yesterday I had an exam, in which I still don't know how to resolve the problem. The question of the exam was: Create a logic circuit with 4 inputs (A,B,C,D) using ONLY AND & OR with 2 INPUTS gates starting from the truth table where:
- When the inputs are all 0, or all 1, the output is indifferent
- The output is 1 if the number of 1 in inputs is odd
- The output is 0 if the number of 1 in inputs is even
So this is the truth table:
And the Karnaugh Map:
So the minimum function is:
ABC + ABD + ACD + BCD + !A!B!C + !A!B!D + !A!C!D + !B!C!D
I think I know how to deal with using AND & OR with 2 inputs, by doing groups like this: (tell me if I'm wrong)
A(BC + BD) + C(AD + BD) + !B(!A!C + !A!D) + !D(!A!C + !B!C)
But I dont understand how it's possible to do this without using a NOT gate, I asked the professor and he clearly said that I can only use AND & OR gates. I tried multiple approaches but they always involved NOT, NOR, NAND gates.
The professor said it's possible to resolve this exercise. Can you please help me?