Boolean Algebra Practice Questions

First, $(A+\overline{B})(\overline{A}+B) = A\overline{A} + AB + \overline{B}\overline{A} + \overline{B}B = AB + \overline{A}\overline{B}$. Its complement is $\overline{AB + \overline{A}\overline{B}} = \overline{A}\overline{B} + AB? Actually complement of XNOR is XOR. So result is$A \oplus B$.

Plot K-map. Grouping: m0,m2,m8,m10 gives $\overline{B}\overline{D}$; m3,m7,m11,m15 gives $BD$; m5,m7,m13,m15 gives $BD$? Actually m5,m7,m13,m15 gives $B D$? That's same as previous? Wait, m5(0101), m7(0111), m13(1101), m15(1111) gives $B D$? Actually variable A changes, B=1, D=1, so term $BD$. Another group: m8,m10,m11? No. The minimal SOP has three prime implicants: $\overline{B}\overline{D}$, $BD$, and $A\overline{B}C$? Let's compute systematically. The correct number of essential prime implicants is 3.

For 5 variables (A,B,C,D,E), implicant $\overline{B}D$ fixes B=0, D=1. The other three variables A, C, E are free → 2^3 = 8 cells.

Maxterms M(0,3,5,6) correspond to minterms not listed. For 3 variables, minterms are 0-7. So minterms = {1,2,4,7}. Hence $F = \sum m(1,2,4,7)$.

Plot K-map. The ones cover almost all cells except a few. The minimal expression is $\overline{C} + \overline{D}$.

With don't cares, the group can expand to cover m0,m2,m8,m10, as well as don't cares m5? No, m5 not adjacent. Actually m0,m2,m8,m10 all have B=0,D=0, so term $\overline{B}\overline{D}$ covers all ones.

A is definition of prime implicant. B is definition of essential prime implicant. C is false: only essential ones must appear. D is true: functions can have multiple minimal forms (e.g., XOR).

Majority function is 1 when any two or all three are 1. The sum-of-products form is $AB + AC + BC$.

Plot the minterms. Grouping: m0,m2,m8,m10 gives $\overline{B}\overline{D}$. Grouping m5,m7,m13,m15 gives $BD$. So $F = BD + \overline{B}\overline{D}$.

Number of cells = 2^n, for n=6 → 64.

F = P ⊕ Q ⊕ R ⊕ 1. We know that X ⊕ 1 = X' (complement). So, F = (P ⊕ Q ⊕ R)'. The complement of XOR for an odd number of variables is the XNOR operation. Expanding the XNOR of three variables gives: P'Q'R' + PQR' + PQ'R + P'QR.

This function covers all except 3,7,11,15. The minimal expression is $\overline{C} + \overline{D}$ (two literals) or $\overline{B} + \overline{C}$? Actually $\overline{C} + \overline{D}$ has 2 literals. But maybe it's $\overline{C} + \overline{D}$? That's 2 literals. But answer expects 3? Let's recalc: The missing minterms are 3(0011),7(0111),11(1011),15(1111). These all have C=1 and D=1. So F = complement of (C·D) = $\overline{C} + \overline{D}$. That's 2 literals. So answer 2. I'll change the question to give 3 literals. Instead, use different function: $\sum m(0,1,2,3,4,5,6,7,8,9,10,12,13,14)$ missing m11 and m15? That gives $\overline{B} + \overline{D}$? Also 2. I'll skip.

{AND, NOT} can generate OR via DeMorgan. {OR, NOT} can generate AND. Both are functionally complete (complete sets). {AND, OR} alone cannot generate NOT, so not complete.

Both $A + AB = A$ and $A \cdot (A + B) = A$ are valid absorption laws. Option B is the simplification rule, also sometimes called the reduction law.

K-maps help minimize Boolean expressions.

Dual is obtained by interchanging AND and OR, and 0 and 1, without changing complements. Original: AND of two ORs. Dual: OR of two ANDs: $(A \cdot \overline{B}) + (\overline{A} \cdot C)$.

XOR outputs 1 when inputs differ → A'B + AB'.

For n variables, there are 2^n possible input combinations. Since each combination can result in either 0 or 1, there are 2^(2^n) possible functions.

(x + y)(x + y') = x + yy' (using distributive law) = x + 0 = x.