CC BY
38
УДК 519.716
Classification of Hyperfunctions of Rank 2 with Respect to Membership in the Maximal Partial Ultraclones
Sergey A. Badmaev*
Institute of Mathematics and Computer Science Buryat State University Smolina, 24a, Ulan-Ude, 670000
Russia
Received 03.04.2019, received in revised form 22.07.2019, accepted 20.09.2019 In this paper, we consider the set of hyperfunctions, which is a subset of the full partial ultraclone of rank 2. For hyperfunctions, the problem of their classification with respect to membership in the the maximal partial ultraclones is solved. The relation of membership in the maximal partial ultraclones is an equivalence relation and generates the corresponding partition into equivalence classes. A complete description of all equivalence classes, the total number of which is 28, is obtained.
Keywords: multifunction, hyperfunction, clone, ultraclone, maximal partial ultraclone, classification of functions.
DOI: 10.17516/1997-1397-2019-12-5-644-651.
Introduction
A sets of multifunctions are considered. A multifunction on a finite set A is a function defined on the set A and taking as its values its subsets. Obviously, superposition in the usual sense does not work when working with multifunctions. Therefore, they need to give a new definition of superposition. Two ways of defining superposition are usually considered: the first is based on the union of subsets of the set A, and in this case the closed sets containing all the projections are called multiclones, and the second is the intersection of the subsets of A, and the closed sets containing all projections are called partial ultraclones. The set of multifunctions on A on the one hand contains all the functions of |A|-valued logic and on the other, is a subset of functions of 2lAl-valued logic with superposition that preserves these subset.
In the theory of functions the problem of classification is interesting. One of the known variants of the classification of functions of fc-valued logic is one in which functions in a closed subset B of a closed set M can be divided according to their membership in the classes that are pre-complete in M. In this paper, the subset of B is the set of all Boolean functions, and the set of M is the set of all multifunctions on the two-element set, and the partial maximal ultraclones are pre-complete classes.
Note that the quality and type of functions are limited to the function The other k-logic logic is used, for example, in the process [1-8].
* badmaevsa@mail.ru © Siberian Federal University. All rights reserved
1. Basic concepts and definitions
Let E = {0,1} h F = {0, {0}, {1}, {0,1}}. We define the following sets of functions: Pl,n = {f \f : En ^ F}, P* = U Pin;
n
P2,n = {f \f g Pin h \f (&)\ = 1 for any 5 G En}, P2 = (JP2,n;
n
P-n = {f \f : En ^ F \ {0}}, P— = (J P-n;
n
Pin = {f\f G Pin and\f(a)\ < 1 for any 5 G En}, P* = (JP*^
n
Functions from P2 are called Boolean functions, functions from P* are called partial functions on E, functions from P— are called hyperfunctions on E, functions from P* are called multifunctions on E.
We believe that the superposition
f (fi (xu... , xm
where f,fi,-..,fn G P*, represents some multifunction g(xi,..xm) on a tuple with elements from the set F, if for any (ai,..., am) G Em
{p| f (fi,.fn) if the intersection is not empty;
I3iefi(ai,... , am)
u f (fi,...,fn) otherwise.
I3iefi(ai,... , am)
On tuples containing 0, the multifunction takes the value 0.
This definition allows you to find the value f (xi,..., xn) for every (ai,..., an) G Fn.
For simplicity we use the following code:
0 ^ *, {0} ^ 0, {1} ^ 1, {0,1} ^ -.
We note that in this paper we will adhere to the terminology adopted in [9], which will allow us not to introduce additional definitions here.
In [10] it is proved that the maximal partial ultraclones of rank 2 are only the following 12 sets:
1) Ki is the set consisting of all multifunctions f such that f (0) G {0, *}.
2) K2 is the set consisting of all multifunctions f such that f (1) G {1, *}.
3) K3 is the set consisting of all multifunctions f for which one of the two conditions is fulfilled:
• f (0) = * or f (1) = *,
• f (0) = 0 and f (1) = 1.
4) K4 is the set consisting of all multifunctions f such that on any binary tuple a one of three conditions is fulfilled:
• f (a) = f (a) = -,
• f (a) = f (a) = *,
f (a) = f (a), where f (a) g {0,1}.
5) K5 is the set consisting of all multifunctions f such that on any binary tuple a one of two conditions is fulfilled:
• f (a) = * or f (a) = *,
• f (â) = f (â), where f (â) G {0,1}.
6) Re = P- U {*}.
7) Rr = P|.
8) R8 is the set of all multifunctions f that simultaneously satisfy three conditions:
• if f (â),f (3), f (y) G {0,1}, then
f Ifll e
a.
where a = (a1,..., an), fl = (fl1,..., fln), 7 = (71,..., jn) are binary tuples such that (aifliYi) e {(000), (001), (010), (111)} for any i e {1,..., n};
• if there is a binary tuple a such that f (a) = —, then for any binary tuple fl true f (fl) = 1;
• let binary tuples a = (a1,..., an), fl = (fl1,..., fln) such that 04 < fli for any i e {1,... ,n} then, if f (a) = *, then f (fl) = *.
9) K9 is the set of all multifunctions f that simultaneously satisfy three conditions:
• if f (a), f (fl), f (a) e{0,1}, then
f (fl)e
where a = (a1,..., an), fl = (fl1,..., fln), a = (j1,..., jn) are binary tuples such that (aifliYi) e {(000), (011), (101), (111)} for any i e {1,..., n};
• if there is a binary tuple a such that f (a) = —, then for any binary tuple fl true f (fl) = 0;
• let binary tuples a = (a1,..., an), fl = (fl1,..., fln) such that ai < fli for any i e {1,... ,n} then, if f (fl) = *, then f (a) = *.
10) K10 is the set of all multifunctions preserve the predicate /0 0 0 0 1 1 1 1 — a\
0 0 1 1 1 1 0 0 — fl 0 10 110 10 — y
V0 1 1 0 1 0 0 1 — sj
which a,fl,Y,S e F are simultaneously satisfy two conditions:
R
10
where (a, ¡3, j, S) are all sorts of columns in
in every column (a, fl, y, S) among a, fl, y, S least two assume the value *;
in every column (a,fl,Y,5)t, if 0 or 1 are found among a,fl,Y,S then all of them are not equal to —.
11) Кц is the set of all multifunctions preserve the predicate
/0 00110 0 - - 01 -****** ДЦ = 100101 0 - 0 - * * * 01 - * * * \0 1 0 0 1 - 0 0 - ****** 01 -
12) K12 is the set of all multifunctions preserve the predicate
/0 01111 1 - - 01 -****** Д12 = (01011 1 - 1 - ** * 01 - ** * \0 1 1 0 1 - 1 1 - ****** 01 -
For any multifunction f we uniquely define vector т(f) = (т1; ... ,712). This vector т(f) is a
0 if f e Ki
vector of membership in the sets K1 — K12 and for every i G {1,..., 12} т\ = ,
[ 1 f f G K
The membership relation in the sets K1 — K12 is an equivalence relation and generates a partition of P2 into equivalence classes. For multifunctions from one class, the membership vectors in the sets K1 — K12 are the same. Since there are 12 maximal partial ultraclones, the largest possible number of equivalence classes is 212 = 4096.
In this paper, we find the number of equivalence classes that consist only of hyperfunctions.
2. The main result
In [1] it is shown that the set of Boolean functions is divided into 15 equivalence classes with respect to membership in the maximal partial ultraclones. Therefore, throughout the paper, we consider only hyperfunctions from the set P- \ P2. Obviously, any hyperfunction belongs to the K6 and does not belong to the K7.
Lemma 1. For any f G P— \ P2 the following statements are true:
1) f G Kb,
2) if f is not the constant hyperfunction —, then f G K10,
3) f G Ki П K2 if and only if f G K3.
Proof. 1) Let f be an arbitrary hyperfunction from the set P- \ P2. There must be a tuple a such that f (à) = —. Moreover, it is obvious that f (à) G {0,1, —}. Therefore, f G K5.
2) Let f be an arbitrary hyperfunction other than the constant hyperfunction —. There are
/-\
/^N
tuples a1 and a2 such that f (à1) = — and f (à2) = Л G {0,1}. Then f
ä1 à2
Л
W
G R
10,
\a2J
where column (ajaja2a2) belongs to the predicate Ri0 for any i G {1,..., n}.
3) The assertion follows directly from the definitions of the Ki, K2, K3. □
Lemma 2. For any f G P— \ P2 the following statements are true:
1)if fGK then f G Kii,
2) if f G Kg then f G Ki2.
Proof. 1) Let f G K8. Suppose that f does not satisfy the first condition in the definition of the K8. There are tuples a1 = (a\,..., aln), where i G {1, 2,3}, such that column (a!a^aj)4 coincides with one of the columns (000)4, (001)4, (010)4, (111)4 for any j and
ai a2
a
à2 I G { I 0 j , ( 1 I , ( 0 j , ( 1 j.If f ( à2 j G { ( 1 I , ( 0 j , ( 1 j, then considering
that (000)t, (001)t, (010)/ (1\1)t e R\ and (011)t, (101)t, (110)t e R11, we get f e K11. If
(a1) (1) (a1) (1)
f I a2 I = I 0 I, then f ( fl I = ( — I e R11, where the tuple fl = (flb ...,fln) is such that
VV W W W
flk = — for those k for which (a\a|a|)t = (010)t, and flk = a| for the remaining k. Therefore,
f e K11.
Now suppose that f does not satisfy the second condition in the definition of the K8. There are tuples a1 and a2 such that f (a1) = — and f(a2) = 1. Consider the value of f on the tuple
a1
consisting only of 1. If f (1) e {0,1}, then f ( 1 ) e 0 ) , ( 1 )e R11. If f (1) = —, then
f | ^J = (—) e R11. Therefore, f e Kn.
2) The proof is similar to the proof of the preceding item due to duality. □
Lemma 3. For any f e P— \ P2 the following statements are true:
1) if f e K1, then f e K9 and f e K12;
2) if f e K2, then f e K8 and f e K11;
3) if feK1f) K2, then feK8[j K9 and f / Kn\J K12.
Proof. 1) Let f e K1. Then f (0) = 0, i.e. there is the tuple on which the value of f is equal to 0. So, taking into account the mandatory existence of a tuple on which the value of the function f is equal to —, we obtain that the hyperfunction f does not satisfy the second condition in the definition of the K9. Therefore, f / K9 and, by the point 2) of Lemma 2, we obtain that
f e K12.
2) The proof is similar to the proof of the preceding item due to duality.
3) The validity of the statement follows from the items 1) and 2) of the present Lemma, as well as the items 1) and 2) of Lemma 2. □
Lemma 4. Let f e P— \ P2. If f e Kx \ K2 or f e K2 \ K1, then f / K4.
Proof. For definiteness, let f e K1 \K2. Then f (0) = 0 and f (1) e {0, —}. We show that in each case the hyperfunction f does not satisfy the conditions in the definition of the K4. If f (1) = 0,
then f (1) = f (0) =0=1= f0) = f (1). If f (1) = —, then f (1) = f (0) = 0 = —.In the case when the hyperfunction f belongs to the set K2 \ K1, the proof is similar. □
Lemma 5. For any f e P— \ P2 the following statements are true:
1) if f e K8f}K9, then f is the constant hyperfunction —;
2) if f e Knn K12, then f is the constant hyperfunction —.
Proof. 1) Suppose f is not the constant hyperfunction —. Then there is a tuple a such that f (a) = A e {0,1}. There is necessarily a tuple in which the value of f is equal to —. So, if A = 0, then f does not satisfy the second condition in the definition of the K9, if A =1, then f does not satisfy the second condition in the definition of the K8. Therefore, either f e K9 or f e K8, which contradicts the fact that f belongs to the set K8 p| K9.
2) Suppose f is not the constant hyperfunction —. From the previous item, we obtain either f e K9 or f e K8. Further, using the assertions of Lemma 2, we obtain that either f e K12 or f e K11, which contradicts the fact that f belongs to the set Kn D K12. □
Lemma 6. Let f e P— \ P2. If f / K1\} K2 and f e K4, then either f is the constant hyperfunction — or f e K8 |J K9 and f e K11 |J K12.
a1
Proof. Since f /KiU K2, then f (0) G {1,-} and f (1) G {0, -}. Considering that f G K4 it is enough to consider cases f (0) = 1, f (1) = 0 and f (0) = f (1) = —. In the case when f (0) = 1, f (1) =0 as well as in the proof of Lemma 3, we get that f G K$U K9 and f G Kn U K12. If f (0) = f (1) = —, then either f is the constant hyperfunction — and the statement of the Lemma holds, or there is a tuple a such that f (a) G {0,1}. Without loss of generality, we can assume that f (a) = 0. Since f G K4, then f (a) = 1. Thus, there are tufles on which the hyperfunction f is equal to 0, 1, and —. Therefore, f G Kg U K9 and f G Kn U K12. □
Lemma 7. Let f G P- \ Pi- If f G KiU K2 and f G K4, then f G K11 U K12.
Proof. Since f G Ki[J K2, then f (0) G {1, —} and /(1) G {0, —}. For the cases when f (0) and
- (0\ i(1\ (1 I (—\\
f (1) are not equal to — simultaneously, we have f |1| G<|0| , ( — I , ( 0 |J G R11 and
„ W U y W w J
f (o| G |Q , (—| , (11 J G R12. Suppose f (0) = f (1) = —. Since f G K, it is not the constant hyperfunction — and, so, there is a tuple a such that f (a) = A G {0,1}. Then f (a | = ( A | G R11 and f ( a | = | A | G R12. □
Theorem 1.1. The set of all hyperfunctions of rank 2 other than Boolean functions generates no more than 13 equivalence classes with respect to membership in the maximal partial ultraclones.
Proof. From the first two points of Lemma 1, it follows that for any of the considered hyperfunctions f the components t5 and r1o of the vector t(f) = (t1t2T3T4T501T8T9T1oT11 t12) are equal to 1, where t(f) is the vector of membership in the K1 — K12. From the third point of the Lemma 1 we get that (t1t2t3) g {(000), (011), (101), (111)}. Consider all these cases.
From the third point of Lemma 3, it follows that the hyperfunctions belonging simultaneously to the K1, K2, K3 are divided into no more than 2 equivalence classes, these classes correspond to the vectors (000010111111), (000110111111).
Now consider the hyperfunctions that either belong to K1 and do not belong to K2, K3, or belong to K2 and do not belong to K1, K3. Using Lemma 2, the first two points of Lemma 3 and Lemma 4, we obtain that the number of equivalence classes for such hyperfunction is no more than 6 and the vectors corresponding to these classes have the form (011110101101), (011110101111), (011110111111), (101110110110), (101110110111), (101110111111).
It remains to consider hyperfunctions that do not belong to any of the K1, K2, K3. It is obvious that among such hyperfunctions there are those that take the value — on each tuple. It is easy to verify that the vector of membership in the K1 — K12 for these hyperfunctions has the form (111010100000). Further we assume that hyperfunctions are not constant. By Lemma 6, we obtain that hyperfunctions belonging to the K4 can generate at most one equivalence class, to which the membership vector (111010111111) corresponds. Further, applying Lemmas 5 and 7, we obtain that hyperfunctions that do not belong to the K4 are divided into no more than 3 equivalence classes, which correspond to the vectors (111110101111), (111110110111), (111110111111). □
Theorem 1.2. The set of all hyperfunctions of rank 2 generates 28 equivalence classes with respect to membership in the maximal partial ultraclones.
Proof. Since the number of classes of Boolean functions is 15 considering the previous theorem, we obtain that all hyperfunctions are divided into no more than 28 equivalence classes.
As a result of computer calculations on hyperfunctions of three variables, 28 different vectors of membership in the K1 — K12 were found. The Tab. 1 shows the vectors of affiliation and the corresponding hyperfunctions. Table 1 shows membership vectors and the corresponding hyperfunctions. Note that at number 23 there is the constant hyperfunction —.
Table 1
N r (f) f (x!,x2,x3) № r (f) f (X1,X2,X3)
1 (000000000000) (00001111) 15 (101110000000) (11111111)
2 (000000011011) (01101001) 16 (101110011011) (10011001)
3 (000000011111) (00010111) 17 (101110011111) (10000001)
4 (000010111111) (000—111) 18 (101110110110) (—1111111)
5 (000110001101) (00000001) 19 (101110110111) (111111 — 1)
6 (000110010110) (00111111) 20 (101110111111) (100000—1)
7 (000110011111) (00000111) 21 (111000011011) (10010110)
8 (000110111111) (000000—1) 22 (111000011111) (10001110)
9 (011110000000) (00000000) 23 (111010100000) —
10 (011110011011) (00111100) 24 (111010111111) (100—110)
11 (011110011111) (00000010) 25 (111110011111) (10000000)
12 (011110101101) (0000000—) 26 (111110101111) (—0000000)
13 (011110101111) (000000—0) 27 (111110110111) (1111111—)
14 (011110111111) (0000001—) 28 (111110111111) (1000000—)
□
This work was supported by RFBR according to the research project no. 18-31-00020.
References
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Классификация гиперфункций ранга 2 относительно принадлежности максимальным частичным ультраклонам
Сергей А. Бадмаев
Институт математики и информатики Бурятский государственный университет Смолина, 24а, Улан-Удэ, 670000 Россия
В данной 'работе рассматривается множество гиперфункций, которое является подмножеством полного частичного ультраклона ранга 2. Для гиперфункций решена задача их классификации относительно принадлежности максимальным частичным ультраклонам. Отношение принадлежности максимальным частичным ультраклонам является отношением эквивалентности и порождает соответствующее разбиение на классы эквивалентности. Получено полное описание всех классов эквивалентности, общее число которых равно 28.
Ключевые слова: мультифункция, гиперфункция, клон, ультраклон, максимальный частичный ультраклон, классификация функций.
