## 3 Network Connectivity Analysis
### Node Non-Isolation Probability  
We consider an arbitrary subarea \(A\) of an infinite plane. In random graph theory [26], connectivity is defined as follows:  
**Definition 2:** A graph is said to be \(k\)-connected if for each pair of nodes there exist at least \(k\) mutually independent paths connecting them. In other words, a graph is \(k\)-connected if and only if there is no set of \(k\)-1 nodes whose removal would disconnect the graph.  
Our focus will first be on one-connectivity. Higher orders of connectivity will be discussed in the future work. We are interested in the critical density \(\rho\), at which the network \(A\) is one-connected a.s. In this section, we derive necessary condition for the network to be one-connected. We denote \(P(I)\) as the isolation probability, which is the probability that a randomly chosen node has no neighbors at all. The probability that none of the nodes in \(A\) is isolated, denoted by \(P(\bar{I})\) can be the upper bound for the probability that the network is one-connected [9; 27], denoted by \(P(C)\). Thus we have:  
\[P(C)\leq P(\overline{I}) \tag{3}\]  
\[\rho(P(C)=p)\geq\rho(P(\overline{I})=p) \tag{4}\]  
Equation (3) is obvious. Equation (4) means under the same network configuration, we need higher node density to reach one-connectivity with probability \(p\) than to reach node non-isolation with the same probability. We assume the isolation of different nodes in A to be independent events and the number of nodes in \(A\) is large, i.e. \(\lambda\geq 100\). The conditional probability of non-isolation is give by:  
\[P(\overline{I}\mid N=k)=(1-P(I))^{k} \tag{5}\]  
Thus the unconditional non-isolation probability can be written as:  
\[\begin{split} P(\overline{I})&=\sum_{k=1}^{\infty }P(\overline{I}\mid N=k)\cdot P(N=k)\\ &=\sum_{k=1}^{\infty}(1-P(I))^{k}\cdot\frac{\lambda^{k}}{k!}e^{- \lambda}\\ &=e^{-P(I)\lambda}\sum_{k=1}^{\infty}\frac{((1-P(I))\lambda)^{k} }{k!}e^{-(1-P(I)\lambda}\\ &=e^{-P(I)\lambda}\end{split} \tag{6}\]  
Since \(P(I)\) is the probability that a randomly chosen node has no neighbors, we denote \(D\) to be the number of neighbors of a node, which is also called node degree. Note that \(D\) also follows Poisson distribution according to (1) and let the expected value of \(D\) be \(D_{0}\). Hence \(P(I)\) can be calculated as:  
\[P(I)=P(D=0)=e^{-D_{0}} \tag{7}\]  
In the following, we calculate the mean node degree \(D_{0}\). Let \(L(i,j)\) denote the event that there is a link between node \(i\) and node \(j\). The distance between the two nodes is known as \(s(i,j)\). Given the distance, the probability for \(L(i,j)\) is denoted by \(P(L|s)\). \(D_{0}\) can be calculated by integrating \(\rho P(L|s)\) over the whole plane:  
\[D_{0}=\rho\int_{0}^{2\pi}\int_{0}^{\infty}P(s)\cdot sds\ d\phi=2\pi\rho\int_{0 }^{\infty}P(s)\cdot sds \tag{8}\]  
Plugging (7) (8) to (6), we have:  
\[P(\overline{I})=\exp(-\rho\big{\|}A\big{\|}\exp(-2\pi\rho\xi)) \tag{9}\]where \(\xi=\int_{0}^{\infty}P(s)sds\). Let the left hand side of (9) be \(p\), then the density \(\rho\) satisfies the following equation:  
\[\rho e^{-2\pi\varphi_{s}^{z}}=\frac{\ln p}{\left\|A\right\|} \tag{10}\]  
The solution of the equation is  
\[\rho(P(\overline{I})=p)=-\frac{1}{2\pi\xi}\mathrm{[PERSON]}(\frac{-2\pi\xi\ln p }{\left\|A\right\|}) \tag{11}\]  
where \(\mathrm{[PERSON]}(x)\) is the solution \(w\) of equation \(w\)-\(\exp(w)\)=\(x\). Finally, the problem has transformed to the analysis of \(\xi\) and \(P(L|s)\). In other words, the study on global network connectivity has turned into the analysis of local link probability.