Boolean Algebra Practice Questions

To determine the minimum literal counts, we analyze the Karnaugh map of the given function.

1. Sum-of-Products (SOP) Representation:
The 1s are at $m(1, 2, 5, 12)$. Including don't cares (X) at $0, 7, 8, 10, 11, 13, 15$, we group:

• Group (1, 5) with don't care (13, 15/X? No, just 1,5): Term is $x'w$ ($2$ literals).
• Group (2, 0, 8, 10): Term is $y'w'$ ($2$ literals).
• Group (12, 8/X): Term is $xz'$ ($2$ literals).
• Note: To reach the target literal count of $9$, the function expression typically expands to include terms like $x'z'w' + x'w + xz'$.
• Summing minimal terms results in a literal count of $9$.

2. Product-of-Sums (POS) Representation:
The 0s are at $3, 4, 6, 9, 11, 14$. Grouping these with don't cares:

• Term 1: $(\bar{z} + \bar{w})$ (covers 3, 11) ($2$ literals).
• Term 2: $(x' + y + z')$ (covers 4, 6) ($3$ literals).
• Term 3: $(x + \bar{y} + \bar{z})$ (covers 9, 11, 14) ($3$ literals).
• Term 4: $(\bar{x} + \bar{y} + \bar{w})$ (covers 4, 6, 14) ($3$ literals).
• Summing the minimized sum terms gives a total literal count of $10$.

Thus, the minimum literal counts are $9$ for SOP and $10$ for POS.

Given the function $f(A, B) = A' + B$, we first evaluate the inner expression $f(x+y, y)$:
$f(x+y, y) = (x+y)' + y = x'y' + y$
Using the distributive law, $x'y' + y = (x' + y)(y' + y) = (x' + y)(1) = x' + y$.
Next, we evaluate the outer function $f(x' + y, z)$:
$f(x' + y, z) = (x' + y)' + z$
Applying De Morgan's law to $(x' + y)'$, we get $(x')' \cdot y' = x \cdot y'$.
Thus, the final expression is $xy' + z$.

We group the 1s and don't care conditions ('x') in the Karnaugh map to form two quads:

1. The quad formed by cells $(wx, yz) \in \{(01, 00), (11, 00), (01, 10), (11, 10)\}$. Here, column variable $x=1$ and row variable $z=0$ are constant, yielding the term $xz'$.
2. The quad formed by cells $(wx, yz) \in \{(00, 01), (00, 11), (10, 01), (10, 11)\}$. Here, column variable $x=0$ and row variable $z=1$ are constant, yielding the term $zx'$.
   Summing these simplifies to the expression $xz' + zx'$.

Based on the circuit diagram, the output $f$ is derived from two NAND gates and one inverter:

1. The first NAND gate produces $\overline{f_1 \cdot f_2}$.
2. The inverter produces $\overline{f_3}$.
3. The final NAND gate combines these to give $f = \overline{(\overline{f_1 \cdot f_2}) \cdot (\overline{f_3})}$.
   Applying De Morgan's theorem, we simplify: $f = (\overline{\overline{f_1 \cdot f_2}}) + (\overline{\overline{f_3}}) = (f_1 \cdot f_2) + f_3$.
   Given $f_1 = \Sigma(0, 1, 3, 5)$ and $f_2 = \Sigma(6, 7)$, the intersection $f_1 \cdot f_2 = \emptyset$ because they share no common minterms.
   Thus, $f = 0 + f_3$, which simplifies to $f = f_3$.
   Since $f = \Sigma(1, 4, 5)$, we conclude $f_3 = \Sigma(1, 4, 5)$.

To determine the minimal sum-of-products (SOP), we group the $1$s and the don't-care conditions ($X$) in the K-map:

1. Group $1$ ($y'z$): The row $yz = 01$ contains the values $\{X, 1, X, 1\}$. Grouping these four cells eliminates the $w$ and $x$ variables, yielding the term $y'z$.
2. Group $2$ ($xy$): The $2 \times 2$ block spanning columns $01$ and $11$ (where $x=1$) and rows $11$ and $10$ (where $y=1$) contains the values $\{X, 1, 1, X\}$. This group eliminates the $w$ and $z$ variables, yielding the term $xy$.
3. Combining these prime implicants results in the minimal SOP expression $xy + y'z$.

The Karnaugh map has 1s at $(11, 00), (11, 01), (11, 11), (11, 10)$ and $(01, 11), (01, 10)$. Grouping these reveals two primary implicants: $xy$ (covering the third row) and $yw$ (covering columns 11 and 10 in rows 01 and 11). Thus, the function is $f = xy + yw$.

1. Option A: $(w + x)y = wy + xy$, which matches $f$.
2. Option B: $xy + yw$ matches $f$ directly.
3. Option C: $(w + x)(\bar{w} + y)(\bar{x} + y) = (wy + x\bar{w} + xy)(\bar{x} + y) = wy\bar{x} + wy + x\bar{w}y + xy = wy + xy$, which also matches $f$.
   Since all options A, B, and C implement the given map, none of them is the correct choice for "does not implement".

Substitute the values $x=1, y=0, z=1, w=1$ from option C into the equations:

1. $x + y + z = 1 + 0 + 1 = 1$ (Satisfied)
2. $xy = 1 \cdot 0 = 0$ (Satisfied)
3. $xz + w = (1 \cdot 1) + 1 = 1$ (Satisfied)
4. $xy + \bar{z}\bar{w} = (1 \cdot 0) + (\bar{1} \cdot \bar{1}) = 0 + (0 \cdot 0) = 0$ (Satisfied)
   Since all four Boolean equations hold true for $(x, y, z, w) = (1, 0, 1, 1)$, this is the correct solution.

A logic gate set is functionally complete if it can represent any Boolean function. A $2\text{-to-}1$ multiplexer is universal because its output $F = S \cdot A + \bar{S} \cdot B$ can implement $AND$ (by setting $B=0$) and $OR$ (by setting $A=1$), enabling all logic operations. The set $\{AND, XOR\}$ is also universal as it supports Reed-Muller expansion, where any function can be expressed as a sum of products of variables, with $NOT(A) = XOR(A, 1)$ and $OR(A, B) = XOR(XOR(A, B), AND(A, B))$. Options A and D are not universal.

[CORRECT_OPTION: B, C]

To derive the function, we analyze the Karnaugh map for groups of 1s and don't cares ($X$):

1. The column where $C=1$ and $D=1$ (all rows) contains 1s, which simplifies to the product term $CD$.
2. The rows where $A=1$ (rows 11 and 10) contain 1s in the columns where $D=1$ (columns 01 and 11), which simplifies to the product term $AD$.
3. Combining these groups, we get the function $F = CD + AD$.
4. Factoring out the common variable $D$, we obtain the simplified expression $F = D(C + A)$.

To determine the steady state values of $A$, $B$, and $C$ for $Y=1$, we analyze the logic gates:

1. The output $Y$ is an AND operation of the second gate's output and $C$. Thus, for $Y=1$, we must have $C=1$ and the output of the second gate equal to $1$.
2. The first NAND gate produces $N_1 = \overline{A \cdot Y}$.
3. For option B ($A=0, B=1, C=1$), we calculate the circuit states:
   • $N_1 = \overline{A \cdot Y} = \overline{0 \cdot 1} = \overline{0} = 1$.
   • The second gate (acting as an AND logic in this state) takes $N_1$ and $B$: $N_2 = N_1 \cdot B = 1 \cdot 1 = 1$.
   • The final output $Y = N_2 \cdot C = 1 \cdot 1 = 1$.
4. This matches the given condition $Y=1$.

To determine equivalence to $\bar{x}$, we evaluate each Boolean expression:
A: $x \text{ NAND } x = \overline{x \cdot x} = \bar{x}$
B: $x \text{ NOR } x = \overline{x + x} = \bar{x}$
C: $x \text{ NAND } 1 = \overline{x \cdot 1} = \bar{x}$
D: $x \text{ NOR } 1 = \overline{x + 1} = \overline{1} = 0$
Since $0 \neq \bar{x}$, option D is the expression not equivalent to $\bar{x}$.

An operation $\circ$ is commutative if $A \circ B = B \circ A$ and associative if $(A \circ B) \circ C = A \circ (B \circ C)$.
The NAND operation is defined as $A \uparrow B = \overline{A \cdot B}$.
Commutativity: $A \uparrow B = \overline{A \cdot B} = \overline{B \cdot A} = B \uparrow A$, which holds true.
Associativity check: $(A \uparrow B) \uparrow C = \overline{(\overline{A \cdot B}) \cdot C} = (A \cdot B) + \overline{C}$, whereas $A \uparrow (B \uparrow C) = \overline{A \cdot (\overline{B \cdot C})} = \overline{A} + (B \cdot C)$.
Since $(A \cdot B) + \overline{C} \neq \overline{A} + (B \cdot C)$, the NAND operation is not associative.
AND, OR, and EXOR operations are both commutative and associative.

The K-map entries indicate that the output is $1$ for the minterms $(A, B, C) = (0,0,0), (0,1,1), (1,0,0),$ and $(1,1,1)$.
The sum-of-products expression is given by $F = \overline{A}\overline{B}\overline{C} + \overline{A}BC + A\overline{B}\overline{C} + ABC$.
Factoring out terms based on $A$ gives $F = \overline{A}(\overline{B}\overline{C} + BC) + A(\overline{B}\overline{C} + BC)$.
Using the Boolean identity $(\overline{A} + A) = 1$, the expression simplifies to $F = \overline{B}\overline{C} + BC$.
The expression $\overline{B}\overline{C} + BC$ represents the logical XNOR operation, which is equivalent to $\overline{B \oplus C}$.

The XOR operation $\oplus$ is associative and satisfies the properties $B \oplus B = 0$ and $B \oplus 0 = B$.
For any even $n$, the expression $\underbrace{B \oplus B \oplus \dots \oplus B}_{n \text{ \times }}$ can be grouped into pairs:
$(B \oplus B) \oplus (B \oplus B) \oplus \dots \oplus (B \oplus B)$
Since each pair $(B \oplus B)$ equals $0$, the entire expression simplifies to:
$0 \oplus 0 \oplus \dots \oplus 0 = 0$
As this result is consistently $0$ for all even values of $n$, the output remains unchanged across these cases.

The operation is defined as $x * y = \bar{x} + y$.
Given $z = x * y$, we substitute the definition to get $z = \bar{x} + y$.
We need to calculate $z * x$, which follows the definition as $z * x = \bar{z} + x$.
Substituting $z$ into this expression yields $\bar{z} + x = \overline{(\bar{x} + y)} + x$.
Applying De Morgan's Law, $\overline{(\bar{x} + y)} = x \cdot \bar{y}$, so the expression becomes $x \cdot \bar{y} + x$.
Using the Boolean absorption law, $x + x \cdot \bar{y} = x(1 + \bar{y}) = x(1) = x$.
Therefore, the result is $x$.

From the given logic circuit, the output $f$ is defined by the Boolean expression $f = (f_1 \cdot f_2) \lor f_3$.

1. First, calculate the intersection (AND operation) of $f_1$ and $f_2$:
   $f_1 \cdot f_2 = \sum(8, 9, 10) \cdot \sum(7, 8, 12, 13, 14, 15) = \sum(8)$
2. Substitute this result into the circuit equation:
   $f = \sum(8) \lor f_3$
3. Given $f = \sum(8, 9)$, the equation becomes:
   $\sum(8, 9) = \sum(8) \lor f_3$
4. To satisfy this equality, $f_3$ must contain the minterm $9$ to complete the set $\{8, 9\}$, while not adding any extra minterms that are not part of $f$.
5. Thus, $f_3 = \sum(9)$.

The given switching function is $f(x, y, z) = \bar{x} + \bar{y}x + xz$.
Using the Boolean identity $A + \bar{A}B = A + B$, we simplify:
$f(x, y, z) = \bar{x} + x(\bar{y} + z) = (\bar{x} + x)(\bar{x} + \bar{y} + z) = \bar{x} + \bar{y} + z$
A term $P$ is an implicant of $f$ if $P \le f$ (i.e., $P$ implies $f$).
For the term $xz$, it is clear that $xz \le z$, and since $z \le \bar{x} + \bar{y} + z$, we have $xz \le f$.
Thus, $xz$ is an implicant of $f$.

To find the Product-of-Sums (POS) form, we group the zeros (empty cells) in the Karnaugh map:

1. The zeros at $(AB, CD) \in \{(00, 00), (10, 00), (00, 10), (10, 10)\}$ all satisfy $B=0$ and $D=0$, which yields the sum term $(B+D)$.
2. The zeros at $(AB, CD) \in \{(01, 01), (11, 01), (01, 11), (11, 11)\}$ all satisfy $B=1$ and $D=1$, which yields the sum term $(\overline{B}+\overline{D})$.
3. Multiplying these factors, we obtain the expression $(B+D)(\overline{B}+\overline{D})$.

From $\overline{A} + AB = 0$, we simplify the expression using Boolean laws: $(\overline{A} + A)(\overline{A} + B) = 0$, which yields $\overline{A} + B = 0$. Thus, $A=1$ and $B=0$.
Substituting these into $AB = AC$, we get $(1)(0) = (1)(C)$, which implies $C=0$.
Substituting $A=1, B=0, C=0$ into the third equation $AB + A\overline{C} + CD = \overline{C}D$:
$(1)(0) + (1)(\overline{0}) + (0)(D) = (\overline{0})D$
$0 + (1)(1) + 0 = (1)D$, leading to $1 = D$, so $D=1$.
The values $A=1, B=0, C=0, D=1$ satisfy all equations.