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Computer Logic and Digital Design
Chapter 2
Henry Hexmoor
AND logic
input
output
AND Gate
B
AB
S1 . S2
S1
off
on
off
on
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S2
off
off
on
on
path?
no
no
no
yes
1
OR logic
input
S1 + S2
S1
S2
S1
off
on
off
on
S2
off
off
on
on
path?
no
yes
yes
yes
output
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2
NOT logic
S
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S'
S
1
0
S
0
1
3
Equivalent Symbols of NAND, NOR Gates
NAND Symbols
Normal Symbol
Alternate NAND Symbol
X
X
(X . Y)’
Y
X’ + Y’
Y
According to DeMorgan’s theorem T13:
(X . Y)’ = X’ + Y’
NOR Symbols
Alternate NOR Symbol
Normal NOR Symbol
X
(X + Y)’
Y
X
X’ . Y’
Y
According to DeMorgan’s theorem T13’:
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(X + Y)’ = X’ . Y’
4
Boolean Algebra
• Literals
(n) and terms
• number of truth table rows = 2n
X
0
0
0
0
1
1
1
1
Y
0
0
1
1
0
0
1
1
Z
0
1
0
1
0
1
0
1
Y’Z
0
1
0
0
0
1
0
0
X+Y’Z
0
1
0
0
1
1
1
1
• See table 2-3
• Using demorgan’s laws OR and AND are interchanged… the
duality
principle
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Hexmoor
5
Switching Algebra Axioms & Theorems
(A1)X = 0 if X ≠ 1
(A2)If X = 0, then X’ = 1
(A3)0 . 0 = 0
(A4)1 . 1 = 1
(A5)0 . 1 = 1 . 0 = 0
(A1’)
(A5’)
X = 1 if X ≠ 0
(A2’)
if X = 1, then, X’ = 0
(A3’)
1+1=1
(A4’)
0+0=0
1+0 =0+1=1
(T1)X + 0 = X
(T1’)
X.1=X
(Identities)
(T2)
X+1 =1
(T2’)
X.0=0
(Null elements)
(T3)
X+X =X
(T3’)
X.X=X
(Idempotency)
(T4)(X’)’ = X
(Involution)
(T5)X + X’ = 1
(T5’)
X . X’ = 0
(Complements)
(T6)X + Y = Y + X
(T6’)
X.Y=Y.X
(Commutativity)
(T7)(X + Y) + Z = X + (Y + Z)
(T7’)
(X . Y) . Z = X . (Y . Z)
(Associativity)
(T8)
X . Y + X . Z = X . (Y + Z)
(T8’)
(X + Y) . (X + Z) = X + Y . Z (Distributivity)
(T9)
X+X.Y=X
(T9’)
X . (X + Y) = X
(Covering)
(T10)
X . Y + X . Y’ = X
(T10’)
(X + Y) . (X + Y’) = X
(Combining)
(T11)
X . Y + X’. Z + Y . Z = X . Y + X’ . Z
(T11’)
(X + Y) . ( X’ + Z) . (Y + Z) = (X + Y) . (X’ + Z)
(Consensus)
(T12)
X+X+ ...+X=X
(T12’)
X . X . . . . . X = X (Generalized idempotency)
(T13)
(X1 . X2 . . . . . Xn)’ = X1’ + X2’ + . . . + Xn’
(T13’)
(X1 + X2 + . . . + Xn)’ = X1’ . X2’ . . . . . Xn’
(DeMorgan’s theorems)
(T14)
[F(X1, X2, . . ., Xn, +, .)]’ = F(X1’, X2’, . . ., Xn’, . , +) (Generalized DeMorgran’s theorem)
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Switching Algebra Axioms
•
First two axioms state that a variable X can only take on only one of two
values:
(A1)
(A1’)
•
X = 0 if X  1
X = 1 if X  0
Not Axioms, formally define X’ (X prime or NOT X):
(A2)
(A2’)
If X = 0, then X’ = 1
if X = 1, then, X’ = 0
Note: Above axioms are stated in pairs with only difference
being the interchange of the symbols 0 and 1.
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Three More Switching Algebra Axioms
• The following three Boolean Algebra axioms state and formally
define the AND, OR operations:
(A3)
(A3’)
0.0 = 0
1+1=1
(A4)
(A4’)
1.1 =1
0+0=0
(A5)
(A5’)
0 . 1 = 1 .0 = 0
1+0 =0+1=1
Axioms A1-A5, A1’-A5’ completely define switching algebra.
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Switching Algebra: Single-Variables Theorems
• Switching-algebra theorems are statements known to be always true
(proven using axioms) that allow us to manipulate algebraic logic
expressions to allow for simpler analysis.
(e.g . X + 0 = X allow us to replace every X +0 with X)
The Theorems: (T1-T5, T1’-T5’)
(T1)
(T2)
(T3)
(T4)
(T5)
X+0=X
X+1 =1
X+X =X
(X’)’ = X
X + X’ = 1
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(T1’) X . 1 = X (Identities)
(T2’) X . 0 = 0 (Null elements)
(T3’) X . X = X (Idempotency)
(Involution)
(T5’) X . X’ = 0 (Complements)
9
Perfect Induction
• Most theorems in switching algebra are simple to prove using
perfect induction:
Since a switching variable can only take the values 0 and 1 we can
prove a theorem involving a single variable X by proving it true for X =
0 and X =1
Example: To prove (T1)
X+0=X
[X = 0]
0 + 0 = 0 true according to axiom A4’
[X = 1]
1 + 0 = 1 true according to axiom A5’
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10
Switching Algebra: Two- and Three-Variable Theorems
(Commutativity)
(T6) X + Y = Y + X
(T6’) X . Y = Y . X
(Associativity)
(T7) (X + Y) + Z = X + (Y + Z)
(T7’) (X . Y) . Z = X . (Y . Z)
T6-T7, T6’ -T7’ are similar to commutative and associative laws for
addition and multiplication of integers and reals.
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Two- and Three-Variable Theorems (Continued)
(Distributivity)
(T8) X . Y + X . Z = X . (Y + Z)
(T8’) (X + Y) . (X + Z) = X + Y . Z
• T8 allows to multiply-out an expression to get sum-of-products
form (distribute logical multiplication over logical addition):
For example:
V . (W + X) . (Y + Z) = V .W . Y + V. W. Z + V. X . Y + V. X . Z
sum-of-products form
• T8’ allows to add-out an expression to get a product-of-sums form
(distribute logical addition over logical multiplication):
For example:
(V . W . X) + (Y . Z ) = (V + Y) . (V + Z) . (W + Y) . (W + Z) . (X + Y) . (X + Z)
product-of-sums form
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Theorem Proof using Truth Table
• Can use truth table to prove T8 by perfect induction.
• i.e, Prove that: X . Y + X . Z = X . (Y + Z)
(i) Construct truth table for both sides of above equality.
x
0
0
0
0
1
1
1
1
y
0
0
1
1
0
0
1
1
z
0
1
0
1
0
1
0
1
y+z
0
1
1
1
0
1
1
1
x.(y + z)
0
0
0
0
0
1
1
1
x.y
0
0
0
0
0
0
1
1
x.z
0
0
0
0
0
1
0
1
x.y + x.z
0
0
0
0
0
1
1
1
(ii) Check that from truth table check that that X . Y + X . Z = X . (Y + Z)
This is satisfied because output column values for X . Y + X . Z and output column values
for X . (Y + Z) are equal for all cases.
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13
Two- and Three-Variable Theorems (Continued)
(Covering)
(T9) X + X . Y = X
(T9’) X . (X + Y) = X
(Combining)
(T10) X . Y + X . Y’ = X
(T10’) (X + Y) . (X + Y’) = X
• T9-T10 used in the minimization of logic functions.
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Two- and Three-Variable Theorems (Continued)
(Consensus)
(T11) X . Y + X’. Z + Y . Z = X . Y + X’ . Z
(T11’) (X + Y) . ( X’ + Z) . (Y + Z) = (X + Y) .
(X’ + Z)
• In T11 the term Y. Z is called the consensus of
the term X . Y and the term X’ . Z:
– If Y . Z = 1, then either X . Y or X’ . Z must also be 1.
– Thus the term Y . Z is redundant and may be dropped.
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15
n-Variable Theorems
(Generalized idempotency)
(T12)
(T12’)
X+X+ ...+X=X
X.X. ... .X=X
(DeMorgan’s theorems)
(T13)
(X1 . X2 . . . . . Xn)’ = X1’ + X2’ + . . . + Xn’
(T13’)
(X1 + X2 + . . . + Xn)’ = X1’ . X2’ . . . . . Xn’
(T13), (T13’) are probably the most commonly
used theorems of switching algebra.
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Examples Using DeMorgan’s theorems
Example: Equivalence of NAND Gate:
A two-input NAND Gate has the output expression
Z = (X . Y)’
using (T13)
Z = (X . Y)’ = (X’ + Y’)
The function of a NAND gate can be achieved with an OR gate with an
inverter at each input.
Example: Equivalence of NOR Gate
A two-input NOR Gate has the output expression Z=(X+Y)’
using (T13’)
Z = (X + Y)’ = X’ . Y’
The function of a NOR gate can be achieved with an AND gate with an
inverter at each input.
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n-Variable Theorems (Continued)
(Generalized DeMorgran’s theorem)
(T14) [F(X1, X2, . . ., Xn, +, .)]’ = F(X1’, X2’, . . ., Xn’, . , +)
• States that given any n-variable logic expression its complement
can be found by swapping + and . and complementing all
variables.
Example:
F(W,X,Y,Z) = (W’.X) + ( X.Y) + (W.(X’ + Z’))
= ((W)’ . X) + (X. Y) + (W.((X)’ + (Z)’))
[F(W,X,Y,Z)]’ = ((W’)’ + X’) .(X’ + Y’).(W’ + ((X’)’.(Z’)’))
Using T4, (X’)’ = X simplifies it to:
[F(W,X,Y,Z)]’ = (W + X’) . (X’ + Y’) . (W’ + (X . Z))
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Logic Function Representation Definitions
•
A literal: is a variable or a complement of a variable
Examples: X, Y, X’, Y’
•
A product term:
Examples: Z’
•
A sum-of-products expression: is a logical sum of product terms.
Example:
Z’ + W.X.Y + X.Y’.Z + W’.Y’.Z
•
A sum term: is a single literal or logical sum of two or more literals
Examples: Z’
W+X+Y
X + Y’ + Z
W’ + Y’ + Z
•
A product-of-sums expression: is a logical product of sum terms.
Example: Z’. (W + X + Y) . (X + Y’ + Z) . (W’ + Y’ + Z)
•
A normal term: is a product or sum term in which no variable appears more
than once
Examples of non-normal terms: W.X.X.Y’
W+W+X’+Y
X.X’.Y
Examples of normal terms:
W . X . Y’
W + X’ + Y
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is a single literal, or a product of two or more literals.
W.Y.Y
X.Y’.Z
W’.Y’.Z
19
Logic Expression Algebraic Manipulation Example
• Prove that the following identity is true using Algebraic expression
Manipulation : (one can also prove it using a truth table)
X .Y + X . Z = ((X’ + Y’) . (X’ + Z’))’
– Starting from the left hand side of the identity:
Let F = X .Y + X . Z
A = X.Y
B = X.Z
Then F = A + B
– Using DeMorgan’s theorem T 13 on F:
F = A + B = (A’ . B’)’
(1)
– Using DeMorgan’s theorem T 13’ on A, B:
A = X . Y = (X’ + Y’)’
(2)
B = X . Z = (X’ + Z’)’
(3)
– Substituting A, B from (2), (3), back in F in (1) gives:
F = (A’ . B’)’ = ((X’ + Y’) . (X’ + Z’))’
Which is equal to the right hand side of the identity.
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Terminology: Minterms
•
•
•
•
A minterm is a special product of literals, in which each input variable appears
exactly once.
A function with n variables has 2n minterms (since each variable can appear
complemented or not)
A three-variable function, such as f(x,y,z), has 23 = 8 minterms:
x’y’z’
x’y’z
x’yz’
x’yz
xy’z’
xy’z
xyz’
xyz
Each minterm is true for exactly one combination of inputs:
Minterm
x’y’z’
x’y’z
x’yz’
x’yz
xy’z’
xy’z
xyz’
xyz
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Is true when…
x=0, y=0, z=0
x=0, y=0, z=1
x=0, y=1, z=0
x=0, y=1, z=1
x=1, y=0, z=0
x=1, y=0, z=1
x=1, y=1, z=0
x=1, y=1, z=1
Shorthand
m0
m1
m2
m3
m4
m5
m6
m7
21
Terminology: Maxterms
•
•
•
A maxterm is a special sum of literals, in which each input variable appears
exactly once.
A function with n variables has 2n maxterms (since each variable can appear
complemented or not)
A three-variable function, such as f(x,y,z), has 23 = 8 maxterms:
Maxterm Shorthand
x’+ y’+ z’
M0
x’+ y’ + z
M1
x’+ y + z’
M2
x’ + y + z M3
x + y’ + z’ M4
x + y’ + z M5
x + y + z’ M6
x+y+z
M7
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Canonical Forms
• Minterms and Maxterms
• Index Representation of Minterms and
Maxterms
• Sum-of-Minterm (SOM) Representations
• Product-of-Maxterm (POM) Representations
• Representation of Complements of
Functions
• Conversions between Representations
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23
Logic Function Representation Definitions
• Minterm
An n-variable minterm is a normal product term with n literals.
There are 2n such products terms.
Example of 4-variable minterms:
W.X’.Y’.Z’
W.X.Y’.Z
W’.X’.Y.Z’
• Maxterm
An n-variable maxterm is a normal sum term with n literals.
There are 2n such sum terms.
Examples of 4-variable maxterms:
W’ + X’ + Y + Z’
W + X’ + Y’ + Z
W’ + X’ + Y + Z
•
A minterm can be defined as as product term that is 1 in exactly one row
of the truth table.
•
A maxterm can similarly be defined as a sum term that is 0 in exactly
one row in the truth table.
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Minterms/Maxterms for A 3-variable function F(X,Y,Z)
Row
0
1
2
3
4
5
6
7
X Y Z
0
0
0
0
1
1
1
1
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0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
F
Minterm
Maxterm
F(0,0,0)
F(0,0,1)
F(0,1,0)
F(0,1,1)
F(1,0,0)
F(1,0,1)
F(1,1,0)
F(1,1,1)
X’.Y’.Z’
X’.Y’.Z
X’.Y.Z’
X’.Y.Z
X.Y’.Z’
X.Y’.Z
X.Y.Z’
X.Y.Z
X+Y+Z
X + Y + Z’
X + Y’ + Z
X + Y’ + Z’
X’ + Y + Z
X’ + Y + Z’
X’ + Y’ + Z
X’ + Y’ + Z’
25
Canonical Sum Example
• The function represented by the truth table:
Row X Y Z
0
0 0 0
1
0 0 1
2
0 1 0
3
0 1 1
4
1 0 0
5
1 0 1
6
1 1 0
7
1 1 1
F
1
0
0
1
1
0
1
1
has the canonical sum representation:
Minterm list using S notation
F = S X,Y,Z m(0, 3, 4, 6, 7)
= X’.Y’.Z’ + X’.Y.Z + X.Y’.Z’ + X.Y’.Z’ + X.Y.Z
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Algebraic canonical sum of minterms
26
Canonical Product Example
• The function represented by the truth table:
Row X Y Z
0
0 0 0
1
0 0 1
2
0 1 0
3
0 1 1
4
1 0 0
5
1 0 1
6
1 1 0
7
1 1 1
F
1
0
0
1
1
0
1
1
has the canonical product representation:
F = P X,Y,Z M(1,2,5)
= (X + Y + Z’) . (X + Y’ + Z) . (X’ + Y + Z’)
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Minterm and Maxterm Relationship
• Review: DeMorgan's Theorem
x · y = x + y and x + y = x  y
• Two-variable example:
M 2 = x + y and m 2 = x·y
M2 is the complement of m2 and vice-versa.
• Since DeMorgan's Theorem holds for n
variables, the above holds for terms of n
variables
• giving:
mi = M i
Mi = mi
and
Thus Mi is the complement of mi.
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Conversion Between Minterm/Maxterm Lists
• To convert between a minterm list and a maxterm list
take the set complement.
Example (page 44, 3rd ed.):
F’(X,Y, Z) = S m(1,3, 4, 6) = m1 + m3 + m4 + m6
F(X,Y, Z) = P M(1,3,4,6) = M1M3M4M6
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Verbal Synthesis Example: An Alarm Circuit
• A verbal logic description:
– The ALARM output is 1 if the panic input is 1, or if the ENABLE input is 1, the
EXISTING input is 0, and the house is not secure.
– The house is secure if the WINDOW, DOOR, GARAGE inputs are all 1
• This can be put in logic expressions as follows:
ALARM = PANIC + ENABLE . EXISTING’ . SECURE’
SECURE = WINDOW. DOOR. GARAGE
ALARM = PANIC + ENABLE . EXISTING’. (WINDOW . DOOR . GARAGE)’
In sum of products form as (by using DeMorgan T13 and multiplying out) :
ALARM = PANIC + ENABLE. EXISTING’ . WINDOW’
+ ENABLE . EXISTING’. DOOR’+ ENABLE. EXISTING’. GARAGE’
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cost criteria
2-4
1. Literal cost
2. Gate input cost: literals + terms
OR Gate
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Combinational Circuit Minimization
•
Canonical sum and product logic expressions do not provide a circuit
realization with the minimum number of gates.
•
Minimization methods reduce the cost of two level AND-OR, NAND-NAND,
OR-AND, NOR-NOR circuits in three ways:
•
1
2
By minimizing the number of first level gates
By minimizing the number of inputs of each first-level gate.
3
Minimizing the inputs of the second level gate
Most minimization methods are based on the combining theorems T10,
T10’:
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32
Karnaugh Maps
•
A Karnaugh Map or (K-map for short) is a graphical representation of
the truth table of a logic function.
•
The K-map for an n-input logic function is an array with 2n cells or
squares, one for each input combination or minterm.
•
The rows and columns are labeled so that the input combination for any
cell is determined from the row and column headings.
•
The row and columns of the map are ordered in such a way that each
cell differs from an adjacent cell in only one input variable:
– Thus for an n-variable K-map, each cell has n adjacent cells.
•
The K-map for a function is filled by putting:
–
–
a ‘1’ in the square corresponding to a minterm
a ‘0’ otherwise (maybe omitted)
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2-Variable K-map
For a 2-variable logic function F(X,Y):
Truth Table:
Row
0
1
2
3
X
0
0
1
1
Y
0
1
0
1
Y
Y
F
Minterm
F(0,0)
X’.Y’
F(0,1)
X’.Y
F(1,0)
X.Y’
F(1,1)
X .Y
0
X
0
K-map
1
0
2
1
3
X
1
Example: For the function F(X,Y) = S X,Y m(1,2,3)
Truth Table:
K-map
Y
Row
0
1
2
3
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X
0
0
1
1
Y
0
1
0
1
F
0
1
1
1
Y
0
X
0
0
1
1
1
2
1
3
1
1
X
34
3 variable Karnaugh map
Textbook convention
YZ
00
0
X
1
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01
11
10
X’ Y Z’ X’ Y Z
X’ Y Z
X’ Y Z
XY’ Z’
XYZ
X Y Z’
X Y’Z
35
3-Variable K-map
For a 3-variable logic function F(X,Y,Z):
Truth Table:
Row
0
1
2
3
4
5
6
7
X Y Z
0
0
0
0
1
1
1
1
0
0
1
1
0
0
1
1
0
1
0
1
0
1
0
1
F
Minterm
F(0,0,0)
F(0,0,1)
F(0,1,0)
F(0,1,1)
F(1,0,0)
F(1,0,1)
F(1,1,0)
F(1,1,1)
X’.Y’.Z’
X’.Y’.Z
X’.Y.Z’
X’.Y.Z
X.Y’.Z’
X.Y’.Z
X.Y.Z’
X.Y.Z
Example: For the function
F(X,Y,Z) = S X,Y,Z (1,4,6,7)
Truth Table:
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Row
0
1
2
3
4
5
6
7
X
0
0
0
0
1
1
1
1
K-map
Y
YZ
00
X
F
0
1
0
0
1
0
1
1
10
0
2
6
4
1
3
7
5
X
1
Y
Y
YZ
00
X
Z
0
1
0
1
0
1
0
1
11
0
K-map
Y
0
0
1
1
0
0
1
1
01
0
01
2
0
10
6
4
7
5
1
3
1
1
11
1
1
1
X
Z
36
4-Variable K-map
For a 4-variable logic function F(W,X,Y,Z):
Truth Table:
K-map
Y
YZ
WX
Row W X Y Z
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
0
0
0
0
0
0
0
0
1
1
1
1
1
1
1
1
0
0
0
0
1
1
1
1
0
0
0
0
1
1
1
1
0
0
1
1
0
0
1
1
0
0
1
1
0
0
1
1
Henry Hexmoor
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
F
F(0,0,0,0)
F(0,0,0,1)
F(0,0,1,0)
F(0,0,1,1)
F(0,1,0,0)
F(0,1,0,1)
F(0,1,1,0)
F(0,1,1,1)
F(1,0,0,0)
F(1,0,0,1)
F(1,0,1,0)
F(1,0,1,1)
F(1,1,0,0)
F(1,1,0,1)
F(1,1,1,0)
F(1,1,1,1)
Minterm
W’.X’.Y’.Z’
W’. X’.Y’.Z
W’. X’.Y.Z’
W’. X’.Y.Z
W’. X.Y’.Z’
W’.X.Y’.Z
W’.X.Y.Z’
W’.X.Y.Z
W.X’.Y’.Z’
W.X’.Y’.Z
W.X’.Y.Z’
W.X’.Y.Z
W.X.Y’.Z’
W.X.Y’.Z
W.X.Y.Z’
W.X.Y.Z
00
00
01
11
10
0
4
12
8
1
5
13
9
3
7
15
11
2
6
14
10
01
W
X
11
10
Z
37
Minimization Using K-maps
• Group or combine as many adjacent 1-cells
as possible:
– The larger the group is, the fewer the number of literals in the
resulting product term.
– Grouping 2 adjacent 1-cells eliminates 1 variable, grouping 4 1cells eliminates 2 variables, grouping 8 1-cells eliminates 3
variables, and so on. In general, grouping 2n squares eliminates
n variables.
• Select as few groups as possible to cover all
the 1-cells (minterms) of the function:
– The fewer the groups, the fewer the number of product terms in
the minimized function.
Henry Hexmoor
38
Karnaugh maps
Example 2-4 (Figure 2-14b, page 54, 3rd ed.)
• 1cell Square overlap is OK.
YZ
X
0
1
00
01
1
1
1
11
000
Z’ 100
010
110
10
1
1
XY’
100
101
F2(X,Y,Z) = m(0,2,4,5,6) = Z’ + XY’
Henry Hexmoor
39
4 variable Karnaugh map
example 2-5 (Figure 2-19, page 57, 3rd ed.)
YZ
00
01
01
1
1
1
1
11
1
1
10
1
1
00
WX
11
10
1
1
1
0000
0001
1100
1101
1000
1001
0000
0001
0100
0101
0010
0110
0100
1100
0110
1110
F(W,X,Y,Z) = S m(0,1,2,4,5,5,8,9,12,13,14) = Y’ + W’Z’ + XZ’
Henry Hexmoor
40
prime/essential implicants
W’Z’ and XZ’ are prime implicants
(each rectangle needs all its
cells)
Y’ is an essential prime implicant…
(the rectangle contains
exclusive (i.e., nonshared) cells
YZ
00
01
01
1
1
1
1
11
1
1
10
1
1
00
WX
11
10
1
1
1
F(X,Y,Z) = m(0,1,2,4,5,5,8,9,12,13,14) = Y’ + W’Z’ + XZ’
Selection rule: minimize overlap among prime implicants
Henry Hexmoor
41
product of sums: Maxterms
YZ
00
01
11
01
1
1
1
1
0
0
11
1
1
0
10
1
1
0
00
WX
10
1
1
1
0
F(X,Y,Z) = m(0,1,2,4,5,5,8,9,12,13,14) = YZ + WX’Y
 (Y’ + Z’) (W’ + X + Y)
Selection rule: minimize overlap among prime implicants
Henry Hexmoor
42
More Examples…
Y
0
0
X
1
1
1
0
0
1
• x’z + y’z + xyz’.
Z
Y
W
1
0
0
1
0
1
1
0
0
0
0
0
1
0
0
1
X
x’z’ + xy’z
Y
Z
W
w’x’y’z’
w’xy’z’
wxy’z’
wx’y’z’
w’x’y’z
w’xy’z
wxy’z
wx’y’z
w’x’yz
w’xyz
wxyz
wx’yz
w’x’yz’
w’xyz’
X
wxyz’
wx’yz’
Z
Henry Hexmoor
43
multilevel optimization
2-6
Manipulate equations algebraically
G = AC’E + AC’F + AD’E + AD’F + BCDE’F’
= A (C’E + C’F + D’E + D’F) + BCDE’F’
= A (C’ + D’)(E + F) + BCDE’F’
OR Gate
Henry Hexmoor
44
other gate types
See figures 2-22 and 2-23 in 4th edition
OR Gate
Henry Hexmoor
45
HW 2
1. Demonstrate by means of truth tables the validity of
the following identity:
(XYZ)’ = X’ + Y’ + Z’
(Q 2-1)
2. Optimize the following Boolean function by means a
three-variable map:
1. F(X,Y,Z) = Sm(1,3,6,7)
(Q 2-14)
OR Gate
Henry Hexmoor
46