Analytical Solution for the Size of the Minimum Dominating Set
in Complex Networks

A graph $G=(V,E)$ is a set of nodes $V$ and edges $E$. A Dominant Set (DS) $S$ is defined to be a subset of $V$, where each node $i\in V$ belongs to $S$ or adjacent to an element of $S$. For each node $i\in V$, we define a binary function $\sigma_{i}$ to be $+1$ if $i\in S$, otherwise $0$. We call the node $i$ occupied (resp. empty) if $\sigma_{i}$ is $+1$ (resp. $0$). The set of the adjacent nodes to node $i$ is denoted by $\partial i$. We can then see that the constraint to define a DS is as follows:

For each node $i\in V$, we define a binary function $I_{i}$ to be $+1$ if and only if $\sigma_{i}+\sum_{j\in\partial i}\sigma_{j}\geq 1$, otherwise 0. A set $S$ is an Minimum Dominating Set (MDS) if the size is smallest among all dominating sets (see Fig. 1).

Let us consider the following Hamiltonian function:

where $\sigma=\{\sigma_{i}|i\in V\}$. Then the partition function is given by

where the summation is taken over all configurations of $\sigma$ and $\beta$ is the inverse temperature.

In what follows, we apply the cavity method to derive the analytic formula for the density of MDS in complex networks. Note that the cavity approach has been used in a similar way to solve the maximal independent set problem dalla . First, we assume the graph $G$ has tree structure and let $\partial i\backslash j$ be the set of $\partial i$ except for node $j$ (see Fig. 2). Note also that the minimum degree of the graph should be two, otherwise some of $\partial i\backslash j$ do not exist. Then, we can write $\sigma_{i\to j}=\{\sigma_{k}|k\in\partial i\backslash j\}$. Next, let $\nu_{i\to j}(\sigma_{i},\sigma_{i\to j})$ be the probability that the node $i$ and $\partial i\backslash j$ take value $\sigma_{i}$ and $\sigma_{i\to j}$, respectively, and constraints $I_{i}$ and $I_{j}$ are not included:

where the first summation is taken over all configurations of $\sigma^{\prime}=\{\sigma_{\beta}|(\beta\neq i)\wedge(\beta\notin\partial i\backslash j)\wedge(\beta\in V)\}$. The product for the constraint $I_{\alpha}$ is taken over all nodes of $V$ except for $i$, $j$ (i.e. $\alpha\in V$ and $\alpha\neq i,j$) and $Z_{i\to j}$ is the normalization constant given by

Eq. (4) can be written in a different way, which will be easier to compute. As shown in Fig. 3, we divide a graph into two subgraphs ($A$ and $B$) by cutting the edge ($i$, $j$) between node $i$ and $j$. Let $A$ be the set of all the nodes which belongs to the subgraph including $i$, and $B$ be the set of all the nodes which belongs to the subgraph including $j$. Then, Eq. (4) can be transformed into

where the first summation is taken over all configurations of $\sigma^{\prime}_{A}=\{\sigma_{\beta}|(\beta\neq i)\wedge(\beta\notin\partial i\backslash j)\wedge(\beta\in A)\}$. The product for the constraint $I_{\alpha}$ is taken over all nodes of $A$ except $i$ (i.e. $\alpha\in A$ and $\alpha\neq i$) and $Z^{\prime}_{i\to j}$ is a normalization constant given by

where the last summation is taken over all configurations of $\sigma_{A}=\{\sigma_{\beta}|(\beta\in A)\}$.

Then the following exact recursive equation can be derived:

This equation can be iteratively solved with its corresponding normalization constant.

Let $\bar{\nu}(\sigma_{i},m)$ be the summation of $\nu_{i\to j}(\sigma_{i},\sigma_{i\to j})$, where the number of occupied neighbors $\sigma_{i\to j}$ is $m$:

Then, let $r_{1}^{i\to j}$ be the probability that the node $i$ is occupied in the cavity graph. Let $r_{00}^{i\to j}$ be the probability that the node $i$ and all the neighbors $\sigma_{i\to j}$ are empty. Let $r_{0}^{i\to j}$ be the probability that the node $i$ is empty and at least one of the neighbors $\sigma_{i\to j}$ are occupied. Then, we can write:

where $k_{i}$ is the degree of node $i$. Here we note that $r_{1}^{i\to j}+r_{00}^{i\to j}+r_{0}^{i\to j}=1$, because of probability normalization.

The above definitions lead to the following iterative equations:

where $N$ is the normalization constant given by

We note that until now we are considering the problem of an DS, which is treated as finite temperature problem ($\beta$ is finite) in the context of statistical mechanics. To address the minimum dominating set (MDS) problem, we have to consider the zero-temperature limit ($\beta\to\infty$) which gives the ground energy state of the Hamiltonian shown in Eq. (2). To solve the equations associated to the zero-temperature limit, we can use an Ultra-Discretization (UD) procedure UD4 .

Discretization is a well-known method to discretize independent variable in continuous theory such as differential equation (see Table 1 from ($i$) to ($ii$)). UD can go further and aims to discretize the dependent variable in the discretized equation (from ($ii$) to ($iii$)). As a result, in the transformed UD equation ($iii$), both independent and dependent variables are discretized (i.e. both are integer variables). The key formula of UD, which transforms the equation type from ($ii$) to ($iii$) is

When the discretized equation ($ii$) has no subtraction operator, we can transform it into the corresponding UD equation ($iii$) by replacing $x=e^{\beta X}$ and $y=e^{\beta Y}$ and taking limit $\beta\to\infty$. Here, $x$ and $y$ belong to discretized equation ($ii$) and $X$ and $Y$ belong to UD equation ($iii$). The operator in the original discretized equation ($ii$) is transformed in the UD equation ($iii$) as follows:

Here we remark that in order to obtain the UD equation ($iii$), it is necessary to be able to remove any minus operator in the original discretized equation ($ii$).

In order to ultra-discretize the previous Eqs. (13), (14) and (15), we make them consist of only plus, multiplication and division operators, avoiding minus operators. By simple computation, we have

and

where $n$ is the number of elements of $\partial i\backslash j$, and $m_{p}=0~\mbox{or}~1$ $(p=1,2,3,\cdots,n)$.

By inserting Eqs. (21) and (22) into Eqs. (13), (14) and (15), we have

Here we remark that the above derived equations consist only of three operators (plus, multiplication and division), which is suitable for the ultra-discretization.

Here, we then replace the following three variables as follows:

where $R_{1}^{i\to j}$, $R_{00}^{i\to j}$ and $R_{0}^{i\to j}$ are variables in UD system.

After inserting (26), (27), (28) into (23), (24), (25) and taking zero-temperature limit $\beta\to\infty$, Eqs. (23), (24) and (25) are transformed into

Eq. (23), (24), (25) (or Eq. (29), (30), (LABEL:eqn:UD_equation_3)) are very difficult to solve, because every edge direction has three variables and three associated equations. Therefore, in order to avoid this high complexity, we use a coarse-grained method.

Let $p_{k}$ be the degree distribution of the network which gives the probability to find nodes with degree $k$ and $\langle k\rangle=\sum_{k}kp_{k}$ be the average degree of the network. The excess degree distribution is given by $q_{k}=(k+1)p_{k+1}/{\langle k\rangle}$, that is the probability to find that a neighbor node has degree k.

Let us assume that the network is enough large. Each $r_{1}^{i\to j}$ and $r_{0}^{i\to j}$ values are assigned to the edge direction from $i$ to $j$. The opposite edge direction gets the values from $r_{1}^{j\to i}$ and $r_{0}^{j\to i}$. $r_{00}^{i\to j}$ is not independent variable because of the normalization. Let $P(r_{1},r_{0})$ be the probability density function of $r_{1}^{i\to j}$, $r_{0}^{i\to j}$.

Then we have the coarse-grained equation (cavity mean field equation) for (23) and (25) as follows:

We transform the probability density function by variables transformation (26) and (28) as follows:

Then, we have

Taking UD limit (zero temperature) $\beta\to\infty$, we have the ultra-discretization version of cavity equation:

Next, we will solve this equation. Eqs. (29), (30), (LABEL:eqn:UD_equation_3) imply that $R_{1}$ and $R_{0}$ takes integer values, since $R_{1}$ and $R_{0}$ at the boundary of network takes integer values. Furthermore, considering the probability conservation, Eqs. (26), (28) imply that $R_{1}$ and $R_{0}$ takes value 0 or negative integer. By considering Eq. (29), firstly, we can see that $R_{1}^{i\to j}$ takes value only $0$ or $-1$. Secondly, if $R_{1}^{i\to j}$ is $-1$, then for each $k\in\partial i\backslash j$, one of $R_{0}^{k\to i}$ or $R_{1}^{k\to i}$ takes value $0$. In this case ($R_{1}^{i\to j}=-1$), from Eq. (LABEL:eqn:UD_equation_3), $R_{0}^{i\to j}$ takes value $0$ or $-1$. Therefore, we can set the distribution $\bar{P}(R_{1},R_{0})$ as follows:

where $a+b+\sum_{n=0}^{\infty}c_{n}=1$.

Inserting (36) into the cavity equation (LABEL:eqn:_cavity_equation_UD), we have

Once the degree distribution $p_{k}$ is given, we can determine $a$, $b$, $c_{n}$ $(n=0,1,2,\cdots)$.

Here, we need the average energy of the Hamiltonian (2), which can be identified as the average number of all DS configurations. After taking zero temperature limit $\beta\to\infty$, we will obtain the analytical formula for the density of MDS.

The average energy can be computed by $<H>=-\frac{\partial}{\partial\beta}\log Z$, where the partition function is shown in Supplementary Information. A generic expression for density is given by

This equation, however, cannot be ultra-discretized because of the minus operator, and without being ultra-discretized, analytical solution cannot be obtained. Therefore, we avoid minus operator by inserting Eqs. (21) and (22), which leads to the following result:

After taking zero temperature limit $\beta\to\infty$, we obtain

By inserting (36) into this equation, we finally obtain

More simply, we transform the previous equation into

Eq. (LABEL:eqn:_analytic_formula2) is our main result. Using this equation, we can analytically compute the density of MDS from any degree distribution $p_{k}$. More concretely, we can summarize the procedure as follows. Once the degree density function $p_{k}$ is given, we can easily compute the excess degree density function $q_{k}$. Then, by solving (37), (38) and (39), we obtain $a$, $b$, $c_{n}$ $(n=0,1,2,\cdots)$. Finally, inserting the value $a$, $b$, $c_{n}$ $(n=0,1,2,\cdots)$ into the above equation (LABEL:eqn:_analytic_formula2), we obtain the density of MDS.

We note that when we consider random network with average degree $z$ as a special case, we can derive the simple expression $\rho=1/z$ by using large average degree limit approximation ($z\to\infty$) (See supplementary Information).