1. Multi-Valued Input Two-Valued Output Functions

2. Multi-Valued Input Slide 2 Example Automobile features 0123 X1X1 TransManAuto X2Doors234 X3ColourSilverRedBlackBlue

3. Multi-Valued Input Slide 3 Example Function Table X1X1 X2X2 X3X3 F 0001 0010 0020 0030 0100 0111 0120 0131 X1X1 X2X2 X3X3 F 0200 0210 0221 0230 1001 1010 1020 1031 X1X1 X2X2 X3X3 F 1100 1111 1120 1131 1200 1210 1221 1230

4. Multi-Valued Input Slide 4 Definition A mapping F: P 1  P 2  P n  is a multi-valued input two-valued output function, P i = {0,1, … p i-1 }, and B = {0,1}. Let X be a variable that takes one value in P = {0,1, …, p-1}. Let S  P. Then X S is a literal of X. When X  S, X S = 1, and when X  S, X S = 0. Let S i  P i, then X 1 S 1 X 2 S 2 … X n S n is a logical product. We can also define minterm and SOP Any binary function can be represented this way Problem: Find the SOP for the function given in the previous slide.

5. Multi-Valued Input Slide 5 Bit Representation Example: Bit representation for the previous example: x1 x2 x3 01-012-0123 11-100-1000 11-010-0101 11-001-0010 01-110-0001

6. Multi-Valued Input Slide 6 Restriction

7. Multi-Valued Input Slide 7 Example

8. Multi-Valued Input Slide 8 Restriction Theorem c F = c F(|c) The restriction is also called cofactor What is the relation Shannon’s expansion?

9. Multi-Valued Input Slide 9 Example

10. Multi-Valued Input Slide 10 Tautology When the logical expression F is equal to 1 for all the input combinations, F is a tautology. The problem of determining whether a given logical expression is a tautology or not is the tautology decision problem. Example. Which are tautologies?

11. Multi-Valued Input Slide 11 Inclusion Relation Let F and G be logic functions. For all the minterms c such that F(c) = 1, if G(c) = 1, then F ≤ G, and G contains F. If a logic function F contains a product c, then c is an implicant of F. Let c be a logical product and F be a logical expression. Then c ≤ F  F(|c)  1.

12. Multi-Valued Input Slide 12 Example When c 2 = (11-010-1101)?

13. Multi-Valued Input Slide 13 Equivalence The following theorem shows that the determination of the equivalence of two SOPs is transformed to the tautology problem:

14. Multi-Valued Input Slide 14 Divide and Conquer Method (9.1)

15. Multi-Valued Input Slide 15 Divide and Conquer Method By using the previous theorem, a given SOP is partitioned into k SOPs. In performing some operation on F, first expand F into the form (9.1). Then, for each F(|c i ), apply the operation independently. Finally, by combining the results appropriately, we have the results for the operation on F. Since, the same operation can be applied to F(|c i ) as to F, this method can be computed by a recursive program. Definition. Let t(F) be the number of products in an SOP F.

16. Multi-Valued Input Slide 16 Divide and Conquer Methods

17. Multi-Valued Input Slide 17 Selection Method for Variables Chose variables that have the maximum number of active columns Among those, chose variables where the total sum of 0’s in the array is maximum Example

18. Multi-Valued Input Slide 18 Complementation of SOPs

19. Multi-Valued Input Slide 19 Complementation of SOPs When n = 10, this method is about 100 times faster than using De Morgan’s theorem

20. Multi-Valued Input Slide 20 Example Find the complement for F

21. Multi-Valued Input Slide 21 Tautology Decision When there is a variable X i and at least one constant a  P i satisfying F(|X i = a) ≤ F(X 1, …, X i, …, X n ), the function F is weakly unate with respect to the variable X i. In an array F, consider the sub-array consisting of cubes that depend on X i. In the variable X i in this array, if all the values in a column are 0, then the SOP F is weakly unate with respect to the variable X i.

22. Multi-Valued Input Slide 22 Example Consider the F. Is F weakly unate? 1111-1110-1110 1111-1101-1101 0110-0110-1101 0101-0111-1101

23. Multi-Valued Input Slide 23 Theorems

24. Multi-Valued Input Slide 24 Tautology Decision

25. Multi-Valued Input Slide 25 Generation of Prime Implicants Let X be a variable that takes a value in P = {0, 1, …, p-1}. If there is a total order (  ) on the values of variable X in function F, such that j  k (j,k  P) implies F(|X=j) ≤ F(|X=k), then the function is strongly unate with respect to X. If F is strongly unate with respect to all variables, then the function F is strongly unate. Assume that F is an SOP. If there is a total order (  ) among the values of a variable X, and if j  k, then each product term of the SOP F(|X=j) is contained by all the product terms of the SOP F(|X=k). In this case the SOP F is strongly unate with respect to X. Lemma. If F is strongly unate with respect to X i, then F is weakly unate with respect to X i.

26. Multi-Valued Input Slide 26 Generation of Prime Implicants

27. Multi-Valued Input Slide 27 Generation of Prime Implicants

28. Multi-Valued Input Slide 28 The Sharp Operation Sharp operation (#) and disjoined sharp operation ( # ) compute F  G. Example. Let a = (11-11-11) and b = (01-01- 01). Find a#b and a # b Example. Let B = {b 1,b 2 }, where b 1 = (11-11- 11) and b 2 = (10-10-10). Let C={c 1,c 2,c 3 }, where c 1 = (10-11-11), c 2 = (11-10-11), and c 3 = (11-11-10). Find a#C and a # C