california state university, northridge boolean algebra
TRANSCRIPT
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
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
,-----------------------·-··-----------·--------·~--------- - -·-----· ---------------------------------------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