Electrical – Is XOR equal to XNOR when odd number of inputs are considered

The standard adder / subtract circuit is given by the figure below.For an explanation of how it works please see my previous answer here : Subtraction using adder circuit

Now notice that if you were to change the input that is tied to '0' to Cin, then the circuit would now perform the operation: A + ~B + (Cin ^ SUB), meaning that:

1) it would be a normal Adder/Subtract circuit when Cin is 0.

2) if Cin = 1 and Sub = 1 it would evaluate to:

A + ~B + (1 ^ 1) = A + ~B = A + ~B + 1 - 1 = A + (~B + 1) - 1 = A - B - 1 ,

This means that the normal Add/Sub circuit can achieve the operation you require, all you need to do is change the input tied to '0' your Cin.

You could also design the desired circuit by using the usual method of writing all your possible inputs into a truth table, then deriving the Boolean equations for them, then simplifying them to get your final circuit equation.

I understand you want to find out which of the two numbers is larger? You can do that by comparing the bits of A and B, starting at the highest ones, let's call those A3 and B3.

There are three cases: they are equal, or otherwise A3 is 1 (and B3 is 0), or B3 is 1 (and A3 is zero). In the first case (equal), which one is larger is determined by the next (lower) bit. If they are not equal, take the A3 bit and it tells you whether A is larger. So for the left most bit:

A_is_larger = ( ( A3 == B3 ) * ( input from lower bits ) ) + ( ( A3 != B3 ) * A3 ).

For the rightmost bit the you can simplify this to A1 (assuming that for equal values you don't care what the outcome is).