The number of essential prime implicants for the Function Y = A'B'C'D + A'BCD' + ABC'D' is given by ______
3
Essential Prime Implicants: Understanding Boolean Function Simplification
To determine the number of essential prime implicants for the given Boolean function \( Y = A'B'C'D + A'BCD' + ABC'D' \), we will use a 4-variable Karnaugh Map (K-map). A K-map is a graphical method used to simplify Boolean expressions. It helps in identifying adjacent minterms (product terms where only one variable changes its state) that can be combined to form larger product terms, known as prime implicants.
Boolean Function Minterm Identification
First, let's identify the minterms represented by the given Boolean expression:
- The term \( A'B'C'D \) corresponds to the minterm \( 0001_2 \), which is \( m_1 \).
- The term \( A'BCD' \) corresponds to the minterm \( 0110_2 \), which is \( m_6 \).
- The term \( ABC'D' \) corresponds to the minterm \( 1100_2 \), which is \( m_{12} \).
So, the function can be written as \( Y(A,B,C,D) = \Sigma (1, 6, 12) \).
K-map Construction and Plotting
Now, let's construct a 4-variable K-map and place '1's in the cells corresponding to these minterms. The K-map is arranged such that adjacent cells differ by only one bit, allowing for grouping.
| CD\AB | AB | |||
|---|---|---|---|---|
| 00 | 01 | 11 | 10 | |
| 00 | 0 (\(m_0\)) | 0 (\(m_4\)) | 1 (\(m_{12}\)) | 0 (\(m_8\)) |
| 01 | 1 (\(m_1\)) | 0 (\(m_5\)) | 0 (\(m_{13}\)) | 0 (\(m_9\)) |
| 11 | 0 (\(m_3\)) | 0 (\(m_7\)) | 0 (\(m_{15}\)) | 0 (\(m_{11}\)) |
| 10 | 0 (\(m_2\)) | 1 (\(m_6\)) | 0 (\(m_{14}\)) | 0 (\(m_{10}\)) |
Prime Implicants Identification
A prime implicant (PI) is a product term obtained by combining the maximum possible number of adjacent '1's in the K-map. These groups must be powers of 2 (1, 2, 4, 8, etc.).
Looking at the plotted K-map:
- The '1' at \( m_1 \) (cell 0001) has no adjacent '1's (horizontally, vertically, or wrapping around). Therefore, it can only be grouped with itself, forming a group of 1. This gives us the prime implicant: \( P_1 = A'B'C'D \)
- The '1' at \( m_6 \) (cell 0110) also has no adjacent '1's. It forms a group of 1. This gives us the prime implicant: \( P_2 = A'BCD' \)
- The '1' at \( m_{12} \) (cell 1100) similarly has no adjacent '1's. It forms a group of 1. This gives us the prime implicant: \( P_3 = ABC'D' \)
In this specific case, all the minterms are isolated, meaning they cannot be combined with any other '1' to form larger groups. Thus, each minterm itself is a prime implicant.
The prime implicants for the function are: \( A'B'C'D \), \( A'BCD' \), and \( ABC'D' \).
Essential Prime Implicants Determination
An essential prime implicant (EPI) is a prime implicant that covers at least one minterm which no other prime implicant covers. To identify EPIs, we check each minterm:
- Minterm \( m_1 \): This minterm is covered exclusively by the prime implicant \( A'B'C'D \). No other prime implicant covers \( m_1 \). Therefore, \( A'B'C'D \) is an essential prime implicant.
- Minterm \( m_6 \): This minterm is covered exclusively by the prime implicant \( A'BCD' \). No other prime implicant covers \( m_6 \). Therefore, \( A'BCD' \) is an essential prime implicant.
- Minterm \( m_{12} \): This minterm is covered exclusively by the prime implicant \( ABC'D' \). No other prime implicant covers \( m_{12} \). Therefore, \( ABC'D' \) is an essential prime implicant.
Since each of the three minterms is covered uniquely by one of the prime implicants, all three identified prime implicants are essential prime implicants.
Final Count of Essential Prime Implicants
Based on our analysis, the essential prime implicants are:
- \( A'B'C'D \)
- \( A'BCD' \)
- \( ABC'D' \)
Therefore, the total number of essential prime implicants for the given function is 3.
Important Questions from Minimization of Boolean Expression
Consider the following four variable Boolean function in sum-of-product form
$F(b_3, b_2, b_1, b_0) = \sum(0, 2, 4, 8, 10, 11, 12)$.
where the value of the function is computed by considering $b_3b_2b_1b_0$ as a 4-bit binary number, where $b_3$ denotes the most significant bit and $b_0$ denotes the least significant bit. Note that there are no don't care terms. Which ONE of the following options is the CORRECT minimized Boolean expression for F?
- Simplified Boolean equation for the following truth table is :
x y z F 0 0 0 0 0 0 1 1 0 1 0 0 0 1 1 1 1 0 0 1 1 0 1 0 1 1 0 1 1 1 1 0 The simplified form of a Boolean equation $(A\overline{B} + A\overline{B}C + AC)(\overline{A}\overline{C} + \overline{B})$ is :
Simplify the following Boolean expression.
E(E + F) + DE + D(E + F)
What is the value of \( \bar{F}\)?
\(F = AB + \bar{C}\bar{D} + \bar{B}D\)
