Neutral Functions | Minterms and Maxterms

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-

  1. Calculate the total number of neutral functions that are possible.
  2. 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,

Watch this Video Lecture

 

Next Article- Self-Dual Functions

 

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Watch video lectures by visiting our YouTube channel LearnVidFun.

Summary
Neutral Functions | Minterms and Maxterms
Article Name
Neutral Functions | Minterms and Maxterms
Description
A neutral function is a function for which number of minterms and number of maxterms are same. Number of neutral functions possible with n Boolean variables = C ( 2n , 2n-1 ).
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Gate Vidyalay
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