Random directed trees and forest – drainage networks with dependence

Consider the d -dimensional lattice Z d where each vertex is ‘open’ or ‘closed’ with probability p or 1 − p respectively. An open vertex v is connected by an edge to the closest open vertex w in the 45 ◦ (downward) light cone generated at v . In case of non-uniqueness of such a vertex w , we choose any one of the closest vertices with equal probability and independently of the other random mechanisms. It is shown that this random graph is a tree almost surely for d = 2 and 3 and it is an inﬁnite collection of distinct trees for d ≥ 4. In addition, for any dimension, we show that there is no bi-inﬁnite path in the tree.


Introduction
During the last two decades there has been a considerable amount of study to understand the structure of random spanning trees. In particular, for the uniform spanning tree model the tree/forest dichotomy according to the dimension of the lattice was established by Pemantle [13]. Also, for the Euclidean minimal weight spanning tree/forest model Alexander [1] showed that the two dimensional structure of the random graph is that of a tree and Newman and Stein [12] through a study of the fractal dimension of the incipient cluster in the Bernoulli bond percolation problem suggest that the random graph is a forest in suitably high dimensions.
Lately there has been an interest in studying these random spanning trees where the edges have a preferred direction of propagation. These studies have been motivated by studies of Alpine drainage patterns (see e.g., Leopold and Langbein [10], Scheidegger [15], Howard [9]). In a survey of such models, Rodriguez-Iturbe and Rinaldo [14] have explored (non-rigorously) power law structures and other physical phenomenon, while Nandi and Manna [11] obtained relations between these 'river networks' and scale-free networks. Also, in connection with the small-world phenomenon, Baccelli and Bordenave [3] and Bordenave [4] have studied the navigation properties of such directed spanning trees.
Gangopadhyay, Roy and Sarkar [8]  In case of non-uniqueness of such a vertex w, any one of the closest open vertices was chosen with equal probability and independent of the other random mechanisms. They established that for d = 2 and 3, the random graph constructed above is a tree, while for d ≥ 4, the graph is a forest (i.e. infinitely many trees). Ferrari, Landim and Thorisson [6] have obtained a similar dichotomy for a continuous version of this model which they termed Poisson trees. In this model, the vertices are Poisson points in d and, given a Poisson point u, it is connected to another Poisson point v by an edge if (i) the first (d − 1) co-ordinates of v lie in a (d − 1) dimensional ball of a fixed radius r centred at the first (d − 1) co-ordinates of u and (ii) if v is the first such point from u in the direction specified by the dth co-ordinate.
Mathematically these models are also attractive by their obvious connection to the Brownian web as described by Fontes, Isopi, Newman and Ravishankar [7]. In particular, Ferrari, Fontes and Wu [5] have shown that, when properly rescaled, Poisson trees converge to the Brownian web. Tóth and Werner [17] considered coalescing oriented random walks on 2 , oriented so as to allow steps only upwards or rightwards. Wilson's method of 'rooted at infinity' associates to this Markov chain a wired random spanning tree on 2 . Tóth and Werner also obtained an invariance principle for the Markov chain they studied.
Motivated by the above, we consider a general class of such models. Here a source of water is connected by an edge to the nearest (see Figure 1) source lying downstream in a 45 degree light cone generating from the source. Like above a uniform choice is adopted in case of non-uniqueness. The random graph obtained by this construction is the object of study in this paper. We establish that for d = 2 and 3, all the tributaries connect to form a single riverine delta, while for d ≥ 4, there are infinitely many riverine deltas, each with its own distinct set of tributaries. Further we also show that there are no bi-infinite paths in this oriented random graph. The angle of the light cone is unimportant here, and our results are valid for any angle.
These results are similar to those obtained by Gangopadhyay, Roy and Sarkar [8] and as in that paper the results are obtained by a combination of a Lyapunov method and a random walk comparison. However, unlike the Gangopadhyay, Roy and Sarkar [8] or the Ferrari, Landim and Thorisson [6], there is no 'longitudinal' or 'lateral' independence in the generalised set-up we study here. As such, the dependence structure brings its own complicacies resulting in more intricate arguments in both the Lyapunov and random walk methods. In particular, the dependence induces a 'history' in the processes, which has to be carefully incorporated to obtain a suitable Markov model for comparison.
In the next subsection we define the model precisely, state our main results and compare it with the existing results in the literature.

Main Results
Before we describe the model we fix some notation which describe special regions in d . For u = (u 1 , . . . , u d ) ∈ d and k ≥ 1 let m k (u) = (u 1 , . . . , We equip Ω = {0, 1} d with the σ-algebra generated by finite-dimensional cylinder sets and a product probability measure P p defined through its marginals as On another probability space (Ξ, , µ) we accommodate the collection {U u,v : v ∈ Λ(u), u ∈ d } of i.i.d. uniform (0, 1) random variables. The random graph, defined on the product space (Ω × Ξ, × , := P p × µ), is given by the vertex set and the (almost surely unique) edge set The graph = ( , ) is the object of our study here. The construction of the edge-set ensures that, almost surely, there is exactly one edge going 'down' and, as such, each connected component of the graph is a tree.
Our first result discusses the structure of the graph and the second result discusses the structure of each connected component of the graph.
and ||v−m k (u)|| L 1 ≤ ak} then also our results hold. In the case a = ∞, this corresponds to the model considered in [8]. The results would also generalise to the model considered in [6].
Using the notation as in (2), we chose the "nearest" vertex at level h uniformly among all open vertices available at that level to connect to the vertex u. One could relax the latter and choose among all open vertices available at level h in any random fashion. If the random fashion is symmetric in nature then our results will still hold.
For proving Theorem 1 we first observe that the river flowing down from any open point u is a random walk on d . The walk jumps downward only in the d-th coordinate and also conditional on this jump the new position in the first d − 1 coordinates are given by a symmetric distribution. Then the broad idea of establishing the results for the case d = 2, 3 is that we show that two random walks starting from two arbitrary open points u and v meet in finite time almost surely. The random walks are dependent and as oppposed to the model considered in [8] they also carry information,(which we call history), as they traverse downwards in d . The second fact makes the problem harder to work with. Consequently, we construct a suitable Markov chain which carries the "history set" along with it and then find a suitable Lyaponov function to establish the Foster's criterion for recurrence (See Lemma 2.2). We also benefit from the observation that this "history set" is always a triangle. The precise definitions and the complete proof of Theorem 1 (i), is presented in Section 2.
For proving Theorem 1 (ii) one shows that two random walks starting from two open points u and v which are far away do not ever meet. Once the starting points are far away one is able to couple these dependent random walks with a system of independent random walks. The coupling has to be done carefully because of the "history set" information in the two walks. To finish the proof one needs estimates regarding intersections of two independent random walks and these may be of intrinsic interest (See Lemma 3.2 and Lemma 3.3).The details are worked out in Section 3.
The proof of Theorem 2 requires a delicate use of the Burton-Keane argument regarding the embedding of trees in an Euclidean space. This is carried out in Section 4.

Dimensions and 3
In this section we prove Theorem 1(i). We present the proof for d = 2 and later outline the modifications required to prove the theorem for d = 3. To begin with we have a collection {U u,v : v ∈ Λ(u, h), h ≥ 1, u ∈ d } of i.i.d. uniform (0, 1) random variables.
Consider two distinct vertices u := (u 1 , u 2 ) and v : This ensures that u ∈ Λ(v, h), v ∈ Λ(u, h) for any h ≥ 1. We will show that given u and v open, they are contained in the same component of with probability 1.
This suffices to prove the theorem, because if two open vertices u and v do not satisfy the condition (3) then, almost surely, we may get open vertices w 1 , . . . , w n such that each of the pairs w i and w i+1 as well as the pairs u and w 1 , and v and w n satisfy the condition (3). This ensures that all the above vertices, and hence both u and v belong to the same component of .
To prove our contention we construct the process dynamically from the two vertices u and v as given in (3). The construction will guarantee that the process obtained is equivalent in law to the marginal distributions of the 'trunk' of the trees as seen from u and v in . Without loss of generality we take u := (0, 0) and v := (l 0 , −1).
Note that all the processes we construct in this section are independent of those constructed in Section 1.1. Before we embark on the formal details of the construction we present the main ideas. From the points u and v we look at the region Λ(u) ∪ Λ(v). On this region we label the vertices open or closed independently and find the vertices u 1 and v 1 to which u and v connect (respectively) according to the mechanism of constructing edges given in Section 1.1. Having found u 1 and v 1 we do the same process again for these vertices. However now we have to remember that we are carrying a history, i.e., there is a region, given by the shaded triangle in Figure 4, whose configuration we know. In case the vertical distance between u 1 and v 1 is much larger than the horizontal distance or the history set is non-empty we move the vertex on the top (see Figures 5 and 6), otherwise we move both the vertices simultaneously (see Figure 7).
Note that in this construction, the history will always be a triangular region. Also, at every move during the construction of the process we will keep a track of the vertical height, the horizontal length and the history. This triplet will form a Markov process whose recurrence properties will be used to establish Theorem 1(i). We now present a formal construction of the model. In Section 3, we will give a simpler construction, however for the purposes of establishing the Markovian structure as stated in the preceding paragraph, the following construction comes in handy.
Now let ω u,v ∈ {0, 1} Λ(u)∪Λ(v) . Let h u := inf{h : ω u,v (w) = 1 for some w ∈ Λ(u, h)} and h v := inf{h : ω u,v (w) = 1 for some w ∈ Λ(v, h)}. Note that under the product measure as defined earlier via the marginals (1) on {0, 1} Λ(u)∪Λ(v) , the quantities h u and h v are finite for -almost all ω u,v . Let Having obtained u k := (u k (1), u k (2)), v k := (v k (1), v k (2)) and H k , L k and V k we consider the following cases (i) if V k = or if V k = and H k /L k ≥ C 0 for a constant C 0 to be specified later in (15) and (a) if u k (2) ≥ v k (2) (see Figure 5 for V k = and Figure 6 for V k = ) then we set Let and let h u k := inf{h : ω u k ,v k (w) = 1 for some w ∈ Λ(u k , h)}. Again under the product measure on Λ(u k ) such a h u k is finite almost surely. Now let u k+1 : Again under the product measure the quantities h u k and h v k are finite for -almost all ω u k ,v k .
We take is empty -the set which is non-empty is necessarily a triangle).
We will now briefly sketch the connection between the above construction and the model as described in Section 1.1. For this with a slight abuse of notation we take ω u k ,v k to be the sample point Thus for every sample path from u and v obtained by our construction we obtain a realisation of our graph with u and v open. Conversely if ω ∈ Ω gives a realisation of our graph with u and v open then we label vertices u i and v i as vertices such that will correspond to the concatenation of the ω u k ,v k we obtained through the construction.  (2)) then we make the following observations about the history region V k .

Observations:
(i) V k is either empty or a triangle (i.e. the shaded region in the figure), (vii) the vertex u k+1 may lie on the base of the triangle, but not anywhere else in the triangle.
While observations (ii) -(vi) are self-evident, the reason for (i) above is that if the history region has two or more triangles, then there must necessarily be a fourth vertex w besides the vertices under consideration, u, v and v k−1 which initiated the second triangle. This vertex w must either be a vertex u j for some j ≤ k − 1 or a vertex v j for some j ≤ k − 2. In the former case, the history region due to w must lie above u k and in the latter case it must lie above v k−1 . In either case it cannot have any non-empty intersection with the region Λ(u).
In Figure 8  In Figure 8 (b) where the vertex v lies on the base of the shaded triangle, the vertex u k+1 may lie anywhere in the triangle subtended by the sides emanating from the vertex u and the horizontal line.
otherwise we obtain v k+1 by considering the region Λ(v) and remembering that in obtaining u k+1 we may have already specified the configuration of a part of the region in Λ(v).
The new history region V k+1 is now determined by the vertices u k , v k , u k+1 and v k+1 .
This justifies our claim that {(H k , L k , V k ) : k ≥ 0} is a Markov process.
We will now show that We use the Foster-Lyapunov criterion (see Asmussen [2], Proposition 5.3, Chapter I) to establish this result.

Proposition 2.1. (Foster-Lyapunov criterion) An irreducible Markov chain with stationary transition
For this we change the Markov process slightly. We define a new Markov process with state space which has the same transition probabilities as {(H k , L k , V k ) : k ≥ 0} except that instead of (0, 0, ) being an absorbing state, we introduce a transition We will now show using Lyapunov's method that this modified Markov chain is recurrent. This will imply that {(H k , L k , V k ) = (0, 0, ) : for some k ≥ 0} = 1. With a slight abuse of notation we let the modified Markov chain be denoted by Define a function g : + → + by : Some properties of g : Thus, using Taylor's expansion and above formulae for derivatives, we have two inequalities: for and Also, we define u : → + by where |V | denotes the cardinality of V .

Lemma 2.2. For all but finitely many
Proof : Before we embark on the proof we observe that using the inequality (7) and the expression (9), we have that Let a 0 = 0 and for k ≥ 1 let b k = 2k + 1 and a k = k i=1 b i . We define two integer valued random variables T and D whose distributions are given by: For Let 0 < C 0 < 1 be small enough so that We shall consider three cases for establishing (11) Case 1: V is non-empty.
We will prove the following inequalities: and where C 1 , C 2 , C 3 and C 4 are positive constants. Putting the inequalities (16), (17) and (18) in (12), for all (h, l, V ) ∈ with V non-empty, we have for all (h, l) such that h + l sufficiently large. Therefore, outside a finite number of choices of (h, l), we have the above inequality. Now, for these finite choices for (h, l), there are only finitely many possible choices of V which is non-empty. This takes care of the case when the history is non-empty. Now, we prove (16), (17) and (18). Define the random variables and (1)).
It is easy to see that the conditional distribution of H n+1 given (H n , L n , V n ) = (h, l, V ) is same as that of the conditional distribution of |T n+1 − h| given u 0 , . . . , u n , v 0 , . . . , v n and h = |u n (2) − v n (2)|.
Note that, for any given non-empty set V , as noted in observations (iv), at least one diagonal line will be unexplored, and consequently the conditional distribution T n+1 given u 0 , . . . , u n , v 0 , . . . , v n , is dominated by a geometric random variable G with parameter p. Thus, for any k ≥ 1 for a suitable choice of C 1 .
Similarly, we have that the conditional distribution of L n+1 given (H n , L n , V n ) = (h, l, V ) is same as that of |D n+1 − l| given u 0 , . . . , u n , v 0 , . . . , v n and l = |u n (1) − v n (1)|. Since D n ≤ 2T n for every n, we have, For the inequality (18), we require the following observations: Further, when T n+1 ≤ h, we have This is seen in the following way. We look at the case when T n+1 = 1 and connect to the point which will always be unexplored. In that case, |V | − |V n+1 | ≥ 1. Note that if both points on the line T n+1 = 1 were available, this probability is at least p(1 − p). If both points were not available, then there are two possible cases, i.e., the history point on the top line is open or the history point on the top line is closed. In the first case, the probability is p/2 while in the second case the probability is p.
Thus, we have, where C 3 and C 4 are positive constants. This completes the proof in the case when history is nonnull.
Case 2: V = , h |l| ≥ C 0 For this case, we use the inequality (12) with |V | = 0. We will show that, for all h large enough, where C 5 , C 6 , C 7 and C 8 are positive constants.
Using the above estimates (22), (23) and (24) in (12) with |V | = 0, we have, Now, we prove the estimates (22), (23) and (24). It is easy to see that the conditional distribution of H n+1 given (H n , L n , V n ) = (h, l, ) is same as that of |T n+1 − h| where T n+1 is as defined in (19). We note here that V = ensures that T n+1 does not depend on the path u 0 , . . . , u n , v 0 , . . . , v n . Therefore, we have Again, we have that the conditional distribution of L n+1 given (H n , L n , V n ) = (h, l, ) is same as that of |D n+1 − l|. Here too we note here that V = ensures that D n+1 does not depend on the path u 0 , . . . , u n , v 0 , . . . , v n . Therefore, we have, for suitable choices of positive constants C 7 and C 8 .
Taking conditional expectation and denoting, ( We want to show that: for all l large enough, where β 1 , β 2 are as defined earlier and C 9 , C 10 , C 11 and C 12 are positive constants. Using the above inequalities in (25), and the definition of C 0 , we have, −48(1 + h 4 + l 4 )l 6 β 1 + C 10 l 9 + C 11 exp(−C 12 l) < 0 for all l large enough, because the leading term in the first inequality is of the order l −2 and it appears with the coefficient [36(β 1 + C 2 0 β 2 ) − 48β 1 ] which is negative by our choice of C 0 given in (15).

Proof of (26) and (27):
Let us assume without loss of generality that v n (1) = u n (1) + l and v n (2) = u n (2) (1) and T u n+1 = u n (2) − u n+1 (2), D u n+1 = u n (1) − u n+1 (1). Using this notation, we have Now, it is easy to observe that if both T u n+1 < (l +h)/2 and T v n+1 < (l −h)/2, both of them will behave independently and the distribution on that set is same as that of T . Further, the tail probabilities of the height distribution T decays exponentially. Thus, for some constant C 13 > 0 and for all j ≥ 1.
Since 0 < C 0 < 1 we have for all j ≥ 1 and suitable constants C 14 , C 15 > 0, where T 1 , T 2 are i.i.d. copies of T and D 1 , D 2 are i.i.d copies of D.
Now, to conclude (26), we just need the observation that all odd moments of the terms T 1 − T 2 and D 1 − D 2 are 0. Thus in the conditional expectation of (30), we see that the terms involving h 3 and l 3 do not contribute, the coefficient of l 2 in the second term contributes 6 (D 1 − D 2 ) 2 , the coefficient of h 2 contributes 6 (T 1 − T 2 ) 2 and all other terms have smaller powers of h and l. From the fact that h/l < C 0 and our choice of C 0 given by (15), we conclude the result.
To show (27), studying R 2 n , we note that there are only three terms which are important, (a) coefficient of l 6 , (b) coefficient of h 6 and (c) coefficient of h 3 l 3 . All other terms, are of type h i l j have i + j < 6 and, since h/l < C 0 , these terms are of order smaller than l 6 . The coefficient of l 6 is 16 Proof of (28): In the expansion of R 3 n , all the terms are of type h i l j with i + j ≤ 9. Thus, using the fact that h/l < C 0 , and all terms have finite expectation, we conclude the result.

Proof of (29) : Finally, the history set
. Again, since the tails probabilities decay exponentially, the above expectations decay exponentially with l.
This completes the proof of Lemma 2.2 and we obtain that holds outside l ≥ l 0 , h ≤ h 0 and |V | ≤ k for some constants l 0 , h 0 and k.
Our Markov chain being irreducible on its state space, by the Foster-Lyapunov criteria (Lemma 2.1) we have that it is recurrent. Therefore, we have For d = 3 we need to consider the 'width', i.e. the displacement in the third dimension. Thus we have now a process (H n , L n , W n , V n ), n ≥ 0 where H n is the displacement in the direction of propagation of the tree (i.e. the third coordinate), and L n and W n are the lateral displacements in the first and the second coordinates respectively. Now the history region V n would be a tetrahedron. Instead of (10) we now consider the Lyapunov function and g(x) = log(1 + x) is as in (6). This yields the required recurrence of the state (0, 0, 0, ) for the Markov process (H n , L n , W n , V n ) thereby completing the proof the first part of Theorem 1.

d ≥ 4
For notational simplicity we present the proof only for d = 4. We first claim that on 4 the graph admits two distinct trees with positive probability, i.e., We start with two distinct open vertices and follow the mechanism described below to generate the trees emanating from these vertices. Given two vertices u and v, we say u ≻ v if u(4) > v(4) or if u(4) = v(4), and u is larger than v in the lexicographic order. Starting with two distinct open vertices, u and v with u ≻ v we set Let R(u) ∈ 4 be the unique open vertex such that 〈u, R(u)〉 ∈ . We set u 1 = max{R(u), v} and v 1 = min{R(u), v} where the maximum and minimum is over vectors and henceforth understood to be with respect to the ordering ≻. We define, The set V 1 is exactly the history set in this case.
Having defined, {T u,v (k) : k = 0, 1, . . . , n} for n ≥ 1, we define T u,v (n + 1) in the same manner. Let R(u n ) ∈ 4 be the unique open vertex such that 〈u n , R(u n )〉 ∈ . Define u n+1 = max{R(u n ), v n }, v n+1 = min{R(u n ), v n } and The process {T u,v (k) : k ≥ 0} tracks the position of trees after the k th step of the algorithm defined above along with the history carried at each stage. Clearly, if u k = v k for some k ≥ 1, the trees emanating from u and v meet while the event that the trees emanating from u and v never meet corresponds to the event that {u k = v k : k ≥ 1}.
A formal construction of the above process is achieved in the following manner. We start with an independent collection of i.i.d. random variables, For ω ∈ Ω, we define k u 0 = k u 0 (ω) as and H (z) : z ∈ V n+1 } as follows: for ω ∈ Ω, we define k u n = k u n (ω) as This construction shows that the chain starting  at (u 0 , v 0 , , ). A formal proof that this Markov chain describes the joint distribution of the trees emanating from the vertices u and v can be given in the same manner as in Lemma 2.1.
We will work towards the proof of Lemma 3.1. Towards that, we introduce an independent version of the above process. In the same probability space, starting with two vertices u and v, and the same set of uniformly distributed random variables, define u The independent version tracks the two trees, emanating from the vertices u and v, with the condition that the trees do not depend on the information (history) carried. The only constraint is that while growing the tree from a vertex, it waits for the tree from the other vertex to catch up, before taking the next step. Note that if the history set is empty, then both constructions match exactly.
We define an event similar to A n,ε but in terms of {(u and two open vertices u, v ∈ and define the event, We will show that the following Lemma holds: Lemma 3.2. For 0 < ε < 1/3 there exist constants C 2 , γ > 0 and n 0 ≥ 1 such that, for all n ≥ n 0 , First we prove Lemma 3.1, assuming Lemma 3.2. Proof of Lemma 3.1: Given 0 < ε < 1/3, fix n 0 ≥ 1 from Lemma 3.2. Now fix n ≥ n 0 and u, v ∈ 4 such that n 1−ε ≤ u − v 1 ≤ n 1+ε and 0 ≤ u(4) − v(4) < log n. Now note that both the events A n,ε (u, v) and B n,ε (u, v) are defined on the same probability space.
We claim that A n,ε (u, v) ⊇ B n,ε (u, v). To prove this, we show that, on the set B n,ε (u, v), we have that This follows easily from the observation that if V k = , for some k ≥ 0, then the two constructions, given before, match exactly.
. Furthermore, on the event B n,ε (u, v), we have that u i+1 − v i+1 1 ≥ log(n 2 ) and 0 ≤ u i+1 (4) − v i+1 (4) < log n 2 . From the definition of the history set, the separation of u and v implies that V i+1 = . Therefore the claim follows by an induction argument. Thus, we have Hence, Lemma 3.1 follows by choosing C 1 = C 2 and β = γ.
Then, the event B n,ε can be restated in terms of S In order to study the event C n,ε (s), we have to look at the steps taken by u The construction clearly shows that each {X k : k ≥ 1} is a sequence of i.i.d. random variables.
The distribution of X k can be easily found. Let a 0 = 0, and for i ≥ 1, a i = |Λ(0, i)| and b i = |H(0, i)|. Define a random variable T on {1, 2, . . . }, given by Now, define D on 3 as follows: (4) for k ≥ 1. We will show that the first three co-ordinates of both the processes have the same norm. In other words, We postpone the proof of Lemma 3.4 for the time being. We define a random walk version of the event C n,ε (s). For n ≥ 1 and 0 < ε < 1/3 and s ≻ 0, define Note that the events E n,ε , F n,ε and G n,ε depend on the random walk part while H n,ε depends on the fourth co-ordinate of S (RW) k . Also note that k j=1 D j is an aperiodic, isotropic, symmetric random walk whose steps are i.i.d. with each step having the same distribution as D where ar(D) = σ 2 I, for some σ > 0 and z∈ 3 z 2 (D = z) < ∞. The events F n,ε and G n,ε are exactly as in Lemma 3.3 of Gangopadhyay, Roy and Sarkar [8]. Hence, we conclude that there exist constants C 3 , C 4 > 0 and α > 0 such that for all n sufficiently large, sup n 1−ε ≤ s 1 ≤n 1+ε , 0≤s(4)<log n F n,ε = sup s (1) ∈∆ n 1+ε \∆ n 1+ε F n,ε ≤ C 3 n −α and sup n 1−ε ≤ s 1 ≤n 1+ε , 0≤s(4)<log n G n,ε = sup The probability of the event E n,ε can be computed in the same fashion as in Lemma 3.3 of Gangopadhyay, Roy and Sarkar [8]. Indeed, we have, E n,ε = r k ≤ 2 log n for some k = 1, 2, . . . , for some suitable positive constant C 5 .
Finally, we are left with the proof of Lemma 3.4.

Proof of Lemma 3.4:
We define an intermediate process on 4 ×{−1, 1} in the following way. Given s ≻ 0 and the steps {(D i , −T i ) : i ≥ 1}, define (S 0 , F 0 ) = (s, 1) and for k ≥ 1, Using induction, it is easy to check that where z (1) is the first three co-ordinates of z. Therefore, we have that for k ≥ 1, S k 1 = s (1) are identical in distribution. Note that, from the definition of S (I) k andS k , we can write, for k ≥ 0, where f : 4 × 3 × → 4 is a suitably defined function. The exact form of f is unimportant, the only observation that is crucial is that we can use the same f for both the cases. This establishes (39) and completes the proof of Lemma 3.4.
Hence we have shown that { is disconnected} > 0 and, by the inherent ergodicity of the process, this implies that A similar argument along with the ergodicity of the random graph, may further be used to establish that for any k ≥ 1 { has at least k trees} = 1.
Consequently, we have that k≥1 { has at least k trees } = 1 and thus { has infinitely many trees} = 1.

Geometry of the graph
We now show that the tree(s) are not bi-infinite almost surely. For this argument, we consider, d = 2. Similar arguments, with minor modifications go through for any dimensions.
Thus, we have that Now, call a vertex x ∈ R t a branching point if there exist distinct points x 1 and x 2 such that x 1 , x 2 ∈ B 1 (x) and B n (x 1 ) = , B n (x 2 ) = for all n ≥ 1, i.e, x has at least two distinct infinite branches of ancestors.
We first show that, if Since (|R 0 | = ∞) > 0, we may fix two vertices x = (x 1 , 0) and y = ( y 1 , 0) such that Suppose that x 1 < y 1 Now, we define C t as the set of all vertices with their second co-ordinates strictly larger than t, each vertex having infinite ancestry and its immediate offspring having the second co-ordinate at most t, i.e., C t = {y = ( y 1 , u) ∈ : u > t, B n (y) = for n ≥ 1, and h(y) = (x 1 , v), with v ≤ t}. Since every open vertex has a unique offspring, for each y ∈ C t , the edge joining y and h(y) intersects the line L t at a single point only, say (I y (t), t). We define, for all n ≥ 1, C t (n) := {y ∈ C t : 0 ≤ I y (t) < n} and r t (n) = |C t (n)|.
We show that (r t (n)) < ∞ and consequently r t (n) is finite almost surely. First, note that C t (n) = ∪ n j=1 {y ∈ C t : j − 1 ≤ I y (t) < j}.
Since the sets on the right hand side of above equality are disjoint, we have r t (n) = n j=1 r ( j) t where r ( j) t = |{y ∈ C t : j − 1 ≤ I y (t) < j}| for 1 ≤ j ≤ n. By the translation invariance of the model, it is clear that the marginal distribution of r ( j) t , are the same for 1 ≤ j ≤ n. Thus, it is enough to show that (r (1) t ) < ∞. Now, we observe that {y ∈ C t : 0 ≤ I y (t) < 1} ⊆ U t := {y = ( y 1 , u) ∈ : u > t, h(y) = (x 1 , v), v ≤ t, 0 ≤ I y (t) < 1}. The second set represents the vertices whose second co-ordinates are strictly larger than t, for each such vertex its immediate offspring having a second co-ordinate at most t and the edge connecting the vertex and its offspring intersecting the line L t at some point between 0 and 1. Note that we have relaxed the condition of having infinite ancestry of the vertices above.
Set a i = 1 + 2i, for i = 1, 2, . . . and s i = i j=1 a j . Here, a i is the number of vertices on the line L t+i which can possibly be included in the set U t . Now, if s i+1 ≥ |U t | > s i , then some vertex x whose second co-ordinate is at least (t + i + 1) will connect to some vertex y whose second co-ordinate is at most t. Thus, the probability of such an event is dominated by the probability of the event that the vertices in the cone Λ(y) up to the level x are closed. Since there are at least i 2 − 1 many vertices in this region, we have Thus, (|U t |) < ∞.
Since every vertex has a unique offspring, we must have, D 1 ∩ D 2 = . Further, by definition of C 1 (n), we have, D 1 ∪ D 2 ⊆ C 1 (n). Also, it is clear that (D 1 ∪ D 2 ) ∩ C (2) 0 (n; 1) = as the offspring of any vertex in D 1 and D 2 lie on the line L 1 while the offspring of any vertex in C (2) 0 (n; 1) lie on L u for u ≤ 0. Thus, we have,