Digital logic design

Question

How many Boolean functions of three variables $\textsf{f(x,y,z)}$ have the property that $\textsf{f(x,y,z)=(f(x’,y’,z’))}$?

• A. $64$
• B. $16$
• C. $256$
• D. $8$

Comments

When we have $3$ variables means $8$ minterms, so no of possible boolean functions are $2^{8} = 256$

But now we have a condition that $f(x,y,z) = f(x',y',z')$ that means these two terms are mutually exclusive

For ex: $000 ==111$ i.e. function will give same value for these two inputs.

Now we have $4$ groups of mutually exclusive terms
$000$ and $111$
$ 001 $ and $110$
$010$ and $101$
$011$ and $100$
No of possible function are $2^{4}$ = 16

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https://gateoverflow.in/68123/ace-page%23-105-q%2329

Answer 1

Your minterms are halved when we have to satisfy the property f(x,y,z) = f(x',y',z').

Hence ans = number of functions from 4 minterms to output set {0,1}

Answer 2

For n variable there is 2^2^n means here for 2^2^3=256

but acc. to question-

f(x,y,z)=f(x′,y′,z′), means output should be same for (0,0,0)and (1,1,1), (0,0,1)) and (1,1,0)(1,1,0) and so on total 4 groups .

So, We have four input cases (pairs) for which output can be 00 or 11. So, we have two cases for each pair.

Total functions possible =2^4=16