Finitely generated maximal partial clones and their intersections

TL;DR

This paper investigates the finite generation properties of certain partial clones on finite sets, revealing differences between small and larger sets and the complexity of their intersections.

Contribution

It demonstrates that on a two-element set, certain partial clones are not finitely generated but are intersections of finitely generated maximal partial clones, and for larger sets, intersections can be non-finitely generated.

Findings

For |A|=2, the selfdual monotone partial functions form a non-finitely generated partial clone.

On larger sets, intersections of finitely generated maximal partial clones can be non-finitely generated.

The structure of partial clones varies significantly with the size of the underlying set.

Abstract

Let A be a finite non-singleton set. For |A|=2 we show that the partial clone consisting of all selfdual monotone partial functions on A is not finitely generated, while it is the intersection of two finitely generated maximal partial clones on A. Moreover for |A| >= 3 we show that there are pairs of finitely generated maximal partial clones whose intersection is a non-finitely generated partial clone on A.

Equations

$$\mathrm{Par}(A):=\bigcup _{n\ge 1}{\mathrm{Par}}^{(n)}(A).$$

$$\mathrm{O}(A):=\bigcup _{n\ge 1}\{f\in {\mathrm{Par}}^{(n)}(A)|\mathrm{dom}(f)=A^n\}.$$

$$\mathrm{dom}(h):=\{\vec{a}\in A^m|\vec{a}\in \bigcap _{i=1}^n\mathrm{dom}(g_i)\text{mbox}and(g_1(\vec{a}),\dots ,g_n(\vec{a}))\in \mathrm{dom}(f)\}$$

$$h({\vec{a}}):=f(g_{1}({\vec{a}}),\dots,g_{n}({\vec{a}})),\quad\mbox{for all ${\vec{a}}\in{\rm dom}\hphantom{.}(h)$.}$$

$${\mathcal{F}}_a=\bigcup _{n\ge 1}\{f\in {\mathrm{Par}}^{(n)}(A)|\text{mbox}if(a,\dots ,a)\in \mathrm{dom}(f),\text{mbox}thenf(a,\dots ,a)=a\}.$$

$$\mathrm{pPol}\rho :=\{f\in \mathrm{Par}(A)|f\mathrm{preserves}\rho \}.$$

$$\begin{gathered} \mathrm{pPol}(\le ):=\{f\in \mathrm{Par}(2)|(a_1,\dots ,a_n)\in \mathrm{dom}(f),(b_1,\dots ,b_n)\in \mathrm{dom}(f)\text{mbox}and \\ {a}_1\le b_1,\dots ,a_n\le b_n⟹f(a_1,\dots ,a_n)\le f(b_1,\dots ,b_n)\} \end{gathered}$$

$$\begin{gathered} \mathrm{pPol}(\ne ):=\{f\in \mathrm{Par}(2)|(a_1,\dots ,a_n)\in \mathrm{dom}(f)\text{mbox}and \\ (1+a_1,\dots ,1+a_n)\in \mathrm{dom}(f)⟹f(1+a_1,\dots ,1+a_n)=1+f(a_1,\dots ,a_n)\}, \end{gathered}$$

$$\text{mbox}if(f_1(\vec{a}),\dots ,f_m(\vec{a}))=(f_1(\vec{b}),\dots ,f_m(\vec{b})),\text{mbox}then\vec{a}=\vec{b}.$$

$$\varrho \in \{\{0\},\{1\},\{(0,1)\},\{(0,1),(1,0)\},\{(0,0),(0,1),(1,1)\}\}.$$

$$\mathrm{maj}(x_1,x_2,x_3)=(x_1\wedge x_2)\vee (x_1\wedge x_3)\vee (x_2\wedge x_3)=(x_1\vee x_2)\wedge (x_1\vee x_3)\wedge (x_2\vee x_3).$$

$$(f_1(\vec{0}),\dots ,f_m(\vec{0}))=(0,\dots ,0).$$

$$(f_1(\vec{a}),\dots ,f_m(\vec{a}))=(0,\dots ,0).$$

$${\vec{e}}_i:=(0,\dots ,0,\underset{i-\mathrm{th}\mathrm{position}}{\underbrace{1}},0,\dots ,0)\in 2^{m+1}.$$

$$(f_1(\vec{0}),\dots ,f_m(\vec{0}))=(f_1({\vec{e}}_t),\dots ,f_m({\vec{e}}_t)),$$

$$(\mathrm{pPol}(\ne )\cap \mathrm{pPol}(\le ))\cap \mathrm{O}(2)=\mathrm{Pol}(\ne )\cap \mathrm{Pol}(\le )$$

$$\pi =(01\dots p-1)(p(p+1)\dots (2p-1))\dots ((k-p)(k-p+1)\dots (k-1)).$$

$${\varrho }_{\pi }:=\{(x,\pi (x),{\pi }^2(x),\dots ,{\pi }^{p-1}(x))|x\in \mathbf{k}\}.$$

$$\begin{array}[]{lll}({\rm pPol}\,(\leq)\cap{\rm pPol}\,(\varrho_{\pi})\cap{\rm O}(A)&=&({\rm pPol}\,(\leq)\cap{\rm O}(A))\cap({\rm pPol}\,(\varrho_{\pi})\cap{\rm O}(A))\\ &=&{\rm Pol}\,(\leq)\cap{\rm Pol}\,(\pi)\\ &=&{\rm J}(A).\end{array}$$

$$(f_1(\vec{0}),\dots ,f_m(\vec{0}))=(0,\dots ,0).$$

$$\vec{a}={\vec{e}}_i:=(0,\dots ,0,\underset{i-\mathrm{th}\mathrm{position}}{\underbrace{1}},0,\dots ,0)\in {\mathbf{k}}^{m+1}.$$

Full text

1. Preliminaries

Let $A$ be a finite set with cardinality $|A|:=k\geq 2$. For a positive integer $n$, an  $n$-ary partial function on $A$ is a map $f:{\rm dom}\hphantom{.}(f)\to A$ where ${\rm dom}\hphantom{.}(f)$ is subset of $A^{n}$ called the domain of $f$. Let ${\rm Par}^{(n)}(A)$ denote the set of all $n$-ary partial functions on $A$ and let

[TABLE]

Moreover, denote by ${\rm O}(A)$ the set of all total or everywhere defined functions, i.e.,

[TABLE]

For $n,m\geq 1$, $f\in{\rm Par}^{(n)}(A)$ and $g_{1},\dots,g_{n}\in{\rm Par}^{(m)}(A)$, the composition of $f$ and $g_{1},\dots,g_{n}$, denoted by $h:=f[g_{1},\dots,g_{n}]\in{\rm Par}^{(m)}(A)$, is the partial function whose domain is

[TABLE]

and defined by

[TABLE]

For every positive integer $n$ and each $1\leq i\leq n$, let $e_{i}^{n}$ denote the $n$-ary i-th projection defined by ${\rm dom}\hphantom{.}(e^{n}_{i})=A^{n}$ and $e^{n}_{i}(a_{1},\dots,a_{n})=a_{i}$ for all $(a_{1},\dots,a_{n})\in A^{n}$. Denote by ${\rm J}(A)$ the set of all projections on $A$.

A partial clone on $A$ is a subset of ${\rm Par}(A)$ closed under composition and containing the set ${\rm J}(A)$ of all projections. A partial clone $C$ is said to be strong if it contains all subfunctions of its functions, i.e., if for every $g\in{\rm Par}(A)$, we have $g\in C$ whenever $g=f|_{{\rm dom}\hphantom{.}(g)}$, for some $f\in C$.

Example 1**.**

Let $a\in A$ and consider the set

[TABLE]

Then $\mathcal{F}_{a}$ is a strong partial clone on $A$.

The idea behind this example is formalized as follows. For $h\geq 1$, let $\rho$ be an $h$-ary relation on $A$ and $f$ be an $n$-ary partial function on $A$. We say that $f~{}preserves~{}\rho$ if for every $h\times n$ matrix $M=[M_{ij}]$, whose columns $M_{*j}\in\rho,~{}(j=1,\ldots n)$ and whose rows $M_{i*}\in{\rm dom}\hphantom{.}(f)$ $(i=1,\ldots,h)$, we have $(f(M_{1*}),\ldots,f(M_{h*}))\in\rho$. Define

[TABLE]

Example 2**.**

Denote by $\leq$ and $\neq$ the two binary relations $\{(0,0),(0,1),(1,1)\}$ and $\{(0,1),(1,0)\}$, respectively, on ${\bf 2}:=\{0,1\}$. Then

[TABLE]

and

[TABLE]

where $+$ denotes the sum modulo 2.

It is well known (see, e.g., [1, 5] and [9], chapter 20) that ${\rm pPol}\,\rho$ is a strong partial clone called the partial clone determined by the relation $\rho$. Let ${\rm Pol}\,(\rho):={\rm pPol}\,(\rho)\cap{\rm O}(A)$ be the partial clone consisting of all total functions preserving the relation $\rho$. The set of all partial clones on $A$, ordered by inclusion, forms an algebraic lattice $\mathcal{L}_{A}$ where the meet coincides with the intersection.

We say that a partial clone $C_{0}$ on $A$ is covered by a partial clone $C_{1}$ on $A$ if there is no partial clone $C$ such that $C_{0}\subset C\subset C_{1}$. A maximal partial clone is a partial clone covered by ${\rm Par}(A)$.

For ${\rm F}\subseteq{\rm Par}(A)$, let $\langle{\rm F}\rangle$ denote the partial clone generated by ${\rm F}$, i.e., the smallest partial clone containing ${\rm F}$ or, equivalently, the intersection of all partial clones on $A$ containing ${\rm F}$. If $C=\langle{\rm F}\rangle$, then we say that ${\rm F}$ is a generating set for $C$. A partial clone $C$ is said to be finitely generated if it admits a finite generating set, i.e., if $C=\langle{\rm F}\rangle$ for some finite set ${\rm F}\subseteq C$.

Generating sets for clones and partial clones have been extensively studied in the literature (see, e.g., surveys in [13] for the total case and [2] for the partial case). For instance, Freivald [4] showed that there are eight maximal partial clones on ${\bf 2}:=\{0,1\}$, and Lau [8] showed that exactly two of them are not finitely generated, namely, the two strong maximal partial clones of Slupecki type (see [5] and [9], section 20, for details).

In this paper we are particularly interested in the two maximal partial clones ${\rm pPol}\,(\leq)$ and ${\rm pPol}\,(\neq)$. As shown in [7], the interval of partial clones $[{{\rm pPol}\,(\leq)}\cap{{\rm pPol}\,(\neq)},{\rm Par}({\bf 2})]$ is infinite. However, it is still unknown if this interval is countably infinite or of continuum cardinality, and this problem seems to be difficult to decide.

This lead the authors to study the partial clone ${{\rm pPol}\,(\leq)}\cap{{\rm pPol}\,(\neq)}$. Here, we show that the partial clone ${{\rm pPol}\,(\leq)}\cap{{\rm pPol}\,(\neq)}$ is not finitely generated. We also generalize this result by showing that, on every non-singleton finite set $A$, there are pairs of finitely generated maximal partial clones whose intersection is a not finitely generated.

We shall make use the concept of “separating clone” as introduced in [1] (see also [2, 5, 6]).

Definition 1**.**

A clone $C$ on $A$ is separating if there exists $m\geq 1$ such that for all $n>m$ and all ${\vec{b}}\in A^{n}$, there are $m$ functions $f_{1},\dots,f_{m}\in C\cap{\rm O}^{(n)}_{A}$ such that for every ${\vec{a}},{\vec{b}}\in A^{n}$,

[TABLE]

For example, ${\rm O}(A)$ is a separating clone on $A$ and it is shown in [1] that ${\rm Pol}\,\varrho$ is a separating clone for every

[TABLE]

Note that if $C\subseteq D$ are two clones and if $C$ is a separating clone, then $D$ is also a separating clone.

The main results of this paper are based on the following criterion established in [1] (see also [2, 5, 6]).

Theorem 1**.**

Let $A$ be a finite set with at least two elements, and let $D$ be a strong partial clone on $A$. If $C:=D\cap{\rm O}(A)$ is not a separating clone, then $D$ is not finitely generated.

2. Intersection of finitely generated maximal partial clones

We start with the two element set ${\bf 2}:=\{0,1\}$. It is shown in [4] that there are eight maximal partial clones on ${\bf 2}$ and it is shown in [8] that exactly two of them are non-finitely generated (they are the two strong maximal partial clones of Slupecki type, see [5] and [9] section 20 for details). In particular, Lau [8] showed the following interesting result.

Lemma 2**.**

([8])* Let $A={\bf 2}$. Then,*

1. (i)

${\rm pPol}\,(\neq)$* is generated by its ternary partial functions, and* 2. (ii)

${\rm pPol}\,(\leq)$* is generated by its binary partial functions.*

Another proof of the fact that both ${\rm pPol}\,(\leq)$ and ${\rm pPol}\,(\neq)$ are finitely generated is given in Proposition 3.10 of [1]. The proof is based on the concept of separating clones and on the fact that every partial function in ${\rm pPol}\,(\leq)$ (in ${\rm pPol}\,(\neq)$) can be extended to a total function in ${\rm Pol}\,(\leq)$ (in ${\rm Pol}\,(\neq)$ respectively). It is noteworthy that the clone ${\rm Pol}\,(\leq)\cap{\rm Pol}\,(\neq)$ on ${\bf 2}$ is generated by the (total) ternary majority function

[TABLE]

Lemma 3**.**

Let $n\geq 3$ and let $f\in{\rm Pol}\,(\leq)\cap{\rm Pol}\,(\neq)$ be an $n$-ary function on ${\bf 2}$. Then,

1. (i)

$f(0,\ldots,0)=0$, and 2. (ii)

if $f(1,0,\ldots,0)=1$, then $f=e^{n}_{1}$ is the $n$-ary first projection function on ${\bf 2}$.

Proof.  To see that (i) holds, suppose, for the sake of a contradiction, that $f(0,\ldots,0)=1$. Then, as $f\in{\rm Pol}\,(\neq)$, we have $f(1,\ldots,1)=0$, and hence $f\not\in{\rm Pol}\,(\leq)$.

Now, we show that (ii) also holds. Since $f\in{\rm Pol}\,(\leq)$ and $f(1,0,\ldots,0)=1$, we have that $f(1,x_{2},\ldots,x_{n})=1$, for all $x_{2},\ldots,x_{n}\in{\bf 2}$. As $f\in{\rm Pol}\,(\neq)$, we have that $f(0,x_{2},\ldots,x_{n})=0$, thus showing that $f=e^{n}_{1}$.

As mentioned above, it was shown in [1] that both ${\rm Pol}\,(\leq)$ and ${\rm Pol}\,(\neq)$ are separating clones on ${\bf 2}$. However, this is not the case for their intersection.

Lemma 4**.**

The clone ${\rm Pol}\,(\leq)\cap{\rm Pol}\,(\neq)$ is not a separating clone on ${\bf 2}$.

Proof.  Let $m\geq 1$ and set $n:=m+1$. Let $f_{1},\ldots,f_{m}\in{\rm Pol}\,(\leq)\cap{\rm Pol}\,(\neq)$ be functions of arity $m+1$ and let ${\vec{b}}={\vec{0}}:=(0,\ldots,0)\in{\bf 2}^{m+1}$. Clearly,

[TABLE]

We show that there is ${\vec{a}}\in{\bf 2}^{m+1}\setminus\{{\vec{0}}\}$ such that

[TABLE]

For $i=1,\ldots,m+1$, let ${\vec{e}}_{i}$ be the vector corresponding to the $i$-th row of the identity $(m+1)\times(m+1)$ matrix, that is,

[TABLE]

Now, for every $j=1,\ldots,m$, there is at most one $i\in\{1,\ldots,m+1\}$ such that $f_{j}({\vec{e}}_{i})=1$. Indeed, by Lemma 3 this is the case only if $f_{j}$ is the $(m+1)$-ary $i$-th projection function on ${\bf 2}$. Therefore, there is at least one $t\in\{1,\ldots,m+1\}$ such that $f_{j}({\vec{e}}_{t})=0$, for all $j\in\{1,\ldots,m\}$. Set ${\vec{a}}={\vec{e}}_{t}$. Then

[TABLE]

thus proving that ${\rm Pol}\,(\leq)\cap{\rm Pol}\,(\neq)$ is not a separating clone on ${\bf 2}$.

Now, it is clear that

[TABLE]

and thus, combining Theorem 1 and Lemma 4, we obtain the following result.

Theorem 5**.**

*The intersection of the two finitely generated maximal partial clones ${\rm pPol}\,(\neq)$ and ${\rm pPol}\,(\leq)$ is not a finitely generated partial clone on ${\bf 2}$.  *

We generalize Theorem 5 to any finite set $A={\bf k}:=\{0,1,\ldots,k-1\}$ with $k\geq 3$. First, we need to recall a construction of [3]. Let $\leq$ be the natural (thus linear and bounded) order on ${\bf k}$, $p$ be a prime divisor of $k$ and $\pi$ be the fixed-point-free permutation defined by

[TABLE]

Note that the permutation $\pi$ consists of $k/p$ cycles of length $p$. It is well known (see e.g., [5, 13] and [9] chapter 5) that both ${\rm Pol}\,(\leq)$ and ${\rm Pol}\,(\pi)$ are maximal clones on ${\bf k}$. Moreover, we have the following result appearing in [3].

Theorem 6**.**

([3])* Let $k\geq 3$. Then ${\rm Pol}\,(\leq)\cap{\rm Pol}\,(\pi)={\rm J}(A)$.*

The partial clone ${\rm pPol}\,\pi$ is not a maximal partial clone (see [5, 6]) but is contained in a maximal partial clone. The following result is stated in [5] and its proof is given in [6].

Theorem 7**.**

([6])*  Let $k\geq 3$, $p$ be a prime divisor of $k$ and $\pi$ be any fixed-point-free permutation with all cycles of length $p$ on ${\bf k}$. Define*

[TABLE]

Then,

1. (i)

${\rm pPol}\,(\pi)\subset{\rm pPol}\,(\varrho_{\pi})$, 2. (ii)

${\rm pPol}\,(\varrho_{\pi})$* is the unique maximal partial clone containing the maximal clone ${\rm Pol}\,(\pi)$ on ${\bf k}$, and* 3. (iii)

${\rm pPol}\,(\varrho_{\pi})$* is generated by its unary and binary partial functions.*

It is known that for $k=8$, there is a bounded partial order $\leq_{8}$ on ${\bf k}$, for which the maximal partial clone ${\rm Pol}\,(\leq_{8})$ is not finitely generated. This was shown by Tardos in [14] for the partial order whose diagram is given below.

[math]

$4$

$3$

$2$

$5$

$6$

$7$

$1$

Also, it is well known that every nontrivial order relation (whether it is bounded or not) determines a maximal partial clone on ${\bf k}$ (see e.g., [5, 6] and [9] chapter 20) and such maximal partial clones are all finitely generated.

Theorem 8**.**

([10, 11])* Let $\leq$ be a non-trivial order relation on ${\bf k}$. Then the maximal partial clone ${\rm pPol}\,(\leq)$ is generated by its binary partial functions on ${\bf k}$.*

Combining Theorems 1, 6, 7 and 8 we obtain the following theorem which asserts that, for every $k\geq 1$, the partial clone ${\rm pPol}\,(\varrho_{\pi})\cap{\rm pPol}\,(\leq)$ on ${\bf k}$ is not finitely generated.

Theorem 9**.**

Let $k\geq 3$, $\varrho_{\pi}$ as in Theorem 7 and let $\leq$ be a linear order on ${\bf k}$. Then the intersection of the two finitely generated maximal partial clones ${\rm pPol}\,(\varrho_{\pi})$ and ${\rm pPol}\,(\leq)$ is not a finitely generated partial clone on ${\bf k}$.

Proof.  Clearly, ${\rm pPol}\,(\leq)\cap{\rm pPol}\,(\varrho_{\pi})$ is a strong partial clone on ${\bf k}$, and we have

[TABLE]

We show that ${\rm J}(A)$ is a not a separating clone on ${\bf k}$. Let $m\geq 1$, set $n:=m+1$ and let $f_{1},\ldots,f_{m}$ be projections on ${\bf k}$ of arity $m+1$. Set ${\vec{b}}={\vec{0}}:=(0,\ldots,0)$. Then

[TABLE]

Since there are $m+1$ different projections on ${\bf k}$ of arity $m+1$, there is $i\in\{1,2,\ldots,m,m+1\}$ such that the projection $e_{i}^{m+1}$ is not among $f_{1},\ldots,f_{m}$. Choose

[TABLE]

Then ${\vec{a}}\neq{\vec{0}}$ and $(f_{1}({{\vec{a}}}),\ldots,f_{m}({{\vec{a}}}))=(0,\ldots,0)$, thus proving that ${\rm J}(A)$ is a not separating clone on ${\bf k}$. By Theorem 1, ${\rm pPol}\,(\leq)\cap{\rm pPol}\,(\varrho_{\pi})$ is not a finitely generated clone on ${\bf k}$.

Note that the partial clone ${\rm pPol}\,(\leq)\cap{\rm pPol}\,(\varrho_{\pi})$ does not consist only of partial projections. Indeed, let $n\geq 3$ and construct the partial $n$-ary function $f$ on ${\bf k}$ by setting ${\rm dom}\hphantom{.}(f):=\{0\}\times{\bf k}^{n-1}$ and $f({\vec{v}})=1$ for all ${\vec{v}}=(0,x_{2},\ldots,x_{n})\in\{0\}\times{\bf k}^{n-1}$. Clearly, $f$ is not a partial projection. Moreover, $f\in{\rm pPol}\,(\leq)$ since it is a constant function, and $f\in{\rm pPol}\,(\varrho_{\pi})$ since there is no $p\times n$ matrix over ${\bf k}$ with columns in $\varrho_{\pi}$ and rows in ${\rm dom}\hphantom{.}(f)$.

The study in this paper yields the following interesting problem:

Problem 1**.**

Let $|A|=k\geq 2$. Describe the strong partial clones on $A$ whose total part is ${\rm J}(A)$.

Note that Str$({\rm J}(A))$, the partial clone consisting of all subfunctions of projections, is one of them. These are not finitely generated partial clones.