1. 4.2 Strategy for minimization So far, we used an intuitive approach to decide how the 1s in the K-map should be grouped together to obtain the minimum-cost impl. –Approach: Find as few (fewer product terms) as possible and as large (fewer number of variables in each term) as possible groups of 1s that cover all cases where the function has a value of 1. –This works well as the K-maps were small. For larger logic functions, which have many variables, this approach is unsuitable. –We must have an organized method for deriving a min-cost impl. –We will introduce a method similar to the one used in CAD tools. 1

2. Literal –Each appearance of a variable in a product term, either un-complemented or complemented. Implicant –a product term that indicates the input valuation for which a given function is equal to 1. Most basic implicants are minterms: For an n-variable function, a minterm is an implicant consisting of n literals Prime implicant –An implicant that it is impossible to delete any literal and still have a valid implicant. 2

3. More terminology Cover –A collection of implicants that account for all valuations for which a given function is equal to 1. Multiple different covers exist for most functions. A set of all minterms for which f = 1 is a cover; a set of all prime implicants is also a cover. Cost –The number of gates plus the total number of inputs to all gates in the circuit. 3

4. Figure 4.9. Three-variable function f (x 1, x 2, x 3 ) =  m(0, 1, 2, 3, 7). An example 4

5. Minimization procedure Lowest-cost implementation is achieved when the cover of a given function consists of prime implicants. –How to determine the minimum-cost subset of prime implicants that will cover the function? –Essential prime implicant If a prime implicant includes a minterm for which f=1 that is not included in any other prime implicant, then it must be included in the cover. 5

6. Process of finding a minimum-cost circuit 1.Generate all prime implicants for the given function f. 2.Find the set of essential prime implicants. 3.If the set of essential prime implicants covers all valuations for which f=1, then this set is the desired cover of f. otherwise, determine the nonessential prime implicants that should be added to form a complete minimum-cost cover. 6

7. Figure 4.10. Four-variable function f ( x 1,…, x 4 ) =  m(2, 3, 5, 6, 7, 10, 11, 13, 14). 7

8. Figure 4.11. The function f ( x 1,…, x 4 ) =  m(0, 4, 8, 10, 11, 12, 13, 15). 8

9. Figure 4.12. The function f ( x 1,…, x 4 ) =  m(0, 2, 4, 5, 10, 11, 13, 15). 9

10. 4.3 Minimization of POS forms 10 Figure 4.13. POS minimization of f (x 1, x 2, x 3 ) =  M(4, 5, 6).

11. 11 x 1 x 2 x 3 x 4 0 00011110 000 0110 1101 1111 00 01 11 10 x 2 x 3 +  x 3 x 4 +  x 1 x 2 x 3 x 4 +++  Figure 4.14. POS minimization of f ( x 1,…, x 4 ) =  M(0, 1, 4, 8, 9, 12, 15).

12. 4.4 Incompletely specified functions 12 x 1 x 2 x 3 x 4 0 00011110 1d0 01d0 00d0 11d1 00 01 11 10 x 2 x 3 x 3 x 4 (a) SOP implementation x 1 x 2 x 3 x 4 0 00011110 1d0 01d0 00d0 11d1 00 01 11 10 x 2 x 3 +  x 3 x 4 +  (b) POS implementation Figure 4.15. Two implementations of the function f ( x 1,…, x 4 ) =  m(2, 4, 5, 6, 10) + D(12, 13, 14, 15). d can represent either 0 or 1.