1. Computer Engineering (Logic Circuits) (Karnaugh Map)
Lec. # 5
(Karnaugh Map)
Dr. Tamer Samy Gaafar
Dept. of Computer & Systems Engineering
Faculty of Engineering
Zagazig University

2. Course Web Page http://www.tsgaafar.faculty.zu.edu.eg

3. Lec. # 5
Karnaugh Map

4. F = X’Y’Z + XY’Z’ + XY’Z + XYZ’ + XYZ
SOP Functions
Take 1’s from Truth Table
Truth Table 1
1 1 1
1 1 0
1 0 1
1 0 0
0 1 1
0 1 0
0 0 1
0 0 0
X Y Z F
F = X’Y’Z + XY’Z’ + XY’Z + XYZ’ + XYZ
Simplify
F = x + y’z

5. F’= x’y’z’ + x’yz’ + x’yz F = (x+y+z).(x+y’+z).(x+y’+z’)
POS Functions
Take 0’s from truth table (F’)
Apply Demorgan (F)
Truth Table 1
1 1 1
1 1 0
1 0 1
1 0 0
0 1 1
0 1 0
0 0 1
0 0 0
X Y Z F
F’= x’y’z’ + x’yz’ + x’yz
F = (x+y+z).(x+y’+z).(x+y’+z’)

6. Universal Gates: NAND and NOR
AND/OR/NOT gates are sufficient for building any Boolean functions.
We call the set {AND, OR, NOT} a complete set of logic.

7. (x.x)' = x' (T1: idempotency)
NAND Gate (1/2)
NAND gate is self-sufficient (can build any logic circuit with it).
Therefore, {NAND} is also a complete set of logic.
Can be used to implement AND/OR/NOT.
Implementing an inverter using NAND gate: x x'
(x.x)' = x' (T1: idempotency)

8. (x.y)= ((x.y)')' idempotency
NAND Gate (2/2)
Implementing AND using NAND gates: x
x.y y
(x.y)'
(x.y)= ((x.y)')' idempotency
= (xy) involution
Implementing OR using NAND gates: x
x+y y x' y'
X+Y = (X+Y)’’
= ( X‘ . Y’) ‘

9. (x+x)' = x' (T1: idempotency)
NOR Gate (1/2)
NOR gate is also self-sufficient.
Therefore, {NOR} is also a complete set of logic
Can be used to implement AND/OR/NOT.
Implementing an inverter using NOR gate: x x'
(x+x)' = x' (T1: idempotency)

10. NOR Gate (2/2) Implementing AND using NOR gates: (XY) = (XY)’’
Implementing OR using NOR gates: x
x+y y
(x+y)'
x+y = ((x+y)')‘
= (x+y)

11. Outline Universal Gates : NAND and NOR NAND Gate NOR Gate
Implementation using NAND Gates
Implementation using NOR Gates

12. Implementation using NAND gates (1/2)
Possible to implement any Boolean expression using NAND gates.
Procedure:
(i) Obtain sum-of-products Boolean expression:
Example: F3 = x.y'+x'.z
(ii) Use DeMorgan theorem to obtain expression using 2-level NAND gates
= (x.y'+x'.z)' ' involution
= ((x.y')' . (x'.z)')' DeMorgan

13. Implementation using NAND gates (2/2)x' z F3
(x'.z)'
(x.y')' x y'
F3 = ((x.y')'.(x'.z)') ' = x.y' + x'.z
NAND gate is used to implement a SOP Function

14. Outline Universal Gates : NAND and NOR NAND Gate NOR Gate
Implementation using NAND Gates
Implementation using NOR Gates

15. Implementation using NOR gates (1/2)
Possible to implement any Boolean expression using NOR gates.
Procedure:
(i) Obtain product-of-sums Boolean expression:
Example: F6 = (x+y').(x'+z)
(ii) Use DeMorgan theorem to obtain expression using 2-level NOR gates.
= ((x+y').(x'+z))' ' involution
= ((x+y')'+(x'+z)')' DeMorgan

16. Implementation using NOR gates (2/2)x' z F6
(x'+z)'
(x+y')' x y'
F6 = ((x+y')'+(x'+z)')' = (x+y').(x'+z)
NOR gate is used to implement a POS Function

17. Implementing a SOP using NOR gates
Procedure:
(i) Obtain sum-of-products Boolean expression:
Example: F3 = x.y'+x'.z
(ii) Take the Complement for both sides
G = F3’= (x.y’+x’.z)’
Then
G = F3’ = (x’+y) . (x+z’)
Note:
G = F3’ > POS ( NOR Implemention)
but needs an Inverter

18. Implementation using NOR gates (2/2)X Z’
(x+z’)'
(x’+y)' X’ Y
G = F3’ F3
G = F3’ = (x’+y) . (x+z’)
G’’ = [(x’+y)’ + (x+z’)’]’
NOR gate is used to implement a POS Function

19. Implementing POS using NAND gates
Possible to implement any Boolean expression using NAND gates.
Procedure:
(i) Obtain product-of-sums Boolean expression:
Example: F7 = (x+y').(x'+z)
(ii) Use DeMorgan theorem to obtain expression using 2-level to represent as SOP
F7’= x’.y+x.z’ DeMorgan

20. Implementation using NAND gates (2/2)X’ Z’
F7’
(X.Z’)'
(X’.Y)' Y F7
F7’ = ((x’.y)'.(x.z’)') ' = x.y' + x.z’
NAND gate is used to implement a POS Function

21. Exclusive-OR (XOR)
Exclusive-OR (XOR) produces a HIGH output whenever the two inputs are at opposite levels.

22. Exclusive-NOR Circuits (XNOR)
Exclusive-NOR (XNOR) produces a HIGH output whenever the two inputs are at the same level.

23. XOR function can also be implemented with AND/OR gates (also NANDs).

24. Positive and Negative Logic
Positive logic AND Negative Logic AND F B A 1 F B A 1

25. Karnaugh Maps (K maps)

26. What are Karnaugh maps?
Karnaugh maps provide an alternative way of simplifying logic circuits.
Instead of using Boolean algebra simplification techniques, you can transfer logic values from a Boolean statement or a truth table into a Karnaugh map.
The arrangement of 0's and 1's within the map helps you to visualize the logic relationships between the variables and leads directly to a simplified Boolean statement.

27. Karnaugh maps A A A A B B B B 00 10 01 11
Karnaugh maps, or K-maps, are often used to simplify logic problems with 2, 3 or 4 variables.
Cell = 2n ,where n is a number of variables
For the case of 2 variables, we form a map consisting of 22=4 cells
as shown in Figure A A A A A A 1 1 B B 00 10 01 11 B B 2 B 1 B 1 1 3
Minterm

28. Karnaugh maps A A C C B B B 3 variables Karnaugh map Cell = 23=8 AB C00 01 11 10 2 6 4 C 1 3 7 5 C 1 B B B

29. Karnaugh maps A A C D D C D B B B 4 variables Karnaugh map AB 00 01 1110 CD 4 12 8 C 00 D 1 5 13 9 01 D 3 7 15 11 11 C 2 6 14 10 10 D B B B

30. Cell numbers are written in the cells. 2-variable K-map
Examples of K-Maps:
Examples:
Cell numbers are written in the cells.
2-variable K-map B B B 1 A A 1 2 3 A 1

31. 3-Variable K-Map: 3-variable K-map B B 00 01 11 10 00 01 11 10 A 1 1 3BC BC A A A 1 1 3 2 4 5 7 6 1 A C C C

32. 4-variable K-map CD AB 00 1 3 2 4 5 7 6 12 13 15 14 8 9 11 10 01 11 10

33. Four-variable K-maps

34. Karnaugh maps
The Karnaugh map is completed by entering a ‘1’ (or ‘0’) in each of the appropriate cells.
Within the map, adjacent cells containing 1's (or 0’s) are grouped together in twos, fours, or eights.

35. Karnaugh Mapping
K-Map Format
Digital Electronics
2.2 Intro to NAND & NOR Logic
Each minterm in a truth table corresponds to a cell in the K- Map.
K-Map cells are labeled such that both horizontal and vertical movement differ only by one variable.
Since the adjacent cells differ by only one variable, they can be grouped to create simpler terms in the sum-of-products expression.
The sum-of-products expression for the logic function can be obtained by OR-ing together the cells or group of cells that contain 1s.
Explain how a K-Map is formatted and how a simplified logic expression can be written from the cell groupings.
Project Lead The Way, Inc.
Copyright 2009

36. Simplification Guidelines for K-maps
Each cell of an n-variable K-map has n logically adjacent cells.
Cells may be combined in groups of 2,4,8,…,2k.
A group of cells can be combined only if all cells in the group have the same value for some set of variables.
Always combine as many cells in a group as possible. This will result in the fewest number of literals in the term that represents the group.
Make as few groupings as possible to cover all minterms. This will result in the fewest product terms.
Always begin with the “loneliest” cells.

37. Minimization with K-maps
Draw a K-map
Combine maximum number of 1’s following rules:
Only adjacent squares can be combined
All 1’s must be covered
Covering rectangles must be of size 1,2,4,8, … 2n
Check if all covering are really needed
Read off the SOP expression

38. 2 variable K-Maps (Adjacency)
In an n-variable k-map, each square is adjacent to exactly n other squares
Q: What if you have 1 in all squares?
Ahmad Almulhem, KFUPM 2009

39. Adjacent Cells = Simplification
Karnaugh Mapping
Adjacent Cells = Simplification
Digital Electronics
2.2 Intro to NAND & NOR Logic V 1 V V 1
Adjacent cells only vary by one variable change. Review the proof that the one variable changing results in it being eliminated algebraically. This is fundamental to how a K-Map works.
Project Lead The Way, Inc.
Copyright 2009

40. Truth Table to K-Map Mapping
Karnaugh Mapping
Truth Table to K-Map Mapping
Digital Electronics
2.2 Intro to NAND & NOR Logic
Two Variable K-Map V 1 2 3 V W X
FWX
Minterm – 0 1
Minterm – 1
Minterm – 2
Minterm – 3 1 1
Demonstrate how the minterms of a two variable truth table are mapped to a K-Map.
Project Lead The Way, Inc.
Copyright 2009

41. Two Variable K-Map Groupings
Karnaugh Mapping
Digital Electronics
2.2 Intro to NAND & NOR Logic
Two Variable K-Map Groupings
Groups of One – 4 V 1 1 1 1
Two variable K-Map, groups of one.
Project Lead The Way, Inc.
Copyright 2009

42. Two Variable K-Map Groupings
Karnaugh Mapping
Digital Electronics
2.2 Intro to NAND & NOR Logic
Two Variable K-Map Groupings
Groups of Two – 4 V 1 1 1 1
Two variable K-Map, groups of two.
Project Lead The Way, Inc.
Copyright 2009

43. Two Variable K-Map Groupings
Karnaugh Mapping
Digital Electronics
2.2 Intro to NAND & NOR Logic
Two Variable K-Map Groupings
Group of Four – 1 V 1
Two variable K-Map, group of four.
Project Lead The Way, Inc.
Copyright 2009

44. Example #1: 2 Variable K-Map
Karnaugh Mapping
Example #1: 2 Variable K-Map
Digital Electronics
2.2 Intro to NAND & NOR Logic
Example:
After labeling and transferring the truth table data into the K-Map, write the simplified sum-of-products (SOP) logic expression for the logic function F1. V J K F1 1
Pause the presentation and allow the student to complete the example. The solution is on the next slide.
Project Lead The Way, Inc.
Copyright 2009

45. Example #1: 2 Variable K-Map
Karnaugh Mapping
Example #1: 2 Variable K-Map
Digital Electronics
2.2 Intro to NAND & NOR Logic
Example:
After labeling and transferring the truth table data into the K-Map, write the simplified sum-of-products (SOP) logic expression for the logic function F1.
Solution: V 1 J K F1 1
Here is the solution. If you print handouts, do not print this page.
Project Lead The Way, Inc.
Copyright 2009

46. Truth Table to K-Map Mapping
Three Variable K-Map

47. 3-variable K-maps
For 3-variable functions, the k-maps are larger and look different.
Total number of minterms that need to be accommodated in the k-map = 8
To maintain adjacency neighbors don’t have more than 1 different bit B BC A 1
m m m3 m2
m m m m6 A C

48. 3-variable K-maps BC A 00 01 11 10 1 m0 m1 m3 m2 m4 m5 m7 m6 1
m m m3 m2
m m m m6
Note: You can only combine a power of 2 adjacent 1-squares. For e.g. 2, 4, 8, 16 squares. You cannot combine 3, 7 or 5 squares BC A 1
m m m3 m2
m m m m6
Minterms mo, m2, m4, m6 can be combined as m0 and m2 are adjacent to each other, m4 and m6 are adjacent to each other
mo and m4 are also adjacent to each other, m2 and m6 are also adjacent to each other

49. Example 1 Simplify F = ∑m(1, 3, 4, 6) using K-map B BC A 00 01 11 10 11 3 2 4 5 7 6 A C

50. Example 1 F = A’C + AC’ Simplify F = ∑m(1, 3, 4, 6) using K-map B BC A00 01 11 10 1 3 2 4 5 7 6
F = A’C + AC’ A C

51. Example 2 Simplify F = ∑m(0,1, 2, 4, 6) using K-map B B A 00 01 11 101 3 2 4 5 7 6 A C

52. Example 2 F = A’ B’ + C’ Simplify F = ∑m(0,1, 2, 4, 6) using K-map B B00 01 11 10 1 3 2 4 5 7 6
F = A’ B’ + C’ A C

53. 3 variable K-Maps (Adjacency)
A 3-variable map has 12 possible groups of 2 minterms
They become product terms with 2 literals 00 01 11 10 1 00 01 11 10 1 00 01 11 10 1

54. 3 variable K-Maps (Adjacency)
A 3-variable map has 6 possible groups of 4 minterms
They become product terms with 1 literals 00 01 11 10 1 00 01 11 10 1 00 01 11 10 1

55. Example F = CD + A’D + BC + AB’C’ Minimize the function
F(A,B,C,D)=∑m(1,3,5,6,7,8,9,11,14,15) C CD AB 00 01 11 10 1 1 1 1 1 B 1 1 A 1 1 1 D
F = CD + A’D + BC + AB’C’

56. Example F(A,B,C,D) = Sm(0,1,2,5,8,9,10) CD AB 00 01 11 10 1 C=1 B=1

57. Example F(A,B,C,D) =Sm(0,1,2,5,8,9,10) F = B’ D’ + B’ C’ + A’ C’ D
Solution:
F = B’ D’ + B’ C’ + A’ C’ D
CD AB 00 01 11 10 1
C=1
B=1
A=1
D=1

58. Example (POS) F(A,B,C,D) = Sm(0,1,2,5,8,9,10)
Write F in the simplified product of sums (POS)
Two methods?
You already know one!
CD AB 00 01 11 10 1
C=1
B=1
A=1
D=1

59. Example (POS) F’’ = F = (A’+B’)(C’+D’)(B’+D)
F(A,B,C,D) = Sm(0,1,2,5,8,9,10)
Write F in the simplified product of sums (POS)
Follow same rule as before but for the
ZEROs
F’ = AB + CD + BD’
Therefore,
F’’ = F = (A’+B’)(C’+D’)(B’+D)
CD AB 00 01 11 10 1
C=1
B=1
A=1
D=1

60. Truth Table to K-Map Mapping
Karnaugh Mapping
Truth Table to K-Map Mapping
Digital Electronics
2.2 Intro to NAND & NOR Logic
Four Variable K-Map W X Y Z
FWXYZ
Minterm – 0
Minterm – 1 1
Minterm – 2
Minterm – 3
Minterm – 4
Minterm – 5
Minterm – 6
Minterm – 7
Minterm – 8
Minterm – 9
Minterm – 10
Minterm – 11
Minterm – 12
Minterm – 13
Minterm – 14
Minterm – 15 1 V 1 3 2 4 5 7 6 12 13 15 14 8 9 11 10 1 1 1
Demonstrate how the minterms of a four variable truth table are mapped to a K-Map.
Project Lead The Way, Inc.
Copyright 2009

61. Four Variable K-Map Groupings
Karnaugh Mapping
Digital Electronics
2.2 Intro to NAND & NOR Logic
Four Variable K-Map Groupings
Groups of One – 16 (not shown)
Groups of Two – 32 (not shown)
Groups of Four – 24 (seven shown) 1 1 1 V 1 1 1 1
Four Variable K-Map:
There are 16 groups of one that are not shown because they are obvious.
There are 32 groups of two that are not shown because they are obvious.
There are 24 groups of four. Seven are show, the other 17 are obvious.
Project Lead The Way, Inc.
Copyright 2009

62. Four Variable K-Map Groupings
Karnaugh Mapping
Digital Electronics
2.2 Intro to NAND & NOR Logic
Four Variable K-Map Groupings
Groups of Eight – 8 (two shown) 1 V 1
Four variable K-Map: There are 8 groups of eight. Two are shown, the other 6 are obvious.
Project Lead The Way, Inc.
Copyright 2009

63. Four Variable K-Map Groupings
Karnaugh Mapping
Digital Electronics
2.2 Intro to NAND & NOR Logic
Four Variable K-Map Groupings
Group of Sixteen – 1 V 1
Four variable K-Map, group of 16.
Project Lead The Way, Inc.
Copyright 2009

64. Example : 4 Variable K-Map
Karnaugh Mapping
Example : 4 Variable K-Map
Digital Electronics
2.2 Intro to NAND & NOR Logic
Example:
After labeling and transferring the truth table data into the K-Map, write the simplified sum-of-products (SOP) logic expression for the logic function F3. R S T U F3 1 V
Pause the presentation and allow students to complete the example. The solution is on the next slide.
Project Lead The Way, Inc.
Copyright 2009

65. Example : 4 Variable K-Map
Karnaugh Mapping
Digital Electronics
2.2 Intro to NAND & NOR Logic
Example:
After labeling and transferring the truth-table data into the K-Map, write the simplified sum-of-products (SOP) logic expression for the logic function F3. R S T U F3 1 V 1
Solution:
Here is the solution. If you print handouts, do not print this page.
Project Lead The Way, Inc.
Copyright 2009

66. Steps of graphical method (continued):
Combine adjacent 1’s into group of 2n each such that
Each group contains only 1’s.
The group is not completely a part of a larger group.
Choose the minimum number of the largest sized groups needed to cover all the 1’s.
Each group is represented by an expression which is an intersection of the minterm in the group.
The simplified solution is a logical OR of the expressions of all the groups chosen in steps 3 above.

67. Definitions An implicant is a product term of a function
Any group of 1’s in a K-Map
A prime implicant is a product term obtained by combining the maximum possible number of adjacent 1’s in a k-map
Biggest groups of 1’s
Not all prime implicants are needed!
If a minterm is covered by exactly one prime implicant then this prime implicant is called an essential prime implicant

68. Example Y=1 Consider F(X,Y,Z) = Sm(1,3,4,5,6)
List all implicants, prime implicants and essential prime implicants
Solution:
Implicants: XY’Z’, XZ’, XY’, XY’Z, X’Y’Z, Y’Z, …
P.Is: XY’, XZ’, Y’Z, X’Z
EPIs: X’Z, XZ’
YZ X 00 01 11 10 1
X=1
Z=1
Y=1
YZ X 00 01 11 10 1
X=1
The simplest expression is NOT unique!
Z=1

69. Finding minimum SOP
Find each essential prime implicant and include it in the solution
If any minterms are not yet covered, find minimum number of prime implicants to cover them (minimize overlap).

70. Example Simplify F(A, B, C, D) = ∑ m(0, 1, 2, 4, 5, 10,11,13, 15)
Note:
Only A’C’ is E.P.I
For the remaining minterms:
Choose 1 and 2 (minimize overlap)
For m2, choose either A’B’D’ or B’CD’
F = A’C’ + ABD + AB’C + A’B’D’

71. Examples to illustrate terms
0 0
1 1
1 0 D A
0 1 B C
5 prime implicants: BD, ABC', ACD, A'BC, A'C'D
essential
minimum cover: 4 essential implicants

72. 4 Variable Maps f(A,B,C,D) = m(0,1,2,3,6,8,9,11,13,14)
f = A C' D + B C D' + B' C ' + B' D + A'B'
C D A B 00 01 11 10 1

73. These are not essential!
Another 4 Variable Map
f(A,B,C,D) = m(0,1,2,5,6,7,8,9,10,13,15)
f = B D + A' B C + B' D' + B' C' or
f = B D + A' B C + B' D' + C' D (there are 2 more)
C D A B 00 01 11 10 1
These are not essential!

74. Terminology
An implicant is a product term in an SOP expression (or a sum term in POS expression)
Implicants are always rectangular in shape and the # of 1's covered is a power of 2
A prime implicant is an implicant that is not fully contained in some other larger implicant
B C A 00 01 11 10 1
red  implicant (but not prime) many more are not shown
blue  prime implicants (only two)

75. Essential Prime Implicants
An essential prime implicant is a prime implicant that contains a 1 not included in any other prime implicant
The minimum Boolean expression must use this term
A cover is a collection of implicants that accounts for all valuations in which the function is “on” (e.g. 1)

76. essential prime implicants
Find the Essential Prime Implicants
C D A B 00 01 11 10 1
C D A B 00 01 11 10 1
essential prime implicants
f = C'D' + A'B + B'D' + ACD

77. 1 square = a term with 4 literals 2 square = a term with 3 literal
16 square = a function equal to 1 Y YZ WX 00 01 X 11 W 10 Z

78. Simplification of kmap
Generate all PIs
Find EPIs
If EPIs can cover all minterms, then it is answer. Otherwise choose some non-essential PIs (which has less cost) such that all minterms are cover.

79. Step 1 Generate PIs Blue circle are PIs
They are the largest circle you can drawn on kmap

80. Step 2 Find EPIs Red circle are EPIs
Minterm 5, 14, 11 can only be cover by these 3 red circle

81. Step 3 EPIs cannot cover minterm 7
Choose between green/blue circle to cover minterm 7
Green is chosen as it is larger
Less cost

82. Final
Final result is obtained
x3x4’ + x2’x3 + x1’x3 + x2x3’x4