Prime Implicant

Unit 6
Quine-McCluskey
Quine McCluskey
Method

Quine-McCluskey
Quine McCluskey Method
Quine-McCluskey
McCluskey Mathod
Quine
6.1 Determination of Prime Implicants
6.2 The Prime Implicant Chart
6.3 Petrick’s Petrick Method
6.4 Simplification of Incompletely Specified
Functions
6.5 Simplification Using Map-Entered
Map Entered
Variables
6.6 Conclusions
Unit 06 2

Quine-McCluskey
Quine McCluskey Method
Cover: A switching function f(x (x1,x ,x2,…,xn) ) is said to
cover another function g(x (x1,x ,x2,…,xn), ), if f assumes the
value 1 whenever g does.
Implicant : Given a function F of n variables, a
product term P is an implicant of F iff for every
combination of values of the n variables for which
P=1 , F is also equal 1.That is, P=1 implies F=1. F=1
Cover:
Implicant
Prime Implicant: Implicant A
Prime
A prime implicant of a
function F is a product term implicart which is no
longer an implicant if any literal is deleted from it.
Essential Prime Implicant: Implicant If
If a minterm is
covered by only one prime implicant, implicant,
then that prime
implicant is called an essential prime implicant.
implicant
Unit 06 3

Quine-McCluskey
Quine McCluskey Method
On a Karnaugh Map
Any single 1 or any group of 1’s 1 s (2 k 1’s, s, k=0,1,2,…) k=0,1,2, )
which can be combined together on a map of the
function F represents a product term which is
called an implicant of F.
A product term implicant is called a prime implicant
if it cannot be combined with another term to
eliminate a variable.
If a minterm is covered by only one prime implicant, implicant,
then that prime implicant is called an essential
prime implicant.
implicant
Unit 06 4

Quine-McCluskey
Quine McCluskey Method
A systematic simplification procedure
Input: Input:
minterm expansion
Output: Output:
a minimum sum of products
Method: Method
1. Generate all prime implicants
Eliminate as many literals as possible from each
term by systematically applying the theorem
XY+XY’=X XY+XY =X
2. Find the minimum solution
Use a prime implicant chart to select a minimum
set of prime implicants which contain a minimum
number of literals.
1.
2.
Unit 06 5

Quine-McCluskey
Quine McCluskey Method
Example: : F(a,b,c)= F(a,b,c)=a’b’c’+ab
+ab’c’+ab +ab’c+abc c+abc
All Implicants: Implicants:
a’b’c’,ab ,ab’c’,ab ,ab’c,abc c,abc
ab’,b ab ,b’c’,ac ,ac
Prime Implicants: Implicants
ab’,b ab ,b’c’,ac ,ac
Essential Prime Implicants: Implicants
b’c’,ac ,ac
Minimum sum of products: products
F(a,b,c)= F(a,b,c)=b’c’+ac
+ac
Example
Unit 06 6

Quine-McCluskey
Quine McCluskey Method
Quine-McCluskey
Quine McCluskey Method: Method
1. Find all of the prime implicants of the
function.
2. Construct the prime implicant table and find
the essential prime implicants of the function.
3. Include the essential prime implicants in the
minimal sum.
Unit 06 7

Quine-McCluskey
Quine McCluskey Method
Quine-McCluskey
Quine McCluskey Method: Method
4. Manipulate the prime implicant table.
4-1. . Delete all essential prime implicant from the
prime implicant table,
4-2. 2. Determine and delete the dominated rows
and dominating columns in the table,
4-3. 3. Find the (secondary ( secondary) ) essential implicants. implicants
5. Repeat Steps 3 and 4 as many times as they
are applicable until a minimal cover of the
function is found.
Unit 06 8

Determination of Prime Implicants
Find all of the prime implicants
Find
(1). Represent each minterm by a binary code. code
(2). Find the decimal number for each binary code.
(3). Define the number of 1’s 1 s in binary number as
the index of the number.
(3 (3-1). 1). Group all the binary numbers of the same
index into a group.
(3 (3-2). 2). List all the groups in a column in the index
ascending order. order
(3 (3-3). 3). Within each group, the binary number are
listed in the ascending order of their decimal-number
decimal number
equivalent.
equivalent
Unit 06 9

Determination of Prime Implicants
(4). Start with the terms in the set of lowest
index; index;
compare them with those, if any, in
the set whose index is 1 greater, greater,
and
eliminate all redundant variables by
XY+XY’=X. XY+XY =X.
(5). Check off all the terms that entered into
the combinations.
combinations.
The ones that are left are
prime implicants. implicants
(6). Repeat Steps (4) and (5) until no further
reduction is possible.
Unit 06 10

Determination of Prime Implicants
Example: Find
Example: Find all of the prime implicants
f
( a,
b,
c,
d)
= ∑ m(
0,
1,
2,
5,
6,
7,
8,
9,
10,
14)
implicants of the function
Unit 06 11

Determination of Prime Implicants
Example: Find
Example: Find all of the prime implicants
f
( a,
b,
c,
d)
= ∑ m(
0,
1,
2,
5,
6,
7,
8,
9,
10,
14)
implicants of the function
Unit 06 12

Determination of Prime Implicants
Example (Cont.):
Minimum form ???
Unit 06 13

The Prime Implicant Chart
Construct the Prime Implicant Table
(Chart) and find the Essential Prime
Implicants of the function
Construct
(1) Construct the prime implicant table
(1)
(1 (1-1) 1) Each column carries a decimal number
at the top which correspond to the one of
the minterm in the given function,
(1 (1-2) 2) The column are assigned by such a
number in ascending order, order
(1 (1-3) 3) Each row corresponds to one of the
prime implicants, implicants,
P1,P2,….
P1,P2,
Unit 06 14

The Prime Implicant Chart
(2) Make a cross under each decimal
number that is a term contained in the
prime implicant represented by that row.
(3) Find all the columns that contain a
single cross and circle them; place an
asterisk * at the left of those rows in which
you circle a cross.
The rows marked with an asterisk are the
essential prime implicants.
implicants
Unit 06 15

The Prime Implicant Chart
Example:
Unit 06 16

The Prime Implicant Chart
Example:
Unit 06 17

The Prime Implicant Chart
Example with a cyclic prime implicant
table
F
= ∑ m(
0,
1,
2,
5,
6,
7)
Sol: Sol:
Find all of the prime implicants
Unit 06 18

The Prime Implicant Chart
Sol(cont.): Sol(cont.):
Select P1 first.
The minimum sum of products
F =
a′
b′
+ bc′
+
ac
Unit 06 19

The Prime Implicant Chart
Sol(cont.): Sol(cont.):
Select P2 first.
The minimum sum of products
The minimum sum of product is not unique. unique
F =
a′
c′
+ b′
c +
ab
Unit 06 20

Petrick’s Petrick Method
A technique for determining all minimum
sum-of sum of-products products solutions from a prime
implicant table.
Before applying Petrick’s Petrick method, all
essential prime implicants and minterms
they cover should be removed from the
table.
Unit 06 21

Petrick’s Petrick Method
Petrick’s Method
Petrick
1. Label the rows of the table, P 1,P ,P2 ,…
2. Form a logic function P(P 1,P ,P2 ,…), ), which
is true when all of the minterms in the table
have been covered.
3. Reduce P to a minimum sum of products
using (X+Y)(X+Z)=X+YZ and X+XY=X.
4. Select one solution that has minimum
number of prime implicant, implicant,
minimum
number of literals.
Unit 06 22

Petrick’s Petrick Method
Example: Example:
F
= ∑ m(
0,
1,
2,
5,
6,
7)
Unit 06 23

Petrick’s Petrick Method
Example (cont.):
In order to cover minterm 0, we must choose
P1 or P 2 ; therefore, the expression P 1+P +P2 must be true.
cover 0 P1+P +P2 1 P1+P +P3 2 P2+P +P4 5 P3+P +P5 6 P4+P +P6 7 P5+P +P 6
Unit 06 24

Petrick’s Petrick Method
Example (cont.):
P = P + P )( P + P)(
P + P )( P + P )( P + P )( P + P ) = 1
( 1 2 1 3 2 4 3 5 4 6 5 6
Using (
X + Y )( X + Z)
= X + YZ and the
distributive law
Unit 06 25

Petrick’s Petrick Method
Example (cont.):
Use to delete redundant terms from P
P = PP
P + PP
P P + P PP
P + P PP
P
1
1
1
4
4
3
5
+ PP
P P
= PP
P
5
4
6
1
1
2
2
3implicants
Two minimum solutions:
F
F
=
=
X + XY =
P
1
P
2
1
5
5
2
4
6
+ P P PP
+ PP
P P
+
+
P
4
P
3
+
+
6
P
P
3
X
6
2
2
3
3
2
4
4
4
3
5
+ P PP
+ P PP
P
5
6
5
4
P
6
2
1
3
3
5
2
4
4
3
6
+ P PP
+ PP
P P
= a′
b′
+ bc′
+ ac
= a′
c′
+ b′
c +
6
ab
6
+ P PP
2
3
6
3
Unit 06 26

Simplification of Incompletely
Specified Functions
Modify the Quine-McCluskey
Quine McCluskey procedure
Finding the Prime Implicants
Treat Treat the don’t don t care terms as if they were
required minterms. minterms
Forming the Prime Implicant Table
The The don’t don t cares are not list at the top of the
table.
Unit 06 27

Simplification of Incompletely
Specified Functions
Example: Simplify
F
A,
B,
C,
D)
∑ m(
2,
3,
7,
9,
11,
13)
+
Sol: Treat the don’t don t cares (1,10,15) as
required minterms
( d
= ∑ ( 1,
10,
15)
Unit 06 28

Simplification of Incompletely
Specified Functions
Sol(cont.): .): The don’t don t cares are not list
at the top of the table
Sol(cont
Unit 06 29

Simplification Using Map-Entered
Map Entered
Variables
Map-Entered Map Entered Variables –
A modification of the Karnaugh Map
Quine-McCluskey
Quine McCluskey Method
Can be used with functions with a fairly
large number of variables. variables
Not very efficient for functions that have
many variables and relatively few terms
Unit 06 30

Simplification Using Map-Entered
Map Entered
Variables
Map-Entered Entered Variables
Map
Example:
F ( A,
B,
C,
D)
= A′
B′
C + A′
BC + A′
BC′
D + ABCD+
( AB′
C)
where (AB’C) (AB C) is don’t don t care. D appears in only two terms,
choose D as a map-entered map entered variable. variable.
F(
A,
B,
C,
D)
=
A′
B′
C+
A′
BC+
A′
BC′
D+
ABCD+
( AB′
C)
= A′
B′
C+
A′
BC+
( A′
BC′
) D+
( ABC)
D+
( AB′
C)
=
m
1
+
m
3
+
m
2
D
+
( m
Unit 06 31
m
7
D
+
5
)

Simplification Using Map-Entered
Map Entered
Variables
Example (cont.):
Unit 06 32

Simplification Using Map-Entered
Map Entered
Variables
Map-Entered Map Entered Variables
Example:
G(
A,
B,
C,
D,
E,
F)
=
m
0
+ m
+ ( m ) + ( m
1
2
+ m
3
10
+ Em
) + ( m
5
13
+ Em
)
Unit 06 33
7
+ Fm
9
+ m
11
+ m
15

Simplification Using Map-Entered
Map Entered
Variables
General Method of Simplifying Functions
Using Map-entered Map entered Variables
If a variable Pi is placed in square mj of a map of
function F, , this means that F=1 =1 when Pi=1 =1 and the
variable are chosen so that mj=1. =1.
F =
MS
MS
i
:
0
+ PMS
1
Minimum
Sum
+ ...
Unit 06 34
1
+
P
2
MS
2

Simplification Using Map-Entered
Map Entered
Variables
MS
0 is the minimum sum obtained by
setting P1
= P2
= ... = . 0
MS is the minimum sum obtained by setting
1
setting
P1 = 1,
Pj
= 0(
j ≠ 1)
and replacing all 1’s 1 s on the map
with don’t don t cares. cares
MS is the minimum sum obtained by setting
i
Pi =
1, Pj
= 0(
j ≠ i)
and replacing all 1’s 1 s on the map
with don’t don t cares. cares
The resulting expression for F is
F
= MS + PMS
+ P MS
0
1
1
+ ...
The expression will not generally be minimum if the
variables P 1,P ,P2,… are dependent (e.g. P 1=E, =E, P 2=E =E’,…). ).
2
2
Unit 06 35

Simplification Using Map-Entered
Map Entered
Variables
Map-Entered Entered Variables
Map
Example: Use Map-entered Map entered Variables to simplify
the function
F ( A,
B,
C,
D)
= A′
B′
C + A′
BC + A′
BC′
D + ABCD+
( AB′
C)
where (AB’C) (AB C) is don’t don t care.
Sol: D appears in only two terms, choose D as a map-entered
map entered
variable. variable.
Sol:
F(
A,
B,
C,
D)
=
A′
B′
C+
A′
BC+
A′
BC′
D+
ABCD+
( AB′
C)
= A′
B′
C+
A′
BC+
( A′
BC′
) D+
( ABC)
D+
( AB′
C)
=
m
1
+
m
3
+
m
2
D
+
( m
Unit 06 36
m
7
D
+
5
)

Simplification Using Map-Entered
Map Entered
Variables
Sol (cont.):
Unit 06 37

Simplification Using Map-Entered
Map Entered
Variables
Sol(cont.): Sol(cont
.): F ( A,
B,
C,
D)
= A′
B′
C + A′
BC + A′
BC′
D + ABCD + ( AB′
C)
F = A′
C + D(
C + A′
B)
= A′
C + CD + A′
BD
Unit 06 38

Simplification Using Map-Entered
Map Entered
Variables
Example: : Use the Map-Entered Map Entered Variables method
to simplify the function
Example
G(
A,
B,
C,
D,
E,
F)
= m
Let P1=E, =E, P 2=F =F. .
0
+ m
+ ( m ) + ( m
1
2
+ m
3
10
+ Em
) + ( m
5
13
+ Em
)
7
+ Fm
9
+ m
Unit 06 39
11
+ m
15

Simplification Using Map-Entered
Map Entered
Variables
Example (cont.): (cont.)
MS 0 : The minimum inimum sum um obtained by setting P 1=P =P2=0 =0 (E=F=0).
MS 1 : The minimum inimum sum um obtained by setting P 1=1, =1, P 2=0 =0 (E=1. F=0).
MS2
: The minimum inimum sum um obtained by setting P 1=0, =0, P 2=1 =1 (E=0, F=1).
MS 2
G
=
MS
0
+ PMS
1
1
+ P MS
= A′
B′
+ ACD + EA′
D + FAD
2
2
= ( A′
B′
+ ACD)
+ E(
A′
D)
+ F(
AD)
Unit 06 40