Minterms and Maxterms-
| For any function consisting of n Boolean variables,
Number of minterms possible = Number of maxterms possible = 2n |
Minterms and Maxterms Examples-
The examples of minterms and maxterms are-
Example-01:
For any function consisting of 2 Boolean variables A and B, we have-
- Number of minterms possible = 22 = 4.
- Number of maxterms possible = 22 = 4.
The following table shows the minterms and maxterms-
| A | B | Minterms | Maxterms |
| 0 | 0 | A’B’ | A + B |
| 0 | 1 | A’B | A + B’ |
| 1 | 0 | AB’ | A’ + B |
| 1 | 1 | AB | A’ + B’ |
Example-02:
For any function consisting of 3 Boolean variables A, B and C, we have-
- Number of minterms possible = 23 = 8.
- Number of maxterms possible = 23 = 8.
The following table shows the minterms and maxterms-
| A | B | C | Minterms | Maxterms |
| 0 | 0 | 0 | A’B’C’ | A + B + C |
| 0 | 0 | 1 | A’B’C | A + B + C’ |
| 0 | 1 | 0 | A’BC’ | A + B’ + C |
| 0 | 1 | 1 | A’BC | A + B’ + C’ |
| 1 | 0 | 0 | AB’C’ | A’ + B + C |
| 1 | 0 | 1 | AB’C | A’ + B + C’ |
| 1 | 1 | 0 | ABC’ | A’ + B’ + C |
| 1 | 1 | 1 | ABC | A’ + B’ + C’ |
Neutral Functions-
| A neutral function is a function for which number of minterms and number of maxterms are same. |
Example-
The following function is an example of a neutral function-
f( A, B) = A ⊕ B
This is because this function has equal number of minterms and maxterms as shown by the following table-
| A | B | A ⊕ B |
| 0 | 0 | 0 |
| 0 | 1 | 1 |
| 1 | 0 | 1 |
| 1 | 1 | 0 |
Number Of Neutral Functions Possible-
Here n = Number of Boolean variables in the function.
Explanation-
- For n variables, total number of terms possible = number of combinations of n variables = 2n.
- Since maximum number of terms possible = 2n, so we choose half of the terms i.e 2n / 2 = 2n-1.
- We assign them the output logic ‘1’.
- We assign ‘0’ to rest half of the terms.
- Thus, number of neutral functions possible with n Boolean variables = C ( 2n , 2n-1 ).
PRACTICE PROBLEM BASED ON NEUTRAL FUNCTIONS-
Problem-
Consider any function consisting of 2 Boolean variables A and B-
- Calculate the total number of neutral functions that are possible.
- Write all the neutral functions.
Solution-
We know, number of neutral functions possible with n Boolean variables = C ( 2n , 2n-1 ).
Substituting n = 2, we get-
Number of neutral functions possible with 2 Boolean variables
= C ( 22 , 22-1 )
= C ( 4 , 2 )
= 6
Thus, total 6 Boolean functions are possible with 2 Boolean variables.
The 6 Boolean Functions are-
- f ( A , B ) = A
- f ( A , B ) = A’
- f ( A , B ) = B
- f ( A , B ) = B’
- f ( A , B ) = A ⊙ B
- f ( A , B ) = A ⊕ B
| Variables | Boolean Functions | ||||||
| A | B | A | A’ | B | B’ | A⊙B | A⊕B |
| 0 | 0 | 0 | 1 | 0 | 1 | 1 | 0 |
| 0 | 1 | 0 | 1 | 1 | 0 | 0 | 1 |
| 1 | 0 | 1 | 0 | 0 | 1 | 0 | 1 |
| 1 | 1 | 1 | 0 | 1 | 0 | 1 | 0 |
The above table clearly shows that for each function, number of minterms = number of maxterms.
So, they all are neutral functions.
To gain better understanding about Neutral Functions,
Next Article- Self-Dual Functions
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