An Implementation of the Python Code for Computing the Independent Number of Genus Graphs

Abstract

The computation of the independent number is a fundamental problem in graph theory. It has many applications, including route planning, computer graphics, network analysis, computational biology, and network architecture. Particularly in computational biology, it provides a powerful mathematical tool for modeling tumor robustness and optimizing cancer treatment. In this article, an algorithmic method for computing the independent number of genus graphs is presented. Moreover, a Python (version 3.10.5) implementation has been provided for constructing genus graphs and computing their independent number.

1. Introduction

A fundamental aspect of graph theory and combinatorial optimization is the study of independent sets in graphs, which offers extensive insights into the structural characteristics of graphs. Fundamentally, it is NP-hard to find an independent set of maximal cardinality [1]. The independent number, which is the size of the largest subset of vertices, such that no two vertices are adjacent, holds significance for a variety of applications, including route planning [2], computer graphics [3,4], network analysis [5], computational biology [6,7,8], network architecture, and resource allocation. Enabling sensors to conserve power while maintaining network coverage, it helps wireless sensor network designers create energy-efficient communication protocols [9,10]. By preventing conflicting tasks from being scheduled concurrently, it aids in the optimization of timetables and test schedules for graph coloring and scheduling. Independent sets, which are effective for maximizing influence and advertising, are non-connected groups in social network analysis. Additionally, in coding theory, it stands in for distinct codewords in error correction. Also, they assist in locating security rules or filters that do not conflict in network security [11,12]. DNA methods are presented by Zhaocai Wang et al. [13] to solve the maximal independent set problem using biological processes in the Adleman–Lipton model.

Studying the interaction between algebraic structures, such as cofinite submonoids and graph theory, has received more attention in the last few years. The extensively explored concepts within this domain include the zero divisor graph [14], nilpotent graphs [15], and Cayley graph [16]. In [17], Cameron explored the recent surge in research on graphs defined on algebraic structures, focusing on how these studies benefit both graph theory and algebra, particularly through three types of interactions between graphs and groups, with examples, and raises several open problems. Recently, different mathematicians assign graphs to the numerical semigroups and studied many invariants such as planarity, clique, and metric dimension [18,19,20,21,22,23].

Let ${\mathbb{N}}_0$ be the set of non-negative integers; then, a numerical semigroup is a subset $\mathfrak{S}\subseteq {\mathbb{N}}_0$ that is closed under addition, contains the zero element, and has a finite complement in ${\mathbb{N}}_0$. The elements of $\mathfrak{G}={\mathbb{N}}_0\setminus \mathfrak{S}$ are called gaps, and their total number is the genus $g\left(\mathfrak{S}\right)$ of the semigroup. The Frobenius number F is the largest integer not in $\mathfrak{S}$, and the multiplicity is the smallest nonzero element of $\mathfrak{S}$, denoted by $\omega$. Given a nonempty set $A\subseteq N_0$, the submonoid generated by A, denoted $\langle A\rangle$, is a numerical semigroup if and only if $gcd\left(A\right)=1$. Every numerical semigroup has a unique minimal system of generators, and the number of these generators defines the embedding dimension $e\left(\mathfrak{S}\right)$. A symmetric numerical semigroup is one where F is odd and for every $x\in \mathbb{Z}\setminus \mathfrak{S}$, $F-x$ belongs to $\mathfrak{S}$.

The rest of the paper is organized as follows: Section 2 provides a theoretical approach in the form of a sequence of results to compute the independent number of genus graphs. In Section 2, we present an algorithm and its implementation to compute the independent number of a genus graph.

2. Independent Number of Genus Graph of Symmetric Cofinite Submonoid

It is well known that the computation of the independent number is an NP-hard problem, and therefore, difficult to solve for general graphs. So, researchers continue to solve this problem by considering the special classes of graphs. We consider the genus graphs of symmetric cofinite submonoid

$$\mathfrak{S}=\langle \omega ,\omega +1,q\omega +2q+2,\dots ,q\omega +(\omega -1)\rangle ,q\in \mathbb{N},\omega \ge 2q+7,e=\omega -2q$$

and give an algorithmic way to compute the independent number of these graphs. Details of the class $\mathfrak{S}$ can be found in [24]. The genus graph is a simple, undirected graph denoted by $G_{\mathfrak{S}}$, having the set of vertices

$$V\left(G_{\mathfrak{S}}\right)=\{v_i:i\in {\mathbb{N}}_0\setminus \mathfrak{S}\}$$

and the set of edges

$$E\left(G_{\mathfrak{S}}\right)=\{v_i{v}_j\iff i+j\in \mathfrak{S}\}.$$

Consider the sets

$$A=\{i+1,0\le i\le \lfloor {\displaystyle \frac{\omega }{2}}\rfloor -1\},$$

$$B=\{\tau (\omega +1)+z+1,1\le \tau \le \lceil {\displaystyle \frac{q}{2}}\rceil -1,0\le z\le \lceil {\displaystyle \frac{\omega }{2}}\rceil -\tau -2\},$$

If $q\ge 2$ is an even integer, then

$$C=\{{\displaystyle \frac{q}{2}}(\omega +1)+t+1,0\le t\le {\displaystyle \frac{q}{2}}\},$$

and define

$$V_A=\{v_x:x\in A\},V_B=\{v_y:y\in B\},V_C=\{v_z:z\in C\}.$$

In the following lemma, we show that $V_A,V_B$ and $V_C$ are contained in $V\left(G_{\mathfrak{S}}\right)$ by showing that $A,B$ and C consist on gaps of $\mathfrak{S}$. Note that the Frobenius number F for $\mathfrak{S}$ is $F=2q\omega +2q+1$.

Lemma 1. 

Let $G_{\mathfrak{S}}$ be a genus graph, then $V_A,V_B,V_C\subseteq V\left(G_{\mathfrak{S}}\right)$.

Proof. 

For $0\le i\le \lfloor {\displaystyle \frac{\omega }{2}}\rfloor -1,0\le z\le \lceil {\displaystyle \frac{\omega }{2}}\rceil -\tau -2,0\le t\le {\displaystyle \frac{q}{2}}$, we have

$$i+1, z+\tau (\omega +1)>0,$$

and for $q\ge 2$ (even)

$${\displaystyle \frac{q}{2}}(\omega +1)+t+1>0.$$

Since $1\le 1+i\le \lfloor {\displaystyle \frac{\omega }{2}}\rfloor$, therefore $i+1\in g\left(\mathfrak{S}\right)$. Note that

$$\tau (\omega +1)+z+1=F-\left(\right(2q-\tau -z\left)\right(\omega +1)+z\omega )$$

and

$${\displaystyle \frac{q}{2}}(\omega +1)+t+1=F-(({\displaystyle \frac{3q}{2}}-t)(\omega +1)+t\omega ).$$

If $2q-\tau \ge z$, then $\left(\right(2q-\tau -z\left)\right(\omega +1)+z\omega )\in \mathfrak{S}$, and by symmetric property $F-\left(\right(2q-\tau -z\left)\right(\omega +1)+z\omega )\in g\left(\mathfrak{S}\right)$. If $2q-\tau <z$, then

$$(2q-\tau -z)(\omega +1)+z\omega =q\omega +(\omega +2q-\tau -z)+(q-x-1)\omega .$$

Since $2q-\tau -z<0$, therefore

$$q\omega +(\omega +2q-\tau -z)\in \{q\omega +(\omega -1),q\omega +(\omega -2),\dots ,q\omega +2q+2\}.$$

Also for $q\ge 3$, $(q-\tau -1)>0$, this gives $(q-\tau -1)\omega \in \mathfrak{S}$. Hence,

$$F-\left(\right(2q-\tau -z\left)\right(\omega +1)+z\omega )\in g\left(\mathfrak{S}\right).$$

Since ${\displaystyle \frac{3q}{2}}>{\displaystyle \frac{q}{2}}$, therefore $({\displaystyle \frac{3q}{2}}-t)((\omega +1)+t\omega )\in \mathfrak{S}$. This implies

$$F-(({\displaystyle \frac{3q}{2}}-t)(\omega +1)+t\omega )\in g\left(\mathfrak{S}\right).$$

□

The proof of the following lemma can be found in [25] (see Lemma 1).

Lemma 2. 

Let $\mathfrak{S}=\langle a,a+1,\dots ,a+x\rangle$ be a numerical semigroup with $1\le x<a$. Then $n\in \mathfrak{S}$ if and only if $n=qa+i$ with $q\in \mathbb{N}$ and $i\in \{0,\dots ,qx\}$.

Now, in the following lemma, we show that union of $V_A,V_B$ and $V_C$ form an independent set for the genus graph.

Lemma 3. 

Let $G_{\mathfrak{S}}$ be a graph, then $V_A\cup V_B\cup V_C$ is an independent set for the graph $G_{\mathfrak{S}}$.

Proof. 

To prove $V_A,V_B$ and $V_C$ are independent sets, we have the following cases:

Case-I: If ${\lambda }_1,{\lambda }_2\in A$ then

$${\lambda }_1+{\lambda }_2=i_1+1+i_2+1=i_1+i_2+2\in g\left(\mathfrak{S}\right).$$

Since $0\le i_1\ne i_2\le \lfloor {\displaystyle \frac{\omega }{2}}\rfloor -1$, therefore $3<i_1+i_2+2<\omega$. This implies

$${\lambda }_1+{\lambda }_2\in g\left(\mathfrak{S}\right).$$

Case-II: If ${\lambda }_1,{\lambda }_2\in B$ then

$${\lambda }_1+{\lambda }_2=z_1+1+{\tau }_1(\omega +1)+z_2+1+(\omega +1){\tau }_2=({\tau }_1+{\tau }_2)(\omega +1)+2+z_1+z_2,$$

where $z_1\ne z_2$. Note that

$$F-(({\tau }_1+{\tau }_2)(\omega +1)+2+z_1+z_2)=(2q-({\tau }_1+{\tau }_2))\omega +((2q-({\tau }_1+{\tau }_2))-1-z_1-z_2).$$

We can see that $2q\ge ({\tau }_1+{\tau }_2)$, since $1\le {\tau }_1\ne {\tau }_2\le \lceil {\displaystyle \frac{q}{2}}\rceil -1$. If

$$2q-({\tau }_1+{\tau }_2)\ge (z_1+z_2+1),$$

then by Lemma 2, we have

$$(2q-2\tau )\omega +(2q-2\tau -1-z_1-z_2)\in \mathfrak{S}.$$

If $2q-({\tau }_1+{\tau }_2)<(1+z_1+z_2)$ then

$$\begin{array}{cc}\hfill & (2q-({\tau }_1+{\tau }_2)\omega +((2q-({\tau }_1+{\tau }_2))-1-z_1-z_2)\hfill \\ \hfill =& (q-({\tau }_1+{\tau }_2)-1)\omega +q\omega +(\omega +(2q-({\tau }_1+{\tau }_2)-z_1-z_2-1)\in \mathfrak{S}.\hfill \end{array}$$

This implies $({\tau }_1+{\tau }_2)(\omega +1)+2+z_1+z_2\in g\left(\mathfrak{S}\right)$. Hence, ${\lambda }_1+{\lambda }_2\in g\left(\mathfrak{S}\right)$.

Case-III: If ${\lambda }_1,{\lambda }_2\in C$ then

$${\lambda }_1+{\lambda }_2={\displaystyle \frac{q}{2}}(\omega +1)+t_1+1+{\displaystyle \frac{q}{2}}(\omega +1)+t_2+1=q(\omega +1)+2+(t_1+t_2),$$

where $t_1\ne t_2$. Note that

$$F-(q\omega +q+2+(t_1+t_2))=q\omega +(q-(t_1+t_2+1))$$

and $2\le t_1+t_2+1\le q$; therefore, $q-(t_1+t_2+1)\ge 0)$. Again by Lemma 2, $q\omega +q+2+(t_1+t_2)\in g\left(\mathfrak{S}\right)$. Hence, ${\lambda }_1+{\lambda }_2\in g\left(\mathfrak{S}\right)$.

Case-IV: If ${\lambda }_1\in A$ and ${\lambda }_2\in B$ then

$${\lambda }_1+{\lambda }_2=i+1+\tau (\omega +1)+z+1=\tau (\omega +1)+z+i+2.$$

Note that

$$F-\left(\tau \right(\omega +1)+z+i+2)=(2q-\tau )\omega +(2q-\tau -1-z-i).$$

Also, $2q\ge \tau$, as $1\le \tau \le \lceil {\displaystyle \frac{q}{2}}\rceil -1$. If $2q-\tau \ge (z+i+1)$, then

$$(2q-\tau )\omega +(2q-\tau -1-z-i)\in \mathfrak{S}.$$

If $2q-\tau <(1+z+i)$, then

$$(2q-\tau )\omega +(2q-\tau -1-i-z)=(q-\tau -1)\omega +q\omega +(\omega +2q-\tau -z-i-1).$$

As $q\ge 2$, therefore $(q-\tau -1)\omega \in \mathfrak{S}$ and

$$q\omega +(\omega +2q-\tau -z-i-1)\in \{q\omega +(\omega -1),q\omega +(\omega -2),\dots ,q\omega +2q+2\}.$$

This implies $\tau (\omega +1)+2+z+i\in g\left(\mathfrak{S}\right)$. Consequently, ${\lambda }_1+{\lambda }_2\in g\left(\mathfrak{S}\right)$.

Case-V: If ${\lambda }_1\in A$ and ${\lambda }_2\in C$ then

$${\lambda }_1+{\lambda }_2=i+1+{\displaystyle \frac{q}{2}}(\omega +1)+t+1={\displaystyle \frac{q}{2}}(\omega +1)+t+i+2.$$

Here,

$$F-({\displaystyle \frac{q}{2}}(\omega +1)+t+i+2=\left({\displaystyle \frac{3q}{2}}\right)\omega +({\displaystyle \frac{3q}{2}}-1-t-i).$$

If ${\displaystyle \frac{3q}{2}}\ge (t+i+1)$, then $\left({\displaystyle \frac{3q}{2}}\right)\omega +({\displaystyle \frac{3q}{2}}-1-t-i)\in \mathfrak{S}$. If ${\displaystyle \frac{3q}{2}}<(1+t+i)$, then

$$\left({\displaystyle \frac{3q}{2}}\right)\omega +({\displaystyle \frac{3q}{2}}-1-i-t)=({\displaystyle \frac{q}{2}}-1)\omega +q\omega +(\omega +{\displaystyle \frac{3q}{2}}-t-i-1)\in \mathfrak{S}.$$

This implies ${\displaystyle \frac{q}{2}}(\omega +1)+2+t+i\in g\left(\mathfrak{S}\right)$. Consequently, ${\lambda }_1+{\lambda }_2\in g\left(\mathfrak{S}\right)$.

Case-VI: If ${\lambda }_1\in B$ and ${\lambda }_2\in C$ then

$${\lambda }_1+{\lambda }_2=\tau (\omega +1)+z+1+{\displaystyle \frac{q}{2}}(\omega +1)+t+1=({\displaystyle \frac{q}{2}}+\tau )(\omega +1)+t+z+2.$$

Note that

$$F-(({\displaystyle \frac{q}{2}}+\tau )(\omega +1)+t+z+2=({\displaystyle \frac{3q}{2}}-\tau )\omega +({\displaystyle \frac{3q}{2}}-\tau -1-t-z).$$

If ${\displaystyle \frac{3q}{2}}-\tau \ge (t+z+1)$, then $({\displaystyle \frac{3q}{2}}-\tau )\omega +({\displaystyle \frac{3q}{2}}-\tau -1-t-z)\in \mathfrak{S}$.

Since, $1\le \tau \le \lceil {\displaystyle \frac{q}{2}}\rceil -1<{\displaystyle \frac{3q}{2}}$, therefore ${\displaystyle \frac{3q}{2}}-\tau )\omega \in \mathfrak{S}$. If ${\displaystyle \frac{3q}{2}}-\tau <(1+t+z)$, then

$$({\displaystyle \frac{3q}{2}}-\tau )\omega +({\displaystyle \frac{3q}{2}}-\tau -1-z-t)=({\displaystyle \frac{q}{2}}-\tau -1)\omega +q\omega +(\omega +{\displaystyle \frac{3q}{2}}-t-z-\tau -1)\in \mathfrak{S}.$$

This implies $({\displaystyle \frac{q}{2}}+\tau )(\omega +1)+2+t+z\in g\left(\mathfrak{S}\right)$. Consequently, ${\lambda }_1+{\lambda }_2\in g\left(\mathfrak{S}\right)$. □

The following results give us a maximum independent set and the independent number of genus graph $G_{\mathfrak{S}}$.

Proposition 1. 

Let $G_{\mathfrak{S}}$ be a graph associated with symmetric cofinite submonoid of arbitrary multiplicity and embedding dimension.

1. 

If $q\ge 1$ (odd) then $A\cup B$ is a maximal independent set.

2. 

If $q\ge 2$ (even) then $A\cup B\cup C$ is a maximal independent set.

Proof. 

Let $F-(u\omega +v(\omega +1)+w{k}_i)\in g\left(\mathfrak{S}\right)\setminus A\cup B\cup C$, where $u,v,w\ge 0$ and $k_i\in \{q\omega +2q+2,\dots ,q\omega +(\omega -1)\}$.

Case 1: If $u=w=0$, and $v=2q+1-\lfloor {\displaystyle \frac{\omega }{2}}\rfloor -x^{\prime }$, where $1\le x^{\prime }\le 2q-\lfloor {\displaystyle \frac{\omega }{2}}\rfloor$, then for $F-((2q-\lfloor {\displaystyle \frac{\omega }{2}}\rfloor +1)(\omega +1)+(\lfloor {\displaystyle \frac{\omega }{2}}\rfloor -1)\omega )\in A$, we have

$$F-(2q+1-\lfloor {\displaystyle \frac{\omega }{2}}\rfloor -x^{\prime })(\omega +1))+F-((2q-\lfloor {\displaystyle \frac{\omega }{2}}\rfloor +1)(\omega +1)+(\lfloor {\displaystyle \frac{\omega }{2}}\rfloor -1)\omega )=\omega (\lfloor {\displaystyle \frac{\omega }{2}}\rfloor +x^{\prime })+(x^{\prime }-1)\in \mathfrak{S},$$

because $1\le x^{\prime }\le 2q-\lfloor {\displaystyle \frac{\omega }{2}}\rfloor$.

Case 2: If $v=w=0$, and $u=\lfloor {\displaystyle \frac{\omega }{2}}\rfloor +x^{\prime }$, where $0\le x^{\prime }\le 2q-\lfloor {\displaystyle \frac{\omega }{2}}\rfloor$ then $F\setminus ((2q-\lfloor {\displaystyle \frac{\omega }{2}}\rfloor +1+s)(\omega +1)+(\lfloor {\displaystyle \frac{\omega }{2}}\rfloor -1-s)\omega )\in A$, where $0\le s\le \lfloor {\displaystyle \frac{\omega }{2}}\rfloor -1$, we have

$$\begin{gathered} F-\left((\lfloor {\displaystyle \frac{\omega }{2}}\rfloor +x^{\prime })\omega \right)+F-((2q-\lfloor {\displaystyle \frac{\omega }{2}}\rfloor +1+s)(\omega +1)+(\lfloor {\displaystyle \frac{\omega }{2}}\rfloor -1-s)\omega )=\omega (2q+1-\lfloor {\displaystyle \frac{\omega }{2}}\rfloor -x^{\prime }) \\ +2q+1-\omega -s+\lfloor {\displaystyle \frac{\omega }{2}}\rfloor . \end{gathered}$$

Since $2q+1-\lfloor {\displaystyle \frac{\omega }{2}}\rfloor -x^{\prime }-(2q+1-\omega -s+\lfloor {\displaystyle \frac{\omega }{2}}\rfloor )=\omega +s-x^{\prime }>0$, also

$$2q+1+\lfloor {\displaystyle \frac{\omega }{2}}\rfloor -(\omega +s)>0,$$

therefore by Lemma 2, $\omega (2q+1-\lfloor {\displaystyle \frac{\omega }{2}}\rfloor -x^{\prime })+2q+1-\omega -s+\lfloor {\displaystyle \frac{\omega }{2}}\rfloor \in \mathfrak{S}$.

Case 3: If $u=v=0$, $w=1$, and $k_i=q\omega +\omega -\lfloor {\displaystyle \frac{\omega }{2}}\rfloor +2q+1-f$, $1\le f\le \omega -\lfloor {\displaystyle \frac{\omega }{2}}\rfloor -1$ then

$$F-(q\omega +\omega -\lfloor {\displaystyle \frac{\omega }{2}}\rfloor +2q+1-f)+F-((2q-\lceil {\displaystyle \frac{\omega }{2}}\rceil +2)(\omega +1)+(\lceil {\displaystyle \frac{\omega }{2}}\rceil -{\displaystyle \frac{q}{2}}-1)\omega =({\displaystyle \frac{3q}{2}}-1)\omega +f-1\in \mathfrak{S},$$

because $F-({\displaystyle \frac{3q}{2}}-1)\omega +f-1=\omega (q-{\displaystyle \frac{q-2}{2}}+1)+2q+2-f\in g\left(\mathfrak{S}\right)$.

Case 4: If $w=0,u=\lfloor {\displaystyle \frac{\omega }{2}}\rfloor -1+y,v=2q+1-\lfloor {\displaystyle \frac{\omega }{2}}\rfloor -t$, with $1\le y\le t$, $\lfloor {\displaystyle \frac{\omega }{2}}\rfloor <2q$, and $1\le t\le 2q-\lfloor {\displaystyle \frac{\omega }{2}}\rfloor$, then

$$\begin{gathered} F-((\lfloor {\displaystyle \frac{\omega }{2}}\rfloor -1+y)\omega +(2q-\lfloor {\displaystyle \frac{\omega }{2}}\rfloor +1-t)(\omega +1))+F-((2q-\lfloor {\displaystyle \frac{\omega }{2}}\rfloor +1+s)(\omega +1)+(\lfloor {\displaystyle \frac{\omega }{2}}\rfloor -1-s)\omega ) \\ =(2\lfloor {\displaystyle \frac{\omega }{2}}\rfloor -y+t)\omega +t-s\in \mathfrak{S}, \end{gathered}$$

because $t+1-y\ge t-s$. □

Theorem 1. 

Let $G_{\mathfrak{S}}$ be a graph associated with symmetric cofinite submonoid of arbitrary multiplicity and embedding dimension. Then the independent number of $G_{\mathfrak{S}}$ is:

1. 

If $q\ge 1$ (odd) then

$$\alpha \left(G_{\mathfrak{S}}\right)=\lfloor {\displaystyle \frac{\omega }{2}}\rfloor +\sum _{\tau =1}^{\lceil {\displaystyle \frac{q}{2}}\rceil -1}(\lceil {\displaystyle \frac{\omega }{2}}\rceil -\tau -1).$$

2. 

If $q\ge 2$ (even) then

$$\alpha \left(G_{\mathfrak{S}}\right)=\lfloor {\displaystyle \frac{\omega }{2}}\rfloor +\sum _{\tau =1}^{{\displaystyle \frac{q}{2}}-1}(\lceil {\displaystyle \frac{\omega }{2}}\rceil -\tau -1)+{\displaystyle \frac{q}{2}}+1.$$

Proof. 

To prove the independent number we have to show that $A,B$ and C are disjoint. For this we discuss following cases:

Case-I: Let $(2q-i)(\omega +1)+i\omega =(2q-\tau -z)(\omega +1)+z\omega$, then $i=\tau (\omega +1)+z>\omega$. This is a contradiction, since from the definition of set A, we have $i<\omega$. This implies $A\cap B=\varnothing$.

Case-II: Let $(2q-i)(\omega +1)+i\omega =(2q-{\displaystyle \frac{q}{2}}-t)(\omega +1)+t\omega$, then $i={\displaystyle \frac{q}{2}}(\omega +1)+t>\omega$. This is again a contradiction, since from the definition of set A, we have $i<\omega$. This gives $A\cap C=\varnothing$.

Case-III: Let $({\displaystyle \frac{3q}{2}}-t)(\omega +1)+t\omega =(2q-\tau -z)(\omega +1)+z\omega$, then $z={\displaystyle \frac{q}{2}}(\omega +1)+t>\omega$. This is a contradiction, since from the definition of set B, we have $z<\omega$. This implies $B\cap C=\varnothing$.

So, for $q\ge 1$ (odd), we have

$$\alpha \left(G_{\mathfrak{S}}\right)=\left|A\right|+\left|B\right|=\lfloor {\displaystyle \frac{\omega }{2}}\rfloor +\sum _{\tau =1}^{{\displaystyle \frac{q}{2}}-1}(\lceil {\displaystyle \frac{\omega }{2}}\rceil -\tau -1)$$

and for $q\ge 2$ (even)

$$\begin{gathered} \alpha \left(G_{\mathfrak{S}}\right)=\left|A\right|+\left|B\right|+\left|C\right|. \\ =\lfloor {\displaystyle \frac{\omega }{2}}\rfloor +\sum _{\tau =1}^{{\displaystyle \frac{q}{2}}-1}(\lceil {\displaystyle \frac{\omega }{2}}\rceil -\tau -1)+{\displaystyle \frac{q}{2}}+1. \end{gathered}$$

 □

3. Algorithmic Framework and Implementation

In this section, we present the Python-based implementation of a greedy algorithm used to compute the independent number of genus graph associated with cofinite submonoid $\mathfrak{S}$ (see Algorithm 1).

Algorithm 1 Computation of the Genus Graph and Its Independent Number

Input:  $T=\{t_1,t_2,\dots ,t_k\}$, is the minimal system of generator of cofinite submonoid $\mathfrak{S}$ Output:  Genus Graph $G_{\mathfrak{S}}$ and the independent number of $G_{\mathfrak{S}}$   1: $S\leftarrow \varnothing$, frontier $\leftarrow \left[0\right]$, max_limit $\leftarrow 200$   2: while frontier not empty do   3:     current ← frontier.pop(0)   4:     if current > max_limit then continue   5:     end if   6:     if current $\notin S$ then   7:         $S\leftarrow S\cup \left\{current\right\}$   8:         for all $t\in \mathrm{generators}$ do   9:            frontier.append(current $+t$) 10:         end for 11:     end if 12: end while 13: $gaps\leftarrow \{x\mid x<max(S),x\notin S\}$ 14: if $\left|gaps\right|\ge 2$ then 15:     $max\_check\leftarrow$ sum of two largest gaps 16: else 17:     $max\_check\leftarrow max\left(\mathrm{generators}\right)\times 3$ 18: end if 19: Reset $S\leftarrow \varnothing$, frontier $\leftarrow \left[0\right]$ 20: while frontier not empty do 21:     current ← frontier.pop(0) 22:     if current > max_check then continue 23:     end if 24:     if current $\notin S$ then 25:         $S\leftarrow S\cup \left\{current\right\}$ 26:         for all $t\in \mathrm{generators}$ do 27:            frontier.append(current $+t$) 28:         end for 29:     end if 30: end while 31: $gaps\leftarrow \{x\mid x<max\_check,x\notin S\}$ 32: Create graph G with nodes = gaps 33: $edge\_list\leftarrow \varnothing$ 34: for all distinct pairs $(i,j)$ from gaps do 35:     if $i+j\in S$ then 36:         Add edge $(i,j)$ to G and $edge\_list$ 37:     end if 38: end for 39: return G, gaps, edge_list 40: Use greedy strategy to build independent set 41: return independent set and its size 42: while True do 43:     Prompt user: “Enter generators (comma-separated) or ‘q’ to quit:” 44:     if user enters ‘q’ then break 45:     end if 46:     Parse input ⇒ generators 47:     $(G,gaps,edges)\leftarrow$ numerical_semigroup_graph(generators) 48:     Print number of nodes, edges, and list of gaps 49:     if edges not empty then 50:         Print all edge pairs where $i+j\in S$ 51:         Call find_maximum_independent_set(G) and print result 52:     end if 53:     Visualize G using circular layout (optional) 54: end while

To compute an independent set for the $G_{\mathfrak{S}}$, we employ a greedy approximation algorithm. The algorithm proceeds as follows: Initialization: Create an empty independent set I. Select a vertex v in G with minimum degree, then add to independent set I. Remove the vertex v and all its neighbors from G and repeat this process. The greedy algorithm employs a minimum-degree vertex ordering rule. In each iteratation, it selects the vertex with the smallest number of connections in the graph. This heuristic aims to minimize the number of neighboring vertices removed in the subsequent step, thereby preserving more options for expanding the independent set. In case of ties, the algorithm selects the vertex with smallest numerical value. The algoritthm iterates until the vertex set of the graph is empty.

The space and time complexity of the purposed algorithm is quadratic $O\left(k^2\right)$, where k is the genus. Python has emerged as a versatile computational platform, distinguished by its efficiency, extensibility, and extensive library support, which collectively enable the seamless implementation of complex mathematical algorithms. Its adaptability and scalability render it an indispensable tool for the rigorous verification of theoretical results and for the exploration of large-scale computational structures arising in algebraic and combinatorial research.

To further enhance accessibility, we provide an online interface through which users can directly validate their results, thereby eliminating the necessity of installing Python or auxiliary software environments.

To support the theoretical results, we provide a Python implementation that constructs the graph associated with a cofinite submonoid and computes its independent number. Interested readers can access via the following link: https://independent-number.onrender.com/ (accessed on 27 July 2025).

Example 1. 

Let $\mathrm{S}=\langle 15,16,70,71,72,73,74\rangle$, then we can use the following Python code (Listing 1) to compute the genus graph $G_{\mathfrak{S}}$ and its independent number $\alpha \left(G_{\mathfrak{S}}\right)$

By Lemma 1 and Theorem 1, we have $A=\{1,2,3,4,5,6,7\},B=\{17,18,19,20,21,22\}$, and $C=\{33,34,35\}$. So, $\alpha \left(G_{\mathfrak{S}}\right)=16$ (see Figure 1).

Listing 1. Python Console Output.

Example 2. 

Let $G_{\mathfrak{S}}$ be a graph associated with $\mathfrak{S}=\langle 13,14,47,48,49,50,51\rangle$, then we can use the following Python code (Listing 2) to compute the genus graph $G_{\mathfrak{S}}$ and its independent number $\alpha \left(G_{\mathfrak{S}}\right)$.

Listing 2. Python Console Output.

By Lemma 1 and Theorem 1, we have $A=\{1,2,3,4,5,6,\}$, and $B=\{15,16,17,18,19\}$. So, $\alpha \left(G_{\mathfrak{S}}\right)=11$ (see Figure 2).

Example 3. 

Let $q=5,m=13$, then $\mathfrak{S}=\langle 13,14,77\rangle$. From Lemma 1, we have

$$A=\{1,2,3,4,5,6\},$$

$$B=\{15,16,17,18,19,29,30,31,32\}.$$

For $q\ge 1$ an odd integer, Theorem 1 gives maximum independent set for $G_{\mathfrak{S}}$ is $A\cup B$. This implies $\alpha \left(G_{\mathfrak{S}}\right)=15$, which is not correct for the numerical semigroup $\mathfrak{S}=\langle 13,14,77\rangle$ shown as follows (Listing 3).

Listing 3. Python Console Output.

4. Application and Experiments

The application of an independent set: Cancer develops when multiple genes become misregulated, causing disruptions in the pathways that normally regulate healthy cell growth and leading to the uncontrolled accumulation of abnormal cells, while some abnormal gene expressions can be naturally compensated by other genes, many remain uncompensated and act as persistent drivers of tumor progression. In the framework of numerical semigroups, these abnormal gene states are represented by the gaps, and the uncompensated ones are captured by the independent set. This connection allows us to identify precisely which gene expressions must be targeted in therapy. Tumors with larger independent sets are biologically more aggressive, since they contain multiple uncompensated drivers that require combination therapy to suppress. By focusing on the members of the independent set, treatment strategies can be designed to block all critical pathways simultaneously, thereby reducing the risk of cancer survival and drug resistance. Moreover, the independent number itself serves as a measure of tumor complexity and can act as a prognostic marker, guiding both therapeutic design and clinical decision making. Thus, the concept of the independent set in numerical semigroup theory offers a powerful mathematical lens to model tumor robustness and optimize cancer treatment (see

Table 1. Comparison of theoretical and exact independence numbers for randomly sampled semigroups.

Sample	(m,q)	$\mathit{\alpha }\left({\mathit{G}}_{\mathfrak{S}}\right)\left(\mathit{formula}\right)$	$\mathit{\alpha }\left({\mathit{G}}_{\mathfrak{S}}\right)\left(\mathit{exact}\right)$	Match	Time (s)	Genus
1	(9,1)	4	4	Yes	0.10	11
2	(10,1)	5	5	Yes	0.2	12
3	(11,1)	5	5	Yes	0.3	13
4	(12,1)	6	6	Yes	0.5	14
5	(11,2)	7	7	Yes	2.5	25

).

5. Conclusions

In this research work, we find a maximum independent set and the independent number of graphs associated with symmetric cofinite submonoid of arbitrary multiplicity and embedding dimension. This work opens up new avenues for future study, such as generalizing our methods to non-symmetric cofinite submonoid, exploring other graph invariants, and applying these results to more extensive context in combinatorics and algebra. The problem of figuring out the independent number, in general, remains unsolved. Also, we give a Python code to find the independent number of the genus graph $G_{\mathfrak{S}}$.