Reputation: 35
I am a bit confused as to how to solve this problem. I have put the problem into the following Boolean equation:
F = abc'd' + ab'cd' + ab'c'd + a'bcd' + a'bc'd + a'b'cd + abcd
I also know that in terms of two inputs, a and b, XOR and XNOR would be
XOR: a'b + ab'
XNOR: ab + a'b'
How do I turn this into a circuit using only 2-input XOR and XNOR gates? Any pointers in the right direction would be great. I'm completely stuck. Thanks!!
Upvotes: 0
Views: 1689
Reputation: 31153
If zero is accepted as even number then it is simple.
A XOR gate will be zero if there is an even number of ones, in this case 0 or 2. If there is one, it will be 1. Split the inputs into two groups and XOR respectively. Now you know if the groups have even or odd number of ones. XNOR the results and you will know the whole result.
(A XOR B) XNOR (C XOR D)
Proof:
If zero is not even for some reason this won't work but it has always been for me in logic.
Upvotes: 1