1. Lecture 4 Nand, Nor Gates, CS147 Circuit Minimization and
Karnaugh Maps
Prof. Sin-Min Lee
Department of Computer Science

2. Boolean Algebra to Logic Gates
Logic circuits are built from components called logic gates.
The logic gates correspond to Boolean operations +, *, ’.
Binary operations have two inputs, unary has one OR +
AND *
NOT ’

3. Logic Circuits ≡ Boolean Expressions
All logic circuits are equivalent to Boolean expressions and any boolean expression can be rendered as a logic circuit.
AND-OR logic circuits are equivalent to sum-of-products form.
Consider the following circuits: A B C A
Y=AB*+B*C B
ABC C A B C Y
AB*C Y
A*B
y=ABC+AB*C+A*B

4. NAND and NOR Gates
NAND and NOR gates can greatly simplify circuit diagrams. As we will see, can you use these gates wherever you could use AND, OR, and NOT. A B
AB 1
NAND A B
AB 1
NOR

5. XOR and XNOR Gates
XOR is used to choose between two mutually exclusive inputs. Unlike OR, XOR is true only when one input or the other is true, not both. A B
AB 1 A B
A B 1
XOR
XNOR

6. NAND and NOR as Universal Logic Gates
Any logic circuit can be built using only NAND gates, or only NOR gates. They are the only logic gate needed.
Here are the NAND equivalents:

7. NAND and NOR as Universal Logic Gates (cont)
Here are the NOR equivalents:
NAND and NOR can be used to reduce the number of required gates in a circuit.

8. Example Problem
A hall light is controlled by two light switches, one at each end. Find (a) a truth function, (b) a Boolean expression, and (c) a logic network that allows the light to be switched on or off by either switch.
Let x and y be the switches: x y
f(x,y) 1
(What kind of gate has this truth table?

9. x y
f(x,y) 1
Example (cont)
One possible equation is the complete sum-of-products form:
f(X,Y) = XY* + X*Y
Use The Most Complex Machine xLogicCircuit Module to implement the
equation.

10. Hint 2 : Use 2 NAND gates to build 2 NOT gates
How to use NAND gates to build an OR gate?
Truth Table A B C D Q A B C D Q 1
Hint 1 : Use 3 NAND gates
Hint 2 : Use 2 NAND gates to build 2 NOT gates
Hint 3 : Put the 3rd NAND gate after the 2 “NOT” gates

11. Hint 2 : Use 3 NAND gates to build an OR gate
How to use NAND gates to build a NOR gate?
Truth Table A B C D Q E A B C D E Q 1
Hint 1 : Use 4 NAND gates
Hint 2 : Use 3 NAND gates to build an OR gate
Hint 3 : Use a NOR gate to build a NOT gate
Hint 4 : Put the “NOT” gate after “OR” gate

15. Since functions can be represented by a sum of product form
of minterms, any function can be shown in the map by
placing each a 1 in each square which represents a minterm
in the function.
EXAMPLE: Draw the map for f = X + Y
1. Determine which minterms are needed by using truth table.
X Y f = X + Y minterm m0 m1 m2 m3

16. 2. Draw the map and place a 1 in each square for required minterms. Y X 1 1 X 1
f = X + Y = m1 + m2 + m3 = X´ Y + X Y´ + X Y

17. Use the 2-Variable Karnaugh Map
for Minimization
Example: Given f (X,Y) =  (0, 2)
Find: Simplified sum of products Y Y X m0 m1 X m2 m3

18. 2-Variable Karnaugh Map
1. Place the 1’s corresponding to the minterms on
the map.
f (X,Y) =  (0, 2) Y Y X 1 X 1

19. 2-Variable Karnaugh Map
2. Now group the 1’s into columns or rows; in this case
we can group them in the first column.
f (X,Y) =  (0, 2) Y Y X 1 X 1
3. This column is Y´ so the simplified function f (X,Y) = Y´

20. 2-Variable Karnaugh Map
Example 2: Given f (X,Y) =  (0, 2, 3)
Find: Simplified sum of products Y Y X m0 m1 X m2 m3

21. 2-Variable Karnaugh Map
Example 3: Given f (X,Y) =  (0, 1)
Find: Simplified sum of products Y Y X m0 m1 X m2 m3

22. Three Variable Map m0 m1 m3 m2 m4 m5 m7 m6 Y YZ X 0 0 0 1 1 1 10
X´ Y´ Z´ X´ Y´ Z X´ Y Z X´ YZ´ 1
X Y´ Z´ XY´Z X Y Z X Y Z´ X Z

23. Three Variable Map There are 2N = 23 = 8 squares.
The minterms are arranged so that only one variable
changes from 0 to 1 or from 1 to 0 as you move from
square to square in the vertical or horizontal direction.
For any two adjacent squares, only one literal changes
from complemented to non-complemented (normal).
From this property the left and right ends of the map
are “adjacent.”

24. Y YZ X
X´Y´Z´ X´ Y´ Z X´ Y Z X´ Y Z´ X 1
X Y´Z´ X Y´ Z X Y Z X YZ´ Z
Using the distributive and complement properties, any two
adjacent minterms can be combined and simplified to a single
term with one less literal.

25. Combining terms on the Karnaugh Map simplifies Boolean
functions.
Example: combine adjacent squares for X´ Y´ Z and XY´Z
X´Y´ Z + XY´Z = Y´ Z ( X´ + X) = Y´ Z · 1 = Y´ Z
Example: Combine adjacent squares for X Y´Z´ and X YZ´
XY´Z´ + X YZ´ = X Z´ ( Y´ + Y) = X Z´ · 1 = X Z´ Y YZ X
X´Y´Z´ X´ Y´ Z X´ Y Z X´ Y Z´ X 1
X Y´Z´ X Y´ Z X Y Z X YZ´ Z

26. Example: Simplify
f = X´Y´ Z´ + X Y Z + X´Y´ Z + X´ Y Z Y YZ X 1 1 1 1 X 1 Z
f = X´Y´ + Y Z

27. Even if the function is not in its simplest form, we can still
use the map to simplify it further.
Example: Simplify f = X´ Y´ + Y´ Z + X´ Z + X Y Z Y YZ X 1 1 1 1 1 X 1 Z
f = Z + X´ Y´ (by further grouping of minterms)

28. 3-Variable Karnaugh Map
Example: Given f (X,Y,Z) =  (0, 2, 3, 4, 7)
Find: Simplified sum of products YZ Y X m0 m1 m3 m2 m5 m7 m6 m4 X Z

29. 3-Variable Karnaugh Map
Example: Given f (X,Y,Z) =  (0, 2, 3, 4, 7)
Simplified sum of products: f = YZ + YZ + XY YZ Y X 1 1 1 1 1 X Z

30. Truth Table
f (X,Y,Z) =  (0, 2, 3, 4, 7)
Simplified sum of products: f = Y´Z´ + YZ + X´Y
The truth table for f:
X Y Z f

31. f (X,Y,Z) =  (0, 2, 3, 4, 7)
= Y´Z´ + YZ + X´Y Y Z Y f Z X Y

32. 3-Variable Karnaugh Map
Example 2: Given f (X,Y,Z) =  (2, 3, 4, 5)
Find: Simplified sum of products YZ Y X m0 m1 m3 m2 m5 m7 m6 m4 X Z

33. 3-Variable Karnaugh Map
Example 2: Given f (X,Y,Z) =  (1, 2, 5, 6, 7)
Find: Simplified sum of products YZ Y X m0 m1 m3 m2 m5 m7 m6 m4 X Z

34. 3-Variable Karnaugh Map
Example 3: Given f (X,Y,Z) = X´Z + X´Y + XY´Z + YZ
Find: “Sum of minterms” expression YZ Y X m0 m1 m3 m2 m5 m7 m6 m4 X Z

35. Exclusive OR XOR
f = X´Y + XY´ = X  Y
X Y f

36. Exclusive OR
f (X,Y) =  (1, 2) = X  Y Y Y X 1 1 X

37. XOR Truth Table
f (X,Y,Z) = X  Y  Z =  (1, 2, 4, 7)
The truth table for f:
X Y Z f

38. Exclusive OR X Y
f (X,Y,Z) = X  Y  Z Z

39. Four Variable Map Y YZ WX 0 0 0 1 1 1 10 m0 m1 m3 m2 m4 m5 m7 m6 00 01
m m m3 m2
m4 m5 m7 m6
W x´y´z 00 01
m m m m14
m m m11 m10 X 11
w x y´z W 10 Z

40. Four Variable Map N = 4 variables 2N = 24 = 16 square (minterms)
Row and column are numbered using a reflected-code
sequence. The minterm number can be obtained by
concatenation of the row and column number .
Example: Row 4 = 10, Column 2 = 01 giving
1001 = 9 decimal for W X´ Y´ Z.

41. 1 Y YZ 0 0 0 1 1 1 10 wx´y´z wx´y´z´ wx´y´z´ w´x´y z´ 00 01 X 11 W 10
wx´y´z 1
wx´y´z´
wx´y´z´
w´x´y z´ 00 01 X 11 W 10 Z
Notice that top and bottom edges and right and left edges are
“adjacent.”

42. 1 square = a term with 4 literals
16 square = a function equal to 1 Y YZ WX 00 01 X 11 W 10 Z

43. Simplify: f (W, X, Y, Z) = W X´ Z + W X Z + W´ Y Z
f = W Z + Y Z
f = Z ( W + Y) Y YZ WX 00 01 X 11 W 10 Z

44. Simplify: f (W, X, Y, Z) =  ( 0, 1, 4, 5, 6, 8, 9, 12, 13, 14)
f = Y´ + WZ´ + XZ´ Y YZ WX 00 01 X 11 W 10 Z
8 square square
4 square

45. Simplify: f = W´ X´ Y´ + X´Y Z´ + W´ X Y Z´ + W X´ Y´
f = X´ Z´ + X´ Y´ + W´ Y Z´ Y YZ WX 00 01 X 11 10 Z

46. 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.

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

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

49. 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

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

51. Complement and Product of Sums
On the K Map a 1 was placed in the squares which represented
minterms for a function. The squares not used represent the
complement of the function. Mark these squares with 0,
combine them and read the complement as a sum of products
function, f´. Using De Morgan’s Law on f´ will give f, the original
Function. It will be in a product of sums form.
Example: Determine the Product of sums form
for f ( W, X, Y, Z) =  (0, 1, 2, 5, 8, 9, 10)
f´ = W X + YZ + X Z´ from the map
f = (W´ + X´) ( Y´ + Z´) ( X´ + Z) by De Morgan’s

52. Complement and Product of Sums
Also, can determine the product of sums form directly from
The map by recognizing that the 0’s on the map represent
maxterms. Remember that a 1 on the edge of the map
Represents a complemented literal for maxterms.
Thus, for f ( W, X, Y, Z) =  (0, 1, 2, 5, 8, 9, 10)…….
(combine 0’s on k-map)…
f = ( Y´ + Z´) (W´ + X´) ( X´ + Z)

53. Y YZ WX 00 01 11 W 10 Z
f = ( Y´ + Z´ ) ( W´ + X´ ) ( X´ + Z)