# Spectrum of Superhypergraphs via Flows

Authors: Mohammad Hamidi; Florentin Smarandache; Elham Davneshvar
Journal: Journal of Mathematics
 (2022)

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## Abstract

For anyn∈ℕ and given nonempty subset V, the concept of n-superhypergraphs is introduced by Florentin Smarandache based on PnV (n-th power set of V). In this paper, we present the novel concepts supervertices, superedges, and superhypergraph via the concept of flow. This study computes the number of superedges of any given superhypergraphs, and based on the numbers of superedges and partitions of an underlying set of superhypergraph, we obtain the number of all superhypergraphs on any nonempty set. As a main result of the research, this paper is introducing the incidence matrix of superhypergraph and computing the characteristic polynomial for the incidence matrix of superhypergraph, so we obtain the spectrum of superhypergraphs. The flow of superedges plays the main role in computing of spectrum of superhypergraphs, so we compute the spectrum of superhypergraphs in some types such as regular flow, regular reversed flow, and regular two-sided flow. The new conception of superhypergraph and computation of the spectrum of superhypergraphs are introduced firstly in this paper.


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## Body

## 1. Introduction and Preliminaries

The theory of graph is a main and important theory for modeling the real problem in the world, and this theory extends in past years in this regard. The disadvantage of a graph is that it cannot connect more than two elements, so this problem causes weakness in this theory. Berge generalized the theory of graphs to the mathematical concept of theory of hypergraphs with the motivation that hypergraphs solve the conflicts, defects, and shortcomings of graph theory around 1960 [1]. Hypergraphs have some applications in other sciences and the real-world, one of the applications of hypergraphs is a simulation for complex hypernetworks. Today, hypergraphs have a vital role and important performances, so are used in complex hypernetworks such as computer science, wireless sensor hypernetwork, and social hypernetworks. In this regards there has been a lot of research about using hypergraphs to problems in real-world such as hypergraph matching via game-theoretic hypergraph clustering [2], hypergraph matching via game-theoretic hypergraph clustering [3], hypergraph-based centrality metrics for maritime container service networks, a worldwide application [4], clustering ensemble via structured hypergraph learning [5], and hypergraph neural network for skeleton-based action recognition [6]. There is some main connection between graphs and hypergraphs via the mathematical computational tools and basic theorems in which these connections facilitate the modeling of other sciences with mathematics. Further materials regarding graphs and hypergraphs are available such as extending factorizations of complete uniform hypergraphs [7], finding perfect matchings in bipartite hypergraphs [8], graphs and hypergraphs [1], resilient hypergraphs with fixed matching number [9], on the spectrum of hypergraphs [10], on the distance spectrum of minimal cages and associated distance biregular graphs [11], on the spectrum of the perfect matching derangement graph [12], and probabilistic refinement of the asymptotic spectrum of graphs [13]. Recently, Hamidi and Saeid computed eigenvalues of discrete complete hypergraphs and partitioned hypergraphs. They defined positive equivalence relation on hypergraphs that establishes a connection between hypergraphs and graphs, and it makes a connection between a spectrum of graphs and a spectrum of the quotient of any hypergraphs. They studied the construct spectrum of path trees via the quotient of partitioned hypergraphs [14]. A hypergraph on any given set considers a relationship between elements and the set as objects or hyper vertices and describes this relationship if it is a weighted hypergraph. It is an ideal condition if proper weights are known, but in most situations, the weights may not be known, and the relationships are hesitant in a natural sense. With the advent of the fuzzy graph, the importance of this theory increased and fuzzy graph as a generalization of a graph provides more information in real-life problems. Based on Zadeh’s fuzzy relations [15], the notion of hypergraph has been extended in the fuzzy theory and the concept of fuzzy hypergraph was provided by Kaufmann [16]. Recently, some researchers investigated the concept of fuzzy hypergraphs and applications such as fuzzy hypergraphs and related extensions [17], an algorithm to compute the strength of competing interactions in the Bering sea based on Pythagorean fuzzy hypergraphs [18] and bipolar fuzzy soft information applied to hypergraphs [19]. Recently, Smarandache introduced a new the concept as a generalization of hypergraphs to n-superhypergraph, plithogenic n-superhypergraph {with supervertices (that are groups of vertices) and hyperedges {defined on power set of power set…} that is the most general form of a graph as today}, which have several properties and are connected with the real-world [20]. Indeed, n-superhypergraphs are a generalization of hypergraphs, with the advantage that they can communicate between the hyperedges.Regarding these points, we consider a nonempty set and make a partition of the given set into some subsets, and relate this subset together with some maps. Indeed, subsets will call supervertices, the mapping between them will call superedges or flows and the system with supervertices and superedges will call a quasi superhypergraph. The main motivation of this the concept is a generalization of graphs to hypernetworks such that all elements be related together. In hypergraph theory, any hypergraph can relate a set of elements, while without any details that it makes some conflicts, defects, and shortcomings in the hypergraph theory. Thus, by introducing superhypergraph, we try to eliminate defects of graph (sometimes graph structures give very limited information about complex networks) structures and hypergraph structures (although the hypergraph structures are for covering graph defects in the applications but in hypergraphs, the relation between of vertices cannot be described in full details). As a main result of the research, this paper is introducing the incidence matrix of superhypergraph and computing the characteristic polynomial for the incidence matrix of superhypergraph, so we obtain the spectrum of superhypergraphs. Indeed, we computed the number of superedges of any given superhypergraphs and based on superedges and partitions of an underlying set of superhypergraph, we obtained the number of all superhypergraphs on any nonempty set. The flow of superedges plays the main role in computing of spectrum of superhypergraphs, so we computed the spectrum of superhypergraphs in some types regular flow, regular reversed flow, and regular two-sided flow.Definition 1.
[1] Let X be a finite set. A hypergraph on X is a pair H=X,Eii=1m such that for all 1≤i≤m,∅≠Ei⊆X and ∪i=1mEi=X. The elements x1,x2,…,xn of X are called vertices, and the sets E1,E2,…,Em are called the hyperedges of the hypergraph H. In hypergraphs, hyperedges can contain an element loop two elements edge or more than three elements. A hypergraph H=X,Eii=1m is called a complete hypergraph, if for any x,y∈X there is 1≤i≤m such that x,y⊆Ei. A hypergraph H=X,Eii=1n is called as a joint complete hypergraph, if X=n for all 1≤i≤n,Ei=i and Ei⊆Ei+1 element loop. If for all 1≤k≤m, Ek=2, the hypergraph becomes an ordinary undirected graph and n rows representing the vertices x1,x2,…,xn, where for all 1≤i≤n and for all 1≤j≤m, we have mij=1 if xi∈Ej and mij=0 if xi∉Ej.Definition 2.
[20] Let m∈ℕ and V=v1,v2,…,vm be a set of vertices, that contains single vertices the classic alones, indeterminate vertices unclear,vague,unknown and null vertices unknown,empty. Consider PV as the power set of V, P2V=PPV…, and PnV=PPn−1V be the n-power set of the set V. Then, the n-superhypergraph n−SHG is an ordered pair n-SHG=Gn,En, where for any n∈ℕ, Gn⊆PnV is the set of vertices and En⊆PnV is the set of edges. The set Gn contains some type of vertices, such as single vertices the classical ones, indeterminate vertices unclear,vague,partially unknown, null vertices totallyunknown,empty, and supervertices or subset vertex, i.e., two or more single,indeterminate,or null vertices together as a group organization. An n-supervertex is a collection of many vertices such that at least one is a n−1-supervertex and all other supervertices in to the collection if any have the order r≤n−1. The set of edges En contains some type of edges such as single edges the classic alones, indeterminate unclear,vague,partiallyunknown, null-edge empty,totally unknown, hyperedge containig three or moresingle vertices, superedge containing two vertices atleastone of them beinga super vertex, n-superedge containing twovertices,atleast one being ann−supervertex andthe otherof orderr−super vertex withr≤n, superhyperedge (containing three or more vertices, at least one being a supervertex, n-superhyperedge containing threeor more vertices,atleast onebeing ann−super vertexand the otherr−super vertices withr≤n, multiedges two ormore edges connecting the same two vertices, and loop an edge that connects an element.


## 2. On (Quasi) Superhypergraph

In this section, we introduce the concepts of supervertex, superedge, superhypergraph, and investigate their properties. For any given superhypergraph, the lower and upper bound of the set of their superedges is computed and proved in a theorem. Also, we computed and proved the number of all superhypergraphs constructed on any given nonempty set. In the following, for any nonempty setX, will denote P∗X=Y|∅≠Y⊆X.In what follows, based on the concept ofn-superhypergraph [20], recall, define, and investigate a special case in n-superhypergraphs as the notation of quasi superhypergraph.Definition 3.
LetX be a nonempty set. Then,(i)
H=X,Sii=1k,φi,ji,j is called a quasi superhypergraph, if φi,ji,j≠∅ and X=∪i=1nSi, where k≥2,(ii)
for all1≤i≤k,Si∈P∗X, is called a supervertex and for any i≠j, the map φi,j:Si⟶Sj (say Si links to Sj) is called a superedge,(iii)
the quasi superhypergraphH=X,Sii=1k,φi,ji,j is called a superhypergraph, if for any Si∈P∗X, there exists at least one Sj∈P∗X such that Si links to Sjit is not necessary all super vertices be linked.(iv)
The superhypergraphH=X,Sii=1k,φi,ji,j is called a trivial superhypergraph, if k=1S1can't link to itself.Example 1.
LetX=xii=19. Then, H=X,Sii=14,φ1,2,φ2,1,φ2,3,φ2,4 is a quasi superhypergraph there is no any link between ofS4andS3 in Figure 1, where(1)φ1,2=x1,x4,x2,x5,x3,x4,φ2,1=x4,x3,x5,x2,φ2,3=x4,x6,x5,x7,φ2,4=x5,x9,x4,x9.
LetH=X,Sii=1k,φi,ji,j be a superhypergraph. We will denote ΦH=φi,j|i,j≥1 by the set of all superedges of superhypergraph H. In what follows, compute and prove the lower bound and upper bound of ΦH, as set of all superedges of superhypergraph.Figure 1
SuperhypergraphH=X,Sii=14,φ1,2,φ2,1,φ2,3,φ2,4.Theorem 1.
LetH=X,Sii=1k,φi,ji,j be a superhypergraph. Then,(2)k−1≤ΦH≤∑1≤i≠j≤nSiSj.Proof.
LetH=X,Sii=1k,φi,ji,j be a superhypergraph. Since, by definition, for any Si∈P∗X, there exists at least one Sj∈P∗X such that Si links to Sj, we have k−1 superedges. In addition, let ΦSi,Sj=φi,j:Si⟶Sj|i,j. For all 1≤i≠j≤n, in one case, if Si∩Sj=∅, then, ΦSi,Sj=SiSj+SjSi. Hence, if ΦS1,S2,…,Sk−1=∑1≤i≠j≤k−1SiSj, then, for all 1≤i≠j≤n,Si∩Sj=∅ implies that(3)ΦS1,S2,…,Sk−1,Sk=ΦS1,S2,…,Sk−1,Sk=ΦS1,S2,…,Sk−1+∑1≤i≤k−1SiSk+SkSi=∑1≤i≠j≤nSiSj.
In another case, if there exists some1≤i≠j≤n such that Si∩Sj≠∅, since for all 1≤i≠j≤n, Si∩Sj⊆Si, for all 1≤t≤n, we get that ΦSi∩Sj,St⊆ΦSi,St and ΦSi∩Sj,St⊆ΦSj,St. So, in any cases, Φ≤∑1≤i≠j≤nSiSj.Example 2.
LetX=xii=14 and H=X,Sii=13,φi,ji,j be a superhypergraph, where S1=2,S2=1,S3=1. Then, by Theorem 1, 2≤ΦH≤8=21+21+12+11+12+11, where(4)φ1,2=x1,x3,x2,x3,φ1,3=x1,x4,x2,x4,φ2,1=x3,x1,φ2,1′=x3,x2,φ3,1=x4,x1,φ3,1′=x4,x2,φ2,3=x3,x4,φ3,2=x4,x3.
The superhypergraphH=X,Sii=13,φ1,23,2 with minimum superedges is shown in Figure 2.
LetX be a nonempty set, SH=H|His a superhypergraph onX and SHn1,n2,…,nk=X,Sii=1k,φi,ji,j∈SH|Si=niand for alli≠j,Si∩Sj=∅. In what follows, compute and prove the number of SHn1,n2,…,nk, as the set of all superhypergraphs based on any nonempty set X, where X=n.Figure 2
SuperhypergraphH=X,Sii=13,φ1,2,φ3,2 with ΦH=2.Theorem 2.
LetX be a nonempty set, r,n∈ℕ and X=n.(i)
SHn=1.(ii)
If∑i=1rni=n and m=i|ni=nj, then, SHn1,n2,…,nr=1/m!∏i=1rn−∑j=1i−1njni∑1≤i≠j≤rninj.Proof.
(i)
By definitions is clear.(ii)
LetX=n. Since SHn1,n2,…,nk=X,Sii=1k,φi,ji,j∈SH|Si=niand for alli≠j,Si∩Sj=∅, we get that S1,S2,…,Sk∈PX is a partition of set X, where Si=ni. Consider S1, then, the numbers of selected vertices in S1 is equal to nn1. Because for any 1≤i≤k,Si∩Sj=∅, the number of selected vertices in S2 is equal to n−n1n2. It follows that the numbers of ways to chosen the selected vertices between S1,S2 is equal to nn1n−n1n2 for the case n1≠n2 and is equal to 1/2!nn1n−n1n2, for the case n1=n2. Thus, in the process of doing so and by induction, for all 1≤i≤k, PX=1/m!∏i=1kn−∑j=1i−1njni, where m=i|ni=nj. In addition, by Theorem 1, we have ΦH≤∑1≤i≠j≤nSiSj, so SHn1,n2,…,nr=1/m!∏i=1rn−∑j=1i−1njni∑1≤i≠j≤rninj.Theorem 3.
LetX be a nonempty set and X=n. If α=n1,n2,…,nr|∑i=1rni=n,ni,r∈ℕ and m=i|ni=nj, then,(5)SHX=∑α1m!∏i=1rn−∑j=1i−1njni∑1≤i≠j≤rninj.Proof.
It is clear by Theorem2.Example 3.
LetX be an arbitrary set and X=4. Then,(6)SHX=SH4+SH3,1+SH2,2+SH2,1,1+SH1,1,1,1=1+431131+13+12!422222+22+12!42211121+21+12+11+12+11+14!4131211111+11+114=101.


## 3. Incidence Matrix of Superhypergraphs

In this section, we introduce a square matrix as incidence matrix associate with any given superhypergraph with sign function. Indeed, in the incidence matrix associate to any given superhypergraph the domain and range of any map determine the sign function.LetH=X,Sii=1k,φi,ji,j be a superhypergraph and ΨH⊆ΦH. Then, we have the following concepts.Definition 4.
LetH=X,Sii=1k,φi,ji,j be a superhypergraph and X=n. Define AS1,S2,…,Sk=aijn×k+ΨH as incidence matrix of H with k+ΨH columns representing the supervertices S1,S2,…,Sk, superedges φi,j and n rows representing the vertices x1,x2,…,xn, where n=k+ΨH and(7)aij=1,ifxi∈Siorxi∈Domainφi,j,−1,ifxi∈Rangeφi,j,0,otherwise..Example 4.
LetX=xii=15. Then, H=X,Sii=13,φ1,3,φ2,3 is a superhypergraph in Figure 3, where φ1,3=x1,x5,x2,x5 and φ2,3=x3,x5,x4,x5.
Then,A2,2,1 is the incidence matrix of H.(8)Figure 3
SuperhypergraphH=X,Sii=13,φ1,3,φ2,3.


### 3.1. Characteristic Polynomial for the Incidence Matrix of Superhypergraph

In this subsection, we compute the characteristic polynomial of the incidence matrix of any given superhypergraph. LetH=X,Sii=1k,φi,ji,j be a superhypergraph and so investigate the spectrum of the superhypergraph.From now on, letPS1,S2,…,Skx be the characteristic polynomial of the incidence matrix AS1,S2,…,Sk corresponding to superhypergraph H and EAS1,S2,…,Sk=x|is an eigenvalue ofAS1,S2,…,Sk. In addition, for any Si,Sj, if ΦSi,Sj=φi,j|φi,j:Si⟶Sj,i,j≥1, will say Si flows to Sj and will denote by Si⇝Sj. In this case, will denote AS1,S2,…,Sk by AS1,S2,…,Sk⇝ and PS1,S2,…,Skx by PS1,S2,…,Sk⇝x.Theorem 4.
LetS1=1,S2=n, n≥2 and S1⇝S2. Then,(i)
P1,n⇝x=−xn−1x2+n−3x−n−2.(ii)
SpecA1,n⇝=012−nn−111.Proof.
(i)
Letn=2. It is easy to see that P1,2⇝x=−xx2−x. Suppose that k≥3 and P1,k−1⇝x=−xk−2x2+k−4x−k−3. Then, A1,k⇝=c1,c2,…,ck,ck+1, where c3=⋯=ck+1=1,−1,…,−1︸k−timest. It follows that(9)P1,k⇝x=detA1,k⇝−Ik+1×k+1X=detB1,k⇝,such thatB1,k⇝=bijk+1×k+1, where B1,k⇝=c1′,c2′,…,ck′,ck+1′ that for all i∈3,…,k+1,ci′=1,−1,…,−1︸i−2−times,−1−x,−1,…,−1︸k−i+1−timest. Now, consider D1,k⇝=dijk+1×k+1, where(10)dij=bij,j≠k+1,−ck′+ck+1′,j=k+1.Thus,D1,k⇝=c1′,c2′,…,ck−1′,ck′,ck+1″, where ck+1″=0,…,0︸k−1−times,x,−xt=x0,…,0︸k−1−times,1,−1t. Based on induction assumption and computations of determinant based on column ck+1″ in matrix D1,k⇝, have(11)Pk,1⇝x=detD1,k⇝−Ik+1×k+1X=x−P1,k−1⇝x+−xk−1+−xk−2=−xk−1x2+k−3x−k−2.(ii)
It is clear by item (i).Theorem 5.
LetS1=n,S2=1, n≥2 and S1⇝S2. Then, Pn,1⇝x=−xn−1x2−n−2x−2.Proof.
Letn=2. It is easy to see that P2,1⇝x=−xx2−2. Suppose that k≥3 and Pk−1,1⇝x=−xk−2x2−k−3x−2. Then, Ak,1⇝=c1,c2,…,ck,ck+1, where c3=⋯=ck+1=1,1,…,1︸k−times,−1t. It follows that(12)Pk,1⇝x=detAk,1⇝−Ik+1×k+1X=detBk,1⇝,such that Bk,1⇝=bijk+1×k+1, where Bk,1⇝=c1′,c2′,…,ck′,ck+1′ that for all i∈3,4,…,k,ci′=1,1,…,1︸i−1−times,1−x,1,1,…,1︸k−i−times,−1t and ck+1′=1,1,…,1︸k−times,−1−xt. Consider Dk,1⇝=dijk+1×k+1, where(13)dij=bij,j≠k,ck′−ck+1′,j=k.
Thus,Dk,1⇝=c1′,c2′,…,ck−1′,ck′′,ck+1′, where ck′′=0,…,0︸k−1−times,−x,xt=−x0,…,0︸k−1−times,1,−1t. Based on induction assumption and computations of determinant based on column ck′′ in matrix Dk,1⇝, have(14)Pk,1⇝x=detDk,1⇝−Ik+1×k+1X=−xPk−1,1⇝x+−xk.Theorem 6.
LetS1=n,S2=2, n≥2 and S1⇝S2. Then, Pn,2⇝x=−xnx2−n−3x−4.Proof.
Letn=2. It is easy to see that P2,1⇝x=x2x2+x−4. Suppose that k≥3 and Pk−1,1⇝x=−xk−1x2−k−4x−4.Then, Ak,2⇝=c1,c2,…,ck,ck+1,ck+2, where c3=⋯=ck+2=1,1,…,1︸k−times,−1,−1t. It follows that(15)Pk,2⇝x=detAk,2⇝−Ik+2×k+2X=detBk,2⇝,such that Bk,2⇝=bijk+2×k+2, where Bk,2⇝=c1′,c2′,…,ck′,ck+1′,ck+2′ that for all i∈3,4,…,kci′=1,1,…,1︸i−1−times,1−x,1,1,…,1︸k−i−times,−1,−1t,ck+1′=1,1,…,1︸k−times,−1−x,,−1t and ck+2′=1,1,…,1︸k−times,−1,−1−xt. In addition, Dk,2⇝=dijk+2×k+2, where(16)dij=bij,j≠k,ck′−ck+2′,j=k.
Thus,Dk,2⇝=c1′,c2′,…,ck′′,ck+1′,ck+2′, where ck′′=0,…,0︸k−1−times,−x,0,xt=−x0,…,0︸k−1−times,1,0,−1t. Based on induction assumption and computations of determinant based on column ck′′ in matrix Dk,2⇝ show that(17)Pk,2⇝x=detDk,2⇝−Ik+2×k+2X=−xPk−1,2⇝x+−xk+1.Theorem 7.
LetS1=n,S2=3, n≥2 and S1⇝S2. Then, Pn,3⇝x=−xn+1x2−n−4x−6.Proof.
It is similar to Theorem6.Corollary 1.
LetS1=n,S2=m, n≥2,m≥1 and S1⇝S2. Then,(i)
Pn,m⇝x=−xn+m−2x2−n−m−1x−2m.(ii)
SpecAn,m⇝=0α+α2+8m/2α−α2+8m/2n+m−211, where α=n−m−1.(iii)
∑x∈EAn,m⇝x=0 if and only if n=m+1.Example 5.
LetX=x1,x2,x3. Then, H=X,Sii=12,φ1,2 is a superhypergraph as shown in Figure 4 and incidence matrix of A2,1⇝ as follows.(18)
Thus, by Theorem5, we have P2,1⇝x=−xx2−2 and so SpecA2,1⇝=02−2111.Figure 4
SuperhypergraphH=X,Sii=12,φ1,2.


### 3.2. Characteristic Polynomial for the Incidence Matrix of Superhypergraph with Reverse Flows

In this subsection, we compute the characteristic polynomial and spectrum of the superhypergraph in the reverse flows to the previous section.For anySi,Sj, if ΦSj,Si=φi,j|φi,j:Sj⟶Si,i,j≥1, will say Sj flows to Si and will denote by Si←Sjreverse flows toSi⇝Sj and so AS1,S2,…,Sk by AS1,S2,…,Sk← and PS1,S2,…,Skx by PS1,S2,…,Sk←x.Theorem 8.
LetS1=1,S2=n, n≥2 and S1←S2. Then,(i)
P1,n←x=−xn−1x2−n+1x+n.(ii)
SpecA1,n←=01nn−111.Proof.
(i)
Letn=2. It is easy to see that P1,2←x=−xx2−3x+2. Suppose that k≥3 and P1,k−1←x=−xk−2x2−kx+k−1. Then, A1,k←=c1,c2,…,ck,ck+1, where c3=⋯=ck+1=−1,1,…,1︸k−timest. It follows that(19)P1,k←x=detA1,k←−Ik+1×k+1X=detB1,k←,such thatB1,k←=bijk+1×k+1, where B1,k←=c1′,c2′,…,ck′,ck+1′ that for all i∈3,4,…,k+1,ci′=−1,1,…,1︸i−2−times,1−x︸i−th,1,…,1︸k−i+1−timest. Now, consider D1,k←=dijk+1×k+1, where(20)dij=bij,j≠k+1,−ck′+ck+1′,j=k+1.Thus,D1,k←=c1′,c2′,…,ck−1′,ck′,ck+1″, where ck+1″=0,…,0︸k−1−times,x,−xt=x0,…,0︸k−1−times,1,−1t. Based on induction assumption and computations of determinant based on column ck+1′′ in matrix D1,k←, have(21)P1,k←x=detD1,k←−Ik+1×k+1X=x−P1,k−1←x−−xk−1−−xk−2=−xk−1x2−k+1x+k.(ii)
It is clear by item (i).Theorem 9.
LetS1=n,S2=1, n≥2 and S1←S2. Then, Pn,1←x=−xn−1x2+n−4x+2.Proof.
Letn=2. It easy is to see that P2,1←x=−xx2−2x+2. Suppose that k≥3 and Pk−1,1←x=−xk−2x2+k−5x+2.Then, Ak,1←=c1,c2,…,ck,ck+1, where c3=⋯=ck+1=−1,−1,…,−1︸ktimes,1t. It follows that(22)Pk,1←x=detAk,1←−Ik+1×k+1X=detBk,1←,such that Bk,1←=bijk+1×k+1, where Bk,1←=c1′,c2′,…,ck′,ck+1′ that for all i∈3,…,k,ci′=−1,…,−1︸i−1−times,−1−x︸ith,−1,…,−1︸k−i−times,1t and ck+1′=−1,−1,…,−1︸ktimes,1−xt. Moreover, consider Dk,1←=dijk+1×k+1, where(23)dij=bij,j≠k,ck′−ck+1′,j=k.
Thus,Dk,1←=c1′,c2′,…,ck−1′,ck′′,ck+1′, where ck′′=0,…,0︸k−1times,−x,xt=−x0,…,0︸k−1times,1,−1t. Based on induction assumption and computations of determinant based on column ck′′ in matrix Dk,1←, have(24)Pk,1←x=detDk,1←−Ik+1×k+1X=−xPk−1,1←x+−1k+1xk.Theorem 10.
LetS1=n,S2=2,n≥2 and S1←S2. Then, Pn,2←x=−xnx2+n−5x+4.Proof.
Letn=2. It is easy to see that P2,1←x=x2x2−3x+4. Suppose that k≥3 and Pk−1,1←x=−xk−1x2+k−6x+4.Then, Ak,2←=c1,c2,…,ck,ck+1,ck+2, where c3=⋯=ck+2=−1,−1,…,−1︸ktimes,1,1t. It follows that(25)Pk,2←x=detAk,2←−Ik+2×k+2X=detBk,2←,such that Bk,2←=bijk+2×k+2, where Bk,2←=c1′,c2′,…,ck′,ck+1′,ck+2′ that for all i∈3,…,k,ci′=−1,…,−1︸i−1−times,−1−x︸ith,−1,…,−1︸k−i−times,1,1t,ck+1′=−1,…,−1︸k−times,1−x,1t, and ck+2′=−1,…,−1︸k−times,1,1−xt. Consider Dk,2←=dijk+2×k+2 where(26)dij=bij,j≠k,ck′−ck+1′,j=k.
Thus,Dk,2←=c1′,c2′,…,ck′′,ck+1′,ck+2′, where ck′′=0,…,0︸k−1−times,−x,x,0t=−x0,…,0︸k−1−times,1,−1,0t. Based on induction assumption and computations of determinant based on column ck′′ in matrix Dk,2← show that(27)Pk,2←x=detDk,2←−Ik+2×k+2X=−xPk−1,2←x+−1kxk+1.Theorem 11.
LetS1=n,S2=3, n≥2 and S1←S2. Then, Pn,3x=−xn+1x2+n−6x+6.Proof.
It is similar to Theorem6.Corollary 2.
LetS1=n,S2=m, n≥2,m≥1 and S1←S2. Then,(i)
Pn,m←x=−xn+m−2x2+n−m−3x+2m.(ii)
SpecAn,m←=0−α+α2−8m/2−α−α2−8m/2n+m−211, where α=n−m−3.(iii)
∑x∈EAn,mx=0 if and only if n=m+3.Example 6.
LetX=x1,x2,x3. Then, H=X,Sii=12,φ2,1 is a superhypergraph as shown in Figure 5 and incidence matrix of A2,1 as follows.(28)
Thus, by Theorem9, we have P2,1x=−xx2−2x+2 and so Spec A2,1←=01+i1−i111.Figure 5
SuperhypergraphH=X,Sii=12,φ2,1.


### 3.3. Characteristic Polynomial for the Incidence Matrix of Superhypergraph with Two-Sided Flows

In this subsection, we compute the characteristic polynomial and spectrum of the superhypergraph with two-sided flows.For anySi,Sj, if Φ=φi,j|φi,j:Si↔Sj,i,j≥1, will say Si flows to Sj by two-sided and will denote by Si↭Sj and so AS1,S2,…,Sk by AS1,S2,…,Sk↭ and PS1,S2,…,Skx by PS1,S2,…,Sk↭x. When we will show that Si↭Sj of type +,−,+,−,…, it means that the first map flows from Si to Sj, the second map flows from Sj to Si, the third map flows from Si to Sj, etc., respectively. We will show it in Figure 6, where, n is an odd.Figure 6
H=X,Si,Sj,φi,j1,φj,i2,…,φj,im+n−2.Theorem 12.
LetS1=1,S2=n, n≥3 and S1↭S2 of type +,−,+,−,…. Then,(i)
P1,n↭x=−x2kx−1n=2kx2kx−12n=2k+1.(ii)
ifn is an even, then Spec A1,n↭=01n1.(iii)
ifn is an odd, then Spec A1,n↭=01n−12.Proof.
(i) Letn=3 and n=4. It is easy to see that P1,3↭x=x2x−12 and P1,4↭x=−x4x−1. Suppose that k≥3 is an odd and P1,k−1↭x=x2k−2x−12. Then, A1,k↭=c1,c2,…,ck,ck+1, where for all i∈3,5,7,…,k,ci=1,−1,…,−1︸k−timest and where for all i∈4,8,…,k+1,ci=−1,1,…,1︸k−timest. It follows that(29)P1,k↭x=detA1,k↭−Ik+1×k+1X=detB1,k↭,such that B1,k↭=bijk+1×k+1, where B1,k↭=c1′,c2′,…,ck′,ck+1′, where for all i∈3,5,7,…,k,ci′=1,−1,…,−1︸i−2−times,−1−x︸ith,−1,…,−1︸k−i+1−timest and for all i∈4,8,…,k+1,ci′=−1,1,…,1︸i−2−times,1−x︸ith,1,…,1︸k−i+1−timest. Now, consider D1,k↭=dijk+1×k+1, where(30)dij=bij,1≤j≤k−1,ck′+ck−1′,j=k,ck+1′−ck−1′,j=k+1.
Thus,D1,k↭=c1′,c2′,…,ck−1′,ck″,ck+1″, where ck″=0,…,0︸k−2−times,−x,−x,0t=−x0,…,0︸k−2−times,1,1,0t and ck+1″=0,…,0︸k−2−times,x,0,−xt=−x0,…,0︸k−2−times,−1,0,1t. Based on induction assumption and computations of determinant based on column ck+1′′ in matrix D1,k↭, have(31)P1,k↭x=detD1,k↭−Ik+1×k+1X=x2P1,k−2↭x=x2kx−12.
In a similar way, ifn is an even, we get that P1,n↭x=−xnx−1.
(ii) and (iii) They are clear by item (i) and (ii).Theorem 13.
LetS1=n,S2=1, n≥3 and S1↭S2 of type +,−,+,−,…. Then,(32)Pn,1↭x=−x2k−1x2−2,n=2k,x2kx2−3x+2,n=2k+1.Proof.
Letn=3. It is easy to see that P3,1↭x=x2x2−3x+2. Suppose that k≥3 is an odd and Pk−1,1↭x=x2k−2x2−3x+2. Then, Ak,1↭=c1,c2,…,ck,ck+1, where for all i∈3,5,…,k,ci=1,…,1︸k−times,−1t and where for all i∈4,6,…,k+1,ci=−1,…,−1︸k−times,1t. It follows that(33)Pk,1↭x=detAk,1↭−Ik+1×k+1X=detBk,1↭,such that Bk,1↭=bijk+1×k+1, where Bk,1↭=c1′,c2′,…,ck′,ck+1′, where for all i∈3,5,…,k,ci′=1,…,1︸i−1−times,1−x︸ith,1,…,1︸k−i−times,−1t and for all i∈4,8,…,k−1,ci′=−1,…,−1︸i−1−times,−1−x︸ith,−1,…,−1︸k−i−times,1t and for i=k+1,ci′=−1,…,−1︸k−times,1−xt. Now, consider Dk,1↭=dijk+1×k+1, where(34)dij=bij,j=1,…,k−2,k+1,ck′+ck−1′,j=k−1,ck′+ck+1′,j=k.
Thus,Dk,1↭=c1′,c2′,…,ck−2′,ck−1″,ck″,ck+1′, where ck−1″=0,…,0︸k−2−times,−x,−x,0t=−x0,…,0︸k−2−times,1,1,0t and ck″=0,…,0︸k−1−times,−x,−xt=−x0,…,0︸k−1−times,1,1t. Based on induction assumption and computations of determinant based on column ck′′ in matrix Dk,1↭, we have(35)Pk,1↭x=detDk,1↭−Ik+1×k+1X=x2Pk−2,1↭x=x2kx2−3x+2.
In a similar way, ifk is an even, we get that Pk,1↭x=−x2k−1x2−2.
(ii) and (iii) They are clear by item (i) and (ii).Theorem 14.
LetS1=n,S2=2, n≥3 and S1↭S2 of type +,−,+,−,…. Then,(36)Pn,2↭x=x2k+1x−1,n=2k,−x2k+2x−2,n=2k+1.Proof.
Letn=3. It is easy to see that P3,1↭x=−x4x−2. Suppose that k≥3 is an odd and Pk−1,2↭x=−x2kx−2. Then, Ak,2↭=c1,c2,…,ck+1,ck+2, where for all i∈3,5,…,k+2,ci=1,…,1︸k−times,−1,−1t and where for all i∈4,6,…,k+1,ci=−1,…,−1︸k−times,1,1t. It follows that(37)Pk,2↭x=detAk,2↭−Ik+2×k+2X=detBk,2↭,such that Bk,2↭=bijk+2×k+2, where Bk,2↭=c1′,c2′,…,ck+1′,ck+2′, where for all i∈3,5,…,k,ci′=1,…,1︸i−1−times,1−x︸ith,1,…,1︸k−i−times,−1,−1t and for i=k+2,ci′=1,…,1︸k−2−times,−1,−1−xt and for all i∈4,8,…,k+1,ci′=−1,…,−1︸i−1−times,−1−x︸ith,−1,…,−1︸k−i−times,1,1t. Now, consider Dk,2↭=dijk+2×k+2, where(38)dij=bij,j=1,…,k−2,k+1,k+2,ck′+ck−1′,j=k−1,ck′+ck+1′,j=k.
Thus,Dk,2↭=c1′,c2′,…,ck−2′,ck−1″,ck″,ck+1′,ck+2′, where ck−1″=0,…,0︸k−2−times,−x,−x,0,0t=−x0,…,0︸k−2−times,1,1,0,0t and ck″=0,…,0︸k−1−times,−x,−x,0t=−x0,…,0︸k−1−times,1,1,0t. Based on induction assumption and computations of determinant based on column ck′′ in matrix Dk,2↭, we have(39)Pk,2↭x=detDk,2↭−Ik+2×k+2X=x2Pk−2,2↭x=−x2k+2x−2.
In a similar way, ifk is an even, we get that Pk,2↭x=x2k+1x−1.
(ii) and (iii) They are clear by item (i) and (ii).Theorem 15.
LetS1=n,S2=3, n≥2 and S1↭S2 of type +,−,+,−,…. Then,(40)Pn,3↭x=−x2k+1x2−2,n=2kk≥1,−x2k+2x2−3x+2,n=2k+1k≥1.Proof.
It is similar to Theorem14.Corollary 3.
LetS1=n,S2=m, n≥2,m≥1 and S1↭S2 of type +,−,+,−,…. If m is an odd, then,(i)
Pn,m↭x=−x2k+m−2x2−2n=2kk≥1x2k+m−1x2−3x+2n=2k+1k≥1,(ii)
Ifn is an odd, then, Spec An,m↭=012n+m−211,(iii)
Ifn is an even, then Spec An,m↭=0−22n+m−211,(iv)
∑x∈EAn,mx≠0.Corollary 4.
LetS1=n,S2=m, n≥2,m≥1 and S1↭S2 of type +,−,+,−,…. If m is an even, then,(i)
ifn is an even, then, Pn,m↭x=−x2k+m−1x−1n=2kk≥1−x2k+mx−2n=2k+1k≥1,(ii)
Ifn is an odd, then, Spec An,m↭=02n+m−11,(iii)
Ifn is an even, then, Spec An,m↭=01n+m−11,(iv)
∑x∈EAn,mx≠0.When we will show thatSi↭Sj of type −,+,−,+,…, it means that the first map flows from Sj to Si, the second map flows from Si to Sj, the third map flows from Sj to Si, etc., respectively. We will show it in Figure 7, where, n is an odd.Figure 7
H=X,Si,Sj,φi,j1,φj,i2,…,φj,im+n−2.Theorem 16.
LetS1=1,S2=n, n≥3 and S1↭S2 of type −,+,−,+,…. Then,(i)
P1,n↭x=−x2k−1x2−3x+2n=2kx2kx−12n=2k+1,(ii)
ifn is an even, then, Spec A1,n↭=012n−111,(iii)
ifn is an odd, then, Spec A1,n↭=01n−12.Proof.
(i) Letn=3 and n=4. It is easy to see that P1,3↭x=x2x−12 and P1,4↭x=−x3x2−3x+2. Suppose that k≥3 is an even and P1,k−1↭x=−x2k−3x2−3x+2. Then, A1,k↭=c1,c2,…,ck,ck+1, where for all i∈3,5,7,…,k,ci=−1,1,…,1︸k−timest and where for all i∈4,6,8,…,k,ci=1,−1,…,−1︸k−timest. It follows that(41)P1,k↭x=detA1,k↭−Ik+1×k+1X=detB1,k↭,such that B1,k↭=bijk+1×k+1, where B1,k↭=c1′,c2′,…,ck′,ck+1′, where for all i∈4,6,8,…,k,ci′=1,−1,…,−1︸i−2−times,−1−x︸ith,−1,…,−1︸k+1−i−timest and for all i∈3,5,7,…,k,ci′=−1,1,…,1︸i−2−times,1−x︸ith,1,…,1︸k+1−i−timest. Now, consider D1,k↭=dijk+1×k+1, where(42)dij=bij,1≤j≤k−1,ck′+ck−1′,j=k,ck+1′−ck−1′,j=k+1.
Thus,D1,k↭=c1′,c2′,…,ck−1′,ck″,ck+1″, where ck″=0,…,0︸k−2−times,−x,−x,0t=−x0,…,0︸k−2−times,1,1,0t and ck+1″=0,…,0︸k−2−times,x,0,−xt=−x0,…,0︸k−2−times,−1,0,1t. Based on induction assumption and computations of determinant based on column ck+1′′ in matrix D1,k↭, we have(43)P1,k↭x=detD1,k↭−Ik+1×k+1X=x2P1,k−2↭x=−x2k−1x2−3x+2.
In a similar way, ifn is an odd, we get that P1,n↭x=xn−1x−12.
(ii) and (iii) They are clear by item (i).Theorem 17.
LetS1=n,S2=1, n≥3 and S1↭S2 of type −,+,−,+,…. Then,(44)Pn,1↭x=−x2k−1x2−2x+2,n=2k,−x2kx2+x−2,n=2k+1.Proof.
Letn=3,n=4. It is easy to see that P3,1↭x=x2x2+x−2 and P4,1↭x=−x3x2−2x+2 Suppose that k≥3 is an even and Pk−1,1↭x=−x2k−3x2−2x+2. Then, Ak,1↭=c1,c2,…,ck,ck+1, where for all i∈3,5,…,k+1,ci=−1,…,−1︸k−times,1t and where for all i∈4,6,…,k,ci=1,…,1︸k−times,−1t. It follows that(45)Pk,1↭x=detAk,1↭−Ik+1×k+1X=detBk,1↭,such that Bk,1↭=bijk+1×k+1, where Bk,1↭=c1′,c2′,…,ck′,ck+1′, where for all i∈4,6,…,k,ci′=1,…,1︸i−1−times,1−x︸ith,1,…,1︸k−i−times,−1t and for all i∈3,5,…,k−1,ci′=−1,…,−1︸i−1−times,−1−x,−1,…,−1︸k−i−times,1 and for i=k+1,ci′=−1,…,−1︸k−times,1−xt. Now, consider Dk,1↭=dijk+1×k+1, where(46)dij=bij,j=1,…,k−2,k+1,ck′+ck−1′,j=k−1,ck′+ck+1′,j=k.
Thus,Dk,1↭=c1′,c2′,…,ck−2′,ck−1″,ck″,ck+1′, where ck−1″=0,…,0︸k−2−times,−x,−x,0t=−x0,…,0︸k−2−times,1,1,0t and ck″=0,…,0︸k−1−times,−x,−xt=−x0,…,0︸k−1−times,1,1t. Based on induction assumption and computations of determinant based on column ck′′ in matrix Dk,1↭, we have(47)Pk,1↭x=detDk,1↭−Ik+1×k+1X=x2Pk−2,1↭x=−x2k−1x2−2x+2.
In a similar way, ifk is an odd, we get that Pk,1↭x=−x2kx2+x−2.
(ii) and (iii) They are clear by item (i) and (ii).Theorem 18.
LetS1=n,S2=2, n≥3 and S1↭S2 of type −,+,−,+,…. Then,(48)Pn,2↭x=x2k+1x−1,n=2k,−x2k+3,n=2k+1.Proof.
Letn=3. It is easy to see that P3,2↭x=−x5 and for n=4,P4,2↭x=x5x−1. Suppose that k≥3 is an odd and Pk−1,2↭x=−x2k+1. Then, Ak,2↭=c1,c2,…,ck+1,ck+2, where for all i∈3,5,…,k+2,ci=−1,…,−1︸k−times,1,1t and where for all i∈4,6,…,k+1,ci=1,…,1︸k−times,−1,−1t. It follows that(49)Pk,2↭x=detAk,2↭−Ik+2×k+2X=detBk,2↭,such that Bk,2↭=bijk+2×k+2, where Bk,2↭=c1′,c2′,…,ck+1′,ck+2′, where for all i∈3,5,…,k,ci′=−1,…,−1︸i−1−times,−1−x︸ith,−1,…,−1︸k−i−times,1,1t and for i=k+2,ci′=−1,…,−1︸k−times,1,1−xt and for all i∈4,8,…,k−1,ci′=1,…,1︸k−2−times,1−x︸ith,1,…,1︸k−i−times,−1,−1t and for i=k+1,ci′=1,…,1︸k−times,−1−x,−1t. Now, consider Dk,2↭=dijk+2×k+2, where(50)dij=bij,j=1,…,k−2,k+1,k+2,ck′+ck−1′,j=k−1,ck′+ck+1′,j=k.
Thus,Dk,2↭=c1′,c2′,…,ck−2′,ck−1″,ck″,ck+1′,ck+2′, where ck−1″=0,…,0︸k−2−times,−x,−x,0,0t=−x0,…,0︸k−2−times,1,1,0,0t and ck″=0,…,0︸k−1−times,−x,−x,0t=−x0,…,0︸k−1−times,1,1,0t. Based on induction assumption and computations of determinant based on column ck′′ in matrix Dk,2↭, we have(51)Pk,2↭x=detDk,2↭−Ik+2×k+2X=x2Pk−2,2↭x=−x2k+3.
In a similar way, ifk is an even, we get that Pk,2↭x=x2k+1x−1.
(ii) and (iii) They are clear by item (i) and (ii).Theorem 19.
LetS1=n,S2=3, n≥2 and S1↭S2 of type −,+,−,+,…. Then,(52)Pn,3↭x=−x2k+1x2−2x+2,n=2k,x2k+2x2+x−2,n=2k+1.Proof.
It is similar to Theorem18.Corollary 5.
LetS1=n,S2=m, n≥2,m≥1 and S1↭S2 of type −,+,−,+,…. If m is an odd, then,(i)
Pn,m↭x=−x2k+m−2x2−2x+2n=2kk≥1x2k+m−1x2+x−2n=2k+1k≥1,(ii)
Ifn is an odd, then, Spec An,m↭=0−21n+m−211,(iii)
Ifn is an even, then, Spec An,m↭=01+i1−in+m−211,(iv)
∑x∈EAn,mx≠0.Corollary 6.
LetS1=n,S2=m, n≥2,m≥1 and S1↭S2 of type −,+,−,+,…. If m is an even, then,(i)
Pn,m↭x=x2k+m−1x−1n=2kk≥1−x2k+m+1n=2k+1k≥1,(ii)
Ifn is an odd, then, Spec An,m↭=0n+m,(iii)
Ifn is an even, then, Spec An,m↭=01n+m−11,(iv)
∑x∈EAn,mx≠0.


## 3.1. Characteristic Polynomial for the Incidence Matrix of Superhypergraph

In this subsection, we compute the characteristic polynomial of the incidence matrix of any given superhypergraph. LetH=X,Sii=1k,φi,ji,j be a superhypergraph and so investigate the spectrum of the superhypergraph.From now on, letPS1,S2,…,Skx be the characteristic polynomial of the incidence matrix AS1,S2,…,Sk corresponding to superhypergraph H and EAS1,S2,…,Sk=x|is an eigenvalue ofAS1,S2,…,Sk. In addition, for any Si,Sj, if ΦSi,Sj=φi,j|φi,j:Si⟶Sj,i,j≥1, will say Si flows to Sj and will denote by Si⇝Sj. In this case, will denote AS1,S2,…,Sk by AS1,S2,…,Sk⇝ and PS1,S2,…,Skx by PS1,S2,…,Sk⇝x.Theorem 4.
LetS1=1,S2=n, n≥2 and S1⇝S2. Then,(i)
P1,n⇝x=−xn−1x2+n−3x−n−2.(ii)
SpecA1,n⇝=012−nn−111.Proof.
(i)
Letn=2. It is easy to see that P1,2⇝x=−xx2−x. Suppose that k≥3 and P1,k−1⇝x=−xk−2x2+k−4x−k−3. Then, A1,k⇝=c1,c2,…,ck,ck+1, where c3=⋯=ck+1=1,−1,…,−1︸k−timest. It follows that(9)P1,k⇝x=detA1,k⇝−Ik+1×k+1X=detB1,k⇝,such thatB1,k⇝=bijk+1×k+1, where B1,k⇝=c1′,c2′,…,ck′,ck+1′ that for all i∈3,…,k+1,ci′=1,−1,…,−1︸i−2−times,−1−x,−1,…,−1︸k−i+1−timest. Now, consider D1,k⇝=dijk+1×k+1, where(10)dij=bij,j≠k+1,−ck′+ck+1′,j=k+1.Thus,D1,k⇝=c1′,c2′,…,ck−1′,ck′,ck+1″, where ck+1″=0,…,0︸k−1−times,x,−xt=x0,…,0︸k−1−times,1,−1t. Based on induction assumption and computations of determinant based on column ck+1″ in matrix D1,k⇝, have(11)Pk,1⇝x=detD1,k⇝−Ik+1×k+1X=x−P1,k−1⇝x+−xk−1+−xk−2=−xk−1x2+k−3x−k−2.(ii)
It is clear by item (i).Theorem 5.
LetS1=n,S2=1, n≥2 and S1⇝S2. Then, Pn,1⇝x=−xn−1x2−n−2x−2.Proof.
Letn=2. It is easy to see that P2,1⇝x=−xx2−2. Suppose that k≥3 and Pk−1,1⇝x=−xk−2x2−k−3x−2. Then, Ak,1⇝=c1,c2,…,ck,ck+1, where c3=⋯=ck+1=1,1,…,1︸k−times,−1t. It follows that(12)Pk,1⇝x=detAk,1⇝−Ik+1×k+1X=detBk,1⇝,such that Bk,1⇝=bijk+1×k+1, where Bk,1⇝=c1′,c2′,…,ck′,ck+1′ that for all i∈3,4,…,k,ci′=1,1,…,1︸i−1−times,1−x,1,1,…,1︸k−i−times,−1t and ck+1′=1,1,…,1︸k−times,−1−xt. Consider Dk,1⇝=dijk+1×k+1, where(13)dij=bij,j≠k,ck′−ck+1′,j=k.
Thus,Dk,1⇝=c1′,c2′,…,ck−1′,ck′′,ck+1′, where ck′′=0,…,0︸k−1−times,−x,xt=−x0,…,0︸k−1−times,1,−1t. Based on induction assumption and computations of determinant based on column ck′′ in matrix Dk,1⇝, have(14)Pk,1⇝x=detDk,1⇝−Ik+1×k+1X=−xPk−1,1⇝x+−xk.Theorem 6.
LetS1=n,S2=2, n≥2 and S1⇝S2. Then, Pn,2⇝x=−xnx2−n−3x−4.Proof.
Letn=2. It is easy to see that P2,1⇝x=x2x2+x−4. Suppose that k≥3 and Pk−1,1⇝x=−xk−1x2−k−4x−4.Then, Ak,2⇝=c1,c2,…,ck,ck+1,ck+2, where c3=⋯=ck+2=1,1,…,1︸k−times,−1,−1t. It follows that(15)Pk,2⇝x=detAk,2⇝−Ik+2×k+2X=detBk,2⇝,such that Bk,2⇝=bijk+2×k+2, where Bk,2⇝=c1′,c2′,…,ck′,ck+1′,ck+2′ that for all i∈3,4,…,kci′=1,1,…,1︸i−1−times,1−x,1,1,…,1︸k−i−times,−1,−1t,ck+1′=1,1,…,1︸k−times,−1−x,,−1t and ck+2′=1,1,…,1︸k−times,−1,−1−xt. In addition, Dk,2⇝=dijk+2×k+2, where(16)dij=bij,j≠k,ck′−ck+2′,j=k.
Thus,Dk,2⇝=c1′,c2′,…,ck′′,ck+1′,ck+2′, where ck′′=0,…,0︸k−1−times,−x,0,xt=−x0,…,0︸k−1−times,1,0,−1t. Based on induction assumption and computations of determinant based on column ck′′ in matrix Dk,2⇝ show that(17)Pk,2⇝x=detDk,2⇝−Ik+2×k+2X=−xPk−1,2⇝x+−xk+1.Theorem 7.
LetS1=n,S2=3, n≥2 and S1⇝S2. Then, Pn,3⇝x=−xn+1x2−n−4x−6.Proof.
It is similar to Theorem6.Corollary 1.
LetS1=n,S2=m, n≥2,m≥1 and S1⇝S2. Then,(i)
Pn,m⇝x=−xn+m−2x2−n−m−1x−2m.(ii)
SpecAn,m⇝=0α+α2+8m/2α−α2+8m/2n+m−211, where α=n−m−1.(iii)
∑x∈EAn,m⇝x=0 if and only if n=m+1.Example 5.
LetX=x1,x2,x3. Then, H=X,Sii=12,φ1,2 is a superhypergraph as shown in Figure 4 and incidence matrix of A2,1⇝ as follows.(18)
Thus, by Theorem5, we have P2,1⇝x=−xx2−2 and so SpecA2,1⇝=02−2111.Figure 4
SuperhypergraphH=X,Sii=12,φ1,2.


## 3.2. Characteristic Polynomial for the Incidence Matrix of Superhypergraph with Reverse Flows

In this subsection, we compute the characteristic polynomial and spectrum of the superhypergraph in the reverse flows to the previous section.For anySi,Sj, if ΦSj,Si=φi,j|φi,j:Sj⟶Si,i,j≥1, will say Sj flows to Si and will denote by Si←Sjreverse flows toSi⇝Sj and so AS1,S2,…,Sk by AS1,S2,…,Sk← and PS1,S2,…,Skx by PS1,S2,…,Sk←x.Theorem 8.
LetS1=1,S2=n, n≥2 and S1←S2. Then,(i)
P1,n←x=−xn−1x2−n+1x+n.(ii)
SpecA1,n←=01nn−111.Proof.
(i)
Letn=2. It is easy to see that P1,2←x=−xx2−3x+2. Suppose that k≥3 and P1,k−1←x=−xk−2x2−kx+k−1. Then, A1,k←=c1,c2,…,ck,ck+1, where c3=⋯=ck+1=−1,1,…,1︸k−timest. It follows that(19)P1,k←x=detA1,k←−Ik+1×k+1X=detB1,k←,such thatB1,k←=bijk+1×k+1, where B1,k←=c1′,c2′,…,ck′,ck+1′ that for all i∈3,4,…,k+1,ci′=−1,1,…,1︸i−2−times,1−x︸i−th,1,…,1︸k−i+1−timest. Now, consider D1,k←=dijk+1×k+1, where(20)dij=bij,j≠k+1,−ck′+ck+1′,j=k+1.Thus,D1,k←=c1′,c2′,…,ck−1′,ck′,ck+1″, where ck+1″=0,…,0︸k−1−times,x,−xt=x0,…,0︸k−1−times,1,−1t. Based on induction assumption and computations of determinant based on column ck+1′′ in matrix D1,k←, have(21)P1,k←x=detD1,k←−Ik+1×k+1X=x−P1,k−1←x−−xk−1−−xk−2=−xk−1x2−k+1x+k.(ii)
It is clear by item (i).Theorem 9.
LetS1=n,S2=1, n≥2 and S1←S2. Then, Pn,1←x=−xn−1x2+n−4x+2.Proof.
Letn=2. It easy is to see that P2,1←x=−xx2−2x+2. Suppose that k≥3 and Pk−1,1←x=−xk−2x2+k−5x+2.Then, Ak,1←=c1,c2,…,ck,ck+1, where c3=⋯=ck+1=−1,−1,…,−1︸ktimes,1t. It follows that(22)Pk,1←x=detAk,1←−Ik+1×k+1X=detBk,1←,such that Bk,1←=bijk+1×k+1, where Bk,1←=c1′,c2′,…,ck′,ck+1′ that for all i∈3,…,k,ci′=−1,…,−1︸i−1−times,−1−x︸ith,−1,…,−1︸k−i−times,1t and ck+1′=−1,−1,…,−1︸ktimes,1−xt. Moreover, consider Dk,1←=dijk+1×k+1, where(23)dij=bij,j≠k,ck′−ck+1′,j=k.
Thus,Dk,1←=c1′,c2′,…,ck−1′,ck′′,ck+1′, where ck′′=0,…,0︸k−1times,−x,xt=−x0,…,0︸k−1times,1,−1t. Based on induction assumption and computations of determinant based on column ck′′ in matrix Dk,1←, have(24)Pk,1←x=detDk,1←−Ik+1×k+1X=−xPk−1,1←x+−1k+1xk.Theorem 10.
LetS1=n,S2=2,n≥2 and S1←S2. Then, Pn,2←x=−xnx2+n−5x+4.Proof.
Letn=2. It is easy to see that P2,1←x=x2x2−3x+4. Suppose that k≥3 and Pk−1,1←x=−xk−1x2+k−6x+4.Then, Ak,2←=c1,c2,…,ck,ck+1,ck+2, where c3=⋯=ck+2=−1,−1,…,−1︸ktimes,1,1t. It follows that(25)Pk,2←x=detAk,2←−Ik+2×k+2X=detBk,2←,such that Bk,2←=bijk+2×k+2, where Bk,2←=c1′,c2′,…,ck′,ck+1′,ck+2′ that for all i∈3,…,k,ci′=−1,…,−1︸i−1−times,−1−x︸ith,−1,…,−1︸k−i−times,1,1t,ck+1′=−1,…,−1︸k−times,1−x,1t, and ck+2′=−1,…,−1︸k−times,1,1−xt. Consider Dk,2←=dijk+2×k+2 where(26)dij=bij,j≠k,ck′−ck+1′,j=k.
Thus,Dk,2←=c1′,c2′,…,ck′′,ck+1′,ck+2′, where ck′′=0,…,0︸k−1−times,−x,x,0t=−x0,…,0︸k−1−times,1,−1,0t. Based on induction assumption and computations of determinant based on column ck′′ in matrix Dk,2← show that(27)Pk,2←x=detDk,2←−Ik+2×k+2X=−xPk−1,2←x+−1kxk+1.Theorem 11.
LetS1=n,S2=3, n≥2 and S1←S2. Then, Pn,3x=−xn+1x2+n−6x+6.Proof.
It is similar to Theorem6.Corollary 2.
LetS1=n,S2=m, n≥2,m≥1 and S1←S2. Then,(i)
Pn,m←x=−xn+m−2x2+n−m−3x+2m.(ii)
SpecAn,m←=0−α+α2−8m/2−α−α2−8m/2n+m−211, where α=n−m−3.(iii)
∑x∈EAn,mx=0 if and only if n=m+3.Example 6.
LetX=x1,x2,x3. Then, H=X,Sii=12,φ2,1 is a superhypergraph as shown in Figure 5 and incidence matrix of A2,1 as follows.(28)
Thus, by Theorem9, we have P2,1x=−xx2−2x+2 and so Spec A2,1←=01+i1−i111.Figure 5
SuperhypergraphH=X,Sii=12,φ2,1.


## 3.3. Characteristic Polynomial for the Incidence Matrix of Superhypergraph with Two-Sided Flows

In this subsection, we compute the characteristic polynomial and spectrum of the superhypergraph with two-sided flows.For anySi,Sj, if Φ=φi,j|φi,j:Si↔Sj,i,j≥1, will say Si flows to Sj by two-sided and will denote by Si↭Sj and so AS1,S2,…,Sk by AS1,S2,…,Sk↭ and PS1,S2,…,Skx by PS1,S2,…,Sk↭x. When we will show that Si↭Sj of type +,−,+,−,…, it means that the first map flows from Si to Sj, the second map flows from Sj to Si, the third map flows from Si to Sj, etc., respectively. We will show it in Figure 6, where, n is an odd.Figure 6
H=X,Si,Sj,φi,j1,φj,i2,…,φj,im+n−2.Theorem 12.
LetS1=1,S2=n, n≥3 and S1↭S2 of type +,−,+,−,…. Then,(i)
P1,n↭x=−x2kx−1n=2kx2kx−12n=2k+1.(ii)
ifn is an even, then Spec A1,n↭=01n1.(iii)
ifn is an odd, then Spec A1,n↭=01n−12.Proof.
(i) Letn=3 and n=4. It is easy to see that P1,3↭x=x2x−12 and P1,4↭x=−x4x−1. Suppose that k≥3 is an odd and P1,k−1↭x=x2k−2x−12. Then, A1,k↭=c1,c2,…,ck,ck+1, where for all i∈3,5,7,…,k,ci=1,−1,…,−1︸k−timest and where for all i∈4,8,…,k+1,ci=−1,1,…,1︸k−timest. It follows that(29)P1,k↭x=detA1,k↭−Ik+1×k+1X=detB1,k↭,such that B1,k↭=bijk+1×k+1, where B1,k↭=c1′,c2′,…,ck′,ck+1′, where for all i∈3,5,7,…,k,ci′=1,−1,…,−1︸i−2−times,−1−x︸ith,−1,…,−1︸k−i+1−timest and for all i∈4,8,…,k+1,ci′=−1,1,…,1︸i−2−times,1−x︸ith,1,…,1︸k−i+1−timest. Now, consider D1,k↭=dijk+1×k+1, where(30)dij=bij,1≤j≤k−1,ck′+ck−1′,j=k,ck+1′−ck−1′,j=k+1.
Thus,D1,k↭=c1′,c2′,…,ck−1′,ck″,ck+1″, where ck″=0,…,0︸k−2−times,−x,−x,0t=−x0,…,0︸k−2−times,1,1,0t and ck+1″=0,…,0︸k−2−times,x,0,−xt=−x0,…,0︸k−2−times,−1,0,1t. Based on induction assumption and computations of determinant based on column ck+1′′ in matrix D1,k↭, have(31)P1,k↭x=detD1,k↭−Ik+1×k+1X=x2P1,k−2↭x=x2kx−12.
In a similar way, ifn is an even, we get that P1,n↭x=−xnx−1.
(ii) and (iii) They are clear by item (i) and (ii).Theorem 13.
LetS1=n,S2=1, n≥3 and S1↭S2 of type +,−,+,−,…. Then,(32)Pn,1↭x=−x2k−1x2−2,n=2k,x2kx2−3x+2,n=2k+1.Proof.
Letn=3. It is easy to see that P3,1↭x=x2x2−3x+2. Suppose that k≥3 is an odd and Pk−1,1↭x=x2k−2x2−3x+2. Then, Ak,1↭=c1,c2,…,ck,ck+1, where for all i∈3,5,…,k,ci=1,…,1︸k−times,−1t and where for all i∈4,6,…,k+1,ci=−1,…,−1︸k−times,1t. It follows that(33)Pk,1↭x=detAk,1↭−Ik+1×k+1X=detBk,1↭,such that Bk,1↭=bijk+1×k+1, where Bk,1↭=c1′,c2′,…,ck′,ck+1′, where for all i∈3,5,…,k,ci′=1,…,1︸i−1−times,1−x︸ith,1,…,1︸k−i−times,−1t and for all i∈4,8,…,k−1,ci′=−1,…,−1︸i−1−times,−1−x︸ith,−1,…,−1︸k−i−times,1t and for i=k+1,ci′=−1,…,−1︸k−times,1−xt. Now, consider Dk,1↭=dijk+1×k+1, where(34)dij=bij,j=1,…,k−2,k+1,ck′+ck−1′,j=k−1,ck′+ck+1′,j=k.
Thus,Dk,1↭=c1′,c2′,…,ck−2′,ck−1″,ck″,ck+1′, where ck−1″=0,…,0︸k−2−times,−x,−x,0t=−x0,…,0︸k−2−times,1,1,0t and ck″=0,…,0︸k−1−times,−x,−xt=−x0,…,0︸k−1−times,1,1t. Based on induction assumption and computations of determinant based on column ck′′ in matrix Dk,1↭, we have(35)Pk,1↭x=detDk,1↭−Ik+1×k+1X=x2Pk−2,1↭x=x2kx2−3x+2.
In a similar way, ifk is an even, we get that Pk,1↭x=−x2k−1x2−2.
(ii) and (iii) They are clear by item (i) and (ii).Theorem 14.
LetS1=n,S2=2, n≥3 and S1↭S2 of type +,−,+,−,…. Then,(36)Pn,2↭x=x2k+1x−1,n=2k,−x2k+2x−2,n=2k+1.Proof.
Letn=3. It is easy to see that P3,1↭x=−x4x−2. Suppose that k≥3 is an odd and Pk−1,2↭x=−x2kx−2. Then, Ak,2↭=c1,c2,…,ck+1,ck+2, where for all i∈3,5,…,k+2,ci=1,…,1︸k−times,−1,−1t and where for all i∈4,6,…,k+1,ci=−1,…,−1︸k−times,1,1t. It follows that(37)Pk,2↭x=detAk,2↭−Ik+2×k+2X=detBk,2↭,such that Bk,2↭=bijk+2×k+2, where Bk,2↭=c1′,c2′,…,ck+1′,ck+2′, where for all i∈3,5,…,k,ci′=1,…,1︸i−1−times,1−x︸ith,1,…,1︸k−i−times,−1,−1t and for i=k+2,ci′=1,…,1︸k−2−times,−1,−1−xt and for all i∈4,8,…,k+1,ci′=−1,…,−1︸i−1−times,−1−x︸ith,−1,…,−1︸k−i−times,1,1t. Now, consider Dk,2↭=dijk+2×k+2, where(38)dij=bij,j=1,…,k−2,k+1,k+2,ck′+ck−1′,j=k−1,ck′+ck+1′,j=k.
Thus,Dk,2↭=c1′,c2′,…,ck−2′,ck−1″,ck″,ck+1′,ck+2′, where ck−1″=0,…,0︸k−2−times,−x,−x,0,0t=−x0,…,0︸k−2−times,1,1,0,0t and ck″=0,…,0︸k−1−times,−x,−x,0t=−x0,…,0︸k−1−times,1,1,0t. Based on induction assumption and computations of determinant based on column ck′′ in matrix Dk,2↭, we have(39)Pk,2↭x=detDk,2↭−Ik+2×k+2X=x2Pk−2,2↭x=−x2k+2x−2.
In a similar way, ifk is an even, we get that Pk,2↭x=x2k+1x−1.
(ii) and (iii) They are clear by item (i) and (ii).Theorem 15.
LetS1=n,S2=3, n≥2 and S1↭S2 of type +,−,+,−,…. Then,(40)Pn,3↭x=−x2k+1x2−2,n=2kk≥1,−x2k+2x2−3x+2,n=2k+1k≥1.Proof.
It is similar to Theorem14.Corollary 3.
LetS1=n,S2=m, n≥2,m≥1 and S1↭S2 of type +,−,+,−,…. If m is an odd, then,(i)
Pn,m↭x=−x2k+m−2x2−2n=2kk≥1x2k+m−1x2−3x+2n=2k+1k≥1,(ii)
Ifn is an odd, then, Spec An,m↭=012n+m−211,(iii)
Ifn is an even, then Spec An,m↭=0−22n+m−211,(iv)
∑x∈EAn,mx≠0.Corollary 4.
LetS1=n,S2=m, n≥2,m≥1 and S1↭S2 of type +,−,+,−,…. If m is an even, then,(i)
ifn is an even, then, Pn,m↭x=−x2k+m−1x−1n=2kk≥1−x2k+mx−2n=2k+1k≥1,(ii)
Ifn is an odd, then, Spec An,m↭=02n+m−11,(iii)
Ifn is an even, then, Spec An,m↭=01n+m−11,(iv)
∑x∈EAn,mx≠0.When we will show thatSi↭Sj of type −,+,−,+,…, it means that the first map flows from Sj to Si, the second map flows from Si to Sj, the third map flows from Sj to Si, etc., respectively. We will show it in Figure 7, where, n is an odd.Figure 7
H=X,Si,Sj,φi,j1,φj,i2,…,φj,im+n−2.Theorem 16.
LetS1=1,S2=n, n≥3 and S1↭S2 of type −,+,−,+,…. Then,(i)
P1,n↭x=−x2k−1x2−3x+2n=2kx2kx−12n=2k+1,(ii)
ifn is an even, then, Spec A1,n↭=012n−111,(iii)
ifn is an odd, then, Spec A1,n↭=01n−12.Proof.
(i) Letn=3 and n=4. It is easy to see that P1,3↭x=x2x−12 and P1,4↭x=−x3x2−3x+2. Suppose that k≥3 is an even and P1,k−1↭x=−x2k−3x2−3x+2. Then, A1,k↭=c1,c2,…,ck,ck+1, where for all i∈3,5,7,…,k,ci=−1,1,…,1︸k−timest and where for all i∈4,6,8,…,k,ci=1,−1,…,−1︸k−timest. It follows that(41)P1,k↭x=detA1,k↭−Ik+1×k+1X=detB1,k↭,such that B1,k↭=bijk+1×k+1, where B1,k↭=c1′,c2′,…,ck′,ck+1′, where for all i∈4,6,8,…,k,ci′=1,−1,…,−1︸i−2−times,−1−x︸ith,−1,…,−1︸k+1−i−timest and for all i∈3,5,7,…,k,ci′=−1,1,…,1︸i−2−times,1−x︸ith,1,…,1︸k+1−i−timest. Now, consider D1,k↭=dijk+1×k+1, where(42)dij=bij,1≤j≤k−1,ck′+ck−1′,j=k,ck+1′−ck−1′,j=k+1.
Thus,D1,k↭=c1′,c2′,…,ck−1′,ck″,ck+1″, where ck″=0,…,0︸k−2−times,−x,−x,0t=−x0,…,0︸k−2−times,1,1,0t and ck+1″=0,…,0︸k−2−times,x,0,−xt=−x0,…,0︸k−2−times,−1,0,1t. Based on induction assumption and computations of determinant based on column ck+1′′ in matrix D1,k↭, we have(43)P1,k↭x=detD1,k↭−Ik+1×k+1X=x2P1,k−2↭x=−x2k−1x2−3x+2.
In a similar way, ifn is an odd, we get that P1,n↭x=xn−1x−12.
(ii) and (iii) They are clear by item (i).Theorem 17.
LetS1=n,S2=1, n≥3 and S1↭S2 of type −,+,−,+,…. Then,(44)Pn,1↭x=−x2k−1x2−2x+2,n=2k,−x2kx2+x−2,n=2k+1.Proof.
Letn=3,n=4. It is easy to see that P3,1↭x=x2x2+x−2 and P4,1↭x=−x3x2−2x+2 Suppose that k≥3 is an even and Pk−1,1↭x=−x2k−3x2−2x+2. Then, Ak,1↭=c1,c2,…,ck,ck+1, where for all i∈3,5,…,k+1,ci=−1,…,−1︸k−times,1t and where for all i∈4,6,…,k,ci=1,…,1︸k−times,−1t. It follows that(45)Pk,1↭x=detAk,1↭−Ik+1×k+1X=detBk,1↭,such that Bk,1↭=bijk+1×k+1, where Bk,1↭=c1′,c2′,…,ck′,ck+1′, where for all i∈4,6,…,k,ci′=1,…,1︸i−1−times,1−x︸ith,1,…,1︸k−i−times,−1t and for all i∈3,5,…,k−1,ci′=−1,…,−1︸i−1−times,−1−x,−1,…,−1︸k−i−times,1 and for i=k+1,ci′=−1,…,−1︸k−times,1−xt. Now, consider Dk,1↭=dijk+1×k+1, where(46)dij=bij,j=1,…,k−2,k+1,ck′+ck−1′,j=k−1,ck′+ck+1′,j=k.
Thus,Dk,1↭=c1′,c2′,…,ck−2′,ck−1″,ck″,ck+1′, where ck−1″=0,…,0︸k−2−times,−x,−x,0t=−x0,…,0︸k−2−times,1,1,0t and ck″=0,…,0︸k−1−times,−x,−xt=−x0,…,0︸k−1−times,1,1t. Based on induction assumption and computations of determinant based on column ck′′ in matrix Dk,1↭, we have(47)Pk,1↭x=detDk,1↭−Ik+1×k+1X=x2Pk−2,1↭x=−x2k−1x2−2x+2.
In a similar way, ifk is an odd, we get that Pk,1↭x=−x2kx2+x−2.
(ii) and (iii) They are clear by item (i) and (ii).Theorem 18.
LetS1=n,S2=2, n≥3 and S1↭S2 of type −,+,−,+,…. Then,(48)Pn,2↭x=x2k+1x−1,n=2k,−x2k+3,n=2k+1.Proof.
Letn=3. It is easy to see that P3,2↭x=−x5 and for n=4,P4,2↭x=x5x−1. Suppose that k≥3 is an odd and Pk−1,2↭x=−x2k+1. Then, Ak,2↭=c1,c2,…,ck+1,ck+2, where for all i∈3,5,…,k+2,ci=−1,…,−1︸k−times,1,1t and where for all i∈4,6,…,k+1,ci=1,…,1︸k−times,−1,−1t. It follows that(49)Pk,2↭x=detAk,2↭−Ik+2×k+2X=detBk,2↭,such that Bk,2↭=bijk+2×k+2, where Bk,2↭=c1′,c2′,…,ck+1′,ck+2′, where for all i∈3,5,…,k,ci′=−1,…,−1︸i−1−times,−1−x︸ith,−1,…,−1︸k−i−times,1,1t and for i=k+2,ci′=−1,…,−1︸k−times,1,1−xt and for all i∈4,8,…,k−1,ci′=1,…,1︸k−2−times,1−x︸ith,1,…,1︸k−i−times,−1,−1t and for i=k+1,ci′=1,…,1︸k−times,−1−x,−1t. Now, consider Dk,2↭=dijk+2×k+2, where(50)dij=bij,j=1,…,k−2,k+1,k+2,ck′+ck−1′,j=k−1,ck′+ck+1′,j=k.
Thus,Dk,2↭=c1′,c2′,…,ck−2′,ck−1″,ck″,ck+1′,ck+2′, where ck−1″=0,…,0︸k−2−times,−x,−x,0,0t=−x0,…,0︸k−2−times,1,1,0,0t and ck″=0,…,0︸k−1−times,−x,−x,0t=−x0,…,0︸k−1−times,1,1,0t. Based on induction assumption and computations of determinant based on column ck′′ in matrix Dk,2↭, we have(51)Pk,2↭x=detDk,2↭−Ik+2×k+2X=x2Pk−2,2↭x=−x2k+3.
In a similar way, ifk is an even, we get that Pk,2↭x=x2k+1x−1.
(ii) and (iii) They are clear by item (i) and (ii).Theorem 19.
LetS1=n,S2=3, n≥2 and S1↭S2 of type −,+,−,+,…. Then,(52)Pn,3↭x=−x2k+1x2−2x+2,n=2k,x2k+2x2+x−2,n=2k+1.Proof.
It is similar to Theorem18.Corollary 5.
LetS1=n,S2=m, n≥2,m≥1 and S1↭S2 of type −,+,−,+,…. If m is an odd, then,(i)
Pn,m↭x=−x2k+m−2x2−2x+2n=2kk≥1x2k+m−1x2+x−2n=2k+1k≥1,(ii)
Ifn is an odd, then, Spec An,m↭=0−21n+m−211,(iii)
Ifn is an even, then, Spec An,m↭=01+i1−in+m−211,(iv)
∑x∈EAn,mx≠0.Corollary 6.
LetS1=n,S2=m, n≥2,m≥1 and S1↭S2 of type −,+,−,+,…. If m is an even, then,(i)
Pn,m↭x=x2k+m−1x−1n=2kk≥1−x2k+m+1n=2k+1k≥1,(ii)
Ifn is an odd, then, Spec An,m↭=0n+m,(iii)
Ifn is an even, then, Spec An,m↭=01n+m−11,(iv)
∑x∈EAn,mx≠0.


## 4. Conclusions and Future Works

The current paper has introduced a novel concept of superhypergraphs as a generalization of graphs. The advantage of the notation of superhypergraphs is that it considers the relationship between a set of elements separately and as a whole, and this helps to eliminate the defects of graphs and superhypergraphs. The notation of superhypergraphs can be useful tools in modeling the real issues in engineering sciences and other sciences, especially network-related issues. For any given superhypergraph, the lower and upper bound of the number of the set of their superedges is computed and so it is computed and proved the number of all superhypergraphs constructed on any given nonempty set. Polynomial characteristics and eigenvalues of a matrix that represents a superhypergraph can provide useful information about the superhypergraph. The concept of the incidence matrix of superhypergraphs is presented and the characteristic polynomial of the incidence matrix of superhypergraphs and spectrum of superhypergraphs is analyzed and computed. It is shown that the spectrum of superhypergraphs depended on to flows of their maps between supervertices and the spectrum of superhypergraphs varies with the change of direction of flows. We presented and computed the spectrum of superhypergraphs with some types of flows such as one-sided flows, left to the right flows, right to left flows, and two-sided reverse flows. We hope that these results are helpful for further studies in the theory of graphs, hypergraphs, and superhypergraphs. In our future studies, we hope to obtain more results regarding domination sets and domination numbers of superhypergraphs, fuzzy superhypergraphs, and obtain some results in this regard and their applications in the real-world.



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