Question

Consider the following 4-variable Boolean function
𝐹(𝐴, 𝐵, 𝐶, 𝐷) = Σ𝑚(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:

All eight given minterms have \(B=0\), while \(A\), \(C\), \(D\) vary freely, so the function simplifies directly to the single literal \(B'\).

Answer 1

The Correct Option is B

Step 1: List the minterms in binary using \(A\) (MSB), \(B\), \(C\), \(D\) (LSB): \(m0=0000\), \(m1=0001\), \(m2=0010\), \(m3=0011\), \(m8=1000\), \(m9=1001\), \(m10=1010\), \(m11=1011\).
Step 2: Notice that in every one of these minterms, the bit \(B\) is always \(0\): for \(m0\)-\(m3\), \(A=0,B=0\); for \(m8\)-\(m11\), \(A=1,B=0\). Meanwhile \(C\) and \(D\) take all four possible combinations (\(00,01,10,11\)) in both groups, and \(A\) takes both \(0\) and \(1\).
Step 3: This means the function value is \(1\) precisely when \(B=0\), regardless of the values of \(A\), \(C\), and \(D\), since all combinations of \(A\), \(C\), \(D\) appear whenever \(B=0\), and no minterm with \(B=1\) (i.e., \(m4,m5,m6,m7,m12,m13,m14,m15\)) is included in the sum.
Step 4: Using a 4-variable Karnaugh map, the two groups of 4 minterms \(\{0,1,2,3\}\) and \(\{8,9,10,11\}\) each form a group of four adjacent cells, and together they combine into a single group of eight cells covering the entire \(B=0\) half of the map, which simplifies to the single literal \(B'\).
Step 5: This is the minimal sum-of-products form since a single literal (group of 8 minterms) is the largest possible grouping and cannot be reduced further.
Final Answer: \(\boxed{B'}\)