Consider a Boolean function f(w, x, y, z) such that 

f(w, 0, 0, z) = 1

f(1, x, 1, z) = x + z

f(w, 1, y, z) = wz + y

The number of literals in the minimal sum-of-products expression of f is ______

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GATE CS 2021 Official Paper: Shift 2
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Answer (Detailed Solution Below) 6

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Detailed Solution

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Explanation :

Given:

f(w, 0, 0, z)= 1

f(1, x, 1, z)= x+z

f(w, 1, y, z)= wz+y

Only thing makes this Question complicated is how you fill the K map cells.

1. f(w, 0, 0, z) = 1 ; means whenever x = 0 and y = 0 (total 4 cells) then put 1 in k-map cell.

2. f(1, x, 1, z) = x+z ; means whenever w=1 and y=1 (total 4 cells) then evaluate the expression “x+z” as per each cell (only those where w=1 and y=1) and fill the result in that particular cell.

3. f(w, 1, y, z) = wz+y ; means whenever x=1( total 8 cells) then evaluate the expression “wz+y” as per each cell (only those where x=1) and fill the result in that particular cell.

 

1. In the Below K map first we mark x = 0 and y = 0

  1001, 0000 , 0001 , 1000 these 4 cells will be marked as 1 .

  (wx’y’z , w’x’y’z’ , w’x’y’z , wx’y’z’) likewise we can fill the case where w=1 and y=1.

 

2. Now x=1( total 8 cells)

 

                     0100,     0101,           0110,          0111,          1100,        1101,        1110,        1111

                     w’xy’z’,      w’xy’z,      w’xyz’,        w’xyz,        wxy’z’,      wxy’z,        wxyz’,       wxyz

“wz+y”

0

0

1

1

0

1

1

1

Likewise we can fill the K map as given in Image.

F1 Shraddha Raju 17.03.2021 D1

So after minimum sum of product expression will be x’y’ + xy + wz ; so total 6 literals. Answer will be 6. 

F1 Shraddha Raju 17.03.2021 D2

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