The random planar graph process
Stefanie Gerke
∗
Dirk Schlatter
†
Angelika Steger
∗
Anusch Taraz
‡
Abstract
We consider the following variant of the classical random graph process introduced by
Erdős and Rényi. Starting with an empty graph on n vertices, choose the next edge uniformly
at random among all edges not yet considered, but only insert it if the graph remains planar.
We show that for all ε > 0, with high probability, θ(n2 ) edges have to be tested before the
number of edges in the graph reaches (1 + ε)n. At this point, the graph is connected with
high probability and contains a linear number of induced copies of any fixed connected planar
graph, the first property being in contrast and the second one in accordance with the uniform
random planar graph model.
1
Introduction
N
The study of the random graph
process (Gn,t )t=0 , where one starts with an empty graph on n
n
vertices and adds all N := 2 edges in a random order, was initiated by Erdős and Rényi in
a series of papers almost 50 years ago. During the past decades, there has been a wealth of
fascinating results in the area, and although some problems still remain unsolved, the model in
general seems to be well understood. But comparatively little is known about variants of this
process, where extra conditions have to be satisfied when inserting the edges. These conditions
distort the randomness in such a way that the methods and tools employed for the original case
are of little use.
A constrained random graph process (Pn,t )N
t=0 is a random graph process equipped with an
additional acceptance test: after we have randomly chosen the edge to be inserted, we check
whether the present graph together with this edge preserves a certain (usually structural) property.
If so, we take it, otherwise we reject it (and never look at it again).
Special cases which have been considered include the properties triangle-freeness and cyclefreeness. In both cases, the outcome of these random graph processes differ significantly from
the corresponding uniform models, where each graph of the respective class is equally likely.
Erdős, Suen, and Winkler [7] have shown that the outcome of the random triangle-free graph
process has with high probability only O(n3/2 log n) edges, whereas a well-known result by Erdős,
Kleitman, and Rothschild [5] states that a uniformly chosen random triangle-free graph is with
high probability bipartite and has θ(n2 ) edges. Aldous [1] investigated the random cycle-free graph
process (Tn,t )N
t=0 and showed amongst other results that the number of leaves in the resulting tree
is concentrated around approximately 0.406n. By noting that the leaves of a tree are precisely
those vertices whose label does not appear in a Prüfer code, it is easy to see that the number of
leaves in a tree chosen uniformly at random is concentrated around n/e ≈ 0.368n.
In this paper, our requirement is planarity, and we are mainly interested in the evolution of
this constrained random process. It will become crucial to understand how the following two
different parametrizations of the process are related. The first one, Pn,t0 , denotes the random
planar graph obtained after t0 edges have been considered. Pn,m=m0 , on the other hand, describes
∗ ETH
Zurich, Institute of Theoretical Computer Science, 8092 Zurich, Switzerland, {sgerke,steger}@inf.ethz.ch
University Berlin, Institute of Computer Science, Unter den Linden 6, 10099 Berlin, Germany,
[email protected]. Research supported by the Deutsche Forschungsgemeinschaft within the European graduate program “Combinatorics, Geometry, and Computation” (No. GRK 588/2)
‡ Technical University München, Centre for Mathematical Sciences, Boltzmannstr. 3, 85747 Garching bei
München, Germany, [email protected]
† Humboldt
1
the random planar graph after m0 edges have been accepted. As edges between vertices in different
components are always accepted, it is obvious that Tn,t ⊆ Pn,t ⊆ Gn,t for all t = 0, . . . , N . Thus,
after the connectivity threshold for Gn,t —which lies at t = n log n/2—Pn,t must have at least n−1
edges with high probability. The following theorem, which states that we have to consider Ω(n2 )
edges before (1 + ε)n edges have been accepted, may thus seem somewhat surprising.
Theorem 1.1. For every ε > 0, there exists δ > 0 such that
P e Pn,δn2 ≥ (1 + ε)n < e−n .
The uniform model of random planar graphs has found considerable attention in the literature
over the past decade [3, 4, 9, 10, 11, 14, 15]. Recently, Giménez and Noy [12] gave rather precise
asymptotic expressions for both the number of simple labelled planar graphs with n vertices and
dn edges, and the number of those which are connected. These results yield an analytic expression
for the probability that a uniform random planar graph with dn edges is connected. As it turns
out, this probability is bounded away from 0 and 1 for every 1 < d < 3. From Theorem 1.1, we
can immediately infer that this is not true for Pn,m=dn .
Theorem 1.2. For every 1 < d < 3,
P [Pn,m=dn is connected ] −→ 1 as n −→ ∞.
Before we state our next result, let us introduce some terminology. A copy of a labelled graph
H with vertex set [h] := {1, . . . , h} in a labelled graph G with vertex set [n] is a subgraph H 0
of G which is isomorphic to H such that the corresponding bijection between the vertex sets is
increasing. We note that in the literature the last condition is usually not required but for us it is
more convenient to work with this slightly stronger definition. An induced copy of H in G is then
a copy of H in G which is induced as a subgraph in G. If H is connected, we define an H-leaf
in G to be an induced copy H 0 of H which is connected to the rest of G only via a single edge
incident with the smallest vertex in H 0 . Observe that two H-leaves are either vertex-disjoint or
together they form a component.
Gerke, McDiarmid, Steger, and Weißl [10] have shown the following result about the containment of a fixed connected planar graph H in a graph P̂n,m=dn which is chosen uniformly at random
from the class of all simple labelled planar graphs with n vertices and dn edges:
h
i
P P̂n,m=dn contains at most αn pairwise vertex-disjoint H-leaves < e−αn ,
for every 1 < d < 3 and a positive constant α = α(H, d).
In this respect, the two models do agree: the following analogue is our second main result.
Theorem 1.3. Let H be a connected planar graph. For every 1 < d < 3, there exists α =
α(H, d) > 0 such that
P [Pn,m=dn contains at most αn pairwise vertex-disjoint H-leaves ] < e−αn .
2
Sketch of proof
In this section we give a rough description of our proof strategy. In Section 3 we show that a
statement similar to Theorem 1.3 is true for labelled random trees. More precisely, we show that
for any fixed tree B the tree generated by the random tree process (Tn,t )N
t=0 contains with high
probability a linear number of (necessarily pairwise vertex-disjoint) B-leaves.
In Section 4 we first show that the fact that there are at most γ n n! different labelled planar
graphs for some γ implies that the planar graph Pn,n log n obtained by the process after having seen
the first n log n edges contains with high probability only n + o(n) edges. From this Theorem 1.2
follows easily. We shall need a (weakend) version of Theorem 1.2 for the proof of Theorem 1.1.
This is why we prove it even though it follows from Theorem 1.1. As a next step we again use
2
that there exist at most γ n n! different labelled planar graphs to show that with high probability
there exist at most Γn different labelled planar graphs that contain the graph Pn,n log n . Having
shown these results, Theorem 1.1 will follow by an easy counting argument.
In Section 5 we prove Theorem 1.3. The strategy is roughly as follows. Let B be an arbitrary
spanning tree in H. Consider the graph Pn,n log n obtained by the process after having seen the
first n log n edges. From the results on the tree process we know that the tree Tn,n log n obtained
by the corresponding random tree process is with high probability connected and contains a linear
number of pairwise vertex-disjoint B-leaves. As we also know that Pn,n log n contains only n + o(n)
edges, the edges that are contained in Pn,n log n but not in Tn,n log n can destroy at most o(n) of
these B-leaves. That is, we know that Pn,n log n also contains with high probability a linear number
of pairwise vertex-disjoint B-leaves. Now we use the results of Section 4 to see what happens to
them while the next δn2 edges are added. Theorem 1.1 tells us that from these δn2 edges with high
probability at most εn edges will be accepted by the planar graph process. That is, by choosing
ε > 0 small enough we know that the graph Pn,δn2 still contains with high probability a linear
number of these B-leaves. Moreover, the fact that at most εn out of δn2 edges have been accepted,
implies that with high probability there are only a linear number of places where one can still add
an edge to the graph in such a way that the graph remains planar. We call such places addable
edges (despite the fact that these are not yet edges in the graph and might even never become
edges of the graph). By a simple averaging argument it follows that from the linear number of
pairwise vertex-disjoint B-leaves a constant fraction is such that their vertices are incident to only
a constant number of addable edges. On the other hand we know from the definition of a B-leaf
that the edges extending it to an H-leaf are among the addable edges. If there are only a constant
number of such edges, then there is a constant probability that, if we wait until we have added
another εn edges, the edges extending a B-leaf to an H-leaf appear but the other addable edges
do not. Thus, for each of these B-leaves we have a constant probability that it is extended to
an H-leaf. Since they are pairwise vertex-disjoint, it follows that with high probability a linear
number of them will become H-leaves, which proves Theorem 1.3.
Let us introduce some further notation and conventions. Throughout this paper we will only
be concerned with labelled graphs, typically on n vertices, in which case we will assume the vertex
set to be [n], and our interest lies in the asymptotic behaviour of these graphs, i.e. what happens in
the limit as n −→ ∞. Some statements, in particular inequalities, will only hold if n is sufficiently
large and in this case we will assume so, sometimes implicitly. Furthermore, we will not introduce
floors and ceilings but disregard rounding issues, as this will not affect the asymptotic behaviour
of the respective quantities.
In general, the number of vertices in a graph G is v(G), the number of its edges is e(G), and
the subgraph of G which is induced by a subset U of its vertices is G[U ]. With Γ(v) or ΓG (v) we
denote the set of neighbours of a vertex v. We write the falling factorial n · (n − 1) · · · · · (n − k + 1)
as nk . We denote the set of all trees on [n] by Tn and the set of all planar graphs on [n] by Pn .
If the probability of an event An which depends on n tends to 1 as n tends to infinity, this is
often described by saying that An occurs with high probability (w.h.p.). As indicated by our main
result, Theorem 1.3, we shall often want faster convergence: An is said to occur with exponentially
high
probability (w.e.h.p.), if the probability of the complement decreases exponentially, i.e. if
P A < e−Ω(n) . Note that in the proof strategy outlined above we did not distinguish between
w.h.p. and w.e.h.p. We will do so when making the above strategy precise.
3
The random tree process
In this section we have a closer look at Tn = Tn,N , the final outcome of the random tree process
(Tn,t )N
t=0 introduced in Section 1. Let B be an arbitrary but fixed tree on the vertex set [v(B)].
Our main goal is to prove that, w.e.h.p., Tn will contain a linear number of B-leaves. We start
with some definitions.
Let T 0 be a subtree of a tree T . A vertex v ∈ V (T 0 ) is called an anchor in T 0 , if there exists a
vertex u ∈
/ V (T 0 ) such that uv ∈ E(T ). The set of anchors of T 0 is denoted by A(T 0 ). A proper
vertex subset U ⊂ V (T ) is called k-autonomous if T [U ] is a subtree and has at most k anchors.
3
..
.
.
..
1
u
B
3
4
a1
T
v=2
2
5
8
14
a2
..
.
5
U
7
B
Figure 1: a B-leaf B 0 and a 2-autonomous set U in a tree T
Thus, if B 0 is a B-leaf in T , then V (B 0 ) is a 1-autonomous set with the smallest vertex as its
anchor.
By SN we denote the set of all permutations on the edge set of the complete graph Kn on n
vertices. Let σ ∈ SN . If an edge e appears in the ith position we write σ(e) = i, and we say that
an edge e1 is smaller than an edge e2 (with respect to σ) if σ(e1 ) < σ(e2 ).
For any time t, i.e. an integer between 0 and N , σ|t will be the initial part of length t of
the permutation σ. The forest T (σ|t ) associated with such an initial sequence is constructed as
follows: start with the empty graph on n vertices, and consider the t edges of σ|t in this order,
accepting an edge into the present graph if and only if it does not close a cycle. We will also say
that T (σ|t ) is (the output) generated by (the input) σ|t .
Let T ∈ Tn and set S(T ) := {σ ∈ SN |T (σ) = T }. Then
P [Tn = T ] =
|S(T )|
,
N!
(1)
and we call w(T) := |S(T)| the weight of the tree T.
3.1
Local rearrangements
In this subsection we introduce two local tree operations. Their main purpose is to modify a given
tree while controlling the number of permutations generating it. The first operation increases the
number of leaves by 1, without decreasing the weight of the tree.
Lemma 3.1. Let T ∈ Tn and xy ∈ E(T ), such that x, y are not leaves, and set {x1 , . . . , xk } :=
Γ(x) \ {y}. Construct a tree T 0 by
E (T 0 ) := (E(T ) \ {xx1 , . . . , xxk }) ∪ {yx1 , . . . , yxk } ,
see Figure 2. Then w(T ) ≤ w(T 0 ).
Proof. Consider T \ {xy} and let U be the vertices in the component containing y, and W be the
union of the vertices in the other component and {y}, so that U ∪ W = V (T ) and U ∩ W = {y}
(cf. Figure 2). For easier reference, we colour an edge of the underlying complete graph blue if
both endpoints lie in U , green if both endpoints lie in W , and red otherwise. For an arbitrary σ
in S(T ), we define

 σ(xz) : e = yz, z ∈ W \ {x, y},
σ(yz) : e = xz, z ∈ W \ {x, y},
σ 0 (e) =

σ(e)
: otherwise.
4
x1
...
...
y
x
y
U
x
xk
x1
W
...
...
T
T
U
xk
W
Figure 2: the local tree operation in Lemma 3.1
In other words, we obtain σ 0 from σ as follows: for every vertex z in a subtree pending from one
of the xi , we transpose the positions of xz and yz.
We now claim that σ 0 ∈ S(T 0 ). As we can reobtain σ from σ 0 without ambiguity, this implies
|S(T )| ≤ |S(T 0 )| and thus completes the proof.
To prove this claim, let φ : T [W ] −→ T 0 [W ] be the isomorphism given by interchanging x and
y, i.e.

 x : w = y,
y : w = x,
φ(w) =
and
φ(w1 w2 ) = φ(w1 )φ(w2 ).

w : otherwise,
We want to show that σ 0 generates T 0 , or equivalently, that it generates T 0 [W ] = φ(T [W ]) on W
and T 0 [U ] = T [U ] on U . More precisely, we prove the following statements h1 (t), h2 (t), and h3 (t)
for all 0 ≤ t ≤ N by induction:
h1 (t) : T (σ 0 |t ) contains no red edge,
h2 (t) : T (σ 0 |t )[U ] = T (σ|t )[U ],
h3 (t) : T (σ 0 |t )[W ] = φ (T (σ|t )[W ]) .
Observe that these statements trivially hold for t = 0, assume that all three statements hold for
t0 < N , and let ab be the edge at the (t0 + 1)st position of σ 0 .
Suppose that ab is green. As U and W overlap only in y and, according to h1 (t0 ), no red edges
have been accepted by σ 0 so far, the acceptance of ab by σ 0 depends only on the presence of a
green path between a and b in T (σ 0 |t0 ). By h3 (t0 ), there is such a path if and only if there is a
green path between φ−1 (a) and φ−1 (b) in T (σ|t0 ). As σ never accepts red edges, we know that
any path between φ−1 (a) and φ−1 (b) in T (σ|t0 ) must be green, and by our construction of σ 0 we
know that the (t0 + 1)st edge in σ is φ−1 (a)φ−1 (b). Thus we conclude that ab is accepted by σ 0
if and only if φ−1 (a)φ−1 (b) is accepted by σ in the (t0 + 1)st step. Hence h3 (t0 + 1) holds, as do
h1 (t0 + 1) and h2 (t0 + 1) trivially.
The case that ab is blue can be dealt with similarly. In fact, this case is easier as σ and σ 0
agree on the blue edges.
So finally assume that ab is red and, say, a lies in W \ {y} and b in U \ {y}. The statements
h2 (t0 + 1) and h3 (t0 + 1) follow trivially from h2 (t0 ) and h3 (t0 ). It thus suffices to check that σ 0
will not accept ab. Observe that since b ∈ U \ {y}, the edge ab has the same position in σ as in σ 0 .
In σ, it is rejected and hence there must be a path P between a and b in T (σ|t0 ). We distinguish
two cases. First assume that a = x. From the structure of T we infer that xy is the first edge
on P , and that all other edges of P lie in U . By h3 (t0 ) and h2 (t0 ), the edges of the path P must
5
T
T
P
y
<
.
z .
..
..
..
.
<
P
y
x
z .
..
x
Figure 3: the local tree operation in Lemma 3.3
exist in T (σ 0 |t0 ) too, and hence xb will be rejected by σ 0 . Now assume that a 6= x. Again we infer
from the structure of T that P consists of a path from a to x, the edge xy, and then a path from
y to b. Thus T (σ 0 |t0 ) contains a path from a to y (due to h3 (t0 )) and a path from y to b (due to
h2 (t0 )), and hence ab will be rejected by σ 0 as well.
If we apply Lemma 3.1 iteratively until there is only one non-leaf left, we arrive at a star
K1,n−1 , which thus maximizes w(T ) over T ∈ Tn . From [1] we know that
P [Tn = K1,n−1 ] =
(n − 1)!2n−1
.
(2n − 2)!
(2)
Using Stirling’s formula, we note that, for sufficiently large n,
(n − 1)!2n−1
2n e n
(e/2)n
∼√
n−2 .
(2n − 2)!
n
2e 2n
Lemma 3.1 and (2) thus imply
Corollary 3.2.
max w(T ) <
T ∈Tn
e n N !
.
2
nn−2
By (1), this upper bound on the weight of the heaviest tree also provides an upper bound on
how much the distribution of Tn deviates from the uniform distribution on Tn .
We now describe our second local tree operation.
Lemma 3.3. Let T ∈ Tn , x be a leaf, y its neighbour, z another vertex, and P the path between y
and z. Suppose an order ρ on an edge set E with xy ∪ E(P ) ⊆ E ⊆ E(T ) is given, such that each
edge in E(P ) is smaller than xy, and denote by S(E, ρ) the set of all edge permutations which
induce ρ on E. Replace xy by xz in T , E, and ρ to get T 0 , E 0 , and ρ0 , respectively (see Figure 3).
Then
|S(T ) ∩ S(E, ρ)| = |S(T 0 ) ∩ S(E 0 , ρ0 )|.
Proof. By symmetry, it suffices to prove that
|S(T ) ∩ S(E, ρ)| ≤ |S(T 0 ) ∩ S(E 0 , ρ0 )|.
6
a
a
f (v1 )
T1
v1
a
a
f (v2 )
v2
f (v3 )
v1
v3
v2
T3
T2
v4
v1
v3
v2
v1
T4
Figure 4: transforming a tree into a star as in the proof of Lemma 3.4
Consider some arbitrary σ ∈ S(T ) ∩ S(E, ρ) and

 σ(xz) :
σ(xy) :
σ 0 (e) =

σ(e)
:
define
e = xy,
e = xz,
otherwise.
We proceed to show that σ 0 ∈ S(T 0 ) ∩ S(E 0 , ρ0 ). As we can reobtain σ from σ 0 , this implies
|S(T ) ∩ S(E, ρ)| ≤ |S(T 0 ) ∩ S(E 0 , ρ0 )| as required. It is easy to see that σ 0 lies in S(E 0 , ρ0 ), so we
are left to prove that σ 0 generates T 0 .
We prove the following statement by induction on t. For each 0 ≤ t ≤ N ,
h(t) :
T (σ|t ) and T (σ 0 |t ) have the same connected components
(by which we mean that there is a u-v-path in T (σ|t ) if and only if there is one in T (σ 0 |t ), for
all vertices u, v). Observe that this will complete the proof, as for each 0 < t ≤ N , h(t − 1) and
h(t) imply that σ 0 accepts the edge at its t-th position if and only if σ accepts the edge at its t-th
position, which yields T (σ 0 ) = T 0 because of T (σ) = T and the way σ 0 was constructed.
For the induction, we see that h(0) holds trivially and assume that h(t0 − 1) is true for 0 ≤
t0 − 1 < N . Set t1 := σ(xy) = σ 0 (xz) and t2 := σ(xz) = σ 0 (xy), and observe that t1 < t2 . For
t0 ∈
/ {t1 , t2 }, σ and σ 0 agree in the t0 -th position and h(t0 ) follows easily from h(t0 − 1).
If t0 = t1 , then xy ∈ E(T ) tells us that σ merges the two components containing x and y by
accepting xy. On the other hand, P ⊆ T (σ|t1 −1 ) shows that z is contained in the same component
as y at time t1 − 1, and hence we know that z cannot be in the same component as x. By h(t1 − 1),
σ 0 will therefore accept xz, joining the two components in the same way as σ and thus proving
h(t1 ). The same arguments apply for the case t0 = t2 , except that this time the edges are not
accepted.
Again, we will apply this local tree operation iteratively to transform a tree into a star.
Lemma 3.4. Let T 0 be a subtree of T ∈ Tn with exactly one anchor a. Let T ∗ be the tree obtained
from T by deleting all edges in T 0 and attaching all non-anchor vertices in T 0 to a as leaves. Then
w(T ) ≥
w(T ∗ )
.
(v(T 0 ) − 1)!
Proof. For two vertices u, v in T 0 , we denote by dist(u, v) the length of the path from u to v in T 0 .
Order the non-anchor vertices of T 0 such that dist(a, vi ) ≥ dist(a, vj ) for all 1 ≤ i < j ≤ v(T 0 ) − 1,
7
a
a
T 
T
C
b
C
b
T0
T
Figure 5: transforming a subtree with two anchors into a different subtree with one anchor as in
Corollary 3.5
and set l := max{1 ≤ i ≤ v(T 0 ) − 1 : d(a, vi ) > 1}. For 1 ≤ i ≤ l, let f (vi ) be the neighbour of vi
on the path between vi and a in T 0 .
We construct a sequence of trees T = T1 , . . . , Tl+1 as follows: get Ti+1 from Ti by setting
E(Ti+1 ) := (E(Ti ) \ {f (vi )vi }) ∪ {avi } (cf. Figure 4). It is easy to see that Tl+1 = T ∗ . Set
Ei := E(Ti [V (T 0 )]) for 1 ≤ i ≤ l + 1. Furthermore, let ρl+1 be defined by avv(T 0 )−1 < · · · < av1
and for 1 ≤ i ≤ l, get ρi from ρi+1 by replacing avi with f (vi )vi . Observe that in this way, for
each 1 ≤ i ≤ l, we ensure that in Ti the edges on the path from f (vi ) to a are smaller than f (vi )vi
with respect to ρi .
We can therefore apply Lemma 3.3 for each 1 ≤ i ≤ l with T = Ti , x = vi , y = f (vi ), z = a,
E = Ei , and ρ = ρi . Note that for this setting, T 0 = Ti+1 , E 0 = Ei+1 , and ρ0 = ρi+1 , so that
|S(T1 ) ∩ S(E1 , ρ1 )| =|S(T2 ) ∩ S(E2 , ρ2 )|
..
.
=|S(Tl+1 ) ∩ S(El+1 , ρl+1 )|.
As |S(Tl+1 ) ∩ S(El+1 , ρl+1 )| = w(T ∗ )/(v(T 0 ) − 1)! by the symmetry of the star, the claimed
inequality follows from w(T ) ≥ |S(T1 ) ∩ S(E1 , ρ1 )|.
We will now combine Lemma 3.1 and Lemma 3.4 in order to change a subtree T 0 which lives
on a 2-autonomous set into any required subtree T 00 on the same vertex set but with only one
anchor.
Corollary 3.5. Let T 0 be a subtree of T ∈ Tn with anchor set A(T 0 ) = {a, b} and T 00 be another
tree on V (T 0 ). Set C := {c ∈
/ V (T 0 ) : bc ∈ E(T )} and construct a tree T0 from T by replacing T 0
00
with T and bc with ac, for each c ∈ C (see Figure 5). Then
w (T0 ) ≥
w(T )
.
(v(T 0 ) − 1)!
Proof. If a 6= b, let a = a0 , . . . , al = b be the vertices of the path between a and b. Apply
Lemma 3.1 iteratively l times: first on T with (x, y) = (al , al−1 ), then on the resulting tree with
(x, y) = (al−1 , al−2 ), and so on. This will result in a tree T1 for which V (T 0 ) is 1-autonomous,
ΓT (a) ∪ C ⊆ ΓT1 (a), and w(T1 ) ≥ w(T ). So we may assume, without loss of generality, that
a = b. Now use Lemma 3.1 iteratively to attach all non-anchor vertices to the anchor. Then apply
Lemma 3.4 in ‘reverse’ to build a copy of T 00 from that substar. In the first of these two phases,
8
the weight of the corresponding tree will not decrease at all, and in the second phase only by a
factor of 1/(v(T 0 ) − 1)!.
3.2
Partial covering by small subtrees
We now describe an algorithm which covers a constant fraction of the vertices in any T ∈ Tn
with small 2-autonomous sets that pairwise overlap in at most one vertex. Observe that a path
of length n − 1 shows that we cannot hope for a linear number of 1-autonomous sets. We have
chosen an algorithmic approach for proving this, as this allows us to argue that even if we change
the tree slightly (and appropriately) the algorithm will find exactly the same 2-autonomous sets.
This fact will turn out to be very helpful later.
In the description of the algorithm we will claim that its output satisfies a series of properties.
In Lemma 3.6 we will prove that this is indeed the case. In Lemma 3.7 we will then show that
appropriate local rearragements of the tree will lead to the same output of the algorithm.
SearchCandidates
Input: a tree T on [n], v ≥ 2
n
e subsets C1 , . . . , Ck ⊆ [n] such that, ∀1 ≤ i < j ≤ k,
Output: a sequence of k := d 10v
(i) v ≤ |Ci | ≤ 2v − 2,
(ii) Ci is 2-autonomous,
(iii) |Ci ∩ Cj | ≤ 1, i.e. T [Ci ] and T [Cj ] are edge disjoint.
1. Root the tree T at 1. For each vertex x, define T (x) to be the subtree rooted at x, that is,
the tree induced by all vertices whose path to vertex 1 contains x.
2. Say a vertex x is of type 0 if v(T (x)) < v. Denote by T 0 the subtree of T that is induced by
all vertices that are not of type 0.
3. Say that a non-root vertex in T 0 is of type 1, 2, or 3, if its degree in T 0 is 1, 2, or at least 3,
respectively. We do not classify the root because if the root were of type 1 we would have
to treat type 1 vertices that are not the root differently to the root. In T 0 , we contract each
maximal path the inner vertices of which are of type 2 into a single edge, and denote the
resulting tree (on the vertices of type 1 and 3 plus the root) by T 00 .
4. We then construct an order ρ on V (T 00 ) ∪ E(T 00 ) via a breadth-first search, starting at the
root: if we are at the stage where we have considered all vertices of at most some distance
d from the root, let x1 < · · · < xl be the vertices at distance d + 1 in the order given by
the vertex labels. Then we first add the edges to x1 , . . . , xl and then the vertices themselves
(both in precisely this order) to ρ.
5. Let (C1 , . . . , Ck ) be a sequence of vertex sets, all of them initially empty. Consider these
subsets as bins, which will be packed sequentially such that
• a bin is either empty, active (if 0 < |Ci | < v), or loaded (if |Ci | ≥ v),
• in each step of the packing process, the loaded bins come first (in this sequence), followed
by at most one active bin, and then the empty bins,
• after each round (see below), there is no active bin.
6. The packing process proceeds in rounds according to ρ, each round corresponding to either
a vertex x or an edge e of T 00 , and we distinguish these two cases:
Case 1 Suppose we are at the beginning of a round that corresponds to a vertex x ∈ V (T 00 ),
which has type 0 neighbours x1 , . . . , xl ordered by their labels. We describe the packing
in this round by the following pseudo-code:
9
smallest empty bin ∪{x}
= 1 to l
:= C ∪ V (T (xj ))
|C| ≥ v
C := next bin ∪{x}
if |C| < v
C := ∅
C :=
for j
C
if
In other words, we try to pack, one after another, the subtrees pending from the type
0 neighbours of x into a bin initially containing only x. If a bin becomes loaded, we
take the next empty bin, fill it with x, and proceed. In the end, when we have placed
all these subtrees, we empty the last bin we have used if it is not yet loaded.
Case 2: Suppose we are at the beginning of a round that corresponds to an edge e ∈ E(T 00 ).
If e also lies in E(T 0 ), we do nothing. Otherwise, e was created by contracting a maximal
path in T 0 whose inner vertices x1 , . . . , xm are of type 2, where dist(1, xi ) < dist(1, xj )
for 1 ≤ i < j ≤ m. Let them have l1 , . . . , lm neighbours xh,j (1 ≤ h ≤ m and 1 ≤ j ≤ lh )
of type 0, respectively, where for a fixed h, the xh,j are ordered by their labels. We
describe the packing in this round by the following pseudo-code:
C := smallest empty bin
for h = 1 to m
C := C ∪ {xh }
if |C| ≥ v
C := next bin ∪{xh }
for j = 1 to lh
C := C ∪ V (T (xh,j ))
if |C| ≥ v
C := next bin ∪{xh }
if |C| < v
C := ∅
In other words, we start as in the first case with x1 in place of x. However, once we
have placed all subtrees pending from type 0 neighbours of x1 , we also place x2 into
the active bin, followed by the subtrees pending from its neighbours, and so on. If a
bin becomes loaded, we take the next empty bin, fill it with the type 2 vertex xh most
recently considered, and proceed.
Note that, unless we are in a round corresponding to a type 1 vertex, there may well be no
type 0 neighbours of a vertex x or the vertices x1 , . . . , xm . But typically, a round comprises
several packing steps in which a bin receives some vertices, plus possibly one unpacking step
at the end.
7. We stop the packing process if either all Ci are loaded, or all rounds for w ∈ V (T 00 ) ∪ E(T 00 )
have been completed in order ρ.
Note that our choice of the root and the manner in which we process the vertices and edges of
T 00 is somewhat arbitrary. We only need and use that this order is prescribed and fixed in some
way. We will now prove that the bins have the desired properties.
Lemma 3.6. The algorithm SearchCandidates works correctly.
Proof. We first show that for all bins which are loaded in the end, the three properties claimed in
the algorithm are fulfilled:
(i) Clearly |Ci | ≥ v by the definition for being loaded. In each step where a non-empty bin
receives some vertices, i.e. in Case 1, Line (3) and Case 2, Line (3) and (7), there are at most
10
v − 1 such new vertices, as only the vertices in the subtree of a type 0 vertex or a single type
2 vertex are added. As we check immediately afterwards whether the currently active bin
has become loaded, it is always true that |Ci | ≤ 2v − 2.
(ii) For all bins which are loaded in Case 1, the second property is obvious, as each bin is the
union of the vertex sets of subtrees that all intersect in a vertex (and thus induces a subtree),
which is moreover the only vertex in the bin adjacent to a vertex outside of it, and hence the
bin is even 1-autonomous.
In Case 2, it is easy to see that the type 2 vertices contained in some bin induce a subpath
P 0 of T 0 [{x1 , . . . , xm }]. As everything else in this bin lies in subtrees of type 0 neighbours of
these vertices, the bin indeed induces a subtree. Moreover, for the non-leaves in P 0 , all type
0 neighbours must lie in this bin, so only the leaves of P 0 are adjacent to a vertex outside of
it, and hence the bin is 2-autonomous.
(iii) The third property can now be easily deduced from our arguments for the second one.
So we are left to show that once the packing process terminates, all bins are loaded. Observe
that, if all rounds are completed, every vertex was considered in exactly one round. Setting
n00 := v(T 00 ), we also observe that there are 2n00 − 1 rounds, and that in each round, less than v of
the vertices considered are not covered, i.e. placed in some bin which will be emptied in the last
step. Therefore,
αn := |{vertices not covered}| < (2n00 − 1)v.
(3)
On the other hand, we know that in each round corresponding to a type 1 vertex x, we will load
at least one bin, as x is a leaf in T 00 and there are at least v vertices of type 0 in T (x). Summing
degrees in T 00 and writing r for the number of type 1 vertices, we get
2(n00 − 1) ≥ 1 + r + 3(n00 − r − 1),
where the first 1 is caused by the root. Hence r ≥ n00 /2. Therefore,
βn := |{vertices covered}| ≥ rv ≥
n00 v
.
2
(4)
As α + β = 1, we can combine (3) and (4) to get
βn
n00 v/2
1
>
= ,
(1 − β)n
2n00 v
4
n
which tells us that at least n/5 vertices are covered. Therefore, there have to be at least d 10v
e
loaded bins in the end.
3.3
Constructing B-leaves
From now on, we consider an arbitrary but fixed tree B. Recall that a B-leaf in a tree is a copy
of B that is connected to the rest of the tree only via an edge incident with the smallest vertex in
the B-leaf.
For v := 3(v(B) + 1) and T ∈ Tn , we denote by sc(T ) = (C1 , . . . , Ck ) the partial vertex
covering we get when we run SearchCandidates. We now use Corollary 3.5 to show that, for
each 1 ≤ i ≤ k, we can modify T [Ci ] in such a way that the resulting overall tree Ti has the same
partial vertex covering and Ti [Ci ] contains a B-leaf.
We emphasise that the main technical difficulty with our construction of a B-leaf on a subset
of Ci is that we wish to preserve the partial vertex covering, and here the labels of the vertices
are becoming important.
n
Lemma 3.7. Let B be a tree and T ∈ Tn and set v := 3(v(B) + 1) and k := d 10v
e. For every
1 ≤ i ≤ k, there exists a tree Ti ∈ Tn such that
11
(i) sc(Ti ) = sc(T ) =: (C1 , . . . , Ck ),
(ii) T [([n] \ Ci ) ∪ A(T [Ci ])] = Ti [([n] \ Ci ) ∪ A(T [Ci ])],
(iii) Ti [Ci ] contains a B-leaf,
(iv) w(Ti ) ≥
w(T )
(v−2)! .
Proof. Fix 1 ≤ i ≤ k arbitrarily. We consider the two cases which SearchCandidates might
have used in order to load bin Ci .
Case 1: Ci was loaded in a round R corresponding to some vertex x ∈ V (T 00 ). Let x01 < · · · < x0s
be the neighbours of x which are contained in Ci . Note that x01 , . . . , x0s are all of type 0. If a
tree rooted at one of the x0j has at least v(B) + 1 vertices, then we transform T (x0j ) into the
following tree. We arbitrarily select a subset V ⊆ V (T (x0j )) \ {x0j } of size v(B) and construct
the (unique) copy B 0 of B on V . Then we select the smallest vertex v in V and connect it to
x0j . Finally, we attach all the vertices in the (possibly empty) set V (T (x0j )) \ (V ∪ {x0j }) to x0j .
By Corollary 3.5 we can perform this transformation such that for the new overall tree Ti we
have w(Ti ) ≥ w(T )/(v − 2)!, as v(T (x0j )) < v, and so Condition (iv) is fulfilled. Observe that
Condition (i) is also satisfied since the sizes of the trees with roots x01 , . . . , x0s have not changed
and in particular vertices in these trees remain type 0 vertices. We have not modified anything
in T [([n] \ Ci ) ∪ A(T [Ci ])] and thus Condition (ii) is fulfilled. By construction, Ti [Ci ] contains a
B-leaf (namely B 0 ) and thus Condition (iii) is true as well.
Now assume that all these subtrees have less than v(B) + 1 vertices. Let λ be the largest
index such that the trees rooted at x0λ , . . . , x0s have together at least v(B) + 1 vertices. Again,
we use Corollary 3.5, this time to transform these trees together with x and the edges between x
and x0λ , . . . , x0s into the following tree. Build a copy of B on a set of v(B) arbitrary vertices not
containing x or x0s , select the smallest vertex v in B 0 , and connect v, as well as x and all remaining
vertices to x0s . Let Ti denote the resulting overall tree. Note that by the maximality of λ and our
assumption on the sizes of the subtrees, we know that at most 2v(B) + 1 vertices lie in the part
which has been modified, and hence Condition (iv) is satisfied. This also shows, that all vertices
in Ci \ {x} remain of type 0. When T and the set Ci were considered, the algorithm added x0s to
Ci which implies that Ci without the vertices of the tree rooted at x0s has size less than v. Hence,
when considering Ti , the algorithm adds first the trees rooted at x01 , . . . , x0λ−1 , followed by the new
subtree rooted at x0s , which proves Condition (i). Condition (ii) and Condition (iii) are clearly
satisfied by construction.
Case 2: Ci was loaded in a round R corresponding to some edge e ∈ E(T 00 ) that was obtained
by contracting a maximal path in T 0 whose inner vertices were of type 2. Let x01 , . . . , x0s denote
the subpath of this path which consists of the vertices that are contained in Ci . Assume that
dist(1, x01 ) ≤ . . . ≤ dist(1, x0s ). If the type 0 neighbours (in Ci ) of either x01 or x0s contain at least
v(B) + 1 vertices in the trees rooted at them, then we can do the same operations as in Case 1
with x replaced by x01 or x0s , respectively. Thus, if we consider the set M consisting of x02 , . . . , x0s−1
together with the vertices of the trees rooted at their type 0 neighbours, we may assume that
it consists of at least v − 2(v(B) + 1) = v(B) + 1 vertices. On the other hand, the fact that
the algorithm also added x0s to Ci shows that |M | < v − 1. Using Corollary 3.5, this time with
x02 and x0s−1 as anchors of T [M ], we construct a tree on T [M ] consisting of x02 connected to the
smallest vertex in a copy B 0 of B, and all vertices in M \ (V (B 0 ) ∪ {x02 }). Note that Corollary 3.5
implies that x02 is now also connected to x01 and x0s , and that for the new overall tree Ti we have
w(T )
w(T )
w(Ti ) ≥ (|M
|−1)! ≥ (v−3)! . It is easy to verify that Conditions (i)–(iv) are satisfied.
3.4
Random viewpoint
We now use the results of the previous sections in a probabilistic setting in order to show that
w.e.h.p. a constant fraction of the linear number of subtrees induced by the 2-autonomous sets in
sc(Tn ) contain B-leaves. To do so, we need the following definition.
12
For every 1 ≤ i ≤ k, we define an equivalence relation ∼i on SN by
σ ∼i π if and only if sc(T (σ)) = sc(T (π)) =: (C1 , . . . , Ck )
h
i
h
i
and T (σ) ([n] \ Ci ) ∪ A(T (σ)[Ci ]) = T (π) ([n] \ Ci ) ∪ A(T (π)[Ci ]) .
We can now state and prove the following lemma.
n
e, and for 1 ≤ i ≤ k let [σ]i
Lemma 3.8. Let B be a tree, set v := 3(v(B) + 1) and k := d 10v
be an equivalence class of ∼i . Furthermore, let sc(T (σ)) = (C1 , . . . , Ck ) denote the partial vertex
covering common to all trees generated by permutations in [σ]i . Then
|[σ]i | ≤ (3v(B) + 1)! · (6v(B) + 4)6v(B)+2 · |{σ ∈ [σ]i : T (σ)[Ci ] contains a B-leaf}|.
Proof. Let T be the set of trees generated by permutations in [σ]i . Lemma 3.7 tells us that
for any tree T ∈ T , there is a tree Ti ∈ T such that Ti [Ci ] contains a B-leaf and such that
w(T ) ≤ (3v(B) + 1)! · w(Ti ). Hence, we get
w(T ) ≤ (3v(B) + 1)! · |{σ ∈ [σ]i : T (σ)[Ci ] contains a B-leaf}|
and therefore also
X
|[σ]i | =
w(T ) ≤ |T | · (3v(B) + 1)! · |{σ ∈ [σ]i : T (σ)[Ci ] contains a B-leaf}|.
T ∈T
As |Ci | ≤ 2v − 2 = 6v(B) + 4, there are at most (6v(B) + 4)6v(B)+2 different trees on Ci and thus
|T | ≤ (6v(B) + 4)6v(B)+2 .
We are now able to prove that for any fixed tree B, the tree Tn generated by the random tree
process will contain with exponentially high probability a linear number of B-leaves.
Theorem 3.9. Let B be a tree and Y be the maximum number of B-leaves in Tn . Then there
exists β > 0 such that
P [Y ≤ βn] < e−βn .
Before proving Theorem 3.9 we first recall an easy fact from elementary probability theory
that will be used several times in the remainder of the paper.
Proposition 3.10. Let E1 , . . . , Ek be pairwise
h S disjoint
i events. Then any event F that satisfies
k
P [F |Ei ] ≥ q for all i ∈ [k], also satisfies P F i=1 Ei ≥ q.
Proof of Theorem 3.9. Set p := 1/((3v(B) + 1)! · (6v(B) + 4)6v(B)+2 ), v := 3(v(B) + 1), and
n
k := d 10v
e. Choose 0 < β < 1/(20v) such that 1 + log(k/(βn)) − p(k/(βn) − 1) < −1. Let τ be
a permutation chosen uniformly at random from SN , and let (C1 , . . . , Ck ) := sc(T (τ )) denote the
output of SearchCandidates. For 1 ≤ i ≤ k, define the event
Ai : T (τ )[Ci ] contains a B-leaf
Pk
and let Xi be the corresponding indicator variable. Set X := i=1 Xi . Clearly X ≤ Y . It thus
suffices to show that P [X ≤ βn] ≤ e−βn for the constant β > 0 chosen above.
Observe that Lemma 3.8 implies that
∀i ∈ [k] ∀σ ∈ SN :
P [Ai |[σ]i ] ≥ p.
(5)

\
P Ai Aj  ≥ p.
j∈J
(6)
We claim that this implies that in fact

∀i ∈ [k] ∀J ⊆ [k] \ {i} :
13
To see this recall that by definition any two permutations σ and π that are in the same equivalence
class of ∼i generate trees that are identical outside of Ci . As any two sets Cj and Ck share at
most one vertex, this implies that for any σ ∈ SN the equivalence class [σ]i is either
T contained
in AjSor completely disjoint from Aj (for all j 6= i). We can therefore write A := j∈J Aj as
A = [σ]i ⊆A [σ]i and (6) thus follows from Proposition 3.10 and (5).
The following calculation shows that (6) implies the theorem. For notational convenience, we
denote in the following equations the complement of a subset I ⊆ [k] of size s by {i1 , . . . , ik−s }.
j−1
"
#
"
#
#
"
βn X
βn X
βn X k−s
\
X
\
\
X
\
X
Y
Ai ≤
Ai =
Ail
P [X ≤ βn] =
P
Ai ∩
P
P Aij s=0 I⊆[k]
|I|=s
(6)
≤
βn X
X
i∈I
k−s
(1 − p)
=
s=0 I⊆[k]
|I|=s
<
ek
βn
βn
s=0 I⊆[k]
|I|=s
i∈I
/
βn X
k
s
s=0
−p(k−βn)
e
k−s
(1 − p)
s=0 I⊆[k] j=1
|I|=s
i∈I
/
l=1
k
< (βn + 1)
(1 − p)k−βn
βn
k
= exp βn 1 + log
−p
βn
k
−1
,
βn
which is smaller than e−βn by our choice of β.
4
Connectedness and containment
In this section we prove Theorems 1.1 and 1.2. We start by establishing an upper bound on
the number of edges in Pn,n log n . We will use the well-known result that there exists a positive
constant γ ≥ 1 such that the number |Pn | of simple labelled planar graphs on the vertex set [n]
satisfies |Pn | ≤ γ n n!. This follows for example from the work of Tutte [16]. We note that a much
stronger result was recently proven by Giménez and Noy [11, 12].
Proposition 4.1.
2 log log n
P e(Pn,n log n ) ≥ n 1 +
< e−3n .
log n
Proof. Let f (n) := 2n log log n/ log n and fix an arbitrary planar graph P with e(P ) = n + f (n)
edges. The probability that P is contained in Pn,n log n can be bounded from above by counting
the permutations where the edges of P are among the first n log n edges. Hence,
e(P )−1
Y n log n − i
(n log n)e(P ) (N − e(P ))!
P [P ⊆ Pn,n log n ] ≤
=
N!
N −i
i=0
e(P ) e(P )
n log n
3 log n
<
≤
.
N
n
Therefore,
X
P [e (Pn,n log n ) ≥ n + f (n)] ≤
P [P ⊆ Pn,n log n ] < γ n nn
P ∈Pn
e(P )=n+f (n)
3 log n
n
n+f (n)
n+f (n)
(3γ log n)
nf (n)
−3n
<e
,
= exp ((n + f (n)) log(3γ log n) − f (n) log n)
<
by our choice of f (n) and for sufficiently large n.
Theorem 1.2 follows now easily.
14
Proof of Theorem 1.2. Due to the connectivity threshold of random graphs [6] and Proposition 4.1,
we know that, as n −→ ∞, P [Pn,n log n is connected] = P [Gn,n log n is connected] −→ 1 and
P [e(Pn,n log n ) = n + o(n)] −→ 1, respectively.
As the probability that any fixed vertex remains isolated in Gn,t (and hence in Pn,t ) becomes
exponentially small only when t = θ(n2 ), the proof of Theorem 1.2 cannot be strengthened to give
an exponentially high probability. However, it is a well-known result from random graph theory
that there is a large component in Gn,n log n with exponentially high probability. The following
proposition is easily proven by the first moment method. We will use it in the proofs of Lemma 4.3,
5.1, and 5.2.
Proposition 4.2. Let X be the size of a largest component in Gn,n log n . Then
2n
< e−3n .
P X ≤n−
log n
Next, we want to give a bound on the number of planar graphs that contain Pn,n log n . Essentially, we will prove that at this time, the spanning tree of the largest component is w.e.h.p.
contained in only exponentially many planar supergraphs.
Lemma 4.3. There exists Γ > 1
P [Pn,n log n is contained in more than Γn planar graphs ] < e−2n .
Proof. Let Fn be the set of all forests on [n] and let Fn0 ⊂ Fn denote the set of forests with
one component of size larger than n − 2n/ log n and all other vertices isolated. We construct the
˙ n and let {F, P } ∈ E(G) if and only if
following auxiliary bipartite graph G. Set V (G) := Fn0 ∪P
F ⊆ P.
Set Γ := 8γe6 , where γ is a constant such that |Pn | ≤ γ n n!, and define Fn00 := {F ∈ Fn0 :
degG (F ) > Γn }. Clearly,
P
degG (P )
00
|Fn | < P ∈Pn n
.
Γ
As
2n
log n −1 X
e(P )
2n 3n
degG (P ) ≤
<
< n23n ,
log
n
n
−
1
−
i
n
i=0
we obtain, for sufficiently large n,
|Fn00 |
|Pn |n23n
<
<
Γn
8γ
Γ
n
nn+1 .
(7)
Define a function f : SN −→ Fn by assigning to a permutation σ the forest obtained by deleting
all edges of T (σ|n log n ) which are not in a largest component. Fix some F ∈ Fn0 , denote the vertex
set of the non-trivial component by U , and set u := |U |. Clearly, σ induces a permutation of the
set of all edges with both endvertices in U , which we denote by σ|U . Observe that for σ ∈ f −1 (F ),
it is easy to see that T (σ|U ) = F [U ].
By Corollary 3.2, we know that there are less than u2 !(e/2)u /uu−2 permutations of the set
of all edges with both endvertices in U which
produce the tree F [U ]. By symmetry, any such
permutation is contained in exactly N !/ u2 ! permutations in SN , and thus we get, for sufficiently
large n,
n
e u
−1
N ! 2e
f (F ) < N ! 2 < n−2n/ log n−2
uu−2
2n
n − log
n
n
N ! 2e n2n/ log n+2
N !e3n
<
= n−2 .
(8)
n n
n
2
Let τ be a permutation chosen uniformly at random in the planar graph process and note that
f (τ ) ⊆ Pn,n log n . Define the following events:
15
A: Pn,n log n is contained in more than Γn planar graphs,
B: no component of Pn,n log n has size larger than n − 2n/ log n,
C: f (τ ) ∈ Fn00 .
We claim that A ⊆ B ∪ C. To see this, suppose that Pn,n log n has a component of size larger than
n − 2n/ log n. If Pn,n log n is contained in more than Γn planar graphs, then f (τ ) is also contained
in more than Γn planar graphs. Hence A ∩ B ⊆ C.
Therefore, Proposition 4.2 and Equations (7) and (8) imply that
−1
P
(F )
F ∈F 00 f
−3n
P [A] ≤ P [B] + P [C] < e
+
|SN |
n
8γ n n+1 N !e3n
n
8γe3
nn−2
< e−3n + Γ
n3 < e−2n ,
= e−3n +
N!
Γ
by our choice of Γ and for sufficiently large n.
The notion T (σ|t ), which we have introduced in Section 3 for the tree process, can easily be
extended to the planar graph process. So, for σ ∈ SN and 0 ≤ t ≤ N , the planar graph P (σ|t )
associated with σ|t is constructed as follows: start with the empty graph on [n] and consider the
t edges of σ|t in this order, accepting an edge into the present graph if and only if its addition
preserves planarity. We are now ready to prove Theorem 1.1.
Proof of Theorem 1.1. Define the following events:
A: e Pn,δn2 ≥ (1 + ε)n,
B: e (Pn,n log n ) ≤ 1 + 2ε n and Pn,n log n is contained in at most Γn planar graphs on
[n], where Γ is the constant from Lemma 4.3.
Proposition 4.1 and Lemma 4.3 imply that P B < e−3n + e−2n . As P [A] ≤ P [A|B] + P B , it
suffices to show that P [A|B] ≤ e−2n .
We define an equivalence relation ∼ on SN by writing σ ∼ π if and only if σ|n log n = π|n log n .
From the definition of B it follows that for any σ ∈ SS
N the equivalence class [σ] is either contained
in or disjoint from B. Hence, we trivially have B = [σ]⊆B [σ]. By Proposition 3.10, it suffices to
show that
∀[σ] ⊆ B :
P [A|[σ]] < e−2n
(9)
to obtain
 
[
P [A|B] = P A [σ]  ≤ e−2n ,
[σ]⊆B
as desired.
In order to show (9), fix an arbitrary σ in B. Set P0 := P (σ|n log n ) and note that for any
π ∈ [σ], we have P (π|n log n ) = P0 . As σ ∈ B, we know that P0 is contained in at most Γn planar
graphs with (1 + ε)n edges, say P1 , . . . , Pk . For a fixed Pi , 1 ≤ i ≤ k, we obtain in a similar way
as in Proposition 4.1
(δn2 − n log n)e(Pi )−e(P0 ) (N − n log n − (e(Pi ) − e(P0 ))!
P Pi ⊆ Pn,δn2 [σ] ≤
(N − n log n)!
! 2ε n
2
e(Pi )−e(P0 )
ε
δn − n log n
δn2
<
<
= (4δ) 2 n .
n2
N − n log n
4
Hence, we get
P [A|[σ]] ≤
k
X
(10)
ε
P Pi ⊆ Pn,t=δn2 [σ] < Γn (4δ) 2 n < e−2n ,
i=1
by choosing δ sufficiently small.
16
(10)
5
The add-function and copies of a planar graph
In this section we prove Theorem 1.3. We start by defining addable edges of a planar graph on
[n]. For P ∈ Pn , an edge e ∈ E(Kn ) \ E(P ) is called addable if (V (P ), E(P ) ∪ {e}) ∈ Pn . We
denote the set of all addable edges of P by Add(P ), and its size by add(P ). More generally, if P 0
is a subgraph of P , then we denote the set of all addable edges of P which are incident to at least
one vertex in P 0 by Add(P 0 , P ), and its size by add(P 0 , P ). At the beginning, i.e. when t = 0, add(Pn,t ) is simply n2 . On the other end, it is clear that
as long as the number of edges in the planar graph is less than (3 − ε)n, its add value is at least
linear. For the purpose of proving our main result about the containment of a fixed planar graph,
it will be crucial to know that w.e.h.p. the add-function is already linear when Pn,t has (1 + ε)n
edges.
Recall that Pn,m=m0 denotes the random planar graph after m0 edges have been accepted.
Analogously, for a permutation σ, we denote by σ|m=m0 the shortest initial sequence σ|t such that
e(P (σ|t )) = m0 .
Lemma 5.1. For every ε > 0, there exists ∆ > 0 such that
n
P add Pn,m=(1+ε)n ≥ ∆n < e− 2 .
Proof. Let δ be given by Theorem 1.1. We may assume that 0 ≤ (1−ε)n ≤ (1−2/ log n)n. Denote
by X the number of edges accepted in Pn,t during the time interval (n log n, δn2 ] and define the
following events:
A: add Pn,m=(1+ε)n ≥ ∆n,
B: e (Pn,n log n ) > (1 − ε)n,
C: e Pn,δn2 < (1 + ε)n,
D: add Pn,δn2 ≥ ∆n,
E: X < 2εn.
Observe that, by definition, A ∩ C ⊆ D and B ∩ C ⊆ E. Therefore, we get
P [A] ≤ P [A ∩ B ∩ C] + P B + P C ≤ P [D ∩ E] + P B + P C .
As P B < e−3n by Proposition 4.2 and P C < e−n by Theorem 1.1, it thus suffices to prove
P [D ∩ E] < e−n .
In order to show this, we visualize the random graph process in the following auxiliary rooted
tree R. In R, vertices at level i have N − i outgoing edges (the root vertex is assumed to have level
0). Clearly, there is a natural way to label the edges of R with edges of Kn in such a way that the
leaves of R correspond exactly to the permutations in SN . To visualize the random planar graph
process, colour an edge green if it is accepted by the process and black otherwise. Moreover, we
colour a leaf red, if the corresponding permutation σ is contained in D ∩ E.
Note that the path from a red leaf to the root has the following two properties:
(i) the vertex at level δn2 has at least ∆n outgoing green edges (and hence the same is true for
all vertices at level < δn2 , too),
(ii) the subpath of this path between the vertex at level n log n and the vertex at level δn2
contains less than 2εn green edges.
Colour a vertex at level δn2 red, if it has property (i) and its path to the root has property (ii).
Observe that
{red vertices at level δn2 } · QN −1 2 (N − i)
|{red leaves in R}|
i=δn
P [D ∩ E] =
≤
.
(11)
N!
N!
17
To make the counting in the last term easier, we recolour the edges of R as follows: if a vertex
has k > ∆n outgoing green edges, recolour k − ∆n of these outgoing green edges black, and if a
vertex has k < ∆n outgoing green edges, recolour ∆n − k of the outgoing black edges green.
Note that by property (i), the edges of the subpaths between a red vertex at level δn2 and the
root can only change from green to black and not vice versa. Hence, if we now apply the definition
of colouring a vertex at level δn2 red with respect to the new colouring, we can only gain but not
lose red vertices.
Since we now know that every vertex has exactly ∆n green outgoing edges, we can count red
vertices at level δn2 as follows. Set I := (n log n, δn2 ] and let S be an arbitrary subset of I. Then
the number of paths from the root to a vertex at level δn2 with a green edge between the vertex
at level i − 1 and the vertex at level i if and only if i is in S is exactly
n log
Yn−1
(N − i) ·
i=0
Y
Y
∆n ·
i∈S
(N − i + 1 − ∆n).
i∈I\S
Hence, by (11),
P [D ∩ E] ≤
2εn−1
1 X X
N ! s=0
n log
Yn−1
2εn−1
X
Y N − i + 1 − ∆n
∆n
·
N −i+1
N −i+1
i=0
S⊆I
|S|=s
=
s=0
X Y
S⊆I i∈S
|S|=s
Y
(N − i) ·
i∈S
∆n ·
Y
(N − i + 1 − ∆n) ·
i∈I\S
N
−1
Y
(N − i)
i=δn2
i∈I\S
|I|−s 2εn−1
s X δn2 4∆ s ∆n δn2
|I|
∆n
∆n
<
e− N 2
1
−
2
N
−
δn
N
n
s
s
s=0
s=0
!n
2ε
eδ4∆
< 2εn
e−∆δ
< e−n ,
2ε
<
2εn−1
X by choosing ∆ sufficiently large.
We need two more definitions. A B-leaf B 0 in P is called ∆-bounded if add(B 0 , P ) ≤ ∆.
Furthermore, two B-leaves B 0 , B 00 are called independent if Add(B 0 , P ) and Add(B 00 , P ) are
disjoint.
We now have enough information about Pn,m=(1+ε)n to prepare the ground for the proof of
Theorem 1.3. We already know w.e.h.p. that we have a linear sized component at t = n log n and
that this component contains a linear number of (pairwise vertex-disjoint) B-leaves. If we choose
ε small enough, we can certainly ensure that at least half of them survive until Pn,m=(1+ε)n . But
there we know w.e.h.p. that the add-function has become linear, which allows us to show that a
positive fraction of our B-leaves must have constant add-value at this stage. The following lemma
makes these arguments precise.
Lemma 5.2. Let B be a tree. For every 1 < d < 3, there exist positive constants ε < d − 1, ∆,
∆0 , and α0 , such that, if X denotes the size of a maximum family B of pairwise vertex-disjoint
∆0 -bounded, independent B-leaves in Pn,m=(1+ε)n , then
0
P (X ≤ α0 n) ∨ add Pn,m=(1+ε)n ≥ ∆n < e−α n .
Proof. Recall that, if τ is the permutation chosen uniformly in the random planar graph process,
Tn = T (τ ) is always contained in Pn,N . Let β be given by Theorem 3.9 and choose some 0 < ε <
d − 1 such that 2ε + 6/ log n ≤ β/2. Let ∆ be given by Lemma 5.1, set ∆0 := 8∆/β, and choose
some 0 < α0 < min{β/(4(∆0 + 1)), β, 1/2}. If in the following we speak of a collection of disjoint
B-leaves, then we mean that they should be pairwise vertex-disjoint. Define the following events:
18
A: Tn contains more than βn B-leaves,
C: e(Pn,n log n ) < (1 + ε)n,
D: Pn,n log n has a component of size larger than n −
E: Pn,m=(1+ε)n contains more than
F : add Pn,m=(1+ε)n < ∆n,
βn
2
2n
log n ,
disjoint B-leaves,
G: Pn,m=(1+ε)n contains more than βn/4 disjoint, ∆0 -bounded B-leaves,
H: Pn,m=(1+ε)n contains more than α0 n disjoint, ∆0 -bounded, independent B-leaves.
We will prove that A ∩ C ∩ D ⊆ E, E ∩ F ⊆ G, and G ⊆ H. Note that this is sufficient to establish
the lemma as it implies
P H ∪ F ≤P G∩F ≤P E∩F ≤P A∩C ∩D∩F
0
n
≤ P A + P C + P D + P F < e−βn + e−3n + e−3n + e− 2 < e−α n ,
using Theorem 3.9, Proposition 4.2, Proposition 4.1, and Lemma 5.1.
To see that A ∩ C ∩ D ⊆ E, consider σ ∈ A ∩ C ∩ D, let B 0 be a family of more than βn B-leaves
in T (σ), set P0 := P (σ|n log n ) and P1 := P (σ|m=(1+ε)n ), and let K be the largest component in
P0 . As σ ∈ C, we know that T (σ)[K] ⊆ P0 ⊆ P1 . As we have mentioned earlier, two B-leaves
in a large enough tree have to be vertex-disjoint, and therefore less than 2n/ log n elements of B 0
can contain vertices outside K. Moreover, each edge in E(P1 ) \ E(T (σ)[K]) is clearly incident to
vertices in at most two different B-leaves, so we deduce that P1 still contains a family B 00 of more
than
2n
2n
2n
βn −
− 2 (|E(P1 )| − |E(T (σ)[K])|) ≥ βn −
− 2 (1 + ε)n − n −
log n
log n
log n
6
βn
= β − 2ε +
n≥
log n
2
disjoint B-leaves, i.e. σ ∈ E.
Next, consider σ ∈ E ∩ F and let B 00 be a family of more than βn/2 disjoint B-leaves in
P1 := P (σ|m=(1+ε)n ). Suppose that at least βn/4 of these B-leaves are not ∆0 -bounded. As
the endvertices of any edge can lie in at most two different B-leaves, there are thus at least
1/2 · βn/4 · ∆0 = ∆n addable edges for P1 , contradicting σ ∈ F . Hence, more than half of the
elements in B 00 are ∆0 -bounded, so that σ ∈ G.
Finally, consider σ ∈ G and let B 000 be a family of more than βn/4 disjoint, ∆0 -bounded Bleaves in P1 := P (σ|m=(1+ε)n ). Construct an auxiliary graph G0 with these B-leaves as vertices
and an edge between two B-leaves B 0 , B 00 if and only if Add(P1 ) contains an edge between two
vertices in B 0 and B 00 . As the B-leaves are ∆0 -bounded, the maximum degree of G0 is less than
∆0 , and hence there is an independent set B in G0 of size at least
v(G0 )
βn
>
> α0 n,
0
maxdeg(G ) + 1
4(∆0 + 1)
which shows that σ ∈ H.
We are finally in the position to prove our second main result:
Proof of Theorem 1.3. Denote by Y the size of a maximum family of pairwise vertex-disjoint Hleaves in Pn,m=dn and fix a spanning tree B in H. Let ε < d − 1, ∆, ∆0 , and α0 be given
by Lemma 5.2, and let X denote the size of a maximum family of pairwise vertex-disjoint, ∆0 bounded, independent B-leaves in Pn,m=(1+ε)n . Set ε0 := d − 1 − ε, choose some 0 < ε00 <
0
min{3 − d, ε0 }, set β := min{α0 , ε00 /(2∆0 )}, and p := (ε00 /(3∆))∆ , choose 0 < α < β/2 such that
19
1 + log(β/α) − p(β/α − 1) < −2, and let A be the event (X > α0 n) ∧ (add(Pn,m=(1+ε)n ) < ∆n).
Then Lemma 5.2 yields
0
P [Y ≤ αn] ≤ P [Y ≤ αn|A] + P A < P [Y ≤ αn|A] + e−α n .
(12)
Define an equivalence relation ∼ on SN by writing σ ∼ π if and only if σ|m=(1+ε)n = π|m=(1+ε)n .
Then for any σ S
∈ SN , the equivalence class [σ] is either contained in or disjoint from A, which
shows that A = [σ]⊆A [σ]. We now claim that
∀[σ] ⊆ A :
P [Y ≤ αn|[σ]] < e−2αn .
(13)
Observe that this is enough to complete the proof, as

[
0
[σ]  + e−α n
= P Y ≤ αn [σ]⊆A

(12)
0
P [Y ≤ αn] < P [Y ≤ αn|A] + e−α n
3.10,(13)
<
0
e−2αn + e−α n < 2e−2αn < e−αn .
In order to prove (13), we fix an arbitrary equivalence class [σ] ⊆ A and consider a family {Bi : 1 ≤
i ≤ βn} of βn pairwise vertex-disjoint, ∆0 -bounded, independent B-leaves in P0 := P (σ|m=(1+ε)n )
which exists according to Lemma 5.2 and the definition of A and β. Denote by Ei a set of edges
in Add(Bi , P0 ) such that (V (Bi ), E(Bi ) ∪ Ei ) is an H-leaf. Note that the permutation chosen
uniformly in the planar graph process induces a linear order on Add(P0 ), as well as on each
Add(Bi , P0 ), which allows us to regard them as ordered sets. For each 1 ≤ i ≤ βn, define the
following event:
Ci :
all edges in Ei are among the first ε0 n edges in Add(P0 )
and all edges in Add(Bi , P0 ) \ Ei are among the last ε00 n edges in Add(P0 ).
Observe that, if Pn,m=(1+ε)n = P0 , then all the first ε0 n edges and none of the last ε00 n edges in
Add(P0 ) will be considered until Pn,m=dn , since 1 + ε < 1 + ε + ε0 = d < 3 − ε00 and, trivially,
add(Pn,m=dn ) ≥ 3n − 6 − dn > ε00 n. Therefore, if Ci ∩ [σ] holds, then Pn,m=(1+ε)n = P0 and every
edge in Ei , but no edge in Add(Bi , P0 ) \ Ei lies in E(Pn,m=dn ), so that Pn,m=dn [V (Bi )] will be an
H-leaf. Denoting the indicator variable of Ci by Zi , we thus arrive at
#
" βn
X
P [Y ≤ αn|[σ]] ≤ P
Zi ≤ αn [σ] .
(14)
i=1
We now claim that, for sufficiently large n,

\
Cj ∩ [σ]  ≥ p.
P Ci j∈J

∀i ∈ [βn] ∀J ⊆ [βn] \ {i} :
(15)
T
Add(P0 ) \
For
S any fixed i ∈ [βn] and J ⊆ [βn] \ {i}, set C := j∈J Cj ∩ [σ] and E :=
0
Add(B
,
P
).
Define
an
equivalence
relation
∼
on
S
by
writing
π
∼
π
if
and only
j
0
C
N
C
j∈J
if π(e) = π 0 (e) for all e not in E. Again, each equivalence class [π]C is either contained in or
disjoint from C, so that by Proposition 3.10 it is sufficient to prove P [Ci |[π]C ] ≥ p for an arbitrary
[π]C ⊆ C.
It is easy to see that |[π]C | = |E|!, and we get a lower bound for |Ci ∩ [π]C | by constructing
permutations as follows. First, place all edges not in E at the positions determined by [π]C . Place
then the edges in Ei within the first ε0 n/2 free positions. Next, place the edges in Add(Bi , P0 ) \ Ei
within the last ε00 n/2 free positions. Finally, assign all other edges in E to the remaining free
S
positions. Observe that by our choice of β, we know that j∈J Add(Bj , P0 ) < βn · ∆0 ≤
ε00 · n/2 < ε0 · n/2, and hence Ci holds. So there are at least
0 |Ei | 00 |Add(Bi ,P0 )\Ei |
ε n
εn
|E \ Add(Bi , P0 )|!
2
2
20
permutations in Ci ∩ [π]C . Therefore, using add(Bi , P0 ) ≤ ∆0 and |E| ≤ add(P0 ) < ∆n, we get
P [Ci |[π]C ] =
>
ε0 n
2
|Ei | ε00 n
2
|Add(Bi ,P0 )\Ei |
|Ci ∩ [π]C |
≥
|Add(Bi ,P0 )|
|[π]C |
|E|
0
|Ei | 00
|Add(Bi ,P0 )\Ei |
εn
ε n
0
0
−
∆
−
∆
2
2
|Add(Bi ,P0 )|
|E|
>
ε00 n
2
− ∆0
∆n
!∆0
>
ε00
3∆
∆ 0
= p,
for sufficiently large n, which proves our second claim in (15).
We are left to show that this settles the first claim in (13). The following computation is
analogous to the one in the proof of Theorem 3.9. Again, we will denote the complement of an
index subset I ⊆ [βn] of size s by {i1 , . . . , iβn−s }, for notational convenience.
#
#
" βn
"
αn X
X
X
\
\
(14)
Zi ≤ αn [σ] =
P
P [Y ≤ αn|[σ]] ≤ P
Ci ∩
Ci [σ]
s=0 I⊆[βn]
|I|=s
i=0
≤
"
αn X
X
P
s=0 I⊆[βn]
|I|=s
αn
(15) X
≤
<
X
\
i∈I
/
#
αn X
X
Ci [σ] =
α
i∈I
/
βn−s
Y
"
P Cij
s=0 I⊆[βn] j=1
|I|=s
βn−s
(1 − p)
s=0 I⊆[βn]
|I|=s
αn
eβ
−p(β−α)n
e
i∈I
=
αn X
βn
s
s=0
j−1
#
\
Cik ∩ [σ]
k=1
βn−s
(1 − p)
βn
< (αn + 1)
(1 − p)(β−α)n
αn
β
β
−1
,
= exp αn 1 + log − p
α
α
which is smaller than e−2αn by our choice of α.
6
Open Problems
The fact that Pn,m=αn is w.h.p. connected for 1 < α < 3, whereas the uniform random planar graph
of the same density is not, may seem to be a relatively weak distinction criterion between these
two models of random planar graphs — what if, for example, we condition on the uniform random
planar graph being connected? We conjecture that the two models for random planar graphs
differ also in this case. This is related to the following question: how different is a triangulation
generated by the random process from the one generated uniformly at random?
Another open problem is what happens with the add-function between the rejection of the first
edge and the time when the density of our random planar graph process becomes asymptotically
larger than 1. We know that w.h.p.
Ω(n2 ) : α < 1/2,
add(Pn,m=αn ) =
O(n) : α > 1,
but when and how fast does it decrease? A question intimately related to the previous one is, of
course, how the two time scales are related in this phase: how many edges do we have to consider
for, say, Pn,m=3n/4 ?
It is a well-known fact [2] that the first appearance of a graph H as a subgraph in Gn,t depends
on the maximum density over all subgraphs of H, so that, e.g., K4 appears around t = n4/3 . But
our method can guarantee a high probability for the appearance of K4 only after Ω(n2 ) edges have
been considered. Thus one may ask for the the smallest t such that Pn,t contains a copy of K4
with high probability.
A somewhat related question is the following one raised by Luczak [13]: Is the probability that
one planar subgraph H1 appears in Pn,t before another planar subgraph H2 always bounded away
21
from 0? Observe that in Gn,t , H1 appears w.h.p. before H2 if it has smaller maximum subgraph
density.
Acknowledgements We would like to thank the referees for their helpful comments.
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22