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Dr. Heman Pathak, Associate Professor
Dept. of Comp. Sc., Kanya Gurukul DehradunChapter -3
Canonical Forms
2
Any Boolean function is said to be in its
canonical form if it is expressed either as
 Sum of minterms or as
 Product of maxterms
Karnaugh Map
 Karnaugh introduced a method for simplification of
Boolean functions in an easy way.
 This method is known as Karnaugh map method or
K-map method.
 It is a graphical method, which consists of 2n cells for
‘n’ variables. The adjacent cells are differed only in
single bit position.
2 Variable K-Map
 The number of cells in 2 variable K-map is four, since
the number of variables is two. The adjacent cells are
differed only in single bit position.
x y minterm
0 0 x’y’ m0
0 1 x’y m1
1 0 xy’ m2
1 1 xy m3
2 Variable K-Map
 F1 = m0+m1=x’y’ + x’y = x’(y’ + y) = x’
 F2 = m1+m3=x’y + xy = y(x’ + x) = y
 F3 = m3+m2=xy + xy’ = x(y + y’) = x
 F4 = m0+m2=x’y’ + xy’ = y’(x’ + x)= y’
 F =m0+m1+m2+m3
 F = x’y’ + x’y + xy’ + xy
= x’(y’+y) + x(y’ + y)
= x’ + x
= 1
x y minterm
0 0 x’y’ m0
0 1 x’y m1
1 0 xy’ m2
1 1 xy m3
2 Variable K-Map
x y x + y
0 0 0 m0
0 1 1 m1
1 0 1 m2
1 1 1 m3
x y xy
0 0 0 m0
0 1 0 m1
1 0 0 m2
1 1 1 m3
3 Variable K-Map
 The number of cells in 3 variable K-map is 23=eight
x y z minterm
0 0 0 x’y’z’ m0
0 0 1 x’y’z m1
0 1 0 xy’z’ m2
0 1 1 x’yz m3
1 0 0 xy’z’ m4
1 0 1 xy’z m5
1 1 0 xyz’ m6
1 1 1 xyz m7
3 Variable K-Map
 The possible combinations of grouping 2 adjacent min
terms are {(m0, m1), (m1, m3), (m3, m2), (m2, m0), (m4,
m5), (m5, m7), (m7, m6), (m6, m4), (m0, m4), (m1,
m5), (m3, m7) and (m2, m6)}.
x y z minterm
0 0 0 x’y’z’ m0
0 0 1 x’y’z m1
0 1 0 xy’z’ m2
0 1 1 x’yz m3
1 0 0 xy’z’ m4
1 0 1 xy’z m5
1 1 0 xyz’ m6
1 1 1 xyz m7
3 Variable K-Map
 The possible combinations of grouping 4 adjacent min
terms are {(m0, m1, m3, m2), (m4, m5, m7, m6), (m0, m1,
m4, m5), (m1, m3, m5, m7), (m3, m2, m7, m6) and (m2, m0,
m6, m4)}.
x y z minterm
0 0 0 x’y’z’ m0
0 0 1 x’y’z m1
0 1 0 xy’z’ m2
0 1 1 x’yz m3
1 0 0 xy’z’ m4
1 0 1 xy’z m5
1 1 0 xyz’ m6
1 1 1 xyz m7
3 Variable K-Map
YZ 00 01 11 10
0
0 0 1 1
1
1 1 0 0
x
F = x’y +xy’
3 Variable K-Map
YZ 00 01 11 10
0
0 0 1 0
1
1 0 1 1
x
3 Variable K-Map
 F = A’C(B+B’) + A’B(C+C’) + AB’C + BC(A+A’)
 F = A’BC+A’B’C+A’BC+A’BC’+AB’C+ABC+A’BC
 F = A’BC+A’B’C+A’BC’+AB’C+ABC
3 Variable K-Map
4 Variable K-Map
 The number of cells in 4 variable K-map is sixteen.
4 Variable K-Map
4 Variable K-Map
4 Variable K-Map
5 Variable K-Map
 The number of cells in 5 variable K-map is thirty-
two.
6 Variable K-Map
 The number of cells in 6 variable K-map is 64.
Product of Sum Simplification
CD
00 01 11 10
00 1 1 0 1
01 0 1 0 0
11 0 0 0 0
10 1 1 0 1
AB
F = B’C’ + B’D’ + A’C’D
Product of Sum Simplification
CD
00 01 11 10
00 0 0 1 0
01 1 0 1 1
11 1 1 1 1
10 0 0 1 0
AB
F’ = AB + CD +BD’
F’ =(3,4,6,7,11,12,13,14,15)
F = (A’+B’).(C’+D’).(B’+D)
Product of Sum Simplification
F’ = AB + CD +BD’
F = (A’+B’).(C’+D’).(B’+D)
Product of Sum Simplification
Simplification of Boolean Function
NAND Implementation
NAND Implementation
NOR Implementation
NAND/NOR Implementation
NAND/NOR Implementation
NAND/NOR Implementation
DON’T CARE Conditions
DON’T CARE Conditions
DON’T CARE Conditions
DON’T CARE Conditions

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Simplification of Boolean Function

  • 1. Dr. Heman Pathak, Associate Professor Dept. of Comp. Sc., Kanya Gurukul DehradunChapter -3
  • 2. Canonical Forms 2 Any Boolean function is said to be in its canonical form if it is expressed either as  Sum of minterms or as  Product of maxterms
  • 3. Karnaugh Map  Karnaugh introduced a method for simplification of Boolean functions in an easy way.  This method is known as Karnaugh map method or K-map method.  It is a graphical method, which consists of 2n cells for ‘n’ variables. The adjacent cells are differed only in single bit position.
  • 4. 2 Variable K-Map  The number of cells in 2 variable K-map is four, since the number of variables is two. The adjacent cells are differed only in single bit position. x y minterm 0 0 x’y’ m0 0 1 x’y m1 1 0 xy’ m2 1 1 xy m3
  • 5. 2 Variable K-Map  F1 = m0+m1=x’y’ + x’y = x’(y’ + y) = x’  F2 = m1+m3=x’y + xy = y(x’ + x) = y  F3 = m3+m2=xy + xy’ = x(y + y’) = x  F4 = m0+m2=x’y’ + xy’ = y’(x’ + x)= y’  F =m0+m1+m2+m3  F = x’y’ + x’y + xy’ + xy = x’(y’+y) + x(y’ + y) = x’ + x = 1 x y minterm 0 0 x’y’ m0 0 1 x’y m1 1 0 xy’ m2 1 1 xy m3
  • 6. 2 Variable K-Map x y x + y 0 0 0 m0 0 1 1 m1 1 0 1 m2 1 1 1 m3 x y xy 0 0 0 m0 0 1 0 m1 1 0 0 m2 1 1 1 m3
  • 7. 3 Variable K-Map  The number of cells in 3 variable K-map is 23=eight x y z minterm 0 0 0 x’y’z’ m0 0 0 1 x’y’z m1 0 1 0 xy’z’ m2 0 1 1 x’yz m3 1 0 0 xy’z’ m4 1 0 1 xy’z m5 1 1 0 xyz’ m6 1 1 1 xyz m7
  • 8. 3 Variable K-Map  The possible combinations of grouping 2 adjacent min terms are {(m0, m1), (m1, m3), (m3, m2), (m2, m0), (m4, m5), (m5, m7), (m7, m6), (m6, m4), (m0, m4), (m1, m5), (m3, m7) and (m2, m6)}. x y z minterm 0 0 0 x’y’z’ m0 0 0 1 x’y’z m1 0 1 0 xy’z’ m2 0 1 1 x’yz m3 1 0 0 xy’z’ m4 1 0 1 xy’z m5 1 1 0 xyz’ m6 1 1 1 xyz m7
  • 9. 3 Variable K-Map  The possible combinations of grouping 4 adjacent min terms are {(m0, m1, m3, m2), (m4, m5, m7, m6), (m0, m1, m4, m5), (m1, m3, m5, m7), (m3, m2, m7, m6) and (m2, m0, m6, m4)}. x y z minterm 0 0 0 x’y’z’ m0 0 0 1 x’y’z m1 0 1 0 xy’z’ m2 0 1 1 x’yz m3 1 0 0 xy’z’ m4 1 0 1 xy’z m5 1 1 0 xyz’ m6 1 1 1 xyz m7
  • 10. 3 Variable K-Map YZ 00 01 11 10 0 0 0 1 1 1 1 1 0 0 x F = x’y +xy’
  • 11. 3 Variable K-Map YZ 00 01 11 10 0 0 0 1 0 1 1 0 1 1 x
  • 12. 3 Variable K-Map  F = A’C(B+B’) + A’B(C+C’) + AB’C + BC(A+A’)  F = A’BC+A’B’C+A’BC+A’BC’+AB’C+ABC+A’BC  F = A’BC+A’B’C+A’BC’+AB’C+ABC
  • 14. 4 Variable K-Map  The number of cells in 4 variable K-map is sixteen.
  • 18. 5 Variable K-Map  The number of cells in 5 variable K-map is thirty- two.
  • 19. 6 Variable K-Map  The number of cells in 6 variable K-map is 64.
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  • 21. Product of Sum Simplification CD 00 01 11 10 00 1 1 0 1 01 0 1 0 0 11 0 0 0 0 10 1 1 0 1 AB F = B’C’ + B’D’ + A’C’D
  • 22. Product of Sum Simplification CD 00 01 11 10 00 0 0 1 0 01 1 0 1 1 11 1 1 1 1 10 0 0 1 0 AB F’ = AB + CD +BD’ F’ =(3,4,6,7,11,12,13,14,15) F = (A’+B’).(C’+D’).(B’+D)
  • 23. Product of Sum Simplification F’ = AB + CD +BD’ F = (A’+B’).(C’+D’).(B’+D)
  • 24. Product of Sum Simplification
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