GATE Breaker

Given $f_1$, $f_3$ and $f$ in canonical sum of products form (in decimal) for the circuit  $f_1 = \sum m(4, 5, 6, 7, 8)$,  $f_3 = \sum m(1, 6, 15)$,  $f = \sum m(1, 6, 8, 15)$,  then $f_2$ is  (A) $\sum m(4, 6)$  (B) $\sum m(4, 8)$  (C) $\sum m(6, 8)$  (D) $\sum m(4, 6, 8)$  (GATE 2008)  Answer: (C) $\sum m(6, 8)$  Explanantion:  $f = (f_1 \cap f_2) \cup f_3$  Check option (C):  $(f_1 \cap f_2) \cup f_3$  $= \big(\sum m(4, 5, 6, 7, 8) \cap \sum m(6, 8)\big) \cup \sum m(1, 6, 15)$  $= \sum m(6, 8) \cup \sum m(1, 6, 15)$  $= \sum m(1, 6, 8, 15)$   $= f$  Previous Post Next Post

The literal count of a Boolean expression is the sum of the number of times each literal appears in the expression. For example, the literal count of (xy + xz′) is 4. What are the minimum possible literal counts of the product-of-sum and sum-of-product representations, respectively, of the function given by the following Karnaugh map? Here, X denotes “don’t care”.  (A) (11, 9)  (B) (9, 13)  (C) (9, 10)  (D) (11, 11)  (GATE 2003)  Answer: (C) (9, 10)  Explanation:  Sum-of-product (SOP) = WY + W'Y' + Z'WX' + XYZ'  Therefore, literal count = 10  Product-of-sum (POS) = (Y' + Z')(W' + Z')(Z' + Y)(X + Z + W)   Therefore, linteral count = 9  Previous Post Next Post