Question about parallel FULL ADDER


Joined Mar 31, 2012
The answer given is correct.

Start with being sure how to combine and dissect the values and their bits.

If a = a2a1a0, then mathematically

a = (2^2)·a2 + (2^1)·a1 + (2^0)·a0
a = 4·a2 + 2·a1 + 1·a0

and similar for the other two.

Now, given 'a', how could you pick off each of its bits using div (integer division) and mod (remainder after integer division) operations?

Once you have that, then you can either piece together the equation they have, or simply brute force the result from the circuit.

Perhaps it might help to group and slightly rewrite the result as

S = [b - (b mod2)] + [a - {[(a div 2) mod 2[ * 2}] + [3 + 1]