Ace Page# 105, Q#29

Question

How many Boolean functions of the type $f(x,y,z)=f(\bar{x}, \bar{y}, \bar{z})$ are available with three variables?

• A. 4
• B. 8
• C. 32
• D. 16

Answer 1

answer is D.

Comments

shouldn't be 256?

whe have  f(x,y,z)=f(x¯,y¯,z¯)  which is equal to f(a,b,c) {a=x', b=y', c=z'}

now its a normal function with 3 variable.

Answer 2

Answer is (D).

$f(x,y,z) = f(x', y',z')$, means output should be same for $(0,0,0)$ and $(1,1,1)$, $(0,0,1)$ and $(1,1,0)$, $(0,1,0)$ and $(1,0,1)$, $(0,1,1)$ and $(1,0,0)$ and the other input combinations repeat after this.

So, We have four input cases (pairs) for which output can be $0$ or $1$. So, we have two cases for each pair.

Total functions possible $= 2*2*2*2 = 16$

Comments

till four input cases, i have understood, but why we need to consider 0 or 1 for each case. In the question it mentioned we need to consider f(x,y,z) =f(x',y',z'), with this case at anytime we will get only 1 right?
please clarify me