1. Gate-level Minimization
Chapter 3
Gate-level Minimization

2. Chapter3: Gate-level Minimization
3-2 The Map Method
Two-variable map and Three-variable map
3-3 Four-Variable Map
3-4 Five-variable Map
3-5 Product of Sums Simplification
3-6 Don’t-care Conditions
3-7 NAND and NOR Implementation
3-8 Other Two-LeveI Implementations
3-9 Exclusive-OR Function

3. 3-2 The Map Method • Simplification of Boolean Expression • Methods:
– Minimum # of terms, minimum # of literals
– To reduce complexity of digital logic gates
– The simplest expression is not unique
• Methods:
– Algebraic minimization  lack of specific rules
• Section 2.4
– Karnaugh map or K-map
• Combination of 2, 4, … adjacent squares
Logic circuit  Boolean function  Truth table  K-map
 Canonical form (sum of minterms, product of maxterms)
 (Simplifier) standard form (sum of products, product of sums)

4. 3-2 The Map Method
A Karnaugh map is a graphical tool for assisting in the general simplification procedure.

5. Two-Variable Maps
2 variables  4 minterms  4 squares. Y’ Y X’ X

6. Rules for K-Maps We can reduce functions by circling 1’s in the K-map
Each circle represents minterm reduction
Following circling, we can deduce minimized and-or form.
Rules to consider
Every cell containing a 1 must be included at least once.
The largest possible “power of 2 rectangle” must be enclosed.
The 1’s must be enclosed in the smallest possible number of rectangles.

7. Two-Variable Maps (Cont.)1 1 b b 1 1 1 1 1 1
g = b'
f = a
F=AB+A’B A 1 B

8. Y X

9. 2-Variable Map Simplification Example (1)
Example: F(X,Y) = XY’ + XY
From the map, we see that
F (X,Y) = X.
Note: There are implied 0s in other boxes.
This can be justified using algebraic manipulations: F(X,Y) = XY’ + XY
= X(Y’ +Y)
= X.1
= X
1 1 X

10. 2-Variable Map Simplification Example (2)
G(x,y) = m1 + m2 + m3 Y
G(x,y) = m1 + m2 + m3
= X’Y + XY’ + XY
From the map, we can see that G = X + Y 1 X

11. 2-Variable Map Simplification Example (3)
F = Σ(m0,m1)
Using algebraic manipulations
F = Σ(m0,m1)
= x’y + x’y’
= x’ (y+y’)
= x’
x y F
0 0 1
0 1 1
1 0 0
1 1 0 1 y x
x’y’
xy’ xy
x’y 1 y x X’

12. Three-Variable Maps 3 variables 8 minterms (m0 – m7).
How can we locate a minterm square on the map?
 Use figure (a) OR  use column # and row # from figure (b) E.g. m5 is in row 1 column 01 (5 10 = 101 2)
Q. Show the area representing X’? Y’? Z’?

13. • 8 minterms for 3 binary variables
• Any two adjacent squares differ by only
one variable yz yz

14. Three-Variable Maps (Cont.)
By combining squares in powers of 2, we reduce number of literals in a product term, reducing the literal cost, thereby reducing the other two cost criteria.
On a 3-variable K-Map:
One square represents a minterm with three variables
Two adjacent squares represent a product term with two variables
Four “adjacent” terms represent a product term with one variables
Eight “adjacent” terms is the function of all ones (logic 1).

15. 3-Variable Map Simplification Example (1)
Adjacent Squares
m0+m2 = X’Y’Z’ + X’YZ’ = X’Z’(Y’+Y) = X’Z’
m4+m6 = XY’Z’ + XYZ’ = XZ’(Y’+Y) = XZ’

16. 3-Variable Map Simplification Example (2)
F = X’Y’Z’ + X’YZ’ + XY’Z’ + XYZ’
= Z’ (X’Y’ + X’Y + XY’ + XY)
= Z’ (X’ (Y’+Y) + X (Y’+Y))
= Z’ (X’+ X)
= Z’

17. 3-Variable Map Simplification Example (3)
F=AB’C’ +ABC +ABC +ABC + A’B’C + A’BC’ A 1 00 01 BC 11 10
F=A+BC +BC

18. 3-Variable Map Simplification Example (4)00 10 11 01 1 X Z Y YZ
Example:
F (x, y, z)= Σm (2, 3, 6, 7)
Applying the Minimization Theorem three times:
Thus the four terms that form a 2 × 2 square correspond to the term "y". 1 Y y = z yz + x ) , ( F

19. 3-Variable Map Simplification Example (5)
Example: Simplify
F (x, y, z)= Σm (2, 3, 4, 5) 00 10 11 01 1 X Z Y YZ
X’Y 1
XY’
F(X,Y,Z) = X’Y + XY’

20. 3-Variable Map Simplification Example (6)
Example: Simplify
G (a, b, c)= Σm (3, 4, 6, 7) 00 10 11 01 1 a c b X bc bc a 1
ac’
G (a,b,c) = bc + ac’

21. 3-Variable Map Simplification Example (7)
Example: Simplify
F(X, Y, Z) = X’Z + X’Y + XY’Z + YZ
F(X, Y, Z) = Σm (1, 2, 3, 5, 7) z y x + 00 10 11 01 1 X Z Y YZ
In general, as more squares are combined, we obtain a product term with fewer literals.
Overlap is allowed. 1
F(x, y, z) =

22. Examples 3-1 and 3-2

23. Examples 3-3 and 3-4 One square represents one minterm, giving a
term of three literals
• Two adjacent squares
represent a term of two
literals
• Four adjacent squares
represent a term of
one literal

24. 3-3 Four-Variable Map
• Two adjacent squares represent a term of three literals
• Four adjacent squares represent a term of two literals
• Eight adjacent squares represent a term of one literal
The larger the number of squares combined, the smaller the
number of literals in the term

25. Flat Map Vs Torus

26. Examples 3-5 and 3-6 x x

27. 4-Variable Map Simplification Example
F = m(0,1,2,4,5,6,8,9,12,13,14)
F = Y’ + XZ’ + W’Z’

28. 4-Variable Map Simplification Example
F = m(0,2,4,5,6,7,10,13,15)
Do it and show it to me next time!

29. Simplification Using Prime Implicants
A Prime Implicant is a product term obtained by combining the maximum possible number of adjacent squares in the map into a rectangle with the number of squares a power of 2.
A prime implicant is called an Essential Prime Implicant if it is the only prime implicant that covers (includes) one or more minterms.
Prime Implicants and Essential Prime Implicants can be determined by inspection of a K-Map. 29

30. 1 30

31. Example of Prime Implicants
Find ALL Prime Implicants CD
ESSENTIAL Prime Implicants C 1 B D A 1 B C D A D B D B C B
Minterms covered by single prime implicant BD BD AD B A

32. Prime Implicant Practice
Find all prime implicants for:
F (W,X,Y,Z) = (0,2,3,8,9,10,11,12,13,14,15) X Y Z W 1 1 1
Prime implicants are: W, X‘Y, and X‘Z‘
Note that all of these prime implicants are essential. 1 1 1 1 1 1 1 1
Prime implicants are: W, X‘Y, and X‘Z‘
Note that all of these prime implicants are essential. 32

33. Another Example Find all prime implicants for:
G(W,X,Y,Z) = (0,2,3,4,7,12,13,14,15)
Hint: There are seven prime implicants! X Y Z W 1 1 1
Prime Implicants are WX, XY’Z', W’Y’Z', W’X’Z', W’X’Y, W’YZ, XYZ.
There is only one essential prime implicant: WX. 1 1 1 1 1 1
Prime Implicants are WX, XY’Z', W’Y’Z', W’X’Z', W’X’Y, W’YZ, XYZ.
There is only one essential prime implicant: WX. 33

34. 3-4 Five-Variable Map 34

35. 35

36. Example 3-7 36

37. 3-5 Product-of-Sums Simplification
E.g. Simplify the following Boolean function in product-of-sums form
F(A,B,C,D)= m(0,1,2,5,8,9,10)
Mark with 1’s the minterms of F.
Mark the remaining squares with 0’s. These represent F’.
Find the simplified F’. F’ = AB + CD + BD’
Complement 3 to obtain a simplified F in product-of-sums form F = (A’ + B’)(C’ + D’) (B’ + D)

38. Product-of-Sums Simplification
Simplify : F= (0,1,2,5,8,9,10) in Product-of-Sums Form
Mark with 0’s the Maxterms of F
Combine 0’s to obtain a simplified F’ in SOP.
Complement 2 to obtain a simplified F in POS. A D C B
F’ = AB + CD + BD’
F = (A’+B’)(C’+D’)(B’+D)
F’ = AB + CD + BD’
F = (A’+B’)(C’+D’)(B’+D) 38

39. Example 3-8 Simply Boolean function F(A, B, C, D) =(0, 1, 2,5, 8, 9,
10) in (a) sum of products and (b) product of sums
(b)
(a) F = B’D’
1. Obtain simplified complemented function:
+ B’C‘
F’ = AB + CD+BD’
+ A’C’D
2. Applying DeMorgan’s theorem to obtain F
F = (A’ + B’) (C’+D’) (B’ + D) 39

40. Gate Implementation for Example 3-840

41. F(x, y, z) = (1, 3, 4, 6) = (0, 2, 5, 7)
F(x, y, z) = (1, 3, 4, 6) = (0, 2, 5,7)
F = x’z + xz‘
F’ = xz + x’z’  F = (x’ + z’)(x + z) 41

42. 3.6 Don't Cares Condition
Sometimes a function table or map contains entries for which it is known:
the input values for the minterm will never occur, or
The output value for the minterm is not used
In these cases, the output value need not be defined
Instead, the output value is defined as a “don't care”
By placing “don't cares” ( an “x” entry) in the function table or map, the cost of the logic circuit may be lowered.
Example 1: A logic function having the binary codes for the BCD digits as its inputs. Only the codes for 0 through 9 are used. The six codes, 1010 through never occur, so the output values for these codes are “x” to represent “don’t cares.”

43. Example 3-9
product of sums? 43

44. 3-7 NAND and NOR Implementation
Digital circuits are frequently constructed with NAND
or NOR gates rather than with AND and OR gates 44

45. NAND Circuits NAND gate: a universal gate
– Any digital system can be implemented with it
• including AND, OR and complement = 45

46. Two-Level Implementation with NAND
sum of product expression and its equivalent
NAND implementation
F = AB + CD
= [(AB + CD)’]’
= [(AB)’*(CD)’]’ 46

47. Example 3-10 Implement the following Boolean function with NAND gates47

48. Procedures of Implementation with two levels of NAND gates
1. Express simplified function in sum of products
2. Draw a NAND gate for each product term that
has at least two literals to constitute a group of
first-level gates
3. Draw a single gate using AND-invert or invert-
OR in the second level
4. A term with a single literal requires an inverter in
the first level 48

49. Multilevel NAND Circuits
1. Convert all AND gates to NAND gates with AND-invert graphic symbols
2. Convert all OR gates to NAND gates with invert-OR graphic symbols
3. Check all the bubbles in the diagrams. For a single bubble, invert an
inverter (one-input NAND gate) or complement the input literal ‘ 49

50. Figure 3-23 Implementing F=(AB’ + A’B)(C+D’)50

51. NOR Circuite •The NOR operation is the dualof the NAND operation
•The NOR gate is anothar universal gate to implement any
Boolean function = 51

52. NOR Implementation Transformation from OR-AND diagram to NOR diagram
•OR gates => OR-invert
•AND gate => invert-AND 52

53. 3-8 Other Two-level Implementations
wired logic: some NAND or NOR gates allow a cirect wire
connection between the outputs of two gates to provide a
specific logic function
– The wired-AND gate or wired-OR gate is not a physical second-level
gate, but only a symbol 53

54. Nondegeneratd Forms
• 16 possible combinations of two-level forms with 4
types of gates: AND, OR, NAND, and NOR
– 8 degenerate forms: degenerate to a single operation
– 8 generate forms
• NAND-AND = AND-NOR = AND-OR-INVERT
OR-NAND = NOR-OR = OR-AND-INVERT
1 \ 2 st nd
AND OR
NAND
NOR
AND
AND 3-4 NAND = OR
3-4 OR x NOR
NAND
= NAND 3-6 AND
NOR
NOR x OR 3-6 54

55. AND-OR-INVERT Implementation
AND-NOR = NAND-AND = AND-OR-INVERT
F = (AB + CD + E)’
Similar to AND-OR, AND-OR-INVERT requires an
expression in sum of products
Given F, we can implement F’ with AND-OR-INVERT 55

56. OR-AND-INVERT Implementation
OR-NAND = NOR-OR = OR-AND-INVERT
F = [ (A+B) (C+D) E ] ’
Similar to OR-AND, OR-AND-INVERT requires an
expression in products of sum
Given F, we can implement F’ with OR-AND-INVERT 56

57. Exxmple 3-11 Other Two-level Implementations F = x’y’z’ + xyz’
F’ = x’y + xy’ + z
F = (x’)’ = (x’y + xy’ + z)’ 57

58. 3-8 Exclusive-OR (XOR) Function
XOR: x  y = xy’ + x’y
Exclusive-NOR = equivalenxe
(x  y)’ = (xy’ + x’y)’
= (x’ + y)(x + y’) = x’y’ + xy
Communtative: A  B = B  A
Associative: (A  B)  C = A 
(B  C) = A  B  C
Only a limited number of Boolean
functions can be expressed in terms of
XOR operations, but it is particularly
useful in arithmetic operations and
error-detection and correction circuits 58

59. Odd Function
• The 3-variable XOR function is equal to 1 if only one variable
is equal to 1 or if all three variables are equal to 1
• Multiple-variable exclusive OR operation = odd function : odd
number of variables be equal to 1 59

60. map for a 4-variable XOR Function60

61. Parity Generation and Checking
parity bit: an extra bit included with a binary message
to make the number of 1’s either odd or even
3-bit even-parity-generator
P=xyz 61

62. 4-bit Even-Parity-Checker
C = xy  zP
P = 0 4-bit even parity checker
= 3-bit even parity generator 62