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 1 :{0,1,..., } {0,1, } i F p X
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 { } - PowerPoint PPT Presentation
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MVI Function Review• Input X is p-valued variable. Each Input can have Value in Set {0, 1, 2, ..., pi-1}
• literal over X corresponds to subset of values of S {0, 1, ... , p-1} denoted by XS or X{j} where j is the logic value
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 =01 – 100 – 110011 – 111 – 0010
11 – 110 – 111011 – 110 – 000111 – 001 – 1111
F2 =
Z0 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.
F G F(|gj) 1 (j = 1, . . , q) and G(|fj) 1 (i = 1, . . , p)
Example: F = xy + y and G = x + xy
i = 1 j = 1
p q
F(|x) 1, F(|xy) 1, G(|xy) 1, and G(|y) 1, Thus F G
Divide and Divide and Conquer Conquer MethodMethod
Divide and Conquer Method
Let F be a SOP and ci (i = 1, 2,. . , k) be the cubes satisfying the following conditions:
F = ci 1 and ci cj = 0 (i j ).
Then, can partition SOP into k SOPs
F = ci F(|ci)
Operations can be done on each F(|ci) independently and then combined to get result on F
i = 1
k
i = 1
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(|ci) )
and thus the number of products.Partition Example:
k = 2, c1 = XjS
A , c2 = XjS
B
SA SB = Pj and SA SB =
i = 1
k
Divide and Conquer MethodUsing 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 MethodSelection Method:1. Chose all the variables with the
maximum number of active columns 12. 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 MethodExample:
X2 and X3 have the largest number of active columns.
Problems to remember and solveProblems to remember and solve
1. Multi-output Multi-valued prime implicants2. Covering for MV functions.3. Cofactors of MV functions.4. Visualization of MV functions5. Complementation of MV functions.6. Decision trees and Decision Diagrams for MV functions7. Cube Calculus operations and Algorithms for MV functions.