Scaling of average receiving time and average shortest path on claw network

Complex networks are a new research field in recent years and many scholars have also obtained many research results. Complex networks have been applied to many interdisciplinary fields since they show many superiority, such as communication network, airport network, biological network and so on. As a part of complex networks, deterministic networks are widely studied including the structure of network [1–3], spectra of the network[4–7], the applications of spectra [8–10] and so on.Furthermore, some dynamical processes are also hot fields in complex networks, like random walks [11–13], coherence problem [14, 15], epidemic spreading [16], Lotka-Volterra model [17], evolutionary game [18, 19], and so on.

Meanwhile, the trapping problem on network has also attracted the attention of many scholars. Usually scholars will fix a node as a trap to study the trapping problem [20, 21]. Two important quantities about trapping problem are the mean first-passage time (MFPT) which is the time from a node to the trap node [22, 23] and the average receiving time (ART) which is an average of the mean first-passage times. The hot topic issue in the study of random walks is how the ART scales with the size of network.

In addition, the average shortest path (ASP) is also a basic measure of complex networks. It is related to many other structural characteristics of networks, for example, degree distribution [24, 25], centrality [26], fractal [27–29]. Besides, ASP has a strong efficiency on various dynamical processes, such as synchronization [30]. Because of its importance and practicability, ASP has attracted considerable attention. However, the most of networks they considered are special networks, which the whole network conforms to some certain laws and the edges weights of most networks they considered are related to the generation of the network.

Motivated by the previous study, we construct the claw network with a new method. The claw network is a network proposed by the authors and related to the weight of the edge in weighted network. It is different from the previous deterministic network generated by the substitution rules [31, 32] and valuable to study the dynamic process of such network. The claw network has two features, one is the global feature which is a common feature of the networks, another is the local feature which is a unique feature of claw network. However, local feature may make it impossible to study the degree distribution and strength distribution of the claw network. In addition, the weight of edge in next generation is related to the strength of the node in former generation. Then, we calculate the exact expression of ART after dividing claw network into four blocks. Because of the particularity of network, the previous methods of calculating ASP are not suitable for claw network, we propose a novel method to calculate ASP. The step of this method is to divide the whole claw network into four blocks, and then divide each block network into smaller blocks to calculate ASP.

The rest of the paper is arranged as follows: in section 2, claw network is constructed using a new method, which has a special topological structure different from previous network. In section 3, we divide the claw network into four block, then we work out the sum of MFPTs, finally we obtain the accurate relation of ART, which implies that ART grows exponentially with the network order as a power-law function. In section 4, we first divide every block network into small blocks after dividing the whole claw network into four blocks. Based on this operation, we find the sums of shortest paths of every block network and the whole claw network. Eventually, we obtain the result which shows that ASP increases sublinearly with the size of claw network. In section 5, we give the conclusion about ART and ASP after analyzing the expressions of ART and ASP respectively.

The process of network construction is described in this section. Let G_n denote the n-th graph and N_n be the set of nodes in G_n. The strength s_i of node i represents the sum of the weights of all edges connected to node i and the initial value of s_i depends on the sum. Our initial graphs are G₀ and h. G₀ is a non-clique connected graph consisting of three nodes and two edges of unit weight. h is a weighted graph with two nodes and a unit weighted edge. Given the n-th network G_n, G_n+1 is obtained by the following two steps.

Step 1: For every node i with strength s_i in G_n, we take s_i copies of h, and link them to node i by unit weighted edge.

Step 2: The weight of every edge in G_n is increased from w to $2w$.

Figure 1 shows the construction processes of claw network from n = 0 to n = 2.

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In order to describe the topological structure of the network better and study conveniently, the claw network is divided into four blocks: two blocks (${G}_{n}^{(1)}$ and ${G}_{n}^{(2)}$) are connected to node 1, and two blocks (${G}_{n}^{(3)}$ and ${G}_{n}^{(4)}$) are connected to nodes 2 and 3 respectively. It is obvious that the four blocks of G_n have the same structure. For G₁, let node 1 be the hub node of ${G}_{n}^{(1)}$ and ${G}_{n}^{(2)}$, nodes 2 and 3 be hub nodes of ${G}_{n}^{(3)}$ and ${G}_{n}^{(4)}$ respectively, nodes 4, 5, 6, 7, 8, 9, 10 and 11 in G₁ be sub-hub nodes, which are defined as the nodes whose strength are second only to the hub node in each block network. Since the four blocks of G_n have the same structure, we take ${G}_{n+1}^{(1)}$ as an example to show the process of network generation. For n > 0, G_n+1 is obtained from the four blocks of G_n when G_n and G₁ are given. Firstly, four copies of ${G}_{n}^{(1)}$ are taken and connected to node 1 of G₁, nextly two copies of ${G}_{n}^{(1)}$ are taken and connected to node 4 of G₁, and they are also taken and connected to node 8 of G₁ similarly, finally ${G}_{n+1}^{(1)}$ is got. By repeating above operation, we can obtain ${G}_{n+1}^{(2)}$, ${G}_{n+1}^{(3)}$ and ${G}_{n+1}^{(4)}$, then G_n+1 is got.

Take n = 1 as an example, G₂ is obtained from the four blocks of G₁ when G₁ is given. Firstly, we take four copies of ${G}_{1}^{(1)}$ and connect these copies to node 1 , nextly we take two copies of ${G}_{1}^{(1)}$ and connect these copies to node 4, then take two copies of ${G}_{1}^{(1)}$ and connect these copies to node 8, finally we get ${G}_{2}^{(1)}$. By repeating the above operation, ${G}_{2}^{(2)}$, ${G}_{2}^{(3)}$ and ${G}_{3}^{(4)}$ can be obtained, eventually we can get G₂. Figure 2 shows the construction process of claw network.

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From the construction of claw network, we can easily get

$$\begin{eqnarray}&&{N}_{n}=\displaystyle \frac{{8}^{n+1}+13}{7}\end{eqnarray} \tag{ 1 }$$

This section gives the exact expression of average receiving time and shows how it scales with the size of network. On claw network, we consider that the method of walking is weight-dependent walk, it means that every node moves from its current location to any of its neighbors according to probability proportional to the weight of the edge. The transition probability is

$$\begin{eqnarray}&&{p}_{i\to j}^{w}=\displaystyle \frac{{w}_{{ij}}}{{s}_{i}}=\displaystyle \frac{{w}_{{ij}}}{{\displaystyle \sum }_{j\in h(i)}{w}_{{ij}}},\end{eqnarray} \tag{ 2 }$$

Where h(i) represents the neighbors of node i, and s_i is the strength of node i.

In this paper, we consider the trap node is node 1. ${F}_{i\to j}(n)$ and F_i(n) are defined as the mean first-passage time (MFPT) for an agent walking from node i to node j and the MFPT from node i to the trap respectively, where n represents the n-th generation network. Let $\langle T{\rangle }_{n}$ be the average receiving time (ART), which is an average of F_i(n) of all nodes except the trap. Based on the definition of ART, we have

$$\begin{eqnarray}&&\langle T{\rangle }_{n}=\displaystyle \frac{1}{{N}_{n}-1}\displaystyle \sum _{i=2}^{{N}_{n}}{F}_{i}(n)\end{eqnarray} \tag{ 3 }$$

Let ${T}_{{tot}}(n)={\sum }_{i=2}^{{N}_{n}}{F}_{i}(n)$, in order to determine $\langle T{\rangle }_{n}$, we should determine T_tot(n) firstly. Combining the division of G_n, we can get

$$\begin{eqnarray}&&{T}_{{tot}}(n)={T}_{{tot}}^{(1)}(n)+{T}_{{tot}}^{(2)}(n)+{T}_{{tot}}^{(3)}(n)+{T}_{{tot}}^{(4)}(n)\end{eqnarray} \tag{ 4 }$$

Because ${G}_{n}^{(1)}$ and ${G}_{n}^{(2)}$, ${G}_{n}^{(3)}$ and ${G}_{n}^{(4)}$ are symmetrical, we can simplify equation (4).

$$\begin{eqnarray}&&{T}_{{tot}}(n)=2({T}_{{tot}}^{(1)}(n)+{T}_{{tot}}^{(3)}(n))\end{eqnarray} \tag{ 5 }$$

Then we need to determine ${T}_{{tot}}^{(1)}(n)$ and ${T}_{{tot}}^{(3)}(n)$. Based on the special structure of claw network, we have the following relationships.

For ${G}_{n}^{(1)}$:

$$\begin{eqnarray}&&{T}_{{tot}}^{(1)}(n)=8{T}_{{tot}}^{(1)}(n-1)+{F}_{4}(n)\left(\displaystyle \frac{{N}_{n-1}-3}{4}\times 4+2\right)\end{eqnarray} \tag{ 6 }$$

Where ${F}_{4}(n)$ is MFPT from node 4 to node 1 in G_n.

For ${G}_{n}^{(3)}$:

$$\begin{eqnarray}\begin{array}{rcl}{T}_{{tot}}^{(3)}(n) & = & 8[{T}_{{tot}}^{(3)}(n-1)-\left(\displaystyle \frac{{N}_{n-1}-3}{4}+1\right){F}_{2}(n-1)]\\ & & +\,{F}_{6\to 2}(n)({N}_{n-1}-1)+\left(\displaystyle \frac{{N}_{n-1}-3}{4}+1\right){F}_{2}(n)\end{array}\end{eqnarray} \tag{ 7 }$$

Where ${F}_{2}(n)$ is MFPT from node 2 to the trap in G_n, ${F}_{6\to 2}(n)$ is MFPT from node 6 to node 2 in G_n.

Note that ${F}_{4}(n)={F}_{6\to 2}(n)$, thus we need to find the ${F}_{6\to 2}(n)$ and ${F}_{2}(n)$. Let X be the MFPT from node i to any of its old neighbors in G_n, and let Y be the MFPT from a new neighbor of node j generated in G_n+1 to an old neighbor of node i in G_n. Let S be the strength of node i in G_n, then the strength of node i in G_n+1 is $4S$. According to the method of walking and the construction of claw network, we have the following equations:

$$\begin{eqnarray*}&&\left\{\begin{array}{l}X=\tfrac{2S}{4S}+\tfrac{2S}{4S}(1+Y)\\ Y=\tfrac{1}{2}(1+X)+\tfrac{1}{2}(1+Y)\end{array}\right.\end{eqnarray*}$$

It is apparent that X = 4 holds, so we know that if the MFPT from node i to node j in G_n is T, then the MFPT from node i to node j in G_n+1 is $4T$. Combining initial condition ${F}_{2}(0)=1$, we obtain

$$\begin{eqnarray}&&{F}_{2}(n)={4}^{n}\end{eqnarray} \tag{ 8 }$$

For ${F}_{6\to 2}(1)$, we have

$$\begin{eqnarray*}&&{F}_{6\to 2}(1)=\displaystyle \frac{1}{2}+\displaystyle \frac{1}{2}(1+{F}_{10\to 2}(1))=1+\displaystyle \frac{1}{2}{F}_{6\to 2}(1)\end{eqnarray*}$$

From above equations, we can get ${F}_{6\to 2}(1)=2$, so we know

$$\begin{eqnarray}&&{F}_{6\to 2}(n)=2\times {4}^{n-1}\end{eqnarray} \tag{ 9 }$$

By introducing equations (1), (8) and (9) into equations (6) and (7) and combining the initial conditions ${T}_{{tot}}^{(1)}(1)=4$ and ${T}_{{tot}}^{(3)}(1)=16$, we can obtain

$$\begin{eqnarray}&&{T}_{{tot}}^{(1)}(n)=\displaystyle \frac{2}{21}\times {32}^{n}-\displaystyle \frac{3}{7}\times {4}^{n}+\displaystyle \frac{1}{3}\times {8}^{n}\end{eqnarray} \tag{ 10 }$$

$$\begin{eqnarray}&&{T}_{{tot}}^{(3)}(n)=\displaystyle \frac{8}{21}\times {32}^{n}+\displaystyle \frac{2}{7}\times {4}^{n}+\displaystyle \frac{1}{3}\times {8}^{n}\end{eqnarray} \tag{ 11 }$$

Based on equations (5), (10) and (11), we can get

$$\begin{eqnarray}&&{T}_{{tot}}(n)=\displaystyle \frac{20}{21}\times {32}^{n}-\displaystyle \frac{2}{7}\times {4}^{n}+\displaystyle \frac{4}{3}\times {8}^{n}\end{eqnarray} \tag{ 12 }$$

Based on equations (1), (3) and (12), we have

$$\begin{eqnarray}\begin{array}{rcl}\langle T{\rangle }_{n} & = & \displaystyle \frac{20}{3\times ({8}^{n+1}+6)}\times {32}^{n}-\displaystyle \frac{2}{{8}^{n+1}+6}\\ & & \times \,{4}^{n}+\displaystyle \frac{28}{3\times ({8}^{n+1}+6)}\times {8}^{n}\end{array}\end{eqnarray} \tag{ 13 }$$

From equation (1), we note that

$$\begin{eqnarray}&&n={{log}}_{8}(7{N}_{n}-13)-1\end{eqnarray} \tag{ 14 }$$

Considering equations (13) and (14), we can get

$$\begin{eqnarray}\begin{array}{rcl}\langle T{\rangle }_{n} & = & \displaystyle \frac{5}{168({N}_{n}-1)}\times {\left(7{N}_{n}-13\right)}^{\tfrac{5}{3}}-\displaystyle \frac{1}{14({N}_{n}-1)}\\ & & \times \,{\left(7{N}_{n}-13\right)}^{\tfrac{2}{3}}+\displaystyle \frac{7{N}_{n}-13}{12({N}_{n}-1)}\end{array}\end{eqnarray} \tag{ 15 }$$

When the size of network is large enough, we obtain

$$\begin{eqnarray}&&\langle T{\rangle }_{n}\approx 0.76{\left({N}_{n}-2\right)}^{\tfrac{2}{3}}\end{eqnarray} \tag{ 16 }$$

The numerical formula equation (13) and the analytical formula equation (16) are compared for $1\leqslant n\leqslant 6$ (see figure 3).

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As shown in figure 3, the numerical result is slightly greater than the analysis result when $0\leqslant n\leqslant 3$ and the same as the analysis result when n > 3 which shows that the analysis result is accurate when the network size tends to infinity. Equation (16) illustrates that ART increases exponentially with the network order as a power-law function of $\theta =\tfrac{2}{3}$, which implies that ART grows sublinearly with the size of network.

This section we find the accurate expression of average shortest path (ASP). Based on the definition of ASP [31, 32] , we have the following equations.

$$\begin{eqnarray}&&{\lambda }_{n}=\displaystyle \frac{2}{{N}_{n}({N}_{n}-1)}{D}_{{tot}}(n)\end{eqnarray} \tag{ 17 }$$

$$\begin{eqnarray}&&{D}_{{tot}}(n)=\displaystyle \sum _{i\ne j,i,j\in {G}_{n}}{d}_{{ij}}(n)\end{eqnarray} \tag{ 18 }$$

Where ${d}_{{ij}}(n)$ is the shortest path between node i and node j in G_n.

Applying the previous generation method (substitution rule) of the network, the exact value of ASP can be obtained by dividing the network once. However, the network generated by the new method is related to the weight of the edge, so the exact solution of ASP cannot be obtained in previous method. Encouraged by the previous research, we find the exact solution of ASP by proposing a new method of dividing the network twice. Because claw network is different from the previous network, we need divide each ${G}_{n}^{(i)}$ into smaller block when we find the exact expression of ASP. Figure 4 shows the partitioning of network taking ${G}_{n}^{(1)}$ as an example.

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As can be seen from the figure 4, ${D}_{{tot}}^{(1)}(n)$ has the following recursive relationship.

$$\begin{eqnarray}&&{D}_{{tot}}^{(1)}(n)=8{D}_{{tot}}^{(1)}(n-1)+{2}^{n-1}{D}_{{tot}}^{(1)}(1)+{{\rm{\Omega }}}_{n}^{(1)}\end{eqnarray} \tag{ 19 }$$

Where ${{\rm{\Omega }}}_{n}^{(1)}$ is the sum of all shortest paths of two nodes which belong to different blocks. Detailed derivation of ${{\rm{\Omega }}}_{n}^{(1)}$ is shown in appendix A. The result is as follows.

$$\begin{eqnarray}&&{{\rm{\Omega }}}_{n}^{(1)}=\displaystyle \frac{208}{49}\times {128}^{n-1}-\displaystyle \frac{12}{49}\times {16}^{n}-\displaystyle \frac{8}{49}\times {2}^{n-1}\end{eqnarray} \tag{ 20 }$$

Substituting equation (20) into equation (19) and combining the initial condition ${D}_{{tot}}^{(1)}(1)=3$, we can get

$$\begin{eqnarray}\begin{array}{rcl}{D}_{{tot}}^{(1)}(n) & = & 8{D}_{{tot}}^{(1)}(n-1)+\displaystyle \frac{208}{49}\\ & & \times \,{128}^{n-1}-\displaystyle \frac{12}{49}\times {16}^{n}+\displaystyle \frac{131}{49}\times {2}^{n-1}\end{array}\end{eqnarray} \tag{ 21 }$$

By calculating equation (21), we can obtain

$$\begin{eqnarray}\begin{array}{rcl}{D}_{{tot}}^{(1)}(n) & = & \displaystyle \frac{3328}{735}\times {128}^{n-1}-\displaystyle \frac{24}{49}\\ & & \times \,{16}^{n}-\displaystyle \frac{131}{147}\times {2}^{n-1}+\displaystyle \frac{36}{5}\times {8}^{n-1}\end{array}\end{eqnarray} \tag{ 22 }$$

Next, we continue to calculate D_tot(n). Based on partitioning of the claw network, D_tot(n) can be expressed as the following equation.

$$\begin{eqnarray}&&{D}_{{tot}}(n)=4{D}_{{tot}}^{(1)}(n)+{{\rm{\Omega }}}_{n}\end{eqnarray} \tag{ 23 }$$

Where Ω_n is the sum of all shortest paths of two nodes which are in different ${G}_{n}^{(i)}$. See appendix B for detailed deducing. The conclusion of Ω_n is

$$\begin{eqnarray}&&{{\rm{\Omega }}}_{n}=\displaystyle \frac{48}{49}\times {128}^{n}+\displaystyle \frac{100}{49}\times {16}^{n}+\displaystyle \frac{48}{49}\times {2}^{n}\end{eqnarray} \tag{ 24 }$$

Considering equations (22), (23) and (24), we can get

$$\begin{eqnarray}\begin{array}{rcl}{D}_{{tot}}(n) & = & \displaystyle \frac{824}{735}\times {128}^{n}+\displaystyle \frac{4}{49}\\ & & \times \,{16}^{n}-\displaystyle \frac{118}{147}\times {2}^{n}+\displaystyle \frac{18}{5}\times {8}^{n}\end{array}\end{eqnarray} \tag{ 25 }$$

Combining equations (1), (17) and (25), we have

$$\begin{eqnarray}\begin{array}{rcl}\lambda (n) & = & \displaystyle \frac{824}{15(32\times {64}^{n}+76\times {8}^{n}+39)}\times \,{128}^{n}\\ & & +\,\displaystyle \frac{882}{5(32\times {64}^{n}+76\times {8}^{n}+39)}\times {8}^{n}\\ & & +\,\displaystyle \frac{4}{32\times {64}^{n}+76\times {8}^{n}+39}\times {16}^{n}\\ & & -\,\displaystyle \frac{118}{3(32\times {64}^{n}+76\times {8}^{n}+39)}\times {2}^{n}\end{array}\end{eqnarray} \tag{ 26 }$$

Substituting equation (14) into (26), we can obtain

$$\begin{eqnarray}\begin{array}{rcl}\lambda (n) & = & \displaystyle \frac{103}{5880{N}_{n}({N}_{n}-1)}\times {\left(7{N}_{n}-13\right)}^{\tfrac{7}{3}}\\ & & +\,\displaystyle \frac{63{N}_{n}-117}{10{N}_{n}({N}_{n}-1)}\\ & & +\,\displaystyle \frac{1}{98{N}_{n}({N}_{n}-1)}\times {\left(7{N}_{n}-13\right)}^{\tfrac{4}{3}}\\ & & -\,\displaystyle \frac{118}{147{N}_{n}({N}_{n}-1)}\times {\left(7{N}_{n}-13\right)}^{\tfrac{1}{3}}\end{array}\end{eqnarray} \tag{ 27 }$$

When N_n is large enough, we have

$$\begin{eqnarray}&&\lambda (n)\approx 1.64{\left({N}_{n}-2\right)}^{\tfrac{1}{3}}\end{eqnarray} \tag{ 28 }$$

The numerical formula equation (26) and the analytical formula equation (28) are checked for 1 ≤ n ≤ 8 (see figure 5).

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As can be seen from figure 5, the numerical result is slightly less than the analysis result when 0 ≤ n ≤ 4 and the same as the analysis result when n > 4, which shows that the analysis result equation (28) is desirable when the network tends to infinity. As shown in figure 5 and equation (28), ASP grows as a power-law function of the network order with the exponent, denoted by $\gamma =\tfrac{1}{3}$.

In this paper, a new method of generating network is introduced to construct the claw network. Moreover, we determine the exact solution of ART and ASP on claw network. The exact expression of ART is obtained after finding out the results of MFPTS by block method, then we test the analysis result and the numerical result of ART. From figure 3, we can see that before the fourth generation of claw network, the numerical result is slightly larger than the analysis result, after the fourth generation of the network, the analysis result is the same as the numerical result. This proves that our analysis about ART is reasonable, and the analysis result of ART implies that ART increases exponentially with the size of the network.

The accurate expression of ASP is obtained by twice-blocking method. Similarly, we examine the analysis result and the numerical result of ASP. From figure 5, we can see that before the fourth generation of the claw network, the numerical result is slightly smaller than the analysis result, and after the fourth generation of the claw network, the analysis result is the same as the numerical result. This shows that our analysis about ASP is also appropriate, and the analysis result of ASP displays that ASP increases sublinearly with the size of the network.

Research is supported by National Natural Science Foundation of China (Nos. 11 501 255, 11 671 172, 11 801 225), University Science Research Project of Jiangsu Province (No.18KJB110005) and Foundation for advanced talents of Jiangsu University (5503000039).

Let ${{\rm{\Omega }}}_{n}^{(\alpha \beta )}$ be the shortest path of two nodes which are in different blocks. From figure 4, we can get the following equation.

$$\begin{eqnarray*}&&{{\rm{\Omega }}}_{n}^{(1)}=8{{\rm{\Omega }}}_{n}^{({AB})}+20{{\rm{\Omega }}}_{n}^{({AC})}+8{{\rm{\Omega }}}_{n}^{({AD})}\end{eqnarray*}$$

In order to obtain ${{\rm{\Omega }}}_{n}^{(\alpha \beta )}$, we define

$$\begin{eqnarray*}&&{{\rm{\Delta }}}_{n}^{(1)}=\displaystyle \sum _{i\in {G}_{n}^{(1)},i\ne 1}{d}_{i1}\end{eqnarray*}$$

Where d_i1 is the shortest path of any node in ${G}_{n}^{(1)}$ to node 1. According to the topological structure of claw network, we have the recursive relation about ${{\rm{\Delta }}}_{n}^{(1)}$.

$$\begin{eqnarray*}\begin{array}{rcl}{{\rm{\Delta }}}_{n}^{(1)} & = & 4{{\rm{\Delta }}}_{n-1}^{(1)}\\ & & +\,(2{{\rm{\Delta }}}_{n-1}^{(1)}+\displaystyle \frac{{N}_{n-1}-3}{4}\times 2\times {2}^{n-1})\times 2+{2}^{n}\end{array}\end{eqnarray*}$$

The second term of RHS (right-hand side) of equation is the sum of shortest paths from nodes connected to sub-hub nodes (nodes 4 and 8) to node 1, The third term of RHS (right-hand side) of equation is the sum of shortest paths from nodes 4 and 8 to node 1. Simplified the equation, we have

$$\begin{eqnarray*}&&{{\rm{\Delta }}}_{n}^{(1)}=8{{\rm{\Delta }}}_{n-1}^{(1)}+\displaystyle \frac{1}{14}\times {16}^{n}+\displaystyle \frac{3}{7}\times {2}^{n}\end{eqnarray*}$$

Combining the initial condition ${{\rm{\Delta }}}_{1}^{(1)}=2$, we can get

$$\begin{eqnarray*}&&{{\rm{\Delta }}}_{n}^{(1)}=\displaystyle \frac{1}{7}\times ({16}^{n}-{2}^{n})\end{eqnarray*}$$

After getting ${{\rm{\Delta }}}_{n}^{(1)}$, we have

$$\begin{eqnarray*}\begin{array}{rcl}{{\rm{\Omega }}}_{n}^{({AB})} & = & \displaystyle \sum _{i\in {G}_{n-1}^{(A)},j\in {G}_{n-1}^{(B)},i,j\ne 4}{d}_{{ij}}\\ & = & \displaystyle \sum _{i\in {G}_{n-1}^{(A)},j\in {G}_{n-1}^{(B)},i,j\ne 4}({d}_{i4}+{d}_{j4})\\ & = & \displaystyle \frac{{N}_{n-1}-3}{4}{{\rm{\Delta }}}_{n-1}^{(1)}+\displaystyle \frac{{N}_{n-1}-3}{4}{{\rm{\Delta }}}_{n-1}^{(1)}\\ & = & \displaystyle \frac{{N}_{n-1}-3}{2}{{\rm{\Delta }}}_{n-1}^{(1)}\end{array}\end{eqnarray*}$$

and

$$\begin{eqnarray*}\begin{array}{rcl}{{\rm{\Omega }}}_{n}^{({AC})} & = & \displaystyle \sum _{i\in {G}_{n-1}^{(A)},j\in {G}_{n-1}^{(C)},i\ne 4,j\ne 8}{d}_{{ij}}\\ & = & \displaystyle \sum _{i\in {G}_{n-1}^{(A)},j\in {G}_{n-1}^{(C)},i\ne 4,j\ne 8}({d}_{i4}+{d}_{48}+{d}_{j8})\\ & = & \displaystyle \frac{{N}_{n-1}-3}{4}{{\rm{\Delta }}}_{n-1}^{(1)}+{2}^{n-1}{\left(\displaystyle \frac{{N}_{n-1}-3}{4}\right)}^{2}\\ & & +\displaystyle \frac{{N}_{n-1}-3}{4}{{\rm{\Delta }}}_{n-1}^{(1)}=\displaystyle \frac{{N}_{n-1}-3}{2}{{\rm{\Delta }}}_{n-1}^{(1)}\\ & & +{2}^{n-1}{\left(\displaystyle \frac{{N}_{n-1}-3}{4}\right)}^{2}\end{array}\end{eqnarray*}$$

$$\begin{eqnarray*}\begin{array}{rcl}{{\rm{\Omega }}}_{n}^{({AD})} & = & \displaystyle \sum _{i\in {G}_{n-1}^{(A)},j\in {G}_{n-1}^{(D)},i,j\ne 4}{d}_{{ij}}\\ & = & \displaystyle \sum _{i\in {G}_{n-1}^{(A)},j\in {G}_{n-1}^{(D)},i,j\ne 4}({d}_{i4}+{d}_{j4})\\ & = & \displaystyle \frac{{N}_{n-1}-3}{4}{{\rm{\Delta }}}_{n-1}^{(1)}+{2}^{n-1}\displaystyle \frac{{N}_{n-1}-3}{4}{{\rm{\Delta }}}_{1}^{(1)}\end{array}\end{eqnarray*}$$

By calculating,we can get

$$\begin{eqnarray*}&&{{\rm{\Omega }}}_{n}^{(1)}=\displaystyle \frac{208}{49}\times {128}^{n-1}-\displaystyle \frac{12}{49}\times {16}^{n}-\displaystyle \frac{8}{49}\times {2}^{n-1}\end{eqnarray*}$$

Let ${{\rm{\Omega }}}_{n}^{({xx})}$ be the shortest path of two nodes which belong to different blocks, for example, ${{\rm{\Omega }}}_{n}^{(12)}$ denotes the shortest path of two nodes belonging to ${G}_{n}^{(1)}$ and ${G}_{n}^{(2)}$ respectively. From figure 2, we can get the following equation.

$$\begin{eqnarray*}\begin{array}{rcl}{{\rm{\Omega }}}_{n} & = & {{\rm{\Omega }}}_{n}^{(12)}+4{{\rm{\Omega }}}_{n}^{(13)}+{{\rm{\Omega }}}_{n}^{(34)}+{2}^{n}{D}_{{tot}}(0)\\ & & +\,6\left({{\rm{\Delta }}}_{n}^{(1)}+{2}^{n}\left(\displaystyle \frac{{N}_{n-1}-3}{4}\right)\right)\\ & & +\,2\left({{\rm{\Delta }}}_{n}^{(1)}+{2}^{n+1}\left(\displaystyle \frac{{N}_{n-1}-3}{4}\right)\right)\end{array}\end{eqnarray*}$$

Because of the particularity of the initial graph, the expressions of Ω_n and ${{\rm{\Omega }}}_{n}^{(1)}$ are not similar.

$$\begin{eqnarray*}\begin{array}{rcl}{{\rm{\Omega }}}_{n}^{(12)} & = & \displaystyle \sum _{i\in {G}_{n}^{(1)},j\in {G}_{n}^{(2)},i,j\ne 1}{d}_{{ij}}\\ & = & \displaystyle \sum _{i\in {G}_{n}^{(1)},j\in {G}_{n}^{(2)},i,j\ne 1}({d}_{i1}+{d}_{j1})\\ & = & \displaystyle \frac{{N}_{n}-3}{4}{{\rm{\Delta }}}_{n}^{(1)}+\displaystyle \frac{{N}_{n}-3}{4}{{\rm{\Delta }}}_{n}^{(1)}\\ & = & \displaystyle \frac{{N}_{n}-3}{2}{{\rm{\Delta }}}_{n}^{(1)}\end{array}\end{eqnarray*}$$

and

$$\begin{eqnarray*}\begin{array}{rcl}{{\rm{\Omega }}}_{n}^{(13)} & = & \displaystyle \sum _{i\in {G}_{n}^{(1)},j\in {G}_{n}^{(3)},i\ne 1,j\ne 2}{d}_{{ij}}\\ & = & \displaystyle \sum _{i\in {G}_{n}^{(1)},j\in {G}_{n}^{(3)},i\ne 1,j\ne 2}({d}_{i1}+{d}_{12}+{d}_{j2})\\ & = & \displaystyle \frac{{N}_{n}-3}{4}{{\rm{\Delta }}}_{n}^{(1)}+{2}^{n}{\left(\displaystyle \frac{{N}_{n-1}-3}{4}\right)}^{2}+\displaystyle \frac{{N}_{n}-3}{4}{{\rm{\Delta }}}_{n}^{(1)}\\ & = & \displaystyle \frac{{N}_{n}-3}{2}{{\rm{\Delta }}}_{n}^{(1)}+{2}^{n}{\left(\displaystyle \frac{{N}_{n}-3}{4}\right)}^{2}\end{array}\end{eqnarray*}$$

$$\begin{eqnarray*}\begin{array}{rcl}{{\rm{\Omega }}}_{n}^{(34)} & = & \displaystyle \sum _{i\in {G}_{n}^{(3)},j\in {G}_{n}^{(4)},i\ne 2,j\ne 3}{d}_{{ij}}\\ & = & \displaystyle \sum _{i\in {G}_{n-1}^{(A)},j\in {G}_{n-1}^{(D)},i\ne 2,j\ne 3}({d}_{i2}+{d}_{23}+{d}_{j3})\\ & = & \displaystyle \frac{{N}_{n}-3}{2}{{\rm{\Delta }}}_{n}^{(1)}+{2}^{n+1}{\left(\displaystyle \frac{{N}_{n}-3}{4}\right)}^{2}\end{array}\end{eqnarray*}$$

By calculating, we can get

$$\begin{eqnarray*}&&{{\rm{\Omega }}}_{n}=\displaystyle \frac{48}{49}\times {128}^{n}+\displaystyle \frac{100}{49}\times {16}^{n}+\displaystyle \frac{48}{49}\times {2}^{n}\end{eqnarray*}$$