A Measure for the Vulnerability of Uniform Hypergraph Networks: Scattering Number

Abstract

The scattering number of a graph G is defined as $s\left(G\right)=max\left\{\omega \right(G-X)-|X|:X\subset V(G),\omega (G-X)>1\}$, where X is a cut set of G, and $\omega (G-X)$ denotes the number of components in $G-X$, which can be used to measure the vulnerability of network G. In this paper, we generalize this parameter to a hypergraph to measure the vulnerability of uniform hypergraph networks. Firstly, some bounds on the scattering number are given. Secondly, the relations of scattering number between a complete k-uniform hypergraph and complete bipartite k-uniform hypergraph are discussed.

1. Introduction

It is a common method to study network problems using graphs to represent network structures and utilizing the concepts of nodes and edges in graph theory to represent relationships between objects in the networks. However, the application of graphs to describe complex network structures may result in incomplete or inaccurate information. For instance, in a research cooperation network, a graph effectively illustrates the collaboration between two authors in a paper, where the connection is represented by an edge. But for the situation in which more than two people collaborate on a paper, it cannot be described clearly by a graph, because the graph can only reflect whether there is cooperation between the two authors instead of many authors. A key limit of graphs is that the edges can only model the pairwise relations among the objects. This limitation becomes particularly noticeable in various real and complex systems.

The concept of a hypergraph addresses the limitations inherent in representing networks solely through graphs. Many complex structures, mirroring real-life networks, find representation through hypergraphs, which are the most general structure in discrete mathematics. Hypergraphs, a generalization of graphs, permit an edge to connect varying numbers of vertices instead of being restricted to connecting only two vertices. This flexibility has led to the extensive utilization of hypergraph structures in modeling high-order relations within numerous real and complex systems [1,2,3].

A hypergraph $H=(V,E)$ is a finite set V of elements, referred to as vertices, together with a finite multiset E of subsets of V, which are called hyperedges or simply edges. If $V\ne \varnothing$ and $E=\varnothing$, we say that H is a trivial hypergraph. In the scenario where both V and E are empty sets, H is termed an empty hypergraph. A hypergraph is said to be k-uniform if each edge contains exactly k vertices of H. Notably, ordinary graphs are referred to as two-uniform hypergraphs. A vertex $v_i\in V$ is considered incident to an edge $e_j\in E$ if $v_i\in e_j$. Two vertices, $v_i,v_j$, are adjacent if there exists some edge $e\in E$ such that $v_i\in E$ and $v_j\in E$. Moreover, two edges are said to be adjacent if their intersection is not empty. The degree of a vertex $v_i$, denoted by $d\left(v_i\right)$, is defined as $d\left(v_i\right)=\left|\{e_j:v_i\in e_j\in E\}\right|$. A vertex $v_i$ is called isolated if its degree is zero. The degree of an edge e is the cardinality of edge e denoted by $d\left(e\right)$: that is, $d\left(e\right)=\left|e\right|$.

A hypergraph $H^{\prime }=(V^{\prime },E^{\prime })$ is a subhypergraph of $H=(V,E)$ if $V^{\prime }\subseteq V$ and $E^{\prime }\subseteq E$, which is denoted by $H^{\prime }\subseteq H$. For $X\subseteq V$, we use $H\left[X\right]$ to denote the hypergraph induced by X, where $V\left(H\right[X\left]\right)=X$ and $E\left(H\right[X\left]\right)=\{e\in E(H):e\subseteq X\}$. $H-X$ is the hypergraph induced by $V\setminus X$.

Given a hypergraph H, a walk in H is a finite alternating sequence $v_1{e}_1{v}_2\cdots e_q{v}_{q+1}$ of vertices and edges of H such that $v_i\in V\left(H\right)$ for $i=1,2,\cdots ,q+1$; $e_i\in E\left(H\right)$ for $i=1,2,\cdots ,q$; and $v_i,v_{i+1}\in e_i$ for $i=1,2,\cdots ,q$. A path is a walk with additional restrictions that the vertices are all distinct and the edges are all distinct. A hypergraph H is connected if for any pair of vertices $u,v\in V\left(H\right)$, there is a path connecting u and v. Otherwise, H is disconnected. A component of hypergraph H is a maximal connected subhypergraph of H. A subset $X\subseteq V\left(H\right)$ is called a cut set of H if $H-X$ is disconnected.

A k-uniform hypergraph H is simple if there are no multiple edges in H, meaning all edges in H are distinct. In the subsequent sections of this paper, we assume that all hypergraphs considered are simple and k-uniform with $k\ge 3$.

Vulnerability refers to a network’s ability to sustain functionality when a part of the network is either naturally damaged or targeted for attack. There are various vulnerability parameters, such as toughness [4,5], integrity [6], tenacity [7,8], and scattering number [9,10,11]. There have been many significient results on these parameters on ordinary graphs [12,13,14]. These parameters consist of one or more of the following three quantities: the count of non-functioning elements, the number of remaining connected subnetworks, and the size of the largest remaining group within which mutual communication can persist. Undoubtedly, these quantities play a pivotal role in the robustness of networks. While numerous studies have delved into these parameters, they predominantly focus on networks modeled by ordinary graphs, leaving a research gap in the vulnerability of hypernetworks. Currently, the literature related to vulnerability parameters in hypergraphs can be found in [15,16,17,18]. In this context, we extend the application of the scattering number as a parameter to measure the vulnerability of hypergraphs.

The concept of a scattering number was defined by Jung [19]. In this paper, the scattering number of hypergraph H is defined as

$$s\left(H\right)=max\left\{\omega \right(H-X)-|X|:X\subseteq V(H),\omega (H-X)>1\}$$

where $\omega (H-X)$ stands for the number of components of $H-X$. The score of X is defined as $sc\left(X\right)=\omega (H-X)-\left|X\right|$. A set $X\subseteq V\left(H\right)$ fulfilling $sc\left(X\right)=s\left(H\right)$ is said to be a scattering set of H.

In this paper, we initially address the scattering number problem of complete k-uniform hypergraphs and their dual hypergraphs, presenting the bounds of the scattering number for complete k-uniform hypergraphs in Section 2. Section 3 focuses on determining the bounds of the scattering number for complete bipartite k-uniform hypergraphs. The relationships of scattering numbers between complete k-uniform hypergraphs and complete bipartite k-uniform hypergraphs with the same order are discussed in Section 4. Throughout this paper, by $\lceil x\rceil$, we denote the smallest integer not smaller than x. The sign function is denoted by $sgn\left(x\right)$, which returns either −1, 0 or 1 depending on whether the value of x is negative, zero or positive. Any undefined terminology and notations in this paper can be found in [20,21,22].

2. The Scattering Number of Complete $\mathit{k}$-Uniform Hypergraph

A hypergraph $H=(V,E)$ is complete if $E=P\left(V\right)\setminus \varnothing$, where $P\left(V\right)$ denotes the powerset of V. For $\left|V\right|=n$, a complete k-uniform hypergraph on $n\ge k\ge 3$ vertices is a hypergraph which has all k-subsets of V as hyperedges, i.e., $E=P_k\left(V\right)$, where $P_k\left(V\right)$ is the set of all k-subsets of V.

Lemma 1. 

If X is a scattering set of complete k-uniform hypergraph $H=(V,E)$, then $\left|X\right|=\left|V\right(H\left)\right|-k+1$.

Proof. 

Let X be a scattering set of complete k-uniform hypergraph H. Now, we proceed by induction on $\left|V\right(H\left)\right|$ to show $\left|X\right|=\left|V\right(H\left)\right|-k+1$.

Firstly, if $\left|V\right(H\left)\right|=k$, then H is just a hyperedge. It is clear that a scattering number X of H contains one vertex. That means $\left|X\right|=\left|V\right(H\left)\right|-k+1$ holds.

Secondly, assume that the conclusion holds for $\left|V\right(H\left)\right|<m(m>k)$, which means that the scattering set X of H contains $\left|\right|V\left(H\right)|-k+1$ vertices. Now, we verify the case for $\left|V\right(H\left)\right|=m$. Let u be a vertex of H and $H^{\prime }=H-\left\{u\right\}$. Clearly, $|V\left(H^{\prime }\right)|=|V\left(H\right)|-1$, and suppose $X^{\prime }$ is a scattering set of $H^{\prime }$, then $X=X^{\prime }\cup u$ is a scattering set of H. This means $\left|X\right|=|X^{\prime }|+1$. By the inductive assumption, we know that $|X^{\prime }|=|V\left(H^{\prime }\right)|-k+1.$

Thus, we have

$$\begin{array}{cc}\hfill \left|X\right|& =|X^{\prime }|+1=|V\left(H^{\prime }\right)|-k+1+1\hfill \\ \hfill & =\left|V\right(H\left)\right|-1-k+1+1\hfill \\ \hfill & =\left|V\right(H\left)\right|-k+1.\hfill \end{array}$$

This completes the proof. □

The following corollary follows from the definition of scattering number and Lemma 1 imediately.

Corollary 1. 

Let $H=(V,E)$ be a complete k-uniform hypergraph with $\left|V\right|=n$. Then,

$$s\left(H\right)=2(k-1)-n.$$

By Corollary 1, for any complete k-uniform hypergraph with $\left|V\right|=n(n\ge k\ge 3)$, we obtain the bound of scattering number for H as follows.

Theorem 1. 

Let $H=(V,E)$ be a complete k-uniform hypergraph with $\left|V\right|=n$. Then,

$$4-n\le s\left(H\right)\le n-2.$$

Remark 1. 

The bounds in Theorem 1 are the best possible. The upper bound can meet at a complete k-uniform hypergraph with $k=n$. The lower bound can meet at a complete k-uniform hypergraph with $k=3$.

If H is a trivial hypergraph or an empty hypergraph, we specify that the scattering number of H is the order number of H, which leads to the following corollary:

Corollary 2. 

Let $H=(V,E)$ be a complete k-uniform hypergraph with $\left|V\right|=n$. If $V^{\prime }$ is any subset of V, then

$$s\left(H\right)\ge s(H-V^{\prime })-\left|V^{\prime }\right|.$$

Proof. 

If $H-V^{\prime }$ is connected, $H-V^{\prime }$ is also a complete k-uniform hypergraph and its order is $|V\setminus V^{\prime }|$. Let $S^{\prime }$ be a scattering set of $H-V^{\prime }$; then, $V^{\prime }\cup S^{\prime }$ is a scattering set of H. Hence,

$$\begin{array}{cc}\hfill s(H-V^{\prime })& =\omega [(H-V^{\prime })-S^{\prime }]-\left|S^{\prime }\right|\hfill \\ \hfill & =\omega [H-(V^{\prime }\cup S^{\prime })]-|V^{\prime }\cup S^{\prime }|+|V^{\prime }|\hfill \\ \hfill & =s\left(H\right)+|V^{\prime }|.\hfill \end{array}$$

If $H-V^{\prime }$ is not connected, then $H-V^{\prime }$ be a trivial hypergraph, which shows that $|V^{\prime }|\ge n-(k-1)$. Hence,

$$s(H-V^{\prime })-|V^{\prime }|=n-2|V^{\prime }|\le n-2[n-(k-1)]=2(k-1)-n=s\left(H\right).$$

The proof is completed. □

The incidence graph of a hypergraph $H=(V,E)$ can be considered a general graph with a special structure, and it is denoted as $G\left(H\right)$. It is a bipartite graph with a vertex set $S=V\cup E$, where $v\in V$ and $e\in E$ are adjacent if and only if $v\in E$.

Lemma 2. 

Let $H=(V,E)$ be a hypergraph. Then, ${\sum }_{v\in V}d\left(v\right)={\sum }_{e\in E}d\left(e\right).$

Proof. 

Let $G\left(H\right)$ be the incidence graph of H. We sum the degrees in the part E and in the part V in $G\left(H\right)$. Since the sum of the degrees in these two parts is equal to each other, we obtain the result. □

To each hypergraph $H=(V,E)$ with $V=\{v_1,v_2,\cdots ,v_n\}$ and $E=\{e_1,e_2,\cdots ,e_m\}$, there corresponds a hypergraph $H^{*}=(V^{*},E^{*})$ whose vertices are points $v_1^{*},v_2^{*},\cdots ,v_m^{*}$ (that, respectively, represent $e_1,e_2,\cdots ,e_m$) and whose edges are sets $e_1^{*},e_2^{*},\cdots ,e_n^{*}$ (that, respectively, represent $v_1,v_2,\cdots ,v_n$), where, for all j, $e_j^{*}=\left\{v_i^{*}\right|i\le m,v_j\in e_i\}$. Thus, $e_j^{*}\ne \varnothing$ and ${\bigcup }_j{e}_j^{*}=V^{*}$, and $H^{*}$ is a hypergraph, which is called the dual hypergraph of H.

Lemma 3 

([22]). Let $H=(V,E)$ be a non-empty hypergraph and $H^{*}=(V^{*},E^{*})$ be its dual. Then, incidence graph $G\left(H\right)$ and $G\left(H^{*}\right)$ are isomorphic graphs.

A set $B\subseteq V$ is a transversal if it meets every hyperedge, i.e., for all $e\in E$, $B\cap V\left(e\right)\ne \varnothing$. The minimum cardinality of a transversal is the transversal number. It is denoted by $\gamma \left(H\right)$. A hyperedge cover is a subset of hyperedges: $\{e_j:e_j\in E,{\bigcup }_{e_j\in E}{e}_j=V\}$. The hyperedge covering number $\rho \left(H\right)$ is the minimum cardinality of a hyperedge cover.

Lemma 4 

([22]). Let $H=(V,E)$ be a hypergraph without isolation. Then,

$$\rho \left(H\right)=\gamma \left(H^{*}\right).$$

Lemma 5. 

Let $H=(V,E)$ be a complete k-uniform hypergraph with $\left|V\right|=n$. Then, the hyperedge covering number of H

$$\rho \left(H\right)=\lceil \frac{n}{k}\rceil .$$

Proof. 

Let Y be a hyperedge cover of H. By the definition of hyperedge cover, we have

$$n-\left|Y\right|k\le 0.$$

Thus,

$$\left|Y\right|\ge \frac{n}{k}.$$

Therefore, the minimum cardinality of a hyperedge cover is $\lceil \frac{n}{k}\rceil .$

The proof is completed. □

Lemma 6. 

Let $X^{*}$ be a scattering set of dual hypergraph $H^{*}$ of complete k-uniform hypergraph $H=(V,E)$ with $\left|V\right|=n$. Then,

$$|X^{*}|=\gamma \left(H^{*}\right).$$

Proof. 

If $|X^{*}|<\gamma \left(H^{*}\right)$, then $\omega (H^{*}-X^{*})<\left(\genfrac{}{}{0pt}{}{n}{k}\right)-\left(\genfrac{}{}{0pt}{}{n-1}{k-1}\right)-\left|X^{*}\right|$,

$$sc\left(X^{*}\right)=\omega (H^{*}-X^{*})-|X^{*}|<\left(\genfrac{}{}{0pt}{}{n}{k}\right)-\left(\genfrac{}{}{0pt}{}{n-1}{k-1}\right)-2|X^{*}|<\left(\genfrac{}{}{0pt}{}{n}{k}\right)-2\left|X^{*}\right|;$$

If $|X^{*}|>\gamma \left(H^{*}\right)$, then $\omega (H^{*}-X^{*})<\left(\genfrac{}{}{0pt}{}{n}{k}\right)-\left|X^{*}\right|$,

$$sc\left(X^{*}\right)=\omega (H^{*}-X^{*})-|X^{*}|<\left(\genfrac{}{}{0pt}{}{n}{k}\right)-2\left|X^{*}\right|;$$

If $|X^{*}|=\gamma \left(H^{*}\right)$, then $\omega (H^{*}-X^{*})=\left(\genfrac{}{}{0pt}{}{n}{k}\right)-\left|X^{*}\right|$,

$$sc\left(X^{*}\right)=\omega (H^{*}-X^{*})-|X^{*}|=\left(\genfrac{}{}{0pt}{}{n}{k}\right)-2\left|X^{*}\right|.$$

By the definition of scattering set, we have $|X^{*}|=\gamma \left(H^{*}\right)$.

This completes the proof. □

By the definition of scattering number and Lemmas 4, 5 and 6, we can obtain the scattering number of dual hypergraph $H^{*}$ of complete k-uniform hypergraph $H=(V,E)$.

Theorem 2. 

Let $H^{*}=(V^{*},E^{*})$ be a dual hypergraph of complete k-uniform hypergrah $H=(V,E)$ with $\left|V\right|=n$. Then,

$$s\left(H^{*}\right)=\left(\genfrac{}{}{0pt}{}{n}{k}\right)-2\lceil \frac{n}{k}\rceil .$$

3. The Scattering Number of Complete Bipartite $\mathit{k}$-Uniform Hypergraph

We subsequently consider the scattering number of a complete bipartite k-uniform hypergraph. Let H be a hypergraph with vertex set V. H is called bipartite if V can be partitioned into two subsets $V_1$ and $V_2$ such that $e\cap V_1\ne \varnothing$ and $e\cap V_2\ne \varnothing$ for any $e\in H$. Furthermore, if $\left|e\right|=k$ for any $e\in H$, then we call H a bipartite k-uniform hypergraph, which is written $H=(V_1,V_2;E)$. H is called the complete bipartite k-uniform hypergraph with vertex set $V=V_1\cup V_2$ if $E=\{e:e\subseteq V,|e|=k$ and $e\cap V_i\ne \varnothing$ for $i=1,2\}$.

Lemma 7. 

Let $H=(V_1,V_2;E)$ be a complete bipartite k-uniform hypergraph. Then,

$$\begin{array}{c}\hfill s\left(H\right)=\left\{\begin{array}{cc}\mid |V_1|-|V_2|\mid ,\hfill & \mathrm{if}3\le k\le max\left\{\right|V_1|,|V_2\left|\right\};\hfill \\ 2(k-1)-\left(\right|V_1|+|V_2\left|\right),\hfill & \mathrm{if}max\left\{\right|V_1|,|V_2\left|\right\}<k\le |V_1|+|V_2|.\hfill \end{array}\right.\end{array}$$

Proof. 

Suppose the partite sets of H are $V_1,V_2$; clearly, $k\le |V_1|+|V_2|$, and if not, H is a trival hypergraph. We will complete the proof in two cases.

Case 1. If $k\le max\left\{\right|V_1|,|V_2\left|\right\}$, then $X^{\prime }=V_i$ for $i=1,2$ is a vertex cut of H.

If $X^{\prime }=V_1$, then $\omega (H-X^{\prime })=|V_2|,sc\left(X^{\prime }\right)=|V_2|-|V_1|;$

If $X^{\prime }=V_2$, then $\omega (H-X^{\prime })=|V_1|,sc\left(X^{\prime }\right)=|V_1|-|V_2|.$

Therefore, $s\left(H\right)\ge max\left\{\right|V_2|-|V_1|,|V_1|-|V_2\left|\right\}.$

For any vertex cut $X^{″}\supset V_i$ for $i=1,2$, the following apply.

If $X^{″}\supset V_1$, then

$$\omega (H-X^{″})-|X^{″}|<|V_2|-|V_1|\le max\{|V_2|-|V_1|,|V_1|-|V_2\left|\right\};$$

If $X^{″}\supset V_2$, then

$$\omega (H-X^{″})-|X^{″}|<|V_1|-|V_2|\le max\{|V_2|-|V_1|,|V_1|-|V_2\left|\right\}.$$

For any vertex set $X^{‴}\subset V_i$ for $i=1,2$, according to the structure of a complete bipartite k-uniform hypergraph, we have $\omega (H-X^{‴})=1$.

Thus

$$s\left(H\right)\le max\left\{\right|V_2|-|V_1|,|V_1|-|V_2\left|\right\}.$$

This implies that

$$s\left(H\right)=\left|\right|V_1|-|V_2\left|\right|.$$

Case 2. If $max\left\{\right|V_1|,|V_2\left|\right\}<k\le |V_1|+|V_2|$, then H is a complete k-uniform hypergraph, by Corollary 1,

$$s\left(H\right)=2(k-1)-\left(\right|V_1|+|V_2\left|\right).$$

This completes the proof. □

From the proof of Lemma 7, it is clear that

Theorem 3. 

If $H=(V_1,V_2;E)$ is a complete bipartite k-uniform hypergraph, then

$$0\le s\left(H\right)\le k-2.$$

Remark 2. 

The bounds in Theorem 3 are also the best possible. The upper bound can meet at the complete bipartite k-uniform hypergraph with either $|V_1|+|V_2|=k$. The lower bound can meet at the complete bipartite k-uniform hypergraph with $|V_1|=|V_2|\ge k$.

From the proof of Lemma 7, we can obtain the following result for the scattering set of complete bipartite k-uniform hypergraph $H=(V_1,V_2;E)$.

Corollary 3. 

Let X be a scattering set of complete bipartite k-uniform hypergraph $H=(V_1,V_2;E)$. Then,

$$\begin{array}{c}\hfill \left|X\right|=\left\{\begin{array}{cc}min\left\{\right|V_1|,|V_2\left|\right\},\hfill & if3\le k\le max\left\{\right|V_1|,|V_2\left|\right\};\hfill \\ |V_1|+|V_2|-k+1,\hfill & ifmax\left\{\right|V_1|,|V_2\left|\right\}<k\le |V_1|+|V_2|.\hfill \end{array}\right.\end{array}$$

In the complete bipartite k-uniform hypergraph $H=(V_1,V_2;E)$, it has been known that when $max\left\{\right|V_1|,|V_2\left|\right\}<k\le |V_2|+|V_2|$, H is also a complete k-uniform hypergraph with vertex set $V=V_1\cup V_2$, and for any subset $V^{\prime }$ of V, the relation of scattering number between hypergraph $H-V^{\prime }$ and H is shown in Corollary 2. For the case of $3\le k\le max\left\{\right|V_2|,|V_1\left|\right\}$, the following relationship of scattering number between hypergraphs $H-V^{\prime }$ and H holds.

Theorem 4. 

Let $H=(V_1,V_2;E)$ be a complete bipartite k-uniform hypergraph and $3\le k\le max\left\{\right|V_1|,|V_2\left|\right\}$, $V^{\prime }$ be a any subset of $V_1\cup V_2$. Then

$$s(H-V^{\prime })=sgn\left(x\right)·[s\left(H\right)+|V^{\prime }\cap V_1|-|V^{\prime }\cap V_2\left|\right],$$

where $sgn\left(x\right)$ is sign function, $x=|V_2\setminus (V^{\prime }\cap V_2)|-|V_1\setminus (V^{\prime }\cap V_1)|$.

Proof. 

Let $V^{\prime }$ be a any subset of $V_1\cup V_2$; without loss of generality, we assume $|V_1|\le |V_2|$ and distinguish three cases to complete the proof.

Case 1. If $V^{\prime }=V_1$, then $H-V^{\prime }$ is a trivial hypergraph. Thus,

$$s(H-V^{\prime })=|V_2|=|V_2|-|V_1|+|V_1|=s\left(H\right)+|V_1|=s\left(H\right)+\left|V^{\prime }\right|.$$

Similarly, if $V^{\prime }=V_2$, then

$$s(H-V^{\prime })=|V_1|=-(|V_2|-|V_1\left|\right)+|V_2|=-s\left(H\right)+|V_2|=-s\left(H\right)+\left|V^{\prime }\right|.$$

Case 2. If $V^{\prime }\subset V_i(i=1,2)$, then $H-V^{\prime }$ is also a complete bipartite k-uniform hypergraph. Thus,

if $V^{\prime }\subset V_1$, then $V_1\setminus V^{\prime }$ is scattering set of $H-V^{\prime }$. Thus,

$$s(H-V^{\prime })=|V_2|-|V_1\setminus V|=|V_2|-|V_1|+|V^{\prime }|=s\left(H\right)+\left|V^{\prime }\right|.$$

If $V^{\prime }\subset V_2$, then either $|V_1|<|V_2\setminus V^{\prime }|$ or $|V_1|\ge |V_2\setminus V^{\prime }|$ holds.

If $|V_1|<|V_2\setminus V^{\prime }|$, then

$$\begin{array}{cc}\hfill s(H-V^{\prime })& =|V_2\setminus V^{\prime }|-|V_1|\hfill \\ \hfill & =|V_2|-|V_1|-|V^{\prime }|=s\left(H\right)-|V^{\prime }|.\hfill \end{array}$$

If $|V_1|\ge |V_2\setminus V^{\prime }|$, then

$$\begin{array}{cc}\hfill s(H-V^{\prime })& =|V_1|-|V_2\setminus V^{\prime }|\hfill \\ \hfill & =|V_1|-|V_2|+|V^{\prime }|=-s\left(H\right)+|V^{\prime }|.\hfill \end{array}$$

Case 3. If $V^{\prime }\cap V_1\ne \varnothing$ and $V^{\prime }\cap V_2\ne \varnothing$, we suppose $|V_1\setminus (V_1\cap V^{\prime })|\le |V_2\setminus (V_2\cap V^{\prime })|$, then

$$\begin{array}{cc}\hfill s(H-V^{\prime })& =|V_2\setminus (V_2\cap V^{\prime })|-|V_1\setminus (V_1\cap V^{\prime })|\hfill \\ \hfill & =|V_2|-|V_2\cap V^{\prime }|-|V_1|+|V_1\cap V\prime |\hfill \\ \hfill & =|V_2|-|V_1-|V_2\cap V^{\prime }|+|V_1\cap V\prime |\hfill \\ \hfill & =s\left(H\right)-|(V_2\cap V^{\prime })|+|(V_1\cap V^{\prime })|.\hfill \end{array}$$

if $|V_1\setminus (V_1\cap V^{\prime })|>|V_2\setminus (V_2\cap V^{\prime })|$, then

$$\begin{array}{cc}\hfill s(H-V^{\prime })& =|V_1\setminus (V_1\cap V^{\prime })|-|V_2\setminus (V_2\cap V^{\prime })|\hfill \\ \hfill & =|V_1|-|V_1\cap V^{\prime }|-|V_2|+|V_2\cap V\prime |\hfill \\ \hfill & =|V_1|-|V_2-|V_1\cap V^{\prime }|+|V_2\cap V^{\prime }|\hfill \\ \hfill & =-s\left(H\right)-|(V_1\cap V^{\prime })|+|(V_2\cap V^{\prime })|.\hfill \end{array}$$

We use the sign function $sgn\left(x\right)$ and take $x=|V_2\setminus (V_2\cap V^{\prime })|-|V_1\setminus (V_1\cap V^{\prime })|$.

To sum it up, we have $s(H-V^{\prime })=sgn\left(x\right)·[s\left(H\right)+|V^{\prime }\cap V_1|-|V^{\prime }\cap V_2\left|\right].$

The proof is completed. □

According to the different cases of the subset $V^{\prime }$, we give the following example.

Example 1. 

Let $H=(V_1,V_2;E)$ be a complete bipartite 3-uniform hypergraph. $V_1=\{v_1,v_2,v_3\},V_2=\{u_1,u_2,u_3,u_4,u_5\}$. By Lemma 7, we have $s\left(H\right)=|V_2|-|V_1|=5-3=2.$

1.

If $V^{\prime }=V_i(i=1,2).$

1.1.

$V^{\prime }=V_1$, then $H-V^{\prime }$ is a trivial hypergraph. Thus,

$$s(H-V^{\prime })=|V_2|=5=s\left(H\right)+\left|V^{\prime }\right|;$$

1.2.

$V^{\prime }=V_2$, then $H-V^{\prime }$ is also a trivial hypergraph. Thus,

$$s(H-V^{\prime })=|V_1|=3=-s\left(H\right)+\left|V^{\prime }\right|;$$

2.

If $V^{\prime }\subset V_i(i=1,2).$

2.1.

$V^{\prime }=\{v_1,v_2\}\subset V_1$, then $H-V^{\prime }$ is a complete bipartite 3-uniform hypergraph and $|V_1\setminus V\prime |<|V_2|$. Thus,

$$s(H-V^{\prime })=|V_2|-|V_1\setminus V\prime |=4=s\left(H\right)+\left|V^{\prime }\right|;$$

2.2.

$V^{\prime }=\left\{u_1\right\}\subset V_2$, then $H-V^{\prime }$ is a complete bipartite 3-uniform hypergraph and $|V_2\setminus V\prime |>|V_1|$. Thus,

$$s(H-V^{\prime })=|V_2\setminus V\prime |-|V_1|=1=s\left(H\right)-\left|V^{\prime }\right|;$$

2.3.

$V^{\prime }=\{u_1,u_2,u_3\}\subset V_2$, then $H-V^{\prime }$ is a complete bipartite 3-uniform hypergraph and $|V_2\setminus V\prime |<|V_1|$. Thus,

$$s(H-V^{\prime })=|V_1|-|V_2\setminus V\prime |=1=-s\left(H\right)+\left|V^{\prime }\right|;$$

3.

If $V^{\prime }\cap V_1\ne \varnothing$ and $V^{\prime }\cap V_2\ne \varnothing .$

3.1.

$|V_1\setminus (V_1\cap V\prime |<|V_2\setminus (V_2\cap V\prime |$.

3.1.1.

$V^{\prime }=\{v_1,u_1,u_2\}$, then $H-V^{\prime }$ is a complete bipartite 3-uniform hypergraph. Thus,

$$s(H-V^{\prime })=|V_2\setminus (V_2\cap V^{\prime })|-|V_1\setminus (V_1\cap V^{\prime })|=3-2=1=-s\left(H\right)+\left|V^{\prime }\right|;$$

3.1.2.

$V^{\prime }=\{v_1,v_2,v_3,u_1,u_2\}$, then $H-V^{\prime }$ is a trivial hypergraph. Thus,

$$s(H-V^{\prime })=|V_2\setminus (V_2\cap V^{\prime })|-|V_1\setminus (V_1\cap V^{\prime })|=3-0=-s\left(H\right)+\left|V^{\prime }\right|;$$

3.2.

If $|V_1\setminus (V_1\cap V\prime |>|V_2\setminus (V_2\cap V\prime |.$

3.2.1.

$V^{\prime }=\{v_1,u_1,u_2,u_3,u_4\}$, then $H-V^{\prime }$ is a complete bipartite 3-uniform hypergraph. Thus,

$$\begin{array}{cc}\hfill s(H-V^{\prime })& =|V_1\setminus (V_1\cap V^{\prime })|=|V_1\setminus (V_1\cap V^{\prime })|-0\hfill \\ \hfill & =|V_1\setminus (V_1\cap V^{\prime })|-|V_2\setminus (V_2\cap V^{\prime })|=1\hfill \\ \hfill & =-s\left(H\right)-|(V_1\cap V^{\prime })|+|(V_2\cap V^{\prime })|;\hfill \end{array}$$

3.2.2.

$V^{\prime }=\{v_1,u_1,u_2,u_3,u_4,u_5\}$, then $H-V^{\prime }$ is a trivial hypergraph. Thus,

$$\begin{array}{cc}\hfill s(H-V^{\prime })& =|V_1\setminus (V_1\cap V^{\prime })|=|V_1\setminus (V_1\cap V^{\prime })|-0\hfill \\ \hfill & =|V_1\setminus (V_1\cap V^{\prime })|-|V_2\setminus (V_2\cap V^{\prime })|=2\hfill \\ \hfill & =-s\left(H\right)-|(V_1\cap V^{\prime })|+|(V_2\cap V^{\prime })|.\hfill \end{array}$$

4. The Relations of Scattering Number between Complete $\mathit{k}$-Uniform Hypergraph and Complete Bipartite $\mathit{k}$-Uniform Hypergraph

In this section, we will derive the relations of scattering number between a complete k-uniform hypergraph and complete bipartite k-uniform.

Let $V_1,V_2$ be two non-empty vertex sets and $|V_1|+|V_2|\ge k$, where $V_1\cap V_2=\varnothing$. $H=(V_1\cup V_2,E)$ is a complete k-uniform hypergraph, $H\left[V_1\right]$ and $H\left[V_2\right]$ are two hypergraphs induced by $V_1$ and $V_2$, respectively. $H^{\prime }=(V_1,V_2;E^{\prime })$ is a complete bipartite k-uniform hypergraph.

For the hypergraphs $H=(V_1\cup V_2,E)$ and $H^{\prime }=(V_1,V_2;E^{\prime })$ specified as above, by Corollary 1 and Lemma 7, it can be seen that the following relationship of scattering number exists between H and $H^{\prime }$.

Theorem 5. 

Let H and $H^{\prime }$ be two hypergraphs specified as above. Then,

$$s\left(H\right)\le s\left(H^{\prime }\right).$$

Proof. 

We will complete the proof in three cases. Without loss of generality, we assume $|V_1|\le |V_2|$.

Case 1. If $|V_1|\le |V_2|<k$, then it is obvious that $H=H^{\prime }$. Thus,

$$\begin{array}{c}\hfill s\left(H\right)-s\left(H^{\prime }\right)=0\end{array}$$

Case 2. If $|V_1|<k\le |V_2|$, then

$$\begin{array}{cc}\hfill s\left(H\right)-s\left(H^{\prime }\right)& =2(K-1)-\left(\right|V_1|+|V_2\left|\right)-\left(\right|V_2|-|V_1\left|\right)\hfill \\ \hfill & =2[(k-1)-|V_2\left|\right]<0.\hfill \end{array}$$

Case 3. If $k\le |V_1|\le |V_2|$, then

$$\begin{array}{cc}\hfill s\left(H\right)-s\left(H^{\prime }\right)& =2(K-1)-\left(\right|V_1|+|V_2\left|\right)-\left(\right|V_2|-|V_1\left|\right)\hfill \\ \hfill & =2[(k-1)-|V_2\left|\right]<0.\hfill \end{array}$$

Therefore, $s\left(H\right)\le s\left(H^{\prime }\right).$ □

Theorem 6. 

Let $H\left[V_1\right],H\left[V_2\right]$ and H be three hypergraphs specified as above. Then,

$$s\left(H\left[V_1\right]\right)+s\left(H\left[V_2\right]\right)>s\left(H\right).$$

Proof. 

Similar to the classification when proving Theorem 5, we also complete the proof in three cases.

Case 1. If $|V_1|\le |V_2|<k$, then $H\left[V_1\right]$ and $H\left[V_1\right]$ are both trivial hypergraph. Thus,

$$\begin{array}{c}\hfill s\left(H\left[V_1\right]\right)+s\left(H\left[V_2\right]\right)=|V_1|+|V_2|\ge k>k-2\\ \hfill s\left(H\right)=2(k-1)-\left(\right|V_1|+|V_2\left|\right)\le 2(k-1)-k=k-2\end{array}$$

Therefore, $s\left(H\left[V_1\right]\right)+s\left(H\left[V_2\right]\right)>s\left(H\right).$

Case 2. If $|V_1|<k\le |V_2|$, then

$$\begin{array}{cc}\hfill s\left(H\left[V_1\right]\right)+s\left(H\left[V_2\right]\right)& =|V_1|+2(k-1)-|V_2|\hfill \\ \hfill & =2(k-1)-\left(\right|V_2|+|V_1\left|\right)+2|V_1|\hfill \\ \hfill & =s\left(H\right)+2|V_1|>s\left(H\right)\hfill \end{array}$$

Case 3. If $k\le |V_1|\le |V_2|$, then $H\left[V_1\right]$ and $H\left[V_1\right]$ are both complete k-uniform hypergraphs. Thus,

$$\begin{array}{cc}\hfill s\left(H\left[V_1\right]\right)+s\left(H\left[V_2\right]\right)& =2(k-1)-|V_1|+2(k-1)-|V_2|\hfill \\ \hfill & =4(k-1)-\left(\right|V_1|+|V_2\left|\right)\hfill \\ \hfill & >2(k-1)-\left(\right|V_1|+|V_2\left|\right)\hfill \\ \hfill & =s\left(H\right).\hfill \end{array}$$

□

Similar to the proof of Theorem 6, the following theorem follows from Corollary 1 and Lemma 7 imediately.

Theorem 7. 

Let $H\left[V_1\right],H\left[V_2\right]$ and $H^{\prime }$ be three hypergraphs specified as above. Then,

$$\begin{array}{c}\hfill s\left(H^{\prime }\right)=\left\{\begin{array}{cc}\mid s\left(H\left[V_1\right]\right)-s\left(H\left[V_2\right]\right)\mid ,\hfill & if|V_1|\le |V_2|<k\mathit{and}k\le |V_1|\le |V_2|;\hfill \\ 2(k-1)-[s\left(H\left[V_1\right]\right)+s\left(H\left[V_2\right]\right)],\hfill & if|V_1|<k\le |V_2|.\hfill \end{array}\right.\end{array}$$

For the hypergraphs $H\left[V_1\right],H\left[V_2\right],H^{\prime }=(V_1,V_2;E^{\prime })$ and $H=(V_1\cup V_2,E)$ as defined above, there is relationship between them $H\left[V_1\right]\cup H\left[V_2\right]\cup H^{\prime }=H$. By Theorems 5 and 6, we obtain

Theorem 8. 

Let $H\left[V_1\right],H\left[V_2\right]$, $H^{\prime }$ and H be four hypergraphs specified as above. Then,

$$s\left(H\left[V_1\right]\right)+s\left(H\left[V_2\right]\right)+s\left(H^{\prime }\right)>2s\left(H\right).$$

5. Conclusions

The scattering number is a crucial parameter for quantifying the vulnerability of a network. Despite its significance, there exists limited research on the vulnerability of hypernetworks. This paper addresses this gap by utilizing the scattering number to assess the vulnerability of uniform hypergraphs. Notably, many vulnerability parameters for hypergraphs remain unexplored, presenting an avenue for further investigation, particularly in the realm of non-uniform hypergraphs. Our work aims to stimulate additional research in this field.