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<article xmlns:xlink="http://www.w3.org/1999/xlink">
  <front>
    <journal-meta />
    <article-meta>
      <title-group>
        <article-title>The Minimization Method of Boolean Functions in Polynomial Set-theoretical Format</article-title>
      </title-group>
      <contrib-group>
        <contrib contrib-type="author">
          <string-name>Bohdan Rytsar</string-name>
          <email>bohdanrytsar@gmail.com</email>
          <xref ref-type="aff" rid="aff0">0</xref>
        </contrib>
        <aff id="aff0">
          <label>0</label>
          <institution>National University “L'viv Polytechnic”</institution>
          ,
          <addr-line>S. Bandera Str. , 12, L'viv, 79646, Ukraine</addr-line>
          ,
          <institution>Rzeszow University, Chair of Computer Science</institution>
          ,
          <addr-line>Prof. S. Pigonia Str. 1, 35-310 Rzeszow</addr-line>
          ,
          <country country="PL">Poland</country>
        </aff>
      </contrib-group>
      <fpage>130</fpage>
      <lpage>146</lpage>
      <abstract>
        <p>A generalized of conjuncterms simplification rules in polynomial settheoretical format has been considered. These rules are based on the proposed theorems for different initial conditions transform of pair conjuncterms, Hamming distance between them can be arbitrary. These rules may be useful to minimize in polynomial set-theoretical format of arbitrary logic functions with n variables. Advantages of the proposed rules of simplification are illustrated by several examples.</p>
      </abstract>
      <kwd-group>
        <kwd>Boolean function</kwd>
        <kwd>polynomial set-theoretical format</kwd>
        <kwd>simplification of conjuncterms</kwd>
        <kwd>Hamming distance</kwd>
      </kwd-group>
    </article-meta>
  </front>
  <body>
    <sec id="sec-1">
      <title>-</title>
      <p>Investigations [1–8] have shown that it is economically profitable to build digital
devices such as arithmetic units, coding-error detectors as well as devices with
programmed logic, etc. on logical elements AND-EXOR, which realize polynomial basis
{&amp;, ⊕, 1}, that is AND, EXCLUSIVE OR (EXOR) logical operations and constant 1.
It is easier to test and diagnose [9–11] digital devices on AND-EXOR if compared to
the devices built on AND-OR. However, in spite of the mentioned advantages it is
more difficult to minimize a function in polynomial format, i. e. in ESOP (EXOR
SumOf-Product), than in disjunctive format, i. e. in SOP (Sum-Of-Product). If the merge
operation of adjacent conjuncterms (conjunction of literals) is only applied in the SOP
minimization, than, in addition, the same operations can be applied in the ESOP
minimization [1, 2].</p>
      <p>A precise solutions of a minimization problem in ESOP generally are based on
analytical [2] or on visual transformations [1–3]. Respectively, such methods are suitable
only for functions from small amount of variables [5, 7, 10–13] and only for special
classes for functions with up to 10 variables [14]. Heuristic methods have
comparatively wider practical application [1, 8, 16–23]. Among them there are minimization
method based on a coefficient of generalized canonical Reed-Muller forms using of
matrix transformations [1, 8, 11, 16] and the method based on iterative execution of
operations with conjuncterms of different ranks of the given function. To the last belongs
the algorithm [17], which after transformation of the given function in Positive Polarity
Reed-Muller expression minimizes it on the basis of three operations with
conjuncterms. Better results have been shown by algorithm based on the procedure of so-called
linked product terms [18, 19]. Later, on the basis of this procedure, the algorithms have
been developed and completed with more perfect operations (that is primary xlink,
secondary xlink, unlink, exorlink), which can be used for minimization of a system of
complete and incomplete functions [20, 21].</p>
      <p>However, the mentioned above algorithms have one drawback in common. They
involve the procedure of linking in pairs only conjuncterms of the same rank r ∈
{1, 2, . . . , n}, which differs between each other by binary positions. Correspondingly,
this limits the use of such algorithms to functions given in SOP or ESOP, which can
have triple conjuncterms in the different part. In these cases to conjuncterms that
differ in ranks certain procedures of transformation are applied leading to an increase of
procedural steps and processing time. Besides, the above mentioned operations of
conjuncterms linking and other rules of simplification [24–26] do not have generalized
character as to Hamming distance between any two conjuncterms of different ranks of
a given function that does not guarantee the final minimized result.</p>
      <p>In this paper we consider a new method of minimization of Boolean functions with
n variables in polynomial set-theoretical format (PSTF), based on a procedure of
splitting of conjuncterms [27–29] and on usage of generalized set-theoretical rules of
conjuncterms simplification [30]. The suggested rules guarantee better (as to costs of
realization and number of procedure steps) results of minimization of logic functions
proved by the numerous examples that are borrowed from publications of other authors
for comparison purposes.
2</p>
    </sec>
    <sec id="sec-2">
      <title>Problem Formulation</title>
      <p>Boolean function f (x1, x2, . . . , xn) that undergoes minimization, will be given
by a set of binary minterms or in perfect set-theoretical form (STF) as a
Y 1 = {m1, m2, . . . , mk}1, or in perfect polynomial set-theoretical form (PSTF) as
a Y ⊕ = {m1, m2, . . . , mk}⊕ [30, 31].</p>
      <p>The generalized set-theoretical rules of simplification [30] of a conjuncterm set
of any function f , given in PSTF Y ⊕, are based on iterative process of
simplification of two conjuncterms θ1r1 = (σ1σ2 · · · σn) and θ2r2 = (σ1σ2 · · · σn),
σi ∈{0, 1, –}, r1, r2 ∈{1, 2, . . ., n}, which differ in Hamming difference d = 1, 2, . . ., n
it is number of different in value α, β, γ, δ, . . . ∈ {0, 1, −} of onename positions. Here
the different part α, β, γ, δ, . . . of these conjuncterms may have a different total number
of literals kl. For example, two pairs of conjuncterms 1011––0110 and 1011––1–01 have
d = 3, but the first has kl = 6, and the second kl = 5. Therefore, different initial
conditions of transformation of two conjuncterms are possible.</p>
      <p>We will consider the following conditions:
• when kl = 2d, here two conjuncterms are of the same r-rank θ1r and θ2r but differ in
d onename binary positions α, β, γ, δ, . . . ∈ {0, 1};
• when kl = 2d−1, here one conjuncterm of (r−1)-rank θ1r−1 and the second of r-rank
θ2r differ in d onename positions α, β, γ, δ, . . . ∈ {0, 1, –}, where dash (–) belongs to
θr−1;
1
• when kl = 2(d − 1), here two conjuncterms are of the same (r − 1)-rank θr−1 and
1
θr−1 differ in d onename positions α, β, γ, δ, . . . ∈ {0, 1, –} and each of them has
2
one dash (–).</p>
      <p>As a result of transformation of two conjuncterms in PSTF a transformed PSTF Y ⊕
will be formed, where power kY will depend on distance d. Efficiency of simplification
of two different conjuncterms for the mentioned above initial conditions will be
estimated on the basis of comparison of interrelation k∗/kl∗, obtained on the ground of data
θ
of transformed PSTF Y ⊕, where kθ∗ is number of transformed conjuncterms and kl∗ is
number of their literals, with initial interrelation kθ/kl, where in this case kθ = 2.
3</p>
      <p>Three Theorems about Conjuncterms Transformation
Theorem 1. Two conjuncterms of r-rank θ1r and θ2r, r ∈ {1, 2, . . ., n}, of the function
f (x1, x2, . . ., xn), that differ in values d of onename binary positions α, β, γ, δ, . . . ∈
{0, 1}, in polynomial set-theoretical format form a set of transformed PSTF Y ⊕ of
ipnowdieffrekreYnt=padr!t, tehaecthotoafltnhuemmbceornosfisltisteorfaklsθ∗k=l∗ =d cdo(ndj−un1ct)e.rms of (r −1)-rank and has
Proof. To determine kθ∗/kl∗ and kY let us consider the transformation of conjuncterms
θ1r and θ2r for d = 0, 1, 2, 3, 4. Here it should be mentioned that initial interrelation
kθ/kl = 2/2d.
• If d = 0, than θ1r = θ2r. So, transformed PSTF Y ⊕ = {θ1r, θ2r}⊕ = ∅, that
corresponds to analytical expression a ⊕ a = 0. In this case kθ∗/kl∗ = 0/0; kY = 1.
• Let d = 1. Then θ1r = (σ1 · · · α¯i · · · σn) and θ2r = (σ1 · · · αi · · · σn), αi ∈ {0, 1}.
Respectively, for analytical expression a¯ ⊕ a = 1 we can write:</p>
      <p>
        Y ⊕ = {(σ1 ··· α¯i ··· σn), (σ1 ··· αi ··· σn)}⊕ = (σ1 ··· −i ··· σn),
(
        <xref ref-type="bibr" rid="ref1">1</xref>
        )
where the transformed PSTF Y ⊕ = {(σ1 ··· −i ··· σn)}⊕ = θr−1 is a triple conjuncterm
of (r−1)-rank.
      </p>
      <p>
        For (
        <xref ref-type="bibr" rid="ref1">1</xref>
        ) interrelation kθ∗/kl∗ = 1/0, and as initial interrelation kθ/kl = 2/2, then
it indicates on a result of transformation (
        <xref ref-type="bibr" rid="ref1">1</xref>
        ) the simplification took place due to the
replacement of two conjuncterms of r-rank by one conjuncterm of (r−1)-rank; kY = 1.
      </p>
      <p>
        To simplify the writing of the conjuncterms of the given and transformed PSTF
Y ⊕ will be considered only for their different positions which will be written down in
a column. In (
        <xref ref-type="bibr" rid="ref1">1</xref>
        ) such position is αi, so, simplified writing down (
        <xref ref-type="bibr" rid="ref1">1</xref>
        ) with taking into
account αi ≡ α ∈ {0, 1}, will look like:
      </p>
      <p>
        αα¯ ⇒⊕ (–), (
        <xref ref-type="bibr" rid="ref2">2</xref>
        )
where ⇒⊕ – operator of transftrmation of the conjuncterms θ1r and θ2r in
polynomial format of the function f . In examples of transformation, the same in
meaning onename positions of conjuncterms will be rewritten without any change. For
example, 11––0111 ⇒⊕ (1– –1), that in decimal format corresponds to 191,,1135 ⇒⊕
⊕
⇒ (
        <xref ref-type="bibr" rid="ref11 ref13 ref15 ref9">9, 11, 13, 15</xref>
        ) and in analytical form is x1x¯3x4 ⊕ x1x3x4 = x1x4.
• Let d = 2. Then θ1r = (σ1 ···α¯i ···β¯j ···σn), θ2r = (σ1 ···αi ···βj ···σn), αi, βj ∈ {0, 1},
(a¯ ⊕ b
and respectively to analytical expressions a¯¯b ⊕ ab = a ⊕ ¯b (if αi = βi) and a¯b ⊕ a¯b=
(if αi 6= βi), in simplified way (for αi ≡ α, βi ≡ β, α, β ∈ {0, 1}) we will
(a¯ ⊕ ¯b
      </p>
      <p>
        a ⊕ b
obtain
(
        <xref ref-type="bibr" rid="ref3">3</xref>
        )
α¯β¯ ⊕
αβ ⇒
α¯– α–
–β , –β¯
.
      </p>
      <p>
        For (
        <xref ref-type="bibr" rid="ref3">3</xref>
        ) we have kθ∗/kl∗ = 2/2, that is indicative of simplification of the given
conjuncterms due to reduction of their rank from r to (r − 1), as initial interrelation
kθ/kl = 2/4; kY = 2.
• Let d = 3. Then θ1r = (σ1 ···α¯i ···β¯j ···γ¯k ···σn) and θ2r = (σ1 ···αi ···βj ···γk ···σn),
αi, βj, γk ∈ {0, 1}. For αi ≡ α, βi ≡ β, γi ≡ γ, α, β, γ ∈ {0, 1} we have
α¯β¯–α¯β¯–α¯–γ¯α¯–γ¯–β¯γ¯–β¯γ¯
αα¯ββ¯γγ¯ ⇒⊕α¯–γ,–β¯γ,–βγ¯,α¯β–,αβ¯–,α–γ¯ . (
        <xref ref-type="bibr" rid="ref4">4</xref>
        )
 –βγ α–γ αβ– –βγ α–γ αβ– 
      </p>
      <p>
        For (
        <xref ref-type="bibr" rid="ref4">4</xref>
        ) kθ∗/kl∗ = 3/6 indicates on an increase of power of each transformed PSTF
Y ⊕ and unchangeability of number of literals of their conjuncterms as initial
interrelation kθ/kl = 2/6; kY = 6.
• Let d = 4. Then θ1r = (σ1···α¯i···β¯j···γ¯k···δ¯l···σn) and θ2r = (σ1···αi···βj···γk···δl···σn),
αi, βj, γk, δl ∈ {0, 1}. For αi ≡ α, βi ≡ β, γi ≡ γ, δi ≡ δ, α, β, γ, δ ∈ {0, 1} we have:
α¯β¯γ¯–α¯β¯γ¯–α¯β¯γ¯–α¯β¯γ¯–α¯β¯γ¯–α¯β¯γ¯–α¯β¯–δ¯
α¯β¯γ¯δ¯ ⇒⊕α¯β¯–δ,α¯β¯–δ,α¯–γ¯δ,α¯–γ¯δ,–β¯γ¯δ ,–β¯γ¯δ ,
αβγδ α–¯β–γγδδα–β–¯γγδδα–ββγ¯–δδα–¯ββγ–δδααβ–¯γ–δδααβ–γ¯–δδαα–¯ββ–γγγδδ¯¯–,
α¯β¯–δ¯α¯β¯–δ¯α¯β¯–δ¯α¯β¯–δ¯α¯β¯–δ¯α¯–γ¯δ¯α¯–γ¯δ¯α¯–γ¯δ¯α¯–γ¯δ¯
α¯–γδ¯,–β¯γδ¯,–β¯γδ¯ α¯β¯γ–,α¯β¯γ–,–βγ¯δ¯,–βγ¯δ¯ α¯βγ¯–,α¯βγ¯–
α¯βγ–αβ¯γ–α–γδ¯,α¯–γδ–β¯γδ αβγ¯–αβ–δ¯,α¯β–δ–βγ¯δ ,
         
–βγδ α–γδ αβγ– –βγδ α–γδ αβ–δ αβγ– –βγδ αβ–δ
α¯–γ¯δ¯α¯–γ¯δ¯–β¯γ¯δ¯–β¯γ¯δ¯–β¯γ¯δ¯–β¯γ¯δ¯–β¯γ¯δ¯–β¯γ¯δ¯
α¯β–δ¯ α¯β–δ¯,αβ¯γ¯–,αβ¯γ¯–,αβ¯–δ¯ αβ¯–δ¯,α–γ¯δ¯,α–γ¯δ¯ 
–βγδ¯,α¯βγ–αβ¯–δα–γ¯δα–γδ¯,αβ¯γ–αβγ¯–αβ–δ¯. (
        <xref ref-type="bibr" rid="ref5">5</xref>
        )
αβγ––βγδ α–γδαβ–δαβγ–α–γδαβ–δαβγ–
So, for (
        <xref ref-type="bibr" rid="ref5">5</xref>
        ) kθ∗/kl∗ = 4/12 indicates on an increase of power of transformed PSTF Y ⊕
and the number of literals, as the initial interrelation kθ/kl = 2/8; kY = 24.
      </p>
      <p>In the case of necessity for any pair of conjuncterms of r-rank of a function f , that
have distance d &gt; 4, one can in similar way form a set of d! of transformed PSTF Y ⊕;
d = 1, 2, ..., n.</p>
      <p>Based on the considered above, one can state that two conjuncterms of r-rank θr
1
and θ2r function f , that differ d = 1, 2,..., n in different by values onename binary
positions α, β, γ, δ, ... ∈ {0, 1}, form in polynomial format a set with kY = d! of
transformed PSTF Y ⊕, each of them consisting of different conjuncterms of (r−1)-rank with
kθ∗/kl∗ =d/d(d−1), that is the proof of Theorem 1. ⊔⊓
Theorem 2. Two conjuncterms of the function f(x1, x2, ..., xn), one of which of
(r−1)rank θr−1 differs from another r-rank θ2r in the number of d different in values
one1
name positions α, β, γ, δ, ... ∈ {0, 1, –}, among which the dash (–) belongs to θr−1,
1
r ∈ {1, 2, ..., n}, in polynomial set-theoretical format create kY = (d − 1)! of sets of
transformed PSTF Y ⊕, each of them has power kθ∗ = d and the total number of literals
in different part kl∗ = d(d − 1) − (d − 2), here:
• if d = 1, then α–˜ ⇒⊕ (α¯˜);</p>
      <p>
        α¯– ⊕
• if d = 2, then αβ˜ ⇒
––
αβ˜¯ ,
α¯–––β¯–
• if d = 3, then αα¯ββ¯γ–˜ ⇒⊕α–ββ–γ˜¯,ααβ–γ–˜¯,
–β¯ ⊕ ––
α˜β ⇒ α˜¯β ;
α¯––––γ¯ –β¯–––γ¯
αα¯β–˜γ¯γ ⇒⊕––¯γ ,α–¯–, α–˜ββ¯γ¯γ ⇒⊕α–˜¯β–γγ,α¯˜–ββ–γ, (
        <xref ref-type="bibr" rid="ref9">9</xref>
        ),(
        <xref ref-type="bibr" rid="ref10">10</xref>
        )
 αβ˜γ αβ˜γ 
(
        <xref ref-type="bibr" rid="ref6">6</xref>
        )
(
        <xref ref-type="bibr" rid="ref7">7</xref>
        )
(
        <xref ref-type="bibr" rid="ref8">8</xref>
        )
(
        <xref ref-type="bibr" rid="ref11">11</xref>
        )
(
        <xref ref-type="bibr" rid="ref12">12</xref>
        )
(
        <xref ref-type="bibr" rid="ref13">13</xref>
        )
(
        <xref ref-type="bibr" rid="ref14">14</xref>
        )
• if d = 4, then
      </p>
      <p>α¯β¯––α¯β¯––α¯–γ¯–α¯–γ¯––β¯γ¯––β¯γ¯–
α¯β¯γ¯– ⇒⊕α¯–γ–,–β¯γ–,α¯β––,–βγ¯– αβ¯––,α–γ¯– 
αβγδ˜ –βγ–α–γ––βγ–αβ––,α–γ–αβ–– ,
αβγδ˜¯αβγδ˜¯αβγδ˜¯αβγδ˜¯αβγδ˜¯αβγδ˜¯</p>
      <p>α¯β¯––α¯β¯––α¯––δ¯α¯––δ¯ –β¯–δ¯ –β¯–δ¯
α¯β¯–δ¯ ⇒⊕α¯––δ, –β¯–δ ,α¯β––, –β–δ¯,αβ¯––,α––δ¯ 
αβγ˜δ α–ββ–γ˜¯δδααβ–γ–˜¯δδα–ββ–γ˜¯δδααββ–γ˜¯δ–ααβ–γ–˜¯δδααββ–γ˜¯δ–,
α¯–γ¯–α¯–γ¯–α¯––δ¯α¯––δ¯––γ¯δ¯––γ¯δ¯
α¯–γ¯δ¯ ⇒⊕α¯––δ,––γ¯δ,––γδ¯,α¯–γ–,α–γ¯–,α––δ¯ 
αβ˜γδ α–β–˜¯γγδδααβ–˜¯γ–δδααβ–˜¯γγ–δα–β–˜¯γγδδααβ–˜¯γ–δδααβ–˜¯γγ–δ,
–β¯γ¯––β¯γ¯––β¯–δ¯–β¯–δ¯––γ¯δ¯––γ¯δ¯
–β¯γ¯δ¯ ⇒⊕–β¯–δ¯ ––γ¯δ,–β¯γ–,––γδ¯,–βγ¯–,–β–δ¯ 
α˜βγδ ––γ¯δ¯,–β–δ ––γδ–βγ––β–δ –βγ– ,
α˜¯βγδα˜¯βγδα˜¯βγδα˜¯βγδα˜¯βγδα˜¯βγδ
where α˜, β˜, γ˜, δ˜ are binary positions of any value 0 or 1.</p>
      <p>Proof. In this case the given PSTF Y ⊕ has interrelation kθ/kl = 2/(2d − 1).
• Let d = 1. Then θ1r−1 = (σ1 ··· −i ··· σn), θ2r = (σ1 ··· α˜i ··· σn), α˜i ∈ {0, 1}, and
respectively for the expression 1 ⊕ a˜ = a˜¯, a˜ ∈ {a, a¯}, we can write down such PSTF
catioAnsoinfttehreregliavteionnPkSθ∗T/Fkl∗Y =⊕ d1u/e1,t othreenmcoovmalpoafreodnwecitohnkjuθn/cktler=m;2k/Y1 w=e1h.ave
simplifi• Let d = 2. Then for θ1r−1 = (σ1 ··· α¯i ··· –j ··· σn) and θ2r = (σ1 ···αi ··· β˜j ··· σn) and
θr−1 = (σ1 ··· –i···β¯j ···σn) and θ2r = (σ1···α˜i···βj ···σn), αi, βj∈{0, 1}, we will obtain
1
¯
Y ⊕ = {(σ1···α¯i···–j···σn),(σ1···αi···β˜j···σn)}⊕ = {(σ1···–i···–j···σn),(σ1···αi···β˜j···σn)}⊕
and Y ⊕ = {(σ1···–i···β¯j···σn),(σ1···α˜i···βj···σn)}⊕ = {(σ1···–i···–j···σn),(σ1···α˜¯i···βj···σn)}⊕.</p>
      <p>Comparing the obtained interrelation kθ∗/kl∗ = 2/2 with kθ/kl = 2/3, we see that
transformed PSTF Y ⊕ is simpler than the given PSTF Y ⊕ for one literal. It should be
noted, that a number of transformed PSTF Y ⊕ is determined by a number of binary
positions in different part θ1r−1 and θ2r, that conforms to Theorem 1. Therefore, for
d = 2 we have kY = 1.
• Let d = 3. Then for θ1r−1= (σ1···α¯i···β¯j ···–k ···σn) and θ2r= (σ1···αi···βj ···γ˜k ···σn),
θr−1= (σ1···α¯i···–j···γ¯k···σn) and θ2r= (σ1···αi···β˜j···γk···σn), θ1r−1= (σ1···–i···β¯j···γ¯k···σn)
1
and θ2r= (σ1 ···α˜i ···βj ···γk ···σn), we have:
= ((σσ11·····α·¯–Yii·⊕····=·–β¯{j(j·σ····1·–·–k·k··α¯····i·σ·σ·nn·)β¯,),(j(σ·σ·11··–···k··–α·i·i····σ···nβ–)jj,(··σ····1––k·k·····α···σiσn·n)·,)·(,(βσσj11········γ˜·ααkii········σβ·βnjj·)}····⊕·γ˜¯γ=˜¯kk······σσnn)) ⊕,</p>
      <p>Y ⊕={(σ1 ···α¯i ···–j ···γ¯k ···σn),(σ1 ···αi ···β˜j ···γk ···σ¯n)}⊕= ⊕
=((σ1 ···α¯i ···–j ···–k ···σn),(σ1 ···–i ···–j ···γk ···σn),(σ1 ···αi ···β¯˜j ···γk ···σn)) ,
(σ1 ···–i ···–j ···γ¯k ···σn),(σ1 ···αi ···–j ···–k ···σn),(σ1 ···αi ···β˜j ···γk ···σn)
= ((σσ11······––iYi···⊕···β=¯–j{j(···σ···1–γ¯·kk·····–···iσσ·n·n·)β),¯,((jσσ·1·1···γ¯···k·––·i·i····σ···n–βj),j·(··σ···γ1kk········α˜·σσin·n·),)·,(β(σσj1·1·······γ·α˜¯kα˜¯i·i·······σ·ββnjj)·}··⊕···γ=γkk······σσnn)) ⊕.</p>
      <p>
        The obtained kθ∗/kl∗ = 3/5 indicates on an increase by one conjuncterm, since
kθ/kl = 2/5; kY = 2.
• Let d = 4. Based on the considered above and taking into account the rule (
        <xref ref-type="bibr" rid="ref4">4</xref>
        ) of
Theorem 1 for d = 3 (three positions are common), one can state that for
θr−1 = (σ1 ··· α¯i ··· β¯j ···γ¯k ···–l ···σn) and θ2r = (σ1 ···αi ···βj ···γk ···δ˜l ···σn),
1
θr−1 = (σ1 ···α¯i ···β¯j ···–k ···δ¯l ···σn) and θ2r = (σ1 ···αi ···βj ···γ˜k ···δl ···σn),
1
θr−1 = (σ1 ···α¯i ···–j ···γ¯k ···δ¯l ···σn) and θ2r = (σ1 ···αi ···β˜j ···γk ···δl ···σn),
1
θr−1 = (σ1 ···–i ···β¯j ···γ¯k ···δ¯···σn) and θ2r = (σ1 ···α˜i ···βj ···γk ···δk ···σn),
1
the interrelation kθ∗/kl∗ = 4/10 which, compared with kθ/kl = 2/7, indicates on an
increase of a number of conjuncterms as well as their literals; kY = 6.
      </p>
      <p>Thus, if a conjuncterm of (r − 1)-rank θr−1 differs from a conjuncterm of r-rank
1
θ2r in the number d of different by value onename positions α, β, γ, δ, . . . ∈ {0, 1, –},
here dash (–) belongs to θ1r−1, r ∈ {1, 2, . . ., n}, then (d − 1)! of sets
transformed PSTF Y ⊕ will be formed, each of which having the interrelation
kθ∗/kl∗ = d/(d(d − 1) − (d − 2)), that is the proof of Theorem 2. ⊔⊓
Theorem 3. Two conjuncterms of (r − 1)-rank θr−1 and θr−1, r ∈ {1, 2, . . ., n},
1 2
of the function f (x1, x2, . . ., xn) differ in d onename binary positions
• if d = 2, then
• if d = 3, then
α, β, γ, δ, ... ∈ {0, 1, –}, where each conjuncterm has one (–), in polynomial
settheoretical format starting with d = 2, create kY = (d − 2)! of sets transformed PSTF
Y ⊕, each of which has power kθ∗ = d and the total number of literals in the different
part kl∗ = d(d − 1) − 2(d − 2) and :
– – –
α¯β˜– ⊕ ¯
α–γ˜ ⇒α¯β˜–,
α–γ˜¯</p>
      <p>
        (
        <xref ref-type="bibr" rid="ref15">15</xref>
        )
– – –
α˜–γ¯ ⇒⊕ α˜¯–γ¯ ,
–β˜γ  ¯ 
–β˜γ
(
        <xref ref-type="bibr" rid="ref16">16</xref>
        ),(
        <xref ref-type="bibr" rid="ref17">17</xref>
        ),(
        <xref ref-type="bibr" rid="ref18">18</xref>
        )
α¯– – –α– – –
αα¯β–˜γγ¯δ˜– ⇒⊕α–¯β–¯˜γγ¯––,α–¯β–¯˜γ¯γ¯––,
α–γδ˜¯α–γδ˜¯
(
        <xref ref-type="bibr" rid="ref19">19</xref>
        ),(
        <xref ref-type="bibr" rid="ref20">20</xref>
        )
α˜– ⊕ α˜¯–
–β˜ ⇒ –β˜¯ ,
α˜β¯– ⇒⊕α–˜¯–β¯––,
–βγ˜ –βγ˜¯
      </p>
      <p>–β¯– ––β– –
α˜β¯γ¯– ⇒⊕α¯˜β¯γ¯–,α¯˜β¯γ¯– ,</p>
      <p>– –γ– – –γ¯– 
–βγδ˜</p>
      <p>
        –βγδ˜¯–βγδ˜¯
α¯– – –α– – –
α¯β¯γ˜– ⇒⊕–β– – –β¯– – 
αβ–δ˜ α¯β¯γ˜¯–,α¯β¯γ˜¯– ,
αβ–δ˜¯αβ–δ˜¯
α¯– – –α– – –
αα¯β–˜γ˜–δδ¯ ⇒⊕α–α¯–β–¯˜γ˜¯––δδδ¯,α–α¯–β–¯˜γ˜¯––δδδ¯¯,
–β¯– ––β– –
α˜β¯–δ¯ ⇒⊕– – –δ – – –δ¯ 
–βγ˜¯δ α–˜¯ββ¯γ˜¯–δδ¯,α–˜¯ββ¯γ˜¯–δδ¯, (
        <xref ref-type="bibr" rid="ref21">21</xref>
        ),(
        <xref ref-type="bibr" rid="ref22">22</xref>
        )
      </p>
      <p>
        – –γ¯–– –γ–
α˜–γ¯δ¯ ⇒⊕–α¯˜––γ¯–δδ¯,α¯˜–γ¯δ¯ , (
        <xref ref-type="bibr" rid="ref23">23</xref>
        ),(
        <xref ref-type="bibr" rid="ref24">24</xref>
        )
      </p>
      <p>
– – –δ¯ </p>
      <p>
–β˜γδ –β˜¯γδ–β˜¯γδ
where α˜, β˜, γ˜, δ˜ are binary positions of any value 0 or 1.</p>
      <p>
        Proof. Given PSTF Y ⊕ has the initial interrelation kθ/kl = 2/2(d − 1).
• Let d = 2. Then θ1r = (σ1· · ·α˜i · · · –j · · · σn) and θ2r = (σ1 · · · –i · · · β˜j · · · σn),
¯
α˜i, β˜j ∈ {0, 1}. For f (a, b) respectively to (
        <xref ref-type="bibr" rid="ref15">15</xref>
        ) we have a˜ ⊕ ˜b = (a˜¯ ⊕ 1) ⊕ (˜b ⊕ 1) =
¯
= a˜¯ ⊕ ˜b, that corresponds to PSTF
      </p>
      <p>Y ⊕ = {(σ1 ··· α˜i ··· β¯j ··· –k ··· σn), (σ1 ··· –i ··· βj ··· γ˜k ··· σn)}⊕ =
= {(σ1 ··· –i ··· –j ··· –k ··· σn), (σ1 ··· α˜¯i ··· β¯j ··· –k ··· σn), (σ1 ··· –i ··· βj ··· γ˜¯k ··· σn)}⊕,
Y ⊕ = {(σ1 ··· α¯i ··· β˜j ··· –k ··· σn), (σ1 ··· αi ··· –j ··· γ˜k ··· σn)}⊕ =
= {(σ1 ··· –i ··· –j ··· –k ··· σn), (σ1 ··· α˜¯i ··· –j ··· γ¯k ··· σn), (σ1 ··· –i ··· β˜¯j ··· γk ··· σn)}⊕.</p>
      <p>Compared with kθ/kl = 2/4 here kθ∗/kl∗ = 3/4 means that the transformed PSTF
Y ⊕ has one more conjuncterm, here its rank is (r − 3); kY = 1.
• Let d = 4. Then θ1r = (σ1 · · · α˜i · · · β¯j · · · γ¯k · · · –l · · · σn) and
θ2r= (σ1 ··· –i ··· βj ··· γk ··· δ˜l ··· σn),
θ1r= (σ1 ··· α¯i ··· β˜j ··· γ¯k ··· –l ··· σn) and θ2r= (σ1 ··· αi ··· –j ··· γk ··· δ˜l ··· σn),
θ1r= (σ1 ··· α¯i ··· β¯j ··· γ˜k ··· –l ··· σn) and θ2r= (σ1 ··· αi ··· βj ··· –k ··· δ˜l ··· σn),
θ1r=(σ1 ··· α˜i ··· β¯j ··· –k ··· δ¯l ··· σn) and θ2r= (σ1 ··· –i ··· βj ··· γ˜k ··· δl ··· σn),
θ1r= (σ1 ··· α¯i ··· β˜j ··· –k ··· δ¯l ··· σn) and θ2r= (σ1 ··· αi ··· –j ··· γ˜k ··· δl ··· σn),
θ1r= (σ1 ··· α˜i ··· –j ··· γ¯k ··· δ¯l ··· σn) and θ2r= (σ1 ··· –i ··· β˜j ··· γk ··· δl ··· σn).
Here, for d = 4 the interrelation kθ∗/kl∗ = 4/8 is greater than the initial one kθ/kl =
2/6; kY = 2.</p>
      <p>So, if two conjuncterms of (r−1)-rank θ1r−1 and θ2r−1, r ∈ {1, 2, . . ., n}, of the
function f (x1, x2, . . ., xn) differ in d different by values onename positions α, β, γ, δ, . . . ∈
{0, 1, –}, among which each of these conjuncterms has one dash (–), then in
polynomial set-theoretical format, starting with d = 2, they form kY = (d − 2)! of the sets
transformed PSTF Y ⊕, each of them has interrelation kθ∗/kl∗ = d/(d(d−1)−2(d−2)),
that is the proof of Theorem 3. ⊔⊓
Example 1. To apply Theorems 1, 2 and 3 to the function f (x1, x2, x3, x4), that is
given by perfect STF Y 1 = {0, 3, 5, 6, 7, 8, 9, 10, 12, 15}1, which is minimized in
polynomial format by K-maps method to the expression f = x1 ⊕ x2x3 ⊕ x2x4⊕
⊕x3x4 ⊕ x1x2x3x4 ⊕ x¯1x¯2x¯3x¯4 [32, p. 97].</p>
      <p>
        Solution. This function has PSTF Y ⊕ = {(1– – –),(–11–),(–1–1),(– –11),(1111),
(0000)}⊕. To the pair (1111) and (0000), that has d = 4, we will apply, for
example, the fourth PSTF from the rule (
        <xref ref-type="bibr" rid="ref5">5</xref>
        ):
 000–⊕
 0–01 .
      </p>
      <p>Y ⊕ = (1– – –), (–11–), (–1–1), (– –11), 01–1</p>
      <p>
         –111
Applying the rule (
        <xref ref-type="bibr" rid="ref6">6</xref>
        ) of Theorem 2 to the underlined pairs that have d = 1, namely
–1–1 ⇒⊕ (11–1), – –11 ⇒⊕ (–011), and the rule (
        <xref ref-type="bibr" rid="ref7">7</xref>
        ), in the formed set namely
01–1 –111
–011 ⊕ – –1– , we will obtain PSTF Y ⊕={(1– – –),(– –1–),(–010),(000–),(11–1),
–11– ⇒ –010
(
        <xref ref-type="bibr" rid="ref1">0–01</xref>
        )}⊕. Doing further transformations and according to the rules (
        <xref ref-type="bibr" rid="ref16">16</xref>
        ) and (
        <xref ref-type="bibr" rid="ref17">17</xref>
        ) of
–0– – – –0–
Theorem 3, namely –000100– ⇒⊕ 1–00101–, 010–00–1 ⇒⊕ 01–1000–, we’ll obtain the final
minimal PSTF Y ⊕ = {(1– – –),(– –1–),(–0– –),(– –0–),(–011),(0–00),(110–),(11–1)}⊕ ⇒
⇒ {(1– – –),(–1– –),(–011),(0–00),(110–),(11–1)}⊕. Here the cost of realization of the
minimized function f = x1 ⊕ x2 ⊕ x¯2x3x4 ⊕ x¯1x¯3x¯4 ⊕ x1x2x¯3 ⊕ x1x2x4 is equal to
kθ∗/kl∗ = 6/14 that is a better result if compared to [32], where kθ/kl = 6/15.
4
      </p>
      <p>Minimization of Complete and Incomplete Functions
The proposed method of Boolean functions minimization in the polynomial
settheoretical format is based on the idea of minterms splitting of a given function
f (x1, x2, . . ., xn) in the disjunctive format [27–29].</p>
      <p>The algorithm of minimization of a function f in the polynomial set-theoretical
format is realized on two stages:</p>
      <p>1-st stage: the procedure of splitting of minterms of a given function f and the
obtaining of a set of covering of a matrix of splitting;</p>
      <p>2-nd stage: the procedure of iterative simplification of conjuncterms of a set of
covering (obtained on the 1-st stage) on the basis of generalized rules of Theorems 1, 2 and
3 and formation of a minimal PSTF Y ⊕ of a given function f .</p>
      <p>The 1-st stage is realized by sequence of such steps:</p>
      <p>Step 1: the given binary minterms m1, m2, . . .mk of the perfect PSTF
Y ⊕ = {m1, m2, . . .mk}⊕ of the function f are split (operator ⇒S) by using the
matrixcolumn of the masks of literals of rank r ≥ n − log2 k, r = 1, 2, . . ., n, as a result of
n!
this a matrix of splitting Mnr of Cnr × k dimension is formed, where Cnr = (n−r)!r! ;
for example, let n = 5; if the number k of minterms is 8 ≤ k &lt; 16, then we use the
matrix of masks of rank r = 2, and as a result the matrix M52 of the dimension C52 × k
is formed;</p>
      <p>Step 2: in the matrix Mnr (in our example M52) for execution of the procedure
of covering (operator ⇒C) the conjuncterms-copies of r-rank, the number of which
2n−r−1 &lt; kr ≤ 2n−r (4 &lt; kr ≤ 8) are highlighted by underlining; priority is given
to conjuncterms-copies and their number is kr = 2n−r (kr = 8); if k = kr, then the
matrix is covered with a conjuncterms-copy of r-rank; if kr &lt; 2n−r (kr &lt; 8), then the
covering of the matrix will be made of the conjuncterms-copies the number of which
2n−r−1 &lt; kr &lt; 2n−r, and if there are not enough of them, then together with
generating minterms of the matrix Mnr; if kr &lt; 2n−r−1, then the transition to step 1 is done
for realization of similar procedures with application of the matrix of masks of the rank
r = 3 and etc. until to getting in the covering of the matrix Mnr of the minterms
splitting of which provides its covering, if such minterms &gt; 2, then the transition to step 1
is done.</p>
      <p>The 1-st stage of algorithm is completed when there are not only minterms in the
set of covering of the matrix Mnr or when the split elements do not provide its covering.</p>
      <p>The 2-nd stage of the minimization algorithm is the procedure of iterative
simplification. It is done with the conjuncterms of the set of the covering in sequence of the
following steps:</p>
      <p>
        Step 1: for every pair with d = 1 (pairs with d = 0 are not taken into account) either
the rule (
        <xref ref-type="bibr" rid="ref2">2</xref>
        ) of Theorem 1, or the rule (
        <xref ref-type="bibr" rid="ref6">6</xref>
        ) of Theorem 2; are applied; after respective
replacement the transition to the 1-st is done, if there are not such pairs, then go to the
step 2;
      </p>
      <p>
        Step 2: for every pair with d = 2 we apply either one from the sets of the rule (
        <xref ref-type="bibr" rid="ref3">3</xref>
        ) of
Theorem 1, or the rule (
        <xref ref-type="bibr" rid="ref7">7</xref>
        ) of Theorem 2; after respective replacement the transition to
the 1-st step is done and if there are not such pairs, then to the 3-rd step;
      </p>
      <p>
        Step 3: for every pair with d = 3 we apply either one from the sets of the rule (
        <xref ref-type="bibr" rid="ref4">4</xref>
        ) of
Theorem 1, or one from the sets of the rules (
        <xref ref-type="bibr" rid="ref8">8</xref>
        ), (
        <xref ref-type="bibr" rid="ref9">9</xref>
        ) or (
        <xref ref-type="bibr" rid="ref10">10</xref>
        ) of Theorem 2, or one from
the rules (
        <xref ref-type="bibr" rid="ref15">15</xref>
        ), (
        <xref ref-type="bibr" rid="ref16">16</xref>
        ) or (
        <xref ref-type="bibr" rid="ref17">17</xref>
        ) of Theorem 3; after respective replacement the transition to
the 1-st step is done and if there are not such pairs, then go to the 4-th step;
      </p>
      <p>
        Step 4: for every pair with d = 4 we apply one from the sets of the rule (
        <xref ref-type="bibr" rid="ref5">5</xref>
        ) of
Theorem 1, or one from the rules (
        <xref ref-type="bibr" rid="ref8">8</xref>
        ), (
        <xref ref-type="bibr" rid="ref9">9</xref>
        ) or (
        <xref ref-type="bibr" rid="ref10">10</xref>
        ) of Theorem 2, or one from the sets of
the rules (
        <xref ref-type="bibr" rid="ref18">18</xref>
        )–(
        <xref ref-type="bibr" rid="ref23">23</xref>
        ) of Theorem 3; after respective replacement the transition to the 1-st
step is done and if there are not such pairs, then go to the 5-th step;
      </p>
      <p>Step 5: if further transformation does not lead to the simplification of the set of
conjuncterms, then this set is the found minimal PSTF Y ⊕ of the function f , the cost
of realization of which is determined by the interrelation kθ∗/kl .
∗
Example 2. To minimize the function f (x1, x2, x3, x4) in the polynomial format by
using the splitting method. This function has perfect STF Y 1 = {0, 6, 14, 15}1 (this
function is borrowed from [21, p. 28]).</p>
      <sec id="sec-2-1">
        <title>Solution.</title>
        <p>000– 000– 001– 011– 010– 011–
00101010 ⇒00–1,01–1,00–0,00–0,01–1,01–0 .</p>
        <p>
           0–11 0–01 0–11 0–10 0–00 0–00 
After replacement of minterms (0000) and (0111) by the underlined sets in the set of
covering, we obtain two equal as to the realization cost of solutions of minimization of
the given function which is reflected by the minimal PSTF:
⊕
(
Y ⊕ = {(–11–),(0000),(0111)}⊕⇒ (–11–),(011–),
(1. (00–0),(
          <xref ref-type="bibr" rid="ref1 ref10 ref2 ref3 ref4 ref5 ref6 ref7 ref8 ref9">0–10</xref>
          ))
2. (01–0),(0–00)
⊕
⇒
(
⇒ (111–),
(1. (00–0),(
          <xref ref-type="bibr" rid="ref1 ref10 ref2 ref3 ref4 ref5 ref6 ref7 ref8 ref9">0–10</xref>
          ))
2. (01–0),(0–00)
⊕
.
        </p>
        <p>Answer. The cost of realization of the minimized function determines the
interrelation kθ∗/kl∗ = 3/9 that is a better result than in [21], where the PSTF Y ⊕ =
{(−11−), (0000), (0111)}⊕ that is equal to 3/10.</p>
        <p>In [27, 28], the incomplete function f (x1, x2, . . ., xn) can be given by the perfect
STF {Y 1, Y ∼}, where Y 1 and Y ∼ there are subsets of the complete set E2n, on which
the function f takes the value respectively 1 and ∼ (so-called “don’t-care”). In the
polynomial set-theoretical format the sets Y ⊕ and Y ⊕˜ correspond to the sets Y 1 and
Y ∼, the elements of which are binary minterms. Thus, incomplete function f can be
given by the perfect PSTF {Y ⊕, Y ⊕˜}.</p>
        <p>Similarly, [27, 29] the procedure of splitting of conjuncterms of incomplete function
.
f is realized by the matrix of splitting Mnr, which is designated as M r⊕..M r ⊕˜, where
.</p>
        <p>M r⊕ that is basic submatrix, M r ⊕˜ that is additional submatrix, and .. that is a symbol
of separation of the matrix Mnr. As a result of covering of the matrix Mnr a set of the
.
splitting conjuncterms Y ⊕..Y ⊕˜ is be obtained.</p>
        <p>An algorithm of minimization of an incomplete function in the polynomial
settheoretical format is realized in the same way as for a complete function in two stages.
On the 1-st stage the minterms of the perfect PSTF {Y ⊕, Y ⊕˜} are split by using of the
matrix Mnr, where the main role in its cover is played by the conjuncterms-copies of
the basic submatrix M r⊕. Whereas the 2-nd stage of the algorithm of minimization of
an incomplete function is realized in similar way as for a complete function.
Example 3. To minimize incomplete function f (x1, x2, x3, x4) in the
polynomial format by using splitting method. This function is has perfect STF
(Y 1= {3, 5, 6, 9, 12, 15}1</p>
        <p>(this function is borrowed from [33, p. 460]).</p>
        <p>Y ∼= {1, 2, 8, 11}∼</p>
      </sec>
      <sec id="sec-2-2">
        <title>Solution.</title>
        <p>(Y ⊕ = {(0011), (0101), (0110), (1001), (1100), (1111)}⊕ S
⇒</p>
        <p>Y ⊕˜ = {(0001), (0010), (1000), (1011)} ⊕˜
l l – – 00– – 01– – 01– – 10– – 11– – 11– – 00– – 00– – 10– – 10– –
l – l – 0–1– 0–0– 0–1– 1–0– 1–0– 1–1– 0–0– 0–1– 1–0– 1–1–
=0– –1 0– –1 0– –0 1– –1 1– –0 1– –1 0– –1 0– –0 1– –0 1– –1⇒C
⇒S l– –l 
 –ll–  –01– –00– –11– –00– –10– –11– –00– –01– –00– –01–
 –l–l  –0–1 –1–1 –1–0 –0–1 –1–0 –1–1 –0–1 –0–0 –0–0 –0–1
– –ll – –11 – –01 – –10 – –01 – –00 – –11 – –01 – –10 – –00 – –11
. ⊕
⇒C {l–l–} = n((0–1–),(0111)), (1–0–),(1101) ,(0101),(1111)..(0001),(1011)o ⇒
⇒ {(0–1–),(–111),(1–0–),(–101)}⊕⇒ {(0–1–),(–1–1),(1–0–)}⊕.</p>
        <p>
          After the transformation of the pair of highlighted conjunñterms by the rule
(
          <xref ref-type="bibr" rid="ref3">3</xref>
          ) of the Theorem 1, i. e. 01––10–– ⇒⊕ 1––––1–– , we will obtain the final
minimal PSTF Y ⊕ ={(– –1–), (1– – –), (–1–1)}⊕, to which coresponds the minimal PSTF
Y ⊕ = {(
          <xref ref-type="bibr" rid="ref10 ref11 ref14 ref15 ref2 ref3 ref6 ref7">2, 3, 6, 7, 10, 11, 14, 15</xref>
          ),(
          <xref ref-type="bibr" rid="ref10 ref11 ref12 ref13 ref14 ref15 ref8 ref9">8, 9, 10, 11, 12, 13, 14, 15</xref>
          ),(
          <xref ref-type="bibr" rid="ref13 ref15 ref5 ref7">5, 7, 13, 15</xref>
          )}⊕={2, 3, 5, 6,
8, 9, 12, 15}⊕, where the highlighted in bold font elements belong to set Y ∼.
Answer. The cost of realization of the given function is equal to kθ∗/kl∗ = 3/4. If
compared with [33] it is a better result, where Y ⊕ = {(–11–), (11– –), (– – –1)}⊕ and the
cost of realization is equal to kθ∗/kl∗ = 3/5.
        </p>
        <p>Minimization of System of Complete and Incomplete Functions
In general, the system F (X ), X = {x1, x2, . . ., xn}, of Boolean functions, fi(X ),
i = 1, 2, . . ., s can be represented in the polynomial set-theoretical format as a perfect
PSTF {Yi⊕, Yi⊕∗} [27, 29]:</p>
        <p>
          Y1⊕={m11, m12, . . ., m1k1 }⊕ , Y1⊕∗={mk1+1, mk1+2, . . ., m2n−k1−ν1 }⊕∗ ,
F (X ) = Y2⊕={m21, m22, . . ., m2k2 }⊕ , Y2⊕∗={mk2+1, mk2+2, . . ., m2n−k2−ν2 }⊕∗ ,
 Y.s.⊕. =..{. m..s.1., .m. .s.2., .. ....,.m. .s.k.s.}.⊕. ., .Y.s⊕..∗.=.{. m..k.s.+. 1. ,. m.. k.s. +..2., .. ....,. m..2.n.−.k.s.−. ν.s. }.⊕..∗.,. .
(
          <xref ref-type="bibr" rid="ref25">25</xref>
          )
where νi &lt; 2n −ki; mij are binary minterms of functions fi(X ), j = 1, 2, . . ., ki; while
if F (X ) it is a system of the completely specified functions, then Yi⊕∗ ≡ ∅ and we have
perfect PSTF Yi⊕ , and if F (X ) is a system of the incompletely specified functions,
then Yi⊕∗ ≡ Yi⊕˜ and we have perfect PSTF, where symbol ∼ represents incomplete
(“don’t care”) values of functions fi of system F (X ).
        </p>
        <p>
          Similarly as in SOP form [27, 29] compatible minimization of the system of
PSTF {Yi⊕, Yi⊕∗} (
          <xref ref-type="bibr" rid="ref25">25</xref>
          ) is performed by splitting method with the system minterms
(m)1,2,. . .,s′ , s′ ∈ {1, 2, . . ., s}, formed from minterms of the system F (X ).
        </p>
        <p>The algorithm of compatible minimization of the system F (X ) of complete
functions given by the perfect PSTF Yi⊕ , Yi∗ ≡ ∅, is realized in the following way. On the
first stage the system minterms (m)1,2,. . .,s′ of the set YI⊕ , I ∈ {1, 2, . . ., s}, are split
by using the matrix Mnr creating a system conjuncterms of (n−1)-rank (θin−1)1,2,. . .,s′ .
The minimal covering of the matrix Mnr is done in similar way [27, 29] by the identical
system conjuncterms-copies of (n−1)-rank. But among them, a decisive role for
realization of compatible minimization of the given system will be played by those ones the
indices of which contain the greatest quantity of numbers with the set {1, 2, . . ., s}.
Herewith, if (θir−1)1,2,. . .,s′ and (θjr−1)1,2,. . .,s′′ , s′, s′′ ∈ {1, 2, . . ., s}, these are identical
system conjuncterms-copies of (r−1)-rank of the matrix Mnr, r = 1, 2, . . ., n−1, then they
can be elements of its covering if the indices of their generative form (θir)1,2,. . .,s′ and
(θjr)1,2,. . .,s′′ will form a not empty intersection, i. e. {1, 2, . . ., s′} T{1, 2, . . ., s′′} 6= ∅.
For example, let the system minterms (100)1,2,4, (110)1,3, (010)1,2,3 be generative
elements of the matrix Mnn−1. For the mask {l–l} the identical system
conjunctermscopies will be (1–0)1 and (1–0)1, the index of which determines the intersection
{1, 2, 4} T {1, 3} = {1}, and for the mask {–ll} will be (–10)1,3 and (−10)1,3
because {1, 3} T {1, 2, 3} = {1, 3}. So, in this case for covering of the matrix Mnn−1 it is
recommended to choose the pair of the mask {–ll} because the power |{1, 3}| &gt; |{1}|.
Example 4. In the polynomial set-theoretical format to minimize the system F (X ) of
complete functions fi(x1, x2, x3), i = 1, 2, 3, by the splitting method. This has
perY11= {(000),(010),(101),(110)}1

fect STF Y21= {(001),(011),(101)}1 . This example is borrowed from [21, p. 35]
Y31= {(000),(001),(010),(011)}1
where the author illustrates efficiency of xlinking method.
Solution. Having transformed the given system of the perfect STF {Y11,2,3} into the
system of the perfect PSTF {Y1⊕,2,3} and, having formed from it a set of system minterms,
we will execute the splitting procedure by using the matrix M31 and its covering
procedure:</p>
        <p>Y1⊕,2,3 = (000)1,3, (001)2,3, (010)1,3, (011)2,3, (101)1,2, (110)1
⊕ S
⇒
⇒Sl––l––= 0––0––1,3 0––0––2,3 0––1––1,3 0––1––2,3 1––0––1,2 1––1––1⇒C
– –l – –0 – –1 – –0 – –1 – –1 – –0
⊕
⇒Cn (0– –)1,3, (001)1,3, (011)1,3 , (0– –)2,3, (000)2,3, (010)2,3 , (101)1,2, (110)1o .
We will do the splitting procedure with system minterms of the formed set by using the
matrix M32 and its covering procedure:
ll–
S
(001)1,3, (011)1,3, (000)2,3, (010)2,3, (101)1,2, (110)1 ⇒ l–l =
–ll
01–</p>
        <p>
          00–
=0–11,3 0–11,3 0–02,3 0–02,3 1–1 1–0 ⇒C{(
          <xref ref-type="bibr" rid="ref1">0–1</xref>
          )1,3,(0–0)2,3,(101)1,2,(110)1}⊕.
        </p>
        <p>
          –01 –11 –00 –10 –01 –10
Having distributed system conjuncterms in the functions we obtain the system of the
PSTF {Y1⊕,2,3}, with the underlined elements of which we will do step by step the
transformations according to the rules (
          <xref ref-type="bibr" rid="ref2">2</xref>
          ), (
          <xref ref-type="bibr" rid="ref3">3</xref>
          ) and (
          <xref ref-type="bibr" rid="ref7">7</xref>
          ) of the theorems:
Y ⊕
 1 = {(0– –),(
          <xref ref-type="bibr" rid="ref1">0–1</xref>
          ),(101),(110)}⊕= {(0–0),(
          <xref ref-type="bibr" rid="ref1">1–1</xref>
          ),(11–)}⊕= {(0– –),(– –1),(11–)}⊕
Y2⊕ = {(0– –), (0–0), (101)}⊕ = {(
          <xref ref-type="bibr" rid="ref1">0–1</xref>
          ), (101)}⊕ = {(– –1), (111)}⊕ .
Y3⊕ = {(
          <xref ref-type="bibr" rid="ref1">0–1</xref>
          ), (0–0)}⊕ = {(0– –)}⊕
Y ⊕ = {(0– –), (– –1), (11–)}⊕
 1
Answer. Minimal system of the PSTF {Y1⊕,2,3}: Y2⊕ = {(– –1), (111)}⊕ . Cost
Y3⊕ = {(0– –)}⊕
of its realization reflects the interrelation k∗/kl∗= 4/7. If compared with [21] it is a better
result, where this system is compatibly minθimized by xlinking method with kθ∗/kl∗= 4/9,
Y ⊕ = {(0– –), (
          <xref ref-type="bibr" rid="ref1">0–1</xref>
          ), (110), (101)}⊕
 1
namely Y2⊕ = {(
          <xref ref-type="bibr" rid="ref1">0–1</xref>
          ), (101)}⊕ .
        </p>
        <p>
          Y3⊕ = {(0– –)}⊕
In case of the system of incomplete functions (
          <xref ref-type="bibr" rid="ref25">25</xref>
          ) the set of system minterms
. .
will consist of two subsets separated by the symbol .. and reflected as {YI⊕..YI⊕˜},
I ∈ {1, 2, . . ., s}. Here the system minterms undergo the splitting procedure by
using the matrix Mnr, the elements of which are the conjuncterms of r-rank. It should be
noted, that in the course of covering the matrix Mnr two procedures are realized at a
time: making the matrix compatible as its elements are used to maximum extent with
higher capacity of the set I, and making it more predetermined in which the elements
of the submatrix YI⊕˜ are used. After distribution of the last ones in the functions of the
system we obtain {YI⊕...YI⊕˜}, the elements of which for every function further undergo
the simplification procedure according to the rules of the respective theorems in
Section 3, selecting out of possible variants of transformation those which will provide the
compatible minimization of the given system F (X) in the best way.
        </p>
        <p>I a b c f1 f2
0 0 0 0 1 0
1 0 0 1 ∼ ∼
2 0 1 0 0 1
3 0 1 1 1 ∼
4 1 0 0 ∼ ∼
5 1 0 1 ∼ 0
6 1 1 0 1 0
7 1 1 1 0 1
, with which we will do the
Example 5. [37, p. 228, example 5.1] To minimize the system
F (X) of incomplete functions f1(a, b, c) and f2(a, b, c), given
by the truth table (see right) with the help of splitting method in
the polynomial set-theoretical format.</p>
        <p>Solution. The given system F (X) has the perfect PSTF
(Y ⊕ 1˜= {(001), (100), (101)} ⊕˜
1 = {(000), (011), (110)}⊕, Y ⊕</p>
        <p>2˜= {(001), (011), (100)} ⊕˜
Y2⊕= {(010), (111)}⊕, Y ⊕
. ˜
We will form a set of system minterms i.e. Y1⊕,2..Y1⊕,2
splitting procedure by using the matrix M31 and the procedure of its covering, for
example, for the mask {–l–}:
Y1⊕,2...Y1⊕˜,2 = {(000)1,(010)2,(011)1,(110)1,(111)2...(001)1,2,(011)2,(100)1,2,(101)1}⊕ ⇒S
l– –  0– – 0– – 0– – 1– – 1– – 0– – 0– – 1– – 1– – 
⇒S–l–= –0–1 –1–2 –1–1 –1–1 –1–2 –0–1,2 –1–2 –0–1,2 –0–1  ⇒C {–l–} =
– –l – –0 – –0 – –1 – –0 – –1 – –1 – –1 – –0 – –1
. ⊕
=n(000)1, (–1–)2,(011)2,(110)2 , (–1–)1,(010)1,(111)1 ..(001)1,2,(011)2,(100)1,2,(101)1o .
After removal of the system minterm (011)2 the set of covering will look like
. ˜ . ⊕
Y1⊕,2..Y1⊕,2=n(000)1, (–1–)2,(110)2 , (–1–)1,(010)1,(111)1 ..(001)1,2,(011)2,(100)1,2,(101)1o .
Having distributed the system conjuncterms on the functions, we obtain the system
 . .</p>
        <p>PSTF Y1⊕..Y1⊕˜ = {(–1–), (000), (010), (111)..(001), (100), (101)}⊕</p>
        <p>. . .</p>
        <p>Y2⊕..Y2⊕˜ = {(–1–), (110) .. (001), (011), (100)}⊕</p>
        <p>We will do the splitting procedure with the minterms of the PSTF of the function f1
by using the matrixM31 and the procedure of its covering:</p>
        <p>. l– – 0– – 0– – 1– – 0– – 1– – 1– –
{(000),(010),(111)..(001),(100),(101)}⊕ ⇒S –l–=–0– –1– –1– –0– –0– –0–⇒C
– –l – –0 – –0 – –1 – –1 – –0 – –1
⇒C n (0– –), (011) , (– –1), (011) o</p>
        <p>
          ⇒ {(0– –), (– –1)}⊕.
⊕
Having taken into account the rule (
          <xref ref-type="bibr" rid="ref3">3</xref>
          )
of the minimal PSTF Y1⊕ = (–1–),
⊕
        </p>
        <p>
          We will do similar procedures for the minterms of the PSTF of the function f2 by
applying the matrix M32 for their splitting:
{(110)...(001), (011), (100)}⊕ ⇒S lll–−l  =  11–10– 000––1 001––1 11–00–  ⇒C {(1–0), (
          <xref ref-type="bibr" rid="ref1">0–1</xref>
          )}⊕.
        </p>
        <p>
          –ll –10 –01 –11 –00
After the transformation according to the rule (
          <xref ref-type="bibr" rid="ref3">3</xref>
          )
obtain two solutions of the minimal PSTF Y2⊕ = (–1–),
f2(a, b, c) = b ⊕ c ⊕ a
;
        </p>
        <p>(f1(a, b, c) = b ⊕ c¯ ⊕ a
f2(a, b, c) = b ⊕ c¯ ⊕ a¯
.</p>
        <p>Cost of realization of the system for the both solutions is equal to kθ∗/kl∗ = 4/4. If
compared with [37] it is a better result, where cost of realization is equal to kθ∗/kl∗ =
4/7, namely:
6</p>
      </sec>
    </sec>
    <sec id="sec-3">
      <title>Conclusions</title>
      <p>(f1(a, b, c) = b ⊕ c¯ ⊕ ab</p>
      <p>.</p>
      <p>f2(a, b, c) = b ⊕ abc¯
A new minimization method in the polynomial set-theoretical format of complete and
incomplete logic functions with n variables and their system has been presented. It
consists in the splitting procedure of given minterms and iterative simplification of
conjuncterms on the based set-theoretical rules. The method’s efficiency has been proved by
numerous examples borrowed from well-known publications (see References) related
to different minimization methods. The vast majority of functions and their systems
minimized by the proposed method showed better results. This is due to the fact that
in the process of transformation are involved the conjuncterms with Hamming distance
d ≥ 3, the transformed PSTF of which may have elements for which in the given set
there will be a pair with smaller d. The search procedure of such elements has s
combinative character: after each replacement of a chosen pair of conjuncterms of the given
PSTF Y ⊕ for certain set of the transformed PSTF Y ⊕ we obtain a new set where it
is necessary to determine distance d between new pairs and, having chosen from them
the elements with minimal d, to apply the rules of respective theorem and build again
a new set and so on. As a result, the probability of effective simplification of
conjuncterms set increases through the use of appropriate transformation rules that reduce the
implementation cost of realization of minimized function.</p>
    </sec>
  </body>
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