#  On Ideal Secret-Sharing Schemes for $k$-homogeneous access structures

Younjin Kim Extremal Combinatorics and Probability Group (ECOPRO), Institute
for Basic Science (IBS), Daejeon, South Korea. Email<EMAIL_ADDRESS>Y.K. was supported by the Institute for Basic Science (IBS-R029-C4). Jihye
Kwon Department of Mathematics, Ewha Womans University, Seoul, South Korea.
Email<EMAIL_ADDRESS>J.K. was supported by the National Research
Foundation of Korea(NRF) grant funded by the Ministry of Education (No.
2019R1A6A1A11051177) Hyang-Sook Lee Department of Mathematics, Ewha Womans
University, Seoul, South Korea. Email<EMAIL_ADDRESS>H.L. was supported by
the National Research Foundation of Korea(NRF) grant funded by the Ministry of
Education (No. 2019R1A6A1A11051177) and partially by the Korea
Government(MSIT) (No. NRF-2021R1A2C1094821).

###### Abstract

A $k$-uniform hypergraph is a hypergraph where each $k$-hyperedge has exactly
$k$ vertices. A $k$-homogeneous access structure is represented by a
$k$-uniform hypergraph $\mathcal{H}$, in which the participants correspond to
the vertices of hypergraph $\mathcal{H}$. A set of vertices can reconstruct
the secret value from their shares if they are connected by a $k$-hyperedge,
while a set of non-adjacent vertices does not obtain any information about the
secret. One parameter for measuring the efficiency of a secret sharing scheme
is the information rate, defined as the ratio between the length of the secret
and the maximum length of the shares given to the participants. Secret sharing
schemes with an information rate equal to one are called ideal secret sharing
schemes. An access structure is considered ideal if an ideal secret sharing
scheme can realize it. Characterizing ideal access structures is one of the
important problems in secret sharing schemes. The characterization of ideal
access structures has been studied by many authors [5, 6, 7, 8, 10, 14, 15].
In this paper, we characterize ideal $k$-homogeneous access structures using
the independent sequence method. In particular, we prove that the reduced
access structure of $\Gamma$ is an $(k,n)$-threshold access structure when the
optimal information rate of $\Gamma$ is larger than $\frac{k-1}{k}$, where
$\Gamma$ is a $k$-homogeneous access structure satisfying specific criteria.

##  1 Introduction

A secret sharing scheme is a tool utilized in numerous cryptographic
protocols. It involves a dealer who possesses a secret, a set of $n$
participants, and a collection $\mathcal{F}$ of subsets of participants
defined as the access structure. A secret sharing scheme for $\mathcal{F}$ is
a method in which the dealer distributes shares of a secret value $k$ among
the $n$ participants so that any subset within $\mathcal{F}$ can reconstruct
the secret value $k$ from their shares, while any subset not in $\mathcal{F}$
cannot reveal any information about the secret value $k$. The qualified
subsets in the secret sharing scheme are defined as the subsets of
participants capable of reconstructing the secret value $k$ from their shares.
A collection of qualified subsets of participants is referred to as the access
structure of the secret sharing scheme. In contrast, the unqualified subsets
or forbidden subsets in the secret sharing scheme are defined as the subsets
of participants who cannot obtain any information about the secret value $k$
from their shares.

In 1979, Shamir [13] introduced a $(t,n)$-threshold secret sharing scheme as
the pioneering work in secret sharing. In this scheme, the qualified subsets
consist of all subsets with at least $t$ participants from a set of $n$
participants, and the size of each share is equal to the size of the secret.
This implies that the $(t,n)$-threshold secret sharing scheme is determined by
the basis consisting of all subsets with exactly $t$ distinct participants
from a set of $n$ participants.

A hypergraph is a generalization of a graph in which hyperedges may connect
more than two vertices. A $k$-uniform hypergraph is one in which each
hyperedge has exactly $k$ vertices. An access structure is a $k$-uniform
hypergraph access structure, denoted by $\mathcal{H}$, if the set of vertices
connected by a $k$-hyperedge can reconstruct the secret, and the set of non-
adjacent vertices in $\mathcal{H}$ does not reveal any information about the
secret. Access structures of this type are also called $k$-homogeneous. A
$k$-homogeneous access structure is determined by a family of minimal
qualified subsets, each consisting of exactly $k$ different participants, or
$k$-uniform hypergraphs $\mathcal{H}(V,E)$, where $V$ is a vertex set and
$E\subseteq 2^{V}$ is the edge set of hyperedges of cardinality $k$. Several
authors have constructed secret sharing schemes for $k$-homogeneous access
structures using various techniques.

One way to measure the efficiency of a secret sharing scheme is to use the
information rate, defined as the ratio between the length of the secret and
the maximum length of the shares given to the participants. Since the length
of any share is greater than or equal to the length of the secret in a secret
sharing scheme, the information rate cannot exceed one. Secret sharing schemes
with an information rate equal to one are called ideal secret sharing schemes.
For example, Shamir’s threshold secret sharing scheme is ideal.

An access structure is ideal if there exists an ideal secret sharing scheme to
implement it. The characterization of ideal access structures is an important
issue in secret sharing schemes, which has been studied by numerous authors
[5, 6, 7, 8, 10, 11, 14, 15]. In 1992, Stinson [14] exactly characterized all
ideal $2$-homogeneous access structures. In 2007, Martí-Farré and Padró [11]
characterized all ideal $3$-homogeneous access structures in which the number
of minimal qualified subsets contained in any set of four participants is not
equal to three. Later, in 2009, Martí-Farré and Padró [10] also characterized
ideal rank-three access structures in some cases. Also, in 2009, Tassa and Dyn
[15] studied an ideal secret sharing scheme that realizes compartmented access
structures using bivariate interpolation. Recently, in 2021, Janbaz and
Bagherpour [7] characterized ideal graph-based $3$-homogeneous access
structures. In this paper, we characterize the ideal $k$-homogeneous access
structures by utilizing the independent sequence method.

######  Theorem 1.1.

Suppose $\Gamma$ is a $k$-homogeneous access structure on a set of
participants $\mathcal{P}$ such that the number of minimal qualified subsets
contained in any set of $k+1$ participants is not equal to $1$ and $k$,
containing $k+1$. Then the following conditions are equivalent.

  * (1)

$\Gamma$ is a vector space access structure

  * (2)

$\Gamma$ is an ideal access structure

  * (3)

$\rho^{*}(\Gamma)>\frac{k-1}{k}$

  * (4)

The reduced access structure of $\Gamma$ is an $(k,n)$-threshold access
structure.

Our paper is organized as follows. In Section 2, we introduce the definitions
of an Ideal Secret Sharing Scheme and the Independent Sequence Method. In
Section 3, we present several access structures related to this paper. In
Section 4, we present the necessary results and lemmas required for proving
our main theorem. Finally, in Section 5, we provide the proof of Theorem 1.1.

##  2 Secret Sharing Scheme

A secret sharing scheme consists of a dealer who possesses a secret, a set of
$n$ participants, and a collection $\mathcal{F}$ of subsets of participants
defined as the access structure. A secret sharing scheme for $\mathcal{F}$ is
a method by which the dealer distributes shares of a secret value $k$ to the
set of $n$ participants such that any qualified subset in $\mathcal{F}$ can
reconstruct the secret value $k$ from their shares, while any unqualified
subset not in $\mathcal{F}$ cannot reveal any information about the secret
value $k$.

###  2.1 Ideal Secret Sharing Schemes

One parameter used to measure the efficiency of a secret sharing scheme is the
information rate, which is defined as the ratio between the length of the
secret and the maximum length of the shares given to the participants, as
follows.

######  Definition 2.1 (information rate).

Let $\mathcal{P}$ be the set of all participants, $\mathcal{S}$ be the set of
all secret keys, and $K(p)$ be the set of all possible shares given to a
participant $p\in\mathcal{P}$. In the secret sharing scheme $\mathcal{F}$, the
information rate, denoted by $\rho(\mathcal{F})$, is defined as

$\rho(\mathcal{F})=\frac{\log|S|}{\max_{p\in\mathcal{P}}\log|K(p)|}\ .$

Since the length of any share is greater than or equal to the length of the
secret in the secret sharing scheme, the information rate cannot be greater
than one. Therefore, $\rho(\mathcal{F})=1$ is the optimal situation. When
designing a secret sharing scheme for the given access structure $\Gamma$, we
may try to maximize the information rate, as defined below.

######  Definition 2.2 (optimal information rate).

In the secret sharing scheme $\mathcal{F}$, the optimal information rate of
the access structure $\Gamma$, denoted by $\rho^{*}(\Gamma)$, is defined as

$\rho^{*}(\Gamma)=\sup(\rho(\mathcal{F}))$

where the supremum is taken over all possible secret sharing schemes
$\mathcal{F}$ with access structure $\Gamma$. Of course, the optimal
information rate of an ideal access structure is equal to one.

Secret sharing schemes with an information rate equal to one are called ideal
secret sharing schemes. An access structure is ideal if there exists an ideal
secret sharing scheme that realizes it. Characterizing ideal access structures
and providing bounds on the optimal information rate are two important
problems in the secret sharing schemes. These problems have been studied
extensively in several particular families of access structures by many
authors [5, 6, 7, 8, 10, 11, 14, 15].

###  2.2 Independent Sequence Methods

To prove the main theorem, we will utilize the independent sequence method,
which we introduce in this section. In 1997, Blundo, Santis, Simone, and
Vaccaro [4] presented the independent sequence method as a way to find upper
bounds on the optimal information rate. Later, in 2000, Padró and Sáez [12]
introduced a slight generalization of this method, as follows.

######  Definition 2.3 (Independent Sequence).

Let $\Gamma$ be an access structure on a set of participants $\mathcal{P}$. A
sequence of non-empty sets $B_{1},B_{2},\cdots,B_{m}$, where

$\emptyset\neq B_{1}\subset B_{2}\subset\cdots\subset
B_{m}\subset\mathcal{P},$

is called independent if

  * 1.

$B_{m}\not\in\Gamma$

  * 2.

there exist $X_{1},X_{2},\cdots,X_{m}\subset\mathcal{P}$ such that $B_{i}\cup
X_{i}\in\Gamma$ and $B_{i-1}\cup X_{i}\not\in\Gamma$ for all $i=1,2,\cdots,m$,
where $B_{0}=\emptyset$.

######  Theorem 2.4 (Independent Sequence Method [12]).

Let $\Gamma$ be an access structure on a set of participants $\mathcal{P}$.
Suppose that $\emptyset\neq B_{1}\subset B_{2}\subset\cdots\subset
B_{m}\subset\mathcal{P}$ is an independent sequence, and let
$A\subset\mathcal{P}$ be a minimum set such that
$\bigcup^{m}_{i=1}X_{i}\subset A$ which makes this sequence independent. Then,
we have

  * (i)

$\rho^{*}(\Gamma)\leq\frac{|A|}{m+1}$ if $A\in\Gamma$

  * (ii)

$\rho^{*}(\Gamma)\leq\frac{|A|}{m}$ if $A\not\in\Gamma$.

##  3 Access Structures

In a secret sharing scheme, the qualified subsets are the sets of participants
who can reconstruct the secret value from their shares. The collection of
qualified subsets of participants is called the access structure of the secret
sharing scheme. Conversely, the unqualified subsets are the sets of
participants who cannot obtain any information about the secret value from
their shares. In any secret sharing scheme, the access structure is said to be
monotone if the superset of any qualified subset is also a qualified subset,
and it is determined by the family of minimal qualified subsets of
participants. The collection of minimal qualified subsets of participants is
called the basis of the access structure. Moreover, in a secret sharing
scheme, every participant must belong to at least one minimal qualified
subset.

###  3.1 $k$-Homogeneous Access Structures

A hypergraph is a generalization of a graph in which hyperedges may connect
more than two vertices. A $k$-uniform hypergraph (or $k$-hypergraph) is a
hypergraph in which each hyperedge has exactly $k$ vertices, or is a
$k$-hyperedge. In particular, the complete $k$-uniform hypergraph on $n$
vertices has all $k$-subsets of $\\{1,2,\dots,n\\}$ as $k$-hyperedges. A
$k$-hypergraph access structure is represented by a $k$-uniform hypergraph
$\mathcal{H}$ in which the participants correspond to the vertices of
$\mathcal{H}$. A set of vertices can reconstruct the secret value from their
shares if they are connected by a $k$-hyperedge, while the set of non-adjacent
vertices receives no information on the secret. A $k$-hypergraph access
structure is also called a $k$-homogeneous access structure.

A $k$-homogeneous access structure is determined by the family of minimal
qualified subsets, each consisting of exactly $k$ participants. Recall that in
the $(k,n)$-threshold secret sharing scheme, the qualified subsets are formed
by all subsets with at least $k$ participants among the set of $n$
participants. For example, consider an access structure on a set of five
participants $P=\\{p_{1},p_{2},p_{3},p_{4},p_{5}\\}$ with minimal qualified
subsets $A_{1}=\\{p_{1},p_{2},p_{3}\\},A_{2}=\\{p_{2},p_{3},p_{4}\\}$, and
$A_{3}=\\{p_{3},p_{4},p_{5}\\}$. This access structure is not
$(3,n)$-threshold but is $3$-homogeneous. Note that there is a one-to-one
correspondence between $k$-uniform hypergraphs and $k$-homogeneous access
structures. Furthermore, complete $k$-uniform hypergraphs correspond to
$(k,n)$-threshold access structures. Many authors [1, 2, 3] have constructed
secret sharing schemes for $k$-homogeneous access structures using various
techniques.

###  3.2 Vector Space Access Structures

Let $\Gamma$ be an access structure defined on a set of participant
$\mathcal{P}$, and let $D\not\in\mathcal{P}$ be a dealer. We say that an
access structure $\Gamma$ is a vector space access structure if there exists a
function

$f:\mathcal{P}\cup{D}\rightarrow E\backslash\\{0\\}$

where $E$ is a vector space over a finite field $K$, such that if
$A\subseteq\mathcal{P}$ then $A\in\mathcal{P}$ if and only if the vector
$f(D)$ can be expressed as a linear combination of the vectors in the set
$f(A)=\\{f(p)|\ p\in A\\}$. An example of a vector space access structure is
the $(k,n)$-threshold access structure, which consists of all subsets with at
least $k$ participants among the set of $n$ participants. (See [14]). The
relationship between vector space access structures and ideal access
structures is as follows.

######  Theorem 3.1.

[9, 10] The vector space access structures are ideal.

###### Proof.

Let $\Gamma$ be a vector space access structure defined on a set of
participant $\mathcal{P}$, and let $D\not\in\mathcal{P}$ be a dealer. Then,
there exists a function $f:\mathcal{P}\cup{D}\rightarrow E\backslash\\{0\\}$,
where $E$ is a vector space over a finite field $K$, such that if
$A\subseteq\mathcal{P}$ then $A\in\mathcal{P}$ if and only if the vector
$f(D)$ can be expressed as a linear combination of the vectors in the set
$f(A)=\\{f(p)|\ p\in A\\}$. Given a secret value $k\in K$, the dealer $D$
selects an element $v\in E$ at random such that $v\cdot f(D)=k$, where $f(D)$
is a linear combination of the vectors in the set $\\{f(p)|\ p\in D\\}$. The
dealer then distributes shares $s_{p}=v\cdot f(p)$ to each participant
$p\in\mathcal{P}$. Note that the ratio between the length of the secret and
the maximum length of the shares given to the participants is one. Therefore,
we conclude that $\Gamma$ is ideal. ∎

###  3.3 Reduced Access Structures

Let $\Gamma$ be an access structure defined on a set of participant
$\mathcal{P}$. We say that an access structure $\Gamma$ is a reduced access
structure if there are no pairs of distinct participants that are equivalent.
In relation to the access structure $\Gamma$, we define the equivalence
relation $\sim$ on $\mathcal{P}$ as follows.

######  Definition 3.2 (equivalence relation).

The two participants $a,b\in\mathcal{P}$ are said to be equivalent, denoted by
$a\sim b$, if either $a=b$ or $a\neq b$ and the following two conditions are
satisfied:

  * (i)

$\\{a,b\\}\nsubseteq A$ if $A\in\Gamma_{0}$

  * (ii)

if $A\subset\mathcal{P}\backslash\\{a,b\\}$, then $A\cup\\{a\\}\in\Gamma_{0}$
if and only if $A\cup\\{b\\}\in\Gamma_{0}$

where $\Gamma_{0}$ is a family of minimal qualified subsets.

Let us define the equivalence classes, induced by $\sim$, on the set of
participants $\mathcal{P}$ as
$\mathcal{P}/\sim=\\{[a_{1}],[a_{2}],\cdots,[a_{m}]\\}$. Then, an access
structure $\Gamma_{\sim}$ on $\mathcal{P}/\sim$ can be obtained naturally from
$\Gamma$ by identifying equivalent participants. The reduced access structure
of $\Gamma$ is denoted as $\Gamma_{\sim}$, and it is isomorphic to
$\Gamma(\\{a_{1},a_{2},\cdots,a_{m}\\})$. Therefore, we have
$\rho^{*}(\Gamma_{\sim})\geq\rho^{*}(\Gamma)$. Furthermore, $\Gamma$ is a
vector space access structure if and only if $\Gamma_{\sim}$ is as well.

##  4 Lemmas

In this section, we present several lemmas to prove the main result. The
following lemma corresponds to the $s=2$ case of Lemma 4.2, and we omit its
proof. Let us define $w(Q,\Gamma)$ as the number of minimal qualified subsets
contained in the set $Q$. Let $\Omega(m,\Gamma)$ be the set of possible values
of $w(Q,\Gamma)$ with $|Q|=m$. This implies that $\Omega(m,\Gamma)$ collects
the numbers of minimal qualified subsets contained in any set of $m$
participants.

######  Lemma 4.1.

Let $\Gamma$ be a $k$-homogeneous access structure defined on a set of
participants $\mathcal{P}$ such that the number of minimal qualified subsets
contained in any set of $k+1$ participants is not equal to $k$, and
$\rho^{*}(\Gamma)>\frac{k-1}{k}$. Let us consider any $k+2$ distinct
participants $u_{1},u_{2},\cdots,u_{k},u_{k+1},v\in\mathcal{P}$ such that

$\\{u_{1},u_{2},u_{l_{1}},u_{l_{2}},\cdots,u_{l_{k-2}}\\}\in\Gamma$ (4.1)

for all $3\leq l_{1},l_{2},\cdots,l_{k-2}\leq k+1$.

Then either $w(\\{u_{1},u_{2},u_{3},\cdots,u_{k},u_{k+1}\\},\ \Gamma)=k+1$ or
$\\{u_{3},u_{4},u_{5},\cdots,u_{k},u_{k+1},v\\}\not\in\Gamma$.

######  Lemma 4.2 (General version of Lemma 4.1).

Suppose that $\Gamma$ is a $k$-homogeneous access structure defined on a set
of participants $\mathcal{P}$, such that the number of minimal qualified
subsets contained in any set of $k+1$ participants is not equal to $k$, and
$\rho^{*}(\Gamma)>\frac{k-1}{k}$. Let us consider any $k+2$ distinct
participants $u_{1},u_{2},\cdots,u_{k},u_{k+1},v\in\mathcal{P}$ such that

$\\{u_{1},u_{2},\cdots,u_{s},u_{l_{1}},u_{l_{2}},\cdots,u_{l_{k-s}}\\}\in\Gamma$
(4.2)

for all $s+1\leq l_{1},l_{2},\cdots,l_{k-s}\leq k+1$, where $s\geq 2$.

Then there exist two participants $u_{i}$ and $u_{j}$, where $1\leq i,j\leq
k+1$, such that either

$w(\\{u_{1},u_{2},u_{3},\cdots,u_{k},u_{k+1}\\},\ \Gamma)=k+1$

or

$\\{u_{m_{1}},u_{m_{2}},u_{m_{3}},\cdots,u_{m_{k-1}},v\\}\not\in\Gamma,$

where $i,j\not\in\\{m_{1},m_{2},m_{3},\cdots,m_{k-1}\\}$ and $1\leq
m_{1},m_{2},m_{3},\cdots,m_{k-1}\leq k+1$.

###### Proof.

Let us assume that
$w(\\{u_{1},u_{2},u_{3},\cdots,u_{k},u_{k+1}\\},\>\Gamma)\neq k+1$ holds. This
implies that the number of minimal qualified subsets contained in the set of
$k+1$ participants, $\\{u_{1},u_{2},u_{3},\cdots,u_{k},u_{k+1}\\}$, is not
equal to $k+1$. Also, by using the condition $k\not\in\Omega(k+1,\Gamma)$,
which means that the number of minimal qualified subsets contained in any set
of $k+1$ participants is not equal to $k$, we can conclude that there exist
$i$ and $j$ such that

$\displaystyle\\{u_{i},u_{m_{1}},u_{m_{2}},u_{m_{3}},\cdots,u_{m_{k-1}}\\}\notin\Gamma{\text{
\ and \ \ }}\\{u_{j},u_{m_{1}},u_{m_{2}},\cdots,u_{m_{k-1}}\\}\notin\Gamma.$
(4.3)

where $i,j\not\in\\{m_{1},m_{2},\cdots,m_{k-1}\\}$, $1\leq
i,j,m_{1},\cdots,m_{s-2}\leq s$, $s+1\leq m_{s-1},\cdots,m_{k-1}\leq k+1$.

Note that
$\\{u_{1},u_{2},u_{3},\cdots,u_{k},u_{k+1}\\}=\\{u_{i},u_{j},u_{m_{1}},u_{m_{2}},u_{m_{3}},\cdots,u_{m_{k-1}}\\}$.
Now we need to show that
$\\{u_{m_{1}},u_{m_{2}},u_{m_{3}},\cdots,u_{m_{k-1}},v\\}\notin\Gamma$. Let us
consider the two cases as follows:
$\\{u_{i},u_{m_{1}},u_{m_{2}},u_{m_{3}},\cdots,u_{m_{k-2}},v\\}\notin\Gamma$
or $\\{u_{i},u_{m_{1}},u_{m_{2}},u_{m_{3}},\cdots,u_{m_{k-2}},v\\}\in\Gamma$.

Case I: $\\{u_{i},u_{m_{1}},u_{m_{2}},\cdots,u_{m_{k-2}},v\\}\notin\Gamma$.

In this case, let us first prove that
$\\{u_{i},u_{m_{1}},\cdots,u_{m_{k-2}},u_{m_{k-1}},v\\}\notin\Gamma$. Let us
assume otherwise, that is, the set of $k+1$ different participants
$\\{u_{i},u_{m_{1}},\cdots,u_{m_{k-2}},u_{m_{k-1}},v\\}$ is in $\Gamma$. We
can consider the following subsets of participants $\mathcal{P}$:
$B_{1}=\\{u_{i}\\}$,$\>B_{2}=\\{u_{i},u_{m_{1}}\\}$,
$\>B_{3}=\\{u_{i},u_{m_{1}},u_{m_{2}}\\},\cdots,B_{k-1}=\\{u_{i},u_{m_{1}},u_{m_{2}},\cdots,u_{m_{k-2}}\\},$
and $\>B_{k}=\\{u_{i},u_{m_{1}},u_{m_{2}}\cdots,u_{m_{k-2}},v\\}$. From the
condition of Case I, we have $\emptyset\neq B_{1}\subset
B_{2}\subset\cdots\subset
B_{k}=\\{u_{i},u_{m_{1}},u_{m_{2}},\cdots,u_{m_{k-2}},v\\}\not\in\Gamma$. Let
us consider a subset
$A=\\{u_{j},u_{m_{1}},u_{m_{2}},\cdots,u_{m_{k-3}},u_{m_{k-1}}\\}\subseteq\mathcal{P}$
consisting of $k-1$ participants, where $1\leq j,m_{1},\cdots,m_{s-2}\leq s$,
and $s+1\leq m_{s-1},\cdots,m_{k-1}\leq k+1$. Note that $A\not\in\Gamma$. We
can now consider the subsets of $A$ as follows:
$X_{1}=\\{u_{j},u_{m_{1}},u_{m_{2}},\cdots,u_{m_{k-3}},u_{m_{k-1}}\\},\>X_{2}=\\{u_{j},u_{m_{2}},u_{m_{3}},\cdots,u_{m_{k-3}},u_{m_{k-1}}\\},\>X_{3}=\\{u_{j},u_{m_{3}},\cdots,u_{m_{k-3}},u_{m_{k-1}}\\},\cdots,X_{k-2}=\\{u_{j},u_{m_{k-1}}\\}$,
$\>X_{k-1}=\\{u_{j}\\}$, and $\>X_{k}=\\{u_{m_{k-1}}\\}$.

By using the condition (4.2), we derive that $B_{1}\cup
X_{1}=\\{u_{i},u_{j},u_{m_{1}},\cdots,u_{m_{k-3}},u_{m_{k-1}}\\}\in\Gamma$,
$\cdots,$ $B_{k-2}\cup
X_{k-2}=\\{u_{i},u_{j},u_{m_{1}},\cdots,u_{m_{k-3}},u_{m_{k-1}}\\}\in\Gamma,$
$B_{k-1}\cup
X_{k-1}=\\{u_{i},u_{j},u_{m_{1}},\cdots,u_{m_{k-3}},u_{m_{k-2}}\\}\in\Gamma$.
From the assumption, we also observe that $B_{k}\cup
X_{k}=\\{u_{i},u_{m_{1}},\cdots,u_{m_{k-2}},u_{m_{k-1}},v\\}\in\Gamma$. Since
the set of $k-1$ participants can not be in $\Gamma$, we derive that
$B_{0}\cup
X_{1}=\\{u_{j},u_{m_{1}},u_{m_{2}},\cdots,u_{m_{k-3}},u_{m_{k-1}}\\}\notin\Gamma,B_{1}\cup
X_{2}=\\{u_{i},u_{j},u_{m_{2}},\cdots,u_{m_{k-3}},u_{m_{k-1}}\\}\notin\Gamma,\cdots,B_{k-2}\cup
X_{k-1}=\\{u_{i},u_{j},u_{m_{1}},\cdots,u_{m_{k-3}}\\}\notin\Gamma$. By using
(4.3), we can also derive that $B_{k-1}\cup
X_{k}=\\{u_{i},u_{m_{1}},u_{m_{2}},\cdots,u_{m_{k-2}},u_{m_{k-1}}\\}\notin\Gamma$.

Then the sequence $\phi\neq B_{1}\subseteq B_{2}\subseteq\cdots\subseteq
B_{k}\notin\Gamma$ is made independent by the set
$A=\\{u_{j},u_{m_{1}},\cdots,u_{m_{k-3}},u_{m_{k-1}}\\}\notin\Gamma$.
Therefore, using the independent sequence method, we can conclude that
$\rho^{*}(\Gamma)\leq\frac{|A|}{k}=\frac{k-1}{k}$, which leads to a
contradiction. Hence, we conclude that
$\\{u_{i},u_{m_{1}},u_{m_{2}},\cdots,u_{m_{k-2}},u_{m_{k-1}},v\\}\notin\Gamma$,
which implies that
$\\{u_{m_{1}},u_{m_{2}},\cdots,u_{m_{k-2}},u_{m_{k-1}},v\\}\notin\Gamma$.

Case II: $\\{u_{i},u_{m_{1}},u_{m_{2}},\cdots,u_{m_{k-2}},v\\}\in\Gamma$.

In this case, we need to prove
$\\{u_{m_{1}},u_{m_{2}},\cdots,u_{m_{k-1}},v\\}\notin\Gamma$. Let us assume
the opposite, that is,
$\\{u_{m_{1}},u_{m_{2}},\cdots,u_{m_{k-1}},v\\}\in\Gamma$. We can consider the
following subsets of participants $\mathcal{P}$:
$B_{1}=\\{u_{m_{1}}\\},\>B_{2}=\\{u_{m_{1}},u_{m_{2}}\\},\cdots,B_{k-1}=\\{u_{m_{1}},u_{m_{2}}\cdots,u_{m_{k-1}}\\}$,
and $B_{k}=\\{u_{j},u_{m_{1}},\cdots,u_{m_{k-1}}\\}$. By using (4.3), we have
$\emptyset\neq B_{1}\subset B_{2}\subset\cdots\subset
B_{k}=\\{u_{j},u_{m_{1}},u_{m_{2}}\cdots,u_{m_{k-2}},u_{m_{k-1}}\\}\not\in\Gamma$.
Let us consider a subset
$A=\\{u_{i},u_{m_{2}},u_{m_{3}},\cdots,u_{m_{k-3}},u_{m_{k-2}},v\\}\subseteq\mathcal{P}$
consisting of $k-1$ participants. Note that $A\not\in\Gamma$. Now we can
consider the subsets of $A$ as follows:
$X_{1}=\\{u_{i},u_{m_{2}},\cdots,u_{m_{k-2}},v\\}$,$\>X_{2}=\\{u_{i},u_{m_{3}},\cdots,u_{m_{k-2}},v\\}$,$\>X_{3}=\\{u_{i},u_{m_{4}},\cdots,u_{m_{k-2}},v\\}$,
$\cdots,X_{k-3}=\\{u_{i},u_{m_{k-2}},v\\}$,$\>X_{k-2}=\\{u_{i},v\\}$,$\>X_{k-1}=\\{v\\}$,$\>X_{k}=\\{u_{i}\\}$.

Since $\\{u_{i},u_{m_{1}},u_{m_{2}},\cdots,u_{m_{k-2}},v\\}\in\Gamma$, we can
observe that $B_{1}\cup X_{1}=$
$\\{u_{i},u_{m_{1}},u_{m_{2}},\cdots,u_{m_{k-2}},v\\}$ $\in\Gamma$, $\
B_{2}\cup X_{2}=\\{u_{i},u_{m_{1}},u_{m_{2}},\cdots,u_{m_{k-2}},v\\}$
$\in\Gamma$, $\cdots,\ B_{k-2}\cup X_{k-2}$
$=\\{u_{i},u_{m_{1}},u_{m_{2}},\cdots,u_{m_{k-2}},v\\}\in\Gamma$. Moreover,
since the set of $k-1$ participants can not be in $\Gamma$, we can also
observe that $B_{0}\cup
X_{1}=\\{u_{i},u_{m_{2}},\cdots,u_{m_{k-2}},v\\}\notin\Gamma,\ B_{1}\cup
X_{2}=\\{u_{i},u_{m_{1}},u_{m_{3}},\cdots,u_{m_{k-2}},v\\}\notin\Gamma,\cdots,B_{k-3}\cup
X_{k-2}=\\{u_{i},u_{m_{1}},\cdots,u_{m_{k-3}},v\\}\notin\Gamma,\>B_{k-2}\cup
X_{k-1}=\\{u_{m_{1}},\cdots,u_{m_{k-2}},v\\}\notin\Gamma$. From equation
(4.3), we can see that $B_{k-1}\cup
X_{k}=\\{u_{i},u_{m_{1}},u_{m_{2}},\cdots,u_{m_{k-2}},u_{m_{k-1}}\\}\notin\Gamma$
and
$B_{k}=\\{u_{j},u_{m_{1}},u_{m_{2}},\cdots,u_{m_{k-2}},u_{m_{k-1}}\\}\notin\Gamma$.
Additionally, from the assumption, we have $B_{k-1}\cup
X_{k-1}=\\{u_{m_{1}},u_{m_{2}},\cdots,u_{m_{k-1}},v\\}\in\Gamma$. By using the
condition (4.2), we derive that
$\\{u_{i},u_{j},u_{m_{1}},\cdots,u_{m_{k-3}},u_{m_{k-1}}\\}\in\Gamma$, which
implies that $B_{k}\cup
X_{k}=\\{u_{i},u_{j},u_{m_{1}},\cdots,u_{m_{k-2}},u_{m_{k-1}}\\}\in\Gamma$.

Then the sequence $\phi\neq B_{1}\subseteq B_{2}\subseteq\cdots\subseteq
B_{k}\notin\Gamma$ is made independent by the set
$A=\\{u_{i},u_{m_{2}},u_{m_{3}},\cdots,u_{m_{k-3}},u_{m_{k-2}},v\\}\notin\Gamma$.
Hence, using the independent sequence method, we can conclude that
$\rho^{*}(\Gamma)\leq\frac{|A|}{k}=\frac{k-1}{k}$, which leads to a
contradiction. Therefore, we conclude that
$\\{u_{m_{1}},u_{m_{2}},\cdots,u_{m_{k-2}},u_{m_{k-1}},v\\}\notin\Gamma$.

∎

######  Lemma 4.3.

Let $\Gamma$ be a $k$-homogeneous access structure on a set of participants
$\mathcal{P}$ such that $\rho^{*}(\Gamma)>\frac{k-1}{k}$. Let
$a_{1},a_{2},\cdots,a_{k},a_{k+1},b_{1},b_{2}\in\mathcal{P}$ be $k+3$
different participants satisfying the following two conditions:

$\displaystyle w(\\{a_{1},a_{2},\cdots,a_{k},a_{k+1}\\},\Gamma)=k+1$ (4.4)

and

$\displaystyle\\{a_{1},a_{i_{2}},a_{i_{3}},\cdots,a_{i_{k-2}},b_{1},b_{2}\\}\in\Gamma,$
(4.5)

where $2\leq i_{2},i_{3},\cdots,i_{k-2}\leq k+1.$

Then there exist $j_{1},j_{2},\cdots,j_{k-1}$, where $1\leq
j_{1},j_{2},\cdots,j_{k-1}\leq k+1$, such that either

$\\{a_{j_{1}},a_{j_{2}},\cdots,a_{j_{k-1}},b_{1}\\}\in\Gamma$ or
$\\{a_{j_{1}},a_{j_{2}},\cdots,a_{j_{k-1}},b_{2}\\}\in\Gamma$.

###### Proof.

Let us assume otherwise, that is,

$\displaystyle\\{a_{j_{1}},\cdots,a_{j_{k-1}},b_{1}\\}\notin\Gamma\ \
{\text{and}}\ \ \\{a_{j_{1}},\cdots,a_{j_{k-1}},b_{2}\\}\notin\Gamma\ \
{\text{for\ all}}\ \ 1\leq j_{1},\cdots,j_{k-1}\leq k+1.$ (4.6)

We can consider the following subsets of participants $\mathcal{P}$:
$B_{1}=\\{a_{1}\\}$, $B_{2}=\\{a_{1},a_{4}\\}$,
$B_{3}=\\{a_{1},a_{4},a_{5}\\}$, $\cdots$,
$B_{k-2}=\\{a_{1},a_{4},\cdots,a_{k}\\}$,
$B_{k-1}=\\{a_{1},a_{4},\cdots,a_{k},b_{1}\\},$ and
$B_{k}=\\{a_{1},a_{4},\cdots,a_{k},a_{k+1},b_{1}\\}$. Note that $\emptyset\neq
B_{1}\subset B_{2}\subset\cdots\subset
B_{k}=\\{a_{1},a_{4},a_{5}\cdots,a_{k},a_{k+1},b_{1}\\}$. Let us consider a
subset
$A=\\{a_{2},a_{3},a_{4},\cdots,a_{k-2},b_{1},b_{2}\\}\subseteq\mathcal{P}$
consisting of $k-1$ participants. Note that $A\not\in\Gamma$. We can now
consider the subsets of $A$ as follows:
$X_{1}=\\{a_{2},a_{3},a_{4},\cdots,a_{k-2},b_{1},b_{2}\\},\>X_{2}=\\{a_{2},a_{3},a_{5},\cdots,a_{k-2},b_{1},b_{2}\\},\>X_{3}=\\{a_{2},a_{3},a_{6},\cdots,a_{k-2},b_{1},b_{2}\\},\cdots,X_{k-3}=\\{a_{2},b_{1},b_{2}\\},\>X_{k-2}=\\{b_{1},b_{2}\\},\>X_{k-1}=\\{b_{2}\\},\>X_{k}=\\{a_{3}\\}$.

By using the condition (4.5), we can derive that $B_{1}\cup
X_{1}=\\{a_{1},a_{2},a_{3},\cdots,a_{k-2},b_{1},b_{2}\\}\in\Gamma$, $B_{2}\cup
X_{2}=\\{a_{1},a_{2},a_{3},\cdots,a_{k-2},b_{1},b_{2}\\}\in\Gamma,$ $\cdots,$
$B_{k-2}\cup
X_{k-2}=\\{a_{1},a_{4},a_{5},\cdots,a_{k},b_{1},b_{2}\\}\in\Gamma.$ Moreover,
since the set of $k-1$ participants can not be in $\Gamma$, we also observe
that $B_{0}\cup
X_{1}=\\{a_{2},a_{3},\cdots,a_{k-2},b_{1},b_{2}\\}\notin\Gamma$, $B_{1}\cup
X_{2}=\\{a_{1},a_{2},a_{3},a_{5},\cdots,a_{k-2},b_{1},b_{2}\\}\notin\Gamma$,$\cdots$,
$B_{k-2}\cup X_{k-1}=\\{a_{1},a_{4},\cdots,a_{k},b_{2}\\}\notin\Gamma$.
Furthermore, using the assumption (4.6), we obtain $B_{k-1}\cup
X_{k}=\\{a_{1},a_{3},\cdots,a_{k},b_{1}\\}\notin\Gamma$. Additionally, using
the condition (4.5), we have $B_{k-1}\cup
X_{k-1}=\\{a_{1},a_{4},a_{5},\cdots,a_{k},b_{1},b_{2}\\}\in\Gamma$. By using
(4.4), we derive that $\\{a_{1},a_{3},a_{4},\cdots,a_{k},a_{k+1}\\}\in\Gamma$,
which implies that $B_{k}\cup
X_{k}=\\{a_{1},a_{3},a_{4},\cdots,a_{k},a_{k+1},b_{1}\\}\in\Gamma$.

Then the sequence $\phi\neq B_{1}\subseteq B_{2}\subseteq\cdots\subseteq
B_{k}\notin\Gamma$ is made independent by the set
$A=\\{a_{2},a_{3},a_{4},\cdots,a_{k-2},b_{1},b_{2}\\}$, which is also not in
$\Gamma$. Therefore, using the independent sequence method, we can conclude
that $\rho^{*}(\Gamma)\leq\frac{|A|}{k}=\frac{k-1}{k}$. This leads to a
contradiction. Hence, we conclude that there exist
$j_{1},j_{2},\cdots,j_{k-1}$, where $1\leq j_{1},j_{2},\cdots,j_{k-1}\leq
k+1$, such that either

$\displaystyle\\{a_{j_{1}},\cdots,a_{j_{k-1}},b_{1}\\}\in\Gamma\ \
{\text{or}}\ \ \\{a_{j_{1}},\cdots,a_{j_{k-1}},b_{2}\\}\in\Gamma.$

This completes the proof of Lemma 4.3.

∎

######  Lemma 4.4.

Assume that $\Gamma$ is a $k$-homogeneous access structure on a set of
participants $\mathcal{P}$ such that the number of minimal qualified subsets
contained in any set of $k+1$ participants is not equal to $k$, and
$\rho^{*}(\Gamma)>\frac{k-1}{k}$. Let us consider $k+2$ different participants
$a_{1},a_{2},\cdots,a_{k},a_{k+1},b\in\mathcal{P}$ satisfying the following
three conditions:

$\displaystyle w(\\{a_{1},a_{2},\cdots,a_{k},a_{k+1}\\},\Gamma)=k+1,$ (4.7)

and

$\displaystyle\\{a_{1},a_{2},a_{i_{1}},a_{i_{2}},\cdots,a_{i_{k-3}},b\\}\in\Gamma\>\>\text{where}\>\>3\leq
i_{1},i_{2},\cdots,i_{k-3}\leq k+1,$ (4.8)

and

$\displaystyle\\{a_{t_{1}},a_{t_{2}},\cdots,a_{t_{k-3}},a_{k},a_{k+1},b\\}\in\Gamma\>\>\text{where}\>\>1\leq
t_{1},t_{2},\cdots,t_{k-3}\leq k-1.$ (4.9)

Then, the induced access structure
$\Gamma(\\{a_{1},a_{2},\cdots,a_{k},a_{k+1},b\\})$ is the $(k,k+2)$-threshold
access structure.

###### Proof.

To prove this, we need to show that
$w(\\{a_{p_{1}},a_{p_{2}},a_{p_{3}},\cdots,a_{p_{k}},b\\},\Gamma)=k+1$, where
$1\leq p_{1},p_{2},\cdots,p_{k}\leq k+1$. We will prove it by showing the
following two claims.

######  Claim 0.1.

$w(\\{a_{1},a_{2},a_{q_{1}},a_{q_{2}},\cdots,a_{q_{k-2}},b\\},\Gamma)=k+1$,
where $3\leq q_{1},q_{2},\cdots,q_{k-2}\leq k+1$.

###### Proof of Claim 0.1.

Using the condition (4.7), we obtain

$\displaystyle\\{a_{1},a_{2},a_{q_{1}},a_{q_{2}},\cdots,a_{q_{k-2}}\\}\in\Gamma,$
(4.10)

where $3\leq q_{1},q_{2},\cdots,q_{k-2}\leq k+1$.

Moreover, using the condition (4.8), we also obtain

$\displaystyle\\{a_{1},a_{2},a_{i_{1}},a_{i_{2}},\cdots,a_{i_{k-3}},b\\}\in\Gamma$
(4.11)

where $i_{1},i_{2},\cdots,i_{k-3}\in\\{q_{1},q_{2},\cdots,q_{k-2}\\}$ with
$3\leq q_{1},q_{2},\cdots,q_{k-2}\leq k+1$.

Thus, we need to show that
$\\{a_{2},a_{q_{1}},a_{q_{2}},\cdots,a_{q_{k-2}},b\\}\in\Gamma$ or
$\\{a_{1},a_{q_{1}},a_{q_{2}},\cdots,a_{q_{k-2}},b\\}\in\Gamma$, where $3\leq
q_{1},q_{2},\cdots,q_{k-2}\leq k+1$. Suppose, for the sake of contradiction,
that $\\{a_{2},a_{q_{1}},a_{q_{2}},\cdots,a_{q_{k-2}},b\\}\not\in\Gamma$ and
$\\{a_{1},a_{q_{1}},a_{q_{2}},\cdots,a_{q_{k-2}},b\\}\not\in\Gamma$, where
$3\leq q_{1},q_{2},\cdots,q_{k-2}\leq k+1$. Now we consider the subsets of
participants $\mathcal{P}$ as follows:
$B_{1}=\\{a_{q_{k-2}}\\},\>B_{2}=\\{a_{q_{k-2}},b\\}$,
$\>B_{3}=\\{a_{q_{k-3}},a_{q_{k-2}},b\\},\cdots,\>B_{k-1}=\\{a_{q_{1}},a_{q_{2}},\cdots,a_{q_{k-2}},b\\},$
and $\>B_{k}=\\{a_{2},a_{q_{1}},a_{q_{2}},a_{q_{3}},\cdots,a_{q_{k-2}},b\\}$.
From the given assumption, we have a sequence of sets $\emptyset\neq
B_{1}\subset
B_{2}\subset\cdots\subset\>B_{k}=\\{a_{2},a_{q_{1}},a_{q_{2}},\cdots,a_{q_{k-2}},b\\}\not\in\Gamma$.
Let us choose a participant, denoted by $a_{q_{k-1}}$, from the set of $k+1$
participants $a_{1},a_{2},\cdots,a_{k+1}$ who is not in the set
$\\{a_{1},a_{2},a_{q_{1}},a_{q_{2}},\cdots,a_{q_{k-2}}\\}$. Next, we consider
a subset
$A=\\{a_{1},a_{2},a_{q_{2}},\cdots,a_{q_{k-3}},a_{q_{k-1}}\\}\subseteq\mathcal{P}$
consisting of $k-1$ participants. Note that $A\not\in\Gamma$. Now, we consider
the subsets of $A$ as follows:
$X_{1}=\\{a_{1},a_{2},a_{q_{2}},a_{q_{3}},\cdots,a_{q_{k-3}},a_{q_{k-1}}\\},\>X_{2}=\\{a_{1},a_{2},a_{q_{2}},a_{q_{3}},\cdots,a_{q_{k-4}},a_{q_{k-1}}\\},\cdots,X_{k-3}=\\{a_{1},a_{2},a_{q_{k-1}}\\},X_{k-2}=\\{a_{1},a_{2}\\}$,
$\>X_{k-1}=\\{a_{q_{k-1}}\\}$, and $\>X_{k}=\\{a_{1}\\}$.

Using condition (4.7), we observe $B_{1}\cup
X_{1}=\\{a_{1},a_{2},a_{q_{2}}\cdots,a_{q_{k-3}},a_{q_{k-2}},a_{q_{k-1}}\\}\in\Gamma$.
Additionally, using condition (4.8), we obtain that $\ B_{2}\cup X_{2}$
$=\\{a_{1},a_{2},a_{q_{2}}\cdots,a_{q_{k-4}},a_{q_{k-2}},a_{q_{k-1}},b\\}\in\Gamma$,
$\cdots$, $\ B_{k-2}\cup
X_{k-2}=\\{a_{1},a_{2},a_{q_{2}}\cdots,a_{q_{k-3}},a_{q_{k-2}},b\\}\in\Gamma$.
Also, we have $\ B_{k-1}\cup X_{k-1}=$
$\\{a_{q_{1}},a_{q_{2}}\cdots,a_{q_{k-2}},a_{q_{k-1}},b\\}=\\{a_{3},a_{4},a_{5},\cdots,a_{k},a_{k+1},b\\}\in\Gamma$.
By using the condition (4.8), we also deduce that
$\\{a_{1},a_{2},a_{r_{1}},a_{r_{2}}\cdots,a_{r_{k-3}},b\\}\in\Gamma$, where
$a_{r_{1}},a_{r_{2}},\cdots,a_{r_{k-3}}\in\\{a_{q_{1}},a_{q_{2}}\cdots,a_{q_{k-2}}\\}$,
which implies that $\ B_{k}\cup
X_{k}=\\{a_{1},a_{2},a_{q_{1}},a_{q_{2}}\cdots,a_{q_{k-2}},b\\}\in\Gamma$.

Since the set of $k-1$ participants can not be in $\Gamma$, we derive that
$B_{0}\cup
X_{1}=\\{a_{1},a_{2},a_{q_{2}},a_{q_{3}},\cdots,a_{q_{k-3}},a_{q_{k-1}}\\}\notin\Gamma,$
$\ B_{1}\cup
X_{2}=\\{a_{1},a_{2},a_{q_{2}},\cdots,a_{q_{k-4}},a_{q_{k-2}},a_{q_{k-1}}\\}\notin\Gamma,$
$\cdots,B_{k-2}\cup
X_{k-1}=\\{a_{q_{2}},\cdots,a_{q_{k-2}},a_{q_{k-1}},b\\}\notin\Gamma$. From
the given assumption, we get $B_{k-1}\cup
X_{k}=\\{a_{1},a_{q_{2}},\cdots,a_{q_{k-2}},b\\}\notin\Gamma$.

Then the sequence $\phi\neq B_{1}\subseteq B_{2}\subseteq\cdots\subseteq
B_{k}\notin\Gamma$ is made independent by the set
$A=\\{a_{1},a_{2},a_{q_{2}},\cdots,a_{q_{k-3}},a_{q_{k-1}}\\}$. Therefore,
using the independent sequence method, we can obtain that
$\rho^{*}(\Gamma)\leq\frac{|A|}{k}=\frac{k-1}{k}$, which leads to a
contradiction. Hence, we conclude that
$\\{a_{2},a_{q_{1}},a_{q_{2}},\cdots,a_{q_{k-2}},b\\}\in\Gamma$ or
$\\{a_{1},a_{q_{1}},a_{q_{2}},\cdots,a_{q_{k-2}},b\\}\in\Gamma$, where $3\leq
q_{1},q_{2},\cdots,q_{k-2}\leq k+1$. By using the condition
$k\not\in\Omega(k+1,\Gamma)$, we derive that
$\\{a_{2},a_{q_{1}},a_{q_{2}},\cdots,a_{q_{k-2}},b\\}\in\Gamma$ and
$\\{a_{1},a_{q_{1}},a_{q_{2}},\cdots,a_{q_{k-2}},b\\}\in\Gamma$, where $3\leq
q_{1},q_{2},\cdots,q_{k-2}\leq k+1$. Therefore, we conclude that
$w(\\{a_{1},a_{2},a_{q_{1}},a_{q_{2}},\cdots,a_{q_{k-2}},b\\},\Gamma)=k+1$,
where $3\leq q_{1},q_{2},\cdots,q_{k-2}\leq k+1$.
∎

######  Claim 0.2.

$w(\\{a_{u_{1}},a_{3},a_{4},a_{5},\cdots,a_{k},a_{k+1},b\\},\Gamma)=k+1$,
where $u_{1}=1$ or $2$.

###### Proof of Claim 0.2.

By using the condition (4.9), we derive

$\displaystyle\\{a_{j_{1}},a_{j_{2}},\cdots,a_{j_{k-3}},a_{k},a_{k+1},b\\}\in\Gamma$
(4.12)

where $j_{1},i_{2},\cdots,j_{k-3}\in\\{u_{1},3,4,5,\cdots,k-1\\}$ with
$u_{1}=1$ or $2$.

Thus, we need to prove that $\\{a_{u_{1}},a_{3},\cdots,a_{k-1},a_{k+1},b\\}$
$\in\Gamma$ or $\\{a_{u_{1}},a_{3},\cdots,a_{k-1},a_{k},b\\}\in\Gamma$, where
$u_{1}=1$ or $2$. Suppose, for the sake of contradiction, that
$\\{a_{u_{1}},a_{3},a_{4},\cdots,a_{k-1},a_{k+1},b\\}\not\in\Gamma$ and
$\\{a_{u_{1}},a_{3},a_{4},\cdots,a_{k-1},a_{k},b\\}\not\in\Gamma$, where
$u_{1}=1{\text{\ or }}2$. Let us consider the following subsets of
participants $\mathcal{P}$: $B_{1}=\\{a_{u_{1}}\\},\>B_{2}=\\{a_{u_{1}},b\\}$,
$\>B_{3}=\\{a_{u_{1}},a_{3},b\\},\cdots,B_{k-2}=\\{a_{u_{1}},a_{3},\cdots,a_{k-2},b\\},\>B_{k-1}=\\{a_{u_{1}},a_{3},\cdots,a_{k-1},b\\},$
and $\>B_{k}=\\{a_{u_{1}},a_{3},\cdots,a_{k-1},a_{k+1},b\\}$, where
$u_{1}=1{\text{\ or }}2$. From the given assumption, we have $\emptyset\neq
B_{1}\subset
B_{2}\subset\cdots\subset\>B_{k}=\\{a_{u_{1}},a_{3},\cdots,a_{k-1},a_{k+1},b\\}\not\in\Gamma$,
where $u_{1}=1{\text{\ or }}2$.

Let us choose a participant, denoted by $u_{2}$, from the set of two
participants $a_{1},a_{2}$ who is not equal to $u_{1}$. Next, we consider a
subset
$A=\\{a_{u_{2}},a_{4},a_{5},\cdots,a_{k},a_{k+1}\\}\subseteq\mathcal{P}$
consisting of $k-1$ participants. Note that $A\not\in\Gamma$. We can now
consider the subsets of $A$ as follows:
$X_{1}=\\{a_{u_{2}},a_{4},a_{5},\cdots,a_{k},a_{k+1}\\},\>X_{2}=\\{a_{u_{2}},a_{5},\cdots,a_{k},a_{k+1}\\},\cdots,X_{k-3}=\\{a_{u_{2}},a_{k},a_{k+1}\\},X_{k-2}=\\{a_{k},a_{k+1}\\}$,
$\>X_{k-1}=\\{a_{u_{2}}\\}$, and $\>X_{k}=\\{a_{k}\\}$. By using the condition
(4.7), we observe that $B_{1}\cup
X_{1}=\\{a_{u_{1}},a_{u_{2}},a_{4},a_{5},\cdots,a_{k},a_{k+1}\\}=\\{a_{1},a_{2},a_{4},a_{5},\cdots,a_{k},a_{k+1}\\}\in\Gamma$.
Moreover, using the condition (4.8) and (4.9), we also derive that $\
B_{2}\cup
X_{2}=\\{a_{u_{1}},a_{u_{2}},a_{5},\cdots,a_{k},a_{k+1},b\\}\in\Gamma$,
$\cdots$, $\ B_{k-2}\cup
X_{k-2}=\\{a_{u_{1}},a_{3},a_{4},\cdots,a_{k-2},a_{k},a_{k+1},b\\}\in\Gamma$,
and $\ B_{k-1}\cup
X_{k-1}=\\{a_{u_{1}},a_{u_{2}},a_{3},\cdots,a_{k-1},b\\}=\\{a_{1},a_{2},a_{3},\cdots,a_{k-1},b\\}\in\Gamma$.
Additionally, using (4.9), we can derive that
$\\{a_{u_{1}},a_{r_{1}},a_{r_{2}}\cdots,a_{r_{k-4}},a_{k},a_{k+1},b\\}\in\Gamma$,
where
$a_{r_{1}},a_{r_{2}},\cdots,a_{r_{k-4}}\in\\{a_{3},a_{4}\cdots,a_{k-1}\\}$,
which implies that $\ B_{k}\cup
X_{k}=\\{a_{u_{1}},a_{3},a_{4},\cdots,a_{k-1},a_{k},a_{k+1},b\\}\in\Gamma$.
Since the set of $k-1$ participants can not be in $\Gamma$, we deduce that
$B_{0}\cup
X_{1}=\\{a_{u_{2}},a_{4},a_{5},\cdots,a_{k},a_{k+1}\\}\notin\Gamma,\ B_{1}\cup
X_{2}=\\{a_{u_{1}},a_{u_{2}},a_{5},\cdots,a_{k},a_{k+1}\\}=\\{a_{1},a_{2},a_{5},\cdots,a_{k},a_{k+1}\\}\notin\Gamma,\cdots,B_{k-2}\cup
X_{k-1}=\\{a_{u_{1}},a_{u_{2}},a_{3},\cdots,a_{k-2},b\\}=\\{a_{1},a_{2},a_{3},\cdots,a_{k-2},b\\}\notin\Gamma$.
From the given assumption, we have $B_{k-1}\cup
X_{k}=\\{a_{u_{1}},a_{3},\cdots,a_{k-1},a_{k},b\\}\notin\Gamma$.

Then the sequence $\phi\neq B_{1}\subseteq B_{2}\subseteq\cdots\subseteq
B_{k}\notin\Gamma$ is made independent by the set
$A=\\{a_{u_{2}},a_{4},a_{5},\cdots,a_{k},a_{k+1}\\}$. Hence, using the
independent sequence method, we can conclude that
$\rho^{*}(\Gamma)\leq\frac{|A|}{k}=\frac{k-1}{k}$, which leads to a
contradiction. Therefore, we conclude that
$\\{a_{u_{1}},a_{3},a_{4},\cdots,a_{k-1},a_{k+1},b\\}\in\Gamma$ or
$\\{a_{u_{1}},a_{3},a_{4},\cdots,a_{k-1},a_{k},b\\}\in\Gamma$, where $u_{1}=1$
or $2$. By using the condition $k\not\in\Omega(k+1,\Gamma)$, we derive that
$\\{a_{u_{1}},a_{3},a_{4},\cdots,a_{k-1},a_{k+1},b\\}\in\Gamma$ and
$\\{a_{u_{1}},a_{3},a_{4},\cdots,a_{k-1},a_{k},b\\}\in\Gamma$, where $u_{1}=1$
or $2$. Therefore, we conclude that
$w(\\{a_{u_{1}},a_{3},a_{4},a_{5},\cdots,a_{k},a_{k+1},b\\},\Gamma)=k+1$,
where $u_{1}=1$ or $2$. ∎

Using Claim 0.1 and Claim 0.2, we can conclude that

$w(\\{a_{p_{1}},a_{p_{2}},a_{p_{3}},\cdots,a_{p_{k}},b\\},\Gamma)=k+1,$

where $1\leq p_{1},p_{2},\cdots,p_{k}\leq k+1$.

Thus, the induced access structure
$\Gamma(\\{a_{1},a_{2},\cdots,a_{k},a_{k+1},b\\})$ is the $(k,k+2)$-threshold
access structure. This completes the proof of Lemma 4.4.

∎

Before stating the next lemma, we need to introduce the following notation.
Let $p$ and $q$ be two participants from the set of all participants
$\mathcal{P}$. We say that $p$ and $q$ are equivalent if either (i) $p=q$ or
(ii) $p\neq q$ and the following two conditions are satisfied: (1)
$\\{p,q\\}\not\subset A$ if $A\in\Gamma_{0}$, and (2) if
$A\subset\mathcal{P}\backslash\\{p,q\\}$, then $A\cup\\{p\\}\in\Gamma_{0}$ if
and only if $A\cup\\{q\\}\in\Gamma_{0}$, where $\Gamma_{0}$ is the collection
of minimal qualified subsets.

######  Lemma 4.5.

Let $\Gamma$ be a $k$-homogeneous access structure on a set of participants
$\mathcal{P}$, such that the number of minimal qualified subsets contained in
any set of $k+1$ participants is not equal to $1$ and $k$, and
$\rho^{*}(\Gamma)>\frac{k-1}{k}$. Let us consider $k+2$ different participants
$a_{1},a_{2},\cdots,a_{k},a_{k+1},b\in\mathcal{P}$ satisfying the following
conditions:

$\displaystyle(i)\ \ \ w(\\{a_{1},a_{2},\cdots,a_{k},a_{k+1}\\},\Gamma)=k+1$
(4.13) $\displaystyle(ii)\ \ \ \\{a_{1},a_{2},\cdots,a_{k-1},b\\}\in\Gamma.$
(4.14) $\displaystyle(iii)\ \ \ a_{k}{\text{\ }is\ not\ equivalent\ to\
}a_{k+1}$ (4.15)

Moreover, $k+2$ participants $a_{1},a_{2},\cdots,a_{k},a_{k+1},b$ satisfy the
one condition among the following three statements.

$\displaystyle(iv)$ $\displaystyle\ \ \
\\{a_{t_{1}},a_{t_{2}},a_{t_{3}},\cdots,a_{t_{k-3}},a_{k},a_{k+1},b\\}\not\in\Gamma\
\ \ where\ \ 1\leq t_{1},t_{2},\cdots,t_{k-3}\leq k-1.$ (4.16) $\displaystyle
or\
\\{a_{l_{1}},a_{l_{2}},a_{l_{3}},\cdots,a_{l_{k-2}},a_{k+1},b\\}\not\in\Gamma\
\ \ where\ \ 1\leq l_{1},l_{2},\cdots,l_{k-2}\leq k-1.$ (4.17) $\displaystyle
or\
\\{a_{s_{1}},a_{s_{2}},a_{s_{3}},\cdots,a_{s_{k-2}},a_{k},b\\}\not\in\Gamma\ \
\ where\ \ 1\leq s_{1},s_{2},\cdots,s_{k-2}\leq k-1.$ (4.18)

Then, either $b$ is equivalent to $a_{k+1}$ or $b$ is equivalent to $a_{k}$.

###### Proof.

First, let us consider $k+2$ participants in $\mathcal{P}$, denoted as
$a_{1},a_{2},\cdots,a_{k},a_{k+1}$, and $b$, who satisfy conditions (4.13),
(4.14), (4.15), and (4.16). Now, let us focus on $k+1$ participants
$a_{\Delta_{1}},a_{\Delta_{2}},\cdots a_{\Delta_{k-2}},a_{k},a_{k+1},b$, where
$1\leq\Delta_{1},\cdots,\Delta_{k-2}\leq k-1$. Using (4.13), we obtain

$\displaystyle\\{a_{\Delta_{1}},a_{\Delta_{2}},\cdots
a_{\Delta_{k-2}},a_{k},a_{k+1}\\}\in\Gamma.$ (4.19)

Moreover, by (4.16), we derive that

$\displaystyle\\{a_{\beta_{1}},a_{\beta_{2}},\cdots
a_{\beta_{k-3}},a_{k},a_{k+1},b\\}\not\in\Gamma,$ (4.20)

where, $a_{\beta_{1}},a_{\beta_{2}},\cdots
a_{\beta_{k-3}}\in\\{a_{\Delta_{1}},a_{\Delta_{2}},\cdots
a_{\Delta_{k-2}}\\}.$

By using the conditions $1\not\in\Omega(k+1,\Gamma)$ and $a_{k}$ is not
equivalent to $a_{k+1}$, we obtain that

$\displaystyle\\{a_{\Delta_{1}},a_{\Delta_{2}},\cdots
a_{\Delta_{k-2}},a_{k+1},b\\}\not\in\Gamma\ \ \ \ \ or$
$\displaystyle\\{a_{\Delta_{1}},a_{\Delta_{2}},\cdots
a_{\Delta_{k-2}},a_{k},b\\}\not\in\Gamma,$ (4.21)

where $1\leq\Delta_{1},\Delta_{2}\cdots,\Delta_{k-2}\leq k-1$.

Case I: $\\{a_{\Delta_{1}},a_{\Delta_{2}},\cdots
a_{\Delta_{k-2}},a_{k+1},b\\}\not\in\Gamma$, where
$1\leq\Delta_{1},\Delta_{2},\cdots,\Delta_{k-2}\leq k-1.$

In this case, we aim to demonstrate the equivalence of $a_{k+1}$ and $b$.
Given that $a_{k+1}$ and $b$ represent distinct participants, it is necessary
to establish that:

  * (1)

$\\{a_{k+1},b\\}\not\subset A$ for any $A\in\Gamma_{0}$,

  * (2)

if $A\subset\mathcal{P}\backslash\\{a_{k+1},b\\}$, then
$A\cup\\{a_{k+1}\\}\in\Gamma$ if and only if $A\cup\\{b\\}\in\Gamma$,

where $\Gamma_{0}$ is the collection of minimal qualified subsets.

Based on our assumption of Case I, we observe that

$\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-2}},a_{k+1},b\\}\not\in\Gamma,$ (4.22)

where $1\leq l_{1},l_{2},\cdots,l_{k-2}\leq k-1$.

Since $a_{k}$ is not equivalent to $a_{k+1}$, we can also deduce that

$\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-2}},a_{k},b\\}\in\Gamma,$ (4.23)

where $1\leq l_{1},l_{2},\cdots,l_{k-2}\leq k-1$.

Using the property (4.19), we also obtain that

$\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-2}},a_{k},a_{k+1}\\}\in\Gamma,$ (4.24)

where $1\leq l_{1},l_{2},\cdots,l_{k-2}\leq k-1$.

To prove the property $(1)$, we will first establish the following claims.

######  Claim 0.1.

$\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-3}},x,a_{k+1},b\\}\not\in\Gamma$ and
$\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-3}},x,a_{k},b\\}\in\Gamma$, where
$1\leq l_{1},l_{2},\cdots,l_{k-3}\leq k-1$ and
$x\in\mathcal{P}\backslash\\{a_{1},a_{2},\cdots,a_{k},a_{k+1},b\\}$.

###### Proof of Claim 0.1.

Assuming the contrary, let us consider
$\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-3}},x,a_{k+1},b\\}\in\Gamma$, where
$1\leq l_{1},l_{2},\cdots,l_{k-3}\leq k-1$ and
$x\in\mathcal{P}\backslash\\{a_{1},a_{2},\cdots,a_{k},a_{k+1},b\\}$.

Next, let us select a participant from $\\{a_{1},a_{2},\cdots,a_{k-1}\\}$ who
is not part of $\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-3}}\\}$ and denote this
participant as $a_{l_{k-2}}$. We will now consider the following subsets of
participants in $\mathcal{P}$:
$B_{1}=\\{a_{l_{1}}\\},\>B_{2}=\\{a_{l_{1}},b\\}$,
$\>B_{3}=\\{a_{l_{1}},a_{l_{2}},b\\},\cdots,B_{k-2}=\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-3}},b\\},\>B_{k-1}=\\{a_{l_{1}},a_{l_{2}}\cdots,a_{l_{k-2}},b\\},$
and $\>B_{k}=\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-2}},b,x\\}$. Note that
$\emptyset\neq B_{1}\subset
B_{2}\subset\cdots\subset\>B_{k}=\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-2}},b,x\\}$.

Now, let us consider a subset
$A=\\{a_{l_{2}},a_{l_{3}},\cdots,a_{l_{k-2}},a_{k},a_{k+1}\\}\subseteq\mathcal{P}$
consisting of $k-1$ participants. Note that $A\not\in\Gamma$. We can then
consider the subsets of $A$ as follows:
$X_{1}=\\{a_{l_{2}},a_{l_{3}},\cdots,a_{l_{k-2}},a_{k},a_{k+1}\\}$,
$X_{2}=\\{a_{l_{2}},a_{l_{3}},\cdots,a_{l_{k-2}},a_{k}\\},\>X_{3}=\\{a_{l_{3}},a_{l_{4}},\cdots,a_{l_{k-2}},a_{k}\\},\cdots,$
$X_{k-3}=\\{a_{l_{k-3}},a_{l_{k-2}},a_{k}\\},$
$X_{k-2}=\\{a_{l_{k-2}},a_{k}\\}$, $\>X_{k-1}=\\{a_{k}\\}$, and
$\>X_{k}=\\{a_{k+1}\\}$.

Using 4.24, we observe that $B_{1}\cup
X_{1}=\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-2}},a_{k},a_{k+1}\\}\in\Gamma$.
Using 4.23, we also derive that $\ B_{2}\cup
X_{2}=\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-2}},a_{k},b\\}\in\Gamma,$ and so
on up to $\ B_{k-1}\cup
X_{k-1}=\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-2}},a_{k},b\\}\in\Gamma$. Based
on the assumption, we have
$\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-3}},a_{k+1},b,x\\}\in\Gamma$, which
implies that $\ B_{k}\cup
X_{k}=\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-2}},a_{k+1},b,x\\}\in\Gamma$.
Since the set of $k-1$ participants can not be in $\Gamma$, we derive that
$B_{0}\cup X_{1}\notin\Gamma,\ B_{1}\cup X_{2}\notin\Gamma,\cdots,B_{k-2}\cup
X_{k-1}\notin\Gamma$. Using 4.22, we also derive that $B_{k-1}\cup
X_{k}=\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-2}},a_{k+1},b\\}\not\in\Gamma$.

If
$\>B_{k}=\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-3}},a_{l_{k-2}},b,x\\}\not\in\Gamma$,
then the sequence $\phi\neq B_{1}\subseteq B_{2}\subseteq\cdots\subseteq
B_{k}\notin\Gamma$ is made independent by the set
$A=\\{a_{l_{2}},a_{l_{3}},\cdots,a_{l_{k-2}},a_{k},a_{k+1}\\}$. Hence, using
the independent sequence method, we can conclude that
$\rho^{*}(\Gamma)\leq\frac{|A|}{k}=\frac{k-1}{k}$, which leads to a
contradiction. Therefore, we can conclude that

$\displaystyle\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-3}},a_{l_{k-2}},b,x\\}\in\Gamma,$
(4.25)

where $1\leq l_{1},l_{2},l_{3},\cdots,l_{k-3},l_{k-2}\leq k-1$.

Let us consider $k+2$ distinct participants:
$a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-2}},a_{k},a_{k+1},x,b\in\mathcal{P}$,
where $1\leq l_{1},l_{2},\cdots,l_{k-2}\leq k-1$ and
$x\in\mathcal{P}\backslash\\{a_{1},a_{2},\cdots,a_{k},a_{k+1},b\\}$. Next, we
can apply Lemma 4.1 with $u_{1}=b,u_{2}=x$, and $v=a_{k}$. Then either

$w(\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-2}},a_{k+1},b,x\\},\
\Gamma)=k+1,{\text{\ where\ }}1\leq l_{1},l_{2},\cdots,l_{k-2}\leq k-1,$

or

$\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-2}},a_{k+1},a_{k}\\}\not\in\Gamma,{\text{\
for \
}}a_{k}\in\mathcal{P}\backslash\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-2}},a_{k+1},b\\}.$

Since $\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-2}},a_{k+1},b\\}\not\in\Gamma$,
where $1\leq l_{1},l_{2},\cdots,l_{k-2}\leq k-1$, we can conclude that

$w(\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-2}},a_{k+1},b,x\\},\ \Gamma)\neq
k+1,{\text{\ where\ }}1\leq l_{1},l_{2},l_{3},\cdots,l_{k-3},l_{k-2}\leq k-1.$

Let us consider $k+1$ participants:
$a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-2}},a_{k},a_{k+1},b\in\mathcal{P}$, where
$1\leq l_{1},l_{2},\cdots,l_{k-2}\leq k-1$. Since the number of minimal
qualified subsets contained in any set of $k+1$ participants is not equal to
$1$, we must have

$\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-2}},a_{k+1},a_{k}\\}\in\Gamma,{\text{\
for \
}}a_{k}\in\mathcal{P}\backslash\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-2}},a_{k+1},b\\},$

which leads to a contradiction. Therefore, we can conclude that

$\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-3}},x,a_{k+1},b\\}\not\in\Gamma,$

where $1\leq l_{1},l_{2},\cdots,l_{k-3}\leq k-1$ and
$x\in\mathcal{P}\backslash\\{a_{1},a_{2},\cdots,a_{k},a_{k+1},b\\}.$

Since $a_{k}$ is not equivalent to $a_{k+1}$, we also conclude that

$\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-3}},x,a_{k},b\\}\in\Gamma,$

where $1\leq l_{1},l_{2},\cdots,l_{k-3}\leq k-1$ and
$x\in\mathcal{P}\backslash\\{a_{1},a_{2},\cdots,a_{k},a_{k+1},b\\}$.

This completes the proof of Claim 0.1. ∎

######  Claim 0.2.

$\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-4}},x,a_{k},a_{k+1},b\\}\not\in\Gamma,$

where $1\leq l_{1},l_{2},\cdots,l_{k-4}\leq k-1$ and
$x\in\mathcal{P}\backslash\\{a_{1},a_{2},\cdots,a_{k},a_{k+1},b\\}.$

###### Proof of Claim 0.2.

Let us assume otherwise, that is,
$\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-4}},x,a_{k},a_{k+1},b\\}\in\Gamma$,
where $1\leq l_{1},l_{2},\cdots,l_{k-4}\leq k-1$ and
$x\in\mathcal{P}\backslash\\{a_{1},a_{2},\cdots,a_{k},a_{k+1},b\\}$.

Next, we can consider two participants from $\\{a_{1},a_{2},\cdots,a_{k-1}\\}$
who are not in $\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-4}}\\}$ and denote them
as $a_{l_{k-3}}$ and $a_{l_{k-2}}$ . We will now consider the following
subsets of participants in $\mathcal{P}$:
$B_{1}=\\{b\\},\>B_{2}=\\{a_{l_{1}},b\\}$,
$\>B_{3}=\\{a_{l_{1}},a_{l_{2}},b\\},\cdots$,
$B_{k-2}=\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-4}},a_{k},b\\},$ $B_{k-1}=$
$\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-4}},a_{k},a_{k+1},b\\},$ and
$\>B_{k}=\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-4}},a_{l_{k-3}},a_{k},a_{k+1},b\\}$.
Note that $\emptyset\neq B_{1}\subset
B_{2}\subset\cdots\subset\>B_{k}=\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-3}},a_{k},a_{k+1},b\\}\not\in\Gamma$.
Next, we consider a subset
$A=\\{a_{l_{2}},a_{l_{3}},\cdots,a_{l_{k-4}},a_{l_{k-2}},a_{k},a_{k+1},x\\}\subseteq\mathcal{P}$
consisting of $k-1$ participants. Note that $A\not\in\Gamma$. We now consider
the subsets of $A$ as follows:
$X_{1}=\\{a_{l_{2}},a_{l_{3}},\cdots,a_{l_{k-4}},a_{l_{k-2}},a_{k},a_{k+1},x\\}$,
$X_{2}=\\{a_{l_{2}},a_{l_{3}},\cdots,a_{l_{k-4}},a_{l_{k-2}},a_{k},x\\},$
$X_{3}=$ $\\{a_{l_{3}},a_{l_{4}},\cdots,a_{l_{k-4}},a_{l_{k-2}},a_{k},x\\},$
$\cdots$, $X_{k-2}=\\{a_{l_{k-2}},x\\}$, $\>X_{k-1}=\\{x\\}$, and
$\>X_{k}=\\{a_{l_{k-2}}\\}$.

Based on the assumption, we observe $B_{1}\cup
X_{1}=\\{a_{l_{2}},a_{l_{3}},\cdots,a_{l_{k-4}},a_{l_{k-2}},x,a_{k},a_{k+1},b\\}\in\Gamma$.
Additionally, using Claim 0.1, we obtain that

$\ B_{2}\cup
X_{2}=\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-4}},a_{l_{k-2}},x,a_{k},b\\}\in\Gamma,$

and so on up to $\ B_{k-2}\cup
X_{k-2}=\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-4}},a_{l_{k-2}},x,a_{k},b\\}\in\Gamma$.
Based on the assumption, we have $\ B_{k-1}\cup
X_{k-1}=\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-4}},x,a_{k},a_{k+1},b\\}\in\Gamma$.
Using 4.24, we obtain that
$\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-2}},a_{k},a_{k+1}\\}\in\Gamma$, which
implies $\ B_{k}\cup
X_{k}=\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-2}},a_{k},a_{k+1},b\\}\in\Gamma$.

Since the set of $k-1$ participants can not be in $\Gamma$, we derive that
$B_{0}\cup X_{1}\notin\Gamma,$ $\ B_{1}\cup X_{2}\notin\Gamma$, $\cdots,$
$B_{k-2}\cup X_{k-1}\not\in\Gamma$. Based on the assumption, we also deduce
that

$B_{k-1}\cup
X_{k}=\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-4}},a_{l_{k-2}},a_{k},a_{k+1},b\\}\not\in\Gamma.$

Then the sequence $\phi\neq B_{1}\subseteq B_{2}\subseteq\cdots\subseteq
B_{k}\notin\Gamma$ is made independent by the set
$A=\\{a_{l_{2}},a_{l_{3}},\cdots,a_{l_{k-4}},a_{l_{k-2}},a_{k},a_{k+1},x\\}\subseteq\mathcal{P}$.
Hence, using the independent sequence method, we can conclude that
$\rho^{*}(\Gamma)\leq\frac{|A|}{k}=\frac{k-1}{k}$, which leads to a
contradiction. Therefore, we can conclude that

$\displaystyle\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-4}},x,a_{k},a_{k+1},b\\}\not\in\Gamma$
(4.26)

where $1\leq l_{1},l_{2},\cdots,l_{k-4}\leq k-1$ and
$x\in\mathcal{P}\backslash\\{a_{1},a_{2},\cdots,a_{k},a_{k+1},b\\}$.

This completes the proof of Claim 0.2.
∎

######  Claim 0.3.

Suppose that the following three statements hold for $1\leq t\leq k-2$.

  1. (i)

$\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-t-1}},x_{1},x_{2},\cdots,x_{t-1},a_{k+1},b\\}\not\in\Gamma$,

  2. (ii)

$\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-t-1}},x_{1},x_{2},\cdots,x_{t-1},a_{k},b\\}\in\Gamma$,

  3. (iii)

$\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-t-2}},x_{1},x_{2},\cdots,x_{t-1},a_{k},a_{k+1},b\\}\not\in\Gamma$,

where $1\leq l_{1},l_{2},\cdots,l_{k-t-1}\leq k-1$ and
$x_{1},x_{2},\cdots,x_{t-1}\in\mathcal{P}\backslash\\{a_{1},a_{2},\cdots,a_{k},a_{k+1},b\\}$.

Then for $1\leq t\leq k-2$, we have

$\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-t-2}},x_{1},x_{2},\cdots,x_{t},a_{k+1},b\\}\not\in\Gamma,$

and

$\\{a_{l_{1}},a_{l_{2}},\cdots,a_{l_{k-t-2}},x_{1},x_{2},\cdots,x_{t},a_{k},b\\}\in\Gamma,$

where $1\leq l_{1},l_{2},\cdots,l_{k-t-2}\leq k-1$ and
$x_{1},x_{2},\cdots,x_{t}$
$\in\mathcal{P}\backslash\\{a_{1},a_{2},\cdots,a_{k},a_{k+1},b\\}$.

###### Proof of Claim 0.3.

Let us consider $k+2$ different participants in $\mathcal{P}$, denoted as
$a_{l_{1}},\cdots,a_{l_{k-t-1}},$ $x_{1},\cdots,x_{t},a_{k},a_{k+1},b$, where
$1\leq l_{1},\cdots,l_{k-t-1}\leq k-1$ and $x_{1},\cdots,x_{t}$
$\in\mathcal{P}\backslash\\{a_{1},a_{2},\cdots,a_{k},a_{k+1},b\\}$.

Using the properties $(i),(ii),(iii)$, now we apply Lemma 4.2 with
$u_{i}=a_{l_{k-t-1}},u_{j}=a_{k},v=x_{t}$. Then either

$w(\\{a_{l_{1}},\cdots,a_{l_{k-t-1}},x_{1},\cdots,x_{t-1},a_{k},a_{k+1},b\\},\
\Gamma)=k+1$

or

$\\{a_{l_{1}},\cdots,a_{l_{k-t-2}},x_{1},\cdots,x_{t-1},x_{t},a_{k+1},b\\}\not\in\Gamma,$

where $1\leq l_{1},\cdots,l_{k-t-1}\leq k-1$ and
$x_{1},\cdots,x_{t}\in\mathcal{P}\backslash\\{a_{1},\cdots,a_{k},a_{k+1},b\\}$.

From $(i)$ and $(iii)$, we observe that
$w(\\{a_{l_{1}},\cdots,a_{l_{k-t-1}},x_{1},\cdots,x_{t-1},a_{k},a_{k+1},b\\},\
\Gamma)\neq k+1$. Therefore, we can conclude that

$\\{a_{l_{1}},\cdots,a_{l_{k-t-2}},x_{1},\cdots,x_{t},a_{k+1},b\\}\not\in\Gamma,$

where $1\leq l_{1},\cdots,l_{k-t-1}\leq k-1$ and
$x_{1},\cdots,x_{t}\in\mathcal{P}\backslash\\{a_{1},\cdots,a_{k},a_{k+1},b\\}$.

Since $a_{k}$ is not equivalent to $a_{k+1}$, for $1\leq t\leq k-2$, we also
conclude that

$\\{a_{l_{1}},\cdots,a_{l_{k-t-2}},x_{1},\cdots,x_{t},a_{k},b\\}\in\Gamma,$

where $1\leq l_{1},\cdots,l_{k-t-2}\leq k-1$ and
$x_{1},\cdots,x_{t}\in\mathcal{P}\backslash\\{a_{1},\cdots,a_{k},a_{k+1},b\\}$.
This completes the proof of Claim 0.3.
∎

Using Claim 0.3, we can conclude that

$\\{a_{k+1},b\\}\not\subset A{{\text{\ }\ \ {for\ any}\ \ }}A\in\Gamma_{0},$

where $\Gamma_{0}$ is the collection of minimal qualified subsets.

Now we are prepared to demonstrate the property $(2)$ by using the property
$(1)$. To accomplish this, we will begin by establishing the following claims.

######  Claim 0.4.

Let $A$ be a subset in $\mathcal{P}\backslash\\{a_{k+1},b\\}$. If
$A\cup\\{a_{k+1}\\}\in\Gamma$, then $A\cup\\{b\\}\in\Gamma$.

###### Proof of Claim 0.4.

Let us consider $k+2$ distinct participants in $\mathcal{P}$, denoted as
$a_{p_{1}},a_{p_{2}},\cdots,a_{p_{k-1}},$ $a_{k},a_{k+1},b$, where
$\\{a_{p_{1}},\cdots,a_{p_{k-1}}\\}\in\mathcal{P}\backslash\\{a_{k},a_{k+1},b\\}$.
First, let $A=\\{a_{p_{1}},,\cdots,a_{p_{k-2}},a_{k}\\}$ be a subset in
$\mathcal{P}\backslash\\{a_{k+1},b\\}$. Suppose that
$A\cup\\{a_{k+1}\\}=\\{a_{p_{1}},\cdots,a_{p_{k-2}},a_{k},a_{k+1}\\}\in\Gamma$.
Using the property $(1)$, we clearly deduce that
$A\cup\\{b\\}=\\{a_{p_{1}},\cdots,a_{p_{k-2}},a_{k},b\\}\in\Gamma$, where
$a_{p_{1}},\cdots,a_{p_{k-2}}$ is in
$\mathcal{P}\backslash\\{a_{k},a_{k+1},b\\}$. Next, let
$A=\\{a_{p_{1}},\cdots,a_{p_{k-1}}\\}$ be a subset in
$\mathcal{P}\backslash\\{a_{k},a_{k+1},b\\}$. Suppose that
$A\cup\\{a_{k+1}\\}=\\{a_{p_{1}},\cdots,a_{p_{k-1}},a_{k+1}\\}\in\Gamma$,
where $a_{p_{1}},\cdots,a_{p_{k-1}}$ is in
$\mathcal{P}\backslash\\{a_{k},a_{k+1},b\\}$. Assuming the contrary, i.e.,
$A\cup\\{b\\}=\\{a_{p_{1}},\cdots,a_{p_{k-1}},b\\}\not\in\Gamma$, where
$a_{p_{1}},\cdots,a_{p_{k-1}}$ are in
$\mathcal{P}\backslash\\{a_{k},a_{k+1},b\\}$, we can proceed to prove the
desired result.

Let us first define the following subsets of participants in $\mathcal{P}$:
$B_{1}=\\{b\\},\>B_{2}=\\{a_{p_{2}},b\\}$,
$\>B_{3}=\\{a_{p_{2}},a_{p_{3}},b\\},\cdots,B_{k-2}=\\{a_{p_{2}},\cdots,a_{p_{k-3}},a_{p_{k-2}},b\\},\>B_{k-1}=\\{a_{p_{1}},a_{p_{2}},\cdots,a_{p_{k-2}},b\\},$
and $\>B_{k}=\\{a_{p_{1}},a_{p_{2}},\cdots,a_{p_{k-2}},a_{k+1},b\\}$. Based on
the property $(1)$, we observe that $\emptyset\neq B_{1}\subset
B_{2}\subset\cdots\subset\>B_{k}=\\{a_{p_{1}},\cdots,a_{p_{k-2}},a_{k+1},b\\}\not\in\Gamma$.
Next, we consider a subset
$S=\\{a_{p_{2}},\cdots,a_{p_{k-1}},a_{k}\\}\subseteq\mathcal{P}$ consisting of
$k-1$ participants. Note that $S\not\in\Gamma$. We now define the following
subsets of $S$: $X_{1}=\\{a_{p_{2}},a_{p_{3}},\cdots,a_{p_{k-1}},a_{k}\\},$
$\>X_{2}=\\{a_{p_{3}},\cdots,a_{p_{k-1}},a_{k}\\},\>X_{3}=\\{a_{p_{4}},\cdots,a_{p_{k-1}},a_{k}\\}$
$,\cdots,X_{k-3}=\\{a_{p_{k-2}},a_{p_{k-1}},a_{k}\\},$
$X_{k-2}=\\{a_{p_{k-1}},a_{k}\\}$, $\>X_{k-1}=\\{a_{k}\\}$, and
$\>X_{k}=\\{a_{p_{k-1}}\\}$.

Using the property $(1)$, we observe that $B_{1}\cup
X_{1}=\\{a_{p_{2}},a_{p_{3}},\cdots,a_{p_{k-1}},a_{k},b\\}\in\Gamma$, and so
on up to $B_{k-2}\cup
X_{k-2}=\\{a_{p_{2}},a_{p_{3}},\cdots,a_{p_{k-1}},a_{k},b\\}\in\Gamma$.
Additionally, we obtain that $B_{k-1}\cup
X_{k-1}=\\{a_{p_{1}},a_{p_{2}},\cdots,a_{p_{k-2}},a_{k},b\\}\in\Gamma$. Based
on the assumption, we also have $\\{a_{p_{1}},\cdots,a_{p_{k-1}},a_{k+1}\\}$
$\in\Gamma$, which implies that $B_{k}\cup
X_{k}=\\{a_{p_{1}},\cdots,a_{p_{k-1}},a_{k+1},b\\}\in\Gamma$. Since the set of
$k-1$ participants can not be in $\Gamma$, we derive that $B_{0}\cup
X_{1}\not\in\Gamma$, $\cdots$, $B_{k-2}\cup X_{k-1}\not\in\Gamma$. Based on
the assumption, we also have $B_{k-1}\cup
X_{k}=\\{a_{p_{1}},a_{p_{2}},\cdots,a_{p_{k-1}},b\\}\not\in\Gamma$.

Therefore, we can apply the independent sequence method to the sequence
$\phi\neq B_{1}\subseteq B_{2}\subseteq\cdots\subseteq B_{k}\notin\Gamma$ with
the set $S=\\{a_{p_{2}},a_{p_{3}},\cdots,a_{p_{k-1}},a_{k}\\}$, to obtain
$\rho^{*}(\Gamma)\leq\frac{|S|}{k}=\frac{k-1}{k}$, which leads to a
contradiction. Thus, we can conclude that
$A\cup\\{b\\}=\\{a_{p_{1}},a_{p_{2}},\cdots,a_{p_{k-1}},b\\}\in\Gamma$, where
$a_{p_{1}},\cdots,a_{p_{k-1}}$ is in
$\mathcal{P}\backslash\\{a_{k},a_{k+1},b\\}$. This completes the proof of
Claim 0.4. ∎

######  Claim 0.5.

Let $A$ be a subset in $\mathcal{P}\backslash\\{a_{k+1},b\\}$. If
$A\cup\\{b\\}\in\Gamma$, then we have $A\cup\\{a_{k+1}\\}\in\Gamma$.

###### Proof of Claim 0.5.

Let us consider $k+2$ distinct participants in $\mathcal{P}$, denoted as
$a_{p_{1}},a_{p_{2}},\cdots,a_{p_{k-1}},$ $a_{k},a_{k+1},b$, where
$a_{p_{1}},\cdots,a_{p_{k-1}}\in\mathcal{P}\backslash\\{a_{k},a_{k+1},b\\}$.
First, let $A=\\{a_{p_{1}},\cdots,a_{p_{k-2}},a_{k}\\}$ be a subset in
$\mathcal{P}\backslash\\{a_{k+1},b\\}$. Suppose that
$A\cup\\{b\\}=\\{a_{p_{1}},\cdots,a_{p_{k-2}},a_{k},b\\}\in\Gamma$, where
$a_{p_{1}},\cdots,a_{p_{k-2}}$ are in
$\mathcal{P}\backslash\\{a_{k},a_{k+1},b\\}$. Assuming the contrary, i.e.,
$A\cup\\{a_{k+1}\\}=\\{a_{p_{1}},\cdots,a_{p_{k-2}},a_{k},a_{k+1}\\}\not\in\Gamma$,
where $a_{p_{1}},\cdots,a_{p_{k-2}}$ are in
$\mathcal{P}\backslash\\{a_{k},a_{k+1},b\\}$. Using the property $(1)$, we
obtain $\\{a_{p_{1}},\cdots,a_{p_{k-2}},a_{k+1},b\\}$ $\not\in\Gamma$ and
$\\{a_{s_{1}},\cdots,a_{s_{k-3}},a_{k},a_{k+1},b\\}$ $\not\in\Gamma$, where
$s_{1},s_{2},\cdots s_{k-3}\in\\{p_{1},p_{2},\cdots,p_{k-2}\\}$ and
$a_{p_{1}},\cdots,a_{p_{k-2}}$ are in
$\mathcal{P}\backslash\\{a_{k},a_{k+1},b\\}$. Then, this contradicts the fact
that $1\not\in\Omega(k+1,\Gamma)$. Therefore, we can conclude that
$A\cup\\{a_{k+1}\\}=\\{a_{p_{1}},\cdots,a_{p_{k-2}},a_{k},a_{k+1}\\}\in\Gamma$,
where $a_{p_{1}},\cdots,a_{p_{k-2}}$ are in
$\mathcal{P}\backslash\\{a_{k},a_{k+1},b\\}$.

Next, let $A=\\{a_{p_{1}},a_{p_{2}},\cdots,a_{p_{k-1}}\\}$ be a subset in
$\mathcal{P}\backslash\\{a_{k},a_{k+1},b\\}$. Suppose that
$A\cup\\{b\\}=\\{a_{p_{1}},\cdots,a_{p_{k-1}},b\\}\in\Gamma$, where
$a_{p_{1}},\cdots,a_{p_{k-1}}$ are in
$\mathcal{P}\backslash\\{a_{k},a_{k+1},b\\}$. Assuming the contrary, i.e.,
$A\cup\\{a_{k+1}\\}=\\{a_{p_{1}},\cdots,a_{p_{k-1}},a_{k+1}\\}\not\in\Gamma$,
where $a_{p_{1}},\cdots,a_{p_{k-1}}$ are in
$\mathcal{P}\backslash\\{a_{k},a_{k+1},b\\}$. Moreover, using the property
$(1)$, we derive that

$\displaystyle\\{a_{\beta_{1}},\cdots
a_{\beta_{k-2}},a_{k+1},b\\}\not\in\Gamma,$ (4.27)

where, $a_{\beta_{1}},\cdots a_{\beta_{k-2}}\in\\{a_{p_{1}},a_{p_{2}},\cdots
a_{p_{k-1}}\\}.$

Then, let us consider $k+1$ participants
$a_{p_{1}},a_{p_{2}},\cdots,a_{p_{k-1}},a_{k+1},b$, then the above results
lead to a contradiction, since the number of minimal qualified subsets
contained in any set of $k+1$ participants is not equal to $1$. Thus, we can
conclude that
$A\cup\\{a_{k+1}\\}=\\{a_{p_{1}},\cdots,a_{p_{k-1}},a_{k+1}\\}\in\Gamma$,
where $a_{p_{1}},\cdots,a_{p_{k-1}}$ are in
$\mathcal{P}\backslash\\{a_{k},a_{k+1},b\\}$. This completes the proof of
Claim 0.5. ∎

Using Claim 0.4 and 0.5, we can conclude that

if $A\subset\mathcal{P}\backslash\\{a_{k+1},b\\}$, then
$A\cup\\{a_{k+1}\\}\in\Gamma$ if and only if $A\cup\\{b\\}\in\Gamma$,

where $\Gamma_{0}$ is the collection of minimal qualified subsets.

Using the property $(1)$ and $(2)$, we conclude that $b$ is equivalent to
$a_{k+1}$.

Case II: $\\{a_{\Delta_{1}},a_{\Delta_{2}},\cdots
a_{\Delta_{k-2}},a_{k},b\\}\not\in\Gamma$, where
$1\leq\Delta_{1},\Delta_{2},\cdots\Delta_{k-2}\leq k-1$.

In this case, we aim to demonstrate the equivalence of $a_{k}$ and $b$. Given
that $a_{k}$ and $b$ represent distinct participants, it is necessary to
establish that:

  * (1)

$\\{a_{k},b\\}\not\subset A$ for any $A\in\Gamma_{0}$,

  * (2)

if $A\subset\mathcal{P}\backslash\\{a_{k},b\\}$, then
$A\cup\\{a_{k}\\}\in\Gamma$ if and only if $A\cup\\{b\\}\in\Gamma$,

where $\Gamma_{0}$ is the collection of minimal qualified subsets.

The proof of Case II is omitted since it is similar to the proof of Case I by
replacing $a_{k}$ with $a_{k+1}$.
∎

##  5 Proof of Theorem 1.1

Now, we are ready to establish Theorem 1.1. In this section, we we will prove
Theorem 1.1 by utilizing Lemma 4.3, 4.4, and 4.5.

###### Proof of Theorem 1.1.

(1) $\Rightarrow$ (2) : Since the vector space access structures are ideal
using Theorem 3.1, the proof is complete.
(2) $\Rightarrow$ (3) : As the optimal information rate of an ideal access
structure equals one, the proof is complete.
(3) $\Rightarrow$ (4) : Let $\Gamma_{\sim}$ be a reduced access structure of
$\Gamma$ on a set of participants $\mathcal{P}$. Since
$k+1\in\Omega(k+1,\Gamma_{\sim})$, there exist $k+1$ distinct participants
$a_{1},a_{2},\cdots,a_{k},a_{k+1}\in\mathcal{P}$, such that
$\omega(\\{a_{1},a_{2},\cdots,a_{k},a_{k+1}\\},\Gamma_{\sim})=k+1$.
Consequently, we can conclude that the induced access structure
$\Gamma_{\sim}(\\{a_{1},a_{2},\cdots,a_{k},a_{k+1}\\})$ is a
$(k,k+1)$-threshold access structure.

Let us consider $\mathcal{P}^{\prime}\subset\mathcal{P}$, consisting of $m$
participants, where $m\geq k+1$ and
$\\{a_{1},a_{2},\dots,a_{k+1}\\}\subset\mathcal{P}^{\prime}$, such that the
induced structure $\Gamma_{\sim}(\mathcal{P}^{\prime})$ is the
$(k,m)$-threshold access structure, while
$\Gamma(\mathcal{P}^{\prime}\cup\\{b\\})$ is not $(k,m+1)$-threshold access
structure for $b\in\mathcal{P}\setminus\mathcal{P}^{\prime}$. We now assert
that $\mathcal{P}=\mathcal{P}^{\prime}$. To demonstrate this, let us assume
the contrary, that is, $\mathcal{P}\neq\mathcal{P}^{\prime}$. Then, we claim
that there must exist $a_{1},a_{2},\dots,a_{k+1}\in\mathcal{P}^{\prime}$ and
$b\in\mathcal{P}\setminus\mathcal{P}^{\prime}$ such that

  * (i)

$\\{a_{1},a_{2},\dots,a_{k-1},b\\}\in\Gamma$

  * (ii)

$\\{a_{t_{1}},a_{t_{2}},\dots,a_{t_{k-3}},a_{k},a_{k+1},b\\}\notin\Gamma$,
where $1\leq t_{1},t_{2},\dots,t_{k-3}\leq k-1$

or $\\{a_{l_{1}},a_{l_{2}},\dots,a_{l_{k-2}},a_{k},b\\}\notin\Gamma$, where
$1\leq l_{1},l_{2},\dots,l_{k-2}\leq k-1$

or $\\{a_{s_{1}},a_{s_{2}},\dots,a_{s_{k-2}},a_{k+1},b\\}\notin\Gamma$, where
$1\leq s_{1},s_{2},\dots,s_{k-2}\leq k-1$.

To begin, let us establish property $(i)$. Given that
$\mathcal{P}\neq\mathcal{P}^{\prime}$, there exists $A\in{\Gamma_{\sim}}_{0}$
such that $A\cap\mathcal{P}^{\prime}\neq\phi$, and
$A\not\subset\mathcal{P^{\prime}}$. Now, let us consider $k+3$ participants,
denoted as $a_{1},a_{2},\dots,a_{k+1},b_{1},b_{2}\in\mathcal{P}$, where
$a_{1},a_{2},\dots,a_{k+1}\in\mathcal{P}^{\prime}$, and
$b_{1},b_{2}\in\mathcal{P}\setminus\mathcal{P}^{\prime}$. We have two cases to
consider. First, if $|A\cap\mathcal{P}^{\prime}|=k-1$, then we can conclude
that $A=\\{a_{1},a_{2},\dots,a_{k-1},b_{i}\\}\in\Gamma$ for some
$a_{1},a_{2},\dots,a_{k-1}\in\mathcal{P}^{\prime}$ and
$b_{i}\in\mathcal{P}\setminus\mathcal{P}^{\prime}$, where $i=1,2$. Second, if
$|A\cap\mathcal{P}^{\prime}|\neq k-1$, then we deduce that
$A=\\{x_{1},x_{2},\dots,x_{k-2},b_{1},b_{2}\\}\in\Gamma$, where
$x_{1},x_{2},\dots,x_{k-2}\in\mathcal{P}^{\prime}$ and
$b_{1},b_{2}\in\mathcal{P}\setminus\mathcal{P}^{\prime}$.

Let us consider three distinct participants, denoted as
$x_{k-1},x_{k},x_{k+1}\in\mathcal{P}^{\prime}\setminus\\{x_{1},x_{2},\cdots,x_{k-2}\\}$.
Since $x_{1},x_{2},\cdots,x_{k-1},x_{k},x_{k+1}\in\mathcal{P}^{\prime}$ and
$\Gamma_{\sim}(\mathcal{P}^{\prime})$ is a $(k,m)$-threshold access structure,
we have $\omega(\\{x_{1},x_{2},\cdots,x_{k},x_{k+1}\\},\Gamma_{\sim})=k+1$,
which satisfies the condition of Lemma 4.3. Therefore, by utilizing Lemma 4.3,
there exist $j_{1},j_{2},\dots,j_{k-1}$ such that either

$\\{x_{j_{1}},x_{j_{2}},\dots,x_{j_{k-1}},b_{1}\\}\in\Gamma$

or

$\\{x_{j_{1}},x_{j_{2}},\dots,x_{j_{k-1}},b_{2}\\}\in\Gamma.$

Now, property $(i)$ holds for the second case with
$x_{j_{1}}=a_{1},x_{j_{2}}=a_{2},\cdots,x_{j_{k-1}}=a_{k-1}$, and $b_{1}=b$.

Now, we establish property $(ii)$. Since $\Gamma_{\sim}(\mathcal{P}^{\prime})$
is a $(k,m)$-threshold access structure and
$\Gamma_{\sim}(\mathcal{P}^{\prime}\cup\\{b\\})$ is not a $(k,m+1)$-threshold
access structure, where $b\notin\mathcal{P}^{\prime}$, there must exist two
participants, denoted as $y_{k},y_{k+1}\in\mathcal{P}^{\prime}$, such that at
least one of the following conditions holds:

$\displaystyle\\{a_{t_{1}},a_{t_{2}},\dots,a_{t_{k-3}},y_{k},y_{k+1},b\\}\notin\Gamma,\>\text{where}\>\>1\leq
t_{1},t_{2},\dots,t_{k-3}\leq k-1$ (5.1)
$\displaystyle\text{or}\>\>\>\\{a_{l_{1}},a_{l_{2}},\dots,a_{l_{k-2}},y_{k},b\\}\notin\Gamma,\>\text{where}\>\>1\leq
l_{1},l_{2},\dots,l_{k-2}\leq k-1$ (5.2)
$\displaystyle\text{or}\>\>\>\\{a_{s_{1}},a_{s_{2}},\dots,a_{s_{k-2}},y_{k+1},b\\}\notin\Gamma,\>\text{where}\>\>1\leq
s_{1},s_{2},\dots,s_{k-2}\leq k-1,$ (5.3)

where $b\in\mathcal{P}\setminus\mathcal{P}^{\prime}$.

From property $(i)$, we can conclude that there exist
$a_{1},a_{2},\dots,a_{k+1}\in\mathcal{P}^{\prime}$ and
$b\in\mathcal{P}\setminus\mathcal{P}^{\prime}$ such that
$\\{a_{1},a_{2},\dots,a_{k-1},b\\}\in\Gamma$. We must now consider two cases.

First, if $\\{a_{1},a_{2},\dots,a_{k-1}\\}\cap\\{y_{k},y_{k+1}\\}=\phi$, then
property $(ii)$ holds with $y_{k}=a_{k}$ and $y_{k+1}=a_{k+1}$. Second,
without loss of generality, we assume that $a_{1},a_{2},\dots,a_{k-1},y_{k}$
are distinct, and $y_{k+1}=a_{1}$. Since
$|\mathcal{P}^{\prime}|=m^{\prime}\geq k+1$, there exists $z_{k+1}$ such that
$z_{k+1}\in\mathcal{P}^{\prime}\setminus\\{a_{1},a_{2},\dots,a_{k-1},y_{k}\\}$.
Consequently, we can deduce that
$\Gamma_{\sim}(\\{a_{1},a_{2},\dots,a_{k-1},y_{k},z_{k+1},b\\})$ is not a
$(k,k+2)$-threshold access structure by utilizing Equations
($\leavevmode\nobreak\ \ref{tt1}),(\ref{tt2}),(\ref{tt3}).$ Now, we assert
that

$\\{a_{t_{1}},a_{t_{2}},\dots,a_{t_{k-3}},y_{k},z_{k+1},b\\}\notin\Gamma,\>\text{where}\>\>1\leq
t_{1},t_{2},\dots,t_{k-3}\leq k-1$
$\text{or}\>\>\>\\{a_{l_{1}},a_{l_{2}},\dots,a_{l_{k-2}},y_{k},b\\}\notin\Gamma,\>\text{where}\>\>1\leq
l_{1},l_{2},\dots,l_{k-2}\leq k-1$
$\text{or}\>\>\>\\{a_{s_{1}},a_{s_{2}},\dots,a_{s_{k-2}},z_{k+1},b\\}\notin\Gamma,\>\text{where}\>\>1\leq
s_{1},s_{2},\dots,s_{k-2}\leq k-1.$

To accomplish this, let us assume otherwise, that is, suppose that

$\\{a_{t_{1}},a_{t_{2}},\dots,a_{t_{k-3}},y_{k},z_{k+1},b\\}\in\Gamma,\>\text{where}\>\>1\leq
t_{1},t_{2},\dots,t_{k-3}\leq k-1$
$\text{and}\>\>\>\\{a_{l_{1}},a_{l_{2}},\dots,a_{l_{k-2}},y_{k},b\\}\in\Gamma,\>\text{where}\>\>1\leq
l_{1},l_{2},\dots,l_{k-2}\leq k-1$
$\text{and}\>\>\>\\{a_{s_{1}},a_{s_{2}},\dots,a_{s_{k-2}},z_{k+1},b\\}\in\Gamma,\>\text{where}\>\>1\leq
s_{1},s_{2},\dots,s_{k-2}\leq k-1.$

Using property $(i)$, we can find $a_{1},a_{2},\dots,a_{k-1}$ such that
$\\{a_{1},a_{2},\dots,a_{k-1},b\\}\in\Gamma$, as required by Lemma 4.4.
Consequently, $\Gamma_{\sim}(\\{a_{1},a_{2},\dots,a_{k-1},y_{k},z_{k+1},b\\})$
is a $(k,k+2)$-threshold access structure, which creates a contradiction.
Therefore, we can conclude that property $(ii)$ holds with $y_{k}=a_{k}$ and
$z_{k+1}=a_{k+1}$. Since $a_{1},a_{2},\cdots,a_{k+1}\in\mathcal{P}^{\prime}$
and $\Gamma_{\sim}(\mathcal{P}^{\prime})$ is $(k,m)$-threshold access
structure, we have
$\omega(\\{a_{1},a_{2},\dots,a_{k+1}\\},\Gamma_{\sim})=k+1$. Utilizing Lemma
4.5, we can conclude that either $b$ is equivalent to $a_{k}$ or $b$ is
equivalent to $a_{k+1}$. This contradicts the fact that $\Gamma_{\sim}$ is a
reduced access structure. Therefore, we can deduce that
$\mathcal{P}=\mathcal{P}^{\prime}$. Consequently, $\Gamma_{\sim}(\mathcal{P})$
is also $(t,m)$-threshold access structure.

(4) $\Rightarrow$ (1) : The results from Section $3$ establish that (4)
implies (1). According to the definitions, a $(k,n)$-threshold access
structure is an example of a vector space access structure. Therefore, if the
reduced access structure of $\Gamma$ is a $(k,n)$-threshold access structure,
then it is also a vector space access structure. Furthermore, the reduced
access structure of $\Gamma$ is a vector space access structure if and only if
$\Gamma$ is a vector space access structure.

This completes the proof of Theorem 1.1. ∎

## References

  * [1] A. Beimel and O. Farrás, The share size of Secret-Sharing Schemes for almost all access structures and graphs, Cryptology ePrint Archive (2020) https://eprint.iacr.org/2020/664
  * [2] A. Beimel, O. Farrás, and Y. Mintz, Secret Sharing Schemes for very dense graphs, Journal of Cryptology, 29(2) (2016) 336-362.
  * [3] J. Benaloh and J. Leichter, Generalized Secret Sharing and monotone functions, in: CRYPTO’88, LNCS, vol. 403 (1988) 27-35.
  * [4] C. Blundo, A.D. Santis, R. D. Simone, and U. Vaccaro, Tight bounds on the information rate of secret sharing schemes, Designs Codes and Cryptography, 11(2) (1997) 107-122.
  * [5] E.F. Brickell and D.M. Davenport, On the classification of ideal secret sharing schemes, Journal of Cryptology, 4(73) (1991) 123-134.
  * [6] L. Csirmaz and G. Tardos, Optimal information rate of secret sharing schemes on trees, IEEE Trans. on Information Theory, 59(4) (2013) 2527-2530.
  * [7] S. Janbaz, A. Zaghian, and B. Bagherpour, Ideal secret sharing schemes on graph-based 3-homogeneous access structures, Transactions on Combinatorics, 10(2) (2021) 107-120.
  * [8] J. Martí-Farré and C. Pardró, Secret sharing schemes with three or four minimal qualified subsets, Designs Codes and Cryptography, 34 (2005) 17-34.
  * [9] J. Martí-Farré and C. Padró, Secret sharing schemes on access structures with intersection number equal to one, Discrete Applied Mathematics, 154 (2006) 552-563.
  * [10] J. Martí-Farré and C. Pardró, Ideal secret sharing scheme whose minimal qualified subsets have at most three participants, Designs Codes and Cryptography, 52 (2009) 1-14.
  * [11] J. Martí-Farré, A note on secret sharing schemes with three homogeneous access structures, Information Processing Letters, 102 (2007) 133-137.
  * [12] C. Padró, G. Sáez, Secret sharing schemes with bipartite access structure, IEEE Trans. Inform. Theory 46 (2000) 2596-2604.
  * [13] A. Shamir, How to share a secret, Communications of the ACM, 22 (1979) 612-613.
  * [14] D.R. Stinson, An explanation of secret sharing schemes, Designs Codes and Cryptography, (1992) 157-390.
  * [15] T. Tassa and N. Dyn, Multipartite secret sharing by bivariate interpolation, Journal of Cryptology, 22 (2009) 227-258.


# Comment on “Constraints on Low-Energy Effective Theories from Weak Cosmic
Censorship”

Jie Jiang<EMAIL_ADDRESS>Department of Physics, Beijing Normal
University, Beijing 100875, China  Aofei Sang<EMAIL_ADDRESS>Department of Physics, Beijing Normal University, Beijing 100875, China  Ming
Zhang  Corresponding author<EMAIL_ADDRESS>Department of Physics,
Jiangxi Normal University, Nanchang 330022, China

###### Abstract

Recently, it was argued in [Phys. Rev. Lett. 126, 031102 (2021)] that the WCCC
can serve as a constraint to high-order effective field theories. However, we
find there exists a key error in their approximate black hole solution. After
correcting it, their calculation cannot show the ability of WCCC to constrain
the gravitational theories.

Weak cosmic censorship conjecture (WCCC) Penrose:1969pc  is a basic principle
that guarantees the predictability of spacetime. One critical scientific
question is whether the WCCC can give a new constraint to the gravitational
theory. Recently, an attempt to this question was given in Ref. Chenprl .
Using Sorce-Wald’s method SW  under the first-order approximation, the authors
showed that the WCCC fails for some possible generations; thus they argued
that the WCCC provides a constraint to the high-order low-energy effective
theories(EFT). However, after examining their letter, we found that key errors
occur in their approximate black hole solution. We will clarify this issue and
show that their discussion cannot give the constraint to the high-order
theories.

The Lagrangian of the EFT considered in Chenprl  is given by

$\displaystyle\begin{aligned}
\mathcal{L}&=\frac{1}{2}R-\frac{1}{4}F_{ab}F^{ab}+c_{1}R^{2}+c_{2}R_{ab}R^{ab}+c_{3}R_{abcd}R^{abcd}\\\
&+c_{4}RF_{ab}F^{ab}+c_{5}R_{ab}F^{ac}F^{b}{}_{c}+c_{6}R_{abcd}F^{ab}F^{cd}\\\
&+c_{7}F_{ab}F^{ab}F_{cd}F^{cd}+c_{8}F_{ab}F^{bc}F_{cd}F^{da}\end{aligned}$
(1)

where $c_{i}$’s are some coupling constants which are treated as small
parameters in the calculations. The equation of motion (EOM) is given by

$\displaystyle\begin{aligned}
H^{ab}=0\,,\quad\quad\nabla_{a}E_{F}^{ab}&=0\,,\end{aligned}$ (2)

in which

$H^{ab}=E_{R}^{cde(a}R_{cde}{}^{b)}+2\nabla_{c}\nabla_{d}E_{R}^{(a|c|b)d}-E_{F}^{c(a}F^{b)}{}_{c}-\frac{1}{2}g^{ab}\mathcal{L}\,,$

with $E_{R}^{abcd}=\partial\mathcal{L}/\partial R_{abcd}$ and
$E_{F}^{ab}=\partial\mathcal{L}/\partial F_{ab}$.

First, we reexamine the solution given by Eqs. (6) and (7) in Chenprl . With a
straightforward calculation, it is easy to check

$\displaystyle\begin{aligned}
\nabla_{a}E_{F}^{ab}(dt)_{b}&=\frac{2q^{3}}{r^{7}}[c_{2}+4c_{3}+10c_{4}+3(c_{5}+c_{6})]+\mathcal{O}(c_{i}^{2})\quad\\\
&\neq 0\,,\end{aligned}$ (3)

which means that there are some errors in the solution given by Ref. Chenprl .
We start with the most general spherically symmetric static metric

$\displaystyle\begin{aligned}
ds^{2}=-f(r)dv^{2}+2\mu(r)dvdr+r^{2}(d\theta^{2}+\sin^{2}\theta
d\phi^{2})\,,\end{aligned}$ (4)

and the Maxwell field

$\displaystyle\begin{aligned} \bm{A}=\Psi(r)dv\,.\end{aligned}$ (5)

Under the leading-order correction of $c_{i}$, we can expand the solution to

$\displaystyle\begin{aligned}
&f_{0}(r)=1-\frac{m}{r}+\frac{q^{2}}{2r^{2}}+f_{1}(r)+\mathcal{O}(c_{i}^{2})\,,\\\
\mu_{0}(r)=1+&\mu_{1}(r)+\mathcal{O}(c_{i}^{2})\,,\,\Psi_{0}(r)=-\frac{q}{r}+\Psi_{1}(r)+\mathcal{O}(c_{i}^{2})\,,\end{aligned}$
(6)

where we used the fact that the background spacetime is a Reissner-Nordstrom
black hole solution. $f_{1}(r)$, $\mu_{1}(r)$ and $\Psi_{1}(r)$ are the linear
functions of $c_{i}$.

From the EOM $\nabla_{a}E_{F}^{ab}=0$, we can obtain

$\displaystyle\begin{aligned}
\Psi_{1}(r)=&\frac{2q}{5r^{5}}[c_{5}q^{2}+c_{6}(6q^{2}-5mr)+(8c_{7}+4c_{8})q^{2}]\\\
&+q\int\frac{\mu_{1}(r)}{r^{2}}dr\,.\end{aligned}$ (7)

Substituting the above result to $H^{vv}=0$, it is easy to obtain

$\displaystyle\begin{aligned}
\mu_{1}(r)=\frac{q^{2}}{r^{4}}(c_{2}+4c_{3}+10c_{4}+3c_{5}+3c_{6})\,,\end{aligned}$
(8)

which gives

$\displaystyle\begin{aligned}
\Psi_{1}(r)=&-\frac{q^{3}}{5r^{2}}\left[c_{2}+4c_{3}+10c_{4}+c_{5}\right.\\\
&\left.-\left(9-10mrq^{-2}\right)c_{6}-16c_{7}-8c_{8}\right]\,.\end{aligned}$
(9)

This result shows a different expression to $A_{a}$ given by Eq. (6) of
Chenprl . Finally, using $H^{\theta\theta}=0$, we can find that $f(r)$ shows
the same expression of $g_{tt}$ in Chenprl . Therefore, in the first-order
gedanken experiments, the condition of not destroying an extremal solution is
also given by Eq. (14) in Chenprl , i.e.,

$\displaystyle\begin{aligned} \delta m-\sqrt{2}\delta
q\left(1+\frac{4c_{0}}{5q^{2}}\right)\geq 0\,.\end{aligned}$ (10)

with

$\displaystyle\begin{aligned} c_{0}\equiv
c_{2}+4c_{3}+c_{5}+c_{6}+4c_{7}+2c_{8}\,.\end{aligned}$ (11)

The condition that the test particle can drop into the horizon or the
infalling matter satisfies the null energy condition is given by Eqs. (18) and
(27) of Chenprl , i.e.,

$\displaystyle\begin{aligned} \delta m-\Phi_{H}^{c}\delta q\geq
0\,,\end{aligned}$ (12)

in which $\Phi_{H}^{c}\equiv-\left.\xi^{a}A_{a}\right|_{H}$ with
$\xi^{a}=(\partial/\partial v)^{a}$ is the electric potential of the black
hole. Using our corrected expression (9) of $A_{a}$, for the extremal black
hole, we have

$\displaystyle\begin{aligned}
\Phi_{H}^{\text{ext}}=\sqrt{2}\left(1+\frac{4c_{0}}{5q^{2}}\right)+\mathcal{O}(c_{i}^{2})\,,\end{aligned}$
(13)

which is different from Eq. (11) of Chenprl . Then, the inequality (12) shows
the same expression as inequality (10), which implies that the extremal
charged black hole cannot be destroyed. This result is just contrary to that
shown by Ref. Chenprl  where there are destructions of the extremal black
holes. This implies that after correcting the solution, their letter Chenprl
cannot show the ability of WCCC to constrain the gravitational theories.

## References

  * (1) R. Penrose, Gravitational collapse: The role of general relativity, Riv. Nuovo Cim. 1 , 252-276 (1969).
  * (2) B. Chen, F. L. Lin, B. Ning and Y. Chen, Constraints on low-energy effective theories from weak cosmic censorship, Phys. Rev. Lett. 126, 031102 (2021).
  * (3) J. Sorce and R.M. Wald, Gedanken experiments to destroy a black hole. II. Kerr-Newman black holes cannot be overcharged or overspun, Phys. Rev. D 96, 104014 (2017).