Digital Logic Design Revised

Using Karnaugh map of the required function, we get: $f(x,y,z)=\overline{x}yz+x\overline{y}z+xy\overline{z}$

To get a minimal solution, we need:

• Three different NOT-gates to get $\overline{x}$ , $\overline{y}$ and $\overline{z}$

• Three different 3-input-AND-gates to get $\overline{x}yz$ , $x\overline{y}z$ and $xy\overline{z}$

• One 3-input-OR-gate to get the final expression $f(x,y,z)=\overline{x}yz+x\overline{y}z+xy\overline{z}$ .

Collectively, we need at-least $7$ logic gates to design the function.

Note:

• $\text{xyz is x AND y AND z, and x+y+z is x OR y OR z}$

• Our aim was to minimize the number of gates, that is why we have chosen 3-input OR and AND gates. Otherwise we could, for example, get $\overline{x}yz$ by using two 2-input AND-gates such that $\overline{x}yz= (\overline{x}y)z$

I guessed the answer as 7

Because according to we only need one check:

If (x+y+z==2)

Bool_function(x,y,z)=1

Else

Bool_function(x,y,z)=0

I am pretty confident that i am wrong. Therefore please correct me and it would be nice if any of you can re-explain the question in a layman's language.