The Degree Sequence of Random Apollonian Networks

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📝 Original Info

  • Title: The Degree Sequence of Random Apollonian Networks
  • ArXiv ID: 1106.1940
  • Date: 2011-06-13
  • Authors: Charalampos E. Tsourakakis

📝 Abstract

We analyze the asymptotic behavior of the degree sequence of Random Apollonian Networks \cite{maximal}. For previous weaker results see \cite{comment,maximal}.

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1. Introduction 1.1. Results. Random Apollonian Networks (RANs) is a popular model of planar graphs with power law properties [11], see also Section 1. 3. In this note we analyze the degree sequence of RANs. For earlier, weaker results see [10,11]. Our main result is Theorem 1.

Theorem 1. Let Z k (t) denote the number of vertices of degree k at time t, k ≥ 3. For any k ≥ 3 there exists a constant b k (depending on k) such that for t sufficiently large |E [Z k (t)]b k t| ≤ K, where K = 3.6. Furthermore let λ > 0. For any k ≥ 3

1.2. Related Work. Bollobás, Riordan, Spencer and Tusnády [3] proved rigorously the power law distribution of the Barabási-Albert model [2]. Random Apollonian Networks were introduced in [11]. Their degree sequence was analyzed inaccurately in [11] (see comment in [10]) and subsequently in [10] using physicist’s methodology. Cooper & Uehara [4] and Gao [8] analyzed the degree distribution of random k trees, a closely related model to RANs. In RANs -in contrast to random k trees-the random k clique chosen at each step has never previously been selected. For example in the two dimensional case any chosen triangular face is being subdivided into three new triangular faces by connecting the incoming vertex to the vertices of the boundary. Darrasse and Soria analyzed the degree distribution of random Apollonian network structures in [6].

1.3. Model. For convenience, we summarize the RAN model here, see also [11]. A RAN is generated by starting with a triangular face and doing the following until the network reaches the desired size: pick a triangular face uniformly at random, insert a vertex inside the sampled face and connect it to the vertices of the boundary.

1.4. Prerequisites. In Section 2 we invoke the following lemma.

Lemma 1 (Lemma 3.1, [5]). Suppose that a sequence {a t } satisfies the recurrence

)a t + c t for t ≥ t 0 . Furthermore suppose lim t→+∞ b t = b > 0 and lim t→+∞ c t = c. Then lim t→+∞ at t exists and

We also use in Section 2 the Azuma-Hoeffding inequality 2, see [1,9].

Lemma 2 (Azuma-Hoeffding inequality). Let (X t ) n t=0 be a martingale sequence with |X t+1 -X t | ≤ c for t = 0, . . . , n -1. Also, let λ > 0. Then:

denote the degree of vertex v at time t and let 1(d v (t) = k) be an indicator variable which equals 1 if d v (t) = k, otherwise 0. Then, for any k ≥ 3 we can express the expected number N k (t) of vertices of degree k as a sum of expectations of indicator variables:

We distinguish two cases in the following.

• Case 1 k = 3:

Observe that a vertex of degree 3 is created only by an insertion of a new vertex. The expectation N 3 (t) satisfies the following recurrence1

The basis for Recurrence (2) is N 3 (1) = 4. We prove the following lemma which shows that lim t→+∞ N 3 (t) t = 2 5 . Lemma 3. N 3 (t) satisfies the following inequality:

Proof. We use induction. Assume that N 3 (t) = 2 5 t + e 3 (t). We wish to prove that for all t, |e 3 (t)| ≤ 3.6. The result trivially holds for t = 1. Assume the result holds for some t. We shall show it holds for t + 1.

Hence by induction, Inequality (3) holds for all t ≥ 1.

• Case 2 k ≥ 4:

For k ≥ 4 the following Equation holds for each indicator variable 1(d v (t) = k):

Therefore, substituting in Equation ( 1) the expression from Equation 4we obtain for k ≥ 4

Now, we use induction to show that lim t→+∞ N k (t) t

for k ≥ 4 exists.

Proof. For k = 3 the result holds by Lemma 3. We use induction. Rewrite Recurrence (5) as:

) which shows a power law degree distribution with exponent 3.

Finally consider the proof of Inequality (6). The case k = 3 was proved in the Lemma 3. Assume the result holds for some k ≥ 3, i.e., |e k (t)| ≤ K where K = 3.6. We will show it holds for k + 1 too. Let e k (t) = N k (t)-b k t. Substituting in Recurrence (5) and using the fact that b k-1 (k-1) = b k (k+2) we obtain the following:

Hence by induction, Inequality ( 6) holds for all k ≥ 3.

It’s worth pointing out that Lemma 4 agrees with [7] where it was shown that the maximum degree is Θ( √ t). Finally, the Lemma 5 proves the concentration of Z k (t) around its expected value for k ≥ 3. This lemma applies the Azuma-Hoeffding inequality 2 and completes the proof of Theorem 1.

Lemma 5. Let λ > 0. For any k ≥ 3

Proof. Let (Ω, F , P) be the probability space induced by the construction of a Random Apollonian Network (see Section 1.3) after t insertions. Fix k (k ≥ 3) and let (X i ) i∈{0,1,…,t} be the martingale sequence defined by

where F 0 = {∅, Ω} and F i is the σ-algebra generated by the RAN process after i steps. Notice

be two sequences of face choices differing only at time j. Also, let P , P ′ continue from P j , P ′ j until t. We call the faces Y j , Y ′ j special with respect to P , P ′ . We define a measure preserving map P → P ′ in the following way: for every choice of a non-special face in process P at time l we make the same face choice in P ′ at time l. For every choice of a face inside the special face Y j in

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