Question

Consider the following 4-variable Boolean function

\[ F(A, B, C, D) = \Sigma m(0, 1, 2, 3, 8, 9, 10, 11) \]
Consider \(A\) as MSB, \(D\) as LSB. Which one of the following options represents the minimal sum of products form for the above function?

Note: + is OR operation, . is AND operation, ' is NOT operation

• A. \(A' + B' + C' + D'\)
• B. \(B'\)
• C. \(A'.B' + A.B\)
• D. \(A'\)

Hint:

Write out all eight minterms in binary and look for a single variable that is 0 in every one of them.

Answer 1

The Correct Option is B

A different way to reach the same answer is to build the 4-variable Karnaugh map directly and read off the grouping, instead of spotting the pattern from the binary listing.

Lay out the K-map with rows for AB (00, 01, 11, 10) and columns for CD (00, 01, 11, 10), which is the standard Gray code layout so adjacent cells differ in exactly one variable.

Row AB=00 (minterms 0,1,3,2 in column order 00,01,11,10): all four of 0,1,2,3 are in the given list, so this entire row is 1.
Row AB=01 (minterms 4,5,7,6): none of 4,5,6,7 are in the given list, so this entire row is 0.
Row AB=11 (minterms 12,13,15,14): none of these are in the given list, so this entire row is 0.
Row AB=10 (minterms 8,9,11,10): all four of 8,9,10,11 are in the given list, so this entire row is 1.

So the map has two full rows of 1s: the AB=00 row and the AB=10 row, and two full rows of 0s: the AB=01 row and the AB=11 row. Both 1-rows share the property B=0 (row 00 has A=0,B=0; row 10 has A=1,B=0), and together they form the largest possible group on the map, since a group covering all four columns (all of C, D) and both rows where B=0 covers all 8 ones in a single rectangle of size 8. A group of 8 cells eliminates 3 variables from the term, leaving only 1 literal.

Since the group spans both values of A (rows 00 and 10 differ in A) and both values of C and D (all four columns), those three variables drop out of the term entirely, and only B stays, fixed at 0 throughout the group, giving the single literal $B'$.

Checking that no smaller expression is possible: since the ones occupy exactly half the map (8 out of 16 cells) in a shape that is already the largest legal group size (a group must be a power of 2 in size, and next size up would be 16, which would mean F is always 1, which is false since the AB=01 and AB=11 rows are 0), $B'$, a single literal, is already the smallest possible product term, so it is the minimal SOP.

This K-map reading matches the earlier pattern-spotting approach exactly. \[ \boxed{F = B' \ \text{(Option B)}} \]