**Proposition 2.**


**Theorem 2.** *The only relation σ which results in a quasihypergroup or semihypergroup is the one with M<sup>σ</sup>* = *S*. *Additionally, L<sup>σ</sup> is the total hypergroup.*

It was revealed that with a few lines of the Mathematica program, the results were constructed for the enumeration of the hypergroupoid associated with binary relations of orders 2, 3, 4, and 5 by a significantly simpler procedure [31].

**Definition 3.** *A graph G is a pair G* = (*V*, *E*), *where V is a set of elements described as vertices and E is a set of edges [32]. The two vertices associated with an edge are called endpoints. If x=y, then the edge is considered as a loop. A vertex is isolated if it is incident with no edges. The graph G is simple if it has no loops and no two distinct edges have the same pair of ends. The graph G is called null graph when its edges set is empty. Graph H is named a subgraph of graph G if V*(*H*) ⊆ *V*(*G*), *E*(*H*) ⊆ *E*(*G*), *and the ends of an edge e* ∈ *E*(*H*) *are the same as its ends in G*. *Denote d*(*x*) *as the degree of vertex x as well as the number of edges incident with x*.

*A path in graph G consists of a sequence x*1,*e*1, *x*2,*e*2, ... ,*ek*, *xk that the edges ei are distinct. Furthermore, if x*<sup>1</sup> = *xk then, we call the path a cycle. Consider that d*(*x*, *y*) *is the length of the shortest path between two vertices x and y*. *Note that diam*(*G*) = *sup*{*d*(*a*, *b*)} *for all a and b that are vertices of G*, *which is called the diameter of graph G*. *The graph G is connected if there exists a path from vertex x to vertex y*, *or graph G includes several connected components. A tree is a connected graph that includes no simple cyclic path. Denote kn as a complete graph, where every pair of vertices is adjacent. An Eulerian circuit is a closed path through a graph applying each edge once and an Eulerian graph is a graph that has this property. Furthermore, graph G is called a Hamiltonian graph if it has a cycle that passes each vertex exactly once. If every vertex has the same degree, the graph is regular, or k-regular if* ∀*x* ∈ *V*, *d*(*x*) = *k*.

**Theorem 3.** *A finite graph G without isolated vertices is Eulerian if and only if G is connected and each vertex has an even degree [32].*

**Definition 4.** *The Cartesian product of two graphs G*<sup>1</sup> = &*V*1, *E*1' *and G*<sup>2</sup> = &*V*2, *E*2' *is denoted by G*1*G*2, *that is a graph with vertices set V*<sup>1</sup> × *V*2, *where vertices* (*t*1, *t*2),(*w*1, *w*2) *are adjacent if and only if t*<sup>1</sup> = *w*1,(*t*2, *w*2) ∈ *E*<sup>2</sup> *or t*<sup>2</sup> = *w*2,(*t*1, *w*1) ∈ *E*<sup>1</sup> *for t*1, *w*<sup>1</sup> ∈ *V*1, *t*2, *w*<sup>2</sup> ∈ *V*<sup>2</sup> *[33].*

## **3. Semihypergroup-Based Graph (SBG) Based on Relation** *γ*

Consider an SBG of *G* = &*H*; *E* = (*γn*)*n*∈N', where (*H*, ◦) is a semihypergroup and *γ<sup>n</sup>* is the relation on *H*. The order of *G* is *o*(*G*) =|*H*| . The elements of *H* are represented as vertices and the relations *γ<sup>n</sup>* are appointed as edges. We assign *x* and *y* to be adjacent, if *xγny*. Clearly, for *n* = 1 and *xγ*1*x*, the edge is a loop.

Indeed, *γ<sup>n</sup>* was determined in [10] as follows:

$$\forall x \gamma\_n y \Longleftrightarrow \exists (a\_1, \dots, a\_n) \in H^n, \exists \sigma \in S\_n : \ x \in \prod\_{i=1}^n a\_{i\prime}, \ y \in \prod\_{i=1}^n a\_{\sigma(i)} \tag{2}$$

Consider *γ*<sup>1</sup> = {(*a*, *a*) |*a* ∈ *H*}. Clearly, the relations *γ<sup>n</sup>* have symmetric property and relation *γ* has a reflexive and symmetric property for every *n* ∈ N, where *γ* = *<sup>n</sup>*≥<sup>1</sup> *γn*. Let *γ*<sup>∗</sup> be the transitive closure of *γ*. The class of *H*/*γ*<sup>∗</sup> was addressed as *γ*∗(*z*) = {*w* |*zγ*∗*w*}, for *z*, *w* ∈ *H*. It was proven that for hypergroup *H*, the relation *γ* is transitive and *γ*<sup>∗</sup> has the smallest strongly regular equivalence property that results *H*/*γ*<sup>∗</sup> is an Abelian group (fundamental group).

**Theorem 4.** *Assume that H is a hypergroup. Then, for an SBG of G* = &*H*; *E* = (*γn*)*n*∈N', *the following statements hold:*


**Proof.** Proof of (i): Consider a path from vertex *x* to vertex *y*. Then, there exists a sequence (*a*1, ... , *ak*) ∈ *H<sup>k</sup>* so that *x* = *a*1*γ*1*a*<sup>2</sup> ... *γ<sup>k</sup> ak* = *y*, that is equal to *xγ*∗*y*. Conversely, if *xγ*∗*y*, then ∃(*a*1, ... , *ak*) ∈ *H<sup>k</sup>* such that *x* = *a*1*γ*1*a*<sup>2</sup> ... *γ<sup>k</sup> ak* = *y*. Therefore, there exists a path from vertex *x* to vertex *y*.

Proof of (ii): By applying (i), for *x*, *y* ∈ *H*, a path exists from vertex *x* to vertex *y* if and only if *xγ*∗*y*. Therefore, the SBG of *G* is a connected graph if and only if *γ*<sup>∗</sup> = *H* × *H* (i.e., clearly, *γ*<sup>∗</sup> ⊆ *H* × *H*. Furthermore, for all *x*, *y* ∈ *H*, since (*x*, *y*) ∈ *γ*∗, then *H* × *H* ⊆ *γ*∗). Since *xγ*∗*y*, we have *γ*∗(*x*) = *γ*∗(*y*) which means that the fundamental group *H*/*γ*<sup>∗</sup> = {*γ*∗(*x*)|*x* ∈ *H*} is a singleton, i.e., |*H*/*γ*∗| = 1.

**Theorem 5.** *The connected components SBG of G are precisely the elements of the fundamental group H*/*γ*∗.

**Proof.** Let *x*, *y* be two vertices SBG of *G*. By employing Theorem 4, vertex *x* is connected to vertex *y* if and only if *xγ*∗*y*. Then, for all *a* ∈ *H*, every element of *γ*∗(*a*) is connected. With the equivalence relation of *γ*∗, the elements of *H*/*γ*<sup>∗</sup> would be the connected components SBG of *G*.

**Theorem 6.** *Let H be a semihypergroup. If the SBG of G* = &*H*, *E*' *is complete, then the relation γ is transitive.*

**Proof.** Let *xγy* and *yγz*. For some *n*1, *n*<sup>2</sup> ∈ N, we have *xγn*<sup>1</sup> *y* and *yγn*<sup>2</sup> *z*. Since the SBG of *G* is complete, therefore, for some *n* ∈ N, we have *xγnz* that yields *xγz*.

**Remark 1.** *Note that a loop is not considered an edge. If xγx*, *then for every ai* ∈ *H*, ∃*σ* ∈ *Sn we have x* ∈ ∏*<sup>n</sup> <sup>i</sup>*=<sup>1</sup> *ai and x* ∈ <sup>∏</sup>*<sup>n</sup> <sup>i</sup>*=<sup>1</sup> *<sup>a</sup>σ*(*i*). *Hence,* <sup>∏</sup>*<sup>n</sup> <sup>i</sup>*=<sup>1</sup> *ai* = <sup>∏</sup>*<sup>n</sup> <sup>i</sup>*=<sup>1</sup> *aσ*(*i*).

**Definition 5.** *Let H be a nonvoid set and let γ*<sup>∗</sup> *be the defined relation in Equation* (2)*. Consider the hypercomposition "" on H as follows:*

$$\mathbf{x} \odot \mathbf{y} = \{ w \in H : (\mathbf{x}, w) \in \gamma^\*, \ (w, y) \in \gamma^\* \}\tag{3}$$

*We denote the hypercompositional structure* (*H*, ) *by Hγ*<sup>∗</sup> . *The Hγ*<sup>∗</sup> *is a hypergroupoid if* ∃*w* ∈ *H so that* (*x*, *w*) ∈ *γ*<sup>∗</sup> *and* (*w*, *y*) ∈ *γ*<sup>∗</sup> *for every x*, *y* ∈ *H*. *Since γ*<sup>∗</sup> *is transitive, we have* (*x*, *y*) ∈ *γ*<sup>∗</sup> *for all x*, *y* ∈ *Hγ*<sup>∗</sup> , *then the reproductivity property holds. In fact, for the arbitrary element x* ∈ *Hγ*<sup>∗</sup> , *the reproductivity axiom y* ∈ *x Hγ*<sup>∗</sup> *holds for all y* ∈ *Hγ*<sup>∗</sup> , *as per the transitive property of γ*∗.
