**7. Example of proof of a result on control using combinatorial enumeration**

In [30] I used the combinatorial enumeration methods and probability for showing the applicability of the selection node criteria in a virus spreading control problem in complex networks. The main purpose of this section is to illustrate how the enumerative combinatorics in combination with probability theory can be used for demonstrating mathematically a result in an application field. We can mention that the case of homogeneity in the behavior of the nodes and their interaction cannot be discarded given what has been observed in the reaction of the agents in the context of social networks is that they try to minimize the conflict. Many successful models can, for example [30–33], base their predicting effectiveness on the homogeneity of the behavior of the nodes and their interaction. By the other side, in [30] we have obtained a criteria for selecting the nodes to be controlled, but such criteria fail if we have homogeneity in the behavior of the nodes and, at the same time, the topology of the network is regular. Then, what we want to do here is to justify the applicability of node selection criteria, keeping the homogeneity of the nodes and trying to compare the number of regular graphs with *n* vertices with the total of graphs that can be constructed with *n* vertices. For this end, based on the results on combinatorial graph enumeration mentioned on Theorems 1.9 and 1.10, we can state our main result as follows. First of all, we suppose that our graph is labeled, *G* ¼ ð Þ *V; E* is *r*-regular with *r*≥ 3 constant and *rn* ¼ 2*m*, where *n* ¼ ∣*V*∣ corresponds to the number of vertices and *m* ¼ ∣*E*∣ corresponds to the number of edges. Let *Lr* the number of labeled regular graphs of degree *r* whose asymptotic value is [28]

$$L\_r \sim e^{-\frac{r^2 - 1}{4}} \frac{(2m)!}{2^m m!} (r!)^{n}. \tag{47}$$

Let *Gn* be the number of all possible graphs with *n* vertices whose value is

$$\mathbf{G}\_{\mathfrak{n}} = \mathbf{2}^{\binom{n}{2}}.\tag{48}$$

Theorem 1.11 If *r*≥3 and *nr* ¼ 2*m*, then

$$\lim\_{n \to \infty} \frac{L\_r}{G\_n} = \mathbf{0}.\tag{49}$$

Proof of Theorem 1.11. If *nr* ¼ 2*m* and *r* is constant of value *c*1, then we can deduce that *<sup>m</sup>* <sup>¼</sup> *<sup>r</sup>* <sup>2</sup> *<sup>n</sup>* <sup>¼</sup> *<sup>c</sup>*<sup>1</sup> <sup>2</sup> *n*, and this implies that *m* ¼ *O n*ð Þ so let us say that *m* ¼ *c*2*n*; then,

$$\lim\_{n \to \infty} \frac{L\_r}{G\_n} = \lim\_{n \to \infty} \frac{e^{-\frac{\epsilon\_1^2 - 1}{4} \frac{(2c\_2 n)! (c\_1!)^{-n}}{2^n (c\_2 n)!}}}{2^{\binom{n}{2}}} \tag{50}$$

*Combinatorial Enumeration of Graphs DOI: http://dx.doi.org/10.5772/intechopen.88805*

$$\mathfrak{h} = \lim\_{n \to \infty} \frac{\frac{e^{-\frac{\mathfrak{e}\_1^2 - 1}{\mathfrak{h}}}}{(c\_1!)^{n}} \frac{(2c\_2 n)!}{2^{c\_2 n} (c\_2 n)!}}{\mathfrak{2^{\frac{n(n-1)}{2}}}},\tag{51}$$

applying the approximation Stirling formula *<sup>n</sup>*! � ffiffiffiffiffiffiffiffi <sup>2</sup>*π<sup>n</sup>* <sup>p</sup> *<sup>n</sup> e* � �*<sup>n</sup>*

$$(c\_1!)^n = \left(\sqrt{2\pi c\_1} \left(\frac{c\_1}{\varepsilon}\right)^{c\_1}\right)^n \tag{52}$$

then

*<sup>U</sup>*<sup>Δ</sup> � *<sup>L</sup>*<sup>Δ</sup>

where *<sup>m</sup>* <sup>¼</sup> <sup>Δ</sup>*<sup>n</sup>*

**enumeration**

value is [28]

deduce that *<sup>m</sup>* <sup>¼</sup> *<sup>r</sup>*

then,

**238**

2 .

*Probability, Combinatorics and Control*

*<sup>n</sup>*! � *<sup>e</sup>*

**7. Example of proof of a result on control using combinatorial**

*Lr* � *e*

Theorem 1.11 If *r*≥3 and *nr* ¼ 2*m*, then

<sup>2</sup> *<sup>n</sup>* <sup>¼</sup> *<sup>c</sup>*<sup>1</sup>

lim*n*!∞ *Lr Gn* �*r*2�<sup>1</sup> 4 ð Þ 2*m* ! 2*mm*!

Let *Gn* be the number of all possible graphs with *n* vertices whose value is

*n* 2 

*Gn* ¼ 2

lim*n*!∞ *Lr Gn*

<sup>¼</sup> lim*<sup>n</sup>*!<sup>∞</sup>

Proof of Theorem 1.11. If *nr* ¼ 2*m* and *r* is constant of value *c*1, then we can

*e*� *c*2 1 �1 ð Þ *<sup>r</sup>*! *<sup>n</sup>*

<sup>2</sup> *n*, and this implies that *m* ¼ *O n*ð Þ so let us say that *m* ¼ *c*2*n*;

<sup>4</sup> ð Þ <sup>2</sup>*c*2*<sup>n</sup>* !ð Þ *<sup>c</sup>*1! �*<sup>n</sup>* <sup>2</sup>*c*2*n*ð Þ *<sup>c</sup>*2*<sup>n</sup>* !

> 2 *n* 2

*:* (47)

*:* (48)

¼ 0*:* (49)

(50)

In [30] I used the combinatorial enumeration methods and probability for showing the applicability of the selection node criteria in a virus spreading control problem in complex networks. The main purpose of this section is to illustrate how the enumerative combinatorics in combination with probability theory can be used for demonstrating mathematically a result in an application field. We can mention that the case of homogeneity in the behavior of the nodes and their interaction cannot be discarded given what has been observed in the reaction of the agents in the context of social networks is that they try to minimize the conflict. Many successful models can, for example [30–33], base their predicting effectiveness on the homogeneity of the behavior of the nodes and their interaction. By the other side, in [30] we have obtained a criteria for selecting the nodes to be controlled, but such criteria fail if we have homogeneity in the behavior of the nodes and, at the same time, the topology of the network is regular. Then, what we want to do here is to justify the applicability of node selection criteria, keeping the homogeneity of the nodes and trying to compare the number of regular graphs with *n* vertices with the total of graphs that can be constructed with *n* vertices. For this end, based on the results on combinatorial graph enumeration mentioned on Theorems 1.9 and 1.10, we can state our main result as follows. First of all, we suppose that our graph is labeled, *G* ¼ ð Þ *V; E* is *r*-regular with *r*≥ 3 constant and *rn* ¼ 2*m*, where *n* ¼ ∣*V*∣ corresponds to the number of vertices and *m* ¼ ∣*E*∣ corresponds to the number of edges. Let *Lr* the number of labeled regular graphs of degree *r* whose asymptotic

� <sup>Δ</sup>ð Þ <sup>2</sup>�<sup>1</sup> 4

For the details of the proof of Theorems 1. 9 and 1.10, see [28, 29], respectively.

ð Þ 2*m* ! 2*mm*!

ð Þ <sup>Δ</sup>! �*<sup>n</sup>*

*<sup>n</sup>*! *,* (46)

$$=\lim\_{n\to\infty} \frac{\frac{e^{\frac{\epsilon\_1^2}{4}-1}}{\left(\sqrt{2\pi c\_1}\left(\frac{c\_1}{\epsilon}\right)^{\epsilon\_1}\right)^n \overline{Z^{\epsilon\_1^n}(c\_2n)!}}{\mathcal{Z}^{\frac{n(n-1)}{2}}}\tag{53}$$

simplifying

$$=\lim\_{n\to\infty} \frac{\frac{1}{\left(\sqrt{2\pi c\_1}\right)^{n}(c\_1)^{\epsilon\_1 n}e^{\frac{\epsilon\_1^2 + 4c\_1n + 1}{4}}}{2^{\frac{n(n-1)}{2}}}}{2^{\frac{n(n-1)}{2}}}\tag{54}$$

applying the approximation Stirling formula

$$(2c\_2n)! = \sqrt{2\pi(2c\_2n)} \left(\frac{(2c\_2n)}{\varepsilon}\right)^{(2c\_2n)}\tag{55}$$

and

$$(c\_2 n)! = \sqrt{2\pi (c\_2 n)} \left(\frac{(c\_2 n)}{\varepsilon}\right)^{(c\_2 n)} \tag{56}$$

we get

$$=\lim\_{n\to\infty} \frac{\frac{1}{\left(\sqrt{2\pi c\_1}\right)^{n}\left(c\_1\right)^{c\_1}\frac{\tau\_1^2+4\varsigma\_1n+1}{4}}\frac{\sqrt{2\pi(2c\_2n)}\left(\frac{(2\varsigma\_2n)}{\tau}\right)^{(2\varsigma\_2)}}{2^{\varsigma\_2}\sqrt{2\pi(c\_2n)}\left(\frac{(\varsigma\_2n)}{\tau}\right)^{(\varsigma\_2n)}}},\tag{57}$$

simplifying

$$=\lim\_{n\to\infty} \frac{\frac{\sqrt{2}}{\left(\sqrt{2\pi c\_1}\right)^n c\_1^{c\_1 n} e^{\left(c\_1^{2+4c\_1 n+1}\right)/4}} \left(\frac{c\_2 n}{e}\right)^{c\_2 n}}{\mathcal{Z}^{\frac{n(n-1)}{2}}}\tag{58}$$

given that *r*≥3, for which we assumed that *r* is a constant *c*1, and that *nr* ¼ 2*m*, then we have that *c*<sup>1</sup> ¼ 2*c*2, and replacing that, in (58), we can express it in terms of *c*1, which is the regular degree *r* assumed as fixed; then, we get

$$\lim\_{n\to\infty} \frac{\frac{\sqrt{2}}{\left(\sqrt{2\pi c\_1}\right)^n c\_1^{c\_1 n} \epsilon \left(\epsilon\_1^{2+4c\_1 n+1}\right)/4} \left(\frac{c\_1 n/2}{\epsilon}\right)^{c\_1 n/2}}{2^{\frac{n(n-1)}{2}}},\tag{59}$$

and, replacing *c*<sup>1</sup> by *r* in (59), we get

$$\lim\_{n\to\infty} \frac{\frac{\sqrt{2}}{\left(\sqrt{2\pi}r\right)^{n}r^{m}\epsilon^{\left(r^{2}+4m+1\right)/4}}\left(\frac{m/2}{\epsilon}\right)^{m/2}}{2^{\frac{n(n-1)}{2}}}.\tag{60}$$

Therefore, if the degree *<sup>r</sup>* is constant, the lim*n*!<sup>∞</sup> *Lr Gn* ¼ 0.

Now, our main result can be stated as a consequence of Theorem 1.11.

Theorem 1.12 If we assume that all graphs are uniformly distributed and that the nodes have homogeneous behavior, then the criteria for selecting nodes to be controlled are almost always applicable.

**Proof of Theorem 1.11.** As a consequence of Theorem 11, we know that the probability that a regular graph appears tends to zero as *n* ! ∞. Then, the mentioned criteria are almost always applicable.
