Constructing Q-Ideals for Boolean Semiring Partitioning Using Seeds

Abstract

Semiring partitioning is widely used in mathematics, computer science, and data analysis. The purpose of this paper is to add to the theory of semirings by proposing a novel method to construct Q-ideals for partitioning Boolean semirings. We introduce the set of all seeds—all s-tuples over a particular Boolean algebra—and the notion of their weight and complement. Utilizing this new method for constructing Q-ideals, we develop a nested hierarchical partitioning algorithm based on the weight of selected seeds. Additionally, we determine the maximal semiring homomorphism corresponding to this proposed method.

1. Introduction

A semiring is a generalization of a ring, where it is not required that each element has an additive inverse. The definition of a semiring varies throughout the literature. In this paper, we use the definition of a semiring in [1]. The most common example of a semiring is the set of natural numbers, i.e., the set $\{1,2,3,\dots \}$. Semirings were formally introduced by Vandiver in 1934 (see [2]) but made its first implicit appearance in the work of authors like Dedekind forty years earlier.

Semirings have been studied extensively over the past few decades, either in an attempt to extend the theory of semigroups and rings, or, to broaden its applications in mathematics, computer science, and data analysis; for example, see [3,4,5]. Bourne [6] was the first to investigate semirings in topology, and Ahsan, Saifullah, and Khan [7] initiated the study of semirings in a fuzzy context. The theory of semirings has developed across different branches of pure mathematics, including graph theory, geometry, topology, and linear algebra; for example, see [8,9,10,11]. Semirings are also used in various applications in computer science; for example, see [12,13].

The properties of ideals in semirings do not necessarily coincide with those in rings. For example, it is well known that the quotient structure of a ring modulo an ideal partitions the ring. However, this is not necessarily the case for semirings, making the task of generalizing ring theorems to semirings a non-trivial matter. Authors like Bourne, Henriksen, and LaTorre used equivalence relations to determine quotient structures of semirings using ideals (see [14,15,16]). Using this approach, LaTorre [16] derived multiple analogs of ring theorems for a large class of semirings. Bourne [14] and LaTorre [16] attempted to derive a precise analog of the Fundamental Theorem of Homomorphisms but encountered some difficulty.

Allen [17] remedied this by defining a special class of ideals, namely partitioning ideals (also called Q-ideals). A Q-ideal of a semiring S is an ideal I of S for which there exists a subset $Q\subseteq S$ such that $\{q+I:q\in Q\}$ partitions S. This notion allowed Allen to generalize the Fundamental Theorem of Homomorphisms for a large class of semirings.

The study of Q-ideals continued. Chaudhari and Ingale [18] introduced a partitioning ideal for ternary semirings. Sharma [19] continued the investigation of the partitioning ideals of skew group semirings. Chaudhari, Nemade, and Davvaz [20] extended some theory of partitioning ideals to $(m,n)$ semirings, and could thus generalize the Fundamental Theorem of Homomorphisms to include $(m,n)$ semirings.

Partitioning elements of sets according to their properties is central to mathematics and many applications thereof. For example, in computer science, the k-means clustering method partitions Boolean vectors into k groups. The method works by iteratively assigning data points to the closest cluster centroid and then updating the centroids based on the assigned points. This partitioning method has wide applications in biology, marketing, medicine, and psychology (see [21]). In mathematics, a method called matroid partitioning partitions a set of elements into matroids, combinatorial structures that generalize the concept of linear independence in graphs (for example, see [22]). The origin of the topic of this paper lies in the question of how best to partition site-by-species matrices in ecology. This is related to the partitioning of Boolean vectors, which is prominent in computer science and mathematics.

Boolean rings are closely related to Boolean algebras, in that every Boolean ring can be interpreted as a Boolean algebra by treating its operators as logical operators. Conversely, every Boolean algebra can be interpreted as a Boolean ring by treating its operators ∨ and ∧ as addition and multiplication, respectively. These structures are thus mathematically equivalent, since they describe the same set of elements and have operators that behave in a similar way. Consequently, there is a close connection between ideals of Boolean algebras and ideals of Boolean rings, in that an ideal of a Boolean algebra can be mapped to an ideal of the corresponding Boolean ring.

In this paper, we study the ideals of the Boolean semiring $V_s=(V_s,+,·{,}^{\prime },\mathbf{0},\mathbf{1})$, where $V_s$ is the product of the two-element Boolean algebra $\{0,1\}$ with itself s times. We notice that, since $V_s$ is also a Boolean algebra, studying its ideals means studying the ideals of its corresponding Boolean ring, a well-known topic. We choose to study $V_s$ as a semiring and develop partitions thereof using the language of Q-ideals, thus extending the work of Allen in [17].

In Section 2, we give the preliminary material needed for this paper. We construct the Boolean semiring $V_s$, the set of all s-tuples over a particular Boolean algebra. In Section 3, we propose an alternative construct for the partitioning of $V_s$ using the ideal $I_{\mathbf{v}}$ (see Theorem 1) and call this the Q-ideal partition. We prove that the cells of a Q-ideal partition have equal cardinality and express this cardinality in terms of the number of non-zero elements of its initial Boolean vector, called the initial seed (Lemma 3). We prove that each seed induces a unique Q-ideal partition (Lemma 4). Using this novel Q-ideal, we develop a hierarchical partitioning algorithm (Algorithm 1) and prove that it is nested (Theorem 2). We determine the number of steps after which this algorithm terminates, as well as the number of cells into which $V_s$ is partitioned at each step of the algorithm (Lemma 5).

As Allen introduced maximal semiring homomorphisms to construct Q-ideals that partition semirings (see [1]), we further investigate the relationship between these concepts. While his result is insightful, the method for identifying such homomorphisms remains unclear. To connect the notions of maximal semiring homomorphisms and Q-ideal partitions, we draw inspiration from one of Allen’s examples to determine the maximal semiring homomorphism corresponding to the Q-ideal partition (see Lemma 7).

2. Preliminary Material

We start with the definition of a special type of algebra.

Definition 1

([23]). A Boolean algebra is a tuple $B=(B,+,·{,}^{\prime },0,1)$, where B is a non-empty set, and + and · are two binary operations on B, such that

(a) 

Both + and · are commutative;

(b) 

Each operation is distributive over the other;

(c)  

For all $b\in B$, we have $b+0=b$ and $b·1=b$;

(d) 

For all $b\in B$, there exists $b^{\prime }\in B$, called the complement of b, such that $b+b^{\prime }=1$ and $b·b^{\prime }=0$.

A Boolean matrix is a matrix over a Boolean algebra [23]. The origin of the topic of this paper lies in the question of how best to partition the rows of a subclass of Boolean matrices, called site-by-species matrices. Below, we give the well-known definition of such a matrix.

Definition 2.

A site-by-species matrix of size $n\times s$ is a Boolean matrix

$$\begin{array}{cccc}& s_1& \cdots & s_s\\ L_1& p_{11}& \cdots & p_{1s}\\ ⋮& ⋮& \ddots & ⋮\\ L_n& p_{n1}& \cdots & p_{ns}\end{array}$$

with rows $L_1,\dots ,L_n$ representing sites and columns $s_1,\dots ,s_s$ representing species. For each i and j, we have $p_{ij}=1$ if species $s_j$ is detected at site $L_i$; otherwise, $p_{ij}=0$.

The following example demonstrates Definition 2.

Example 1.

Consider the $3\times 3$ site-by-species matrix

$$\begin{array}{cccc}& s_1& s_2& s_3\\ L_1& 0& 0& 1\\ L_2& 1& 0& 0\\ L_3& 0& 1& 0\end{array}$$

where $L_1$, $L_2$, and $L_3$ represent three distinct ponds in the Western Cape, and $s_1$, $s_2$, and $s_3$ represent three distinct fish species. Since $p_{13}=1$, species $s_3$ is detected at site $L_1$, and since $p_{31}=0$, species $s_1$ is not detected at site $L_3$.

In order to achieve this objective, we express the rows of such a matrix as Boolean vectors

$$\mathbf{v}=(v_1,\dots ,v_s)$$

where, for each i, the element $v_i$ belongs to the well-known two-element Boolean algebra $B_0=(\{0,1\},+,·{,}^{\prime },0,1)$. Thus, the problem of partitioning the rows of a site-by-species matrix using algebraic tools reduces to partitioning a set of Boolean vectors over $B_0$.

The following result is well known.

Lemma 1.

If $(A,+,·)$ is an algebraic structure, $A_1,\dots ,A_k$ is a partition of A and $X\subseteq A$, then $X_1,\dots ,X_k$ is a partition of X, where $X_i=X\cap A_i$ for each $i\in \{1,\dots ,k\}$.

Proof. 

Let $(A,+,·)$ be an algebraic structure, and let $A_1,\dots ,A_k$ be a partition of A. Let $X\subseteq A$, and for each $i\in \{1,\dots ,k\}$, define $X_i=X\cap A_i$. We show that $X_1,\dots ,X_k$ is a partition of X, that is, (1) $X={\bigcup }_{i=1}^k{X}_i$ and (2) $X_i\cap X_j=\varnothing$ for each $i\ne j$.

(1)

If $x\in X$, then since $X\subseteq A$, we have $x\in A$. Since $A={\bigcup }_{i=1}^k{A}_i$, it follows that $x\in A_i$ for some $i\in \{1,\dots ,k\}$. Hence, $x\in X\cap A_i$ for some $i\in \{1,\dots ,k\}$, so that $x\in X_i$ for some $i\in \{1,\dots ,k\}$. By the definition of the set union, we deduce that $x\in {\bigcup }_{i=1}^k{X}_i$. Since x was arbitrary, we conclude that $X\subseteq {\bigcup }_{i=1}^k{X}_i$. Conversely, if $x\in {\bigcup }_{i=1}^k{X}_i$, then $x\in X_i$ for some $i\in \{1,\dots ,k\}$. Since $X_i\subseteq X$ for each $i\in \{1,\dots ,k\}$, it follows that $x\in X$. Since x was arbitrary, we conclude that ${\bigcup }_{i=1}^k{X}_i\subseteq X$. Therefore, $X={\bigcup }_{i=1}^k{X}_i$.

(2)

Suppose $i,j\in \{1,\dots ,k\}$ with $i\ne j$. Suppose, for contradiction, that $X_i\cap X_j\ne \varnothing$. It follows by construction that $(X\cap A_i)\cap (X\cap A_j)=X\cap (A_i\cap A_j)\ne \varnothing$; thus, $A_i\cap A_j\ne \varnothing$ is a contradiction, since $A_1,\dots ,A_k$ is a partition of A. From this, we deduce $X_i\cap X_j=\varnothing$ for each $i\ne j$.

This completes the proof.    □

Our objective is to find partitions of the algebraic structure where the elements are all possible rows $\mathbf{v}=(v_1,\dots ,v_s)$ over $B_0$ (also termed in ecology, the set of all possible realizations), and then, by applying Lemma 1, to find partitions of the rows of a site-by-species matrix.

Definition 3

([23]). Let $V_s$ denote the set of s-tuples $\mathbf{v}=(v_1,\dots ,v_s)$ over $B_0$, called seeds of length s, where the i-th entry of  $\mathbf{v}$ is denoted by $v_i$. The weight of $\mathbf{v}$ , denoted by  $\omega \left(\mathbf{v}\right)$ , is the number of non-zero entries in  $\mathbf{v}$ , while the complement of $\mathbf{v}$ , denoted by ${\mathbf{v}}^{\prime }$ , is the distinct element in $V_s$ such that $v_i^{\prime }=1$ if and only if $v_i=0$. We define two binary operators + and · on $V_s$ by making use of the operators on $B_0$ componentwise. That is,

$$\mathbf{v}+\mathbf{w}=(v_1+w_1,\dots ,v_s+w_s)$$

and

$$\mathbf{v}·\mathbf{w}=(v_1·w_1,\dots ,v_s·w_s)$$

for all $\mathbf{v},\mathbf{w}$ in $V_s$.

Notation 1.

We denote the seeds $(0,\dots ,0)$ and $(1,\dots ,1)$ by $\mathbf{0}$ and $\mathbf{1}$, respectively.

It is clear that $V_s$ is a product of $B_0$ with itself s times. In the next result, we use Definition 1 in order to rigorously prove that $V_s$ is also a Boolean algebra.

Proposition 1.

$V_s=(V_s,+,·{,}^{\prime },\mathbf{0},\mathbf{1})$is a Boolean algebra.

Proof. 

Let $\mathbf{v},\mathbf{w},\mathbf{z}\in V_s$.

(a)

If + and · are commutative in $B_0$, then

$$\mathbf{v}+\mathbf{w}=(v_1+w_1,\dots ,v_s+w_s)=(w_1+v_1,\dots ,w_s+v_s)=\mathbf{w}+\mathbf{v}$$

and

$$\mathbf{v}·\mathbf{w}=(v_1·w_1,\dots ,v_s·w_s)=(w_1·v_1,\dots ,w_s·w_s)=\mathbf{w}·\mathbf{v}.$$

(b)

If + and · are distributive over each other in $B_0$, then

$$\begin{array}{cc}\hfill \mathbf{v}·(\mathbf{w}+\mathbf{z})& =(v_1·(w_1+z_1),\dots ,v_s·(w_s+z_s))\hfill \\ \hfill & =((v_1·w_1)+(v_1·z_1),\dots ,(v_s·w_s)+(v_s·z_s))\hfill \\ \hfill & =(v_1·w_1,\dots ,v_s·w_s)+(v_1·z_1,\dots ,v_s·z_s)\hfill \\ \hfill & =\mathbf{v}·\mathbf{w}+\mathbf{v}·\mathbf{z}\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill \mathbf{v}+(\mathbf{w}·\mathbf{z})& =(v_1+(w_1·z_1),\dots ,v_s+(w_s·z_s))\hfill \\ \hfill & =((v_1+w_1)·(v_1+z_1),\dots ,(v_s+w_s)·(v_s+z_s))\hfill \\ \hfill & =(v_1+w_1,\dots ,v_s+w_s)·(v_1+z_1,\dots ,v_s+z_s)\hfill \\ \hfill & =(\mathbf{v}+\mathbf{w})·(\mathbf{v}+\mathbf{z}).\hfill \end{array}$$

(c)

Since $v+0=v$ and $v·1=v$ for all v in $B_0$, it follows that

$$\mathbf{v}+\mathbf{0}=(v_1+0,\dots ,v_s+0)=(v_1,\dots ,v_s)=\mathbf{v}$$

and

$$\mathbf{v}·\mathbf{1}=(v_1·1,\dots ,v_s·1)=(v_1,\dots ,v_s)=\mathbf{v}.$$

(d)

Since, for each $\mathbf{v}$ in $V_s$, we have that $v_i+v_i^{\prime }=1$ and $v_i·v_i^{\prime }=0$ for all $i\in \{1,\dots ,s\}$, it follows that

$$\mathbf{v}+{\mathbf{v}}^{\prime }=(v_1+v_1^{\prime },\dots ,v_s+v_s^{\prime })=(1,\dots ,1)=\mathbf{1}$$

and

$$\mathbf{v}·{\mathbf{v}}^{\prime }=(v_1·v_1^{\prime },\dots ,v_s·v_s^{\prime })=(0,\dots ,0)=\mathbf{0}.$$

It follows by Definition 1 that $V_s$ is a Boolean algebra.    □

The partitioning of Boolean algebras using ideals is a well-known topic. We make use of a different approach by viewing $V_s$ as a Boolean semiring.

Definition 4

([1]). A semiring is a tuple $(S,+,·)$, where S is a non-empty set, and + and · are two associative binary operators on S, such that

(i) 

+ is commutative;

(ii) 

· distributes over + both from the left and from the right;

(iii) 

There exists $0\in S$ such that $s+0=s$ and $s·0=0$ for each $s\in S$.

A semiring is commutative if the multiplication operator · is commutative. Furthermore, a semiring S is Boolean if for all s in S, the condition

$$s·s=s$$

holds.

There is a distinction between a Boolean semiring and an idempotent semiring. An idempotent semiring is a semiring in which the addition operation has the idempotent property; that is, for each s in S, we have $s+s=s$. Since the additive operation in a Boolean semiring B corresponds to set union, and $A\cup A=A$ for every set A, it follows that for each b in B, we have $b+b=b$. This shows that every Boolean semiring is an idempotent semiring, but the converse is not necessarily true. That is, not every idempotent semiring is a Boolean semiring. This is because the multiplication operation in an idempotent semiring need not have the idempotency property.

We give an example of an idempotent semiring that is not a Boolean semiring.

Example 2.

Consider the set ${\mathbb{R}}^{\ge 0}$ of positive real numbers (including zero) equipped with two binary operators: $r+s=max(r,s)$ and $r·s=r\times s$, where × is the usual multiplication. We verify that $({\mathbb{R}}^{\ge 0},max,\times )$ is a semiring.

(a) 

For each $r,s\in {\mathbb{R}}^{\ge 0}$, we have $r+s=max(r,s)=max(s,r)=s+r$.

(b) 

For each $r,s,t\in {\mathbb{R}}^{\ge 0}$, we have

$$\begin{array}{c}\hfill r·(s+t)=r\times max(s,t).\end{array}$$

Suppose w.l.o.g. that $s\le t$, then $max(s,t)=t$ and $r\times s\le r\times t$, so that

$$r·(s+t)=r\times t=max(r\times s,r\times t)=max(r·s,r·t)=r·s+r·t.$$

A similar method can be used to show that $(r+s)·t=r·t+s·t$.

(c) 

For each $r\in {\mathbb{R}}^{\ge 0}$, we have $r+0=max(r,0)=r$ and $r·0=r\times 0=0$.

It follows by Definition 4 that $({\mathbb{R}}^{\ge 0},max,\times )$ is a semiring. Moreover, since $r+r=max(r,r)=r$ for each r in ${\mathbb{R}}^{\ge 0}$, it follows that $({\mathbb{R}}^{\ge 0},max,\times )$ is an idempotent semiring. However, since × is not idempotent (for example, ${\displaystyle \frac{1}{2}}\times {\displaystyle \frac{1}{2}}={\displaystyle \frac{1}{4}}\ne {\displaystyle \frac{1}{2}}$), $({\mathbb{R}}^{\ge 0},max,\times )$ is not a Boolean semiring.

We give an example of a Boolean semiring.

Example 3

([1]). Let $(S,\le )$ be a well-ordered set where $a+b=max(a,b)$ and $a·b=min(a,b)$ for all $a,b\in S$. It follows that $S=(S,+,·)$ is a semiring. To see why S is Boolean, notice that for every a in S, we have

$$a·a=min(a,a)=a.$$

Since S is a Boolean semiring, it is also an idempotent semiring.

Notation 2.

From here onwards, to indicate the multiplication of two vectors $\mathbf{x}$ and $\mathbf{y}$, we will write $\mathbf{x}\mathbf{y}$ instead of $\mathbf{x}·\mathbf{y}$, and for elements a and b that are not vectors, we will write $ab$ instead of $a·b$. In cases where two numbers are multiplied, for example, 0 and 1, we will write $0·1$ instead of 01 to avoid confusion.

In ring theory, ideals are significant. The same notion exists for semirings.

Definition 5

([1]). An ideal in a commutative semiring $(S,+,·)$ is a subset $I\subseteq S$ such that for any $a,b\in I$ and $c\in S$, we have $a+b\in I$ and $ac\in I$.

It is well known that for any ring R and ideal I of R, the quotient $R/I$ of R with respect to I is a partition of R. We give an example to show that this property cannot be generalized to include semirings.

Example 4.

Consider the ideal $I=\left\{\right(0,0,0),(1,0,0),(0,0,1),(1,0,1\left)\right\}$ in the semiring $V_3$. We have

$$(0,1,0)+I=\left\{\right(0,1,0),(1,1,0),(0,1,1),(1,1,1\left)\right\}$$

and

$$(1,1,1)+I=\left\{\right(1,1,1\left)\right\},$$

so that

$$\left(\right(0,1,0)+I)\cap \left(\right(1,1,1)+I)=\left\{\right(1,1,1\left)\right\}\ne \varnothing .$$

Therefore, $V_s/I$ does not partition $V_3$.

A special class of ideals was introduced by Allen to partition a semiring.

Definition 6

([1]). An ideal I of a semiring $(S,+,·)$ is a Q-ideal if there exists a subset $Q\subseteq S$, such that the following conditions hold:

(i) 

$S=\bigcup \{q+I:q\in Q\}$;

(ii) 

For all $q,q^{*}\in Q$, we have $(q+I)\cap (q^{*}+I)=\varnothing$ if and only if $q\ne q^{*}$.

We give examples of Q-ideals.

Example 5.

Consider the set $\mathbb{N}$ of natural numbers $\{0,1,2,\dots \}$ with usual addition and multiplication. Let $I=\{n\in \mathbb{N}:{\displaystyle \frac{n}{2}}\in \mathbb{N}\}$ and $Q=\{0,1\}$. It can be argued that, since $0+I$ is the set of all even natural numbers (including zero), and $1+I$ is the set of all odd natural numbers, we have $\mathbb{N}=(0+I)\cup (1+I)$ and $(0+I)\cap (1+I)=\varnothing$. Therefore, $\{q+I:q\in \{0,1\left\}\right\}$ is a Q-ideal of $\mathbb{N}$.

Example 6

([1]). Let $(S,\le )$ be a well-ordered set where $a+b=max(a,b)$ and $ab=min(a,b)$ for all $a,b\in S$. Fix $c\in S$. Let $I_c=\{s\in S:s\le c\}$ and $Q=\left\{0\right\}\cup \{s\in S:s\le c\}$. Then, $I_c$ is a Q-ideal of S.

Since $V_s$ is a semiring, partitioning $V_s$ requires constructing Q-ideals of $V_s$. We adopt the same construction used by Allen in Example 6.

We start by defining the following relation on $V_s$.

Definition 7

([23]). For any $\mathbf{a},\mathbf{v}\in V_s$ , we say that $\mathbf{a}$ is related to $\mathbf{v}$ and write $\mathbf{a}\le \mathbf{v}$ , if for all $i\in \{1,\dots ,s\}$ , we have that $a_i=1$ implies that $v_i=1$ . We say that $\mathbf{a}$ is unrelated to $\mathbf{v}$ , and write $\mathbf{a}\le \mathbf{v}$ , if there exists an $i\in \{1,\dots ,s\}$ such that $a_i=1$ and $v_i=0$.

Proposition 2.

The following holds:

(a) 

For all $\mathbf{v}\in V_s$ , we have that ${\mathbf{v}}^{\prime }\le \mathbf{v}$.

(b) 

For all $\mathbf{v},\mathbf{w}\in V_s$ , we have that $\mathbf{v}\le \mathbf{w}$ if and only if ${\mathbf{w}}^{\prime }\le {\mathbf{v}}^{\prime }$.

(c) 

$(V_s,\le )$ is an ordered set with the least element $\mathbf{0}$ and the largest element $\mathbf{1}$.

(d) 

For all $\mathbf{v}\in V_s$, we have that $I_{\mathbf{v}}=\{\mathbf{x}\in V_s:\mathbf{x}\le \mathbf{v}\}$ is an ideal in $V_s$.

Proof. 

(a)

This follows since, for all $i\in \{1,\dots ,s\}$, if $v_i^{\prime }=1$, then $v_i=0$.

(b)

For any $\mathbf{v},\mathbf{w}\in V_s$, suppose that $\mathbf{v}\le \mathbf{w}$, and let $i\in \{1,\dots ,s\}$ be arbitrary. If $w_i^{\prime }=1$, then $w_i=0$, so that $v_i=0$ (since $\mathbf{v}\le \mathbf{w}$), and thus $v_i^{\prime }=1$. Since i was arbitrary, we have that ${\mathbf{w}}^{\prime }\le {\mathbf{v}}^{\prime }$. Hence, ${\mathbf{w}}^{\prime }\le {\mathbf{v}}^{\prime }$ whenever $\mathbf{v}\le \mathbf{w}$. Conversely, suppose that ${\mathbf{w}}^{\prime }\le {\mathbf{v}}^{\prime }$. If $v_i=1$, then $v_i^{\prime }=0$, so that $w_i^{\prime }=0$ (since ${\mathbf{w}}^{\prime }\le {\mathbf{v}}^{\prime }$), and thus $w_i=1$. Since i was arbitrary, we have that $\mathbf{v}\le \mathbf{w}$. Hence, $\mathbf{v}\le \mathbf{w}$ whenever ${\mathbf{w}}^{\prime }\le {\mathbf{v}}^{\prime }$.

(c)

Suppose $\mathbf{v},\mathbf{w},\mathbf{z}\in V_s$ and that $i\in \{1,\dots ,s\}$.

(i)

If $v_i=1$, then $v_i=1$. Hence, $\mathbf{v}\le \mathbf{v}$, so that ≤ is reflexive.

(ii)

Suppose that $\mathbf{v}\le \mathbf{w}$ and that $\mathbf{w}\le \mathbf{z}$. If $v_i=1$, then $w_i=1$ (since $\mathbf{v}\le \mathbf{w}$), and thus $z_i=1$ (since $\mathbf{w}\le \mathbf{z}$). Hence, $\mathbf{v}\le \mathbf{z}$, so that ≤ is transitive.

(iii)

Suppose that $\mathbf{v}\le \mathbf{w}$ and that $\mathbf{w}\le \mathbf{v}$. If $v_i=1$, then $w_i=1$ (since $\mathbf{v}\le \mathbf{w}$). If $v_i=0$, then $w_i=0$ (since $\mathbf{w}\le \mathbf{v}$). Hence, $v_i=w_i$ for all $i\in \{1,\dots ,s\}$, so that $\mathbf{v}=\mathbf{w}$. Therefore, ≤ is antisymmetric.

It follows from (i), (ii), and (iii) that ≤ is an order relation on $V_s$, so that $(V_s,\le )$ is an ordered set. To see why $\mathbf{0}$ is the least element in $V_s$, notice that $v_i=0$ whenever $0_i=0$. Hence, $\mathbf{0}\le \mathbf{v}$. To see why $\mathbf{1}$ is the largest element of $V_s$, notice that $1_i=1$ whenever $v_i=1$. Hence, $\mathbf{v}\le \mathbf{1}$.

(d)

Suppose $\mathbf{x},\mathbf{y}\in I_{\mathbf{v}}$ and $\mathbf{z}\in V_s$. Let $i\in \{1,\dots ,s\}$ be arbitrary. If $x_i+y_i=1$, then either $x_i=1$ or $y_i=1$ (or both). Suppose w.l.o.g. that $x_i=1$, then $v_i=1$ (since $\mathbf{x}\le \mathbf{v}$). Since i was arbitrary, it follows that $\mathbf{x}+\mathbf{y}\le \mathbf{v}$. Hence, $\mathbf{x}+\mathbf{y}\in I_{\mathbf{v}}$. If $x_i{z}_i=1$, then $x_i=1$ (and $z_i=1$), so that $v_i=1$ (since $\mathbf{x}\le \mathbf{v}$). Since i was arbitrary, it follows that $\mathbf{x}\mathbf{z}\le \mathbf{v}$. Hence, $\mathbf{x}\mathbf{z}\in I_{\mathbf{v}}$. It follows that $I_{\mathbf{v}}$ is an ideal in $V_s$.

   □

If we were to use the same Q as Allen did in Example 6, then $I_{\mathbf{v}}$ need not be a Q-ideal of $V_s$. We illustrate this with the following example:

Example 7.

If $\mathbf{v}=(1,0,1)$, then

$$I_{\mathbf{v}}=\{(0,0,0),(1,0,0),(0,0,1),(1,0,1)\}$$

and

$$Q=\left\{\right(0,0,0),(0,1,0),(1,1,0),(0,1,1),(1,1,1\left)\right\}.$$

We have that

$$(1,1,0)+I_{\mathbf{v}}=\{(1,1,0),(1,1,1)\}$$

and

$$\{(0,1,0)+I_{\mathbf{v}}=\{(0,1,0),(1,1,0),(0,1,1),(1,1,1)\},$$

so that

$$((1,1,0)+I_{\mathbf{v}})\cap ((0,1,0)+I_{\mathbf{v}})=\{(1,1,0),(1,1,1)\}\ne \varnothing .$$

Since Definition 6(ii) is not satisfied, $I_{\mathbf{v}}$ is not a Q-ideal of $V_3$.

In the next result, we show under which conditions $I_{\mathbf{v}}$ will be a Q-ideal of $V_s$ for Q as defined in Example 6.

Lemma 2.

Let $\mathbf{v}\in V_s$. It follows that $I_{\mathbf{v}}$ is a Q-ideal of $V_s$, where $Q=\left\{\mathbf{0}\right\}\cup \{\mathbf{x}\in V_s:\mathbf{x}\le \mathbf{v}\}$ , if and only if $\mathbf{v}\in \{\mathbf{0},\mathbf{1}\}$.

Proof. 

Suppose $\mathbf{v}\in \{\mathbf{0},\mathbf{1}\}$. If $\mathbf{v}=\mathbf{0}$, then $\bigcup \{\mathbf{q}+I_{\mathbf{0}}:\mathbf{q}\in Q\}=\bigcup \{\mathbf{q}+\left\{\mathbf{0}\right\}:\mathbf{q}\in V_s\}=V_s.$ Moreover, Definition 6(ii) is satisfied since for all $\mathbf{q},{\mathbf{q}}^{*}\in V_s$, we have $\left\{\mathbf{q}\right\}\cap \left\{{\mathbf{q}}^{*}\right\}=\varnothing$ if and only if $\mathbf{q}\ne {\mathbf{q}}^{*}$. If $\mathbf{v}=1$, then $\bigcup \{\mathbf{q}+I_{\mathbf{1}}:\mathbf{q}\in Q\}=\bigcup \{\mathbf{q}+V_s:\mathbf{q}\in \left\{\mathbf{0}\right\}\}=V_s.$ Moreover, Definition 6(ii) is satisfied, since $Q=\left\{\mathbf{0}\right\}$. Therefore, $I_{\mathbf{v}}$ is a Q-ideal of $V_s$ whenever $\mathbf{v}\in \{\mathbf{0},\mathbf{1}\}$.

Conversely, suppose $\mathbf{v}\notin \{\mathbf{0},\mathbf{1}\}$. Since $\mathbf{1}\in {\mathbf{v}}^{\prime }+I_{\mathbf{v}}$ and $\mathbf{1}\in \left\{\mathbf{1}\right\}=\mathbf{1}+I_{\mathbf{v}}$, it follows that $\mathbf{1}\in ({\mathbf{v}}^{\prime }+I_{\mathbf{v}})\cap (\mathbf{1}+I_{\mathbf{v}}).$ Since ${\mathbf{v}}^{\prime }\le \mathbf{v}$ (by Proposition 2(a)) and $\mathbf{1}\le \mathbf{v}$ (since $\mathbf{1}$ is the largest element and $\mathbf{v}\ne \mathbf{1}$), we have that ${\mathbf{v}}^{\prime },\mathbf{1}\in Q$. Moreover, since ${\mathbf{v}}^{\prime }\ne \mathbf{1}$, Definition 6(ii) is not satisfied. Hence, $I_{\mathbf{v}}$ is not a Q-ideal in $V_s$ whenever $\mathbf{v}\notin \{\mathbf{0},\mathbf{1}\}$.

This concludes the proof.    □

If we plan to use the ideal $I_{\mathbf{v}}$ for partitioning $V_s$, an alternative construction of Q is needed. We address this in the next section.

3. Constructing $\mathit{Q}$-Ideals Using Seeds

Returning to Example 6, we notice that if Q was restricted to $\left\{\right(0,0,0),(0,1,0\left)\right\},$ then $I_{\mathbf{v}}$ would be a Q-ideal of $V_3$. This is used as motivation for the new choice of Q constructed in this section. Before proposing this different choice of Q, we express the cardinality of $I_{\mathbf{v}}$ in terms of the weight of $\mathbf{v}$.

Lemma 3.

Let $\mathbf{v}\in V_s$. For all $\mathbf{q}\in I_{{\mathbf{v}}^{\prime }}$, we have that

$$|I_{\mathbf{v}}|=|\mathbf{q}+I_{\mathbf{v}}|=2^{\omega \left(\mathbf{v}\right)}.$$

(1)

Proof. 

If $\mathbf{q}$ = 0, then the result follows trivially. Suppose $\mathbf{q}\in I_{{\mathbf{v}}^{\prime }}-\left\{\mathbf{0}\right\}$. Define a map $\varphi :I_{\mathbf{v}}\to \mathbf{q}+I_{\mathbf{v}}$ by $\varphi \left(\mathbf{g}\right)=\mathbf{q}+\mathbf{g}$ for all $\mathbf{g}\in I_{\mathbf{v}}$. We verify that $\varphi$ is bijective.

It follows by construction that $\varphi$ is surjective. To show that $\varphi$ is injective, suppose that $\varphi \left(\mathbf{g}\right)=\varphi \left(\mathbf{h}\right)$ for some $\mathbf{g},\mathbf{h}\in I_{\mathbf{v}}$. Then, for all $i\in \{1,\dots ,s\}$, we have that

$$q_i+g_i=q_i+h_i.$$

(2)

We need to prove that $\mathbf{g}=\mathbf{h}$ Suppose by way of contradiction that $\mathbf{g}\ne \mathbf{h}$. Then, for some i ∈ {1, …, s}, we have that g_i ≠ h_i Consider this i-th entry.

Case 1. If $q_i=g_i=0$, then $h_i=1$, so that $q_i+g_i=0\ne 1=q_i+h_i$, contradicting (2).

Case 2. If $q_i=0$ and $g_i=1$, then $h_i=0$, so that $q_i+g_i=1\ne 0=q_i+h_i$, contradicting (2).

Case 3. If $q_i=1$ and $g_i=0$, then, since $\mathbf{q}\le {\mathbf{v}}^{\prime }$, we have that $v_i^{\prime }=1$, and, since $\mathbf{h}$, we have that $h_i=1$, which implies that $v_i=1$. Hence, $v_i=v_i^{\prime }$, a contradiction.

We do not consider the case where $q_i=g_i=1$, since if $q_i=g_i=1$, then, since $\mathbf{q}\le {\mathbf{v}}^{\prime }$, we have that $v_i^{\prime }=1$, and, since $\mathbf{g}\le \mathbf{v}$, we have that $v_i=1$. Hence, $v_i=v_i^{\prime }$, a contradiction.

We deduce that $\mathbf{g}=\mathbf{h}$. Therefore, ϕ is injective. Hence, ϕ is bijective, so that $|I_{\mathbf{v}}|=|\mathbf{q}+I_{\mathbf{v}}|.$

Suppose that $\omega \left(\mathbf{v}\right)=s-j$, where $1\le j\le s$, and w.l.o.g. that $v_1=\cdots =v_j=0$. It follows that the $\mathbf{y}$ in $V_s$ for which $\mathbf{y}\le \mathbf{v}$ are those $\mathbf{y}$ for which $y_1=\cdots =y_j=0$ and $y_i\in \{0,1\}$ for all $i\in \{j+1,\dots ,s\}$. Clearly, there are $2^{s-j}$ such $\mathbf{y}$. If $\omega \left(\mathbf{v}\right)=s$, then $\mathbf{v}=\mathbf{1}$, so that $I_{\mathbf{v}}=V_s$, and thus $|I_{\mathbf{v}}|=|V_s|=2^s=2^{\omega \left(\mathbf{v}\right)}$.

This concludes the proof.    □

We are now in a position to prove one of our main results. The partition constructed will be called the Q-ideal partition.

Theorem 1.

Let $\mathbf{v}\in V_s$ . It follows that $I_{\mathbf{v}}$ is a Q-ideal if and only if $Q=I_{{\mathbf{v}}^{\prime }}$.

Proof. 

Let $\mathbf{v}\in V_s$. Suppose that $I_{\mathbf{v}}$ is a Q-ideal in $V_s$ and by way of contradiction that $Q\ne I_{{\mathbf{v}}^{\prime }}$. Then, either there exists a $\mathbf{q}$ in Q such that $\mathbf{q}\notin I_{{\mathbf{v}}^{\prime }}$, or there exists an $\mathbf{x}$ in $I_{{\mathbf{v}}^{\prime }}$ such that $\mathbf{x}\notin Q$ (or both).

Case 1. Suppose there exists a $\mathbf{q}$ in Q such that $\mathbf{q}\notin I_{{\mathbf{v}}^{\prime }}$. Since $\mathbf{0},{\mathbf{v}}^{\prime }\in I_{{\mathbf{v}}^{\prime }}$, we have that $\mathbf{q}\notin \{\mathbf{0},{\mathbf{v}}^{\prime }\}$. Consider $\mathbf{q}\mathbf{v}$. We notice that $\mathbf{q}\mathbf{v}\le \mathbf{v}$ and $\mathbf{q}\mathbf{v}\le \mathbf{q}$ (since for all $i\in \{1,\dots ,s\}$, if $q_i{v}_i=1$, then $q_i=v_i=1$), and $\mathbf{q}\mathbf{v}\ne \mathbf{0}$ (since $\mathbf{q}\ne \mathbf{0}$ and $\mathbf{q}\ne {\mathbf{v}}^{\prime }$). Thus, we have that $\mathbf{q}+\mathbf{0}=\mathbf{q}=\mathbf{q}+\mathbf{q}\mathbf{v}$, where $\mathbf{0},\mathbf{q}\mathbf{v}\in I_{\mathbf{v}}$ and $\mathbf{0}\ne \mathbf{q}\mathbf{v}$. Hence, $|\mathbf{q}+I_{\mathbf{v}}|\le |I_{\mathbf{v}}|-1<|I_{\mathbf{v}}|$, contradicting (1).

Case 2. If there exists an $\mathbf{x}$ in $I_{{\mathbf{v}}^{\prime }}$ such that $\mathbf{x}\notin Q$, then $\mathbf{x}\notin \mathbf{q}+I_{\mathbf{v}}$ for no $\mathbf{q}$ in Q, so that $\mathbf{x}+\mathbf{v}\notin \mathbf{q}+I_{\mathbf{v}}$ for any $\mathbf{q}$ in Q. Hence, $\mathbf{x}+\mathbf{v}\notin \bigcup \{\mathbf{q}+I_{\mathbf{v}}:\mathbf{q}\in Q\}$, so that $\bigcup \{\mathbf{q}+I_{\mathbf{v}}:\mathbf{q}\in Q\}\subset V_s$, a contradiction.

From this, we deduce that $Q=I_{{\mathbf{v}}^{\prime }}$.

Conversely, suppose that $Q=I_{{\mathbf{v}}^{\prime }}$. It is either the case that $\mathbf{v}=\mathbf{1}$, or that $\mathbf{v}\ne \mathbf{1}$.

Case 1. If $\mathbf{v}=\mathbf{1}$, then $I_{{\mathbf{v}}^{\prime }}=\left\{\mathbf{0}\right\}$ (since ${\mathbf{v}}^{\prime }=\mathbf{0}$), so that $Q=\left\{\mathbf{0}\right\}$, and, $I_{\mathbf{v}}=V_s$. Since $\bigcup \{\mathbf{q}+I_{\mathbf{v}}:\mathbf{q}\in Q\}=\bigcup \{\mathbf{q}+V_s:\mathbf{q}\in \left\{\mathbf{0}\right\}\}=V_s$, Definition 6(i) is satisfied. Also, Definition 6(ii) is satisfied since $Q=\left\{\mathbf{0}\right\}$. Hence, $I_{\mathbf{v}}$ is a Q-ideal of $V_s$ whenever $\mathbf{v}=\mathbf{1}$. Case 2. If $\mathbf{v}\ne \mathbf{1}$, suppose that $\omega \left(\mathbf{v}\right)=s-j$, where $1\le j\le s$, and w.l.o.g. that

$$v_1=\dots =v_j=0.$$

(3)

We will now verify two properties of what it means for the cosets of a Q-ideal to partition $V_s$, namely that (i) their unions equals $V_s$ and (ii) they are pairwise disjoint.

(i)

Their union equals $V_s$:

It is trivial that $\bigcup \{\mathbf{q}+I_{\mathbf{v}}:\mathbf{q}\in I_{{\mathbf{v}}^{\prime }}\}\subseteq V_s$. To verify that $V_s\subseteq \bigcup \{\mathbf{q}+I_{\mathbf{v}}:\mathbf{q}\in I_{{\mathbf{v}}^{\prime }}\}$, suppose that $\mathbf{y}\in V_s$. It is either the case that $\mathbf{y}\in I_{\mathbf{v}}$ or that $\mathbf{y}\notin I_{\mathbf{v}}$.

Subcase 1. If $\mathbf{y}\in I_{\mathbf{v}}$, then, since $I_{\mathbf{v}}=\mathbf{0}+I_{\mathbf{v}}$ and $\mathbf{0}\in I_{{\mathbf{v}}^{\prime }}$, we have that $\mathbf{y}\in \bigcup \{\mathbf{q}+I_{\mathbf{v}}:\mathbf{q}\in I_{{\mathbf{v}}^{\prime }}\}$.

Subcase 2. If $\mathbf{y}\notin I_{\mathbf{v}}$, then there exist at least one $i\in \{1,\dots ,s\}$ such that $y_i=1$ and $v_i=0$. Suppose w.l.o.g. that this is true for $y_1,\dots ,y_k$, where $1\le k\le j$.

For each $i\in \{1,\dots ,s\}$, let

$$\begin{array}{cc}\hfill q_i& =\left\{\begin{array}{cc}1\hfill & \mathrm{if}1\le i\le k,\hfill \\ 0\hfill & \mathrm{otherwise}.\hfill \end{array}\right.\hfill \end{array}$$

(4)

Let $\mathbf{q}=(q_1,\dots ,q_s)$. We verify that $\mathbf{q}\in I_{{\mathbf{v}}^{\prime }}$. Let $i\in \{1,\dots ,s\}$ be arbitrary. If $q_i=1$, then by (4), we have that $1\le i\le k\le j$, i.e., $1\le i\le j$. It follows by (3) that $v_i=0$, so that $v_i^{\prime }=1$. Since i was arbitrary, we have that $\mathbf{q}\in I_{{\mathbf{v}}^{\prime }}$.

For each $i\in \{1,\dots ,s\}$, let

$$\begin{array}{cc}\hfill g_i& =\left\{\begin{array}{cc}0\hfill & \mathrm{if}1\le i\le k\hfill \\ y_i\hfill & \mathrm{otherwise}.\hfill \end{array}\right.\hfill \end{array}$$

(5)

Let $\mathbf{g}=(g_1,\dots ,g_s)$. We verify that $\mathbf{g}\in I_{\mathbf{v}}$. Let $i\in \{1,\dots ,s\}$ be arbitrary. If $g_i=1$, then by (5), we have that $i\notin \{1,\dots ,k\}$ and $g_i=y_i$, so that $y_i=1$. Since, by assumption, $y_i=0$ for all $i\in \{k+1,\dots ,j\}$, and $y_i=1$, we have that $i\notin \{k+1,\dots ,j\}$. Hence, $i\notin \{1,\dots ,j\}$. By (3), we thus have that $v_i=1$. Since i was arbitrary, it follows that $\mathbf{g}\in I_{\mathbf{v}}$. Moreover, we have for each $i\in \{1,\dots ,s\}$ that

$$\begin{array}{cc}\hfill q_i+g_i& =\left\{\begin{array}{cc}1\hfill & \mathrm{if}1\le i\le k,\hfill \\ y_i\hfill & \mathrm{otherwise}.\hfill \end{array}\right.\hfill \end{array}$$

Hence, $\mathbf{y}=\mathbf{q}+\mathbf{g}$. Therefore, $\mathbf{y}\in \bigcup \{\mathbf{q}+I_{\mathbf{v}}:\mathbf{q}\in I_{{\mathbf{v}}^{\prime }}\}$.

Since $\mathbf{y}$ was arbitrary, it follows that $V_s\subseteq \bigcup \{\mathbf{q}+I_{\mathbf{v}}:\mathbf{q}\in I_{{\mathbf{v}}^{\prime }}\}$. Hence, $V_s=\bigcup \{\mathbf{q}+I_{\mathbf{v}}:\mathbf{q}\in I_{{\mathbf{v}}^{\prime }}\}$.

(ii)

They are pairwise disjoint:

Suppose that $(\mathbf{q}+I_{\mathbf{v}})\cap ({\mathbf{q}}^{*}+I_{\mathbf{v}})=\varnothing$. By way of contradiction, suppose that $\mathbf{q}={\mathbf{q}}^{*}$. Then,

$$\mathbf{q}+I_{\mathbf{v}}=(\mathbf{q}+I_{\mathbf{v}})\cap ({\mathbf{q}}^{*}+I_{\mathbf{v}})=\varnothing ,$$

which contradicts the fact that $\mathbf{q}\in \mathbf{q}+I_{\mathbf{v}}$. We deduce that $\mathbf{q}\ne {\mathbf{q}}^{*}$.

Conversely, suppose that $\mathbf{q}\ne {\mathbf{q}}^{*}$. By way of contradiction, suppose that $(\mathbf{q}+I_{\mathbf{v}})\cap ({\mathbf{q}}^{*}+I_{\mathbf{v}})\ne \varnothing$. Then, there exists an $\mathbf{y}$ in $V_s$ such that $\mathbf{y}\in \mathbf{q}+I_{\mathbf{v}}$ and $\mathbf{y}\in {\mathbf{q}}^{*}+I_{\mathbf{v}}$. Let ${\beta }_{{\mathbf{v}}^{\prime }}=I_{{\mathbf{v}}^{\prime }}-\{\mathbf{0},{\mathbf{v}}^{\prime }\}$.

Subcase 1. If $\mathbf{q}=\mathbf{0}$ and ${\mathbf{q}}^{*}={\mathbf{v}}^{\prime }$, then $\mathbf{y}\in I_{\mathbf{v}}$ and $\mathbf{y}\in {\mathbf{v}}^{\prime }+I_{\mathbf{v}}$. Thus, there exists a $\mathbf{g}\in I_{\mathbf{v}}$ such that ${\mathbf{v}}^{\prime }+\mathbf{g}\le \mathbf{v}$, i.e., for all $i\in \{1,\dots ,s\}$, if $v_i=0$, then $v_i^{\prime }+g_i=0$ (a). Since $v_1=0$ (by (3)), we have that $g_1=0$ (since $\mathbf{g}\le \mathbf{v}$) and $v_1^{\prime }=1$, so that $v_1^{\prime }+g_1=1+0=1\ne 0$, contradicting (a).

Subcase 2. If $\mathbf{q}=\mathbf{0}$ and ${\mathbf{q}}^{*}=\mathbf{b}$ for some $\mathbf{b}\in {\beta }_{{\mathbf{v}}^{\prime }}$, then $\mathbf{y}\in I_{\mathbf{v}}$ and $\mathbf{y}\in \mathbf{b}+I_{\mathbf{v}}$ for some $\mathbf{b}\in {\beta }_{{\mathbf{v}}^{\prime }}$. Thus, there exists a $\mathbf{g}\in I_{\mathbf{v}}$ such that $\mathbf{b}+\mathbf{g}\le \mathbf{v}$, i.e., for all $i\in \{1,\dots ,s\}$, if $v_i=0$, then $b_i+g_i=0$. In particular, for all $i\in \{1,\dots ,s\}$, if $v_i=0$, then $b_i=0$ (b). Suppose that $i\in \{1,\dots ,s\}$ and that $v_i=0$. Then, $i\in \{1,\dots ,j\}$ (by (3)) and $v_i^{\prime }=1$. Since $\mathbf{b}\le {\mathbf{v}}^{\prime }$ and $\mathbf{b}\ne \mathbf{0}$, it follows that $b_i=1$ for some $i\in \{1,\dots ,j\}$, contradicting (b).

Subcase 3. If $\mathbf{q}={\mathbf{v}}^{\prime }$ and ${\mathbf{q}}^{*}=\mathbf{b}$ for some $\mathbf{b}\in {\beta }_{{\mathbf{v}}^{\prime }}$, then $\mathbf{y}\in {\mathbf{v}}^{\prime }+I_{\mathbf{v}}$ and $\mathbf{y}\in \mathbf{b}+I_{\mathbf{v}}$ for some $\mathbf{b}\in {\beta }_{{\mathbf{v}}^{\prime }}$. Thus, there exist $\mathbf{g},I_{\mathbf{v}}$ such that ${\mathbf{v}}^{\prime }+\mathbf{g}=\mathbf{b}+\mathbf{h}$, i.e., for all i ∈ {1,…, s}, we have that $v_i^{\prime }$ + g_i = b_i + h_i (c). Since v_i = 0 for all i ∈ {1, …, j} and g, h ∈ I_v, it follows that g_i = h_i = 0 for all i ∈ {1, …, j}. Also, we have that b_i = 0 for an i ∈ {1, …, j} (since b ≠ v′). Consider this i-th entry. We have that $v_i^{\prime }$ = 1, so that + g_i = 1 + 0 = 1 ≠ 0 = 0 + 0 = b_i + h_i, contradicting (c).

Subcase 4. If $\mathbf{q}=\mathbf{b}$ and ${\mathbf{q}}^{*}={\mathbf{b}}^{*}$ for some $\mathbf{b},{\mathbf{b}}^{*}\in {\beta }_{{\mathbf{v}}^{\prime }}$, then $\mathbf{y}\in \mathbf{b}+I_{\mathbf{v}}$ and $\mathbf{y}\in {\mathbf{b}}^{*}+I_{\mathbf{v}}$ for some $\mathbf{b},{\mathbf{b}}^{*}\in {\beta }_{{\mathbf{v}}^{\prime }}$. Thus, there exist $\mathbf{g},I_{\mathbf{v}}$ such that $\mathbf{b}+\mathbf{g}={\mathbf{b}}^{*}+\mathbf{h}$, i.e., for all i ∈ {1,…, s}, we have that b_i + g_i = $b_i^{*}$ + h_i (d). Since v_i = 0 for all i ∈ {1, …, j} and g,I_v, it follows that g_i = h_i = 0 for all i ∈ {1, …, j}. Also, since b ≠ v′, we have that b_i = 0 for some i ∈ {1, …, j}. Consider this i-th entry for which b_i = 0 and = 1. It follows that b_i + g_i = 0 + 0 = 0 ≠ 1 = 1 + 0 = + h_i, contradicting (d).

We deduce that $(\mathbf{q}+I_{\mathbf{v}})\cap ({\mathbf{q}}^{*}+I_{\mathbf{v}})=\varnothing$ whenever $\mathbf{q}\ne {\mathbf{q}}^{*}$.

It follows by Definition 6 that $I_{\mathbf{v}}$ is a Q-ideal of $V_s$ whenever $\mathbf{v}\ne \mathbf{1}$.

This concludes the proof.    □

We illustrate Theorem 1 with the following example.

Example 8.

Let $\mathbf{v}=(1,1,0)$.

(a) 

If $Q=\{(0,0,0),(0,0,1)\}=I_{{\mathbf{v}}^{\prime }}$, then$I_{\mathbf{v}}$is a Q-ideal, because

$$\begin{array}{cc}\hfill (0,0,0)+I_{\mathbf{v}}& =\left\{\right(0,0,0),(1,0,0),(0,1,0),(1,1,0\left)\right\}\hfill \\ \hfill (0,0,1)+I_{\mathbf{v}}& =\left\{\right(0,0,1),(1,0,1),(0,1,1),(1,1,1\left)\right\}\hfill \end{array}$$

is a partition of $V_3$.

(b) 

If $Q=\{(1,0,0),(1,1,0)\}\ne I_{{\mathbf{v}}^{\prime }}$, then $I_{\mathbf{v}}$ is not a Q-ideal, because

$$\begin{array}{cc}\hfill (1,0,0)+I_{\mathbf{v}}& =\left\{\right(1,0,0),(1,1,0\left)\right\}\hfill \\ \hfill (1,1,0)+I_{\mathbf{v}}& =\left\{\right(1,1,0\left)\right\}\hfill \end{array}$$

is not a partition of $V_3$.

The following result assures us that each seed induces a unique partition.

Lemma 4.

If $\mathbf{v},\mathbf{w}\in V_s$ , then $\{\mathbf{q}+I_{\mathbf{v}}:\mathbf{q}\in I_{{\mathbf{v}}^{\prime }}\}=\{\mathbf{q}+I_{\mathbf{w}}:\mathbf{q}\in I_{{\mathbf{w}}^{\prime }}\}$ if and only if $\mathbf{v}=\mathbf{w}$.

Proof. 

Suppose $\mathbf{v},\mathbf{w}\in V_s$. If $\mathbf{v}=\mathbf{w}$, then $I_{\mathbf{v}}=I_{\mathbf{w}}$, and, ${\mathbf{v}}^{\prime }={\mathbf{w}}^{\prime }$, so that $I_{{\mathbf{v}}^{\prime }}=I_{{\mathbf{w}}^{\prime }}$. This implies that $\mathbf{q}+I_{\mathbf{v}}=\mathbf{q}+I_{\mathbf{w}}$ for all $\mathbf{q}\in I_{{\mathbf{v}}^{\prime }}$. Hence, $\{\mathbf{q}+I_{\mathbf{v}}:\mathbf{q}\in I_{{\mathbf{v}}^{\prime }}\}=\{\mathbf{q}+I_{\mathbf{w}}:\mathbf{q}\in I_{{\mathbf{w}}^{\prime }}\}$.

Conversely, if $\{\mathbf{q}+I_{\mathbf{v}}:\mathbf{q}\in I_{{\mathbf{v}}^{\prime }}\}=\{\mathbf{q}+I_{\mathbf{w}}:\mathbf{q}\in I_{{\mathbf{w}}^{\prime }}\}$, then for all $\mathbf{q}$ in $I_{{\mathbf{v}}^{\prime }}$ there exists a unique ${\mathbf{q}}^{*}$ in $I_{{\mathbf{w}}^{\prime }}$ such that $\mathbf{q}+I_{\mathbf{v}}={\mathbf{q}}^{*}+I_{\mathbf{w}}$. That is, if $\mathbf{q}=\mathbf{0}$, then there exists a unique ${\mathbf{q}}^{*}$ in $I_{{\mathbf{w}}^{\prime }}$ such that $(0,0,0)+I_{\mathbf{v}}={\mathbf{q}}^{*}+I_{\mathbf{w}}$. We show that ${\mathbf{q}}^{*}=\mathbf{0}$.

Suppose by way of contradiction that ${\mathbf{q}}^{*}\ne \mathbf{0}$. Then, $\mathbf{0}\notin {\mathbf{q}}^{*}+I_{\mathbf{w}}$, a contradiction, since $\mathbf{0}\in I_{\mathbf{v}}$. Therefore, we deduce that ${\mathbf{q}}^{*}=\mathbf{0}$. Hence, $I_{\mathbf{v}}=I_{\mathbf{w}}$. Thus, we have that $\mathbf{v}=\mathbf{w}$. This concludes the proof.    □

Definition 8.

If $\mathbf{v}$ is such that $\{\mathbf{q}+I_{\mathbf{v}}:\mathbf{q}\in I_{{\mathbf{v}}^{\prime }}\}$ is a Q-ideal partition, then we call $\mathbf{v}$ the initial seed of this Q-ideal partition.

In light of Theorem 1, we construct the following hierarchical partitioning algorithm (Algorithm 1).

Algorithm 1 Algorithm QP(seed).

Step 1. Let seed $=\mathbf{v}$ and $n=\omega \left(\mathbf{v}\right)$.

Step 2. Compute $I_{\mathbf{v}}=\mathbf{OI}\left(\mathbf{v}\right)$.

Step 3. Compute ${\mathbf{v}}^{\prime }$.

Step 4. Compute $I_{{\mathbf{v}}^{\prime }}=\mathbf{OI}\left({\mathbf{v}}^{\prime }\right)$.

Step 5. Return $\mathbf{v}$ and $\{\mathbf{q}+I_{\mathbf{v}}:\mathbf{q}\in I_{{\mathbf{v}}^{\prime }}\}$ as a partition of $V_s$.

Step 6. Choose $\mathbf{w}$ in $I_{\mathbf{v}}$ with $\omega \left(\mathbf{w}\right)=n-1$.

Step 7. If $n-1=0$, terminate; else, let seed $=\mathbf{w}$ and go to Step 1.

Example 9.

Consider $V_3$ with seed $=(1,0,1)$. We have the following steps for demonstration:

Step d1. Let $\mathbf{v}=\mathrm{seed}$and $n=\omega \left(\mathrm{seed}\right)=2$.

Step d2. Thus, $I_{\mathbf{v}}=\{(0,0,0),(1,0,0),(0,0,1),(1,0,1)\}$.

Step d3. We have that ${\mathbf{v}}^{\prime }=(0,1,0)$.

Step d4. Thus, $I_{{\mathbf{v}}^{\prime }}=\{(0,0,0),(0,1,0)\}$.

Step d5. We have that $\mathbf{v}=(1,0,1)$, and

$$\begin{array}{cc}\hfill (0,0,0)+I_{\mathbf{v}}& =\left\{\right(0,0,0),(1,0,0),(0,0,1),(1,0,1\left)\right\}\hfill \\ \hfill (0,1,0)+I_{\mathbf{v}}& =\left\{\right(0,1,0),(1,1,0),(0,1,1),(1,1,1\left)\right\}\hfill \end{array}$$

is a partition of $V_3$.

Step d6. Choose $\mathbf{w}=(1,0,0)$ in $I_{\mathbf{v}}$ with $\omega \left(\mathbf{w}\right)=n-1=1$.

Step d7. Let $\mathrm{seed}=\mathbf{w}$ and go to Step 1.

Step d8. Thus,  $\mathbf{v}=(1,0,1)$ and $n=1$.

Step d9. Thus, $I_{\mathbf{v}}=\{(0,0,0),(1,0,0)\}$.

Step d10. We have that ${\mathbf{v}}^{\prime }=(0,1,1)$.

Step d11. Thus, $I_{{\mathbf{v}}^{\prime }}=\{(0,0,0),(0,1,0),(0,0,1),(0,1,1)\}$.

Step d12. We have that $\mathbf{v}=(1,0,0)$, and

$$\begin{array}{cc}\hfill (0,0,0)+I_{\mathbf{v}}& =\left\{\right(0,0,0),(1,0,0\left)\right\}\hfill \\ \hfill (0,1,0)+I_{\mathbf{v}}& =\left\{\right(0,1,0),(1,1,0\left)\right\}\hfill \\ \hfill (0,0,1)+I_{\mathbf{v}}& =\left\{\right(0,0,1),(1,0,1\left)\right\}\hfill \\ \hfill (0,1,1)+I_{\mathbf{v}}& =\left\{\right(0,1,1),(1,1,1\left)\right\}\hfill \end{array}$$

is a partition of $V_3$.

Step d13. Choose $\mathbf{z}=(0,0,0)$ in $I_{\mathbf{w}}$ with $\omega \left(\mathbf{z}\right)=n-1=0$.

Step d14. Since $n-1=0$, terminate the algorithm.

We take note of the following:

Remark 1.

Consider Example 9.

• The hierarchical partitioning of $V_3$, according to Algorithm 1, terminated at 14 steps.
• The semiring $V_3$ is partitioned into

  -

  $2^1$ cells of equal cardinality  $2^2=2^{\omega \left(\mathbf{v}\right)}$  at Step d5, where  $\mathbf{v}$ is the chosen seed at Step d1;

  -

  $2^2$ cells of equal cardinality $2^1=2^{\omega \left(\mathbf{w}\right)}$ at Step d12, where $\mathbf{w}$ is the chosen seed at Step d6.

We use this observation to deduce the following result:

Lemma 5.

In $V_s$, Algorithm 1 will terminate at $7(s-1)$ steps. Let $j\in \{7m-2:m\in \mathbb{N}\}$. If $\mathbf{v}$ is the chosen seed at Step j, then $m=\omega \left({\mathbf{v}}^{\prime }\right)$ , and $V_s$ is partitioned into $2^m$ cells each of equal cardinality $2^{s-m}$.

Proof. 

The conclusion follows immediately from the construction. □

In set theory, a set A is said to be nested within a set B if $A\subseteq B$. We notice the following:

Remark 2.

In Example 9, we observe that every cell generated at some step is nested within a cell from the previous step.

We use the above observation to deduce the following result:

Theorem 2.

Let $1\le j\le s-1$ and let $\mathbf{v}$ (respectively, $\mathbf{w}$ ) be the chosen seed at the j-th (respectively, ($j+1$ )-th) step. For every $\mathbf{q}$ in $I_{{\mathbf{w}}^{\prime }}$ , there exists an $\mathbf{r}$ in $I_{{\mathbf{v}}^{\prime }}$ such that $\mathbf{q}+I_{\mathbf{w}}$ is nested within $\mathbf{r}+I_{\mathbf{v}}$.

Proof. 

Let $1\le j\le s-1$, and let $\mathbf{v}$ (respectively, $\mathbf{w}$) be the chosen seed at the j-th (respectively, $(j+1)$-th) step. Suppose $\mathbf{q}\in I_{{\mathbf{w}}^{\prime }}$. We need to construct an $\mathbf{r}$ in $I_{{\mathbf{v}}^{\prime }}$ so that $\mathbf{q}+I_{\mathbf{w}}\subseteq \mathbf{r}+I_{\mathbf{v}}$ (*).

Since $\mathbf{w}\in I_{\mathbf{v}}$ (by Algorithm 1), it follows from Proposition 2(b) that ${\mathbf{v}}^{\prime }\in I_{{\mathbf{w}}^{\prime }}$, so that $I_{{\mathbf{v}}^{\prime }}\subseteq I_{{\mathbf{w}}^{\prime }}$. Since $\mathbf{q}\in I_{{\mathbf{w}}^{\prime }}$, it is either the case that $\mathbf{q}\in I_{{\mathbf{v}}^{\prime }}$, or that $\mathbf{q}\notin I_{{\mathbf{v}}^{\prime }}$.

Case 1. If $\mathbf{q}\in I_{{\mathbf{v}}^{\prime }}$, then, since $I_{\mathbf{w}}\subseteq I_{\mathbf{v}}$, we have that $\mathbf{q}+I_{\mathbf{w}}\subseteq \mathbf{q}+I_{\mathbf{v}}$. Thus, (*) holds for $\mathbf{r}=\mathbf{q}$.

Case 2. Suppose that $\mathbf{q}\notin I_{{\mathbf{v}}^{\prime }}$ and let $\mathbf{y}\in \mathbf{q}+I_{\mathbf{w}}$ be arbitrary. We need to construct an $\mathbf{r}\in I_{{\mathbf{v}}^{\prime }}$ and a $\mathbf{g}\in I_{\mathbf{v}}$ so that $\mathbf{y}=\mathbf{r}+\mathbf{g}$.

Since $\mathbf{y}\in \mathbf{q}+I_{\mathbf{w}}$, there exists an $I_{\mathbf{w}}$ such that $\mathbf{y}=\mathbf{q}+\mathbf{h}$. Let g = ${\mathbf{q}}^{*}$ + h, where

$$\begin{array}{cc}\hfill q_i^{*}& =\left\{\begin{array}{cc}1\hfill & \mathrm{if}{v}_i=1\mathrm{and}{w}_i=0\hfill \\ 0\hfill & \mathrm{otherwise}\hfill \end{array}\right.\hfill \end{array}$$

(6)

for all $i\in \{1,\dots ,s\}$. We verify that $\mathbf{g}\in I_{\mathbf{v}}$. It follows by construction that ${\mathbf{q}}^{*}\in I_{\mathbf{v}}$. Since $I_{\mathbf{w}}$ and $I_{\mathbf{w}}\subseteq I_{\mathbf{v}}$, we have that $I_{\mathbf{v}}$. Since $I_{\mathbf{v}}$ is an ideal and ${\mathbf{q}}^{*},I_{\mathbf{v}}$, it follows that ${\mathbf{q}}^{*}+I_{\mathbf{v}}$, i.e., $\mathbf{g}\in I_{\mathbf{v}}$.

Now, for all $i\in \{1,\dots ,s\}$, let

$$\begin{array}{cc}\hfill r_i& =\left\{\begin{array}{cc}1\hfill & \mathrm{if}{q}_i=1\mathrm{and}{q}_i^{*}=0,\hfill \\ 0\hfill & \mathrm{otherwise}.\hfill \end{array}\right.\hfill \end{array}$$

(7)

Let $\mathbf{r}=(r_1,\dots ,r_s)$. We verify that $\mathbf{r}\in I_{{\mathbf{v}}^{\prime }}$. Let $i\in \{1,\dots ,s\}$ be arbitrary. If $r_i=1$, then it follows by (7) that $q_i=1$ and $q_i^{*}=0$ (A). Since $\mathbf{q}\le {\mathbf{w}}^{\prime }$ and $q_i=1$, it follows that $w_i^{\prime }=1$, and hence $w_i=0$. We need to show that $v_i^{\prime }=1$. So, by way of contradiction, suppose that $v_i^{\prime }=0$. Then, $v_i=1$. Since $w_i=0$ and $v_i=1$, it follows by (6) that $q_i^{*}=1$, which contradicts (A). Therefore, we deduce that $v_i^{\prime }=1$. Hence, $v_i^{\prime }=1$ whenever $r_i=1$ for an arbitrary i. From this, we deduce that $\mathbf{r}\in I_{{\mathbf{v}}^{\prime }}$.

All that is left to prove is that $\mathbf{y}=\mathbf{r}+\mathbf{g}$, i.e., $\mathbf{q}+\mathbf{h}=\mathbf{r}+{\mathbf{q}}^{*}+\mathbf{h}$. So, let i ∈ {1,…, s} be arbitrary. We show that q_i + h_i = r_i + $q_i^{*}$ + h_i. We consider all possible cases. (Note that we do not consider the cases where r_i = = 1, since by (7), if r_i = 1, then = 0.)

Subcase 1. If $r_i=q_i^{*}=h_i=0$, then $r_i+q_i^{*}+h_i=0$. Since $r_i=0$ and $q_i^{*}=0$, it follows from (7) that $q_i=0$. Hence, $q_i+h_i=0$, so that $q_i+h_i=r_i+q_i^{*}+h_i$.

Subcase 2. If $r_i=1$ and $q_i^{*}=h_i=0$, then $r_i+q_i^{*}+h_i=1$. Since $r_i=1$, it follows from (7) that $q_i=1$. Hence, $q_i+h_i=1$, so that $q_i+h_i=r_i+q_i^{*}+h_i$.

Subcase 3. If $r_i=h_i=0$ and $q_i^{*}=1$, then $r_i+q_i^{*}+h_i=1$; by (6), $v_i=1$, and thus $v_i^{\prime }=0$. Since $h_i=0$, we need $q_i$ to equal 1 for $q_i+h_i$ to equal 1. Thus, by way of contradiction, suppose that $q_i=0$. Since i is arbitrary, we have that $v_i^{\prime }=0$ implies that $q_i=0$, and thus $\mathbf{q}\in I_{{\mathbf{v}}^{\prime }}$, a contradiction. Therefore, we deduce that $q_i=1$, so that $q_i+h_i=1$. Hence, $q_i+h_i=r_i+q_i^{*}+h_i$.

Subcase 4. If $r_i=q_i^{*}=0$ and $h_i=1$, then $r_i+q_i^{*}+h_i=1$ and $q_i+h_i=1$ (regardless of the value of $q_i$), so that $q_i+h_i=r_i+q_i^{*}+h_i$.

Subcase 5. If $r_i=h_i=1$ and $q_i^{*}=0$, then $r_i+q_i^{*}+h_i=1$, and, by (7), we have that $q_i=1$. Hence, $q_i+h_i=1$, so that $q_i+h_i=r_i+q_i^{*}+h_i$.

Subcase 6. If $r_i=0$ and $q_i^{*}=h_i=1$, then $r_i+q_i^{*}+h_i=1$ and $q_i+h_i=1$ (regardless of the value of $q_i$), so that $q_i+h_i=r_i+q_i^{*}+h_i$.

Therefore, for all $i\in \{1,\dots ,s\}$, we have that $q_i+h_i=r_i+q_i^{*}+h_i$. Hence, $\mathbf{y}=\mathbf{r}+\mathbf{g}$. Since $\mathbf{y}$ was arbitrary, it follows that $\mathbf{q}+I_{\mathbf{w}}\subseteq \mathbf{r}+I_{\mathbf{v}}$. Since $\mathbf{q}$ was arbitrary, it follows that for all $\mathbf{q}$ in $I_{{\mathbf{w}}^{\prime }}$, there exists an $\mathbf{r}$ in $I_{{\mathbf{v}}^{\prime }}$ such that $\mathbf{q}+I_{\mathbf{w}}\subseteq \mathbf{r}+I_{\mathbf{v}}$.

This concludes the proof. □

4. Maximal Semiring Homomorphisms

As previously mentioned, the notion of a maximal semiring homomorphism was introduced by Allen [1] as a means of constructing Q-ideals.

Definition 9

([1]). A semiring homomorphism $\eta :S\to T$ is called maximal if, for every $t\in T$, there exists a unique $C_t\in {\eta }^{-1}\left(\left\{t\right\}\right)$ such that for every $x\in {\eta }^{-1}\left(\left\{t\right\}\right)$, we have $x+ker\eta \subseteq C_t+ker\eta$, where $ker\eta =\{x\in S:\eta (x)=0\}$.

We give an example of a maximal semiring homomorphism.

Example 10

([1]). If R denotes the set of non-negative real numbers with the usual ordering, then $(R,+,·)$, where $a+b=max(a,b)$ and $ab=min(a,b)$ for all $a,b\in R$, forms a commutative semiring. Let $S=\left\{x\in R:0\le x\le {\displaystyle \frac{1}{4}}\right\}\cup T$, where $T=\left\{{\displaystyle \frac{n}{2}}\in R:n=0,1,2,\dots \right\}.$ Let $\eta :S\to T$ be a map defined by

$$\begin{array}{cc}\hfill \eta \left(x\right)& =\left\{\begin{array}{cc}x\hfill & \mathrm{if}x\in T,\hfill \\ 0\hfill & \mathrm{otherwise}.\hfill \end{array}\right.\hfill \end{array}$$

(8)

It can be verified that $\eta :S\to T$ is a maximal semiring homomorphism.

It is true that there exist semiring homomorphisms that are not maximal.

Example 11

([1]). Consider the set $\mathbb{N}$ of natural numbers $\{0,1,2,\dots \}$, which is a well-ordered set under the usual ordering of the natural numbers. Thus, $(\mathbb{N},max,min)$ can be viewed as a semiring as described in Example 6. Define $\eta :\mathbb{N}\to B_0$ by

$$\begin{array}{cc}\hfill \eta \left(x\right)& =\left\{\begin{array}{cc}0\hfill & \mathrm{if}x\le 5,\hfill \\ 1\hfill & \mathrm{otherwise}.\hfill \end{array}\right.\hfill \end{array}$$

We verify that η is a homomorphism from $\mathbb{N}$ onto $B_0$: Let $n,m\in \mathbb{N}$ be arbitrary and suppose w.l.o.g. that $n\le m$. Then, $n+m=max(n,m)=m$ and $n·m=min(n,m)=n$.

Case 1. If $\eta \left(n\right)=\eta \left(m\right)=0$, then $\eta \left(n\right)+\eta \left(m\right)=0$ and $\eta \left(n\right)·\eta \left(m\right)=0$. Since $\eta (n+m)=0$ and $\eta (n·m)=0$, we have

$$\eta (n+m)=0=\eta \left(n\right)+\eta \left(m\right)$$

and

$$\eta (n·m)=0=\eta \left(n\right)·\eta \left(m\right).$$

Case 2. If $\eta \left(n\right)=0$ and $\eta \left(m\right)=1$, then $\eta \left(n\right)+\eta \left(m\right)=1$ and $\eta \left(n\right)·\eta \left(m\right)=0$. Since $\eta (n+m)=1$ and $\eta (n·m)=0$, we have

$$\eta (n+m)=1=\eta \left(n\right)+\eta \left(m\right)$$

and

$$\eta (n·m)=0=\eta \left(n\right)·\eta \left(m\right).$$

Case 3. If $\eta \left(n\right)=\eta \left(m\right)=1$, then $\eta \left(n\right)+\eta \left(m\right)=1$ and $\eta \left(n\right)·\eta \left(m\right)=1$. Since $\eta (n+m)=1$ and $\eta (n·m)=1$, we have

$$\eta (n+m)=1=\eta \left(n\right)+\eta \left(m\right)$$

and

$$\eta (n·m)=1=\eta \left(n\right)·\eta \left(m\right).$$

Hence, η is a homomorphism.

Since $ker\eta =\{x\in \mathbb{N}:x\le 5\}$, we have, for each $y\in {\eta }^{-1}\left(1\right)$, that $y+x=max(y,x)=y$ for each $x\in ker\eta$. Thus, $y+ker\eta =\left\{y\right\}$ for each $y\in {\eta }^{-1}\left(1\right)$. Similarly, $C_1+ker\eta =\left\{C_1\right\}$ if $C_1\in {\eta }^{-1}\left(1\right)$. Since, for all $y\ne C_1$, we have $\left\{y\right\}\subseteq \left\{C_1\right\}$, it follows that $y+ker\eta \subseteq C_1+ker\eta$ for all $y\in \left({\eta }^{-1}\left(1\right)\right)-\left\{C_1\right\}$. Hence, no $C_1$ in ${\eta }^{-1}\left(1\right)$ exists such that $y+ker\eta \subseteq C_1+ker\eta$ for all $y\in {\eta }^{-1}\left(1\right)$. Therefore, η is not maximal.

Using the notion of a maximal semiring homomorphism, Allen introduced a method by which the quotient structure of a semiring with respect to a Q-ideal can be constructed.

Lemma 6

([1]). Suppose $\eta :S\to T$ is a semiring homomorphism. If η is maximal, then $ker\eta$ is a Q-ideal, where $Q={\left\{C_t\right\}}_{t\in T}$.

Even though the above result allowed Allen to generalize the Fundamental Theorem of Homomorphisms for a large class of semirings, the method for identifying these homomorphisms remains unclear.

We imitate the construction of Allen in ([1] Example 12, p. 415) with the aim of constructing a maximal semiring homomorphism corresponding to the Q-ideal partition.

Lemma 7.

Let $\mathbf{v}\in V_s-\left\{\mathbf{1}\right\}$. Suppose that $\omega \left(\mathbf{v}\right)=s-j$, where $1\le j\le s$, and w.l.o.g. that

$$v_1=\cdots =v_j=0.$$

(9)

Let $T=\{t_0,t_1,\cdots ,t_{2^j-1}\}$ be an ordered set with least element $t_0$, where $t_i+t_k=max\{t_i,t_k\}$ and $t_i{t}_k=min\{t_i,t_k\}$ for all $i,k\in \{0,\dots ,2^j-1\}$. Define $\eta :V_s\to T$ by

$$\begin{array}{cc}\hfill \eta \left(\mathbf{y}\right)& =\left\{\begin{array}{cc}{t}_0\hfill & \mathrm{if}{y}_1=y_2=y_3=\cdots =y_j=0\hfill \\ t_1\hfill & \mathrm{if}{y}_1=1;y_2=y_3\cdots =y_j=0\hfill \\ t_2\hfill & \mathrm{if}{y}_2=1;y_1=y_3=\cdots =y_j=0\hfill \\ ⋮\hfill \\ t_j\hfill & \mathrm{if}{y}_j=1;y_1=y_2=y_3=\cdots =y_{j-1}=0\hfill \\ t_{j+1}\hfill & \mathrm{if}{y}_1=y_2=1;y_3=\cdots =y_j=0\hfill \\ t_{j+2}\hfill & \mathrm{if}{y}_1=y_3=1;y_2=\cdots =y_j=0\hfill \\ ⋮\hfill \\ t_{2^j-1}\hfill & \mathrm{if}{y}_1=y_2=y_3=\cdots =y_j=1\hfill \end{array}\right.\hfill \end{array}$$

for all $\mathbf{y}\in V_s$. Then, η is a maximal semiring homomorphism.

Proof. 

It is easy to verify that $\eta$ is a semiring homomorphism. We verify that $\eta$ is maximal.

Suppose $t\in T$. We consider all possible cases.

Case 1. $t=t_0$: Let $C_{t_0}=\mathbf{0}$. Suppose $\mathbf{y}\in {\eta }^{-1}\left(\left\{t_0\right\}\right)$. Then, $\mathbf{y}\in ker\eta$. Suppose $\mathbf{z}\in \mathbf{y}+ker\eta$. We have that $\mathbf{z}=\mathbf{y}+\mathbf{n}$ for some $\mathbf{n}\in ker\eta$, and thus

$$\eta \left(\mathbf{z}\right)=\eta (\mathbf{y}+\mathbf{n})=\eta \left(\mathbf{y}\right)+\eta \left(\mathbf{n}\right)=t_0+t_0=max\{t_0,t_0\}=t_0.$$

Hence,

$$\mathbf{z}\in {\eta }^{-1}\left(\left\{t_0\right\}\right)=ker\eta =\mathbf{0}+ker\eta =C_{t_0}+ker\eta .$$

Since $\mathbf{z}$ was arbitrary, it follows that $\mathbf{y}+ker\eta \subseteq C_{t_0}+ker\eta$. Since $\mathbf{y}$ was arbitrary, it follows that $\mathbf{y}+ker\eta \subseteq C_{t_0}+ker\eta$ for all $\mathbf{y}\in {\eta }^{-1}\left(\left\{t\right\}\right)$. This proves existence.

To prove uniqueness, suppose by way of contradiction that there exists a $C_t\in V_s-\left\{C_{t_0}\right\}$ such that for every $\mathbf{y}\in {\eta }^{-1}\left(\left\{t_0\right\}\right)$, we have that

$$\mathbf{y}+ker\eta \subseteq C_t+ker\eta .$$

(10)

If $\mathbf{y}=\mathbf{0}$, then $\mathbf{y}\in {\eta }^{-1}\left(\left\{t_0\right\}\right)$. It follows from (10) that

$$ker\eta \subseteq C_t+ker\eta .$$

(11)

If $\mathbf{z}\in ker\eta$, then

$$\eta \left(\mathbf{z}\right)=t_0,$$

(12)

and $\mathbf{z}\in C_t+ker\eta$ by (11). Thus, there exists an $\mathbf{n}\in ker\eta$ such that $\mathbf{z}=C_t+\mathbf{n}$, so that

$$\eta \left(\mathbf{z}\right)=\eta (C_t+\mathbf{n})=\eta \left(C_t\right)+\eta \left(\mathbf{n}\right)=t+t_0=max\{t,t_0\}=t\ne t_0,$$

which contradicts (12). This proves uniqueness.

Case 2. $t=t_1$: Let $C_{t_1}$ be the element in $V_s$ with first entry equal to 1, and 0 elsewhere. Suppose $\mathbf{y}\in {\eta }^{-1}\left(\left\{t_1\right\}\right)$. Then, $y_1=1$ and $y_2=\cdots =y_j=0$. Suppose $\mathbf{z}\in \mathbf{y}+ker\eta$. We have that $\mathbf{z}=\mathbf{y}+\mathbf{n}$ for some $\mathbf{n}\in ker\eta$; thus,

$$\eta \left(\mathbf{z}\right)=\eta (\mathbf{y}+\mathbf{n})=\eta \left(\mathbf{y}\right)+\eta \left(\mathbf{n}\right)=t_1+t_0=max\{t_1,t_0\}=t_1.$$

Hence, $\mathbf{z}\in {\eta }^{-1}\left(\left\{t_1\right\}\right)$. Thus, $z_1=1$ and $z_2=\cdots =z_j=0$.

For each $i\in \{1,\dots ,s\}$, let $m_i=0$ whenever $i\in \{1,\dots ,j\}$; otherwise, let $m_i=z_i$. If $\mathbf{m}=(m_1,\dots ,m_s)$, then $\mathbf{m}\in ker\eta$ and $C_{t_1,i}+m_i=1$ if $i=1$; $C_{t_1,i}+m_i=0$ if $i\in \{2,\dots ,j\}$; otherwise, $C_{t_1,i}+m_i=z_i$ for each $i\in \{1,\dots ,s\}$. Hence, $\mathbf{z}=C_{t_1}+\mathbf{m}$, so that $\mathbf{z}\in C_{t_1}+ker\eta$. This proves existence.

To prove uniqueness, suppose by way of contradiction that there exists a $C_t\in V_s-\left\{C_1\right\}$ such that for every $\mathbf{y}\in {\eta }^{-1}\left(\left\{t_1\right\}\right)$, we have that

$$\mathbf{y}+ker\eta \subseteq C_t+ker\eta .$$

(13)

If $\mathbf{y}=C_{t_1}$, then $\mathbf{y}\in {\eta }^{-1}\left(\left\{t_1\right\}\right)$. It follows from (13) that

$$C_{t_1}+ker\eta \subseteq C_t+ker\eta .$$

(14)

If $\mathbf{z}=C_{t_1}$, then

$$\eta \left(\mathbf{z}\right)=t_1,$$

(15)

and $\mathbf{z}\in C_{t_1}+ker\eta$. It follows from (14) that $\mathbf{z}\in C_t+ker\eta$. Thus, there exists an $\mathbf{n}\in ker\eta$ such that $\mathbf{z}=C_t+\mathbf{n}$, so that

$$\eta \left(\mathbf{z}\right)=\eta (C_t+\mathbf{n})=\eta \left(C_t\right)+\eta \left(\mathbf{n}\right)=t+t_0=max\{t,t_0\}=t\ne t_1,$$

which contradicts (15). This proves uniqueness. A similar approach can be used for other values of $t\in T-\left\{t_0\right\}$.

To show that $\eta$ corresponds to the Q-ideal partition, we will prove that (a) $I_{\mathbf{v}}=ker\eta$ and (b) $I_{{\mathbf{v}}^{\prime }}=Q$

(a)

If $\mathbf{y}\in ker\eta$, then $y_1=\dots =y_j=0$. Suppose $i\in \{1,\dots ,s\}$. If $y_i=1$, then $i\notin \{1,\dots ,j\}$. By (9), we have that $v_i=1$. Since i was arbitrary, it follows that $\mathbf{y}\in I_{\mathbf{v}}$. Since $\mathbf{y}$ was arbitrary, it follows that $ker\eta \subseteq I_{\mathbf{v}}$. If $\mathbf{y}\in I_{\mathbf{v}}$, then for all $i\in \{1,\dots ,s\}$, if $v_i=0$, then $y_i=0$. Hence, $y_1=\dots =y_j=0$. Thus, $\mathbf{y}\in ker\eta$. Since $\mathbf{y}$ was arbitrary, it follows that $I_{\mathbf{v}}\subseteq ker\eta$. Therefore, $ker\eta =I_{\mathbf{v}}$.

(b)

Suppose $C_t\in Q$. If $C_t=\mathbf{0}$, then $C_t\in I_{{\mathbf{v}}^{\prime }}$. If $C_t\ne \mathbf{0}$, then $C_{t,i}=1$ for at least one i in $\{1,\dots ,j\}$. Consider this i-th entry. By (9), we have that $v_i=0$, so that $v_i^{\prime }=1$. Since i was arbitrary, it follows that $C_t\in I_{{\mathbf{v}}^{\prime }}$. Since $C_t$ was arbitrary, it follows that $Q\subseteq I_{{\mathbf{v}}^{\prime }}$. Now, suppose $\mathbf{y}\in I_{{\mathbf{v}}^{\prime }}$. If $\mathbf{y}=\mathbf{0}$, then $\mathbf{y}=C_{t_0}\in Q$. If $\mathbf{y}\ne \mathbf{0}$, then there exists at least one i in $\{1,\dots ,j\}$ such that $y_i=1$. Hence, $\mathbf{y}\in Q$. Since $\mathbf{y}$ was arbitrary, it follows that $I_{{\mathbf{v}}^{\prime }}\subseteq Q$. Therefore, $Q=I_{{\mathbf{v}}^{\prime }}$.

This completes the proof. □

Remark 3.

In Lemma 7, we assume that $\mathbf{v}$ has j zero entries amongst its s entries and without loss of generality that these zero entries lie among the first j entries of $\mathbf{v}$. If we do not make this assumption, we assume that the zeros of $\mathbf{v}$ are “scattered” across its s entries. That is,

$$v_{\alpha }=v_{\beta }=v_{\lambda }=\dots =v_{ϵ}=v_{\delta }=0,where\alpha ,\beta ,\lambda ,\dots ,ϵ,\delta \in \{1,\dots ,s\}.$$

We define $\eta :Vs\to T$ by

$$\begin{array}{cc}\hfill \eta \left(\mathbf{y}\right)& =\left\{\begin{array}{cc}{t}_0\hfill & \mathrm{if}{y}_{\alpha }=y_{\beta }=y_{\lambda }=\dots =y_{ϵ}=y_{\delta }=0\hfill \\ t_1\hfill & \mathrm{if}{y}_{\alpha }=1;y_{\beta }=y_{\lambda }=\dots =y_{ϵ}=y_{\delta }=0\hfill \\ t_2\hfill & \mathrm{if}{y}_{\beta }=1;y_{\alpha }=y_{\lambda }=\dots =y_{ϵ}=y_{\delta }=0\hfill \\ ⋮\hfill \\ t_j\hfill & \mathrm{if}{y}_{\delta }=1;y_{\alpha }=y_{\beta }=y_{\lambda }=\dots =y_{ϵ}=0\hfill \\ t_{j+1}\hfill & \mathrm{if}{y}_{\alpha }=y_{\beta }=1;y_{\lambda }=\dots =y_{ϵ}=y_{\delta }=0\hfill \\ t_{j+2}\hfill & \mathrm{if}{y}_{\alpha }=y_{\lambda }=1;y_{\beta }=\dots =y_{ϵ}=y_{\delta }=0\hfill \\ ⋮\hfill \\ t_{2^j-1}\hfill & \mathrm{if}{y}_{\alpha }=y_{\beta }=y_{\lambda }=\dots =y_{ϵ}=y_{\delta }=1\hfill \end{array}\right.\hfill \end{array}$$

for each $\mathbf{y}$ in $V_s$. Proving that η is a maximal semiring homomorphism follows the same outline as the proof of Lemma 7.

Example 12.

Let $T=\{t_0,t_1\}$ be a well-ordered set where $t_0\le t_1$. We define $t_0+t_1=t_1$ and $t_0{t}_1=t_0$. Then, $(T,+,·)$ is a semiring. For each $\mathbf{x}\in V_3$, let

$$\begin{array}{cc}\hfill \eta \left(\mathbf{x}\right)& =\left\{\begin{array}{cc}{t}_0\hfill & \mathrm{if}{x}_3=0,\hfill \\ t_1\hfill & \mathrm{otherwise}.\hfill \end{array}\right.\hfill \end{array}$$

It follows from Lemma 7 that $\eta :V_3\to T$ is a maximal semiring homomorphism. We have that

$$ker\eta =\{(0,0,0),(1,0,0),(0,1,0),(1,1,0)\}=I_{(1,1,0)}$$

and

$$\begin{array}{c}\hfill Q=\{(0,0,0),(0,0,1)\}=I_{(0,0,1)}.\end{array}$$

Hence, $\eta :V_3\to T$ corresponds to the Q-ideal partition

$$\begin{array}{cc}\hfill (0,0,0)+I_{(1,1,0)}& =\left\{\right(0,0,0),(1,0,0),(0,1,0),(1,1,0\left)\right\}\hfill \\ \hfill (0,0,1)+I_{(1,1,0)}& =\left\{\right(0,0,1),(1,0,1),(0,1,1),(1,1,1\left)\right\}\hfill \end{array}$$

of $V_3$.

5. Concluding Remarks

We introduced a novel approach for constructing Q-ideals, resulting in a new class of partitions of a Boolean semiring. This method facilitated the development of a hierarchical partitioning algorithm that uses the weight of selected seeds. We provide corresponding maximal homomorphisms and illustrate the results with low-dimensional examples. While Allen’s [1] method for creating Q-ideals relies on maximal homomorphisms, our alternative technique leverages the weight of selected seeds to predict the cell count of the partition and the size of individual cells. As the primary focus of our manuscript is not on maximal semiring homomorphisms, we decided not to delve deeply into this specific topic. The exploration of Q-ideals in the context of maximal semiring homomorphisms could serve as an interesting direction for future research.

The method proposed in Theorem 1 is designed with theoretical clarity and precision in mind, and we recognize that scalability can be a concern when applied to larger Boolean semirings. In applications, one would have to expand the realized vectors (the rows of a site-by-species matrix) to the power set and then, after partitioning, extract the realized vectors from the partitioned power set. This could potentially limit computational efficiency for larger Boolean semirings, since the expansion and extraction of the realized vectors could increase execution time and memory consumption. For smaller or more manageable semirings, we believe that the proposed method offers valuable insights and practical utility and that its broader applicability to larger systems offers an opportunity for future research.

Since the focus of this paper was to introduce our method and give illustrative low-dimensional examples thereof, comparing its computational efficiency to other methods in this paper is not possible but will rather form part of ongoing research. The focus of our follow-up paper is on the evaluation of the “goodness” of our method by elucidating the relationship between the average within- and between-similarity of the resulting partitions. Our results suggest that our proposed method consistently produces “good” partitions. We believe that comparing its computational complexity to those of other methods is relevant to the scope therein.

The number of ways to partition a set corresponds to the Bell numbers. That is, the number of ways to partition a set of size n equals $B_n={\sum }_{k=0}^{n-1}{C}_{n-1}^k{B}_k$, where $C_{n-1}^k$ denotes the number of ways to choose k elements from a set of size $n-1$, where $n\in \{1,2,3,\dots \}$. For example, the number of ways to partition a set of size 8 equals 4140; the number of Q-ideal partitions of a set of size 8 equals 8 (since each seed produces a unique partition; see Lemma 4). Since hierarchical partitioning typically has exponential time complexity, especially when evaluating all possible partitions of a given set (e.g., agglomerative or divisive hierarchical clustering [O($n^3$)], graph partitioning via minimum bisection [NP-hard], optimal decision tree construction [exponential], K-partitioning of sets via exhaustive search [Bell numbers, that is, super-exponential], and integer linear programming ILP-based partitioning [NP-hard]), we believe that our algorithm mitigates this inefficiency. This is because the number of possible Q-ideal partitions is significantly smaller than the total number of partitions, particularly as n increases.

Throughout this paper, there are no bounds placed on the weight of a selected seed to partition a Boolean semiring, except in Lemma 9, where we assume that the weight of a selected seed of length s is strictly less than s. Although we do not place boundaries on the weight of a selected seed in our proposed method to construct Q-ideals (Theorem 1), we are less interested in the two trivial cases, that is, where the weight equals either 0 or s. This is because the resulting Q-ideal partition will either be the set where each seed belongs to its own cell, or the set where all seeds belong to a single cell, respectively. While we believe that using seed weights can significantly reduce the computational cost compared to evaluating all possible partitions, we considered whether the initial choice of seed weights affects the overall partitioning outcome. This is particularly relevant in situations where the seed weights do not fully reflect the structure of the dataset or if they are selected in a manner that is not optimal (for instance, if a seed of significantly low weight is chosen, but the realized set of vectors all have weights close to s). Identifying the optimal set of seeds for hierarchical partitioning or designing a single seed for “good” partitioning remains an open question, offering opportunities for future research and exploration.