1. Chapter 3
Gate-level Minimization

2. 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.

3. 1

4. 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

5. 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.

6. 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 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
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.

7. 3-4 Five-Variable Map

9. Example 3-7

10. 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)

11. 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)

12. 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)

13. Gate Implementation for Example 3-8

14. 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)

15. 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, through 1111 never occur, so the output values for these codes are “x” to represent “don’t cares.”

16. Example 3-9
product of sums?