chapter 7 - uspsites.poli.usp.br/d/pea5737/e-book/chapter7_4.pdf · considered in the 2nd example...

233

7.4 SECOND INVERSE CONSEQUENT BAF(BAF) OPERATION

As we have seen in Section 7.3 of this Chapter, the inverse of the problem

of the Direct “BAF(BAF)” Operation of the precedent Sections could also be

considered as a problem of Boolean Equations.

Thus, if we consider the Direct “BAF(BAF)” Operation in the classical form,

q2 x r)C(p2 x r

(B) q2 x p)A( =→ , where the second member represents the

result, it is possible to define two “Inverse Operations”, because the corresponding

Direct “BAF(BAF)” Operation, as a Logical Implication, do not have the

mathematical commutative property of the given terms.

Then, in this Second Inverse Consequent BAF(BAF) Operation, the given

data are the corresponding Result C r x q1 6 2 and the Antecedent System A px q1 6 2 .

The Consequent System is the Incognito System of this Inverse Operation.

In terms of Boolean Geometry language, we have the following Fig. 15:

In terms of Boolean Algebra language, we have the following literal

expressions:

...... ...

234

Given the following two systems, C r x q1 6 2 and B r x1 6 2p , in the given order,

( )

( , ,..., )

( , ,..., )

...

( , ,..., )

( )

( , ,..., )

( , ,..., )

...

( , ,..., )

C

Z F X X Xq

Z F X X Xq

Zr Fr X X Xq

A

Y f X X Xq

Y f X X Xq

Yp f p X X Xq

and

1 1 1 2

2 2 1 2

1 2

1 1 1 2

2 2 1 2

1 2

=

=

=

%

&KK

'KK

=

=

=

%

&KK

'KK

, determine the

third System of Boolean Functions, ( )

( , ,..., )

( , ,..., )

...

( , ,..., )

B

Z G Y Y Yp

Z G Y Y Yp

Zr Gr Y Y Yp

1 1 1 2

2 2 1 2

1 2

=

=

=

%

&KK

'KK

, in such way as to

have the following logical equalities:

G f X X X f X X X f X X X F X X X

G f X X X f X X X f X X X F X X X

G f X X X f X X X

q q p q q

q q p q q

r q

1 1 1 2 2 1 2 1 2 1 1 2

2 1 1 2 2 1 2 1 2 2 1 2

1 1 2 2 1 2

, ,..., , , ,..., ,..., , ,..., , ,...,

, ,..., , , ,..., ,..., , ,..., , ,...,

...

, ,..., , , ,...,

3 8 3 8 3 8 3 8

3 8 3 8 3 8 3 8

3 8

=

=

q p q r qf X X X F X X X3 8 3 8 3 8,..., , ,..., , ,...,1 2 1 2=

%

&

KKK

'

KKK

In order to solve this Second Inverse Consequent BAF(BAF) Operation, it

is necessary to determine previously the Inverse Analytical System of the given

Antecedent System, ( )Apx q2

, that is, we determine the system Aq x

−14 9 2p

. Thus,

we reduce this problem to the solution of the following equivalent Direct

“BAF(BAF)” Operation:

1462 2 2

11 6

( ) ( ) ( )Aqx p C

r x q Brx p

− → =

To solve the operation given by (146), we can apply the practical gadget of

TABLES of the correspondent “Discreet Function of Discreet Functions” of the

ordinary Algebra, using the given numerical compact representation.

Then, it will be:

235

(147) y(x) → z(y) = z[y(x)] = z(x),

where:

(148) y(x) = NT A NT Apx qxq p

− − − −%&'

()*

= %&'()*

12

1 1 1

21 6 4 9

(149) z(y) = NT C rx q−1

21 6= B

(150) z(x) = NT B rx p−1

21 6= B

Then, expression (147) becomes the following, that is similar to expression

(146):

1512 2 2

1 1 11 6 NT

-1 ( ) ( ) ( )Aqx p NT C

r x q NT Brx p

− − −%&'

()*

→ %&'()*

= %&'()*

11th EXAMPLE:

The Systems of Boolean Arithmetical Functions (BAFs) are given:

( ) . ;#

152 5 23 2

1

2

3

2 1 22 3A NT

Y

Y

Y

X Xx1 6 ; @=%&K

'K

()K

*K=

%&K

'K 6 1 4

and

( ) . ;#

153 0 22 21

22 1 22 2C NT

Z

ZX Xx1 6 ; @=

%&'

()*

=%&'

2 1 3 , determine the following system

of Boolean Arithmetical Functions (BAFs),

( ) . ;#

154 32 21

27 6 5 4 3 2 1 02 1 2 33 2B NT

Z

Zb b b b b b b b Y Y Yx1 6 ; @=

%&'

()*

=%&'

, so that we have:

( )155 3 2 2 2 2 22 3 2 A1 6 1 6 1 6x x xB C→ =

NOTE: This example corresponds to the Second Inverse Consequent BAF(BAF)

Operation, referring to the 1st EXAMPLE considered in Section 7.1.


236

It is opportune to remember that the Systems of BAFs, in the compact

vertical numerical notation, could be also represented by the TABLES of the

practical gadget. This practical gadget corresponds to the “Discreet Function of

Discreet Function” of the ordinary Algebra.

Then, applying expression (151) we have:

1582 23 2 22 2 23

1 1 11 6 NT

-1 ( ) ( ) ( )Ax

NT Cx

NT Bx

− − −%&'()*

→ %&'()*

= %&'()*

Therefore, firstly we must determine the Inverse Analytical System of the

given Antecedent System of BAFs, that is:

A1 63 22x ( )Ax

−1

2 23

col.k

ak#

col.k

ak#

−1

3 5 7 Φ2

2 6 6 21 1 5 30 4 4 0

3 Φ2

2 Φ2

1 10 Φ2

237

Therefore, we have:

(159) ( )

#

. ;Ax

NTX

XY Y Y− =

%&'K

()*K

=1 2 2 2 2

2 231

2 22 3 1 2 3Φ Φ Φ Φ 2 3 0 1

= B , where

(160) ΦΦΦ

2

2

21

13 2=

%&'

()*

=�

!

"

$##

= 1 0 0

0 1 0 2, 1 or 0

(4)

,#

(or four solutions).

Then, applying the practical gadget of TABLES of the correspondent

“Discreet Function of Discreet Function” of the ordinary Algebra, we have:

(161) ( ) ( ) ( )Ax

Cx

Bx

− → =1

2 23 2 22 2 23

NOTE:

It is suitable to observe that as it is an Inverse Operation, whose respective

Direct Operation does not have the mathematical commutative property of the

given terms. Therefore, this Second Inverse Operation is not UNIQUE either.

Then, there are several solutions and one of them is the same, which was

considered in the 1st EXAMPLE of the Direct Boolean Arithmetical Function of

Boolean Arithmetical function Operation or BAF(BAF) Operation of Section 7.1.

Therefore, we have:

GIVEN DATA RESULTS

( )Ax

−1

2 23( )C

x2 22 ( )B

x2 23

col. k ak#

−1 col. k#

ck col. k bk#

7 Φ2 3 0 7 Φ2

6 2 2 2 6 25 3 1 1 5 04 0 0 3 4 33 Φ2 3 Φ2

2 Φ2 2 Φ2

1 1 1 10 Φ2 0 Φ2

238

This problem offers a result of “44 = 256” full solutions, which in the

numerical vertical compact representation of a system of BAF is the following,

(162) B x NTZ

ZY Y Y1 6 = B2

1

22 2 2 2

223 1 2 3 23 2 0 3 1

(4) (4) (4) (4)

=%&'K

()*K

=�

!

"

$##Φ Φ Φ Φ#

. ; ,

or, as binary Boolean Matrix form,

(163) B x NTZ

ZY Y Y1 6 = B2

1

2

4 4 4 4

22

3 1 2 3 23

* * * *

1 0 1 0

0 0 1 1 =

%&'K

()*K

=

�

!

"

$

######

Φ Φ Φ ΦΦ Φ Φ Φ

( ) ( ) ( ) ( )

#

. ;

(with “44 = 256” full solutions)

NOTE:

In that global numerical expression of all 256 full solutions (i.e. solutions

without any time-restriction) of expression (163), the abscissa components marked

with (*) correspond to the abscissa components of the solution, that is, it is the

same as the data in the 1st EXAMPLE of the Direct BAF(BAF) Operation of Section

7.1. In fact, this data was the following numerical expression (5):

(5) NT }3Y2Y1Y;3.{2#2

2213] 1203[2Z1Z

NT32x2)B( )}y(z{ =

== , or

(164) B x NTZ

ZY Y Y1 6 = B2

1

2

22

3 1 2 3

0 1 1

1 0 0 1 23

* * * *

1 0 1 1 0

0 0 1 1 =

%&'K

()*K

=

�

!

"

$

######

. ; , where the

bit columns not marked with (*) are one of the “44 = 256” possible solutions showed

in (163) numerical expression.

239

12th EXAMPLE:


( ) . ;#

165 1 23 2

1

2

3

2 1 22 3A NT

Y

Y

Y

X Xx1 6 ; @=%&K

'K

()K

*K=

%&K

'K 6 7 3

and

( ) . ;#

166 11 25 2

1

2

3

4

5

2 1 22 5C NT

Z

Z

Z

Z

Z

X Xx1 6 ; @=

%

&

KKK

'

KKK

(

)

KKK

*

KKK

=

%

&

KKK

'

KKK

15 15 29 , determine the following

system of Boolean Arithmetical Functions (BAFs),

( ) . ;#

167 35 2

1

2

3

4

5

7 6 5 4 3 2 1 02 1 2 33 5B NT

Z

Z

Z

Z

Z

b b b b b b b b Y Y Yx1 6 ; @=

%

&

KKK

'

KKK

(

)

KKK

*

KKK

=

%

&

KKK

'

KKK

, so that we have:

( )168 3 2 5 2 5 22 3 2 A1 6 1 6 1 6x x xB C→ =


Operation, referring to the 2nd EXAMPLE considered in Section 7.1.


240





Then, applying expression (151) we have:

1692 23 5 22 5 23

1 1 11 6 NT

-1 ( ) ( ) ( )Ax

NT Cx

NT Bx

− − −%&'()*

→ %&'()*

= %&'()*

Therefore, firstly we must determine the Inverse Analytical System of the

given Antecedent System of BAFs, that is:

A1 63 22x ( )Ax

−1

2 23

col.k

ak#

col.k

ak#

−1

3 1 7 12 6 6 21 7 5 Φ2

0 3 4 Φ2

3 02 Φ2

1 30 Φ2

Therefore, we have:

(170) ( )

#

. ;Ax

NTX

XY Y Y− =

%&'K

()*K

=1 2 2 2 2

2 231

21

22 3 1 2 3 2 0 3

Φ Φ Φ Φ = B , where

(171) ΦΦΦ

2

2

21

13 2=

%&'

()*

=�

!

"

$##

= 1 0 0

0 1 0 2, 1 or 0

(4)

,#

(or four solutions).


“Discreet Function of Discreet Function” of the ordinary Algebra, we have:

(172) ( ) ( ) ( )Ax

Cx

Bx

− → =1

2 23 5 22 5 23

241

NOTE:


Direct Operation does not have the mathematical commutative property of the

given terms. Therefore, this Second Inverse Operation is not UNIQUE either.


considered in the 2nd EXAMPLE of the Direct Boolean Arithmetical Function of

Boolean Arithmetical Function Operation or BAF(BAF) Operation of Section 7.1.

Therefore, we have:

GIVEN DATA RESULTS

( )Ax

−1

2 23( )C

x5 22 ( )B

x5 23

col. k ak#

−1 col. k#

ck col. K bk#

7 1 3 11 7 156 2 2 15 6 155 Φ2 1 15 5 Φ5

4 Φ2 0 29 4 Φ5

3 0 3 292 Φ2 2 Φ5

1 3 1 110 Φ2 0 Φ5

In this TABLE, “)5”, means the following “25 = 32” solutions:

(173) Φ

ΦΦΦΦΦ

5231 5=

%

&

KKK

'

KKK

(

)

KKK

*

KKK

= ,#

30, 29, ... , 2, 1, 0

This problem offers a result of “(25)4 = 220” full solutions, which in the


242

( ) . ;

#

174 15 15 35 2

1

2

3

4

5

21 2 33 5

B NT

Z

Z

Z

Z

Z

Y Y Yx1 6 ; @=

%

&

KKK

'

KKK

(

)

KKK

*

KKK

=

%

&

KKK

'

KKK

Φ Φ Φ Φ5 5 5 5 29 11 ,

or,

with “(25)4 = 220” full solutions”,

NOTE:

In that global numerical expression of all “(25)4 = 220” full solutions (i.e.

solutions without any time-restriction) of expression (175), the abscissa

components marked with (*) correspond to the abscissa components of the

solution, that is, it is the same as the data in the 2nd EXAMPLE of the Direct

BAF(BAF) Operation of Section 7.1. In fact, this data was the following numerical

expression:

( ) . ;

#

176 18 01 25 11 35 2

1

2

3

4

5

21 2 33

5B NT

Z

Z

Z

Z

Z

Y Y Yx1 6 ; @=

%

&

KKK

'

KKK

(

)

KKK

*

KKK

= �!

"$#

%

&

KKK

'

KKK

15 15 29 11 * * * *

or,

( ) . ;

* * * *

177

0 1 0

1 0 1

0 0 0 0

1 1 0 0 1

1 1 0 1 1 1 1 1

35 2

1

2

3

4

52

1 2 33B NT

Z

Z

Z

Z

Z

Y Y Yx1 6 ; @=

%

&

KKK

'

KKK

(

)

KKK

*

KKK

=

�

!

"

$

#######

%

&

KKK

'

KKKK

0 1 0 1 1

1 0 1 1 1

1 1 1 0

1 0 1

,

( ) . ;

( ) ( ) ( ) ( )

* * * *

175

0

1

1

1 1 1 1

35 2

1

2

3

4

5

2 2 2 2

2

1 2 33

5 5 5 5

B NT

Z

Z

Z

Z

Z

Y Y Yx1 6 ; @=

%

&

KKK

'

KKK

(

)

KKK

*

KKK

=

�

!

"

$

########

%

&

KKKK

'

KKKK

0 1 0

1 1 1

1 1 1 0

1 0 1

Φ Φ Φ ΦΦ Φ Φ ΦΦ Φ Φ ΦΦ Φ Φ ΦΦ Φ Φ Φ

243

where the bit columns not marked with (*) are one of the “(25)4 = 220” possible

solutions showed in (175) numerical expression.

13th EXAMPLE:


( ) . ;#

178 0 32 21

22 1 2 33 2A NT

Y

YX X Xx1 6 ; @=

%&'

()*

=%&'

1 2 0 2 0 0 1

and

( ) . ;#

179 6 33 2

1

2

3

2 1 2 33 3C NT

Z

Z

Z

X X Xx1 6 ; @=%&K

'K

()K

*K=

%&K

'K 2 7 5 1 3 0 4 , determine the

following system of Boolean Arithmetical Functions (BAFs),

( ) . ;#

180 53x2

1

2

3

31 30 29 28 3 2 1 02 1 2 1 2 35 3B NT

Z

Z

Z

b b b b b b b b Y Y X X X1 6 ; @=%&K

'K

()K

*K=

%&K

'K ...

, so that we have:

( )1815 2 2 2 3 23 5 3 A2 7 1 6 1 6x x xB C→ = , where, with the Complementary System

referring to the proper variables, that is:

( )

#

. ;1821

2

3

07 23 3 1 2 3 06 05 04 03 02 01 00NT

X

X

X

X X X

%&KK

'KK

()KK

*KK

=

%

&KK

'KK

= B ,

we have:

( ) . ; ,183

0010

35 2

1

2

1

2

3 2

1 2 33

1000

0100 0001

1111 0000

1100 1100

1010 1010

orA NT

Y

Y

X

X

X

X X Xx2 7 ; @=

%

&

KKK

'

KKK

(

)

KKK

*

KKK

=

�

!

"

$

######

244

( ) . ; ,183 07 35 2

1

2

1

2

3

2 1 2 33 14 21 04 19 02 01 08 orA NT

Y

Y

X

X

X

X X Xx2 7 ; @=

%

&

KKK

'

KKK

(

)

KKK

*

KKK

=

Then, according to the definition of the Second Inverse Consequent

BAF(BAF) Operation, we have:

1843 25 3 23 3 25

11 6

( ) ( ) ( )Ax

Cx

Bx

− → = , and therefore:

1853 25 3 23 3 25

1 1 11 6 NT

-1 ( ) ( ) ( )Ax

NT Cx

NT Bx

− − −%&'()*

→ %&'()*

= %&'()*


Operation, referring to the 3rd EXAMPLE considered in Section 7.2.






Therefore, to apply expression (183) we must firstly determine the Inverse

Analytical System of the given Antecedent System of BAFs, that is:

F ig. 18

245

A2 75 23x ( )Ax

−1

3 25

col.k

ak#

col.k

ak#

−1 col.k

ak#

−1

7 07 31 Φ3 15 Φ3

6 14 30 Φ3 14 65 21 29 Φ3 13 Φ3

4 04 28 Φ3 12 Φ3

3 19 27 Φ3 11 Φ3

2 02 26 Φ3 10 Φ3

1 01 25 Φ3 09 Φ3

0 08 24 Φ3 08 0

23 Φ3 07 7

22 Φ3 06 Φ3

21 5 05 Φ3

20 Φ3 04 4

19 3 03 Φ3

18 Φ3 02 2

17 Φ3 01 1

16 Φ3 00 8

Therefore, we have:

( ) ( ) [

] . ;#

1843 25

1

2

3

5

1 3 3 3 3

3 3 32 1 2 1 2 33

5 ( ) 3 ( ) 6 ( ) 0 7

( ) 4 ( ) 2 1 ( )

10 1 4 5

2 1 1

Ax

NT

X

X

X

Y Y X X X

− =

%&K

'K

()K

*K

= Φ Φ Φ Φ

Φ Φ Φ

4 9

; @

, where

(185) ΦΦΦΦ

3

2

2

1

1 7 3 3=%&K

'K

()K

*K=

�

!

"

$

#####

= 1 1 1 0 0 0 0

1 0 0 1 1 0 0

1 0 1 0 1 0 1 0

6, 5, 4, 2, 1 or 0

(8)

, ,#

(or, eight solutions).

246


“Discreet Function of Discreet Function” of the ordinary Algebra to the numerical

expression (182), we have the following TABLES:

NOTE:


Direct Operation do not have the mathematical commutative property of the given

terms. Therefore, this Second Inverse Operation is not UNIQUE either.


considered in the 3rd EXAMPLE of the Direct Boolean Arithmetical Function of

Boolean Arithmetical function Operation or BAF(BAF) Operation of Section 7.1.

Therefore, we have:

247

GIVEN DATA RESULTS

( )Ax

−1

3 25( )C

x3 23 ( )B

x3 25

col. k ak#

−1 Col. k#

ck col. k bk#

31 Φ3 7 6 31 Φ3

30 Φ3 6 2 30 Φ3

29 Φ3 5 7 29 Φ3

28 Φ3 4 5 28 Φ3

27 Φ3 3 1 27 Φ3

26 Φ3 2 3 26 Φ3

25 Φ3 1 0 25 Φ3

24 Φ3 0 4 24 Φ3

23 Φ3 23 Φ3

22 Φ3 22 Φ3

21 5 21 720 Φ3 20 Φ3

19 3 19 118 Φ3 18 Φ3

17 Φ3 17 Φ3

16 Φ3 16 Φ3

15 Φ3 15 Φ3

14 6 14 213 Φ3 13 Φ3

12 Φ3 12 Φ3

11 Φ3 11 Φ3

10 Φ3 10 Φ3

09 Φ3 09 Φ3

08 0 08 407 7 07 606 Φ3 06 Φ3

05 Φ3 05 Φ3

04 4 04 503 Φ3 03 Φ3

02 2 02 301 1 01 000 Φ3 00 Φ3

248

In this TABLE, “)3”, means the following “23 = 8” solutions:

(186) ΦΦΦΦ

3

8)

2

1111

7 3=%&K

'K

()K

*K=

�

!

"

$

#####

(

,#

0000

1100 1100

1010 1010

= 6, 5, 4, 3, 2, 1 and 0

2


numerical vertical compact representation of a system in the BAF is the following,

( ) [

] . ;

*

#

187 4

5

3 2

1

2

3

3 10 3 1 3 4 3 5 3 2

3 1 3 1

2 1 2 1 2 3

5

3

B NT

Z

Z

Z

Y Y X X X

x1 6 4 9 4 9 4 9 4 9 4 9

4 9 4 9 ; @

=%&K

'K

()K

*K=

%&K

'K 7 1 2 6

5 3 0

* * * *

* * *

Φ Φ Φ Φ Φ

Φ Φ

,

NOTE:

In that global numerical expression of all “(23)24 = 272” full solutions (i.e.

solutions without any time-restriction) of expression (187), the abscissa

components marked with (*) corresponds to the abscissa components of the

solution, that is, it is the same as the data in the 3rd EXAMPLE of the Direct

BAF(BAF) Operation of Section 7.2. In fact, this data was the following numerical

expression:

( ) ( ) [

]

#

.{ ; },

233

1

2

3

5476

23 5 1 2 1 2 3

x 25 1032 547 6 1032 3210

7654 6745 2301

* * *

* * * * *

B NT

Z

Z

Z

Y Y X X X

=

%&K

'K

()K

*K

=

where the bit columns not marked with (*) are one of the “(25)4 = 220” possible

solutions showed in (187) numerical expression.

249

14th EXAMPLE:


( ) . ;188 1 21 2 2 1 22A NT Y X Xx1 6 ; @ ; @= = = 0 0 1

and

( ) . ;#

189 2103 42 21

22 1 2 3 44 2C NT

Z

ZX X X Xx1 6 ; @=

%&'

()*

=%&'

2213 2213 2103 , determine

the following system of Boolean Arithmetical Functions (BAFs),

( ) . ;#

190 32 21

27 6 5 4 3 2 1 02 3 43 2B NT

Z

Zb b b b b b b b YX Xx1 6 ; @=

%&'

()*

=%&'

, so that we have:

( )1913 2 2 2 2 24 3 4 A4 9 1 6 1 6x x xB C→ =



1924 23 2 24 2 23

11 6

( ) ( ) ( )Ax

Cx

Bx


1934 23 2 24 2 23

1 1 11 6 NT

-1 ( ) ( ) ( )Ax

NT Cx

NT Bx

− − −%&'()*

→ %&'()*

= %&'()*


Operation, relative to the 4th EXAMPLE considered in Section 7.2.


250

The data given by numerical expression (188), must be prepared to

become “ Ax

4 93 24 ” of expression (191), through the following procedure:

We consider the Complementary System referred to the variables {X3 X4},

extended to the BAFi of ordinality “ω4 3 4 1 2= X X X X; @ ”, that is:

(194) NTX

XX X X X3

4 2

3 4 1 2

11114

%&'

()*

=�!

"$#

1111 0000 0000

1111 0000 1111 0000. ;; @ .

Then, we have:

(195) A NT

Y

X

X

X X X Xx

4 9 ; @3 2 3

4 2

3 4 1 24

1001

1111 4=%&K

'K

()K

*K=

�

!

"

$

###

1001 1001 1001

1111 0000 0000

1111 0000 1111 0000

. ; ,

or, in the numerical vertical compact representation,

(196) A NT

Y

X

X

X X X Xx

4 9 ; @3 2

3

4

2 3 4 1 2437337 4=

%&K

'K

()K

*K= 6226 5115 4004

#

. ;

Applying the properties given in CHAPTER 3 of changes in the

permutation of ordinality to the BAFi “ω4 1 2 3 4= X X X X; @ ”, we find:

(197) A NT

Y

X

X

X X X Xx

4 9 ; @3 2 3

4

2 1 2 3 44 37654 4=%&K

'K

()K

*K= 3210 3210 7654

#

. ;


Analytical System of the given Antecedent System of BAFs, suitably prepared by

expression (197), that is:

251

A4 93 24x

A -14 94 23x

Col.k

ak#

Col.k

ak#

−1

15 7 7 15 0314 6 6 14 0213 5 5 13 0112 4 4 12 0011 3 3 11 0710 2 2 10 0609 1 1 09 0508 0 0 08 0407 306 205 104 003 702 601 500 4

Then, with “28 = 256” full solutions, we have:

( ) [

]

#

. ;

( )198

1

2

315 03 02 01 00

07 06 05 0423 3 3 4

4 2

4

2

3 A = NT 14 13 12

11 10 09 08

-1(2) (2) (2)

(2) (2) (2) (2)

4 9

= B

x

X

X

X

X

YX X

%

&KK

'KK

(

)KK

*KK

=

Then, we have:

252

GIVEN DATA RESULTS

Ax

−1

4 234 9 C x1 62 24 B x1 62 23

Col. k ak#

−1 Col. k ck#

Col. k bk#

7 15 03 15 2 7 26 14 02 14 1 6 15 13 01 13 0 5 04 12 00 12 3 4 33 11 07 11 2 3 22 10 06 10 2 2 21 09 05 09 1 1 10 08 04 08 3 0 3

07 206 205 104 303 202 101 000 3

This example has the following unique solution, which is the same as the

one considered as data given by numerical expression (37), in the 4th EXAMPLE of

the Direct BAF(BAF) Operation of the Section 7.2, that is:

( )

#

. ;199 1

22 22 3 3 2 NT 1 0 3 2 2 1 3

Z

ZYX X

%&'K

()*K

= = B ,

15th EXAMPLE:


( ) . ;200 1 21 2 2 1 22 0 0 1A NT Y X Xx1 6 ; @ ; @= = =

and

253

( ) . ;#

201 5476 76 43 2

1

2

3

2 1 2 3 44 3C NT

Z

Z

Z

X X X Xx1 6 ; @=%&K

'K

()K

*K=

%&K

'K54 6745 1032 , determine

the following system of Boolean Arithmetical Functions (BAFs),

( ) . ;#

202 52 2

1

2

3

31 30 29 28 3 2 1 02 1 2 3 43 3 ... B NT

Z

Z

Z

b b b b b b b b YX X X Xx1 6 ; @=%&K

'K

()K

*K=

%&K

'K, so that

we have:

( )2035 2 3 2 3 24

5 4 A��

�� → =

x x xB C1 6 1 6



2044 25 3 24 3 25

11 6

( ) ( ) ( )Ax

Cx

Bx


2054 25 3 24 3 25

1 1 11 6 NT

-1 ( ) ( ) ( )Ax

NT Cx

NT Bx

− − −%&'()*

→ %&'()*

= %&'()*


Operation, referring to the 5th EXAMPLE considered in Section 7.2.


254

The data given by numerical expression (200), must be prepared to

become the term “ Ax

��

��5 24

” of the expression (203), through the following

procedure:

We consider the Complementary System referred to the variables {X3 X4},

extended to the BAFi of ordinality “ω4 3 4 1 2= X X X X; @ ”, with “24 = 16” solutions,

that is:

(206) NT

X

X

X

X

X X X X

3

4

1

2 2

3 4 1 2

1111

4

%

&KK

'KK

(

)KK

*KK

=

�

!

"

$

####

1111 0000 0000

1111 0000 1111 0000

1100 1100 1100 1100

1010 1010 1010 1010

. ;; @ ,


(207) NT

X

X

X

X

X X X X

3

4

1

2

2 3 4 1 215 44

%

&KK

'KK

(

)KK

*KK

= 14 13 12 ... 03 02 01 00#

. ;; @

Then, we have:

(208) A NT

Y

X

X

X

X

X X X Xx

��

�� =

%

&

KKK

'

KKK

(

)

KKK

*

KKK

=

�

!

"

$

######

5 2

3

4

1

2 2

3 4 1 24

1001

1111

4

1001 1001 1001

1111 0000 0000

1111 0000 1111 0000

1100 1100 1100 1100

1010 1010 1010 1010

. ;; @ ,


(209)

A NT

Y

X

X

X

X

X X X Xx

��

�� =

%

&

KKK

'

KKK

(

)

KKK

*

KKK

=5 2

3

4

1

2

2 3 4 1 24531 4 14 13 28 27 10 09 24 23 06 05 20 19 02 01 16

#

. ;; @

Applying the properties given in Chapter 5 of changes in the permutation of

ordinality to the BAFi “ω4 1 2 3 4= X X X X; @ ”, we find:

255

(210)

A NT

Y

X

X

X

X

X X X Xx

��

�� =

%

&

KKK

'

KKK

(

)

KKK

*

KKK

=5 2

3

4

1

2

2 1 2 3 44531 4 27 23 19 14 10 06 02 13 09 05 01 28 24 20 16

#

. ;; @


Analytical System of the given Antecedent System of BAFs, suitably prepared by

expression (210), that is:

Ax

��

��5 24

( )Ax

−14 25

col.k

ak#

col.k

ak#

−1 col.k

ak#

−1

15 31 31 15 15 Φ4

14 27 30 Φ4 14 11

13 23 29 Φ4 13 07

12 19 28 03 12 Φ4

11 14 27 14 11 Φ4

10 10 26 Φ4 10 10

09 06 25 Φ4 09 06

08 02 24 02 08 Φ4

07 13 23 13 07 Φ4

06 09 22 Φ4 06 09

05 05 21 Φ4 05 05

04 01 20 01 04 Φ4

03 28 19 12 03 Φ4

02 24 18 Φ4 02 08

01 20 17 Φ4 01 04

00 16 16 00 00 Φ4

Then, with “ 2 24 16 644 9 = ” full solutions and “ Φ4 ” is given by (207), we have:

256

( ) [

]

#

. ;

211

1

2

315 01 12

24 5 3 4

4

2 2 2 2

1 2 2 2 11 2

A =NT 03 14 02 13

00 11 07 10 06 09 05 08 04

-1

4x2

4 4 4 4

4 4 4 4 4

5

��

��

%

&KK

'KK

(

)KK

*KK

=

X

X

X

X

YX X X X

Φ Φ Φ Φ

Φ Φ Φ Φ Φ

4 9 4 9 4 9 4 9

4 9 4 9 4 9 4 9 4 9 = B

Then, applying the definition given by expression (204), we have:

2044 25 3 24 3 25

11 6

( ) ( ) ( )Ax

Cx

Bx

− → =

GIVEN DATA RESULTS

( )Ax

−14 25

( )Cx3 24

( )Bx3 25

Col. k ak#

−1 Col. k ak#

−1 Col. k ck#

Col. k bk#

Col. k bk#

31 15 15 Φ4 15 5 31 5 15 Φ3

30 Φ4 14 11 14 4 30 Φ3 14 7

29 Φ4 13 07 13 7 29 Φ3 13 628 03 12 Φ4 12 6 28 1 12 Φ3

27 14 11 Φ4 11 7 27 4 11 Φ3

26 Φ4 10 10 10 6 26 Φ3 10 6

25 Φ4 09 06 09 5 25 Φ3 09 7

24 02 08 Φ4 08 4 24 0 08 Φ3

23 13 07 Φ4 07 6 23 7 07 Φ3

22 Φ4 06 09 06 7 22 Φ3 06 5

21 Φ4 05 05 05 4 21 Φ3 05 4

20 01 04 Φ4 04 5 20 3 04 Φ3

19 12 03 Φ4 03 1 19 6 03 Φ3

18 Φ4 02 08 02 0 18 Φ3 02 4

17 Φ4 01 04 01 3 17 Φ3 01 5

16 00 00 Φ4 00 2 16 2 00 Φ3

where “ Φ4 ” is given by expression (207) and “Φ3” is given by the following

numerical expression:

257

(212) NT

Z

Z

Z

1

2

3

7

%&K

'K

()K

*K= , 6, 5, 4 ,3, 2, 1 or 0



( ) [

. ;

*

#

213 5

5

3 2

1

2

3

3 2 3 2 3 2 3 2 3 1

3 2 3 2 3 32 3 4 1 2

5

3

B NT

Z

Z

Z

YX X X X

x1 6 4 9 4 9 4 9 4 9 4 9

4 9 4 9 4 9 4 9 ; @

=%&K

'K

()K

*K=

%&K

'K 1 4 0 7 3 6 2

7 6 6 7 5 4 4 5 ]

* * * * * * *

* * * * * *

2

* *

1

Φ Φ Φ Φ Φ

Φ Φ Φ Φ

As we have seen in Property (27) in Chapter 3, Section 3.3, we can

determine the new numerical expression in a BAFi, whose cardinality/ordinality is

“ 5 1 2 3 4;YX X X X; @ ” and we find:

( ) ) ) ) .

;

* * * *#

214 5 0 7 6

5

2

3

8 4 4

2

1 2 3 4

3 B = NT

Z

4 7 6 ( 1 3 2 ( 6 5 4 7 4 5 (

.

3x2

1

* * *

3

* * *

3

* * * * * *

351 6

; @

Z

Z

YX X X X

%&K

'K

()K

*K= �

! "$#

Φ Φ Φ

NOTE: As we have seen, this example corresponds to the Second Inverse

Consequent BAF(BAF) Operation, relative to the 5th EXAMPLE considered in

Section 7.2 and the value expected for this system of BAF is given by the

numerical expression:

( )

#

.{ ; },B NT

Z

Z

Z

YX X X X3

1

2

223

5 1 2 3 4 x 25 5476 1032 5476 1032 3210 7654 6745 2301 * * * * * * * * * * * * * * * *

=%&K

'K

()K

*K= �

! "$#

It is clear that expression has, apart from the sixteen common number

marked by (*), other numbers, that is one of those “(23)16 = 248” full solutions of

global numerical expression (214).

chapter 7 - uspsites.poli.usp.br/d/pea5737/e-book/chapter7_4.pdf · considered in the 2nd example...

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