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 = 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 ( 2ⁿ , 2^n-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 , 2^n-1 ).

 

Substituting n = 2, we get-

Number of neutral functions possible with 2 Boolean variables

= C ( 2² , 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,

Watch this Video Lecture

 

Next Article- Self-Dual Functions

 

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