multi-valued input two-valued output functions

Multi-Valued InputTwo-Valued Output

Functions

Multi-Valued Input Slide 2

Example

• Automobile features

0 1 2 3

X1 Trans Man Auto

X2 Doors 2 3 4

X3 Colour Silver Red Black Blue


Example

• Function Table

X1 X2 X3 F

0 0 0 1

0 0 1 0

0 0 2 0

0 0 3 0

0 1 0 0

0 1 1 1

0 1 2 0

0 1 3 1

X1 X2 X3 F

0 2 0 0

0 2 1 0

0 2 2 1

0 2 3 0

1 0 0 1

1 0 1 0

1 0 2 0

1 0 3 1

X1 X2 X3 F

1 1 0 0

1 1 1 1

1 1 2 0

1 1 3 1

1 2 0 0

1 2 1 0

1 2 2 1

1 2 3 0


Definition

• A mapping F: P1 P2 Pn is a multi-valued input two-valued output function, Pi = {0,1, … pi-1}, and B = {0,1}. Let X be a variable that takes one value in P = {0,1, … , p-1}. Let S P. Then XS is a literal of X. When X S, XS = 1, and when X S, XS = 0. Let Si Pi, then X1

S1 X2S2… Xn

Sn 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.


Bit Representation

• Example: Bit representation for the previous example:

x1 x2 x3

01-012-012311-100-100011-010-010111-001-001001-110-0001


Restriction

QuickTime™ and aTIFF (LZW) decompressor

are needed to see this picture.


Example




Restriction

• Theorem c F = c F(|c)• The restriction is also called cofactor• What is the relation Shannon’s expansion?


Example




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?




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.


Example

• When c2 = (11-010-1101)?




Equivalence

• The following theorem shows that the determination of the equivalence of two SOPs is transformed to the tautology problem:






Divide and Conquer Method

(9.1)


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(|ci), 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(|ci) 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.


Divide and Conquer Methods




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




Complementation of SOPs




Complementation of SOPs

• When n = 10, this method is about 100 times faster than using De Morgan’s theorem




Example

• Find the complement for F




Tautology Decision

• When there is a variable Xi and at least one constant a Pi satisfying F(|Xi = a) ≤ F(X1, … , Xi, … , Xn), the function F is weakly unate with respect to the variable Xi.

• In an array F, consider the sub-array consisting of cubes that depend on Xi. In the variable Xi in this array, if all the values in a column are 0, then the SOP F is weakly unate with respect to the variable Xi.


Example

• Consider the F. Is F weakly unate?

1111-1110-11101111-1101-11010110-0110-11010101-0111-1101


Theorems




Tautology Decision




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 Xi, then F is weakly unate with respect to Xi.


Generation of Prime Implicants




The Sharp Operation

• Sharp operation (#) and disjoined sharp operation ( # ) compute FG.

• Example. Let a = (11-11-11) and b = (01-01-01). Find a#b and a # b

• Example. Let B = {b1,b2}, where b1 = (11-11-11) and b2 = (10-10-10). Let C={c1,c2,c3}, where c1 = (10-11-11), c2 = (11-10-11), and c3 = (11-11-10). Find a#C and a # C

multi-valued input two-valued output functions

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