Math Problem Statement
Solution
The problem asks to minimize a Boolean function using a 3-variable Karnaugh map (K-map). The function is given as a sum of minterms:
Step-by-step K-map minimization:
-
Set up the K-map:
- The three variables , , and form a 3-variable K-map with 8 cells corresponding to the possible combinations of these variables. These are the minterms from 0 to 7.
AB \ BC 00 01 11 10 00 0 1 3 2 01 4 5 7 6 -
Plot the minterms: From the function , place 1s in the K-map for the corresponding minterms:
- Minterm 0:
- Minterm 1:
- Minterm 3:
- Minterm 5:
- Minterm 6:
This results in the following K-map:
AB \ BC 00 01 11 10 00 1 1 1 0 01 0 1 0 1 -
Group the 1s:
- We can form two groups:
- A group of 4 covering cells (0, 1, 3, 5): This simplifies to .
- A group of 2 covering cells (1, 3): This simplifies to .
- A group of 2 covering cells (5, 6): This simplifies to .
- We can form two groups:
-
Simplified expression: The minimized Boolean expression is:
Answer:
The correct minimized expression corresponds to Option 3: .
Would you like further details or have any questions? Here are some follow-up questions that you can consider:
- How does the K-map help in minimizing Boolean expressions?
- What are prime implicants and how do they apply in this case?
- Can the minimized expression be verified using Boolean algebra?
- What would the expression look like if additional minterms were added?
- How does grouping in a K-map improve simplification efficiency?
Tip: Always aim to form the largest possible groups of 1s in a K-map to achieve maximum simplification.
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Math Problem Analysis
Mathematical Concepts
Boolean Algebra
Karnaugh Maps (K-map)
Logic Simplification
Formulas
F(A, B, C) = Σm(0, 1, 3, 5, 6)
Minimized Boolean Expression
Theorems
Karnaugh Map Theorem
Grouping of Minterms
Suitable Grade Level
University-level digital logic or computer science students
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