This ppt explains about the Quine Mccluskey (Tabulation method) to minimize the boolean functions

1. Digital Electronics
QUINE MCCLUSKEY METHOD
(TABULATION METHOD)
R.KANMANI
ASSISTANT PROFESSOR(SR.GR)/ECE
SRI RAMAKRISHNA INSTITUTE OF TECHNOLOGY
COIMBATORE.

2. Some facts about Quine Mccluskey
• Developed in 1956
• Otherwise called as tabulation method
• Used for minimization of Boolean functions
• Where Karnaugh map could solve for upto 5
bits, Quine Mcclusky can solve for more than
5 bits
• Has easy algorithm implementation in
computer than Karnaugh

3. Steps
Four Main steps
• Generate Prime implicants
• Construct prime implicants table
• Reduce prime implicants
• Solve prime implicants table

4. Example problem
1. Minimize using Quine Mcclusky method
∑m(0,4,5,7,10,12,13,14,15)
Step 1:
Listing the Binary codes for each number
Minterms Binary representation
0 0000
4 0100
5 0101
7 0111
10 1010
12 1100
13 1101
14 1110
15 1111

5. • Step 2 : Listing Binary numbers according to
their number of 1’s
0 0000
4 0100
5 0101
10 1010
12 1100
7 0111
13 1101
14 1110
15 1111

6. • Step 3 : Making table of duals
0,4 0-00
4,5 010-
4,12 -100
5,7 01-1
5,13 -101
10,14 1-10
12,13 110-
12,14 11-0
7,15 -111
13,15 11-1
14,15 111-

7. • Step 4 : Generating table of Quads
Prime Implicants
0,4 0-00
4,5,12,13 -10-
4,12,5,13 -10-
10,14 1-10
5,13,7,15 -1-1
5,7,13,15 -1-1
12,13,14,15 11--
12,14,13,15 11--

8. • Step 5 : Making table of prime implicants
Prime Implicants m0 m4 m5 m7 m10 m12 m13 m14 m15
0,4 ʘ ʘ
4,5,12,13 ʘ ʘ ʘ ʘ
10,14 ʘ ʘ
5,13,7,15 ʘ ʘ ʘ ʘ
12,13,14,15 ʘ ʘ ʘ ʘ
Check for one dot in column wise m0,m7,m10 so definitely take 0,4 5,13,7,15
10,14 for final output expression .
Check for two dot in column wise m4,m5,m12,m14,m15
m4 already taken,m5 taken, m12 so take 4,5,12,13 for final output expression.
Check for three dots, m13 has three dots. But it is already consider. So
12,13,14,15 is not considered for final output expression.

9. • Step 6 : Generating SOP from prime implicants
Prime Implicants ABCD
0,4 0-00
4,5,12,13 -10-
10,14 1-10
5,13,7,15 -1-1
So the minimized SOP is
A’C’D’ + BC’ + ACD’ + BD

10. 2.Simplify the following Boolean function by using
Quine McCluskey Method.
F(A,B,C,D)=∑m(0,2,3,6,7,8,10,12,13)
3. Minimize the expression using Quine McCluskey
method.
Y= A’BC’D’+A’BC’D+ABC’D’+ABC’D+AB’C’D+A’B’CD’

11. 2.Simplify the following Boolean function by using Quine
McCluskey Method.
F(A,B,C,D)=∑m(0,2,3,6,7,8,10,12,13)
0 – 0000 0 – 0000 0,2 00-0
2 – 0010 2 – 0010 0,8 -000
3 – 0011 8 – 1000 2,3 001-
6 – 0110 3 – 0011 2,6 0-10
7 – 0111 6 – 0110 2,10 -010
8 – 1000 10 – 1010 8,10 10-0
10 – 1010 12 – 1100 3,7 0-11
12 – 1100 7 – 0111 6,7 011-
13 - 1101 13 – 1101 12,13 110-
8,12 1-00

12. 8,12 1-00 PI
12,13 110- PI
0,2,8,10 -0-0
0,8,2,10 -0-0
2,3,6,7 0-1-
2,6,3,7 0-1-

13. PI 0 2 3 6 7 8 10 12 13
8,12
. .
12,13
. .
0,2,8,10
. . . .
2,3,6,7
. . . .
F = ABC’+B’D’+A’C

14. 3.Minimize the expression using Quine McCluskey
method.
Y= A’BC’D’+A’BC’D+ABC’D’+ABC’D+AB’C’D+A’B’CD
Solution
4 – 0100 4 – 0100 4,5 010-
5 – 0101 3 - 0011 4,12 -100
12 – 1100 5 - 0101 5,13 -101
13 – 1101 9 – 1001 9,13 1-01
9 – 1001 12 – 1100 12,13 110-
3 – 0011 13 - 1101

15. • 3 0011 PI
• 4,12,5,13 -10- PI
• 9,13 1-01 PI
PI 3 4 5 9 10 12 13
3
.
9,13
. .
4,12,5,13
. . . .
F = A’B’CD+AC’D+BC’

16. 4.Simplify the function
F(w,x,y,z) = ∑m(2,3,12,13,14,15) using tabulation
method . Implement the simplified function
using gates.
Solution :
2 0010 2,3 001- PI
3 0011 12,13 110-
12 1100 12,14 11-0
13 1101 13,15 11-1
14 1110 14,15 111-
15 1111

17. • 2,3 001-
• 12,13,14,15 11- -
• 12,14,13,15 11- -
PI 2 3 12 13 14 15
2,3 . .
12,13,14,15 . . . .
F = w’x’y+wx

18. 5.Minimize the expression using Quine Mccluskey
(Tabulation) method
F = ∑m(0,1,9,15,24,29,30) + ∑d(8,11,31)
Solution:
Step 1: List the binary code of each number
Minterm Binary representation
0 00000
1 00001
8 01000
9 01001
11 01011
15 01111
24 11000
29 11101
30 11110
31 11111

19. List the binary numbers according to
their number of 1’s
Minterm & don’t care Binary representation
m0 00000
m1 00001
d8 01000
m9 01001
m24 11000
d11 01011
m15 01111
m29 11101
m30 11110
d31 11111
Each color in the table represents arrangement based on
number of 1’s

20. Compare one group with next group
and find one bit difference
minterms Binary representation
0,1 0000-
0,8 0-000
1,9 0-001
8,9 0100-
8,24 -1000
9,11 010-1
11,15 01-11
15,31 -1111
29,31 111-1
30,31 1111-
Each color in the table represents arrangement based on
number of 1’s

21. List the Prime Implicants
Prime Implicants Binary representation
0,1,8,9 0-00-
8,24 -1000
9,11 010-1
11,15 01-11
15,31 -1111
29,31 111-1
30,31 1111-

22. List the Prime Implicants
m0 m1 d8 m9 d11 m15 m24 m29 m30 d31
0,1,8,9 ʘ ʘ . ʘ
8,24 . ʘ
9,11 ʘ .
11,15 . ʘ
15,31 ʘ .
29,31 ʘ .
30,31 ʘ .
First check for one dot in the column : m0,m1,m24,m29,m30 . These minterms
will be definitely taken for final output
0,1,8,9 8,24 29,31 30,31
Second check for two dots in the column (neglect if don’t care has two dot in
the column) m9,m15,
9 already consider but 11 is a don’t care . So no need to take 9,11.
15 already considered and 31 contains 3 dots (don’t care) . So need to take
15,31.

23. List the Prime Implicants
m0 m1 d8 m9 d11 m15 m24 m29 m30 d31
0,1,8,9 ʘ ʘ . ʘ
8,24 . ʘ
9,11 ʘ .
11,15 . ʘ
15,31 ʘ .
29,31 ʘ .
30,31 ʘ .
So the final expression is F = A’C’D’ + BC’D’E’ + A’BDE + ABCE + ABCD
0,1,8,9 0-00-
8,24 -1000
11,15 01-11
29,31 111-1
30,31 1111-