Download CPSC 2105 Lecture 6 - Edward Bosworth, Ph.D.

Boolean Algebra & Digital Logic
Boolean algebra was developed by the Englishman George Boole, who published the
basic principles in the 1854 treatise An Investigation of the Laws of Thought on Which to
Found the Mathematical Theories of Logic and Probabilities.
The applicability to computing machines was discovered by three Americans
Claude Shannon
Symbolic Analysis of Relay and Switching Circuits, 1938.
George Stibitz
An employee of Bell Labs, he developed a binary adder
using mechanical relays in 1937, the model “K 1” adder
because he built it at home on his kitchen table.
John Atanasoff
He was probably the first to use purely electronic relays
(vacuum tubes) to build a binary adder.
Boolean algebra is a two–valued algebra based on the constant values denoted as either
FALSE, TRUE
0, 1
The use of this algebra for computation is based on the fact that binary arithmetic
is based on two values, always called “0” and “1”.
Basic Boolean Operators
Boolean algebra is defined in terms of two constants (defined above), which we
call “0” and “1”. Other courses will call these values “F” and “T”.
Boolean algebra is defined in terms of three basic operators, to which we shall add
a useful fourth operator. The three operators are NOT, AND, & OR.
Each of these three basic operators is implemented by a basic electronic device
called a “logic gate”. We present the gates along with the definition.
NOT
This function takes one input and produces one output. The gate is shown
below. The circle at the right end of the triangle is important.
Algebraically, this function is denoted f(X) = X’ or f(X) = X .
The notation X’ is done for typesetting convenience only; the notation
The evaluation of the function is simple:
0 = 1 and 1 = 0.
X is better.
Basic Boolean Operators (Part 2)
Logic OR
This is a function of two Boolean variables. We denote the logical OR of two Boolean
variables X and Y by “X + Y”. Some logic books will use “X  Y”.
The evaluation of the logical OR function is shown by a truth table
X
0
0
1
1
Y
0
1
0
1
X+Y
0
1
1
1
Basic Boolean Operators (Part 3)
Logic AND
This is a function of two Boolean variables. We denote the logical AND of two Boolean
variables X and Y by “X  Y”. Some logic books will use “X  Y”.
The evaluation of the logical AND function is shown by a truth table
X
0
0
1
1
Y
0
1
0
1
XY
0
0
0
1
Another Boolean Operator
While not a basic Boolean operator, the exclusive OR is very handy.
Logic XOR
This is a function of two Boolean variables. We denote the logical XOR of two Boolean
variables X and Y by “X  Y”. Most logic books seem to ignore this function.
The evaluation of the logical XOR function is shown by a truth table
X
Y
0
0
0
1
1
0
1
1
From this last table, we see immediately that
X  0 = X and X  1 = X
XY
0
1
1
0
Bitwise Operations on Binary Integers
These operations are defined for Boolean values, but can be extended to integer
values viewed as binary sequences
X Y XY
X Y X+Y
X Y XY
0
0
0
0
0
0
0
0
0
0
1
1
0
1
1
0
1
0
1
0
1
1
0
1
1
0
0
1
1
0
1
1
1
1
1
1
As binary integer examples, I propose to use the ASCII codes for alphabetic characters.
‘A’
Upper case ASCII 0100 0001
Pattern for OR
0010 0000
This gives lower case 0110 0001
‘N’
0100 1110
0010 0000
0110 1110
Lower case ASCII 0110 0001
0110 1110
Pattern for AND 1101 1111
1101 1111
Upper case
0100 0001
0100 1110
More Boolean Operations on Integers
Consider the 8–bit binary number
0110 1001
The logical NOT of this number is
1001 0110.
The second value is the one’s–complement of the first value.
We could show more examples. Here are a few with 8–bit binary numbers.
0101 1000
+ 0011 0100
0111 1100
0101 1000
 0011 0100
0001 0000
In programming languages, these set or clear individual bits in a binary value.
One might write the following code in some variant of Jave
lower_case = upper_case | 0x20
The lower case representation of a character is the logical OR of the upper case
representation and binary 0010 0000 (0x20).
Truth Tables
The fact that any Boolean variable can assume only one of two possibly values can be
shown, by induction, to imply the following.
For N > 0, N Boolean variables can take only 2N different combinations of values.
For small values of N, we can use this to specify a function using a truth table with 2N
rows, plus a header row to label the variables and the function.
N
2N
1
2
2
4
3
8
4
16
5
32
6
64
Four–variable truth tables have 17 rows total. This is just manageable.
Five–variable truth tables have 33 rows total. This is excessive.
N–variable truth tables, for N > 5, are almost useless.
Sample Truth Table
A
B
C
F1(A, B, C)
0
0
0
0
0
0
1
1
0
1
0
1
0
1
1
0
1
0
0
1
1
0
1
0
1
1
0
0
1
1
1
1
This truth table for 3 variables has 23 = 8 rows, plus a label row.
This truth table forms a complete definition of the function. We shall later
give it another name, but can base all our discussions on this table.
Another Sample Truth Table
A
B
C
F2(A, B, C)
0
0
0
0
0
0
1
0
0
1
0
0
0
1
1
1
1
0
0
0
1
0
1
1
1
1
0
1
1
1
1
1
Two Truth Tables in One
A
B
C
F1(A, B, C)
F2(A, B, C)
0
0
0
0
0
0
0
1
1
0
0
1
0
1
0
0
1
1
0
1
1
0
0
1
0
1
0
1
0
1
1
1
0
0
1
1
1
1
1
1
Truth tables are often used to show pairs of functions, such as these two,
which will later be shown to be related. This is easier than two complete tables.
Truth tables rarely show more than two functions, just because large truth
tables are “messy” and hard to read.
Labeling Rows in a Truth Table
The row numbers are just labels. They are not really a part of the truth table, but
aid in our discussions and conversions to Boolean expressions.
The row numbers are the decimal equivalents of the variable values viewed as binary
numbers. The first row is always “row 0”.
Row
Number
X
Y
Z
F(X, Y, Z)
0
0
0
0
1
0 = 04 + 02 + 01
1
0
0
1
1
1 = 04 + 02 + 11
2
0
1
0
0
2 = 04 + 12 + 01
3
0
1
1
1
3 = 04 + 12 + 11
4
1
0
0
1
4 = 14 + 02 + 01
5
1
0
1
0
5 = 14 + 02 + 11
6
1
1
0
1
6 = 14 + 12 + 01
7
1
1
1
1
7 = 14 + 12 + 11
The  and  Notations
These can be viewed as shorthand notation for expressing truth tables.
 notation:
Give the row numbers in which the function has value 1.
 notation:
Give the row numbers in which the function has value 0.
Example: The Exclusive OR (XOR) function
Row Number
0
1
2
3
X  Y =  ( 1, 2 )
X
0
0
1
1
Y
0
1
0
1
XY
0
1
1
0
X  Y =  ( 0, 3 )
Exercise:
Convert F(X, Y, Z) =  ( 0, 2, 4, 6 ) to the  notation.
Answer:
The function has three variables, so the truth table has 23 = 8 rows,
numbered 0 through 7. If rows 0, 2, 4, and 6 have ones, then the rows
containing zeroes must be 1, 3, 5, and 7. F(X, Y, Z) =  ( 1, 3, 5, 7 ).
Examples: F1 and F2
Consider the two functions (F1 and F2), which we shall explain later.
Row
A
B
C
F1(A, B, C)
F2(A, B, C)
0
0
0
0
0
0
1
0
0
1
1
0
2
0
1
0
1
0
3
0
1
1
0
1
4
1
0
0
1
0
5
1
0
1
0
1
6
1
1
0
0
1
7
1
1
1
1
1
F1(X, Y, Z) =  ( 1, 2, 4, 7 ) =  ( 0, 3, 5, 6 )
F2(X, Y, Z) =  ( 3, 5, 6, 7 ) =  ( 0, 1, 2, 4 )
Evaluation of Boolean Expressions
The relative precedence of the operators is:
1) NOT
do this first
2) AND
3) OR
do this last
As in the usual algebra, parentheses take precedence.
AB + CD, often written as AB + CD, is read as (AB) + (CD)
A  B  C  D is read as A  B   C  D . The latter is really messy.
AB + CD = 10 + 11 = 0 + 1 = 1
A(B + C)D = 1(0 + 1)1 = 1  1  1 = 1
A  B  C  D = 1 0  11 = 0  0 + 1  0 = 0 + 0 = 0
A B = 1 0 = 0 = 1
A  B = 1 0 = 0  1 = 0
Question: Where to Put the 0’s and 1’s
We now ask how to fill a truth table by evaluating a Boolean expression.
Example: F(X, Y, Z) = X  Y  Y  Z  X  Y  Z
Step 1:
There are three Boolean variables, so the truth table will have 8 rows.
Let’s evaluate this function for all eight possible values of X, Y, Z.
Row 0 X = 0
Y=0
Z=0
F(X, Y, Z) = 00 + 00 + 010 = 0 + 0 + 0 = 0
Row 1 X = 0
Y=0
Z=1
F(X, Y, Z) = 00 + 01 + 011 = 0 + 0 + 0 = 0
Row 2 X = 0
Y=1
Z=0
F(X, Y, Z) = 01 + 10 + 000 = 0 + 0 + 0 = 0
Row 3 X = 0
Y=1
Z=1
F(X, Y, Z) = 01 + 11 + 001 = 0 + 1 + 0 = 1
Row 4 X = 1
Y=0
Z=0
F(X, Y, Z) = 10 + 00 + 110 = 0 + 0 + 0 = 0
Row 5 X = 1
Y=0
Z=1
F(X, Y, Z) = 10 + 01 + 111 = 0 + 0 + 1 = 1
Row 6 X = 1
Y=1
Z=0
F(X, Y, Z) = 11 + 10 + 100 = 1 + 0 + 0 = 1
Row 7 X = 1
Y=1
Z=1
F(X, Y, Z) = 11 + 11 + 101 = 1 + 1 + 0 = 1
This function can be written as F(X, Y, Z)
= (3, 5, 6, 7)
where the 1’s are
= (0, 1, 2, 4)
where the 0’s are
Some Basic Identities of Boolean Algebra
Here are a few identities basic to Boolean algebra, presented in two forms: the
“AND form” and the “OR form”. Some resemble standard algebra and some do not.
Identity Name
Identity Law
Null (Dominance) Law
Idempotent Law
Inverse Law
Commutative Law
Associative Law
Distributive Law
Absorption Law
DeMorgan’s Law
AND form
1X = X
0X = 0
XX = X
XX’ = 0
XY = YX
X(YZ) = (XY)Z
X + (YZ) = (X+Y)(X+Z)
X(X + Y) = X
(XY)’ = X’ + Y’
OR form
0+X = X
1+X = 1
X+X = X
X+X’ = 1
X+Y = Y+X
X+(Y + Z) = (X + Y)+Z
X(Y + Z) = (XY) + (XZ)
X+(XY) = X
(X + Y)’ = X’Y’
It is due to the associative law, one can unambiguously write expressions such as
XYZ and X + Y + Z.
The Basic Unusual Boolean Theorem
Here are two sets of theorems in Boolean algebra.
For all X
For all X
For all X
For all X
0X
1X
0+X
1+X
=
=
=
=
0
X
X
1
OK
OK
OK
What?
Consider the following truth tables
W
0
0
1
1
X
0
1
0
1
W+X
0
1
1
1
From this, we derive the truth table proving the last two theorems.
X
0
1
0+X
0
1
1+X
1
1
The Principle of Duality
If a statement in Boolean algebra is true, so is its dual.
To take the dual of an expression do the following:
change all logical AND to logical OR and all logical OR to logical AND
change all 0 to 1 and 1 to 0.
Postulate
0X =
1X =
0+X =
1+X =
0
X
X
1
An Unexpected Pair:
Dual
1+X
0+X
1X
0X
=
=
=
=
1
X
X
0
Two distributive laws, each the dual of the other.
For all Boolean values of Boolean variables A, B, C: A(B + C) = AB + AC
For all Boolean values of Boolean variables A, B, C: A + BC = (A + B)(A + C)
If A = 1, the statement becomes 1 + BC = (1 + B)(1 + C), or 1 = 11.
If A = 0, the statement becomes 0 + BC = (0 + B)(0 + C), or BC = BC.
Another Look at the Second Distributive Law
Let’s use the truth table approach to proving the equality A + BC = (A + B)(A + C).
A
0
0
0
0
1
1
1
1
B
0
0
1
1
0
0
1
1
C
0
1
0
1
0
1
0
1
(A + B)
0
0
1
1
1
1
1
1
(A + C)
0
1
0
1
1
1
1
1
BC
0
0
0
1
0
0
0
1
A + BC
0
0
0
1
1
1
1
1
(A + B)(A + C)
0
0
0
1
1
1
1
1
Note that the last two columns show values that are identical for all possible
combinations or X, Y, and Z. For that reason, the two functions are identical.
We now show a few tricks for generating truth tables.
Example (Page 1)
Consider the truth table for the Boolean expression A + BC.
We start with the basic truth table, which has eight rows.
A
B
C
0
0
0
0
0
1
0
1
0
0
1
1
1
0
0
1
0
1
1
1
0
1
1
1
BC
A + BC
To fill the column BC, we enter a 0 when B = 0 and the value for C when B = 1.
Example (Page 2)
A
B
C
BC
0
0
0
0
0
0
1
0
0
1
0
0
0
1
1
1
1
0
0
0
1
0
1
0
1
1
0
0
1
1
1
1
A + BC
To fill the column A + BC, we place a 1 when A = 1, and the
value for BC when A = 0.
Example (Page 3)
A
B
C
BC
A + BC
0
0
0
0
0
0
0
1
0
0
0
1
0
0
0
0
1
1
1
1
1
0
0
0
1
1
0
1
0
1
1
1
0
0
1
1
1
1
1
1
Other Logic Gates
The top gate shows the NOR gate and its logical equivalent.
The bottom line shows the NAND gate and its logical equivalent.
In my notes, I call these “derived gates” as they are composites of Boolean gates
that are more basic from the purely theoretical approach.
X Y OR NOR
X Y AND NAND
0 0 0
1
0 0
0
1
0 1 1
0
0 1
0
1
1 0 1
0
1 0
0
1
1 1 1
0
1 1
1
0
AND Gates and OR Gates: The Real Way
In actual fact, the NAND and NOR gates are more primitive than the AND, OR,
and NOT gates in that they are easier to build from transistors.
AND is NOT (NAND)
X
0
0
1
1
Y NAND AND
0
1
0
1
1
0
0
1
0
1
0
1
OR is NOT (NOR)
X
0
0
1
1
Y NOR OR
0
1
0
1
0
1
0
0
1
1
0
1
Multiple–Input Gates
The standard definitions of the AND and OR gates call for two inputs.
3–input and 4–input varieties of these gates are quite common.
Here we give informal, but precise, definitions.
Gate
Number of Inputs
Output
NOT
Exactly 1
0 if input is 1, 1 if input is 0
AND
2 or more
0 if any input is 0
1 if and only if all inputs are 1.
OR
2 or more
1 if any input is 1
0 if and only if all inputs are 0.
NAND 2 or more
1 if any input is 0
0 if and only if all inputs are 1
NOR
0 if any input is 1
1 if and only if all inputs are 0.
2 or more
Example: “Changing the Number of Inputs”
Some lab experiments call for gates with input counts other than what we have.
We begin with two ways to fabricate a 4–input AND gate from 2–input ANDs.
Another Example
We now consider how to take a 4–input AND gate and make it act
as if it were a 2–input AND gate.
There are always multiple solutions. Here are two solutions.
There are many others.