UGC Net Computer Science Paper II Dec 13 Page 4 Solved
31. The dual of a Boolean expression is obtained by interchanging
(A) Boolean sums and Boolean products
(B) Boolean sums and Boolean products or interchanging 0’s and 1’s
(C) Boolean sums and Boolean products and interchanging 0’s & 1’s
(D) Interchanging 0’s and 1’s
Answer C
Explanation :-
Dual of a Boolean expression
The duality principle ensures that "if we exchange every symbol by its dual in a formula, we get the dual result".
(a) 0 . 1 = 0: is a true statement asserting that "false and true evaluates to false"
(b) 1 + 0 = 1: is the dual of (a): it is a true statement asserting that "true or false evaluates true."
(c) 1 . 1 = 1: it is a true statement asserting that "true and true evaluates to true".
(d) 0 + 0 = 0: (d) is the dual of (c): it is a true statement asserting, correctly, that "false or false
More example
F1 = x'yz' + x' y' z
The dual of F1 is : ( x' + y + z' ) ( x' + y' + z )
Complement each literal : F1' = ( x + y' + z ) ( x + y + z' )
F2 = x ( y' z' + yz )
The dual of F2 is x + ( y' + z' ) ( y + z )
Complement each literal: F2' = x' + ( y + z ) ( y' + z' )
Product of Sums (POS)
A boolean expression consisting purely of Maxterms (sum terms) is said to be in canonical product of sums form.
Example
Lets say, we have a boolean function F defined on two variables A and B. So, A and B are the inputs for F and lets say, output of F is true i.e., F = 1 when only one of the input is true or 1.
now we draw the truth table for F
Now we will create a column for the maxterm using the variables A and B. If input is 1 we take the complement of the variable and if input is 0 we take the variable as is.
To get the desired canonical POS expression we will multiply the maxterms (sum terms) for which the output is 0.
F = (A+B) . (A’+B’)
