**Minterms and Maxterms-**

For any function consisting of n Boolean variables, Number of minterms possible = Number of maxterms possible = 2 |

**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 = 2
^{2}= 4. - Number of maxterms possible = 2
^{2}= 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 = 2
^{3}= 8. - Number of maxterms possible = 2
^{3}= 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 = 2
^{n}. - Since maximum number of terms possible = 2
^{n}, so we choose half of the terms i.e 2^{n}/ 2 = 2^{n-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 ( 2
^{n}, 2^{n-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 ( 2^{n} , 2^{n-1} ).

Substituting n = 2, we get-

Number of neutral functions possible with 2 Boolean variables

= C ( 2^{2} , 2^{2-1} )

= 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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