# MVI Function Review Input X is p -valued variable. Each Input can have Value in Set {0, 1, 2,..., p i-1 } literal over X corresponds to subset of values.

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MVI Function Review Input X is p -valued variable. Each Input can have Value in Set {0, 1, 2,..., p i-1 } literal over X corresponds to subset of values of S  {0, 1,..., p-1} denoted by X S or X {j} where j is the logic value Empty Literal: X {  } Full Literal has Values S={0, 1, 2, …, p-1} X {0,1,…,p-1} Equivalent to Don’t Care

SOP Bit Representation X 1 X 2 X 3 01 – 012 – 0123 11 – 100 – 1000 11 – 010 – 0101 11 – 001 – 0010 01 – 110 – 0001 c = X 1 S 1 X 2 S 2... X n S n, S i  P i Cube in an n-dimensional hyper-cube

Cofactor for MV

Restriction (Cofactor) Operation in MV cube calculus Restriction of Two-Valued Output Function F obtained by restricting Domain to D, denoted by F(|D) For SOP, the restriction is defined as follows: Let F be a SOP, and c = X 1 S 1 X 2 S 2... X n S n be a product. Then, the restriction F(|c) of F to c is obtained as follows: (1)For each product term in F, make a logical product with c. Delete the zero terms. (2)Let d = X 1 T 1 X 2 T 2... X n T n be a product obtained in (1). Replace d with X 1 (T 1  S 1 ) X 2 (T 2  S 2 )... X n (T n  S n )

Procedure for Finding F(|c) 11 – 100 – 1000 11 – 010 – 0101 11 – 001 – 0010 01 – 110 – 0001 F =F = c = (01-101-1111) Step 1: Bit-wise AND each product term in F with c 01 – 100 – 1000 01 – 001 – 0010 01 – 100 – 0001 F  c = Step 2: Bit-wise OR each product term in F  c with c F(|c) = 11 – 110 – 1000 11 – 011 – 0010 11 – 110 – 0001 Students: check this in maps and equations

Cofactor Concept By Shannon’s Expansion f (x 1, x 2,.., x n ) = x 1 f (0, x 2,.., x n ) + x 1 f (1, x 2,.., x n ) F(|c 0 ) F(|c 1 ) Where c 0 and c 1 are cubes with x 1 =0 and x 1 =1, respectively. Example: f = xy + yz + zx F =F = x y z 01 – 01 – 11 11 – 01 – 01 01 – 11 – 01 f (0,y,z) c 0 = (10-11-11) F  c 0 = [10- 01- 01] F(|c 0 ) = [11- 01- 01] f (1,y,z) c 1 = (01-11-11) F  c 1 = F(|c 1 ) = 01 – 01 – 11 01 – 01 – 01 01 – 11 – 01 11 – 01 – 11 11 – 01 – 01 11 – 11 – 01 Students: check this in maps and equations

Tautology for MV

When the logical expression F is equal to logical 1 for all the input combinations, F is a tautology. Tautology Decision Problem - determining if logical expression is or is not a tautology F1 =F1 = 01 – 100 – 1100 11 – 111 – 0010 11 – 110 – 1110 11 – 110 – 0001 11 – 001 – 1111 F2 =F2 = Z 0 1 2 3 Example: No Yes Y 0 1 2 0 1 2 X = 0 X = 1 Can confirm with K-Maps

Inclusion Relation for MV 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 F contains a product c then c is an implicant of F. 11 – 100 – 1000 11 – 010 – 0101 11 – 001 – 0010 01 – 110 – 0001 F =F = Example: c 1 = (01- 100 - 1001) F(|c 1 ) = 11 – 111 – 1110 11 – 111 – 0111 F(|c 1 )  1, c 1  F c 2 = (11- 010 - 1101) F(|c 2 ) = 11 – 111 – 0111 01 – 111 – 0011 F(|c 2 )  1, c 2  F

Equivalence Relation Let F =  f j and G =  g j then F  G  F(|g j )  1 ( j = 1,.., q ) and G(|f j )  1 ( i = 1,.., p ) Example: F = xy + y and G = x + xy i = 1j = 1 pq F(|x)  1, F(|xy)  1, G(|xy)  1, and G(|y)  1, Thus F  G

Divide and Conquer Method

Let F be a SOP and c i ( i = 1, 2,.., k ) be the cubes satisfying the following conditions: F =  c i  1 and c i  c j = 0 (i  j ). Then, can partition SOP into k SOPs F =  c i  F(|c i ) Operations can be done on each F(|c i ) independently and then combined to get result on F i = 1 k k This was already illustrated graphically to check SAT, TAUTOLOGY and other similar

Divide and Conquer Method Let t ( F ) be the number of products in an SOP F. We can use Divide and Conquer Theorem to minimize  t ( F(|c i ) ) and thus the number of products. Partition Example: k = 2, c 1 = X j S A, c 2 = X j S B S A  S B = P j and S A  S B =  i = 1 k

Divide and Conquer Method Using Divide and Conquer we use the recursive application of the restriction operation to attempt to get columns of all 0’s or 1’s (they can be ignored). A column with both 0 and 1 is active. Selection Method Selection Method: 1. Chose all the variables with the maximum number of active columns 1 2.Among the variables chosen in step 1, choose variables where the total sum of 0’s in the array is maximum 3.For all variables in step 2, find a column that has the maximum number of 0’s and from among them choose the one with the minimum number of 0’s

Divide and Conquer Method Example: X 2 and X 3 have the largest number of active columns. Choose X 3 and let S A = {0,1} and S B = {2,3} X 1 X 2 X 3 11 – 100 – 1000 11 – 010 – 0100 11 – 001 – 0010 01 – 110 – 0001 F =F = c 1 = (11- 111 - 1100) F(|c 1 ) = 11 – 100 – 1011 11 – 010 – 0111 F(|c 2 ) = 11 – 001 – 1110 01 – 110 – 1101 c 2 = (11- 111 - 0011) We split with respect to some values, like cutting a KMAP in our previous examples

Complementation of SOPS

Let F =  c i  1 and c i  c j = 0 (i  j ). Then, the complement of F is F =  c i  F(|c i ) i = 1 k k

Algorithm for Complementation of SOPS 1. F consist of one product c F = X 1 S 1 X 2 S 2... X n S n Then F = X 1 S 1 + X 1 S 1  X 2 S 2 +... + X 1 S 1  X 2 S 2... X n-1 S n-1  X n S n 2. F consist of more than one product Expand F into F = c 1  F(|c 1 ) + c 2  F(|c 2 ), where c 1 = X j S A, c 2 = X j S, S A  S B = P j and S A  S B =  F = c 1  F(|c 1 ) + c 2  F(|c 2 )

Complementation of SOPS Example 1. Expand F w.r.t. X 3 and let S A = {0,1} and S B = {2,3} X 1 X 2 X 3 11 – 100 – 1000 11 – 010 – 0100 11 – 001 – 0011 01 – 110 – 0001 F =F = c 1 = (11- 111 - 1100) F 1 = F(|c 1 ) = 11 – 100 – 1011 11 – 010 – 0111 F 2 = F(|c 2 ) = 11 – 001 – 1111 01 – 110 – 1101 c 2 = (11- 111 - 0011)

Complementation of SOPS Example (Continued) F 1 = F(|c 1 ) = 11 – 111 – 1011 11 – 110 – 0111 2. Next, expand F 1 variable X 2, F 1 = c 3  F 1 (|c 3 ) + c 4  F 1 (|c 4 ) c 3 = (11- 100 - 1111) F 3 = F(|c 3 ) = F 4 = F(|c 4 ) = c 4 = (11- 011 - 1111) 11 – 100 – 1011 11 – 010 – 0111

Complementation of SOPS Example (Continued) F 2 = F(|c 2 ) = 01 – 111 – 1101 11 – 111 – 1111 3. Next, expand F 2 variable X 2, F 2 = c 5  F 2 (|c 5 ) + c 6  F 2 (|c 6 ) c 5 = (11- 110 - 1111) F 5 = F(|c 5 ) = F 6 = F(|c 6 ) = c 6 = (11- 001 - 1111) 11 – 001 – 1111 01 – 110 – 1101

Complementation of SOPS Example (Continued) 4. F 3 through F 6 are single products so we apply alg. Step 1. 11 – 111 – 0100 F3=F3= 10 – 111 – 1111 01 – 111 - 0010 F5=F5= 11 – 001 – 1111 11 – 110 – 1000 F4=F4= F 6 = 0 11 – 111 – 1011 F3=F3= 01 – 111 – 1101 F5=F5= 11 – 110 – 0111 F4=F4= 11 – 111 – 1111 F6=F6=

Complementation of SOPS Example (Completed) 5. Combining all the products gives: F= 11 – 100 – 0100 11 – 001 – 1100 11 – 010 – 1000 10 – 110 – 0011 01 – 110 - 0010 F = c 1 F 1 + c 2 F 2 = c 1 (c 3 F 3 + c 4 F 4 ) + c 2 (c 5 F 5 + c 6 F 6 ) = c 1 c 3 F 3 + c 1 c 4 F 4 + c 2 c 5 F 5 + c 2 c 6 F 6

Problems to remember and solve 1.Multi-output Multi-valued prime implicants 2.Covering for MV functions. 3.Cofactors of MV functions. 4.Visualization of MV functions 5.Complementation of MV functions. 6.Decision trees and Decision Diagrams for MV functions 7.Cube Calculus operations and Algorithms for MV functions.

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