψ(ᾱ, β̄i , β̄o , c) + O(n−1/2 ).
(5.13.4)
Using the asymptotic behavior in (5.13.3) and (5.13.4), (5.13.2) becomes
T
ϕ̃s (t) = exp it
√ 1 T
f (s/n) − (θ , cθ )
n · exp − t ψt + O(n−.49 ),
2
2
2 T
where ψ is evaluated at (ᾱ, β̄i , β̄o , c).
189
(5.13.5)
However the fluctuation of fj around (ᾱ, β̄i , β̄o , c) is on the order of n−1/2 and
√
we multiply fj by n in (5.13.5), so we must take this variation into account. The
Taylor expansion of f1 around (ᾱ, β̄i , β̄o , c) gives uniformly over s ∈ S(a),
√
n f1 (s/n) − θ2 =
∇f1 (ᾱ, β̄i , β̄o , c) ·
ν − ᾱn νi − β̄i n νo − β̄o n
1
√ , √
,
, √
,0 + O √
n
n
n
n
where ∇f1 is the gradient of f1 . Similarly, we find that
√
n f2 (s/n) − cθ2 =
∇f2 (ᾱ, β̄i , β̄o , c) ·
ν − ᾱn νi − β̄i n νo − β̄o n
1
√ , √
, √
,0 + O √
.
n
n
n
n
Hence uniformly over s ∈ S(a), we have
√ exp itT (f (s/n) − (θ2 , cθ2 )T ) n =
ν − ᾱn
νi − β̄i n
νo − β̄o n
+ iu2 √
+ iu3 √
exp iu1 √
+ O(n−0.49 ), (5.13.6)
n
n
n
where u1 = t1 (f1 )α + t2 (f2 )α , u2 = t1 (f1 )βi + t2 (f2 )βi , u3 = t1 (f1 )βo + t2 (f2 )βo with
these partial derivatives being evaluated at ᾱ, β̄i , β̄o , c . Now by Lemma 5.12.1, in
conjuction with (5.13.3), (5.13.6),
lim
X
ϕ̃s (t)P s(D(n, m = cn)) = s =
n→∞
s∈S(a)
Z
1 T
1
1 T
√
exp − t ψt ×
exp iu · x − x Rx dx. (5.13.7)
3/2 det K
2
2
|x|≤a (2π)
Furthermore, we notice that
Z
1
1 T
1 T
√
exp iu · x − x Rx dx = exp − u Ku .
2
2
(2π)3/2 det K
Letting a → ∞ in (5.13.1) and (5.13.7), we obtain that
1 T
1 T
ϕ̃(t) → exp − t ψt − u Ku .
2
2
190
(5.13.8)
One can see that equation (5.13.8) can be rewritten as
1 T
ϕ̃(t) → exp − t Bt ,
2
(5.13.9)
where B(c) = (Bj,k ) is a 2 × 2 matrix defined by

 (f1 )α (f2 )α
ψ
ψ
(f
)
(f
)
(f
)
1,1
1,2
1
α
1
β
1
β
o
i

 
 
B=
+
K
 (f1 )βi (f2 )βi

ψ2,1 ψ2,2
(f2 )α (f2 )βi (f2 )βo
(f1 )βo (f2 )βo







;


again, all functions above are evaluated at
(ᾱ, β̄i , β̄o , c) = 1 − e−2c , e−c − e−2c , e−c − e−2c , c .
5.14
Asymptotic Distribution of the Strong Giant
In the previous section, we determined the asymptotic distribution of (ν̂, µ̂) starting
from s(D(n, m = cn)). However, we are interested in the number of vertices and arcs
where we allow the deletion process to continue until there are no more vertices left,
i.e. the number of vertices and arcs in the (1, 1)−core. In light of our results from
Appendix A, the proof is almost immediate.
Theorem 5.14.1. Let c > 1. Let V1,1 , A1,1 be the vertices and arcs of the (1, 1)−core
and V1 , A1 be the vertices and arcs in the largest strong component of D(n, m = cn).
Then
(i)
|V1,1 | − θ2 n |A1,1 | − cθ2 n
√
√
,
n
n
D
=⇒ N (0, B),
(5.14.1)
and
(ii)
|V1 | − θ2 n |A1 | − cθ2 n
√
√
,
n
n
D
=⇒ N (0, B),
(5.14.2)
where N (0, B) is a 2−dimensional Gaussian with mean 0 and covariance matrix B,
defined in the proof.
191
Proof. (i). An immediate consequence of Lemma 5.12.1 is that w.h.p. the induced
starting point s from D(n, m = cn) is in S0 , where σ = c and = n−1/3 . By
Lemma 5.10.3, uniformly over s ∈ S0 , processes that start at s, w.h.p. have that
ν̂i + ν̂o < (ln n)2 . Therefore, w.h.p. our terminal point induced from D(n, m = cn)
satisfies ν̂i + ν̂o < (ln n)2 . Now that the number of semi-isolated vertices is at most
(ln n)2 at the end of this stopped process. By Lemma 5.1.3 (which is proved in
the Appendix A), with high probability, if at any time the number of semi-isolated
vertices is less than (ln n)2 , then
0 ≤ ν̂ − |V1,1 | ≤ 2(ln n)4
0 ≤ µ̂ − |A1,1 | ≤ 4(ln n)6 .
Since (ln n)6 n1/2 , the pair
|V1,1 | − θ2 n |A1,1 | − cθ2 n
√
√
,
n
n
converges in distibution to N (0, B), by Lemma 5.13.1.
(ii). By Theorem 5.1.2, w.h.p.
0 ≤ |V1,1 | − |V1 | ≤ 2(ln n)8 ,
0 ≤ |A1,1 | − |A1 | ≤ 4(ln n)10 .
Since (ln n)10 is dwarfed by n1/2 , the pair
|V1 | − θ2 n |A1 | − cθ2 n
√
√
,
n
n
also converges in distribution to N (0, B).
5.15
Asymptotic Behavior of the Covariance Matrix, B(c)
Due to our integral formula for ψ, our expression for B(c) is also in integral form.
We can say considerably more about B(c) for c near but larger than 1. We have
192
previously determined the asymptotic behavior of ψj,k , (5.9.2), as c decreases to 1.
Now we need to find the find the asymptotic behavior of K and the partials of fj .
Computing this behavior is relatively simple, and we find that as c ↓ 1,

2
3e − 5
3e − 5
 2e − e − 5

e−1 
K(c) = 4
3e − 5
2e3 − 6e2 + 7e − 5
−2e + 5
e (2e − 3) 

3e − 5
−2e + 5
2e3 − 6e2 + 7e − 5



,


up to an additive error of order (c − 1) in the Euclidean norm; also, we find that, for
both j = 1 and j = 2,
−8e2 + 16e
(c − 1) + O((c − 1)2 ),
(e − 1)2
−8e
(fj )βi =
(c − 1) + O((c − 1)2 ),
(e − 1)2
−8e
(c − 1) + O((c − 1)2 ),
(fj )βo =
(e − 1)2
(fj )α =
Consequently, as c decreases
T 


 (f1 )α (f2 )α 





 (f )
 1 βi (f2 )βi  K 



(f1 )βo (f2 )βo
to 1, we have that


64
(f1 )α (f2 )α 
  (e − 1)2

(f1 )βi (f2 )βi 
=
64

(e − 1)2
(f1 )βo (f2 )βo

64
(e − 1)2 
 (c − 1)2 ,

64
(e − 1)2
up to an additive error of (c − 1)3 . Therefore as c decreases to 1, the entries of B, up
to an additive error of order (c − 1)3 ln(1/(c − 1)), are
Z 1
64
940
h(z)dz (c − 1)2 ,
+
+4
B1,1 = 40(c − 1) + −
3
(e − 1)2
0
Z 1
712
64
B1,2 = 40(c − 1) + −
+4
h(z)dz (c − 1)2 ,
+
2
3
(e − 1)
0
Z 1
484
64
B2,2 = 40(c − 1) + −
+
+4
h(z)dz (c − 1)2 ,
3
(e − 1)2
0
(5.15.1)
where h(z) analytic for |z| ≤ 1.2, and defined by
32
h(z) = 2z 3 z
− e2z + 3e3z − 3e4z + e5z + z − 4ez z + 5e2z z − 2e3z z
e z (e − 1)3
32
+ 2ez z 2 − 6e2z z 2 + 7e3z z 2 − 3e4z z 2 + e2z z 3 − 2e3z z 3 + 2e4z z 3 − 2 ,
z
193
for z 6= 0 with h(0) = −4/3. Numerically integrating h(z) gives
Z
1
h(z)dz = 0.5025...
0
Consider the asymptotics up to and including (c − 1)2 term of (5.15.1), we notice that the corresponding matrix is non-negative definite, but not positive definite.
Specifically, (−1, 1)T is an eigenvector with 0 eigenvalue of this truncated matrix.
This leads us to consider the excess of the strong giant component, which is the number of arcs minus the number of vertices of this strong giant. As a consequence of
our ultimate result, we have that
excess − (c − 1)θ2 n D
√
=⇒ N (0, σ),
n
where σ 2 (c) = B1,1 (c) + B2,2 (c) − 2B1,2 (c). We notice that (c − 1)θ2 n = (1 + O(c −
1))4(c − 1)3 n, while all we can say about the variance is that
σ (c) = O (c − 1)3 ln
2
1
c−1
.
In summary, for c > 1 fixed (does not depend on n), we have shown that the
number of vertices and arcs of the largest strong component of D(n, m = cn) is
jointly asymptotic Gaussian with mean (θ2 , cθ2 )T n and covariance matrix B(c)n.
Furthermore, we found the asymptotic behavior of the mean and covariance as c
decreases to 1. This is not to be confused with finding the size of the largest strong
component of D(n, m = cn n), where cn decreases to 1. Luczak and Seierstad [43]
investigated the case where cn is allowed to tend to 1 as n tends to ∞. They found
that if cn ↓ 1, but (cn − 1)n1/3 → ∞, then w.h.p. the largest strong component of
D(n, p = cn /n) has size (1 + o(1))4(cn − 1)2 n.
194
CHAPTER 6
EXTENSIONS
6.1
Possible Extensions to Hypergraph Results
Once the threshold for k-connectivity is established, naturally one may wonder when
the hypergraph process develops a non-empty k−connected sub(hyper)graph. Trivially, a k-connected subgraph of a hypergraph must be subgraph which has minimum
degree at least k. But if the known results about k-connected subgraphs in the
random graph models, G(n, m), G(n, p) and G̃, are any indication, determination of
when such a minimum degree k subgraph develops may be sufficient.
The k-core of a hypergraph as the maximal subgraph with minimum degree at
least k, as long as this subgraph is non-empty. Bollobás [7], who introduced the
notion of k−core in graphs, showed that for k ≥ 5 and c = c(k) sufficiently large,
with high probability, the k-core of G(n, m = cn/2) exists and is k-connected. As
previously mentioned, Pittel [50] showed that for c > 1, with high probability, the
2-core exists and has size linear in n. Later, Luczak [38], [40] proved that for all
k ≥ 3, with high probability the graph process G̃ is such that when the k-core
exists, it is k-connected and contains at least 0.0002n vertices. Pittel, Spencer, and
Wormald [54] established the sharp threshold, m(n) = ck n, for existence of a k-core
in G(n, m), as well as finding the concentration of the size of the k-core when it first
becomes present. More recently, Janson and M. Luczak [27] proved that the number
195
of vertices and edges in the k-core in G(n, p) and G(n, m), properly centered and
scaled, is asymptotically Gaussian.
Some of these results for the random graph models have been extended to hypergraphs. Cooper [13] considered k-cores of d-uniform hypergraphs with a given degree
sequence. Molloy [46] partially extended the results of Pittel et al.[54], by finding constants cdk for d ≥ 3, k ≥ 2 such that if c < cdk , then w.h.p. Hd (n, p = c(d − 1)!/nd−1 )
has empty k-core and if c > cdk , then w.h.p. Hd (n, p = c(d − 1)!/nd−1 ) has a k-core
with size linear in n. As an immediate consequence, there are analogous results for
Hd (n, m = cn/d). One particular difference between hypergraphs and graphs is that
the existence of a 2-core is a sharp threshold for hypergraphs and when present is
linear in n, which contrasts with the graph case where the threshold is not sharp (we
discuss the 2-core in graphs more in the following section on digraphs). As in [54],
one may naturally suspect that one can allow c to tend to cdk slowly to obtain similar
results. Extending the result from Janson et al.[27], we suspect the following is true:
Conjecture 6.1.1. Suppose k ≥ 2 and c > cdk . If (N, M ) is the ordered pair of
the number of vertices and hyperedges of the k-core in Hd (n, m = cn/d), then there
are some positive constants ν = ν(d, k, c) and µ = µ(d, k, c) and a 2 × 2 matrix
Γ = Γ(d, k, c) such that
N − νn M − µn
√
, √
n
n
D
=⇒ N (0, Γ) ,
where N (0, Γ) is the Gaussian distribution with mean 0 and covariance matrix Γ.
Furthermore, if Tk is the stopping time for when the hypergraph process, H̃,
develops a non-trivial k-core, then we would suspect that the number of vertices in
the k-core in H̃ at Tk divided by n converges to a constant in probability, as in Pittel
et al.[54]. Moreover, as in Janson et al.[27], we suspect the following conjecture is
true:
196
Conjecture 6.1.2. Let k ≥ 2. There is some σ = σ(d, k) > 0 such that
Tk − cdk n/d D
2
=⇒
N
0,
σ
,
n1/2
where N 0, σ 2 is a normal distribution with mean 0 and variance σ 2 .
In Chapter 3, we proved that w.h.p. the diameter of the hypergraph process at
the moment of connectivity takes on at most nine deterministic values. We also found
a likely range of the diameter of Hd (n, m) for m on the order of n ln n. One obvious
extension would be to determine the likely size of the diameter for m n ln n. For
instance, we would expect that the diameter is concentrated on only a couple values,
similar to the graph cases, G(n, p) and G(n, m) (see Burtin [10] and Bollobás [5]).
For a connected hypergraph, the diameter is the largest distance between a pair of
vertices. It is natural to study a notion of diameter for disconnected hypergraphs. We
define the true diameter of a hypergraph to be the maximum diameter among all its
components; for a connected hypergraph, these two definitions of diameter obviously
coincide. Most authors simply use the term diameter for this new definition, but to
avoid confusion here, we employ the substitute term true diameter.
Research by Korshunov [34], Luczak [41], Chung and Lu [14], Fernholz and Ramachandran [22], Nachmias and Peres [47], Luczak and Seierstad [42], and Riordan
and Wormald [58] have looked at this true diameter of G(n, p), G(n, m) and G̃. We
would expect many of these results to have hypergraph analogues. In particular,
similar to Fernholz et al.[22], we expect:
Conjecture 6.1.3. For c > 1, w.h.p. the true diameter of Hd (n, p = c(d − 1)!/nd−1 )
is the diameter of its largest (giant) component.
One crucial difference between these two definitions of “diameter” is that while
the diameter is (non-strictly) decreasing with the addition of more hyperedges, the
197
true diameter does not behave as nicely (if a new hyperedge connects some components, the true diameter of the hypergraph may increase!). Luczak and Seierstad [42]
considered the true diameter of the graph process G̃ and found that the the maximum
value of the true diameter is on the order of n1/3 and occurs during the phase transition (m ≈ n/2). We would expect to find that the maximum true diameter of the
hypergraph process also occurs at the phase transition, m ≈ n/d. Roughly speaking,
the true diameter of G(n, p) is about ln n/| ln(np)| for wide ranges of the parameter p
(note that the maximum true diameter occurs near np ≈ 1). For comparison, we have
shown that for p = c(d − 1)! ln n/nd−1 , where c > 1 fixed, the diameter of Hd (n, p) is
concentrated around
ln n
ln
.
p
n
d−1
In Chapter 4, we found that with high probability the difference between the
stopping time for the hypergraph process having a weak Hamilton cycle, denoted T ,
and the stopping time for the process having minimum degree at least 1, denoted
τ1 , is o(n). The nature of our argument (incrementally increasing the edge set by
batches) rules out, in our opinion, a possibility to refine our method to reduce o(n)
to 0, which we believe is the likely difference. However, the likely size of the batches
in our argument is on the order of ln n and we conjecture that our proof could be
significantly refined to show that w.h.p. T − τ1 < ω(n) ln n, for any ω → ∞. If so, we
can say that w.h.p. the stopping time for having a weak Hamilton cycle is separated
from τ1 by at most ω(n) ln n.
To proceed, we considered weak Hamilton cycles because this notion of a cycle
allowed repeated hyperedges which in turn allowed us to consider path rotations
and apply Pósa’s Lemma in a manner similar to the proof of likely Hamiltonicity of
G(n, p = (ln n + ω)/n). For instance, if we are only interested in longest distinct
hyperedge paths, Pósa sets need not be non-expanding. As such, our method here
198
can not be refined to find the threshold for the existence of a Hamilton cycle (where
the hyperedges of the cycle are distinct). However, we conjecture that one may be
able to show that a Hamilton cycle appears “pretty soon” after the first moment the
hypergraph process has minimum degree 2 by adding hyperedges; moreover, just as
in Bollobás [7], we suspect that w.h.p. the cycle appears precisely at the first moment
where minimum degree is 2.
Recent research of Hamiltonicity in random hypergraphs has primarily focused on
l-overlapping cycles rather than our cycles and weak cycles. An l-overlapping cycle
is a sequence of vertices in hyperedges which in turn overlap by l vertices. See Figure
6.1 for a clarifying example of an l-overlapping cycle. Of particular interest are loose
cycles (1-overlapping cycle) and tight cycles ((d − 1)-overlapping cycle).
v3
v2
v4
v1
v5
v9
v6
v8
v7
Figure 6.1: A 2-overlapping cycle in a 5-uniform hypergraph
Research by Frieze [25] and Dudek and Frieze [15], [16] have found thresholds
(and in case of tight cycles, sharp thresholds) for the existence of a l-overlapping
Hamilton cycle. This is hardly an exhaustive list of results about l-overlapping cycles,
but are closest to our results here. As in this work, one may wish to find the main
199
barrier to l-overlapping Hamiltonicity in the context that the primary barrier to weak
Hamilonicity is the existence of isolated vertices.
In our proof of our ultimate weak Hamiltonicity result in Chapter 4, we considered the probability, denoted P (a, b), that each of b fixed vertices is in at least one
hyperedge with at least one of a fixed other vertices, where potential hyperedges are
present with probability p. The upper bound that we obtained reminds us of a very
elegant and powerful inequality in random graphs known as Stepanov’s Inequality.
Simply stated, if P (n, p) denotes the probability that the Bernoulli random graph,
G(n, p), is connected, then
P (n, p) ≤ 1 − (1 − p)n−1
n−1
.
(6.1.1)
This inequality was originally shown by Stepanov [59]. More recently, Biskup et
al.[3] proved this inequality in the following argument: the probability that G(n, p) is
connected is equal to the probability that each vertex in the directed random graph
D(n, p) is reachable from a fixed vertex v. This latter probability is less than the
probability that each vertex other than v has in-degree at least 1, which is precisely
the bound in (6.1.1).
We can adapt this method as well as a greedy algorithm similar to the greedy algorithm in the proof of bounding P (a, b) to yield the following Stepanov-like inequality
for random d−uniform hypergraphs.
Lemma 6.1.4. Let Pd (n, p) denote the probability that Hd (n, p) is connected. Then
n−1
d−1
Pd (n, p) ≤ 1 − (1 − p)(2 −1)( d−1 )
d n−1
e
d−1
.
(6.1.2)
Note. For d = 2, this bound is precisely Stepanov’s Inequality.
Proof. This inequality will be established by considering a version of a “directed”
hypergraph. There are a few natural ways to define a directed hypergraph. Here, a
200
(hyper)arc, denoted ~e = (T, H), is a pair of sets of vertices comprising of a tail set,
T , and a head set, H, such that the head and tail are non-empty and disjoint with
|T | + |H| = d. We view the arc as starting at the tail set and ending at the head
set. Let Dd (n, p) be a random directed hypergraph on [n] where each possible arc
is present with probability p independently of each other. This particular defintion
of a directed hypergraph allows us to notice that the probability that Hd (n, p) is
connected is equal to the probability that each vertex in Dd (n, p) is reachable from a
fixed vertex v (one can use a breadth-first search to see this). Furthermore, we notice
that
Pd (n, p) ≤ P (each vertex other than v is in at least one head set in Dd (n, p)) .
For a specific vertex, w, notice that the number of pairs of non-empty disjoint sets,
T and H, such that T, H ⊂ [n] and w ∈ H, is equal to
n−1
d−1
2
−1
,
d−1
because you can choose the other d − 1 elements first, then you can distribute these
d − 1 elements among the tail and head in any way as long as you do not put them
all in the head. Hence
P (w is in a head set in Dd (n, p)) = 1 − q (2
)(n−1
d−1 ) .
d−1 −1
(6.1.3)
For the graph case, the events in (6.1.3) for different vertices are independent which
gives rise to Stepanov’s Inequality. However for d ≥ 3, these events are definitely
not independent. We can deal with the dependence of the events in a similar fashion
to our greedy algorithm in the proof of the inequality for P (a, b). In particular, we
start at a vertex (other than v) and look at possible arcs containing this vertex one
after another until we discover a present arc. Then we move to an unseen vertex and
continue looking for present arcs, and so on. Since each head contains at most d − 1
201
n − 1
d−1
arcs in this greedy algorithm. This in turns gives rise to the inequality in (6.1.2), as
vertices and we wish to cover all n − 1 vertices, we can discover at least
desired.
We suspect that this Stepanov-like Inequality could be improved, specifically that
2d−1 − 1 could be significantly lowered. Perhaps a more algebraic proof of Stepanov’s
Inequality such as the proof by Knuth and Schönhage [32] could be adapted to the
random hypergraph model. Lastly, we suspect that the numerous extensions and generalizations of Stepanov’s Inequality by various authors (for instance, see Lomonosov
and Polesskii [37], Takács [61], Biskup et al.[3], and Pittel [53]) could possibly have
hypergraph analogues.
6.2
Possible Extensions to Digraph Results
For a discussion about extensions of our digraph results, we begin with the definition
of the random digraph process. Similar to the hypergraph process model, the random
n(n−1) digraph process, D̃ or {D(n, m)}m=0 , is a random sequence of digraphs, where
D(n, 0) is the empty digraph on [n], and D(n, m + 1) is obtained from D(n, m) by
inserting an extra arc chosen uniformly at random among all n(n−1)−m non-present
arcs.
The (1, 1)-core of a digraph is the maximal sub(di)graph of minimum in-degree
and out-degree at least 1. This (1, 1)-core mimics some of the behavior of the 2-core
in undirected graphs. For example, these two cores exist if and only if a cycle is
present. When the 2-core is first born in the graph process, G̃, it is precisely the first
cycle. If TG denotes the number of edges present when this first cycle is created, then
for each fixed x ∈ (0, 1), we have that
lim P (TG ≥ xn) =
n→∞
202
√
1 − xex/2+x
2 /4
;
see Flajolet et al. [23]. As a consequence, there can not be a sharp threshold for
existence of a 2-core. Surprisingly, although the size of the first cycle of the graph
process is bounded in probability, Flajolet et al. [23] proved that the expected size of
the first cycle is on the order of n1/6 .
One can show that typically in the digraph process, when a cycle first appears it
is from a single birth (see Figure 6.2 for an unlikely example of a twin birth of first
cycles after the addition of arc ~e). As a consequence, it is likely that the (1, 1)-core
at its birth is simply this first unique cycle.
~e
Figure 6.2: An unlikely birth of a first cycle
Although as in the undirected case, the first cycle is bounded in probability,
Jaworski and Luczak [30] found that the expected size of the first cycle is on the
order of ln n rather than n1/6 . Furthermore, if TD denotes the number of arcs present
when the first cycle is born, then for each x ∈ (0, 1) fixed,
lim P (TD ≥ xn) = (1 − x) ex .
n→∞
So just as for the 2-core, the threshold for the existence of the (1, 1)−core is not
sharp.
Let’s generalize the notion of k-cores to digraphs. For ki , ko ∈ N, we say that the
(ki , ko )-core of a digraph is the maximal subgraph of minimum in-degree at least ki
203
and minimum out-degree at least ko . We expect the question of existence of (ki , ko )core when ki + ko ≥ 3 to be similar to the question of existence of k-core for k ≥ 3.
Specifically as in Pittel et al.[54], we expect a sharp threshold for the existence of
a non-empty (ki , ko )-core, and when the core does exists, it has size linear in n.
Formally, for ki , ko ≥ 1 with ki and ko not both 1, there are constants cki ,ko such that
• If c < cki ,ko , then w.h.p. D(n, m = cn) does not have a (ki , ko )-core.
• If c > cki ,ko , then w.h.p. D(n, m = cn) has a (ki , ko )-core. Furthermore, there
is some positive constant, γ = γ(c), such that the size of the core divided by n
tends to γ in probability.
Our earlier comments rule out the above behavior for the (1, 1)-core. Using a first
moment argument similar to Luczak’s method [40] in the undirected graph models
counting subgraphs in which each vertex has degree at least k, one can show that
w.h.p. if a (ki , ko )-core appears in the first O(n) steps in the digraph process, then
necessarily it must have size linear in n. However, trying to prove existence of a
(ki , ko )-core for m/n large enough (but fixed) requires a much more careful analysis
than this first moment method.
One may naturally wonder if there are digraph analogues to the results that we
have shown for hypergraphs. For (fixed) k ∈ N, Frieze [24] proved that at the first
moment the random digraph process, D̃, has minimum in-degree and minimum outdegree at least k, the process develops k arc-disjoint Hamilton cycles; in fact, he found
a randomized polynomial time algorithm which likely finds these cycles. Notice that
(trivially) k arc-disjoint Hamilton cycles could not be present prior to this moment.
As an obvious consequence, the digraph likely becomes strongly connected at the first
moment that the digraph process has minimum in-degree and minimum out-degree
at least 1.
204
As one may suspect, there is a k-(vertex)connected counterpart for digraphs; we
say that a digraph is k-strongly connected if whichever k − 1 vertices are deleted, the
induced digraph is still strongly connected. Similar to Bollobás-Thomason Theorem
[9] for k−connectivity in the graph process, one may conjecture:
Conjecture 6.2.1. Let k = k(n) ∈ N with k ≤ n − 1. With high probability, at
the first moment that the random digraph process, D̃, has minimum in-degree and
minimum out-degree at least k, this digraph is k-strongly connected.
As mentioned before, determining the strong components is not so much a local
property but a global one. This makes arguments concerning strong connectedness a
little more delicate. We have found a way to prove this conjecture for fixed k, and
the proof carefully considers these source sets and sink sets.
205
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210
Appendix A
FINDING THE (1, 1)−CORE SUFFICES
A.1
Overview
We have two goals in this appendix. First, we want to show that the (1, 1)−core
of D(n, m = cn) is little more than the largest strong component. Second, we wish
to show that we can stop the deletion process early and not be too far off from the
(1, 1)−core. Namely, we prove:
Theorem A.1.1. Fix c > 1. If L1 = (V1 , A1 ) is the largest component of D(n, m =
cn) and L1,1 = (V1,1 , A1,1 ) is the (1, 1)−core, then w.h.p.
0 ≤ |V1,1 | − |V1 | ≤ 2(ln n)8 ,
0 ≤ |A1,1 | − |A1 | ≤ 4(ln n)10 ,
and
Lemma A.1.2. Fix c > 1. With high probablity, D(n, m = cn) is such that whenever
D(t), the digraph after some t deletions, has at most (ln n)2 semi-isolated vertices,
then
0 ≤ |V (D(t))| − |V1,1 | ≤ 2(ln n)4 ,
0 ≤ |A(D(t))| − |A1,1 | ≤ 4(ln n)6 .
We do not try to find the best possible bounds above. The key is that these upper
bounds are dwarfed by the natural fluctuations of the strong giant component with
are on the order of n1/2 .
211
In the proof of these two results, we will utilize a depth-first search on the random digraph D(n, p = c/n). We switch to the Bernoulli digraph model because the
independent nature of potential arcs makes analyzing this search much easier, just
like when we looked at a breadth-first search in random hypergraph in Chapter 3.
At the end, we will use a standard conversion technique to switch our results from
D(n, p = c/n) to D(n, m = cn).
Another reason that we switched to the Bernoulli model is that the following
decomposition will be indispensable. Following the idea of Luczak and Seierstad [43],
in D(n, p), let L = (V, A) be the induced sub(di)graph on the vertices either in cycles
of length at least (ln n)2 or is a descendant of a vertex in such a cycle. Viewed another
way, L is the smallest sink-set containing the cycles of length at least (ln n)2 . The
random model D(n, p) conditioned on L = L0 = (V0 , A0 ) is distributed as
• the set of arcs that have both endpoints in V0 is exactly A0 ,
• the potential arcs from [n]\V0 to V0 are present with probability p independently
of one another,
• there are no arcs from V0 to [n] \ V0 (or else more vertices should be added to
L), and
• the arc set within [n] \ V0 is distributed as D([n] \ V0 , p) conditioned on the
event that there are no cycles of length at least (ln n)2 .
Since no vertex of [n] \ V0 can be reached from V0 , the condition that [n] \ V0 does not
contain cycles of length at least (ln n)2 does not impose any constraints on potential
arcs from [n] \ V0 to V0 .
We will prove that w.h.p. L is, in fact, the largest strong component and its
descendant set; however, our current definition of L allows simpler conditioning on
212
the decomposition above. Now vertices in the (1, 1)−core but outside of L must be
either
• in a cycle within [n] \ V , or
• the descendant of a cycle within [n] \ V .
To be able to show that the number of vertices in the (1, 1)−core but outside L is
small, we establish the concentration of the number of vertices of L, then using the
decomposition, we show that w.h.p. the number of vertices in cycles or a descendant
of a cycle within [n] \ V is at most (ln n)8 . Furthermore, we will show that w.h.p.
maximum in-degree and maximum out-degree are at most (ln n)2 . Combining all
these results, we have that w.h.p. the number of vertices in the (1, 1)−core, but not
in the smallest sink-set containing the largest strong component, is at most (ln n)8 and
the number of arcs on these vertices is at most 2(ln n)10 . By symmetry, we have the
analogous result for the smallest source-set containing the largest strong component.
These two results finish the proof of Theorem A.1.1. The proof of Lemma A.1.2 will
be nearly immediate following the proof of the theorem.
The bulk of the proof here is to show the concentration of number of vertices of
L. Namely, we will show:
Lemma A.1.3. Let c > 1. Then in D(n, p = c/n), quite surely
|V (L)| ∈ [θn − 2n1/2 ln n, θn + 3n1/2 ln n],
where θ = θ(c) is the unique root of 1 − θ = e−cθ .
Note: The 2 and 3 in the above are artifacts from the proof, any positive constants
should work.
The proof of the concentration of |V (L)| will entail performing a directed version
of a depth-first search on the whole digraph. The typical size of |V (L)| will follow from
213
the usual structure of this depth-first search. The next section details the depth-first
search on a digraph.
A.2
The Depth-First Search
We start with an overview of the depth-first search. We start at the lowest index
vertex and look at potential arcs starting at this vertex one after another. If we
find a present arc, we move to this descendant and start looking at potential arcs
from this newly found vertex; where if we find another arc, we again move to the
descendant and continue the search. Whenever we do not find an arc we move back
one vertex and continue searching where we left off last time. Eventually, we will
come back to the initial vertex and discover no more arcs. At this point, we pick
the lowest index vertex among all undiscovered vertices and search over from there.
Our analysis requires a detailed understanding of this search and for the rest of this
section we will give a formal description of the depth-first search.
Let D be a digraph on V = {v1 , v2 , . . . , vn }. We will define vertices w1 , w2 , . . . , wn ,
along with subgraphs L1 , L2 , . . . by doing a depth-first search (furthermore denoted
DFS) on D. These wi0 s are defined in the order in which we discover them in the
search. The subgraph L1 will be the subgraph found starting at v1 , L2 will be
subgraph of starting at the first vertex after leaving L1 , and so on.
First, we add a vertex v0 =: w0 and each arc (v0 , vi ), 1 ≤ i ≤ n to D. We begin
the DFS at v0 . We start with W (0), L1 (0), L2 (0), . . . empty. This W (t) will be the
vertices that we have found during the depth-first search up to step t.
We know that (v0 , v1 ) is a arc, so we define w1 := v1 and we add vertex w1 to
L1 (0) to get L1 (1) and define W (1) = {w1 }. Next, we look at possible arcs of the
form (w1 , vi ) where vi ∈ V \ W one after another (starting at lowest index) until we
either (a) find one is present or (b) find that none exist.
214
For (a), suppose this lowest index vertex is vi . We define w2 := vi and define
W (2) = W (1) ∪ {w2 } and add vertex w2 and arc (w1 , w2 ) to L1 (1) obtaining L1 (2).
And then we continue the search from w2 .
For (b), we define L1 (2) = L1 (1), W (2) = W (1) and we go back one vertex. We
continue the search from v0 to V \ W (2) and we start L2 .
The general step: suppose we have finished with step t, so W (t), L1 (t), L2 (t), . . .
are defined. Note that we have just moved to some wi (by either just defining wi or
returning back to wi ) and that so far we have defined w0 , w1 , . . . , wk and non-empty
L1 (t), . . . , Ll (t). If we returned to wi , then we have previously checked some potential
arcs, E, of the form (wi , ·). If we have just defined wi , let E be empty. We now check
arcs of the form (wi , v), where v ∈ V \ W (t) which are not in E, one after another
(starting at lowest index) until we either (a) find one is present or (b) find that none
exists.
For (a), suppose that the arc that we find is (wi , vj ). We define wk+1 := vj and
let W (t + 1) = W (t) ∪ {wk+1 }. If wi ∈ Ll (t), we add wk+1 and (wi , wk+1 ) to Ll (t)
obtaining Ll (t + 1). Otherwise wi must be v0 and we now start Ll+1 (t + 1) with just
vertex wk+1 with no arcs. For the rest of the process, L1 , L2 , . . . , Ll will stay the
same. If W (t + 1) = V , then we are done with the DFS and we stop.
For (b), necessarily wi 6= v0 and we have seen (wi0 , wi ) is present for some i0 < i.
We define W (t + 1) = W (t), Ll (t + 1) = Ll (t) and move back to wi0 and continue the
search.
At the end of this process, we let DFS = (W, E), where E is the set of present
arcs that we have seen from this process. We can only have arcs of the form (wi , wj ),
for i < j, and necessarily DFS is a subgraph of D. Let E 0 = {e ∈ E(D) : e =
(wi < wj ) for some i < j} and DFS 0 = (W, E 0 ). We can find E 0 by doing the DFS
and then checking arcs of the form (wi , wj ), i < j, which we have not checked yet.
215
Furthermore, let E 00 = {e ∈ E(D) : e = (wj , wi ), i < j}. During the DFS, we never
look at arcs of the form of E 00 . Note that E 0 ∪ E 00 = E(D).
For ease of notation, let desc(w) denote the descendant set of w in the digraph
and s-desc(w) denote the descendant set in the depth-first search. Trivially, we have
that s-desc(w) ⊂ desc(w).
In the next section, we begin with a discussion about the structure of the depthfirst search when we start with D(n, p).
A.3
Concentration of the size of L
If we disregard direction, then DFS in D(n, p) is equal in distribution to the depthfirst search in G(n, p)! This is because for each pair of vertices {v, w}, we check at
most one of (v, w) and (w, v). Moreover, each time we check a potential arc, it is
present with probability p. This sort of correspondence betwen D(n, p) and G(n, p)
has been gainfully used before, particularly in Karp [31] and Biskup et al. [3]. In this
correspondence, the L0i s correspond to the components in G(n, p). This gives rise to
the natural definition of the L0i s as search components or s−components. Similarly,
if we disregard direction, DFS 0 in D(n, p) is equal in distribution to G(n, p). This
correspondence will allow us to convert results from G(n, p) to our depth-first search
on D(n, p).
Unless otherwise stated, the probabilities involved are for the probability space of
D(n, p = c/n) for some fixed c > 1. We say that an event, E, holds quite surely, if
for each fixed a > 0, for all n sufficiently large (depends on a),
P (E) ≥ 1 − n−a .
2
First, note that if P (E) ≥ 1 − e−b(ln n) for some b > 0, then E holds quite surely.
216
Our analysis of the depth-first search will utilize a “gap” theorem on the sizes of
components of G(n, p). Using Stepanov’s Inequality,
P (G(k, p) is connected) ≤ (1 − q k−1 )k−1 ,
q = 1 − p,
one can prove the following (see Pittel [50]),
Proposition A.3.1. Let c > 1. Quite surely, G(n, p = c/n) has exactly one component of size in
[θn − n1/2 ln n, θn + n1/2 ln n]
and every other component has size less than (ln n)2 .
We will use this proposition a few times. Let’s begin by showing that
Claim A.3.2. Quite surely, for all i ≤ (ln n)10 , the size of s-desc(wi ) is either less
than (ln n)2 or in
[θn − 2n1/2 ln n, θn + 2n1/2 ln n].
Proof. Once we define wi (conditioned on the prehistory), the size of the descendant set in the DFS (not to be confused with its descendant set in the digraph!) is
equal in distribution to the size of the component of a specific vertex in G(n−i+1, p).
These events are definitely not independent for different i, but an union bound is
helpful here.
For ease of calculation, both here and later, let
1
1
In = [(ln n)2 , n] \ [θn − n 2 ln n, θn + n 2 ln n]],
Kn = [θn − 2n1/2 ln n, θn + 2n1/2 ln n],
Jn = [(ln n)2 , n] \ Kn .
In particular, we have Jn ⊂ In−i+1 ∪ [n − i + 1, n], for 1 ≤ i ≤ (ln n)10 for n large
enough.
217
For i ≤ (ln n)10 , we have that
P size of s-desc(wi ) ∈ Jn ≤ P size of s-desc(wi ) ∈ In−i+1 .
Now by this correspondence as well as Proposition A.3.1, this latter probability is at
most (n − i + 1)−a for each fixed a > 0. Hence by the union bound,
(ln n)10
P
∪ {size of s-desc(wi ) ∈ Jn } ≤ (ln n)10 (n − (ln n)10 )−a ≤ n−a+1 .
i=1
Claim A.3.3. Quite surely, at least one of w1 , w2 , . . . , w(ln n)5 has descendant set of
size in
Kn = [θn − 2n1/2 ln n, θn + 2n1/2 ln n]
in the depth-first search.
Proof. First let’s show that quite surely, one of the first (ln n)5 vertices of [n] is
in the largest component in G(n, p = c/n). Then we will use the correspondence to
get the typical structure of the depth-first search in D(n, p = c/n).
Let V be the vertex set of the largest component of G(n, p) and let ν = |V|. For
vertex subsets, A and B, of the same size, we have that P V = A = P V = B ; so V
conditioned on the value of ν is uniformly distributed on sets of size ν. In particular,
P V ∩ {v1 , v2 , . . . , vk } = ∅ν = l = P V ⊂ {vk+1 , . . . , vn }ν = l
n−k . n
=
≤ exp (−kl/n) .
l
l
Uniformly over l ∈ Kn , we have that
0
5
P V ∩ {v1 , v2 , . . . , v(ln n)5 } = ∅ν = l ≤ e−b (ln n) ,
for some b0 > 0. As a consequence, we have that
0
5
P V ∩ {v1 , . . . , v(ln n)5 } = ∅, ν ∈ Kn ≤ e−b (ln n) .
218
By Proposition A.3.1, we have that quite surely ν ∈ Kn . As a consequence, for all
a > 0, we have that
0
5
P V ∩ {v1 , . . . , v(ln n)5 = ∅ ≤ e−b (ln n) + n−a .
Hence quite surely at least one of v1 , v2 , . . . , v(ln n)5 is in some giant component of size
in Kn . Now we convert this result about G(n, p) to our case here.
In the undirected depth-first search, the trees found are the spanning trees of
components. This implies that the size of component containing vi in G(n, p) is
equal in distribution to the size of the s−component containing vi . Since whenever we return to v0 in the depth-first search we pick the lowest index, vi is in one
of L1 , L2 , . . . , Li . In particular, {v1 , v2 , . . . , vk } ⊂ L1 ∪ L2 . . . ∪ Lk . Hence quite
surely, one of L1 , L2 , . . . , L(ln n)2 has size in Kn . By Claim A.3.2, quite surely, each
of w1 , w2 , . . . , w(ln n)5 either have a s-descendant set of size less than (ln n)2 or in Kn .
Also, the size of Li is equal to the size of the s-descendant set of the vertex that
starts Li . If wj starts Li and |s-desc(wj )| = k, then wj+k starts Li+1 . By pigeonhole principle, if for 1 ≤ j ≤ (ln n)5 , |s-desc(wj )| < (ln n)2 , then the size of each
Li , 1 ≤ i ≤ (ln n)3 − 1 is less than (ln n)2 . Hence one of w1 , w2 , . . . , w(ln n)5 has a
s-descendant set of size in Kn quite surely.
Claim A.3.4. Quite surely, maximum out-degree of D(n, p = c/n) is at most (ln n)2 .
Proof. By an union bound,
P max out-deg ≥ (ln n)2 ≤ nP out-deg(v1 ) ≥ (ln n)2 .
(A.3.1)
Note that
2
P out-deg(v1 ) ≥ (ln n) ] =
X n − 1
pk q n−1−k ≤
k
2
k≥(ln n)
219
X
k≥(ln n)2
ck
.
k!
For k ≥ (ln n)2 ,
c
c
ck+1 /(k + 1)!
1
=
≤
.
≤
ck /k!
k+1
(ln n)2
2
Hence
2
2
P v1 has out-deg ≥ (ln n)
≤
X
k≥(ln n)2
c(ln n)
((ln n)2 )!
k−(ln n)2
2
1
c(ln n)
.
≤2
2
((ln n)2 )!
So equation (A.3.1) becomes
2
2
P max out-deg ≥ (ln n)
c(ln n)
≤ 2n
≤ 2n
((ln n)2 )!
ec
(ln n)2
(ln n)2
≤ exp −(ln n)2 ,
where in the penultimate inequality, we use Stirling’s formula for ((ln n)2 )!.
Let’s consider the event C = Cn , where
• each of w1 , w2 , . . . , w(ln n)10 have a s-descendant set of size either less than (ln n)2
or in Kn ,
• at least one of w1 , . . . , w(ln n)5 has a s-descendant set size in Kn , and
• maximum out-degree is less than (ln n)2 .
We have just shown that C holds quite surely. The next two results give some implications about the structure of the depth-first search on the likely event C.
Claim A.3.5. On C, if for some i < (ln n)10 − (ln n)4 , the size of s-desc(wi ) is in
Kn , then exactly one of its children has s-descendant set of size in Kn . Moreover this
child, wk , with large descendant set has index k < i + (ln n)4 .
Proof. On the set C, when doing the depth first search, the number of children of
wi is bounded above by the degree of wi in the digraph. Suppose i < (ln n)10 −(ln n)4 ,
with size of descendant set of wi in Kn . There are at most (ln n)2 children of wi . Note
that we have the following equivalence:
|s-desc(wi )| = 1 +
X
wl :children of wi
220
|s-desc(wl )|.
In particular, if the size of s-desc(wi ) and s-desc(wl ) are in Kn , where wl is one of
wi0 s children, then wl is only child of wi with large s-descendant set.
Suppose no child of wi has a s-descendant set with size in Kn . Let wk1 , wk2 , . . .
be the children of wi . Necessarily k1 = i + 1 < (ln n)10 with s-descendant set with
size less than (ln n)2 , so k2 ≤ k1 + (ln n)2 < (ln n)10 . We can keep doing this since
kj < i + j(ln n)2 < (ln n)10 for 1 ≤ j ≤ (ln n)2 . This implies that size of the sdescendant set of wi is at most 1 + (ln n)2 · (ln n)2 , which is a contradiction.
Claim A.3.6. On C, there will be a path of length at least (ln n)3 in depth-first search
where each vertex, along this path, has s-descendant set with size in Kn .
Proof. Let y1 be the lowest index element of {w1 , w2 , . . .} such that the size
of s-desc(w) is in Kn . On C, y1 exists and y1 = wi for some i ≤ (ln n)5 . For
2 ≤ j ≤ (ln n)3 , we define yj recursively; we let yj be the child of yj−1 which has
descendant set in DFS of size in Kn . To finish the proof, we need to establish that
these yj0 s are well-defined on C, which we do by induction.
Suppose that yj is defined. By Claim A.3.5, we can define yj+1 as long as yj = wi
for some i < (ln n)10 − (ln n)4 . Since y1 = wi for some i < (ln n)5 , by the same claim
yj = wk for some
k < (ln n)5 + (j − 1)(ln n)4 ≤ (ln n)5 + (ln n)7 < (ln n)10 − (ln n)4 ,
so we can define yj+1 . In particular y(ln n)3 is defined on C and the size of s-desc(y(ln n)3 )
is in Kn .
Now that we have sufficient information about the structure of the depth-first
search on the event C, we can now say more about the likely structure of the depthfirst search.
Claim A.3.7. Quite surely, vertices defined after leaving y(ln n)3 for the last time have
small s-descendant set size.
221
Proof. By the correspondence between DFS in D(n, p) and G(n, p),
P |s-desc(wi )| ≥ (ln n)2
= P size of component of vertex 1 in G(n − i + 1, p = c/n) ≥ (ln n)2 .
For any i ≥ θn − 2n1/2 ln n, we have
P size of component of vertex 1 in G(n − i + 1, p) ≥ (ln n)2
≤ P size of component of vertex 1 in G((1 − θ)n + 2n1/2 ln n, p) ≥ (ln n)2
For n sufficiently large, we definitely have that
(1 − θ)n + 2n1/2 ln n ≤ n,
so that
P size of descendant of wi in DF S ≥ (ln n)2
≤ P size of component of vertex 1 in G(n0 , p = c0 /n0 ) ≥ (ln n0 )2 ,
where n0 = (1 − θ)n + 2n1/2 ln n and
p=
c
c(n0 /n)
=
.
n
n0
One can show from the definition of θ that c(1 − θ) < 1. So for any α ∈ (c(1 − θ), 1),
and n sufficiently large, we have that the probability that the size of s-desc(wi ) is at
least (ln n)2 is at most
P (size of component of vertex 1 in G(n0 , p = α/n0 ) ≥ (ln n0 )2 ).
By a simple first moment argument on the number of trees with size at least (ln n0 )2 ,
0
0 2
one can show this latter probability is bounded above by e−b (ln n ) for some b0 > 0.
222
Since ln n0 ∼ ln n, for any b00 such that 0 < b00 < b0 , for sufficiently large n, we have
that
0
0 2
00 (ln n)2
P (|s-desc(wi )| ≥ (ln n)2 ) ≤ e−b (ln n ) ≤ e−b
.
By taking a union bound over i in [θn − 2n1/2 ln n, n], we have that


n
[
00
2
2
P
|s-desc(wi )| ≥ (ln n)2  ≤ ne−b (ln n) ≤ e−b(ln n) ,
i=θn−2n1/2
(A.3.2)
ln n
for any 0 < b < b00 and n sufficiently large. On the set C, if there is some vertex, wi ,
defined after leaving s−desc(y(ln n)3 ), then necessarily wi has index i ≥ θn−2n1/2 ln n.
2
By equation A.3.2, this probability is less than e−b(ln n) .
Now we have sufficient information about the depth-first search to complete the
proof of the concentration of the number of vertices of L. Namely,
Lemma A.3.8. Quite surely,
|V (L) ∈ [θn − 2n1/2 ln n, θn + 3n1/2 ln n].
Proof. We wish to augment our event C with the likely behavior discovered in the
last claim. Let D be the event where
• C occurs,
• each of the L0i s either have size less than (ln n)2 or in Kn , and
• each vertex defined after leaving s-desc(y(ln n)3 ) has s-descendant set size less
than (ln n)2 .
Note that we have found that quite surely D occurs. We began by noticing some
consequences of these likely events on the structure of the depth-first search. First,
D forces an upper bound on the number of vertices in L.
223
Claim A.3.9. On D,
|V (L)| ≤ θn + 3n1/2 ln n.
Proof. Consider the set D. Exaclty one of the children of v0 (specifically y1 ) has
s-descendant set of size Kn , while all the other children have s-descendant set sizes
less than (ln n)2 . When running a DFS on a digraph, all strong components (and so
all cycles) must be in the same s−component. So all cycles of length at least (ln n)2
in the digraph must reside in the s−component starting with y1 . Let Lk be this large
s−component; necessarily k < (ln n)2 . The descendant set of Lk in the digraph is Lk
with possibly some vertices from L1 , L2 , . . . , Lk−1 . On D, the number of vertices in
Lj , j < k must be less than (ln n)2 . The number of vertices either in cycles of length
at least (ln n)2 or a descendant of such a cycle has size less than |L1 |+|L2 |+. . .+|Lk |.
In other words, we have that
|V (L)| ≤ θn + 2n1/2 (ln n) + (k − 1)(ln n)2 ≤ θn + 3n1/2 ln n.
Now we turn our attention to finding a lower bound on |V (L)|. Consider the event
that DFS, the digraph after running the depth-first search, is equal to some D0 ∈ D.
Now there is a path from y1 to y(ln n)3 along the yi0 s and that B := s-desc(y(ln n)3 ) has
size at least θn − 2n1/2 ln n. If there is an arc from B to one of {y1 , y2 , . . . , y(ln n)2 },
then there is a cycle of length at least
(ln n)3 − (ln n)2 ≥ (ln n)2 ,
which contains y(ln n)3 and so on the event in question,
|V (L)| > |B| ≥ θn − 2n1/2 ln n.
224
Hence
P |V (L)| < θn − 2n1/2 ln n|DFS = D0 ≤
P no arc from B → {y1 , . . . , y(ln n)2 }|DFS = D0 .
Conditioned on the DFS, potential arcs of the form (wj , wi ), with i < j, are present
with probability p independently of one another. Hence
2
P no arc from B → {y1 , . . . , y(ln n)2 }|DFS = D0 = q |B|(ln n) .
Since |B| ≥ θn − 2n1/2 ln n, we have that
q
|B|(ln n)2
c
1/2
2
≤ exp − θn − 2n ln n (ln n) ≤ exp −b0 (ln n)2 ,
n
for any b0 < c(1 − θ) and n sufficiently large, independent of the choice of D0 in D.
Hence we have that
0
2
P {|V (L)| < θn − 2n1/2 ln n} ∩ D ≤ e−b (ln n) .
Therefore, quite surely,
|V (L)| ≥ θn − 2n1/2 ln n,
as desired.
A.4
Number of vertices in cycles
Now that we have the concentration of the number of vertices of L, we turn our
attention to D(n − |V (L)|, p). As a reminder, the vertices in [n] \ V (L) that are in
the (1, 1)−core must necessarily be in a cycle or be the descendant of a cycle within
[n] \ V (L). Furthermore, conditioned on L, the induced digraph on [n] \ V (L) is
distributed as D(n − |V (L)|, p) conditioned on the event that there is not a cycle of
length at least (ln n)2 , which we will find is a very likely event.
225
For ease of notation, let In := [θn − 2n1/2 ln n, θn + 3n1/2 ln n], which is the likely
range of |V (L)|.
Claim A.4.1. Uniformly over ν ∈ In , quite surely, there are no cycles in D(n−ν, p =
c/n) of length ≥ (ln n)2 .
Proof. We can couple D(n0 , p) as an induced subgraph of D(n00 , p) for n0 < n00 , so
it suffices to find this bound when ν = θn − 2n1/2 ln n (note that our p depends on
n and not on n0 and n00 ). By first moment method, the probability that there exists
a cycle of length at least (ln n)2 is less than the expectation of the number of cycles
of length at least (ln n)2 . Let E1 denote the expectation of cycles of length at least
(ln n)2 in D(n − θn + 2n1/2 ln(n), p).
X n − θn + 2n1/2 ln(n)
E1 =
E[# of cycles of length l] =
(l − 1)!pl
l
l≥(ln n)2
l≥(ln n)2
l
X 1
2 ln(n)
c 1 − θ + 1/2
.
≤
l
n
2
X
l≥(ln n)
As previously noted c(1 − θ) < 1, so there is some = (c) > 0 such that c(1 − θ) ≤
2 ln(n)
1 − . So for n sufficiently large, c 1 − θ + 1/2
≤ 1 − . Hence we have that
n
E1 ≤
X 1
2
1
2
(1 − )l ≤b
(1 − )(ln n) ≤ e−b(ln n) ,
2
l
(ln n)
2
l≥(ln n)
for some b > 0.
Since the event that we condition on is extremely likely, we may as well assume
that we are not conditioning at all. For example, for any event, the difference between
the probability in the conditioned probability space versus the probability in the
unconditioned space is at most O(n−a ), for any fixed a > 0. The next few claims
build up to the likely event that the number of vertices in cycles or descendants of
cycles is relatively small in the (unconditioned) random digraph D(n − |V (L)|, p).
226
Claim A.4.2. There is some M = M (c) > 0, such that for ν ∈ In , with probability
at least 1 − M/n, every strong component in D(n − ν, p = c/n) is either a cycle or
an isolated vertex.
Proof. As in Luczak and Seierstad [43], one can see that any strong component
other than a cycle or single vertex contains a pair of directed cycles whose intersection
is a directed path (even if this directed path is a single vertex). Note that on k vertices
there are at most k 2 k! such possible pairs of cycles. Also, there must be at least k + 1
arcs present in these pairs. Hence
n X
n − ν 2 k+1
P (∃ such a strong component) ≤
k k!p .
k
k=3
Note that we have
ln n
n−ν
≤ 1 − θ + 2 1/2
n
n
and as before
2 ln n
c 1 − θ + 1/2
n
≤1−
for some > 0. So
k
n
n
n X
X
1X 2
n−ν
n − ν 2 k+1
2
k
ck (1 − )k .
k k!p
≤p
np ≤
n
n
k
k=3
k=3
k=3
We choose M =
X
ck 2 (1−)k (which converges) to finish the proof of this claim.
k≥3
As a consequence of this last claim, with sufficiently large probability, any present
cycles must be on distinct vertices. Our next claim shows that there are not many
disjoint small cycles present in D(n − |V (L)|, p).
Claim A.4.3. Uniformly over ν ∈ In , quite surely, there are less than (ln n)2 disjoint
cycles of length k, for k ∈ [2, (ln n)2 ], in D(n − ν, p = c/n).
227
Proof. By first moment method,
2
n−ν
2
P (∃(ln n) distinct cycles of length k) ≤
((k − 1)!)(ln n) pk(ln n)
k, k, . . . , k
k(ln n)2
2 ln n
≤ c 1 − θ + 1/2
n
2
2
≤ (1 − )k(ln n) .
A union bound over k yields the desired result.
Now we combine the previous three claims to show that there are not many vertices
in cycles.
Claim A.4.4. With probability at least 1 − M 0 /n for some M 0 > 0, for any ν ∈ In ,
the number of vertices, in D(n − ν, p), in cycles is less than (ln n)6 .
Proof. By Claim A.4.3, quite surely, for each k ∈ 2, (ln n)2 , there are at most
(ln n)2 disjoint cycles of length k. On this event, there are at most (ln n)6 vertices
M
, any two
in these disjoint cycles. By Claim A.4.2, with probability at least 1 −
n
cycles must be disjoint. By Claim A.4.1, quite surely, there are no cycles of length
≥ (ln n)2 .
A.5
Descendant sets
Now that we have a bound on the number of vertices in cycles in D(n − |V (L)|, p), we
turn our attention to finding likely upper bounds on the descendant sets of vertices.
Claim A.5.1. Uniformly over ν ∈ In , quite surely, any set of vertices, V0 , has total
descendant set of size less than |V0 |(ln n)2 in D(n − ν, p = c/n).
228
Proof. For a specific vertex vi ∈ [n], the descendant size of vi in D(n, p) is equal
in distribution to the size of component of vi in G(n, p). So
P (vi has descendant size ≥ (ln n)2 in D(n − ν, p))
= P (vi in component of size ≥ (ln n)2 in G(n − ν, p)).
As we increase the vertex set (with the same parameter p), we increase the likelihood
that vi is a large component. Namely,
P (vi in component of size ≥ (ln n)2 in G(n − ν, p))
≤ P (vi in component of size ≥ (ln n)2 in G(n − θn + 2n1/2 ln n, p)).
We determined an upper bound on this latter probability in the proof of Claim A.3.7.
We found that this latter probability is bounded by e−b
00 (ln n)2
, for some b00 > 0. A
union bound over all vertices completes the proof.
We now have sufficient information about D(n, p = c/n) to be able to say that the
number of vertices in the (1, 1)−core but not in L is at most polynomial logarithmic.
Lemma A.5.2. With probability at least 1−M/n, for some M (c) > 0, D(n, p = c/n)
is such that the descendant set of cycles in [n] \ L is less than (ln n)8 .
Proof. From Lemma A.3.8, we have that quite surely, V (L) ∈ In . Along with our
decomposition of D(n, p), Claim A.4.4 bounds the number of vertices in cycles, in
[n] \ V (L), by (ln n)6 and then Claim A.5.1 bounds the descendant set size of these
vertices by (ln n)6 · (ln n)2 = (ln n)8 , as desired.
Now before we convert these results to D(n, m = cn), we need to show that
likely L is the largest strong component along with its descendant set as previously
expected. Specifically, it suffices to show that vertices in cycles of length at least
(ln n)2 are in the same strong component with sufficiently high probability.
229
Claim A.5.3. Quite surely, in D(n, p = c/n), vertices with descendant sets at least
(ln n)2 have a directed path to vertices with ancestor set at least (ln n)2 .
Proof. Quite surely, the ancestor set and descendant set of any vertex have size
in [1, (ln n)2 ] ∪ θn − n1/2 ln n, θn + n1/2 ln n . Let v, w be vertices of [n]. Then for
any ν, ν 0 ∈ θn − n1/2 ln(n), θn + n1/2 ln(n) ,
P (v has ν descendants, w has ν 0 ancestors, these sets don’t intersect) =
P (w has ancestor set of size ν 0 in D(n − ν, p))P (v has ν descendants).
2
However this probability is less than e−a(ln n) by the proof in Claim A.5.1. Union
bound over all such pairs v, w gives the desired result. As a consequence, with sufficiently large probability, any vertices in cycles of length at least (ln n)2 are in the
same strong component.
Finally, we can prove our desired result for D(n, p = c/n).
Lemma A.5.4. With probability at least 1 − M/n, for some M = M (c) > 0, the
(1, 1)−core of D(n, p = c/n) contains at most 2(ln n)8 vertices and at most 4(ln n)10
arcs outside of its largest strong component.
Proof. As a consequence of Lemma A.5.2 and Claim A.5.3, with sufficiently large
probability, L is the largest strong component and its descendant set and the number
of vertices outside of L but in the (1, 1)−core is bounded by (ln n)8 . By Claim
A.3.4, each of these vertices have out-degree at most (ln n)2 with sufficiently large
probability. Analogously, each of these vertices have in-degree at most (ln n)2 , so the
total number of arcs on these vertices of the (1, 1)−core outside L is at most 2(ln n)10 .
There is a similar treatment about L∗ , the set of vertices in cycles of length at least
(ln n)2 and its ancestor set. Again, we will find that the number of vertices and arcs
of the (1, 1)−core outside of L∗ is at most (ln n)8 and 2(ln n)10 , respectatively. All
230
that remains is to note that, again with sufficiently large probability, the intersection
of L∗ and L is the largest strong component.
To convert the previous lemma to D(n, m = cn), we use a standard conversion
technique similar to Lemma 2.2.6 for hypergraphs.
Lemma A.5.5. Fix c > 1. Suppose Q is some subset of digraphs such that the
probability that D(n, p = c/n) is in Q is at least 1−o(n−1/2 ), then w.h.p. D(n, m = cn)
also is in Q.
Proof. First, for a digraph, D, we define e(D) to be the number of arcs in D. We
have that
n(n−1)
P (D(n, p) ∈
/ Q) =
X
P (D(n, m) ∈
/ Q)P (e(D(n, p)) = m)
m=0
≥ P (D(n, m = cn) ∈
/ Q)P (e(D(n, p)) = cn).
In particular,
P (D(n, m = cn) ∈
/ Q) ≤
P (D(n, p) ∈
/ Q)
.
P (e(D(n, p)) = cn)
Since each potential arc of D(n, p) is present independently of one another, the number of present arcs, denoted e(D(n, p)), is binomially distributed with n(n − 1) trials
and success probability p. Now we wish to determine asymptotics on P (e(D(n, p)) =
cn) for p = c/n. Notice that
n(n − 1) cn n(n−1)−cn
P (e(D(n, p)) = cn) =
p q
.
cn
Now we have that
cn cn−1 Y
n(n − 1)
i
n(n − 1)
1−
=
cn
cn!
n(n − 1)
i=0
!
2 cn −c cn−1
X
n
e
1
e cn
i
√
= 1+O
exp −
.
n
cn
n(n − 1)
2πcn
i=0
231
Hence
n(n − 1)
ncn e−c e cn −c2 /2
∼√
e
.
cn
2πcn c
We also have that
n(n − 1) − cn − p − p2 /2 + O(n−3 )
= exp −c(n − 1) + c2 − c2 /2 + O(n−1 )
= exp −cn + c + c2 /2 + o(1) .
q n(n−1)−cn = exp
Combining these asymptotics, we have that
1
e−c −c2 /2+c+c2 /2
e
=√
,
P (e(D(n, p)) = cn) ∼ √
2πcn
2πcn
which implies that
P (D(n, m = cn) ∈
/ Q) ≤b
√
nP (D(n, p) ∈
/ Q),
and by assumption, P (D(n, p) ∈
/ Q) = o(n−1/2 ). As a consequence, with high probability, D(n, m = cn) is in Q.
The proof that the (1, 1)−core of D(n, m = cn) is little more than the largest
strong component (Theorem A.1.1) follows from applying the conversion lemma
(Lemma A.5.5) to Lemma A.5.4.
A.6
Can stop the deletion process early
In this section, we will show that likely we can stop the deletion process early and
not be too far off from the (1, 1)−core where we allow the process to continue. For
this, we begin with discussing the deletion process.
We start with a digraph D(0), with some vertices of in-degree zero, Oi (0), and
some vertices of out-degree zero, Oo (0). At each step, we delete either a vertex with
in-degree zero or out-degree zero. Let D(t), Oi (t), Oo (t) be these parameters after the
232
t0 th step (you can delete these according to any rule you desired, unlike the way we
randomly delete them).
Any vertex in L will never have in-degree zero during the process, because either
the vertex is in a cycle of length (ln n)2 and never deleted or it is a descendant of such
a cycle. These descendants can have out-degree zero at some point in the process
and eventually be deleted, but never will it have in-degree zero. In other words, for
any t ≥ 0, we have that
Oi (t) ⊂ [n] \ L.
Claim A.6.1. If t1 < t2 , then Oi (t2 ) is contained in the union of Oi (t1 ) and the
descendant set of Oi (t1 ).
Proof. Note that if B is in the descendant set of A, then the descendant set of B
is contained in the descendant set of A as well. As a consequence, it suffices to show
that for any t, we have Oi (t + 1) is contained in the union of Oi (t) and its descendant
set. We break this into cases into whether we initially delete a vertex in (i) Oi (t) or
in (ii) Oo (t) \ Oi (t).
(i) So we have deleted a vertex of Oi (t). The new vertices that have in-degree
zero must have had an arc coming from the deleted vertex. In particular, these new
in-degree zero vertices are descendants of Oi (t).
(ii) If we delete a vertex of zero out-degree, then the in-degree of each remaining
vertex stays the same. Hence Oi (t + 1) = Oi (t).
Claim A.6.2. Quite surely, D(n, p = c/n) is such that for any order of deletions, if
for some t ≥ 0, |Oi (t)| < (ln n)2 , then at most (ln n)4 vertices and 2(ln n)6 arcs will
be deleted for having in degree zero for the rest of the process.
Proof. Note that for each t ≥ 0, the union of Oi (s) over s ≥ t is exactly the vertices
233
deleted sometime during the process for having out degree zero. The previous claim
states that for all t ≥ 0, we have that
∪ Oi (s) ⊂ Oi (t) ∪ desc(Oi (t)),
s≥t
where desc(Oi (t)) denotes the descendant set of Oi (t) in [n] \ L. By Claim A.5.1,
with sufficiently large probability, any set of vertices in [n] \ L of size at most (ln n)2
has at most (ln n)2 · (ln n)2 = (ln n)4 descendants. As before, with sufficiently high
probability, each vertex has at most 2(ln n)2 arcs incident to it.
The proof of the next lemma is immediate from the previous claim as well as the
corresponding claim about Oo (t).
Lemma A.6.3. Let c > 1. Quite surely, D(n, p = c/n) is such that whenever D(t),
the digraph after some t deletions, has at most (ln n)2 semi-isolated vertices, then at
most 2(ln n)4 vertices and 4(ln n)6 arcs will be deleted in the remainder of the deletion
process finding the (1, 1)−core.
Now the conversion to Lemma A.1.2 follows from the application of Lemma A.5.5.
234
Appendix B
PARTIAL DERIVATIVES FROM DELETION PROCESS
In Chapter 5, we found asymptotic expressions for certain random variables according
to our transition probability. We gave the method to find these on page 164. Each
of the following sums are over a = 1, t = 0, b ≥ 0, r ≥ 0 and k ≥ max{1, b + r}. The
approximations below are up to an additive term of O(n−1 ).
νi
,
νi + νo
X
νo zi zo
ezo
νo zi zo
νi
2
1 + zi
,
b qi ≈
νi + νo ezo − 1 (ezi − 1)µ
(e − 1)µ
X
νi
ezo
νo zi zo
abqi ≈
,
z
z
o
νi + νo e − 1 (e o − 1)µ
X
ezo ν − νi − νo zi zo
νi
,
arqi ≈
νi + νo e z o − 1 e z i − 1
µ
X
atqi = 0,
X
νi
zo ezo
akqi ≈
,
νi + νo e z o − 1
X
ezo
νo zi zo (ν − νi − νo )zi zo
νi
brqi ≈
,
z
z
o
νi + νo e − 1 (e i − 1)µ
(ezi − 1)µ
X
btqi = 0,
X
νi
ezo
νo zi zo
bkqi ≈
(1 + zo ),
z
z
o
νi + νo e − 1 (e i − 1)µ
X
a2 q i ≈
235
X
νi
ezo ν − νi − νo zi zo
r qi ≈
νi + νo ezo − 1 ezi − 1
µ
X
rtqi = 0,
X
2
rkqi ≈
ν − νi − νo zi zo
1+
,
ezi − 1
µ
νi
ezo ν − νi − νo zi zo
(1 + zo ) ,
νi + νo ezo − 1 ezi − 1
µ
X
t2 qi = 0,
X
tkqi = 0,
X
k 2 qi ≈
νi
zo ezo
(1 + zo ) .
νi + νo ezo − 1
One can easily find the analogous sums over qo by switching i and o. This leads
us to the following approximate expectations below:
2
E[a ] :=
E[b2 ] :=
E[ab] :=
E[ar] :=
E[at] :=
E[ak] :=
E[br] :=
E[bt] :=
E[bk] :=
E[r2 ] :=
νi
νo
ezi
νi zi zo
νi zi zo
+
1 + zo
,
νi + νo νi + νo ezi − 1 (ezo − 1)µ
(e − 1)µ
νo
νi
ezo
νo zi zo
νo zi zo
+
1 + zi
,
νi + νo νi + νo ezo − 1 (ezi − 1)µ
(e − 1)µ
νi
νo
νi zi zo
e zo
νo zi zo
ezi
,
+
z
z
z
z
νi + νo e o − 1 (e o − 1)µ νi + νo e i − 1 (e i − 1)µ
νi
ezo ν − νi − νo zi zo
,
νi + νo e z o − 1 e z i − 1
µ
νo
νi zi zo (ν − νi − νo )zi zo
e zi
,
z
z
i
νi + νo e − 1 (e o − 1)µ
(ezo − 1)µ
νi
νi zi zo
zo ezo
νo
e zi
+
(1 + zi ),
z
z
z
o
i
νi + νo e − 1 νi + νo e − 1 (e o − 1)µ
νi
e zo
νo zi zo (ν − νi − νo )zi zo
,
z
z
o
νi + νo e − 1 (e i − 1)µ
(ezi − 1)µ
νo
ezi ν − νi − νo zi zo
,
νi + νo ezi − 1 ezo − 1
µ
νi
e zo
νo zi zo
νo
zi ezi
(1
+
z
)
+
,
o
νi + νo ezo − 1 (ezi − 1)µ
νi + νo e z i − 1
νi
ezo ν − νi − νo zi zo
ν − νi − νo zi zo
1+
,
νi + νo e z o − 1 e z i − 1
µ
ezi − 1
µ
E[rt] := 0,
E[rk] :=
νi
ezo ν − νi − νo zi zo
(1 + zo ) ,
νi + νo e z o − 1 e z i − 1
µ
236
νo
ezi ν − νi − νo zi zo
ν − νi − νo zi zo
E[t ] :=
1+
,
νi + νo ezi − 1 ezo − 1
µ
ezo − 1
µ
νo
ezi ν − νi − νo zi zo
E[tk] :=
(1 + zo ) ,
νi + νo ezi − 1 ezo − 1
µ
zo ezo
νo
zi ezi
νi
(1
+
z
)
+
E[k 2 ] :=
(1 + zi ) .
o
νi + νo ezo − 1
νi + νo ezi − 1
2
Now we wish to find the partial derivatives of f1 and f2 with respect to the
variables α, βi , βo and γ. As a reminder, zi , zo , z ∗ are defined as roots of the equations:
zo ezo
γ
γ
zi ezi
,
=
,
=
z
z
o
i
e −1
α − βi
e −1
α − βo
∗
z ∗ ez
γ(α − βi − βo )
=
,
∗
z
e −1
(α − βi )(α − βo )
and f1 and f2 are defined by
f1 =
γ
∗
z ∗ − z ∗ e−z ,
zi zo
f2 =
γ
(z ∗ )2 .
zi zo
First, we find the partials of zi , zo , and z ∗ with respect to these variables.
(ezi − 1)2
−γ
∂zi
= zi zi
·
,
∂α
e (e − 1 − zi ) (α − βi )2
(ezi − 1)2
γ
∂zi
= zi zi
,
·
∂βi
e (e − 1 − zi ) (α − βi )2
∂zi
= 0,
∂βo
1
∂zi
(ezi − 1)2
= zi zi
·
,
∂γ
e (e − 1 − zi ) α − βi
∂zo
(ezo − 1)2
= zo zo
∂α
e (e − 1 − zo )
∂zo
= 0,
∂βi
∂zo
(ezo − 1)2
= zo zo
∂βo
e (e − 1 − zo )
·
−γ
,
(α − βo )2
γ
,
(α − βo )2
∂zo
(ezo − 1)2
1
= zo zo
·
,
∂γ
e (e − 1 − zo ) α − βo
237
·
and
2
∗
ez − 1
∂z ∗
= z∗ z∗
∂α
e (e − 1 − z ∗ )
2
∗
ez − 1
∂z ∗
= z∗ z∗
∂βi
e (e − 1 − z ∗ )
2
∗
ez − 1
∂z ∗
= z∗ z∗
∂βo
e (e − 1 − z ∗ )
2
∗
ez − 1
∂z ∗
= z∗ z∗
∂γ
e (e − 1 − z ∗ )
γ (α − βi ) (α − βo ) − γ (α − βi − βo ) (2α − βi − βo )
,
(α − βi )2 (α − βo )2
−βo γ
,
·
(α − βo ) (α − βi )2
−βi γ
,
·
(α − βi ) (α − βo )2
γ (α − βi − βo )
·
.
(α − βi ) (α − βo )
·
Now we can determine the partial derivatives of fi in terms of the partials already
found. So
∂f1
−γ ∗
γ
∗ ∂zi
∗
∗ −z ∗ ∂zo
= 2
z − z ∗ e−z
−
z
−
z
e
∂α
zi zo
∂α zi zo2
∂α
∗
γ
∂z
∗
∗
+
,
1 − e−z + z ∗ e−z
zi zo
∂α
∗
−γ ∗
γ
∂f1
∗ ∂zi
∗
∗ ∂z
= 2
+
,
z − z ∗ e−z
1 − e−z + z ∗ e−z
∂βi
zi zo
∂βi zi zo
∂βi
∗
∂f1
−γ ∗
γ
−z ∗
∗ −z ∗ ∂z
∗ −z ∗ ∂zo
=
+
1
−
e
+
z
e
,
z
−
z
e
∂βo
zi zo2
∂βo zi zo
∂βo
∗
z ∗ − z ∗ e−z
γ
γ
∂f1
∗ ∂zi
∗
∗ −z ∗ ∂zo
=
− 2
−
z ∗ − z ∗ e−z
z
−
z
e
∂γ
zi zo
zi zo
∂γ
zi zo2
∂γ
∗
γ
∂z
∗
∗
+
1 − e−z + z ∗ e−z
,
zi zo
∂γ
and
∂f2
∂α
∂f2
∂βi
∂f2
∂βo
∂f2
∂γ
−γ ∗ 2 ∂zi
γ
2γz ∗ ∂z ∗
∗ 2 ∂zo
(z
)
−
(z
)
+
,
zi2 zo
∂α zi zo2
∂α
zi zo ∂α
−γ
∂zi 2γz ∗ ∂z ∗
= 2 (z ∗ )2
+
,
zi zo
∂βi
zi zo ∂βi
−γ ∗ 2 ∂zo 2γz ∗ ∂z ∗
(z )
=
+
,
zi zo2
∂βo
zi zo ∂βo
(z ∗ )2
γ
∂zi
γ
2γz ∗ ∂z ∗
∗ 2 ∂zo
=
− 2 (z ∗ )2
−
(z
)
+
.
zi zo
zi zo
∂γ
zi zo2
∂γ
zi zo ∂γ
=
238
Appendix C
GLOSSARY OF TERMS AND NOTATIONS
In the following glossary, Page refers to the page number where the term is formally
defined and Description is a brief explanation of the term.
Notation
Page
(1, 1)−core
116
Description
Maximal subdigraph that has minimum in-degree
and out-degree at least 1
Ancestor
118
w is an ancestor of v if there is path from w to v
Binomial, Bin
22
Binomial Random Variable
Breadth-First
48
A search algorithm which finds all the vertices at
Search, BFS
distance i from a given vertex
Ceiling, dxe
68
The smallest integer greater or equal to x
Component
2
A maximally connected subgraph
Connected Hyper-
2
A hypergraph where there is a path between any
graph
(Berge) Cycle
two vertices
8
A path whose starting and ending vertex are the
same
Cycle (digraph)
9
A (directed) path whose starting and ending vertex are the same
239
Notation
Page
D(n, p)
10
Description
The random digraph on vertex set [n], where arcs
are present with probability p independently
D(n, m)
10
The random digraph distributed uniformly on all
digraphs with vertex set [n] with m arcs
202
D̃
The random digraph process where arcs are
added one after another
d−uniform hyper-
1
graph
Decreasing
A hypergraph in which each hyperedge has d vertices
prop-
4
erty
A property which is closed under deletion of (hyper)edges
Degree, deg(v)
Descendant
2
118
Number of (hyper)edges incident to a vertex v
w is a descendant of v if there is a path from v
to w
Diameter
7
Largest distance between any two vertices
Digraph
9
A graph with directed edges
Distance
7
Length of shortest path
Es [◦]
138
Expectation for the process starting at s
Esq [◦]
138
“Expectation” for the process over the substochastic transition probability qi + qo .
e(H)
2
Number of (hyper)edges of H
E(n, d)
1
Collection of all subsets of [n] of cardinality d
13
Max number of hyperedges incident to v whose
Essential
ess-deg(v)
Degree,
pairwise intersections are {v}
240
Notation
Page
Description
15
n(n − 1)(n − 2) · · · (n − r + 1)
Floor, bxc
35
Largest integer less than or equal to x
g(s)
123
The number of digraphs with parameter s
Falling
Factorial,
(n)r
Graph
1
A hypergraph in which each hyperedge has cardinality 2
Hd (n, p)
2
The random hypergraph where hyperedges are
present independently with probability p
Hd (n, m)
2
The random hypergraph distributed uniformly on
hypergraphs on [n] with m hyperedges
H̃n , H̃
2
The random hypergraph process where hyper
edges are added one at a time
Hamilton cycle (in
8
Cycle spanning all vertices
1
A ordered pair, (V, E), of a set of vertices, V and
a hypergraph)
Hypergraph
a set of hyperedges, E, which are subsets of V
In-degree,
in-
9
Number of arcs leading into vertex v
deg(v)
In-Neighbor
Increasing property
118
w is an in-neighbor of v if (w, v) is a present arc
4
A property which is closed under addition of (hyper)edges
Isolated Vertex
k−connected
116
A vertex with zero in-degree and zero out-degree
6
After deletion of any k − 1 vertices, the hypergraph is still connected
241
Notation
Page
Description
L1
114
Largest strong component
L1,1
116
The (1, 1)−core of a digraph
[n]
1
The set of positive integers less than or equal to
n
(in)
(out)
NS,S 0 , NS,S 0
120
Number of digraphs, D, which can end in D0 after
one deletion step
Non-expanding Set
77
A set of vertices, A, which is adjacent to less than
2|A| neighbors
o(·), O(·), ≤b , Θ(·)
Oi , Oo
5
116
Landau notation for asymptotic bounds
Set of vertices with zero in-degree, zero outdegree
Out-degree,
out-
9
Number of arcs leaving vertex v
deg(v)
Out-Neighbor
118
w is an out-neighbor of v if (v, w) is a present arc
P ∗ , Es∗ [◦]
142
Probability and Expectation for the process
which stops upon leaving S
P̂ , Ês [◦]
171
Probability and Expectation for the process
which stops upon leaving S
Path (in a digraph)
9
Sequence of arcs which connect a starting vertex
to an ending vertex
Path (in a hyper-
2
graph)
Poisson
Sequence of vertices, whose consecutive vertices
are in hyperedges
15
A Poisson Random Variable
242
Notation
Page
Pósa set
83
Description
Set of endpoints of longest paths formed by rotations
Property
3
Quite surely
216
A subset of (hyper)graphs
With probability at least 1 − n−a , for any fixed
a>0
Semi-isolated
116
A vertex with zero in-degree or zero out-degree
3
The first moment that the hypergraph process,
vertex
Stopping time
H̃, develops some property
Strong Component
9
A maximal strongly connected sub(di)graph
Subgraph (of G)
2
A graph whose vertex and edge set is contained
in the vertex and edge set of some larger graph
G
Θ(·)
5
Notation for “on the order of”
Tk
6
Stopping Time for H̃n being k−connectedness
τk
6
Stopping Time for H̃n having minimum degree at
least k
τk0
14
Stopping Time for H̃n having minimum essential
degree at least k
Threshold,
Sharp
4
Threshold
Parameter value where the limiting probability of
a property jumps from 0 to 1
Trivial Component
2
A component consisting of a single isolated vertex
Trivial
9
A strong component consisting of a single vertex
Strong
Component
243
Notation
u(H)
Page
84
Description
Size of smallest non-expanding set of non-isolated
vertices
Weak cycle
8
A cycle in which hyperedges are allowed to be
repeated
Weak Hamilton cy-
8
A weak cycle spanning all vertices
3
With probability tending to 1
cle
With high probability, w.h.p.
[xk ]
129
Linear operator which returns the coefficient by
xk of a power series
244