CALIFORNIA STATE UNIVERSITY, NORTHRIDGE

BOOLEAN ALGEBRA II.-

and

SWITCHING CIRCUITS

A thesis submitted in.partial satisfaction of the requirements for the degree of Master of Science in

Mathematics

by

Cynthia. Gruber Brown

May, 1975

The thesis of Cynthia Gruber Br.own is approved:

California State University, Northridge

May, 1975

ii

I l.

To MOM and DAD

for supplying the .

· PUSH and PULL,

respectively, respectfully.

iii

The assistance of Professors Potts and Karash is

greatly appreciated.

iv

TABLE OF CONTENTS

Approval. page ••••••.••••••••••••••.•••••..••• ~ • . . • • • ii

Dedication ......•................. ·-· ............... . iii

Aclmowl edgemen t . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . i v

Table of Contents . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v

Abstract

Part I:

Part II:

Part III:

Part IV:

........................................ •-• .. vi

Abstract Boolean Algebra ••••••••••••••••• 1

The Logic of Propositions . . . . . . . . . . . . . . . . 8

Applied Boolean Algebra . . . . . . . . . . . . . . . . . . 12

Minimization . . . . . . . . . . . . . . ·-. . . . . . . . . . . . . . 31

Quine-HCCluskey Method for Finding Prime Implicants ••••••••••••••••••••••••• 36

Concensus Method for Finding Prime Implicants ••• ~ ••••••••••• ~............... 42

Karnaugh Maps Implicants

The Computer

for Finding Prime . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

47

56

Biblio_graphy •...•.....••••••••..•••.....•. -.......... 64

v

ABSTRACT

BOOLEAN ALGEBRA

and

SWITCHING CIRCUITS

by

Cynthia Gruber Brown

Master of Science in Mathematics

May, 1975

The logician, George Boole, concerned with developing

a logic of mathematical symbols not dependent on quantity,

devised an algebra of classes which defined adding, sub­

tracting, multiplying, and dividing identical "classes~ of

objects. In an attempt to develop a universal, unambiguous,

symbolic logic, George Cantor expanded Boole's structure by

defining the operations on pre-defined "sets" of objects

which did not need to be similar. Cantor's work finally

evolved into a propositional logic complete with axioms.

Until the 1930's, Boolean algebra existed as an

abstract structure when it developed into an applied system

by providing a mathematical description of switching

circuits. Equivalent algebraic expressions yield equivalent

and sometimes simpler, therefore, cheaper s~itching

circuits. Through the refinement of Boolean algebra

vi

applications has come processes to minimize the materials

and costs of switching circuits.

vii

,-----------------------·-··-----------·--------·~--------- - -·-----· ---------------------------------------1

-I ABSTRACT BOOLEAN ALGEBRA . ; I· I

Boolean algebra, an abstract mathematical strttcture,

constructed from axioms and theorems also serves as an

excellent model for switching circuits. In its application, l :it is used to construct and redesign circuits using swi tche.s

lin electronic relays and computers. In order to understand

ihow Boolean algebra is applied, we must understand its

!structure in pure form.

1 George Boole (1815-1864), while not directly attribut~d i I

!with developing the algebra used today, evolved a process !

iof symbolic logic concerned with non-quantity objects. He

. I

I I

iwas concerned with the applications of the binary operations I I

!of addition, subtraction, multiplication, and division on l 1distinct groups of objects.

Boole began by designating two groups of objects we I

jwill call 'A' and 'B'. He defined addition to yield a third

:grour containing all the objects in 'A' and all the objects ; 'in 'B' • 'B' subtracted from 'A' is defined as all those

'objects contained in 'A' which are not in 'B'. Mul tiplica-

tion betw~en 'A' and 'B' yields a group of objects which

are in both_ 'A' and in 'B' • If an object is. in 'A' and

not in 'B', then it is not in the product of 'A' and 'B'.

Division is defined as the inverse of multiplication. If

we let 'AB' represent the group of objects formed by 'A'

multiplied by 'B', then 'AB' divided by 'B' yields 'A'.

While Boole had defined his system on various groups,

1

each group contained the same type of object. Logicians

-refer to these groups as "classes."

During the late decades of the nineteenth centruy,

Boole's ideas were expanded into an algebra of classes. As

mathematicians George Cantor, Augustus De Horgan, and

Ernst Schroder became involved in symbolic logic, the word

"class" fell way to "s·et." While "class" denoted objects

of the same type, a mathematician could tak~ a group of

any number of completely dissimilar objects and put them

together into a "set." Only two requirements are necessary:

(1) Each element (object) in the set must be specifically

denoted. If a dog is an element of a set, it must be

described so that there is no mistake as to which particular

dog is in the set-. ( 2) No object is counted more than

once in a set.

Cantor, in following Doole's procedures, defined

specific sets and then described operation "cup'' (v) and

"cap" (()). Using modern notation, we will let brackets

denote a set and commas separate names of elements in the

set. For example

Let A ~ {'a", "f", t, ~ and B ~ {• f" , 5 , *, @} Then:

Ar'lB =

{"a", "f", !, @, ;,_*~---all the elements tha~are in 'A' and all the elements that are in 'B'

~"f" @} ' all the elements which are in both 'A' and in 'B'

By comparing Doole's and Cantor's definitions, we can ;

see that we have two identical operations.

Set theory progressed into a structure primarily

concerned with sets in general, what they contain, relations.

that exist between them, and operations that may be

performed on them. While we are not concerned with set

theory, it is important to understand one definition of its

structure that provides the foundation for our Boolean

algebra: Given two sets, 'C' and 'D', 'C' is a subset

of 'D' if and only if every element of set 'C' is an element

of set 'D'.

Boolean algebra is an abstract form of the algebra of

sets. While there are many postulational treatments of

Boolean algebra, the most commonly acc~pted system is

given by E. V. Huntington:

Boolean algebra is a set B of elements a, b, c, •••

together with two binary operations U and (), called

'cup' and 'cap', respectively, unary operation of complement

('), and two special elements denoted 'u' (the universe)

and 'n' (the null set) satisfying the following postulates.

Axiom 1

Axiom 2

Axiom 3

a: a()b = b()a

b: aub = bUa

a: a()u = a

b: auu = a

a: a()(bUc) =

aU(b(Jc) =

- .. -------------- ··------- -

commutativity

identity

(a()b )v<anc) distributivity

(aub){)(auc)

3

,------··---·--·-·-· ---------- --------·--·--·-----~- -----·----·--- ------ ------- ---- -------··- -··- -···--- -· . ---------- ---··-·- -·· -------~

i Theorem 1: Any theorem of Boolean algebra remains valid if the operations U and() , and the identity elements n and u are interchanged throughout the statement of the theorem.

i i

b

Theorem 2 a: a()a = a idempotent

b: aua = a

proof:

a - aun Axiom 2

= au(a()a•) ·Axiom 4

= .(aua)() {aua') Axiom 3

= {aua)()u Axiom 4

= a(Ja Axiom 2

{Due to Theorem 1 , separate proofs for the

dual of a theorem is unnecessary. We may

prove either expressio~ (a) or (b).)

Theorem 3 a: a(Jn = n

b. -auu = u

proof:

n = ana• Axiom 4

= an(a'un) Axiom 2

= . (an,a' )0 (a()n) Axiom 3

= nu(a()n) Axiom 4

= {a(Jn)un Axiom 1 ,~.

= a()n Axiom 2

Theorem 4 a: a()(aub) = a

b: au(a()b) = absorption

a

proof:

a= aun Axiom 2

4

_,

!

------~·-------~~--·-----------------------·------,

= au(b(\n)

= (aub)()(aun).

= (aub)rla

= an(aub)

Theorem 3

Axiom 3

Axiom 2

Axiom 1

Theorem 5 a: a(J(b(}c) = (a()b)(lc associativity

b:

proof:

Let x = av(buc) and y = (aub)uc

anx = (a()a)u(a()(buc)) Axiom 3

= au(a()(bUc)) Theorem 2

=a Theorem 4

and

a()y = (a()(a(Jb))U(a(Jc) Axiom 3

= au (a()c) Theorem 4

= a Theorem 4

therefore · a(Jx = a(Jy

a'()x = (a'(Ja)tj(a'()(b()c)) Axiom 3

= _ (a()a' )U(a' (J(bUc)) Axiom 1

= nu(a •n (bUc)) Axiom 4

= (a'() (bUc ))un Axiom 1 ·

= a' (){buc) Axiom 2

and

a'()y = (a'()(aub))u(a'()c) Axiom 3

= ((a'()a)v(a'(Jb))u(a'()c) Axiom 3

= ({afJa')U(a':Jb))U(a'()c) · Axiom 1

.-

= (nu(a'()b))u(a'()c) Axiom 4

I '

5

[-----------------'------------·-----·--·-----------·-----~~-----·--------··--~

· · = ((a'(}b)(_;n)u(a'()c) Axiom 3 • i . i = ·(a'(")b)u(a'()c) Axiom 2

I I I

! I I j I I

I i

Theorem 6:

Axiom 3

therefore a'C)x = a'()y

(a(")x)u(a'()x) = (a()y)u(a'()y) substitution

(x()a)(J (x()a') = (y()a)u(y()a')

xn<aua') = y()(aUa')

x(Ju = y()u

x=y

· Axiom 1

Axiom 3

Axiom 4

Axiom 2

There exists a unique inverse (comple-t "'") f 1 t men , or every e emen •

proof:

· Let b and c be elements of B such that aub .= u, a()b = n, a;uc = u, and }~.()c = n

c = c()u

= c()(aUb)

= (c(la)U(c()b)

_ (a()c)U(b(Jc)

= nU(b(")c) ·

= (b()c)Un

= b(')c

and, in the same way

b = c(lb

b = c

·Axiom 2

substitution

Axiom 3

Axiom 1

substitution

Axiom 1

Axiom 2

Axiom 1

·definition of a'

Theorem 7 a: (a()b)'= a'ub'

b: (aub)'= a'(lb' DeMorgan' s laws __

6

Q

r---------:------------··------···--------------------------------------------···--------·----, 1 proof: · . I

. I . ~

I i I !

.I I

and

therefore

= (a'u<aub) )(){ (aub)ub') · · Axiom 1

= ( (a'ua)uh)()(au{bub 1))

Theorem 5

= (bu(aua 1 ))(J(aU(bLJb 1) )"

-Axiom 1

= (buu)() (auu) Axiom 4

= unu Theorem 3

= u Axiom 2

(a •()b 1 )() (aub)

Axiom 1

= ( (a'Clb' )()a)u( (a'()b' )fib) Axiom 3

= (a()(a 1()b' ))L)( (a'()b' )Ob) Axiom 1

- ( (a() at ) n b I ) U (a I () ( b h b ) ) Theorem 5

= (b'()(a()a'))u(a'(J(b0b')) Axiom 1

= (b'()n)U(a'()n) Axiom 4

=nun Theorem 3

=n Axiom 2

(a' n b ' ) = (au b ) ' Theorem 6

The transition of Boolean algebra from Boole's original

(

concepts to its modern structure is much more involved than

what has been presented. However, an in depth understanding.

of Boole's work and set theory is not pertinent to the

application of Boolean algebra on switching circuits.

THE LOGIC OF PROPOSITIONS

Any mathematical theory results from a set of

postulates and logic. Boolean algebra is also an abstrac-

tion of the logic of propositions. In order to discuss the

process of logic, we must be free from all the ambiguities

common to ordinary language. We turn to a symbolic logic

known as the logic of propositions. We will begin by

defining our symbols: A proposition is any statement. 'a'

and 1 b' are proposi ti.ons.

disjunction

conjunction

a/\b

implication

equivalence

negation

a'

( read , ' a or b " )

A true proposition when either 'a' or 'b' are true.

I

(read, "a and b")

A true proposition when both 'a' is true and 'b' is true.

(read, "if a then b")

A true proposition when either 'a' is false or 'b' is true •.

(read, "a, if and only if b")

A true proposition when either both 'a' and 'b' are true or both 'a' and 'b' are false.

(read' "not a")

A true proposition when ·~· is false.

i j . i

r---~-~---~------ -------~ ---------- -------- ·----- ----------------- ·---··· -------------------------------------------------- ------· ---- --- -- ---.---------------------, I In o~der to best apply this logic, it is necessary to I !establish a simple, organized pattern •. This pattern, made : . i

lup of "truth tables," is the logic of propositions and is I . .

jderived from the definitions .of the operations. We will ·' !begin with propositions 'a'· and 'b' and let "T" and "F"

signify the ordinary concept of true and false.

Table #1 Table #2

a b aVb a b a/\b

T T T T T T T F T T F F F T T F T F F F F F F F

Table #3 Table #4

a b ~b a b a~b

T T T T T T T F F T F F F T T F T F F F T F F T

Table #5

This pattern of logic continues by combining two or more

·operations while referring to the original tables for the

outcome. For example

..

Table #6

a b aVb (a Vb)~b

T T T T T F T T F T 'i' T F F F T

Column three is obtained by applying Table #1 to columns

one and two. Column four is obtained·by applying Table #3

to columns two and three.

The number of rows in a table depends upon the number

·of propo~itions being used. Each proposition can be either

true or false, and every possible permutation of the

propositions mrist.be considered. Therefore, a table

coritaining two propositions has 22 or 4 rows. A table of

three propositions has 23 or 8 rows. As we will see later,

a table containing six propostions has 26 or 64 rows.

Because disjunction and conjunction are defined the

same as "cup" and "cap," respectively, it is ·easy to

understand· how· the logic th:tt derived tables one and two

c~n be r~lated to Boolean algebra. However, in o~der to

see the impact of implication and equivalence toBoolean

algebra, it is necessary to prove that these operations can

be re-written in terms of disjunction and conjunction.

Table #7

a .b b' a/\b' (a/\b')'

T T F F T T F T T F F T F F T F F T F T

Ignoring the "scratch work" within the heavy lines, Table

I '-'

,--------------------------------·-·--------- -----------·-·--·· . ···-··· -- ·----- - ..... ··-------- - ··------ ------- ----------------------------------------------- ·----- -----

#7 is .identical to Table #3. Therefore, we may now re-write

'a-4b 1 using only those symbols related to Boolean algebra,

namely, conjunction_ anl negation (cap and complement). -

Table #8

a b (a/\b')i/\ (a'!\b)'

T T T T F F F T F F F T

Table #8 (we have left out the scratch work) is identical .

t.o Table #4, and equivalence may be written as (a/\b')'/\

(a'/\ b)'.

1 1

APPLIED BOOLEAN ALGEBRA

By the 1930's, Bnolean algebra developed not only.into

its significant abstract form, but also into an important

branch of applied mathematics. As the first part dealt

with Boolean algebra as abstract mathematics, this part

will emphasize its application.

The algebra discussed earlier serves as a mathematical

·model for switching {and logic) circuits. While Boolean

algebra cannot represent all facets of its physical system,

it does describe all the factors we consider vital. Our

concern is that given a circuit, how can we minimize the

number of switches in it and place them appropriately with

respect to each other in order to reduce costs and

materials. We will see how a circuit is represented

algebraically and how, through the applications of Boolean

axioms and· the.orems, an equ:iv.alent algebraic expression, I

hence, equivalent physical circuit is derived.

An electrical circuit is any conductor which carries

an electrical .impulse from its one extremity to the other.

A switching ci~cuit is a circuit containing a device, a

switch, which when open prevents the electrical impulse from

continuing. A clo~ed switch permits the completion of the

circuit. ·A~ the switch is either opened or closed, the

circuit is called a two-state switching circuit.

The following example is the type of symbol we will use

to illustrate a circuit with a switch A:

12

r----------------------~---·-----:----~-------.___ __ -----:----------.-:.-=------·~----·--·--------~----, ' '

' . \Circuits may become more elaborate by the placement of more t

!switches. For example,

A~~ ----= circuit 1

' !In order for a current to pass through circuit 1, both 'A'

and 'B' must be closed (for if either one is -ope-n.;--the

current stops at that point).

The following table demonstrates all possible combina­

!tions between switches A and B and their effect on circuit 1

("C" indicates closed and "0" d~notes opened).

·effect on A B Circuit 1

c c c c 0 0 0 c 0 0 0 0

.This table is identical to table #2 (part I) when "T" is

substituted for "C" and "F" for "0". This is the funda-

mental application of Boolean algebra to switching circuits.

As would be expected, mathematically, circuit 1 is called

an "and" circuit, maintaining the continuity between its

truth table and table #2.

The following is an example of an "or" circuit:

..

I I

1..)

----------~----·--------------. -------··---------------------------------------~----

r-----'A

B .__ __ _ circuit 2

!This circuit is complete when either •A' or 'B' are closed ' and is logically represented by table #1.

In circuit 1, the switches are said to be "in series;"

and in circuit 2, they are called "in parallel".

As the circuits become more complicated, the need for

!propositional logic becomes greater. Truth tables allow us

:to readily see the outcome of all switch combinations, and i

!the Boolean algebra provides us with a method of manipulating l . /· 1expressions.which represent the relationships between the

switches. . The (Boolean) algebraic expression A /\B

represents circuit 1, and AVB represents cireuit 2.

Circuits inay contain s~d.tches which are placed both . '

:"in series" and "in parallel" as exemplified by circuit 3.

·such a circuit is called a series-parallel circuit. e.g.

r----~c

~---A'

.__ ___ B

circuit 3

By visually following the wiring, we can see that the circuit

14

would be completed when

( 1 ) 'A' and 'C' are c 1 os ed

or

(2) 'A'' is closed

or

(3) 'B' is closed.

This is represented algebraically by

(A 1\C) v A' v B

( 1 ) or (2) or (3)

Applying Theorem 5, we get (A/\C) V(A'VB) whose truth

table is

A B c (A/\C)V(A'VB)

T T T T T T F T T F T T T F F F F T T T F T F T F F T T F F F T

The only·cbmbination of flWi~ches which will not provide a

c~~pleted circuit is when 'A' is closed and all the other

switches are open.

Beginning again with the original expression for this

circuit, (AI\C)VA'V B,. we can find an equivalent expression:

(A 1\C )VA' VB = (A!\C)VA~ VB = Theorem 5

~' V(A!\C~ VB = Axiom 1

"[A'VA)!\(A' vc] vB = Axiom 3

(Y/\(A'VCJ VB = Axiom 4

(A'VC)VB = Axiom 2

15

A'VCVB Theorem 5

We have been able to reduce our original expression of four

terms to three terms, which means we now hr.:t.ve ?, simpler

circuit containing one less switch. Circuit 3 may.be

' replaced by

---+------' c

....__ __ ~- B

which is represented propositionally by

A' B c (A' VB)V C

T T T T T T F T T F T, T. T F F T

.F T T T F T F T F .F T T F F F F

We check equivalence between circuits by comparing their

truth tables. We can see, again, that the only time this

circuit is not completed is when switches A', B, and Care

all open.

C. E •. shannon developed a postulated system for

switching circuits. Shannon's system (actually presented

here in its duality):

II II • represents switches connected in series.

11 +11 represents switches connected in parallel.

10

,----~-----~-- -----,-----------------------------~-- --- ---·-· -------------~------------- ---------------- ----·- ----------------- ---------- ---- ---: - .

"0" represents an open switch and an open circuit.

"1" repiesents a closed switch and a closed circuit.

Axiom 1 a. 0 • 0 = 0

b •. 1 + 1 = 1

Axiom 2 a. 0 • 1 = 1 • 0 = 0

b. 1 + 0 = 0 + 1 = 1

Axiom 3 a. 1 • 1 = 1

b. 0 + 0 = 0

Axiom 4 "x" and "y" represent switches where at any time, X = 1 or X = 0 and y = 1 or y = o.

,Theorem 1 a. X • y = y • X

·b· X + y - y + X

Theorem 2 a. X • (y • z) = (x • y) • z

·p. X + (y + z) =' (x + y) + z

;Theorem 3 a. X • (y + z) = (x • y) + (x • z)

b •. X + (y • z) = (x + y) • (x + z)

~Theorem 4 a. 1 • X = X

b. 0 + X = X

Theorem 5 a. 0 .. X = 0

b. 1 + X = 1

Theorem 6 (when X = o, x' = 1 ; and when X = 1 ' x' = 0)

a. X • x' = 0

b. ·X + x' = 1

Theorem 7 a. O' = 1

b. 1 ' = 0

1._ _ _-_ _,_ ____ -------- .: ... : ...

' f

,..----·-·

i Theorem 8 I·

(x')' = x -~--·-~-·------·-·-·-·-·····-·-·----·--·----;

I

i :Presented here were only eight of the theorems comprising

this system. -

The following circuit diagrams are visual illustrations

!depicting Shannon's axioms and six of the theorems.

Axiom 1 a.

~~ -·~ b.

I L

l

• •

I I

i I i !Axiom 2 a. : i !

!

~ • --~-b.

,

• •

Axiom 3 a.

• • I •

b •

. '

18

19

r-------------·

'Theorem 1 a.· -------------------·-----------------· ------..,

l

b .. -

r-----x .----Y

.__ __ y

l Th~orem 2 a.

X

y

z

Theorem 3 a.

z

b.

20

r-----------····------·---------------------------·· !Theorem 4 a.

----------------, !

--~

• •

b.

---~

• •

lx 1

I I

Theorem 5 a. ,

-~--····------~lr-==-~~-

b.

•

Theorem 6 a.

-· --IIF-x -+---~

,----·-----------·--·-----·----··---------·---·-·----------·----·----·-------~

! b. . . i I . ;

I i 1 !

•

In order to mathematically justify Boolean algebra

:applications on circuit expressions, it is necessary for

1Huntington's postulated system for Boolean algebra to be

. .

:the same as Shannon's system for circuits. We will show i ithat they are, in fact, equivalent systems by proving that i

I .. !Shannon 1 s axioms can be deduced from Huntington's system;

; i 'and that Huntington 1 s axioms can be deduced from Shannon's I i jsystem. i I

I I ' !PART I: Huntington'. s symbols· are equivalent to Shannon 1 s

symbols.

•

The three series circuits of Axioms 1, 2, and 3 can be

;represented by the table

. rst 2nd switch switch circuit

Axiom 1a 0 0 0 Axiom 2a 0 1 0 Axiom 2a 1 0 0

·Axiom 2a 1 1 1

which corresponds to table #2 when true and false are

substituted for "1" and "0", respectively. Since Shannon

defined " " to indicate switches connected in series and

we have shown the equivalence between the tables for circuit

1 and the table above and the "and" table #2, we can replace

21

r---~~-------~--~~-~~~-~------ ------- ·-----------~--------------~~-----------·~·- --~ ---- -~--"'-------~ ------------- ---------------~ 1" •" with 11 1\ 11 throughout Shannon's system. ~ 1

I i I I i In the same manner, we can show that 11 +11 may be ; l ireplaced with "v:ll

1st 2nd switch switch circuit

Axiom 1b 1 1 1 Axiom 2b 1 0 1 Axiom 2b 0 1 1 Axiom 3b 0 0 0

which is equivalent to table #1.

We have already shown that cup and cap are equivalent I :to "V 11 and " 1\. 11 Therefore, Huntington's operations may ; ;

!also _be replaced with 11 V" and " 1\." i

-!

i :PART II: Specific sets for Huntington's 'A', 'B', and 'C. 1

~1 Let each set contain only one proposition: a statement I I !about the condition of a switch. The switch at any time may.

:be opened or closed.-;

:PART III: Huntington 1 s system yields Shannon is axioms.

Huntington's Theorem 2a: A/\A = A and 2b: A VA = A.:

'Letting 1 A' be open, we have o/\ 0 = 0 and 0 VO = 0 wt.ich are

equivalent to Shannon's Axioms 1a and 3b, respectively. If

'A 1 is closed, we have 1/\1 = 1 and 1 V1 = 1 which are

· equivalent to Shannon's Axioms 3a and 1 b, respectively.

It is necessary to understand the specific denotation

of set complementation on our new sets A, B, and C. Since

sets A, B, and C contain only one element, a proposition

about the condition of a switch, their negations can only

22

contain the other st4tement about the switch. Therefore,

when 'A' contains (the) "open" (stateme-nt), 'A" contains

"closed;" and when 'A' contains "closed," 'A" cont-ains

"open." Which leads us to Huntington's Axiom 4a: A/\A'

and 4b: AVA' = u. For 'A' "open," we have 01\1 which

equals "0" by truth table #2; and accordingly, for I A I

=

"closed," we have 1/\0 = o. Therefore, N = O. (We had not

given a complete list of theorems relative to set theory.

One theorem not listed concerns the transitive property

of equlaity on sets.)

And as expected: for 'A' "open," OV1 = 1 (by table

#f) and for 'A' "closed," 1 VO = 1. Now we have U = 1.

N

... - By replacing "N" and "U" with "0". and_ "1 , " respectively,

1and applying Axiom 1 (commutativity) we can obtain

Shannon's Axioms 2a and 2b from Huntington's Axioms 4a and

4b, respectively. I

Shannon's--Axiom 4 is equivalent to our previous

definition of specific sets A, B, and C.

PART IV: Shannon's system yields Huntington's axioms.

Shannon's Theorem 1ab is equivalent to Huntington's

Axiom 1ab by inspection.

Using our replacement for "N" and "U", we can see,

by inspection, that Shannon's Theorem 5ab is equivalent to

Huntington's Axiom 2ab.

Again, by inspection, Shannori's Theorem Jab is the

same as Huntington's Axiom Jab.

23

And, as. shown above, Shannon's Axiom 2ab is equivalent

to Huntington's Axiom 4ab.

Summary of parts III and IV:

Huntington's

Theorem 2a

Theorem 2b

Theorem 2a

·Theorem 2b

Axiom 4a

Axiom 4b

for 'A' "open"

for·'A' "closed"

Definition of sets A, B, and C

Axiom 1ab

Axiom 2ab

Axiom Jab

Axiom 4ab

·Shannon's

Axiom 1a

Axiom Jb

Axiom Ja

Axiom 1b

Axiom 2a

Axiom 2b

Axiom 4

Theorem 1ab

Theorem 5ab

Theorem Jab

Axiom 2ab

Boolean algebra can be applied to circuits other than

series and/or parallel circuits. The following is an

example of a "bridge circuit," so named because switch C

forms a bridge between series switches A and D and series

switches .B and E.

24

r--------------------------------------------·---·-----------···-. ----------'---------1

'"-_ _;__ ___ A

.___ ___ -~E circuit 4

Following all the paths which wouid yield a completed

!circuit, we can see that circuit 4 is represented by

:Axiom 1

(A/\D)V (B/\E)V(A/\C/\E)V(B/\C/\D) =

(A/\D) V (A/\C/\E) V (B/\E)V (B/\C/\D) =

~1\(DV(C/\E)~ V ~1\(EV(C/\D)] ' ·. i 'Axiom 3

;This last expression is represented by

D

.-----~..________,L.---~-· E

circuit 5

Obviously, since this circuit contains 3 more switches than

·circuit 4(switches C, E, and Dare duplicated), circuit 4

is the preferred circuit.

Series-parallel switches may also be represented by

bridge circuits:

25

~-------------------~;------------------------:---------------------------·-------1

D .___ __ _ ...._ __ B

......__ __ ~.....____ r-------E

.,.____ ~D i land represented algebraically by

l

!which is equivalent to i ! i !

(A/\D) V (A/\F /\E) V (B /\D)V (B/\F /\E) V ( C /\E) V ( C/\F 1\D)

;and again, represented by

t-----A

1----,---D

'------~-----'

.._ ____ ___, c

The applications of Boolean algebra on switching

circuits are varied and far reaching. The procedures for

minimizing circuits will be considered in the next part.

26

' ------------------~---------·------------------~----------,

I Not only. does Boolean algebra provide a mathematical i -

- idescription of switching circuits~ but it alio allows us . '

to construct original circuits based on certain criteria • • / ! '

Given a~roblem, we will first describe its truth

,table, giv~ it~ algebraic description, and then picture l

· ithe circuit.

j(1) A committee of three votes affirmative by pressing a 1

lbu:tton. Construct a switching circuit that will pass i !current only when the majority votes yes. i I 1-,

- iLet A, B, and_C stand for members and T and F denote the i [affirmative and negative vote, respectively. i

A B c I

I T T T T T T F T T F T T T F F F F T T T F T Ii'· F F F T F F F F F

(A/\ B) V (A/\ C) V (B/\ C) = (!/\ (BVC}] V (B/\C) Axiom 3

B

~- I ·c

~-------~~--------------- c

27

(2) Design a circuit so that a light may be independently

controlled by two wall switches sueh that the flicking of

one wall switch changes the condition of the light.

Let A and B denote wall switches and T and F denote switch

is up and down, respectively. T and F will also denote

light on and light off; respectively.

condition A B of li ht

F F F F T T T F T T T F

(AVB)/\ (A/\B) 1 =

(AVB)/\(A 1 VB 1 ) =

(A /\A I) v (B /\A I) v (A /\B I ) v (B /\B I) =

Nv.(B!\A 1 )V(A/\B 1 )VN -·

(B/\A 1 )V (A/\B 1 )

~A r-----

(3) A municipal board consists of the mayor, President of

the City Council, Comptroller, and three Borough Presidents.

The mayor has two-votes and all the others one vote. A

motion obtaining a majority passes except that any motion

opposed by all three Borough Presidents fails. Write a

28

··--~------------------------------.

:a;itching circuit which will indicate passage of a motion. ' . i i i

;Let M indicate mayor; P, p~esident; c, comptroller; and

1B1, B2, and BJ the three borough presidents. T and F denote I !affirmative and negative vote, respectively.

M M p c B1 B2 BJ majority

T T T T T T T T T T T T T T F T T T T T T F T T. T T T T T F F T . • •

* T T T T F F F F • • • F F F F F F F F

!

! :There are 64 permutations of T and F for the six elements I !of this table. The outcome is always T when four or more i ;entries in a row are T and always F when there are three or l i

!less .T entries per row. The exception is the row that is 1 ~-

starred (*). '

: [l.t/\ p/\ C) 1\ (B1 VB2-VB3 ~ V [M/\ (PVC)) /\(B1 V B2 VBJ] V

i [M/\B1) A (B2 VBJ] V (M/\B2/\Bj V [P!\..C /\B1) I) (B2 VBJ]

(P/\CI\.B2 AB3 ) V ~1 /\B2 /\B3 )/\ (PVCJ . . _

•·

29

-1----~1

r--B1

---r----B2

....____BJ

1----ir-----B2

.________,.. BJ

t-------~~BJ

t-----~~B1

1----~ ~~-BJ .....___----!

.--P

'----=~~ BJ ._--i

'----· c

30

MINHIIZAT ION

Part II has demonstrated the method for algebraically

representing a circuit and finding an equivalent one which

-appeared to b~ simpler than·the original. But how do we

know whether or not we have found the simplest circuit?

And, when we begin with more difficult circuits such as

(A 1\ B !\ C 1\ D) V (A!\ B 1\ C ' !\ D) V (A 1\ B '1\ C 1 1\ D) V (AI\ B ' 1\ C 1\D) V

(A /\B 1\ C' 1\ D) V (A' 1\ B 1\c' 1\ D' ) V (A' 1\ Bl\ C 1\D 1 ) V (A 1 1\ B' 1\ C '/\

D), how/where do we begin to ·apply our Boolean axioms and

theorems? This third part will illustrate methods for

recognizing and eliminating those switches which are

unnecessary.

I Disjunctive Normal Forms

The.primary foundation for finding the simplest

equivalent circuit is to express the original

circuit in terms of its "disjunctive normal form."

A disjunctive normal form is nothing more than an

algebraic expression of the circuit which meets

certain requirements.

1. Statement letters and their negations are

called literals. A, B, C, and A' are literals.

2. A literal, by itself or a conjunction ( 1\) ·of

two or more literals, where the conjunction contains

no more than one of a ~tatement letter (includes

negation), is called a fundamental conjunction.

A/\B', A/\B, A'/\ BI\C are fundamental conjunctions.

31

A/\B /\A, B VC, A/\ C/\B/\ C' are not fundamental

.conjunctions.

3. If all the literals of a conjunction are also

in another conjunction, the second is said to

include .the first. A/\B includes A.. A/\B'/\ C

includes A/\C. ·However, B/\C does not include

B'/\C.

4. A single fundamental conjunction or a disjunc­

tion ( V) of two or more fundamental conjunctions,

none of which include·s another, is said to be in

di.sjunctive normal form (abbreviated dnf). A,

BVB', AV(B'/\C), and (A/\B/\C)\/(A'I\B/\C) are in

dnf. B/\ B 1 , A/\ (B VC), and (A 1\B) 1\ (A 1\ B VC) are

not in dnf. ·

The;re are two specific forms of dnf's.

1. If a dnf contains a certain number of letters

· (not literals) and ea'ch disjunction contains all

the letters, the dnf is in full disjunctive normal

form. A' and (A/\B') V(A'/\B) and (A/\B./\C)V

(A'/\B/\C)V(A/\B'/\C) are in full disjunctive

normal form with respect to A, A and B, and A, B,

and c, respectively.

2. The s~cond form of a dnf is concerned with

comparing dnf's which are equivalent to a given

circuit. Given two dnf's, if the first has an

equal or less number of literals and an. equal or

less number of disjunctions than a second dnf,

32

the first dnf is simpler than the s~cond (one of

·the inequalities must be a strict inequality). If

an expression A, for a circuit, is in disjunctive

normal form and there are no other expressions in

disjunctive normal form for the circuit which is

simpler than A, then A is a minimal disjunctive

normal form for that circuit.

II Implicants

If a fundam~ntal conjunction within a dnf logically

implies the expression, it is an implicant of the

dnf.

· III Prime Implicants

A fundamental conjunction which does not contain

any implicants of the given dnf and which logically

implies the dnf, is a prime implicant of the dnf.

For example, as we will see later, A 1\C is a prime

implicant of (A/\B'/\C)V(A'/\B'/\G')V(A/\B/\C')V

(A/\B/\G). By constructing a truth table, we can

see that (A/\C)4(A/\B'/\C)VtA'I\B/\C')V(A/\.B/\C')

V(A 1\B/\ C).

A B C (A/\CH(A/\B'/\C)V(A'/\B'/\C')

T T T T T F T F T T }il F F T T F T F F F T F F F

IV Notation

T T T­T T T T T

. . .

33

To simplify our notation, we will omit conjunction

( 1\) and parenthesis. A 1\B and (A!\B) V (A' 1\ C)

will be written AB. and ABVA'C, respectively. It

is also conventional-not to differentiate between

conjunctions which only differ in the permutations

of literals. AB and B'CD are the same as BA and

CB'D, respectively. In the same manrier, we will

not differentiate between dnf's which differ only

in the commuting of the disjunctions. AB VCD is

the same as CD V AB.

By the definition of dnf and implicant, we can see

that ail conjunctions contained· in a dnf are implicants

of it. While the disjunction of all the prime implicants

of an expression is logically equivalent to the expression,

a prime implicant does not necessarily have to be a

conjunction within the original expression. -' .

Theorem PI1 : There is at least one minimal disjunctive

·normal form which corresponds to a disjunction of prime

iJI!plicants.

Proof: Assume there is a minimal dnf. Let c 1 be one of

the disjunctions of the dnf which is not a prime implicant.

But, c1 must be an implicant. By definition, there must

be some conjunction (:riot necessarily of the dnf) c 2 , which

is a prime implicant such that c 1 contains c 2 • By defini­

tion of contain, c 2 must have fewer literals. Since c2

also implies the dnf, it may be added to the original ·

34

expression. · By the Boolean algebra theorem on absorption,

c 1 may be removed from the dnf leaving an expression.with

the same number of terms but fewer literals.

Theorem PI 2 : No prime implicant of an expression

contains any letters not in the given. expression.

Proof: Let V denote a conjunction all of whose letters

are in t~e given dnf, A denote a conjunction all of whose

.letters are not in the dnf, andUA denote a prime implicant

of the expression. Any n.:.tuple which makes UAassume the

value of "1" will also make u-assume the value of "1."

Since the letters as~ociated with Ado not appear in the

expression, the size of the expression will be k-tuples

where k<n~. The value "of the expression is totally deter­

mined by k-tuples which only assign values to the letters

of V. Thus·, independent of the values assign.ed to the

letters ofA., .any set of n-tuples which makes V equal to

. .-"1" will make the expression equal to "1." That is, V

implies the expression. . Since uA contains U, vA cannot be

a prime implicant, which contradicts the hypotheses.

Theorem PI3 : The set of prime implicants of a given

expression is finite.

Proof: By Theorem PI2 , a prime implicant can only contain

the letters of the expression. Fur~hermore, an implicant

is a fundamental conjunction. Each letter may appear ~t

most once within the cbnjunction. Since there are only a

finite number of fundamental conjunctions that can be

con~tructed from a finite set of letters, the set of prime

. 35

implicants must be finite.

The set of prime implicants of an expression can be

obtained by forming all possible conjunctions involving

the letters of the expression, testing to see if the con-

junctions imply the expression, and checking to see that

it does not contain some other conjunction which also

implies the expression.

Quine-HCCluskey Method for Finding Prime Implicants

Our first method for finding prime implicants of dnf's

is the Quine..:.HCCluskey method. By beginning with a full

dnf, we will compare pairs of disjunctions which are the

same for every literal except one. The one differing

literal must appear negated in one disjunction and unnegated

in the other. We form a new disjunction by writing the

conjunction of the literals appearing the same in both

disjunctions and omitting the differing literal. This

process is repeated, comparing all possible pairings, ~ntil

no two disjunctions can contribute to a new conjunction.

Tbe conjunctions (original and formed) which cannot compare

are our prime implicants.

For the full dnf ABC ' V AB 'C VA' BC VA' B' C VA' B' C' the ' . ' disjunctions are

ABC' AB'C~· ·B'C A'BC A'C A'B'C A'.B' A'B'C'

where A'BC and A'B'C appear the same for every literal-

36

except one: .B is negated in the second disjunction and

unnegated in the first. Therefore, A'BC and A'B'C yield a

new conjunction, A'C. Following the above diagram, we can

see that the. prime implicants are ABC', B'C, A'C, and A'B'.

A more extensive comparison can be seen for ABCD V .

ABC I D v AB I G I D v AB I CD v ABC I D I v A I BC I D I v A I BCD I v A I B I c I D

where the prime implicants are A'BD', B'C'D, ABC', BC'D',

and AD.

ABCD ABC'D AB'C'D AB'CD ABC'D' A'BC'D' A'BCD' A'B'C'D

ABD ·

AC'~-AB'D ACD AD A' BD' .· :AB-B'C'D ABC' BC'D'

All fundamental conjunctions in the dnf imply the dnf.

By· observation, we can see that the application of the

Quine-HCCluskey method to the fundamental conjunctions

yields another fundamental co.njunction which also implies

the dnf. Every prime implicant of the dnf will appear

and be unable to pair with any other fundamental conjunction

of the dnf (because all possible conjunctions which are

contained in other conjunctions of the dnf will eventually

appear) •.. Also, no non-prime implicant will be left

unpaired for every non-prime implicant must include a

fundamental conjunction which is a prime implicant (see

proof of Theorem PI1 ).

The final step·in our process is to form the minimal

dnf from the prime implicants of an expression. What we

37

are ultimately trying to do is to form a disjunction of

prime implicants where every disjunction of the full.dnf

includes a disjunction of the prime.implicants.

A method for selecting the prime implicant is called

a prime implicant table. Let's begin with a relatively

simple expression, a full dnf, AB'C!\A'BC'/\ABC'!\ABC

AB'C} A'BC' ABC' ABC

AC · BC'

AB

whose prime implicants are AC, BC', and AB. The prime

implicant table is formed by constructing a matrix with

columns headed by the di~junctions of the full dnf and

wl.th rows of the prime implicants.

AB'C A'BC' ABC' ABC

AC X X

BC' X X

AB X X

Crosses (x) are placed at the intersections of rows and

columns where the prime implicant is contained in a

disjunction. We now note that columns AB'C and A'BC'

contain only one cross each. The corresponding prime

implicants, AC and BC', will be two of the disjunctions of

our minimal dn:f.

AB'C A'BC' ABC' ABC

~1 ~ X

® X I

AB l X X

38

Since AC and BC' are part of our answer, all other crosses

corresponding to them may be eliminated:

AB'C A'BC' . ABC' ABC

@_ X ---X-

@ X .,X..

AB X X

For every column which has an eliminated cross in it, we

can eliminate the. entire column:

AB'C A'BC'

AC X

BC' X

AB

Now there are no crosses corresponding to AB so that AB is

eliminated and our minimal dnf is AC VBC'.

Proof that AB' C VA 'BC' VABC' V ABC ~(----+)AB VBC':

A B c v ACVBC' AB' CVA'BC' ••• ~AC VBC'

T T T T T T T T F T T T T F T T T T T F F F F T F T T F F T F T F T T T F F T F F T F F F F F T

Not all prime implicant tables are this simple. Many

need a more exterisive operational process. The prime

implicants of ABCDVAB'CDVA'BCD' VAB 1 C1 D1 VA 1 BC'D 1 VABCD 1V

AB 1 CD 1 V AB 1 C 1 D VA 1 BCD VA' BC 'DV A 'B' C 1 D' are B' C 1 D' , A' C 'D' ,

AC, BC, AB', and A'B.

39

ABCD AR'CD A'BCD'' AB'C'D A'BC'D' ABCD' AB'CD' AB'C'D A'BCD A'BC'D A'B'C'D'

ACD ABC BCD AB'C AB'D A'BD' BCD'

. A'BC AB'D' AB'G'

I B t c 'D •I A' BC' '

IA'C'D'I ACD' A'BD

AC 4G­BC ~ AB'

-ABL A'B ~

The table appears as

B'C'D'

A' C 'D'

AC

BC

@)

8

B'C'D'

A'C'D'

AC

BC

ABCD Al 1 CD A 1 lCD' AB 1 ( 'D 1 A' B J 'D' ABCD' AB CD'

~

~

x. .} X

X X

-:l ... -:i L- -:r-

..,; - -:i!fO-

AB'C'D A' CD A'BC'D A'B'C'D'

X

X

As before, (1) we have placed crosses where the prime

implicants are contained in the disjunctions. (2) we

40

circled single column crosses and considered their corres­

ponding prime. implicants AB 1 and A1 B part of our answer.

· {3) We eliminated all other crosses corresponding to AB 1

and A 1 B. { 4) 1ve eliminated all columns containing an

eliminated cross. However, we still have left the four

other prime implicants to consider because they contain

'uneliminated crosses. Our new table is

ABCD ABCD' A'B'C'D 1

B'C'D' X

A'C'D' X

AC X X

BC X X

Because columns ABCD and ABCD' are identical, we may

eliminat• one of them:

ABCD A'B'C'D'.

. B'C'D' X

A'C'D' X

AC X

BC " X

Since it is necessary that each column include some disjunc­

tion o~ the minimal dnf, A'B'C'D' would be satisfied by

choosing either B'CD' or A'C'D': and ABCD would be satisfied

by choosing either AC or BC. The~efore, our minimal dnf

can be any one of the following:

AB' v A' B v B' CD I v AC

AB' v A' B v B' CD I v BC

41

AB 1 VA' B VA' C' D1 V AC

AB 1 VA'BVA 'G'D' VBC

Concensus Method £or Finding Prime Implicants

A second method for finding prime implicants is called

the Concensus method. Like the Quina-MCCluskey method, we

will look for exactly_ one letter which is negated in one

disjunction and unnegated in another. However, this time

we are not concerned with whether or not the rest of the two

disjunctions are similar. For this reason, we do not have

to begin our process with a full dnf.

A concensus of two conjunctions is formed when we

omit the negated-unnegated literal and form a new conjunc­

tion of all the remaining literals without repeating any

literal. For example, the concensus of ABD and AB'C is

ACD. The concensus of AB and AB' is A. There is no

concensus for A'BC and AB'C. -

Theorem C =··· The concensus, A, of !/{ and ljl2 logically

implies Vf.t V ~.

Proof: Consider any truth assignment making A true.· LetT

be the letter occuring negated in 1jl,. and unnegated :i..n 1/12 •

If Tis true, then t/J2 is true. If Tis false, thm!/1 is

true. In either case, lJ'1 V ~ is true.

Corollary. C: If A is the concensus of 1./i and ~,

then iflt V l/!2 is logically equivalent to 1/J.t V l.j!2 v A-

Given a dnf, ABC' V AB 1 CD' V AB' \/ A' BC ' VA 1 B' C 'D, we

find all the prime implicants by (i) eliminating any

42

disjunctions which includes another disjunction and then

(ii) add, as a disjunction, the concensus of any two of

the conjunctions. We will continue to repeat the process

until no conjunctions can be eliminated and no consensus can

be formed.

ABC I v .A,B I CD I v AB' v A I BC' v A' B 'c I D

(AB 1 is included in AB 1 CD') ABC 1 V AB' V A' BC ' \1 A' B ' C ' D

.(ABC' and AB' form AC') ABC' v AB' v A I BC I v A' B' c 'D v AC'

(AC' is included in ABC') AB I v A I BC' v A' B I c 'D v AC'

(A 1 BC 1 and A1 B'C'D form A'C'D) AB 1 V A' BC 1 V A 1 B' C ' D V AC ' V A' C 1 D

(A 1 C1 D is included in A'B'C'D) AB ' v A I BC I v AC I \j A I c ' D .

(AC 1 and A1 C1 D form C'D) AB I VA I BC I v AC' VA' c 'D vc 'D

(C 1 D is included in A1 C'D) AB I v A I BC ' v AC ' v c I D

(A 1 BC 1 and AC 1 form BC') AB' v·A 1 BC' vAc 1 vc 'nvBc'

(BC' is included in A'BC') AB'~ AC', C'D, and BC' are the prime implicants

Justification of the Concensus Hethod

(1) The process must come to an end. Since there are

a finite number of dnf's using the letters of the given

expression, we must show that there can be no cycles in the

application of our process. Once we drop a fundamental

conjunction by (i), then it can never reappear because of

(ii). For, in all future steps, there will always be a

43

fundamental conjunction which is included in the expression.

If there were a cycle, it would consist only of the

applications of ( ii). But· ( ii) increases the number of ·

disjunctions.

(2) Every prime implicant of the expression occurs

as a disjunction in the dnf remaining at the end of the

process. Assume the contrary. There must be a fundamental

conjunction Awhich has the maximum number of literals

among all fundamental disjunctions T such that (a) A

includes T, (b) Tincludes no disjunction of the dnf, and

(c) the letters of T occur in the expression. Notice that

a prime implicant is such a fundamental conjunction. By

(a), A logically imples the expression.. Also, i\. cannot

contain all the letters of the expression. (Otherwise, i~

by (b), Awould logically imply the negation of each

disjunction of the dnf, and therefore would logically imply

"not" the expression. But only contradictions logically

im:ply both the expression and "not" the expression, and no

fundamental conjunction is a contradiction). Let 'A' be a

letter of the e'xpression not in A. By the maximali ty of

A, AA.~nd A~ must lack one of the properth:s (a)-(c).

The only one they can lack is (b). There are disjunctions

IJ!1 and ljf2 of the dnf such that AA includes t/J.t and A •i\.

includes lj;2 • · By property (b) of A, 'A' must be a literal

of l/J1 and A •A must be a literal of !j;2 • Since A;\. includes

(/;; and A 'A includes ~, tj;1 and lj;2 do not h~ve any other

literals which are negations of each other. Then the

44

consensus of ~ and ~ is included in \, and therefore, by

{b), includes no disjunctions of the dnf. An application

of (ii) can be made to tfi.t and tJ!2, contradicting the

assumption that the process has ended.

(3) Every disjunction of the dnf remaining at the

end of the process must be a prime implicant of the expres­

sion, otherwise, the disjunctions would include some prime

implicant. By (2), the prime implicant would be a disjunc­

tion of the final dnf, and operation (i) would still be

applicable.

Again, we must form our minimal dnf from prime

implicants. However, we cannot use prime implicant tables

because we do not have full dnf's to work with. We must

find another method.

Given a dnf, if there is a fundamental conjunction

within the dnf that- can be "dropped," without changing the.

logic value of the dnf, the fundamental conjunction is

called superfluous. In much the same manner, if a literal

can be dropped from a fundamental conjunction and the dnf

containing the fundamental conjunction retains its logic

value, the literal is superfluous in the dnf. A dnf which

contains no superfluous disjunctions or literals is irredun-

~· Our objective is to remove all superfluous literals

and disjunctions from our prime implicants, thus obtaining

an irredundant dnf. 'fe will then compare dnf' s and choose

a minimal one.

45

The following process is attributed to M. J. Ghazala

and begins with a prime implicant table with column and

row headings of the prime implicants:

AB' AC' C'D BC'

( 1 ) AB' X C' C'D F

(2) AC' B' X D B

(3) C'D AB' A X B

(4) BC' F A D X

We do not compare identical prime implicants, hence, the

crosses. Assume the row heading conjunctions are true

(e.g. AB' = T .·.A= T and B' = T). We examine the truth

value of the corresponding columns. If the corresponding

conjunctions are false, we indicate so with au "F.n If

the values are undetermined because of a statement letter

which does not appear in the .row heading, we indicate the

literal. We then f~rm a disj~nction of the results for

each ~ow: (1) ... C'VC'DVF, (2) B'\)DVB, (3) AB'VAVB, .i

·and (4) FVAVD. _These disjunctions provide us w·ith the

outcome based upon the given value of the implicant. We can

see that for (2), B'VDVB, it will always be true. Since

it will always be true, regardless of the values assigned

to·the res~ of the switches, it has no effect on the logic

value of the dnf and· is superfluous. Any superfluous con-

junction may be dropped. 1ve cannot determine the value of

the other prime implicants, their value depends upon the

independent conditions of each switch, the given value of

the implicant was insufficient. Hence, they effect the

46

logic value of the circuit and may not be eliminated. The

minimal dnf is AB ' V d 1 D V BC ' •

Karnaugh Naps for Finding Prime Implicants·

Our third and final method of obtaining minimal dnf's

is a pictorial process, limited to six statement letters,

called Karnaugh Naps.· These ''maps" are grid patterns where

the column and row headings are the literals of each con-

junction. Conjunctions AB, A'B, AB', and A'B' are in

A A'

: • /~----· -----t---1

However, if we have specifically AB VA 1 B 1 , we represent each

disjunction with a check (J) in the appropriate, corres­

ponding square:

B

B'

A A' J

J figure 1

For a three-letter statement, ABCVA'B'C', we use a two-

by-four grid: AB AB' A'B' A'B

c I J I I I C' J figure 2

It is necessary to express disjunctions in terms of all

possible permutations of the letters. Also, they must be

shown in such an ord.er that each column differs by only one

literal from its adjacent columns: AB, AB', A'B', A'B is an

acceptable order while AB, A'B', A'B, AB' is not.

We use a four-by-four grid to illustrate four-letter

statements: ABCD'VABC'DVA'BC'D'

47

CD

CD' - C'D'

C'D

AB AB' A'B' A'B

.J J

~ . f~gure. 3

Notice that all permutations of C, D, C', and D' must also

be stated and in an order that allows only one change.of a

literal at a time~

Five-letter statements are illustrated on two, four-

by-four planes: ABCDE VAB'CD'E vABC'D'E'

E

E'

1 And, as expected, a six-letter statement is expressed I iby four; four-by-four planes and each plane is one of the

'·.

-~permutations of E, F, E', and F' which change one literal !

l at-~~~~~~~BC'DEF' v AB'CDE'F' v ABCD'E'F _ _j

48

49

EF

-.::,

EF'

E 1F'

E1 F

Three dimensions limits us to six statement letter conjunc-

tions.

There is only one more aspect of these maps to under-

stand. Adjacent squares are squares which share a common

side. In figure 1, there are four pairs of adjacent -.

J squares : AB & AB 1 , A' B & A' B ' , AB & A' B, and AB 1 & A' B 1 • 1

lrn figure 2, there are not only the ten pairs which are I l~~j a.<: ~n~ ___ !!! ___ ~~~---~!l_ll1e _ _!1!~~-~_r a~_ f :i,gll,:t;"_~_l_,_ _ _Q1J.j;_ th_~re._ar.e_al s_oj

two more pairs which are adjacent. 'fuen the grid is

curved into ~ cylinder so that the two heavy lines meet,

A'BC and A'BC' bec~me adjacent to ABC and ABC', respective-

ly.

In figure 3, not only is the grid curved into a

cylinder (heavy lines meeting), but the top bends back

·down and the bottom bends back up so that the top row of

squares become adjacent to the bottom row of squares.

Figures 4 and 5 still have one more aspect to consider '

~djacent. C~rresponding (parallel) squares in differnet

planes are adjacent.

Now that we can illustrate our full dnf's, we must

realize minimal dnf's from these maps. ·The purpose for

showing all permutations and in changing, one-literal only ..

order is so_that when adjacent squares are checked, we may.

eliminate the letter which differs in the two squares

leaving us a conjunction which will be one of the disjune-

tions of our minim~! dnf.

(1) Two Statement Letters

A pair -of adjacent checks will yield a one­

letter statement~

B.

. B'

AB'V A'BV A'B'

~ ~ =A'VB'

figure 6

50

..

(2) Three Statement Letters

statement.

statement.

A pair of adjacent checks will yield a two-letter

c C'

ABC v A'BC

AB AB' A1B' A'B

f [)I = BC

Four adjacent checks will yield a one-letter

c C'

ABC v AB 'c v AB' c I v ABC I

AB AB' At B I A I B

t } I ~ I I I =A

ABC I v AB I c t v A ' B ' c t v A t BC t

(3) Four Statement Letters

A pair of adjacent checks will yield a three­

letter statement.

CD

CD'

C'D'

C 1 D

AB I CD v AB' c ' D

= AB'D

51

statement.

sta:t.ement.

Four adjacent checks will yield a two-letter

CD CD' C'D'

.. C'D

ABCD V A'BCD v ABC'D v A'BC'D

AB AB' A I B I A' B JJ IC J

= BD

J' (J

Eight adjacent checks will yield a one-letter

AB'CDVA'B'CDVAB'CD'vA'B'CD'VAB'C'D'vA'B'C'D'VAB'C'DVA'B'C'D

CD CD'.

C'D' C'D

AB AB' J J J J

A'B' A'_B_ J J =B

J J

(4) Five Statement.Letters

statement.

statement.

, statement. I

A pair of adjacent checks yield a four-letter

Four adjacent checks yield a three-letter

Eight adjacent checks yield a two-letter

I Sixteen adjacent checks yield a one-letter I· ! statement.

l J

52

( 5) Six Statemen,t Letters

A pair of adjacent checks yield a five-letter

statement.

Four adjacent checks yield a four-letter

statement.

Eight adjacent checks yield a three-letter

statement.

Sixteen adjacent checks yield a two-letter

statement.

Thirty-two adjacent checks yield a one-letter

statement. . a

When a picture of a full dnf does not have all its

checks adjacent, we associate, by circling, checks which

are adjacent, apply one of our techniques listed above,

include solitaire checks, and find our minimal dnf.

For our final example of this process, we will picture

. the full dnf AB' C v A' BC' v ABC 1 v ABC to which we have al-

ready applied the Quina-MCCluskey method.

c C'

AB AB' A'B' A'B

·1s;:p> I I u I = AC vBC'

If we had chosen to circle ABC with ABC', we would have

' t

I

I

i

'been left with two uncircled checks. The success of our

iprocedure depends upon finding an arrangement of circles

!which yields the greatest number of associations. A check

~a:r: __ ~~---~irc_!_~~ ___ f:r!ore t_~~:r!_on_c_~_i~ order to ~-~~-~ci_!!.:!i~~_!l ____ j

53

uncircled check. But, a check may not be used more than

once if all checks in a circle have been used before. In

the above grid, ABC iL not assoicated with ABC' because

both checks have already been circled.

Justification for Using Karnaugh Haps for Finding

Prime Implicants

Theorem K: The set of all prime implicants of an

expression can be obtained from a Karnaugh map.

Proof: By construction. .Given an n-literal map, if all

2n entries are checked, then the prime implicant is 1 (the

dnf has a value 1 regardless of the values assigned to its

letters). If all 2n entries are not checked, determine all

rectangular groupings of checked cells with diminsions

2a x 2b = 2n-1 (each of these groupings represent a one­

letter conjJ].n~tion). Since no two different groupings can ·

represent the same conjunction, and since all cells in the

grouping are checked cells, they must describe prime

implicants. Next, Form.all rectangular groupings of checked

cells with dimensions 2a x 2b = 2n-i for i = 2, such that

no grouping is totally within a single previously formed . . .

grouping. Each of these groupings represents an i-letter

(fori =·1) conjunction which imples the dnf. Since no

grouping is totally_withina single previouslyformed

grouping, its associated conjunction will not contain any

previously found conjunction and hence be a prime implicant.

Repeat thi~ process fori= 3, 4, 5, ••• ,n. Th~ set of all

conjunctions formed are prime implicants. These are all the

54

prime implicants since any other grouping of checked cells

must be totally contained within at least one of the

groupings.

..

55

THE COHPUTER

Switching circuits are the essence of computer opera­

tiqn. The repeated and extended applications presented in

part III contribute to the development of computer ~ystems.

The system begins with an electrical source which,

when instructed to do so, feeds an electroma.gnet, thereby

magnetizing it. The magnet then attracts the switch "gate"

pulling it towards itself and closing the circuit.

'· .. electrical 1 ~

source ~----------~~--~

data input

Switches which are connected in series are referred

to as and gates in computer circuitry and are represented

by~ in circuit diagrams. However, [I] is a simplified

notation for

~lect.ricalr---____.J~ ~ ~ · source _

data input~--------------~------~

Switches connected in parallel are- referred to as 2!.

gates and are represented by Q. . Q_;s a simplified

notation for

56

electrical source

data input

(Half circles in the wiring indicate wires which.cross over

each other but do not connect.)

"Not" switches, · ~· , are also verY important for

they may be placed anywhere in a circuit to change the

condition of the circuit at that point.

---7) X

electrical source

When a current has reached the point indicated by the "X",

it not only tries t6 continue through the switch, but it

57

magnitizes the electromagnet and opens the switch so that

·it cannot. pass. When there is no .current coming in

through "X," the electrical source supplies the.current

without contacting the electromagnet.

Because. computer circuits use the~' ~' and~ notation, the diagrams are referred to as logic circuits.

Of course, it must be realized that the acttial electronic

and mechanical devices such as diodes, transistors, vacuum

tubes, etc., vary with the state of technology. We are

not concerned with the hardware, only the systems they

comprise.

For an example of a logic circuit,- circuit 3 (part II)

provides us a look at series, parallel, and "not" switches.

Reconstructed as a logic circuit, it appears as

A

B

c

A forerUnner of today's computer is the Jacquard loom

used in the production of fabrics having complicated

patterns. Jacquard's loom, produced in 1804, was in wide

use in France and England by the 1820's. The loom was .·

constructed in a criss-cross of vertical, long tailed -

58

hooks and horizontal bars. The spring anchored bars lifted I just before a. vertical shuttle passed. The rising and J falling of a hook would cause the lifting and lowering ·of

the bars which, in turn, produced a pattern from the. thread .

being 'dropped' by the shuttle. But what caused the hooks

to lower and raise? The base of the books rested horizont- 1

ally on a block. A shallow-holed cylinder traveled the

length of the base while rolling, pushing the block out

of the way. As a hook end came in contac~ with a hole on

the drum, the hook fell and then rose as the drum passed.

To prevent.the hooks·from falling into every hole, cards

were selectively punched and then placed over the drum.

When a hook contacted a hole in the card, it fell. Yet

a hole in the drum covered by part of the card unholed,

prevented the hook from falling. Cards with different

hole arrangements made different parts of the pattern. The

cards were strung together so that they were automatically

feed onto the drum.

Card reading computers of today work on the same

principle. Cards are place<l in the data input section and ~-

either permit a current to make contact with a specific l . I

terminal· through a selectively punched hole or else prevent j

the contact with a nonpunched section of the card. There 1

are only two pieces of information going through input:

I :::::::)~ade, contact not made (hole punched, hole not

L Using a '1 ' and a '0' to designate contact to be made

59

60

and contact not to be made, respectively, we have· the

binary system. All numbers can be coded into the binary

system where place value increases by powers. of two.

0 0 0 0 0 0

1 0 0 0 0 1

2 0 0 0 1 0

3 0 0 0 1 1

4 0 0 1 0 0

5 0 0 1 0 1

6 0 0 1 1 0

7 0 0 1 1 1

8 0 1 0 0 0

9 0 1 0 0 1

Our basic logic circuit is constructed by adding all

possible combinations of two, one-digit numbers, A and B.

A 0 A 0 A 1 A 1

B +0 B +1 B +0 B +1

0 1 1 10

Converting these to a table where ' s ' stands for the sum

in the digits place and f c ' stands for the £.arry over from

the digits, we get

A B s c

0 0 0 0

0 1 1 0

1 0 1 0

1 1 0 1

Columns A, B, and c are a truth table·for "and," yielding

the following logic eircuit

A

B ------------~J __ &_r--~> c

However, columns A, B, and s are slightly mo·re complicated.

They appear similar to an "or" table except·for the last

entry. For this entry, if either A or B (exclusive or)

were negated, we'd have an "or" circuit. By including a

"not" gate and two "and" gates, we obtain

A

& &

B

s

By combining the circuits for c and s we have

A .·-----,--------~---------1

& & c

B

s .·

61

62

The combined circuit for c and s is called a half-adder.

A full adder is obtained by adding all possible combinations

of three, one digit numbers, A. '

B, and c. A B c s c

0 0 0 0 0

0 0 1 1 0

0 1 0 1 0

0 1 1 0 1

1 . b 0 1 0

1 0 1 0 1

1 1 0 0 1

1 1 1 1 1

A.~--.-----~--------------------~

B

c

&

c

Circuit.diagrams are now simplified by replacing all

occurrences of half-adders (HA) and full adders (FA) with

-----"~ HA I : : By adding two, three digit numbers, A1A2A3 and B1B2B3 , we

can construct the following logic circuit.

,I S1 , / /

HA \ ' S_2_ ~ ,

( FA (

' , / S1 \.

' FA /

(. c ' ' /

Computer logic circuits continue to build upon this basic

foundation.

63

BIBLIOGRAPHY

.Eves, Howard, and Carroll V. Newson: An Introduction to the.Fouridations and Fundamental Concepts of Hathe­matics, Holt, Rinehart, and Winston, New York, 1965.

Flegg, H. Graham: Boolean Algebra and its AEplication, John Wiley and Sons, Inc., New York, 19 4.

Ghazala, H. J.: "Irredundant Disjunctive and Conjunctive Normal Forms of a Boolean Function," IBM Journal of Research and Devlopment, Vol. 1, pgs. 171-176, 1957.

Givone, Donald D.: Introduction to Switching Circuit Theory, HCGraw-Hill Book Company, New York, 1970.

Hohn, Franz E.: Applied Boolean Algebra, The MacHillan · Co., New York, 1966~ .

King, Kenneth H., and Lelia D. Chance: "Yes, No ••• One, Zero," American Petroleum Institute, New York.

Mendelson, Elliott: Schaum's Outline Series: Theory and Problems of Boolean Algebra and Switching Circuits, MCGraw-Hill Book Company, New York; 1970.

Shannon, C. E.: "ASyffibolic Analysis of Relay and Switch­ing Circuits," American Institute of Electrical Engineers, Vol •. 57, pgs. 713-723, 1938.

Stoll, Robert T.: Set Theory and Logic, W. H. Freeman and Company, San Francisco, 1963.

Wilkes, M., V.: Automatic Digital Computers, John Wiley and Sons, Inc., New York, 1957.

Williams, Gerald E.: Boolean Algebra with Computer Appli­cations, HCGraw-Hill Book Company, New York, 1970.

Fraenkel, A. A.: Encyclopedia Britannica, William Benton, Chicago~ Vol. 20, pgs. 265-267, 1967.

Harley, R.: Encyclopedia Britannica, William Benton, Chicago, Vol. 3, pgs. 944-945, 1967.

Rescher, Hicholas: Encyclopedia Britannica, William Benton, Chicago, Vol. 14, pgs. 209-237, 1967.

64