Question

Pretty simple, really. I want to negate an integer which is represented in 2’s complement, and to do so, I need to first flip all the bits in the byte. I know this is simple with XOR–just use XOR with a bitmask 11111111. But what about without XOR? (i.e. just AND and OR). Oh, and in this crappy assembly language I’m using, NOT doesn’t exist. So no dice there, either.

Answer 1

You can’t build a NOT gate out of AND and OR gates.

As I was asked to explain, here it is nicely formatted. Let’s say you have any number of AND and OR gates. Your inputs are A, 0 and 1. You have six possibilities as you can make three pairs out of three signals (pick one that’s left out) and two gates. Now:

Operation  Result
A AND A    A
A AND 1    A
A AND 0    0
A OR A     A
A OR 1     1
A OR 0     A

So after you fed any of your signals into the first gate, your new set of signals is still just A, 0 and 1. Therefore any combination of these gates and signals will only get you A, 0 and 1. If your final output is A, then this means that for both values of A it won’t equal !A, if your final output is 0 then A = 0 is such a value that your final value is not !A same for 1.

Edit: that monotony comment is also correct! Let me repeat here: if you change any of the inputs of AND / OR from 0 to 1 then the output won’t decrease. Therefore if you claim to build a NOT gate then I will change your input from 0 to 1 , your output also can’t decrease but it should — that’s a contradiction.