2.1. Regularity conditions

Let $\tau _0\ge 2$. This represents the number of nodes in a building block (a seed). We grow the network by adding a number of copies of the seed at places called latches. At each latch, a designated vertex in the seed (called the hook) is fused with the latch. A formal definition of this process is given in the sequel.

We shall consider adding $k_{n}\ge 1$ copies of the seed to construct the $(n+1)$st network, under the following regularity conditions:

1. (R1) $k_n \le \tau _0 + (\tau _0 -1) \sum _{i=0}^{n-1} k_i$.

2. (R2) $\lim _{n \to \infty } {k_n}/{\sum _{i=0}^n k_i} = a \in [0, 1]$.

3. (R3) $\lim _{n \to \infty } {\tau _0}/{k_n} = b \ge 0$.

A sequence of nonnegative integers $\{k_n\}_{n=0}^\infty$ satisfying (R1)–(R3) is called a building sequence. Condition (R1) is to guarantee the feasibility of choosing latches. At no point in time does the process require more (distinct) latches than the number of nodes existing in the network. Conditions (R2)–(R3) facilitate the existence of limits for properties of interest and expedite finding their values. Note that $a=0$ and $b=0$ are both allowed. For instance, for a constant sequence $k_n = k \in \mathbb {N}$, we have $a=0$, and $b = \tau _0/ k \gt 0$, whereas when $k_n = n+1$, we have both $a= 0$ and $b=0$.

Regularity conditions (R1)–(R3) are not too restrictive and the class covered by the investigation remains very broad. The examples that come up in practice satisfy these regularity conditions. For example, at one extreme, the building sequence $k_n=1$ builds networks of linear growth, including trees. At the other extreme, the case of equality in Condition (R1) builds a deterministic network where the entire vertex set is chosen at each step (a take-all model); such extremal case grows the network exponentially fast.

3.1. Notation

The notation ${\rm Hypergeo}(t, r, s)$ stands for the hypergeometric random variable associated with the random sampling of $s$ objects out of a total of $t$ objects, of which $r$ objects are of a special type. So, the hypergeometric random variable counts the number of special objects in the sample.

It is customary to call the cardinality of the vertex set of a graph the order of the graph and reserve the term size of the graph to the cardinality of the set of edges in the graph. Let $V_n$ be the set of vertices of the graph $G_n$, and ${\mathcal {E}}_n$ be the set of edges of that graph. Thus, the seed $G_0 =(V_0, {\mathcal {E}}_0)$ is a connected graph with the set $V_0$ of vertices and the set ${\mathcal {E}}_0$ of edges.

Let $\tau _n$ be the order of the graph at age $n$. Hence, the cardinality of the vertex set $V_0$ of the seed is $|V_0|=\tau _0$. The $n$th hooking step adds $k_{n-1}$ copies of the seed at $k_{n-1}$ distinct latches chosen uniformly at random from $G_{n-1}$. Each copy contributes $\tau _0 -1$ new vertices to the network. The reason for subtracting 1 is the absorption of the hook. This gives the recurrence

(1) \begin{equation} \tau_n = \tau_{n-1} + k_{n-1} (\tau_0-1) . \end{equation}

Unwinding this recurrence, we obtain

(2) \begin{equation} \tau_n = (\tau_0-1)\sum_{i=1}^n k_{i-1} + \tau_0. \end{equation}

We use the notation $\deg (v)$ to denote the degree of node $v$ in a given graph, and we set $h^* = \deg (h)$.

3.2. Useful limits

By the regularity conditions, we can argue from (2) that

$$\frac {\tau_n} {k_{n-1}} = (\tau_0-1)\frac 1 {k_{n-1}}\sum_{i=0}^{n-1} k_i + \frac {\tau_0} {k_{n-1}} \to \frac {\tau_0-1} a+b =: \gamma.$$

Reorganize (1) as

$$1 = \frac {\tau_{n-1}} {\tau_n} + \frac {k_{n-1}}{\tau_n} (\tau_0-1) .$$

to find the limit

$$\frac {\tau_{n-1}} {\tau_n} = 1 - \frac {k_{n-1}}{\tau_n} (\tau_0-1) \to 1- \frac{(\tau_0-1)} {\gamma}.$$

4.1. Evolution of the degree of a specific node

Suppose a node appears for the first time at step $j$. What will become of its degree at step $n$? At step $j$, several copies are added. To avoid a heavy notation identifying the time of appearance $j$, the copy number, which node within the copy to be tracked, and $n$, we use a simpler notation that needs only $j$ and $n$, for after all nodes of the same degree in the seed have the same distribution over time.

Theorem 4.1. Suppose $\{k_n\}_{n=0}^\infty$ is a building sequence of the family of graphs $\{G_n\}_{n=1}^\infty$. Let $X_{j:n}$ be the degree of a node at time $n$ that had appeared for the first time at step $j$. If initially its degree (in the seed) is $\delta$, then we have

$${\mathbb{E}}[X_{j:n}] =\delta + h^* \sum_{i=j}^{n-1} \frac {k_i} {\tau_i} = \frac {h^*a} {(1-a)(\tau_0 -1) + ab}(n-j) + o(n-j) + O(1).$$

Proof. Suppose a node $v$ appears at time $j$ for the first time. So, it belongs to one of the copies adjoined to the graph at that time. As the graph evolves, in any single step, the degree of $v$ can increase, if it is one of the nodes selected as latches in that step; otherwise its degree stays put, and when it does increase, it goes up by $h^* = \deg (h)$, the degree of the hook in the seed. This gives rise to a recurrence:

$$X_{j:n} = X_{j:n-1} + h^* {\mathbb{I}}_{n-1}(v),$$

where ${\mathbb {I}}_{n-1}(v)$ is an indicator of the event of choosing $v$ among the $k_{n-1}$ latches of that step of growth. On average, we have

$${\mathbb{E}}[X_{j:n}] = {\mathbb{E}}[X_{j:n-1}] + h^* \frac {{\tau_{n-1}-1\choose k_{n-1}-1}} {{\tau_{n-1}\choose k_{n-1}}} = {\mathbb{E}}[X_{j:n-1}] +h^* \frac {k_{n-1}} {\tau_{n-1}}.$$

Unwinding the recurrence, we obtain the exact average:

$${\mathbb{E}}[X_{j:n}] =\delta + h^* \sum_{i=j}^{n-1} \frac {k_i} {\tau_i}.$$

By the limits in Subsection 3.2, we obtain

$${\mathbb{E}}[X_{j:n}] = O(1) + o(n-j) + \frac {h^*a} {(1-a)(\tau_0 -1) + ab}(n-j).$$

Remark 4.3. If $a=0$, we can only assert that ${\mathbb {E}}[X_{j:n}] =o(n-j) + \delta$. In this case, a finer analysis is needed to identify the leading order of the average degree of a node that appears at time $j$. For instance, in the case of a tree grown from the complete graph $K_2$, we have $\delta = 1, h^*=1, k_n =1$, and $a=0$. The exact formula in this case yields

$${\mathbb{E}}[X_{j:n}] =1 + \sum_{i=j}^{n-1} \frac 1 {i+2}=H_{n+1} - H_{j+1} + 1.$$

Whence, we have the phases

$${\mathbb{E}}[X_{j:n}]\sim \begin{cases} \ln n, & \text{if} \ j \ \text{is fixed};\\ \ln \dfrac {n} {j(n)}, & \text{if} \ j(n)\to \infty\ \text{and} \ j(n) = o(n);\\ \ln \dfrac 1 {\rho}, & \text{if} \ j(n) \sim \rho n,\ 0 \lt \rho \lt 1;\\ 1, & \text{if} \ j(n) = n - o(n). \end{cases}$$

4.2. The overall average degree

The main result about the overall average degree in the graph is developed in this section. The result is expressed in terms of $\eta$, the number of edges in the seed graph.

Theorem 4.2. Suppose $\{k_n\}_{n=0}^\infty$ is a building sequence of the family of graphs $\{G_n\}_{n=1}^\infty$. Let $Y_n$ be the degree of a randomly chosen node in the graph $G_n$ at age $n$. We have

$$\lim_{n\to\infty}{\mathbb{E}}[Y_n] =\frac {2\eta}{\tau_0-1}.$$

Proof. Upon hooking $k_{n-1}$ copies of the seed to $k_{n-1}$ distinct nodes of $G_{n-1} = (V_{n-1}, {\mathcal {E}}_{n-1})$, we add $\eta k_{n-1}$ edges to the graph. Therefore, we have

$$|{\mathcal{E}}_n| = |{\mathcal{E}}_{n-1}| + \eta k_{n-1}.$$

This recurrence has the solution

$$|{\mathcal{E}}_n| = \eta\left(1+\sum_{i=0}^{n-1} k_i\right).$$

Using the classical First Theorem of Graph Theory, we obtain

$$\sum_{v\in V_n} \deg (v) = 2{|\mathcal{E}}_n| = 2 \eta\left(1+\sum_{i=0}^{n-1} k_i\right).$$

Scaling the equation by $\tau _n$, we get

$$\frac 1 {\tau_n} \sum_{v\in V_n} \deg (v) = \frac {2\eta} {\tau_n}\left(1+\sum_{i=0}^{n-1} k_i\right).$$

Taking limits, and using Eq. (2), we obtain

$$\lim_{n\to\infty} {\mathbb{E}}[Y_n]= 0+\lim_{n\to\infty}2\eta \frac{\sum_{i=0}^{n-1} k_i} {\tau_n} = \frac {2\eta}{\tau_0-1 }.$$

4.3. Nodes of the smallest degree

We study only the nodes of the smallest degree. Let $d^*$ be the smallest degree in the seed. Note that the smallest admissible degree in the graph is $d^*$. After the network grows, the smallest degree in it may be $d^*$ or higher. Let $X_n$ be the number of nodes of degree $d^*$ at time $n$. Thus, $X_0$ is the number of nodes of degree $d^*$ in the seed. Later graphs can have more nodes of degree $d^*$. The seed in Figure 1 has $d^* = 2$, and $X_0 = 1, X_1 = 2, X_2=3$, and $X_3 =3$.

4.3.1. Stochastic recurrence

In the evolution at step $n$, we hook $k_{n-1}$ copies of the seed to the graph $G_{n-1}$. Let $A_0$ be the event $\deg (h) = d^*$ and $\mathbb {I}_{A_0}$ be an indicator that assumes value 1, if $\deg (h) = d^*$, otherwise, it assumes the value 0. A latch of degree $d^*$ in the sample will have a higher degree (namely, its degree goes up to $d^* + \deg (h)$) in $G_n$. So, we lose such vertices in the count of $X_n$. If the hook degree is $d^*$, every hooked copy contributes only $X_0-1$ vertices of degree $d^*$.

For the case when $k_{n-1} = 1$ and the latch is $\ell$, the change from $X_{n-1}$ to $X_n$ for the four cases can be seen as shown in the table below:

Thus, for any value of $k_{n-1}$, the count $X_n$ therefore satisfies a (conditional) stochastic recurrence:

(3) \begin{align} X_n & = X_{n-1} + (X_0 - 1 -\mathbb{I}_{A_0})\, {\rm Hypergeo}(\tau_{n-1}, X_{n-1}, k_{n-1}) \nonumber\\ & \quad + (X_0 - \mathbb{I}_{A_0})(k_{n-1} -{\rm Hypergeo}(\tau_{n-1} , X_{n-1}, k_{n-1})). \end{align}

4.3.2. The average proportion of nodes of degree $d^*$

Take (conditional) expectation of (3) to get

(4) \begin{align} {\mathbb{E}}[X_n\, \vert \, G_{n-1}] & = X_{n-1} + (X_0 - 1 -\mathbb{I}_{A_0}) \frac {X_{n-1}} {\tau_{n-1}} k_{n-1} \nonumber\\ & \quad + (X_0 - \mathbb{I}_{A_0})\left(k_{n-1} - \frac {X_{n-1}} {\tau_{n-1}} k_{n-1}\right) \nonumber\\ & = \left(1 -\frac {k_{n-1}} {\tau_{n-1}} \right) X_{n-1} + (X_0 - \mathbb{I}_{A_0}) k_{n-1}. \end{align}

Theorem 4.3. Suppose $\{k_n\}_{n=0}^\infty$ is a building sequence of the family of graphs $\{G_n\}_{n=1}^\infty$, starting from a seed with $X_0$ nodes of the smallest degree $d^*$. Let $X_n$ be the number of vertices of this degree in the graph after $n$ steps of evolution according to the building sequence. We have

$${\mathbb{E}}[X_n] = (X_0-\mathbb{I}_{A_0}) \sum_{i=1}^n k_{i-1}\prod_{j=i+1}^n \left(1 - \frac {k_{j-1}} {\tau_{j-1}}\right) +X_0\prod_{j=1}^n \left(1 - \frac {k_{j-1}} {\tau_{j-1}}\right).$$

Subsequently, the average proportion converges to a limit independent of the limits $a$ and $b$; namely we have the convergence

$${\mathbb{E}}\left[ \frac {X_n} {\tau_n}\right] \to \frac {X_0-\mathbb{I}_{A_0}} {\tau_0}.$$

Proof. Taking a double expectation of (4) yields

(5) \begin{equation} {\mathbb{E}}[X_n] = \left(1 - \frac {k_{n-1}} {\tau_{n-1}}\right) {\mathbb{E}}[X_{n-1}] + (X_0 - \mathbb{I}_{A_0}) k_{n-1} . \end{equation}

This recurrence equation is of the standard linear form

(6) \begin{equation} y_n = g_n y_{n-1} + h_n, \end{equation}

with solution

(7) \begin{equation} y_n = \sum_{i=1}^n h_i\prod_{j=i+1}^n g_j + y_0 \prod_{j=1}^n g_j. \end{equation}

So, the sought solution for the average of the number of nodes of degree $d^*$ (for $n\ge 1$) is

$${\mathbb{E}}[X_n] = (X_0-\mathbb{I}_{A_0}) \sum_{i=1}^n k_{i-1} \prod_{j=i+1}^n \left(1 - \frac {k_{j-1}} {\tau_{j-1}}\right) + X_0 \prod_{j=1}^n \left(1 - \frac {k_{j-1}} {\tau_{j-1}}\right).$$

The strategy for the asymptotic part of the statement is two-fold: We prove the existence of a limit (under any building sequence) for the proportion from the exact solution. We then find the value of the limit from the recurrence under the mild regularity conditions imposed on the building sequence.

First, express the expected proportion as

(8) \begin{equation} {\mathbb{E}}\left[\frac {X_n}{\tau_n}\right] = (X_0-\mathbb{I}_{A_0})\sum_{i=1}^n \frac {k_{i-1}} {\tau_n}\prod_{j=i+1}^n \left(1 - \frac {k_{j-1}} {\tau_{j-1}}\right) + \frac {X_0} {\tau_n} \prod_{j=1}^n \left(1 - \frac {k_{j-1}} {\tau_{j-1}}\right); \end{equation}

at $i=n$, the first product does not exist, and is taken to be 1, as usual. Let

$$c_n = \sum_{i=1}^n \frac{k_{i-1}} {\tau_n}\prod_{j=i+1}^n \left( 1 - \frac {k_{j-1}} {\tau_{j-1}}\right) = \sum_{i=1}^{n-1} \frac{k_{i-1}} {\tau_n}\prod_{j=i+1}^n \left( 1 - \frac {k_{j-1}} {\tau_{j-1}}\right)+ \frac {k_{n-1}}{\tau_n};$$

We manipulate this to turn it into a recurrence as follows:

\begin{align*} c_{n+1} & = \sum_{i=1}^n \frac{k_{i-1}} {\tau_{n+1}}\prod_{j=i+1}^{n+1} \left( 1 - \frac {k_{j-1}} {\tau_{j-1}}\right) + \frac {k_n}{\tau_{n+1}}\\ & = \frac {\tau_n}{\tau_{n+1}}\left(1 - \frac {k_n}{\tau_n} \right)\sum_{i=1}^n \frac{k_{i-1}} {\tau_n}\prod_{j=i+1}^n \left( 1 - \frac {k_{j-1}} {\tau_{j-1}}\right) + \frac {k_n}{\tau_{n+1}} \\ & = \left(\frac {\tau_n - k_n}{\tau_{n+1}} \right) c_n+ \frac {k_n}{\tau_{n+1}} \\ & = \left(\frac {\tau_{n+1} - (\tau_0-1)k_n - k_n}{\tau_{n+1}} \right) c_n+ \frac {k_n}{\tau_{n+1}} \\ & = \left(\frac {\tau_{n+1} - \tau_0 k_n }{\tau_{n+1}} \right) c_n+ \frac {k_n}{\tau_{n+1}} \end{align*}

Rearrange the recurrence in the form

$$c_{n+1} - \frac 1 {\tau_0} =c_n - \frac {\tau_0 k_n }{\tau_{n+1}}\, c_n + \frac {k_n}{\tau_{n+1}} -\frac 1 {\tau_0} = \left(c_n - \frac 1 {\tau_0}\right) \left(\frac {\tau_{n+1}-\tau_0 k_n}{\tau_{n+1}}\right),$$

leading to the inequality

\begin{align*} \left|c_{n+1} - \frac 1 {\tau_0} \right| & \le\left|c_n - \frac 1 {\tau_0}\right| \left|\frac {\tau_{n+1}-\tau_0 k_n + k_n}{\tau_{n+1}}\right| \\ & = \left|c_n - \frac 1 {\tau_0} \right| \frac{\tau_n}{\tau_{n+1}}\\ & \le \left|c_{n-1} - \frac 1 {\tau_0} \right| \frac {\tau_n}{\tau_{n+1}} \times \frac {\tau_{n-1}}{\tau_n}\\ & \quad \vdots\\ & \le \left|c_0 - \frac 1 {\tau_0} \right| \frac {\tau_n}{\tau_{n+1}} \times \frac {\tau_{n-1}}{\tau_n} \times \cdots\times \frac {\tau_0}{\tau_{1}}. \end{align*}

Noting that the sum in $c_n$ is empty at $n=0$, we have $c_0 = 0$ and the bounds simplify to $0 \le |c_n - 1 /\tau _0| \le 1/\tau _n$. So, both inferior and superior limits of $|c_n - 1 /\tau _0|$ are equal to $0$, which furnishes the existence of a limit for $c_n$ equal to $1/\tau _0$, too.

As for the remainder part

$$r_n := \frac {X_0}{\tau_n}\prod_{j=1}^n \left( 1 - \frac {k_{j-1}} {\tau_{j-1}}\right),$$

in (8), it clearly converges to 0, as $\tau _n$ is increasing, and the product is bounded from above by 1. Plugging in the limits $\lim _{n\to \infty } c_n= 1/\tau _0$ and $\lim _{n\to \infty } r_n = 0$ in (8), we reach the conclusion that

$$\lim_{n\to \infty} {\mathbb{E}}\left[ \frac {X_n} {\tau_n}\right] = \frac {X_0- \mathbb{I}_{A_0}} {\tau_0}.$$

Remark 4.5. In the case when the hook is not of the smallest degree $d^*$, we have ${\mathbb {E}}[X_n/\tau _n] \to X_0/\tau _0$. The initial proportion of nodes of the smallest degree in the seed is preserved on average in larger subsequent graphs.

Remark 4.6. In the case when the hook is of the smallest degree $d^*$, we have ${\mathbb {E}}[X_n/\tau _n] \to (X_0-1)/\tau _0$. The long-term proportion of nodes of the smallest degree is less than the proportion of nodes of degree $d^*$ in the seed.

Remark 4.7. In the case when the hook is the only node of the smallest degree $d^*$ in the seed, we have $X_0=1$, and ${\mathbb {E}}[X_n/\tau _n] \to 0$, for all $n\ge 1$. Indeed, the degree $d^*$ disappears after the first latching at the initial hook and never reappears.

Remark 4.8. The limit in Theorem 4.3 is more than just an ultimate value in the take-all case. In this case, it is the actual value for each $n\ge 0$, which can be seen from the recurrence. The only term that does not vanish is the last term in sum, yielding $(X_0 - I_{A_0}) k_{n-1}/ \tau _n = (X_0 - I_{A_0}) / \tau _0$.

5.1. Average total path length

As the network grows, at step $n$, a sample of size $k_{n-1}$ latches is chosen from $G_{n-1}$ to grow $G_{n-1}$ into $G_n$. Suppose these latches are $\ell _1, \ldots, \ell _{k_{n -1}}$ at depths $d_1, \ldots, d_{k_{n-1}}$. In view of the absorption of the hooks of the added graphs, a copy's hook fused at the $j$th latch adds $\tau _0-1$ nodes, which appear in $G_n$ at depths equal to their distance from the hook of the copy translated by an additional distance from the latch to the reference vertex. So, collectively, the vertices of the copy hooked to $\ell _j$ increase the total path length by $T_0 + (\tau _0-1) d_j$. We have a conditional recurrence:

$${\mathbb{E}}[T_n \, \vert \, G_{n-1}, d_1, \ldots, d_{k_{n-1}}] = T_{n-1} + \sum_{j=1}^{k_{n-1}} (T_0 + (\tau_0-1) d_j).$$

Averaging over the graphs and the choices of the latches within, we get

(9) \begin{equation} {\mathbb{E}}[T_n] = {\mathbb{E}}[T_{n-1}] + T_0 k_{n-1} + (\tau_0-1) \sum_{j=1}^{k_{n-1}} {\mathbb{E}}[ \Delta_{n-1,\ell_j}]. \end{equation}

Lemma 5.1.

$$\sum_{j=1}^{k_{n-1}} {\mathbb{E}}[ \Delta_{n-1,\ell_j}] = \frac {k_{n-1}} {\tau_{n-1}} \, {\mathbb{E}}[T_{n-1}].$$

Proof. Condition on the event $\ell _1 =i_1, \ldots, \ell _{k_{n-1}} = i_{k_{n-1}}$, to get

\begin{align*} \sum_{j=1}^{k_{n-1}} {\mathbb{E}}[ \Delta_{n-1,\ell_j}] & = \sum_{j=1}^{k_{n-1}} \sum_{1\le i_1 \lt i_2 \lt \cdots \lt i_{k_{n-1}} \le \tau_{n-1}} {\mathbb{E}} [ \Delta_{n-1,\ell_j}\, \vert \, d_1 =i_1, \ldots d_{k_{n-1}} = i_{k_{n-1}}]\\ & \quad \times {\mathbb{P}}(d_1 =i_1, \ldots, d_{k_{n-1}} =i_{k_{n-1}}). \end{align*}

The subsets of size $k_{n-1}$ latches that appear in a sample of vertices from $G_{n-1}$ are all equally likely, and we get

\begin{align*} & \sum_{j=1}^{k_{n-1}} {\mathbb{E}}[\Delta_{n-1,\ell_j}] \\ & \quad =\frac 1 {{{\tau_{n-1}} \choose {k_{n-1}}}}\sum_{j=1}^{k_{n-1}} {\mathbb{E}}\left[ \left.\left(\sum_{1\le i_1 \lt i_2 \lt \cdots \lt i_{k_{n-1}} \le \tau_{n-1}} \Delta_{n-1,\ell_j}\right)\right| \ell_1 =i_1, \ldots \ell_{k_{n-1}}= i_{k_{n-1}}\right]. \end{align*}

Let us write out the inner sum in expanded form:

\begin{align*} & \Delta_{n-1,1} + \Delta_{n-1,2} +\cdots + \Delta_{n-1,k_{n-1}} \\ & \quad + \Delta_{n-1,1} + \Delta_{n-1,2} +\cdots + \Delta_{n-1,k_{n-1}-1} + \Delta_{n-1,k_{n-1}+1}\\ & \quad \enspace \vdots\\ & \quad + \Delta_{n-1,\tau_{n-1} - k_{n-1}+1} + \Delta_{n-1,\tau_{n-1} - k_{n-1}+2}+\cdots + \Delta_{n-1,\tau_{n-1}}. \end{align*}

Upon a reorganization collecting similar terms, we get

$${\tau_{n-1 -1} \choose k_{n-1}-1} (\Delta_{n-1,1} + \Delta_{n-1,2} + \cdots + \Delta_{n-1,\tau_{n-1}}) = {\tau_{n-1 -1} \choose k_{n-1}-1} T_{n-1}.$$

Plugging this expression in the expectation, we proceed to

$$\sum_{j=1}^{k_{n-1}} {\mathbb{E}}[\Delta_{n-1,\ell_j}] = \frac {{\tau_{n-1 -1} \choose k_{n-1}-1}} {{\tau_{n-1} \choose k_{n-1}}} {\mathbb{E}}[T_{n-1}] = \frac {k_{n-1}} {{\tau_{n-1}}} {\mathbb{E}}[T_{n-1}].$$

Theorem 5.1. Suppose $\{k_n\}_{n=0}^\infty$ is a building sequence of the family of graphs $\{G_n\}_{n=1}^\infty$, starting from a seed of total path length $T_0$. Let $T_n$ be the total path length after $n$ steps of evolution according to the building sequence. We have

$${\mathbb{E}}[T_n] = T_0\tau_n\left(\sum_{i=1}^n \frac {k_{i-1}} {\tau_i} + \frac 1 {\tau_0}\right).$$

Proof. By Lemma 5.1 and the recurrence (9), we have a recurrence for the average total path length:

\begin{align*} {\mathbb{E}}[T_n] & = {\mathbb{E}}[T_{n-1}] + T_0 k_{n-1}+ \frac {(\tau_0-1) k_{n-1}} {\tau_{n-1}} {\mathbb{E}}[T_{n-1}] \\ & =\left(1 + \frac {(\tau_0-1) k_{n-1}} {\tau_{n-1}} \right){\mathbb{E}}[T_{n-1}] + T_0 k_{n-1}. \end{align*}

Again, the recurrence is of the standard form (6) with the solution (7). In the specific case at hand, this solution is

$${\mathbb{E}}[T_n] = T_0\sum_{i=1}^n k_{i-1}\prod_{j=i+1}^n \left(1 + \frac {(\tau_0-1){k_{j-1}} } {\tau_{j-1}}\right) +T_0 \prod_{j=1}^n \left(1 + \frac {(\tau_0 -1)k_{j-1}} {\tau_{j-1}}\right).$$

The recurrence (1) on the order of the graph simplifies the solution into telescopic products

$${\mathbb{E}}[T_n] = T_0\sum_{i=1}^n k_{i-1}\prod_{j=i+1}^n \frac {\tau_j} {\tau_{j-1}} +T_0 \prod_{j=1}^n \frac {\tau_j} {\tau_{j-1}} = T_0 \tau_n \left(\sum_{i=1}^n \frac{k_{i-1}}{\tau_i} + \frac 1 {\tau_0}\right).$$

5.2. Average depth

Theorem 5.1 provides a benchmark for the calculation of the average depth. Let the depth of a randomly selected node in the $n$th network be $D_n$.

Corollary 5.1.

$${\mathbb{E}}[D_n] = T_0 \sum_{i=1}^n \frac{k_{i-1}}{\tau_i} + D_0.$$

Proof. Given a specific development leading to $G_n$, the average depth in that graph is

$${\mathbb{E}}[D_n \, \vert \, G_{n}] = \frac {\Delta_{n,1} +\cdots + \Delta_{n,\tau_n}} {\tau_n} = \frac{T_n}{\tau_n}.$$

Upon taking expectation, it follows that ${\mathbb {E}}[D_n] = {\mathbb {E}}[T_n] / \tau _n$. The form given in the statement ensues from Theorem 5.1.

Corollary 5.2. Under the regularity conditions (R1)–(R3), we have the asymptotic equivalent

$${\mathbb{E}}[D_n] = \frac {a T_0}{\tau_0 -1 + ab} n + o(n), \quad \text{as} \ n \to \infty.$$

Remark 5.1. Corollary 5.2 is more useful when

$$\lim_{n\to \infty }\frac {k_n} {\sum_{i=0}^n k_i} = a\not = 0.$$

When $a=0$, as in the case of trees for example, one needs to sharpen the argument to find the leading asymptotic term, as we do in some specific cases below.

5.3. Distances under specific building sequences

At one extreme, there is the sequence of least possible growth ($k_n= 1$). At the other extreme, we have a take-all model ($k_n = \tau _n$) in which all the nodes of $G_n$ are taken as latches for $\tau _{n}$ copies of the seed.

In the case of $k_n = k$, for fixed $k\in \mathbb {N}$, of nearly the least growth, the average depth is

$${\mathbb{E}}[D_n] = T_0 \sum_{i=1}^n \frac k {\tau_i} +D_0.$$

The limit $a$ in regularity condition (R2) is 0, and Corollary 5.2 only tells us that ${\mathbb {E}}[D_n] =o(n)$. However, we can sharpen the asymptotic equivalence from the specific construction of the case.

Here, we have $\tau _i = (\tau _0 - 1) i + \tau _0$, which gives

$${\mathbb{E}}[D_n] = \frac {T_0 k} {\tau_0-1} \sum_{i=1}^n \frac 1{ i + \frac {\tau_0} {\tau_0-1}} + D_0.$$

In terms of the generalized harmonic numbersFootnote ¹

$$H_n(x) = \frac 1 {1+x} + \frac 1 {2+x} +\cdots+\frac 1 {n+x},$$

the depth in the near-least-growth is compactly expressed as

$${\mathbb{E}}[D_n] = \frac {T_0 k} {\tau_0-1} H_n\left(\frac {\tau_0}{\tau_0-1}\right) + D_0 \sim \frac {T_0 k} {\tau_0-1} \ln n, \quad \text{as} \ n\to\infty.$$

Remark 5.2. The case $k=1$ and $\tau _0=2$ grows a recursive tree. The seed is a rooted tree on two vertices, in which $T_0=1$ and $D_0 = \frac 1 2$. In this case, the average depth becomes

$${\mathbb{E}}[D_n] =H_n(2) + \frac 1 2 = H_{n+2} - 1 \sim \ln n, \quad \text{as} \ n\to\infty,$$

which recovers a known result [Reference Smythe and Mahmoud19].

Remark 5.3. At the other end of the spectrum, there is the take-all model, in which $k_i = \tau _i$, leading at step $n$ to a graph of order $\tau _n = \tau _0^{n+1}$. Here, the limit $a$ is $(\tau _0-1)/\tau _0$ and the limit $b$ is 0. According to Corollary 5.2, we have ${\mathbb {E}}[D_n] \sim D_0 n$, as $n\to \infty$. This asymptotic estimate can be sharpened as the case is amenable to exact calculation:

$${\mathbb{E}}[D_n] = T_0\sum_{i=1}^n \frac {\tau_{i-1}} {\tau_i} + D_0 = T_0\sum_{i=1}^n \frac 1 {\tau_0} + D_0= D_0 (n+1) .$$

6.1. Eccentricity of a node in $G_n$

The $k_{n-1}$ nodes selected as latches from the graph $G_{n-1}= (V_{n-1}, {\mathcal {E}}_{n-1})$ are vertices that play a key role in designing the network at stage $n$ and onward and contribute significantly in determining the diameter of the graph at the next stage.

As a node's eccentricity changes over time, its value at step $n$ in $G_n$ may be different from its value at step $n-1$ in $G_{n-1}$. We need an eccentricity notation reflecting the possible change over time. For that we use $C_n(v)$ to speak of the eccentricity of a vertex $v$ in $G_n$.

If $v \in V_n$ is a vertex in a copy of $G_0$ latched at a vertex $\ell _i \in V_{n-1}$, we express that by saying $v \in V_0^{co_i}$, otherwise we say $v \in V_{n-1}$. We now introduce some notation:

1. 1. $\mathbb {L}_n = \{\ell _1, \ell _2, \cdots \ell _{k_n}\}$ is the set of latches selected in the graph $G_n$ to produce the graph $G_{n+1}$.

2. 2. $\widetilde {C}_n(v) = C_n(v) \, \vert \, G_{n-1}, \mathbb {L}_{n-1}$. This is the conditional eccentricity $C_n(v)$ of the node $v$ in the graph $G_n$, given $G_{n-1}$ and the $k_{n-1}$ latches in it.

3. 3. For any $v \in V_{n-1}$, we define $d_v^{\#} = \max _{\ell _j \in \mathbb {L}_{n-1}} d(v, \ell _j)$. So, $d_v^{\#}$ is the maximum distance from $v$ to the nodes in $\mathbb {L}_{n-1}$.

Also, in what follows we use the notation ${\mathbb {I}}_{\mathcal {C}}$ to indicate a predicate (condition) $\mathcal {C}$. So, it is 1, when $\mathcal {C}$ holds, and is 0, otherwise.

Theorem 6.1. Suppose $\{k_n\}_{n=0}^\infty$ is a building sequence of the family of graphs $\{G_n\}_{n=1}^\infty$. Let $v$ be a node in the graph $G_n$. Conditional upon the choice of the latches $\ell _1, \ldots, \ell _{k_{n-1}}$ in $G_{n-1}$, the eccentricity $C_n(v)$ is given by

$$\widetilde{C}_n(v)= \begin{cases} \max\{C_{n-1}(v), d_v^{\#}+C_0(h)\}, & \text{if} \ v\in V_{n-1};\\ \max\{d(v,h) + C_{n-1}(\ell_i), & \\ \quad \max\{C_0(v) , (d(v,h) + d_{\ell_i}^{\#} + C_0(h)){\mathbb{I}}_{\{k_{n-1} \gt 1\}}\}\}, & \text{if} \ v \in V_0^{co_i}. \end{cases}$$

Proof. The graph $G_n$ is obtained by attaching a copy of the seed $G_0$ at each of the latches $\ell _1, \ell _2, \cdots, \ell _{k_{n-1}}$ selected in the graph $G_{n-1}$.

We denote the vertex set of the $r$th copy of the seed, for $r= 1, \ldots, k_{n-1}$, by $V_0^{{\rm co}_r}$. We now compute the distance from a node $v$ to a vertex $u$ in $G_n$ by considering the four cases:

For a given $v\in V_n$, the maximum (over the range of $u$ and $j$) in each block is

The result now follows.

Remark 6.1. Suppose a vertex $\ell$ is chosen as a latch from $G_{n-1}$. From Theorem 6.1, we pick up the top line and write

$$\widetilde C_n(\ell) = \max \{C_{n-1}(\ell), d_\ell^{\#}+ C_0(h)\}.$$

If $k_{n-1} = 1$, then $d_\ell ^{\#}=0$, in which case we have $\widetilde C_n(\ell ) = \max \{C_{n-1}(\ell ), C_0(h)\}$.