Boolean Expression Based on Hypergraphs with Algorithm

2020·0.33 MB·English

by  Mohammad Hamidi, Marzieh Rahmati, Akbar Rezaei;

AlgebraicStructuresandTheirApplicationsVol. .. No. (20.. ) pp.
BOOLEAN EXPRESSION BASED ON HYPERGRAPHS WITH
ALGORITHM
MOHAMMAD HAMIDI∗, MARZIEH RAHMATI AND AKBAR REZAEI
Abstract. This paper introduces a novel concept of Boolean function{based hypergraph
with respect to any given T.B.T(total binary truth table). This study de(cid:12)nes a notation
of kernel set on switching functions and proves that every T.B.T corresponds to a Minimum
BooleanexpressionviakernelsetandpresentssomeconditionsonT.B.TtoobtainaMinimum
irreducibleBooleanexpressionfromswitchingfunctions. Finally,wepresentanalgorithmand
so Python programming(with complete and original codes) such that for any given T.B.T,
introduces a Minimum irreducible switching expression.
1. Introduction
TheconceptofhypergraphhasbeenintroducedbyBergeasageneralizationofgrapharound
1960 and one of the initial concerns was to extend some classical results of graph theory and
the notion of hypergraph has been considered as a useful tool to analyze the structure of a
system. Graphs and hypergraphs can be used to describe the network systems. Today, some
features of hypergraphs are used in computer science, notably in machine learning, and there
has been a lot of research about using hypergraphs in relational databases, which might be
MSC(2010): Primary: 05C65,94C10
Keywords: Switchingfunction,Booleanfunction{basedhypergraph,switchingkernel,T.B.T.
Received: 17May2020,Accepted: 04July2020.
∗Correspondingauthor
⃝c 2020YazdUniversity.
1
2 Alg. Struc. Appl. Vol. .. No. . (20..) ..-...
viewed as a sort of data mining. The reason why hypergraphs seem apt to depict relations in
information systems, social networks, document centered information processing, web infor-
mation systems and computer science, are the relationships among services within a service
oriented architecture[4, 5, 9]. With respect to the classical hypergraph, Smarandache (2019)
added the supervertices (a group of vertices put all together form a supervertex), in order
to form a superhypergraph (SHG). Therefore, each SHG-vertex and each SHG-edge belong
to P(V), where V is the set of vertices, and P(V) means the power set of V. Further on,
since in our world we encounter complex and sophisticated groups of individuals and complex
and sophisticated connections between them, Smarandache extended the superhypergraph to
n-superhypergraph, by extending P(V) to Pn(V) that is the n-power set of the set V (i.e.
power-set of the power-set of the powerset of V, n times)[10, 11]. Further materials regarding
graphs and hypergraphs are available in the literature too [1, 4, 5, 6, 7, 8]. Two major new
works on logic were published by prominent British mathematicians, formal logic by Augustus
DeMorgan(1806{1871)andthemathematicalanalysisofLogicbyGeorgeBoole(1815{1864).
Both authors sought to stretch the boundaries of traditional logic by developing a general
method for representing and manipulating logically valid inferences or to develop mechanical
modes of making transitions. Historically, propositional logic and electrical engineering have
beenthemainnurturing(cid:12)eldsforthedevelopmentofresearchonBooleanfunctions. However,
because they are such fundamental mathematical objects, Boolean functions have also been
used to model a large number of applications in a variety of areas. In most applications, how-
ever, moreinformationisavailableregardingtheprocessthatgeneratesthefunctionofinterest
asillus trated by Electrical engineering, Arti(cid:12)cial neural networks, Reliability theory, Game
theory, Integer programming, Distributed computing systems, etc. In other view, Boolean
expressions have been widely used to represent a decision/predicate and a number of branch
testing and techniques have been reported in the literature[2, 3].
Regarding these points, this paper considers the concepts of, Boolean expression, hyper-
graphs and with respect to combination of these concepts, applies it in computer science. In
this paper we introduce the notation of switching functions and investigates the relation be-
tween hypergraphs and switching functions. This study, for any Boolean function constructs
a hyperdiagram which is called a hyperdiagram based on a given Boolean function and in-
vestigate some condition it be a hypergraph based on a given Boolean function. Also for
all arbitrary hypergraph, is extracted a Boolean function titled Boolean functionable hyper-
graph. There is a natural question that, do it correspond a switching expression from any
given T.B.T. The main our motivation from this paper is extraction an irreducible switching
expression from any T.B.T. So we de(cid:12)ne the concept of Boolean function{based hypergraph
and the notation of unitors set of Boolean functions. In (cid:12)nal, we apply these concepts and
Alg. Struc. Appl. Vol. .. No. .. (20..) ..-... 3
prove that every T.B.T corresponds to a Minimum Boolean expression via kernels set and
presents some conditions on T.B.T to obtain a Minimum irreducible Boolean expression from
switching functions.
2. Preliminaries
In this section, we recall some de(cid:12)nitions and results, which we need in what follows.
Let X be an arbitrary set. Then we denote P∗(X) = P(X)r∅, where P(X) is the power
set of X.
De(cid:12)nition 2.1. [1] Let G = {x ,x ,...,x } be a (cid:12)nite set. A hypergraph on G is a pair
1 2 n
∪m
H = (G,{E }m ) such that for all 1 ≤ i ≤ m,∅ ̸= E ⊆ G and E = G. The ele-
i i=1 i i
i=1
ments x ,x ,...,x of G are called vertices, and the sets E ,E ,...,E are called the edges
1 2 n 1 2 m
(hyperedges) of the hypergraph H. For each 1 ≤ k ≤ m if |E | ≥ 2, then E is represented by
k k
a continuous curve joining its vertices, if |E | = 1 by a cycle on the element (loop). If for all
k
1 ≤ k ≤ m |E | = 2, the hypergraph becomes an ordinary (undirected) graph.
k
De(cid:12)nition 2.2. [4] Let G = {x ,x ,...,x } be a (cid:12)nite set. A hyperdiagram on G is a pair
1 2 n
H = (G,{E }m )suchthatforall1 ≤ k ≤ m,E ⊆ Gand|E | ≥ 1. Clearlyeveryhypergraph
k k=1 k k
is a hyperdiagram, while the converse is not necessarily true.
′
We say that two hyperdiagrams H = (G,{E }m ) and H′ = (G′,{E′}m ) are isomorphic
k k=1 k k=1
if m = m′ and there exists a bijection φ : G → G′ and a permutation τ : {1,2,...,m} →
{1,2,...,m′} such that for all x,y ∈ G, if for some 1 ≤ i ≤ m,x,y ∈ E , then φ(x),φ(y) ∈
i
E , if for all 1 ≤ i ≤ m,x,y ̸∈ E , then φ(x),φ(y) ̸∈ E and if for some 1 ≤ i ≤ m,x ∈ E ,
(cid:28)(i) i (cid:28)(i) i
for all 1 ≤ j ≤ m,y ̸∈ E , then φ(x) ∈ E and φ(y) ̸∈ E . Since every hypergraph is a
j (cid:28)(i) j
hyperdiagram, de(cid:12)ne an isomorphic hypergraphs in a similar a way.
3. Switching expression based on hypergraph
Inthissection,weapplythenotationoftotalbinarytruthtable(T.B.T)onBooleanvariables
and introduce the concept of hypergraphable Boolean functions, Boolean functionable hyper-
graphsandinvestigatesomeoftheirproperties. Weconsiderevery(switching)Boolean function
∏m ∑kj
f : B → B = {0,1} by f(x ,x ,...,x ) = x , where for all 1 ≤ i ≤ n,x is a literal
n 1 2 n i i
j=1i=1
(Boolean variable or the complement of a Boolean variable) and m,j,k ∈ N. Let n ∈ N,m ∈
j
N∗, x ,x ,...,x be arbitrary Boolean variables and for all 0 ≤ j ≤ m, f(m)(x ,x ,...,x )
1 2 n 1 2 n
be Boolean functions. We will denote a total binary truth table(T.B.T) on Boolean variables
x ,x ,...,x by a set T(f(0),f(1),...,f(m),x ,...,x ) = {f(0),f(1),...,f(m),(x ,...,x )},
1 2 n 1 n 1 n
4 Alg. Struc. Appl. Vol. .. No. . (20..) ..-...
Table 1. T. B. T with n variables T(f(0),f(1),...,f(m),x ,x ,...,x )
1 2 n
x x ... x f(0)(x ,...,x ) f(1)(x ,...,x ) ... f(m)(x ,...,x )
1 2 n 1 n 1 n 1 n
(0) (1) (m)
0 0 ... 0 f (x ,...,x ) f (x ,...,x ) ... f (x ,...,x )
1 1 n 1 1 n 1 1 n
(0) (1) (m)
0 0 ... 1 f (x ,...,x ) f (x ,...,x ) ... f (x ,...,x )
2 1 n 2 1 n 2 1 n
. . . . . . .
. . . . . . .
. . . . . . .
(0) (1) (m)
0 0 ... 1 f (x ,...,x ) f (x ,...,x ) ... f (x ,...,x )
2n−1 1 n 2n−1 1 n 2n−1 1 n
(0) (1) (m)
1 1 ... 1 f (x ,...,x ) f (x ,...,x ) ... f (x ,...,x )
2n 1 n 2n 1 n 2n 1 n
Table 2. T.B.T T(f,g,x ,x ,x )
1 2 3
x x x f g f +g f.g c(f)
1 2 3
0 0 0 1 0 1 0 0
0 0 1 0 0 0 0 1
0 1 0 1 1 1 1 0
0 1 1 0 1 1 0 1
1 0 0 0 1 1 0 1
1 0 1 1 1 1 1 0
1 1 0 0 0 0 0 1
1 1 1 1 1 1 1 0
where for all 0 ≤ j ≤ m, f(m)(x ,x ,...,x ), are Boolean functions(see a Table 1) and for
1 2 n
m = 0, we will denote it by T(f,x ,x ,...,x ).
1 2 n
c(f(x ,...,x )) = 1−f(x ,...,x ).
1 n 1 n
Example 3.1. ConsideraT.B.TT(f,g,x ,x ,x )inTable2. Thebinaryoperationstogether
1 2 3
with a unary operation is computed in this Table.
Theorem 3.2. (T(f,f′,x ,...,x ),+,.,c) is a Boolean algebra.
1 n
De(cid:12)nition 3.3. Let n ∈ N and T(f,x ,x ,...,x ) be a T.B.T. For all 1 ≤ j ≤ 2n de(cid:12)ne
1 2 n
Kernel(f ) = {(x ,x ,...,x ) | f (x ,x ,...,x ) = 0} and will denote by Ker(f ), in a
j 1 2 n j 1 2 n j
similar a way Kernel(f) is de(cid:12)ned and it is denoted by Ker(f).
Alg. Struc. Appl. Vol. .. No. .. (20..) ..-... 5
Theorem 3.4. Let n ∈ N,1 ≤ j ≤ 2n and T(f,f′,x ,x ,...,x ) be a T.B.T. Then
1 2 n
∪2n
(i) Ker(f) = Ker(f );
j
j=1
(ii) Ker(f′) = Ker(f) if and only if f ∼ f′;
(iii) If Ker(f) ⊆ Ker(f′), then (f +f′) ∼ f;
(iv) If Ker(f′) ⊆ Ker(f), then (f.f′) ∼ f;
(v) Ker(f +f′) = Ker(f)∩Ker(f′);
(vi) Ker(f.f′) = Ker(f)∪Ker(f′);
(vii) Ker(c(f)) = B \Ker(f).
n
Proof. Let n ∈ N and (x ,...,x ) ∈ B .
1 n n
∏2n
(i) Since Ker(f) = {(x ,...,x ) | f (x ,...,x ) = 0}, we get
1 n j 1 n
j=1
(x ,...,x ) ∈ Ker(f) ⇔ ∃ 1 ≤ j ≤ 2n such that f (x ,...,x ) = 0
1 n j 1 n
∪2n
⇔ for some 1 ≤ j ≤ 2n,(x ,...,x ) ∈ Ker(f ) ⇔ (x ,...,x ) ∈ Ker(f ).
1 n j 1 n j
j=1
(ii) Let (x ,x ,...,x ) ∈ Ker(f). Then f(x ,x ,...,x ) = 0 and because f ∼ f′, we get that
1 2 n 1 2 n
f′(x ,x ,...,x ) = 0. It follows that (x ,x ,...,x ) ∈ Ker(f′) and Ker(f) ⊆ Ker(f′) and in
1 2 n 1 2 n
asimilarawayKer(f′) ⊆ Ker(f). LetKer(f′) = Ker(f). Thenf(x ,x ,...,x ) = 0implies
1 2 n
′ ′
that f (x ,x ,...,x ) = 0 and f(x ,x ,...,x ) = 1 implies that f (x ,x ,...,x ) = 1. So
1 2 n 1 2 n 1 2 n
f ∼ f′.
(iii) If (x ,x ,...,x ) ∈ Ker(f), then (f + f′)(x ,x ,...,x ) = f(x ,x ,...,x ) +
1 2 n 1 2 n 1 2 n
f′(x ,x ,...,x ) = 0 = f(x ,x ,...,x ). If (x ,x ,...,x ) ̸∈ Ker(f), then (f +
1 2 n 1 2 n 1 2 n
′ ′
f )(x ,x ,...,x ) = f(x ,x ,...,x )+f (x ,x ,...,x ) = 1 = f(x ,x ,...,x ).
1 2 n 1 2 n 1 2 n 1 2 n
(iv) If (x ,x ,...,x ) ∈ Ker(f),
1 2 n
′ ′
then (f.f )(x ,x ,...,x ) = f(x ,x ,...,x ).f (x ,x ,...,x ) = 0 = f(x ,x ,...,x ). If
1 2 n 1 2 n 1 2 n 1 2 n
(x ,x ,...,x ) ̸∈ Ker(f), because Ker(f′) ⊆ Ker(f), we get that (x ,...,x )
1 2 n 1 n
̸∈ Ker(f′) and so (f.f′)(x ,x ,...,x ) = f(x ,x ,...,x ).f′(x ,x ,...,x ) = 1 =
1 2 n 1 2 n 1 2 n
f(x ,x ,...,x ).
1 2 n
′
(v) By de(cid:12)nition, (f + f )(x ,x ,...,x ) = 0 if and only if f(x ,x ,...,x ) +
1 2 n 1 2 n
′ ′
f (x ,x ,...,x ) = 0 if and only if f(x ,x ,...,x ) = 0 and f (x ,x ,...,x ) = 0 if and
1 2 n 1 2 n 1 2 n
only if (x ,x ,...,x ) ∈ Ker(f)∩Ker(f′).
1 2 n
′
(vi) By de(cid:12)nition, (f.f )(x ,x ,...,x ) = 0
1 2 n
′
if and only if f(x ,x ,...,x ).f (x ,x ,...,x ) = 0 if and only if f(x ,x ,...,x ) = 0 or
1 2 n 1 2 n 1 2 n
f′(x ,x ,...,x ) = 0 if and only if (x ,x ,...,x ) ∈ Ker(f)∪Ker(f′).
1 2 n 1 2 n
6 Alg. Struc. Appl. Vol. .. No. . (20..) ..-...
(vii) (x ,x ,...,x ) ∈ Ker(c(f)) if and only if c(f)(x ,x ,...,x ) = 0 if and only if
1 2 n 1 2 n
1−f(x ,x ,...,x )
1 2 n
= 0 if and only if f(x ,x ,...,x ) = 1 if and only if (x ,x ,...,x ) ̸∈ Ker(f).
1 2 n 1 2 n
Example 3.5. Consider a T.B.T T(f,g,x ,x ,x ) in Table 2. Computations show that
1 2 3
Ker(f +g) = {(0,0,1),(1,1,0)} = Ker(f)∩Ker(g) = {(0,0,1),(1,0,0),(0,1,1),(1,1,0)}∩
{(0,0,0),(0,0,1),(1,1,0)} and Ker(f.g) = {(0,0,0),(0,0,1),(0,1,1),(1,0,0),(1,1,0)} =
Ker(f)∪Ker(g).
Corollary 3.6. Let n ∈ N,1 ≤ j ≤ n and T(f,f′,x ,x ,...,x ) be a T.B.T. If (f +f′) ∼ f,
1 2 n
then |Ker(f)| ≤ |Ker(f′)|.
Proof. Since (f +f′) ∼ f, by Theorem 3.4, Ker(f +f′) = Ker(f). It follows that Ker(f)∩
Ker(f′) = Ker(f +f′) = Ker(f) and so Ker(f) ⊆ Ker(f′).
Corollary 3.7. Let n ∈ N,1 ≤ j ≤ n and T(f,f′,x ,x ,...,x ) be a T.B.T. If (f.f′) ∼ f,
1 2 n
then |Ker(f′)| ≤ |Ker(f)|.
Proof. Since (f.f′) ∼ f, by Theorem 3.4, Ker(f.f′) = Ker(f). It follows that Ker(f) ∪
Ker(f′) = Ker(f.f′) = Ker(f) and so Ker(f′) ⊆ Ker(f).
Inthissection, consideranyT.B.TandwithrespecttoconceptofKernelsset, trytoextract
associated switching expression to given T.B.T.
Theorem 3.8. Let n ∈ N. Then every T(f ̸≡ 0,x ,x ,...,x ) corresponds to a hypergraph.
1 2 n
Proof. Let x ,x ,...,x be arbitrary Boolean variables. Consider a total binary truth table
1 2 n
(T.B.T)T
(f,x ,x ,...,x ) in Table 1. Suppose that for k ∈ N and for all i ∈ {j ,j ,...,j }, we have
1 2 n 1 2 k
∑n
Ker(f ) ̸= ∅. Since x = 0 if and only if for all 1 ≤ j ≤ n,x = 0, for all j ≤ i ≤ j de(cid:12)ne
i j j 1 k
j=1
∑n
f (x ,x ,...,x ) = x +x +...+x , where x = 0. Now, for all j ≤ i ≤ j , set
ji 1 2 n 1 2 n j 1 k
j=1
∑n
E = {x ,x ,...,x | x = f (x ,x ,...,x )}.
i 1 2 n j j 1 2 n
j=1
∪j
k
Thus it is easy to see that H = (G = E ,{E }jk ) is a hypergraph.
i i i=j1
i=j1
Alg. Struc. Appl. Vol. .. No. .. (20..) ..-... 7
Table 3. T. B. T with 3 variables T(f,x ,x ,x )
1 2 3
x x x f(x ,x ,x )
1 2 3 1 2 3
0 0 0 f (x ,x ,x ) = 1
1 1 2 3
0 0 1 f (x ,x ,x ) = 1
2 1 2 3
0 1 0 f (x ,x ,x ) = 1
3 1 2 3
0 1 1 f (x ,x ,x ) = 0
4 1 2 3
1 0 0 f (x ,x ,x ) = 0
5 1 2 3
1 0 1 f (x ,x ,x ) = 0
6 1 2 3
1 1 0 f (x ,x ,x ) = 1
7 1 2 3
1 1 1 f (x ,x ,x ) = 1
8 1 2 3
We will call the hypergraph H in Theorem 3.8, as Boolean function{based hypergraph and
will denote by (H,T).
Example 3.9. Consider a T.B.T T(f,x ,x ,x ) in Table 3. Computations show that
1 2 3
Ker(f ) = {(x ,x′,x′)},Ker(f ) = {(x′,x ,x )},Ker(f ) = {(x′,x ,x′)}. Now, it is ob-
4 1 2 3 5 1 2 3 6 1 2 3
tained a Boolean function{based hypergraph (H,T) in Figure 1.
xx′′ xx
11 11
xx............. xx xx′′ xx′′
33 22 33 22
Figure 1. Hypergraph (H,T)
De(cid:12)ne a relation ∼ on a T.B.T T(f,f′,x ,x ,...,x ) by f ∼ f′ if and only if for all
1 2 n
(x ,x ,...,x ) ∈ B , we have f(x ,x ,...,x ) = f′(x ,x ,...,x )(f ≡ f′). It is clear that ∼
1 2 n n 1 2 n 1 2 n
is a congruence equivalence relation on T(f,f′,x ,x ,...,x ). For 0 ≤ j,j′ ≤ m, we say that
1 2 n
T(f(j),x ,x ,...,x ) and T′(f(j′),x ,x ,...,x ) are equivalent, if f(j) ∼ f(j′).
1 2 n 1 2 n
Theorem 3.10. Let 0 ≤ j,j′ ≤ m. If T(f(j),x ,...,x ) and T′(f(j′),x ,...,x ) are equiva-
1 n 1 n
lent, then their Boolean function{based hypergraph are isomorphic.
Proof. Let 0 ≤ j,j′ ≤ m. Since T(f(j),x ,...,x ) ∼ T′(f(j′),x ,...,x ), for all 1 ≤ i ≤ 2n
1 n 1 n
we get that Ker(f(j)) = Ker(f(j′)). Using Theorem 3.8, T(f(j),x ,x ,...,x ) corresponds to
i i 1 2 n
8 Alg. Struc. Appl. Vol. .. No. . (20..) ..-...
(H,T) if and only if T(f(j′),x ,x ,...,x ) corresponds to (H′,T′). Hence (H,T) ∼= (H′,T′).
1 2 n
De(cid:12)nition 3.11. Let n ∈ N,m ∈ N∗, 1 ≤ k ≤ n and T(f(0),...,f(m),x ,...,x ) be a T.B.T,
1 n
∏2n
where for 0 ≤ t ≤ m,f(t)(x ,...,x ) = f(t)(x ,x ,...,x ). Then
1 n i 1 2 n
i=1
(i) Z(n,f(t),0) = {j | f(t)(x ,x ,...,x ) = 0, where 1 ≤ j ≤ 2n};
j 1 2 n
∑n ∑k ∑n
(ii) S(k,x ,x ,...,x ,0) = { x | x + x = 0}.
1 2 k i i i
i=1 i=1 i=k+1
Example 3.12. Consider a T.B.T T(f,x ,x ,x ) in Table 3. Simple computations show
1 2 3
Z(n,f,0) = {j | f (x ,x ,x ) = 0, where 1 ≤ j ≤ 8} = 3,S(3,x ,x ,x ,0) = {x + x +
j 1 2 3 1 2 3 1 2
x },S(2,x ,x ,0) = {x +x +x ,x +x +x′}, S(1,x ,0) = {x +x +x ,x +x +x′,x +
3 1 2 1 2 3 1 2 3 1 1 2 3 1 2 3 1
x′ +x ,x +x′ +x′},|S(3,x ,x ,x ,0)| = 1,|S(2,x ,x ,0)| = 2 and |S(1,x ,0)| = 4.
2 3 1 2 3 1 2 3 1 2 1
Theorem 3.13. Let n ∈ N,1 ≤ j ≤ n and T(f,x ,x ,...,x ) be a T.B.T. Then |S(k =
1 2 n
j,x ,x ,...,x
1 2 k
,0)| = 2n−j.
∑j
Proof. Let x ,x ,...,x be j Boolean variables. Then x = 0 if and only if x = x =
1 2 j i 1 2
i=1
∑n ∑n ∑n
... = x = 0. So x = 0 or x = 1. Now, x = 1 if and only if for some
j i i i
i=j+1 i=j+1 i=j+1
∑n
j+1 ≤ t ≤ n we have x = 1, where have 2n−j −1 cases. In addition, x = 0 if and only
t i
i=j+1
if for all j +1 ≤ t ≤ n we have x = 0. So |S(j,x ,x ,...,x ,0)| = 2n−j.
t 1 2 j
Corollary 3.14. Let n ∈ N,m ∈ N∗ and T(f,x ,x ,...,x ) be a T.B.T. Then
1 2 n
(i) if k = 1, then |S(k,x ,x ,...,x ,0)| = 2n−1;
1 2 k
(ii) if k = n−1, then |S(k,x ,x ,...,x ,0)| = 2;
1 2 k
(ii) if k = n−i, then |S(k,x ,x ,...,x ,0)| = 2i;
1 2 k
(i) if k = n, then |S(k,x ,x ,...,x ,0)| = 1.
1 2 k
Let n ∈ N and T(f(0),...,f(m),x ,...,x ) be a T.B.T, 1 ≤ k ≤ n,0 ≤
1 n
t,t′ ≤ m. De(cid:12)ne (f(t)f(t′))(x ,...,x ) = {f(s)(x ,...,x ) | f(s)(x ,...,x ) =
1 n 1 n j 1 n
f(t)(x ,...,x )f(t′)(x ,...,x ) for all 1 ≤ j ≤ 2n} and (f(t) + f(t′))(x ,...,x ) =
j 1 n j 1 n 1 n
{f(s)(x ,...,x ) | f(s)(x ,...,x ) = f(t)(x ,...,x )+f(t′)(x ,...,x ) for all 1 ≤ j ≤ 2n}.
1 n j 1 n j 1 n j 1 n
So we have the following Theorem.
Alg. Struc. Appl. Vol. .. No. .. (20..) ..-... 9
Theorem 3.15. Let n ∈ N and T(f(0),...,f(m),x ,...,x ) be a T.B.T, 1 ≤ k ≤ n,0 ≤ t,t′ ≤
1 n
m. Then
(i) Z(n,x ,0) = 2n−1;
k
(ii) Z(n,0,0) = 2n and Z(n,1,0) = 0;
(iii) Z(n,x .f(t),0) = [Z(n,x ,0)+Z(n,f(t),0)]−Z(n,x +f(t),0).
k k k
Proof. (i) Let 1 ≤ k ≤ n. Then by Corollary 3.14, we obtain that Z(n,x ,0) =
k
|S(1,x ,x ,...,x ,0)| = 2n−1.
1 2 k
(ii) In a similar way it is obtained from Corollary 3.14.
(iii) Let 1 ≤ k ≤ n,1 ≤ j ≤ 2n. Because x ∼ 0 or x ∼ 1, for all 1 ≤ j ≤ 2n, we get that
k k
(x .f(t)) ∼ 0 or (x .f(t)) ∼ f(t). If (x .f(t)) ∼ 0, since for all 1 ≤ j ≤ 2n,0+f(t) = f(t) by
k j k j j k j j j
item (ii), we get that
Z(n,x .f(t),0) = 2n = 2n+Z(n,f(t),0)−Z(n,f(t),0)
k j j j
= [Z(n,x ,0)+Z(n,f(t),0)]−Z(n,x +f(t),0).
k j k j
If (x .f(t)) ∼ f(t), then Z(n,x .f(t),0) = 0+Z(n,f(t),0).
k j j k j j
Theorem 3.16. Let n ∈ N and T(f(0),...,f(m),x ,...,x ) be a T.B.T, 1 ≤ k ≤ n,1 ≤ i ≤
1 n
2n,0 ≤ t,t′ ≤ m. Then
(i) Z(n,f(t).f(t′),0) = [Z(n,f(t),0)+Z(n,f(t′),0)]−Z(n,f(t)+f(t′),0);
(ii) Z(n,f(t)+f(t′),0) ≤ min{Z(n,f(t),0),Z(n,f(t′),0)}.
∑n ∑n
Proof. (i) Because for all 1 ≤ j ≤ 2n and 1 ≤ t ≤ m we have f(t) = x and x ∼
j i i
i=1 i=1
∑n ∑n
0, x ∼ 1 or there exists 1 ≤ j ≤ 2n such that x ∼ x , so by Theorem 3.15(iii), the
i i j
i=1 i=1
proof is obtained.
(ii) By de(cid:12)nition, for all 1 ≤ j,j ≤ 2n,f(t)(x ,x ,...,x )+f(t′)(x ,x ,...,x ) = 0 if and
j 1 2 n j 1 2 n
(t) (t′)
only if f (x ,x ,...,x ) = f (x ,x ,...,x ) = 0, so
j 1 2 n j 1 2 n
Z(n,f(t)+f(t′),0) = {j | f(t)(x ,x ,...,x ) = f(t′)(x ,x ,...,x ) = 0, where 1 ≤ j ≤ 2n}
j 1 2 n j 1 2 n
≤ {j | f(t)(x ,x ,...,x ) = 0, where 1 ≤ j ≤ 2n}
j 1 2 n
and
Z(n,f(t)+f(t′),0) = {j | f(t)(x ,x ,...,x ) = f(t′)(x ,x ,...,x ) = 0, where 1 ≤ j ≤ 2n}
j 1 2 n j 1 2 n
≤ {j | f(t′)(x ,x ,...,x ) = 0, where 1 ≤ j ≤ 2n}.
j 1 2 n
Hence Z(n,f(t)+f(t′),0) ≤ min{Z(n,f(t),0),Z(n,f(t′),0)}.
10 Alg. Struc. Appl. Vol. .. No. . (20..) ..-...
Corollary 3.17. Let n ∈ N and T(f,f′,x ,x ,...,x ) be a T.B.T. Then (f.f′) ∼ f if and
1 2 n
′
only if Z(n,f.f ,0) = Z(n,f,0).
Theorem 3.18. Let n ∈ N,1 ≤ j ≤ n and T(f,f′,x ,x ,...,x ) be a T.B.T. If there exists
1 2 n
∑k
some 1 ≤ i ,i ,...,i ≤ n such that f′ ∼ x , then Z(n,f.f′,0) > 2n−k.
1 2 k ij
j=1
∑k
Proof. Let 1 ≤ i ,i ,...,i ≤ n and f′ ∼ x . By Theorem 3.16,
1 2 k ij
j=1
Z(n,f(t).f(t′),0) = [Z(n,f(t),0)+Z(n,f(t′),0)]−Z(n,f(t)+f(t′),0)
∑k ∑k
= [Z(n,f(t),0)+Z(n, x ,0)]−Z(n,(f(t))+ x ,1)
ij ij
j=1 j=1
> Z(n,f(t),0)+2n−k −Z(n,f(t),0) = 2n−k.
∑k
Theorem 3.19. Let n ∈ N,1 ≤ j ≤ n and T(f,f′,x ,x ,...,x ) be a T.B.T and f′ ∼ x ,
1 2 n ij
j=1
where 1 ≤ i ,i ,...,i ≤ n.
1 2 k
(i) If Z(n,f,0) < 2n−1, then (f.f′) ̸∼ f;
(ii) If Z(n,f′,0) = Z(n,f,0) and Ker(f′) ⊆ Ker(f) imply that f′ ∼ f;
Proof. (i) Let (f.f′) ∼ f. Using Theorem 3.16, Z(n,f.f′,0) = Z(n,f,0) + Z(n,f′,0) −
Z(n,f+f′,0). BecauseZ(n,f+f′,0) ≤ min{Z(n,f,0),Z(n,f′,0)},wegetthatZ(n,f.f′,0) ≥
Z(n,f,0)+2n−k −Z(n,f,0) = 2n−k, which is a contradiction.
(ii) Let f′(x ,...,x ) = 0. Then Ker(f′) ⊆ Ker(f) implies that (x ,...,x ) ∈ Ker(f)
1 n 1 n
and so f(x ,...,x ) = 0. Suppose that f′(x ,...,x ) = 1, then Ker(f′) ⊆ Ker(f) implies
1 n 1 n
that (x ,...,x ) ∈ Ker(f)\Ker(f′) or (x ,...,x ) ̸∈ Ker(f). If (x ,...,x ) ̸∈ Ker(f), then
1 n 1 n 1 n
f(x ,...,x ) = 1 and in this case f ∼ f′. If (x ,...,x ) ∈ Ker(f), then f(x ,...,x ) = 0
1 n 1 n 1 n
′
and it follows that Z(n,f ,0) < Z(n,f,0), which is a contradiction.
Theorem 3.20. Every T.B.T corresponds to a Boolean expression.
Proof. Let n ∈ NandT(f,x ,x ,...,x )beaT.B.T.Ifforall1 ≤ j ≤ 2n,f (x ,x ,...,x ) =
1 2 n j 1 2 n
0, then consider g(x ,x ,...,x ) ≡ 0. In a similar a way that, if for all 1 ≤ j ≤
1 2 n
2n,f (x ,x ,...,x ) = 1, then consider g(x ,x ,...,x ) ≡ 1. Now, if there exist k ∈ N
j 1 2 n 1 2 n
and 1 ≤ j ,j ,...,j ≤ 2n such that f (x ,x ,...,x ) = 0, then consider f (x ,x ,...,x ) =
∑n 1 2 k jk 1 2 n j 1 2 n
x in such a way that for all 1 ≤ i ≤ n,x = 0. Applying Theorem 3.8, H = (G =
i i
i=1