The random planar graph process

Transcription

1 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(n 2 ) 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 The study of the random graph process (G n,t ) N t=0, where one starts with an empty graph on n vertices and adds all N := ( 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 (P n,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(n 3/2 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(n 2 ) edges. Aldous [1] investigated the random cycle-free graph process (T n,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, P n,t0, denotes the random planar graph obtained after t 0 edges have been considered. P n,m=m0, on the other hand, describes ETH Zurich, Institute of Theoretical Computer Science, 8092 Zurich, Switzerland, {sgerke,steger}@inf.ethz.ch Humboldt University Berlin, Institute of Computer Science, Unter den Linden 6, Berlin, Germany, schlatter@informatik.hu-berlin.de. 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, Garching bei München, Germany, taraz@ma.tum.de 1

2 the random planar graph after m 0 edges have been accepted. As edges between vertices in different components are always accepted, it is obvious that T n,t P n,t G n,t for all t = 0,..., N. Thus, after the connectivity threshold for G n,t which lies at t = n log n/2 P n,t must have at least n 1 edges with high probability. The following theorem, which states that we have to consider Ω(n 2 ) 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 ( P n,δn 2) (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 P n,m=dn. Theorem 1.2. For every 1 < d < 3, P [P n,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 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 of H which is connected to the rest of G only via a single edge incident with the smallest vertex in H. 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 ˆP n,m=dn which is chosen uniformly at random from the class of all simple labelled planar graphs with n vertices and dn edges: [ ] P ˆPn,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 [P n,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 (T n,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 P n,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

3 that there exist at most γ n n! different labelled planar graphs to show that with high probability there exist at most Γ n different labelled planar graphs that contain the graph P n,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 P n,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 T n,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 P n,n log n contains only n+o(n) edges, the edges that are contained in P n,n log n but not in T n,n log n can destroy at most o(n) of these B-leaves. That is, we know that P n,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 δn 2 edges are added. Theorem 1.1 tells us that from these δn 2 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 P n,δn 2 still contains with high probability a linear number of these B-leaves. Moreover, the fact that at most εn out of δn 2 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 Γ G (v) we denote the set of neighbours of a vertex v. We write the falling factorial n (n 1) (n k + 1) as n k. We denote the set of all trees on [n] by T n and the set of all planar graphs on [n] by P n. If the probability of an event A n which depends on n tends to 1 as n tends to infinity, this is often described by saying that A n occurs with high probability (w.h.p.). As indicated by our main result, Theorem 1.3, we shall often want faster convergence: A n 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 T n = T n,n, the final outcome of the random tree process (T n,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., T n will contain a linear number of B-leaves. We start with some definitions. Let T be a subtree of a tree T. A vertex v V (T ) is called an anchor in T, if there exists a vertex u / V (T ) such that uv E(T ). The set of anchors of T is denoted by A(T ). A proper vertex subset U V (T ) is called k-autonomous if T [U] is a subtree and has at most k anchors. 3

4 . T a 1 B B a 2 U.. 1 u v = Figure 1: a B-leaf B and a 2-autonomous set U in a tree T Thus, if B is a B-leaf in T, then V (B ) is a 1-autonomous set with the smallest vertex as its anchor. By S N we denote the set of all permutations on the edge set of the complete graph K n on n vertices. Let σ S N. If an edge e appears in the ith position we write σ(e) = i, and we say that an edge e 1 is smaller than an edge e 2 (with respect to σ) if σ(e 1 ) < σ(e 2 ). 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 T n and set S(T ) := {σ S N T (σ) = T }. Then P [T n = T ] = and we call w(t) := S(T) the weight of the tree T. 3.1 Local rearrangements S(T ), (1) N! 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 T n and xy E(T ), such that x, y are not leaves, and set {x 1,..., x k } := Γ(x) \ {y}. Construct a tree T by see Figure 2. Then w(t ) w(t ). E (T ) := (E(T ) \ {xx 1,..., xx k }) {yx 1,..., yx k }, 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 (e) = σ(xz) : e = yz, z W \ {x, y}, σ(yz) : e = xz, z W \ {x, y}, σ(e) : otherwise. 4

5 T T x y... U x y... U x 1... x k x 1... x k W W Figure 2: the local tree operation in Lemma 3.1 In other words, we obtain σ from σ as follows: for every vertex z in a subtree pending from one of the x i, we transpose the positions of xz and yz. We now claim that σ S(T ). As we can reobtain σ from σ without ambiguity, this implies S(T ) S(T ) and thus completes the proof. To prove this claim, let φ : T [W ] T [W ] be the isomorphism given by interchanging x and y, i.e. φ(w) = x : w = y, y : w = x, w : otherwise, and φ(w 1 w 2 ) = φ(w 1 )φ(w 2 ). We want to show that σ generates T, or equivalently, that it generates T [W ] = φ(t [W ]) on W and T [U] = T [U] on U. More precisely, we prove the following statements h 1 (t), h 2 (t), and h 3 (t) for all 0 t N by induction: h 1 (t) : h 2 (t) : T (σ t ) contains no red edge, T (σ t )[U] = T (σ t )[U], h 3 (t) : T (σ t )[W ] = φ (T (σ t )[W ]). Observe that these statements trivially hold for t = 0, assume that all three statements hold for t 0 < N, and let ab be the edge at the (t 0 + 1)st position of σ. Suppose that ab is green. As U and W overlap only in y and, according to h 1 (t 0 ), no red edges have been accepted by σ so far, the acceptance of ab by σ depends only on the presence of a green path between a and b in T (σ t0 ). By h 3 (t 0 ), 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 we know that the (t 0 + 1)st edge in σ is φ 1 (a)φ 1 (b). Thus we conclude that ab is accepted by σ if and only if φ 1 (a)φ 1 (b) is accepted by σ in the (t 0 + 1)st step. Hence h 3 (t 0 + 1) holds, as do h 1 (t 0 + 1) and h 2 (t 0 + 1) trivially. The case that ab is blue can be dealt with similarly. In fact, this case is easier as σ and σ 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 h 2 (t 0 + 1) and h 3 (t 0 + 1) follow trivially from h 2 (t 0 ) and h 3 (t 0 ). It thus suffices to check that σ will not accept ab. Observe that since b U \ {y}, the edge ab has the same position in σ as in σ. 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 h 3 (t 0 ) and h 2 (t 0 ), the edges of the path P must 5

6 T T. y P < z.. y P < z. x x Figure 3: the local tree operation in Lemma 3.3 exist in T (σ t0 ) too, and hence xb will be rejected by σ. Now assume that a 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 t0 ) contains a path from a to y (due to h 3 (t 0 )) and a path from y to b (due to h 2 (t 0 )), and hence ab will be rejected by σ as well. If we apply Lemma 3.1 iteratively until there is only one non-leaf left, we arrive at a star K 1,n 1, which thus maximizes w(t ) over T T n. From [1] we know that P [T n = K 1,n 1 ] = Using Stirling s formula, we note that, for sufficiently large n, (n 1)!2n 1. (2) (2n 2)! (n 1)!2 n 1 (2n 2)! 2n ( e ) n (e/2) n 2e 2n n n 2. Lemma 3.1 and (2) thus imply Corollary 3.2. ( e ) n N! max w(t ) < T T n 2 n n 2. By (1), this upper bound on the weight of the heaviest tree also provides an upper bound on how much the distribution of T n deviates from the uniform distribution on T n. We now describe our second local tree operation. Lemma 3.3. Let T T n, 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, E, and ρ, respectively (see Figure 3). Then S(T ) S(E, ρ) = S(T ) S(E, ρ ). Proof. By symmetry, it suffices to prove that S(T ) S(E, ρ) S(T ) S(E, ρ ). 6

7 a a a a f( v 1 ) v 1 f( v 2 ) v 2 f( v 3 ) v v 4 1 v 1 v 1 v v 2 v 2 3 v 3 T 1 T 2 T 3 T 4 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 define σ(xz) : e = xy, σ (e) = σ(xy) : e = xz, σ(e) : otherwise. We proceed to show that σ S(T ) S(E, ρ ). As we can reobtain σ from σ, this implies S(T ) S(E, ρ) S(T ) S(E, ρ ) as required. It is easy to see that σ lies in S(E, ρ ), so we are left to prove that σ generates T. We prove the following statement by induction on t. For each 0 t N, h(t) : T (σ t ) and T (σ 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 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 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 ) = T because of T (σ) = T and the way σ was constructed. For the induction, we see that h(0) holds trivially and assume that h(t 0 1) is true for 0 t 0 1 < N. Set t 1 := σ(xy) = σ (xz) and t 2 := σ(xz) = σ (xy), and observe that t 1 < t 2. For t 0 / {t 1, t 2 }, σ and σ agree in the t 0 -th position and h(t 0 ) follows easily from h(t 0 1). If t 0 = t 1, 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 t 1 1, and hence we know that z cannot be in the same component as x. By h(t 1 1), σ will therefore accept xz, joining the two components in the same way as σ and thus proving h(t 1 ). The same arguments apply for the case t 0 = t 2, 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 be a subtree of T T n with exactly one anchor a. Let T be the tree obtained from T by deleting all edges in T and attaching all non-anchor vertices in T to a as leaves. Then w(t ) w(t ) (v(t ) 1)!. Proof. For two vertices u, v in T, we denote by dist(u, v) the length of the path from u to v in T. Order the non-anchor vertices of T such that dist(a, v i ) dist(a, v j ) for all 1 i < j v(t ) 1, 7

8 a a T T b C b C T T 0 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 ) 1 : d(a, v i ) > 1}. For 1 i l, let f(v i ) be the neighbour of v i on the path between v i and a in T. We construct a sequence of trees T = T 1,..., T l+1 as follows: get T i+1 from T i by setting E(T i+1 ) := (E(T i ) \ {f(v i )v i }) {av i } (cf. Figure 4). It is easy to see that T l+1 = T. Set E i := E(T i [V (T )]) for 1 i l + 1. Furthermore, let ρ l+1 be defined by av v(t ) 1 < < av 1 and for 1 i l, get ρ i from ρ i+1 by replacing av i with f(v i )v i. Observe that in this way, for each 1 i l, we ensure that in T i the edges on the path from f(v i ) to a are smaller than f(v i )v i with respect to ρ i. We can therefore apply Lemma 3.3 for each 1 i l with T = T i, x = v i, y = f(v i ), z = a, E = E i, and ρ = ρ i. Note that for this setting, T = T i+1, E = E i+1, and ρ = ρ i+1, so that S(T 1 ) S(E 1, ρ 1 ) = S(T 2 ) S(E 2, ρ 2 ). = S(T l+1 ) S(E l+1, ρ l+1 ). As S(T l+1 ) S(E l+1, ρ l+1 ) = w(t )/(v(t ) 1)! by the symmetry of the star, the claimed inequality follows from w(t ) S(T 1 ) S(E 1, ρ 1 ). We will now combine Lemma 3.1 and Lemma 3.4 in order to change a subtree T which lives on a 2-autonomous set into any required subtree T on the same vertex set but with only one anchor. Corollary 3.5. Let T be a subtree of T T n with anchor set A(T ) = {a, b} and T be another tree on V (T ). Set C := {c / V (T ) : bc E(T )} and construct a tree T 0 from T by replacing T with T and bc with ac, for each c C (see Figure 5). Then w (T 0 ) w(t ) (v(t ) 1)!. Proof. If a b, let a = a 0,..., a l = b be the vertices of the path between a and b. Apply Lemma 3.1 iteratively l times: first on T with (x, y) = (a l, a l 1 ), then on the resulting tree with (x, y) = (a l 1, a l 2 ), and so on. This will result in a tree T 1 for which V (T ) is 1-autonomous, Γ T (a) C Γ T1 (a), and w(t 1 ) 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 from that substar. In the first of these two phases, 8

9 the weight of the corresponding tree will not decrease at all, and in the second phase only by a factor of 1/(v(T ) 1)!. 3.2 Partial covering by small subtrees We now describe an algorithm which covers a constant fraction of the vertices in any T T n 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 Output: a sequence of k := n 10v subsets C 1,..., C k [n] such that, 1 i < j k, (i) v C i 2v 2, (ii) C i is 2-autonomous, (iii) C i C j 1, i.e. T [C i ] and T [C j ] 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 the subtree of T that is induced by all vertices that are not of type Say that a non-root vertex in T is of type 1, 2, or 3, if its degree in T 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, 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. 4. We then construct an order ρ on V (T ) E(T ) 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 x 1 < < x l be the vertices at distance d + 1 in the order given by the vertex labels. Then we first add the edges to x 1,..., x l and then the vertices themselves (both in precisely this order) to ρ. 5. Let (C 1,..., C k ) 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 < C i < v), or loaded (if C i 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, 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 ), which has type 0 neighbours x 1,..., x l ordered by their labels. We describe the packing in this round by the following pseudo-code: 9

10 C := smallest empty bin {x} for j = 1 to l C := C V (T (x j )) if C v C := next bin {x} if C < v C := 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 ). If e also lies in E(T ), we do nothing. Otherwise, e was created by contracting a maximal path in T whose inner vertices x 1,..., x m are of type 2, where dist(1, x i ) < dist(1, x j ) for 1 i < j m. Let them have l 1,..., l m neighbours x h,j (1 h m and 1 j l h ) of type 0, respectively, where for a fixed h, the x h,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 {x h } if C v C := next bin {x h } for j = 1 to l h C := C V (T (x h,j )) if C v C := next bin {x h } if C < v C := In other words, we start as in the first case with x 1 in place of x. However, once we have placed all subtrees pending from type 0 neighbours of x 1, we also place x 2 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 x h 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 x 1,..., x m. 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 C i are loaded, or all rounds for w V (T ) E(T ) 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 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 C i 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

11 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 C i 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 of T [{x 1,..., x m }]. 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, all type 0 neighbours must lie in this bin, so only the leaves of P 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 n := v(t ), we also observe that there are 2n 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} < (2n 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 and there are at least v vertices of type 0 in T (x). Summing degrees in T and writing r for the number of type 1 vertices, we get 2(n 1) 1 + r + 3(n r 1), where the first 1 is caused by the root. Hence r n /2. Therefore, As α + β = 1, we can combine (3) and (4) to get βn := {vertices covered} rv n v 2. (4) βn (1 β)n > n v/2 2n v = 1 4, which tells us that at least n/5 vertices are covered. Therefore, there have to be at least n 10v 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 T n, we denote by sc(t ) = (C 1,..., C k ) 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 [C i ] in such a way that the resulting overall tree T i has the same partial vertex covering and T i [C i ] contains a B-leaf. We emphasise that the main technical difficulty with our construction of a B-leaf on a subset of C i is that we wish to preserve the partial vertex covering, and here the labels of the vertices are becoming important. Lemma 3.7. Let B be a tree and T T n and set v := 3(v(B) + 1) and k := n 10v. For every 1 i k, there exists a tree T i T n such that 11

12 (i) sc(t i ) = sc(t ) =: (C 1,..., C k ), (ii) T [([n] \ C i ) A(T [C i ])] = T i [([n] \ C i ) A(T [C i ])], (iii) T i [C i ] contains a B-leaf, (iv) w(t i ) 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 C i. Case 1: C i was loaded in a round R corresponding to some vertex x V (T ). Let x 1 < < x s be the neighbours of x which are contained in C i. Note that x 1,..., x s are all of type 0. If a tree rooted at one of the x j has at least v(b) + 1 vertices, then we transform T (x j ) into the following tree. We arbitrarily select a subset V V (T (x j )) \ {x j } of size v(b) and construct the (unique) copy B of B on V. Then we select the smallest vertex v in V and connect it to x j. Finally, we attach all the vertices in the (possibly empty) set V (T (x j )) \ (V {x j }) to x j. By Corollary 3.5 we can perform this transformation such that for the new overall tree T i we have w(t i ) w(t )/(v 2)!, as v(t (x j )) < v, and so Condition (iv) is fulfilled. Observe that Condition (i) is also satisfied since the sizes of the trees with roots x 1,..., x s have not changed and in particular vertices in these trees remain type 0 vertices. We have not modified anything in T [([n] \ C i ) A(T [C i ])] and thus Condition (ii) is fulfilled. By construction, T i [C i ] contains a B-leaf (namely B ) 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 x λ,..., x s 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 x λ,..., x s into the following tree. Build a copy of B on a set of v(b) arbitrary vertices not containing x or x s, select the smallest vertex v in B, and connect v, as well as x and all remaining vertices to x s. Let T i 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 C i \ {x} remain of type 0. When T and the set C i were considered, the algorithm added x s to C i which implies that C i without the vertices of the tree rooted at x s has size less than v. Hence, when considering T i, the algorithm adds first the trees rooted at x 1,..., x λ 1, followed by the new subtree rooted at x s, which proves Condition (i). Condition (ii) and Condition (iii) are clearly satisfied by construction. Case 2: C i was loaded in a round R corresponding to some edge e E(T ) that was obtained by contracting a maximal path in T whose inner vertices were of type 2. Let x 1,..., x s denote the subpath of this path which consists of the vertices that are contained in C i. Assume that dist(1, x 1)... dist(1, x s). If the type 0 neighbours (in C i ) of either x 1 or x s 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 x 1 or x s, respectively. Thus, if we consider the set M consisting of x 2,..., x s 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 x s to C i shows that M < v 1. Using Corollary 3.5, this time with x 2 and x s 1 as anchors of T [M], we construct a tree on T [M] consisting of x 2 connected to the smallest vertex in a copy B of B, and all vertices in M \ (V (B ) {x 2}). Note that Corollary 3.5 implies that x 2 is now also connected to x 1 and x s, and that for the new overall tree T i we have w(t i ) w(t ) ( M 1)! w(t ) (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(t n ) contain B-leaves. To do so, we need the following definition. 12

13 For every 1 i k, we define an equivalence relation i on S N by σ i π if and only if sc(t (σ)) = sc(t (π)) =: (C 1,..., C k ) [ ] [ ] and T (σ) ([n] \ C i ) A(T (σ)[c i ]) = T (π) ([n] \ C i ) A(T (π)[c i ]). We can now state and prove the following lemma. Lemma 3.8. Let B be a tree, set v := 3(v(B) + 1) and k := n 10v, and for 1 i k let [σ] i be an equivalence class of i. Furthermore, let sc(t (σ)) = (C 1,..., C k ) 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)[c i ] 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 T i T such that T i [C i ] contains a B-leaf and such that w(t ) (3v(B) + 1)! w(t i ). Hence, we get and therefore also w(t ) (3v(B) + 1)! {σ [σ] i : T (σ)[c i ] contains a B-leaf} [σ] i = T T w(t ) T (3v(B) + 1)! {σ [σ] i : T (σ)[c i ] contains a B-leaf}. As C i 2v 2 = 6v(B) + 4, there are at most (6v(B) + 4) 6v(B)+2 different trees on C i and thus T (6v(B) + 4) 6v(B)+2. We are now able to prove that for any fixed tree B, the tree T n 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 T n. 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 Let E 1,..., E k be pairwise [ disjoint events. Then any event F that satisfies P [F E i ] q for all i [k], also satisfies P F ] k q. i=1 E i Proof of Theorem 3.9. Set p := 1/((3v(B) + 1)! (6v(B) + 4) 6v(B)+2 ), v := 3(v(B) + 1), and k := n 10v. 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 S N, and let (C 1,..., C k ) := sc(t (τ)) denote the output of SearchCandidates. For 1 i k, define the event A i : T (τ)[c i ] contains a B-leaf and let X i be the corresponding indicator variable. Set X := k i=1 X i. 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 S N : P [A i [σ] i ] p. (5) We claim that this implies that in fact i [k] J [k] \ {i} : P A i j J A j p. (6) 13

14 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 C i. As any two sets C j and C k share at most one vertex, this implies that for any σ S N the equivalence class [σ] i is either contained in A j or completely disjoint from A j (for all j i). We can therefore write A := j J A j as A = [σ] [σ] i A 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 {i 1,..., i k s }. P [X βn] = βn s=0 I [k] I =s (6) βn < [ P s=0 I [k] I =s ( ek βn i I A i i/ I (1 p) k s = A i ] βn s=0 ) βn e p(k βn) = exp ( βn which is smaller than e βn by our choice of β. βn [ ] P A i = s=0 I [k] I =s i / I ( ) k (1 p) k s < (βn + 1) s 4 Connectedness and containment βn [ k s P s=0 I [k] j=1 I =s A ij ( ) k (1 p) k βn βn ( 1 + log k ( ))) k βn p βn 1, j 1 l=1 A il ] In this section we prove Theorems 1.1 and 1.2. We start by establishing an upper bound on the number of edges in P n,n log n. We will use the well-known result that there exists a positive constant γ 1 such that the number P n of simple labelled planar graphs on the vertex set [n] satisfies P n 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. [ ( P e(p n,n log n ) n 1 + )] 2 log log n < 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 P n,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, Therefore, e(p ) 1 P [P P n,n log n ] (n log n)e(p ) (N e(p ))! = N! i=0 ( ) e(p ) ( ) e(p ) n log n 3 log n <. N n P [e (P n,n log n ) n + f(n)] P P n e(p )=n+f(n) by our choice of f(n) and for sufficiently large n. Theorem 1.2 follows now easily. n log n i N i ( ) n+f(n) 3 log n P [P P n,n log n ] < γ n n n n (3γ log n)n+f(n) < = exp ((n + f(n)) log(3γ log n) f(n) log n) n f(n) < e 3n, 14

15 Proof of Theorem 1.2. Due to the connectivity threshold of random graphs [6] and Proposition 4.1, we know that, as n, P [P n,n log n is connected] = P [G n,n log n is connected] 1 and P [e(p n,n log n ) = n + o(n)] 1, respectively. As the probability that any fixed vertex remains isolated in G n,t (and hence in P n,t ) becomes exponentially small only when t = θ(n 2 ), 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 G n,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 G n,n log n. Then [ P X n 2n ] < e 3n. log n Next, we want to give a bound on the number of planar graphs that contain P n,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 [P n,n log n is contained in more than Γ n planar graphs ] < e 2n. Proof. Let F n be the set of all forests on [n] and let F n F n denote the set of forests with one component of size larger than n 2n/ log n and all other vertices isolated. We construct the following auxiliary bipartite graph G. Set V (G) := F n P n and let {F, P } E(G) if and only if F P. Set Γ := 8γe 6, where γ is a constant such that P n γ n n!, and define F n := {F F n : deg G (F ) > Γ n }. Clearly, As deg G (P ) we obtain, for sufficiently large n, F n < 2n log n 1 i=0 P P n deg G (P ) Γ n. ( ) e(p ) < 2n n 1 i log n F n < P n n2 3n Γ n < ( ) 3n < n2 3n, n ( ) n 8γ n n+1. (7) Γ Define a function f : S N F n 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 F n, 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 ( u 2)!(e/2) u /u u 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!/ ( u 2)! permutations in SN, and thus we get, for sufficiently large n, ) u f 1 (F ) < N! ( e 2 u u 2 < ( n 2n log n ) n 2 ) n 2n/ log n 2 N! ( e < N! ( ) e n 2 n 2n/ log n+2 ( n ) n = N!e3n. (8) nn 2 2 Let τ be a permutation chosen uniformly at random in the planar graph process and note that f(τ) P n,n log n. Define the following events: 15

16 A: P n,n log n is contained in more than Γ n planar graphs, B: no component of P n,n log n has size larger than n 2n/ log n, C: f(τ) F n. We claim that A B C. To see this, suppose that P n,n log n has a component of size larger than n 2n/ log n. If P n,n log n is contained in more than Γ n planar graphs, then f(τ) is also contained in more than Γ n planar graphs. Hence A B C. Therefore, Proposition 4.2 and Equations (7) and (8) imply that P [A] P [B] + P [C] < e 3n F F f 1 (F ) + S N ( 8γ ) n < e 3n Γ n n+1 N!e 3n ( ) n 8γe + n 2 = e 3n 3 n + n 3 < e 2n, 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 S N 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 ( P n,δn 2) (1 + ε)n, B: e (P n,n log n ) ( 1 + ε 2) n and Pn,n log n is contained in at most Γ n 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 S N 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 S 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 P 0 := P (σ n log n ) and note that for any π [σ], we have P (π n log n ) = P 0. As σ B, we know that P 0 is contained in at most Γ n planar graphs with (1 + ε)n edges, say P 1,..., P k. For a fixed P i, 1 i k, we obtain in a similar way as in Proposition 4.1 P [ ] P i P n,δn 2 (δn 2 n log n) e(pi) e(p0) (N n log n (e(p i ) e(p 0 ))! [σ] (N n log n)! Hence, we get P [A [σ]] < by choosing δ sufficiently small. ( δn 2 ) e(pi) e(p n log n 0) < N n log n k i=1 ( δn 2 P [ ] (10) P i P n,t=δn 2 [σ] < Γ n (4δ) ε 2 n < e 2n, n 2 4 ) ε 2 n = (4δ) ε 2 n. (10) 16

17 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 P n, an edge e E(K n ) \ E(P ) is called addable if (V (P ), E(P ) {e}) P n. We denote the set of all addable edges of P by Add(P ), and its size by add(p ). More generally, if P 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 by Add(P, P ), and its size by add(p, P ). At the beginning, i.e. when t = 0, add(p n,t ) is simply ( n 2). 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 P n,t has (1 + ε)n edges. Recall that P n,m=m0 denotes the random planar graph after m 0 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 )) = m 0. Lemma 5.1. For every ε > 0, there exists > 0 such that P [ add ( P n,m=(1+ε)n ) n ] < e n 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 P n,t during the time interval (n log n, δn 2 ] and define the following events: A: add ( P n,m=(1+ε)n ) n, B: e (P n,n log n ) > (1 ε)n, C: e ( P n,δn 2) < (1 + ε)n, D: add ( P n,δn 2) 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 K n in such a way that the leaves of R correspond exactly to the permutations in S N. 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 δn 2 has at least n outgoing green edges (and hence the same is true for all vertices at level < δn 2, too), (ii) the subpath of this path between the vertex at level n log n and the vertex at level δn 2 contains less than 2εn green edges. Colour a vertex at level δn 2 red, if it has property (i) and its path to the root has property (ii). Observe that {red leaves in R} {red vertices at level δn 2 } N 1 i=δn2(n i) P [D E] =. (11) N! N! 17

18 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 δn 2 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 δn 2 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 δn 2 as follows. Set I := (n log n, δn 2 ] and let S be an arbitrary subset of I. Then the number of paths from the root to a vertex at level δn 2 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 Hence, by (11), P [D E] 1 N! = 2εn 1 s=0 2εn 1 s=0 n log n 1 i=0 S I S =s S I i S S =s (N i) n n log n 1 i=0 n N i + 1 i S (N i + 1 n). i I\S (N i) n i I\S i S 2εn 1 ( ) ( I n < s N δn 2 s=0 ( (eδ4 ) ) 2ε n < 2εn e δ < e n, 2ε by choosing sufficiently large. (N i + 1 n) i I\S N i + 1 n N i + 1 ) s ( 1 n ) I s < N 2εn 1 s=0 ( δn 2 s ) ( 4 n N 1 i=δn 2 (N i) ) s e n δn 2 N 2 We need two more definitions. A B-leaf B in P is called -bounded if add(b, P ). Furthermore, two B-leaves B, B are called independent if Add(B, P ) and Add(B, P ) are disjoint. We now have enough information about P n,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 P n,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,,, and α, such that, if X denotes the size of a maximum family B of pairwise vertex-disjoint -bounded, independent B-leaves in P n,m=(1+ε)n, then P [ (X α n) ( add ( P n,m=(1+ε)n ) n )] < e α n. Proof. Recall that, if τ is the permutation chosen uniformly in the random planar graph process, T n = T (τ) is always contained in P n,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 := 8 /β, and choose some 0 < α < min{β/(4( + 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