GATE CS 2021 Set 2 — Question 62
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Digital Logic → Boolean Algebra & Logic Gates → K-Map Minimization
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Question
Consider a Boolean function such that
The number of literals in the minimal sum-of-products expression of is ________.
The number of literals in the minimal sum-of-products expression of is ________.
Correct answer
6 to 6
Solution
To find the minimal sum-of-products (SOP) expression, we first determine the minterms for which the function is 1, 0, or Don't Care.Step 1: Analyze the given conditions
Step 3: Grouping
.
1.
- This corresponds to the case where and .
- Indices (binary ): (0), (1), (8), (9).
- All these minterms are 1.
- This corresponds to and .
- Indices:
- (10). Value: .
- (11). Value: .
- (14). Value: .
- (15). Value: .
- This corresponds to .
- Indices:
- (4). Value: .
- (5). Value: .
- (6). Value: .
- (7). Value: .
- (12). Value: .
- (13). Value: .
- (14). Value: (Consistent).
- (15). Value: (Consistent).
- 1s: 0, 1, 6, 7, 8, 9, 11, 13, 14, 15
- 0s: 4, 5, 10, 12
- Don't Cares: The inputs not covered by the conditions are (Indices 2 and 3). We can treat them as Don't Cares () to minimize the expression, though in this specific case, they don't reduce the literal count further.
| 00 | 01 | 11 | 10 | |
|---|---|---|---|---|
| 00 | 1 | 1 | d | d |
| 01 | 0 | 0 | 1 | 1 |
| 11 | 0 | 1 | 1 | 1 |
| 10 | 1 | 1 | 1 | 0 |
1.Group 1: Corners/Edges for minterms 0, 1, 8, 9.
- Cells: .
- Variable constant: .
- Term: . (2 literals)
- Cells: .
- Variable constant: .
- Term: . (2 literals)
- Cells: .
- Variable constant: .
- Term: . (2 literals)
- covers 0, 1, 8, 9.
- covers 6, 7, 14, 15.
- covers 9, 11, 13, 15.
- All 1s are covered. No redundant groups (removing any group leaves some 1s uncovered).
.
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