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For an n-variable Boolean function, the maximum number of prime implicants is
- a)2(n-1)
- b)n/2
- c)2n
- d)2(n-1)
Correct answer is option 'D'. Can you explain this answer?
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For an n-variable Boolean function, the maximum number of prime implic...
For an n-variable Boolean function, the maximum number of prime implicants
=2(n -1)
=2(n -1)
Most Upvoted Answer
For an n-variable Boolean function, the maximum number of prime implic...
Explanation:
In Boolean algebra, a prime implicant is an implicant (a product term that covers at least one minterm of a function) that cannot be further reduced by eliminating literals (variables) from it. Prime implicants are important in the process of minimizing a Boolean function.
The maximum number of prime implicants for an n-variable Boolean function can be calculated as follows:
1. The number of minterms for an n-variable Boolean function is 2^n. This is because each variable can be either 0 or 1, giving 2 possibilities, and there are n variables, so the total number of possible combinations is 2^n.
2. Each minterm can be represented by a product term consisting of the variables that are 1 in that minterm and their complements that are 0 in that minterm. For example, for a 3-variable function with minterms (0,2,3,5), the corresponding product terms are (A'+B'+C'), (A'+B+C'), (A+B+C'), and (A+B'+C').
3. A prime implicant is a product term that covers at least one minterm of the function, and cannot be further reduced by eliminating literals.
4. To find the maximum number of prime implicants, we need to find the minimum number of product terms that cover all the minterms of the function. This is known as the minimal cover.
5. The minimal cover can be found using various algorithms, such as the Quine-McCluskey algorithm or the Petrick's method.
6. The maximum number of prime implicants is equal to the number of product terms in the minimal cover.
7. The maximum number of product terms in the minimal cover is 2^(n-1), since each variable can be either included or excluded from each product term, giving 2 possibilities, and there are n variables, so the total number of possible combinations is 2^n. However, we cannot have all 2^n combinations, since some of them may be redundant or cover the same minterms. Therefore, we need to eliminate the redundant product terms.
8. The number of redundant product terms that can be eliminated is equal to the number of variables minus 1. This is because each variable can be either included or excluded from a product term, except for the last variable, which must be included in at least one product term to cover all the minterms.
9. Therefore, the maximum number of prime implicants is 2^(n-1) - (n-1).
10. For example, for a 3-variable function, the maximum number of prime implicants is 2^(3-1) - (3-1) = 2^2 - 2 = 2, which can be verified using the Quine-McCluskey algorithm or the Petrick's method.
Conclusion:
Hence, the correct option is D, i.e., 2(n-1).
In Boolean algebra, a prime implicant is an implicant (a product term that covers at least one minterm of a function) that cannot be further reduced by eliminating literals (variables) from it. Prime implicants are important in the process of minimizing a Boolean function.
The maximum number of prime implicants for an n-variable Boolean function can be calculated as follows:
1. The number of minterms for an n-variable Boolean function is 2^n. This is because each variable can be either 0 or 1, giving 2 possibilities, and there are n variables, so the total number of possible combinations is 2^n.
2. Each minterm can be represented by a product term consisting of the variables that are 1 in that minterm and their complements that are 0 in that minterm. For example, for a 3-variable function with minterms (0,2,3,5), the corresponding product terms are (A'+B'+C'), (A'+B+C'), (A+B+C'), and (A+B'+C').
3. A prime implicant is a product term that covers at least one minterm of the function, and cannot be further reduced by eliminating literals.
4. To find the maximum number of prime implicants, we need to find the minimum number of product terms that cover all the minterms of the function. This is known as the minimal cover.
5. The minimal cover can be found using various algorithms, such as the Quine-McCluskey algorithm or the Petrick's method.
6. The maximum number of prime implicants is equal to the number of product terms in the minimal cover.
7. The maximum number of product terms in the minimal cover is 2^(n-1), since each variable can be either included or excluded from each product term, giving 2 possibilities, and there are n variables, so the total number of possible combinations is 2^n. However, we cannot have all 2^n combinations, since some of them may be redundant or cover the same minterms. Therefore, we need to eliminate the redundant product terms.
8. The number of redundant product terms that can be eliminated is equal to the number of variables minus 1. This is because each variable can be either included or excluded from a product term, except for the last variable, which must be included in at least one product term to cover all the minterms.
9. Therefore, the maximum number of prime implicants is 2^(n-1) - (n-1).
10. For example, for a 3-variable function, the maximum number of prime implicants is 2^(3-1) - (3-1) = 2^2 - 2 = 2, which can be verified using the Quine-McCluskey algorithm or the Petrick's method.
Conclusion:
Hence, the correct option is D, i.e., 2(n-1).
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For an n-variable Boolean function, the maximum number of prime implicants isa)2(n-1)b)n/2c)2nd)2(n-1)Correct answer is option 'D'. Can you explain this answer?
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