How to deduce the XOR variant of the Full 1bit adder circuit design?

Question

I'm trying to figure out how to find the short version of Sum in a full adder, from the truth table I got this DNF:

(A && ~B && ~C) || (~A && B && ~C) || (~A && ~B && C) || (A && B && C)

where A = A, B = B, and C = CIn

But according to wikipedia, this is equivalent to:

A XOR B XOR C

Is there a way I can somehow figure out the latter version or do I just need to "see it" in the truth table?

Thanks!

Answer 1

The terms in your DNF have one thing in common: An odd number of inputs is true.

The output line of a full-adder is 1 if and when an odd number (one or three) of its input lines are 1. If zero inputs are 1 (= all are 0), the output is 0. If two inputs are 1, carry-out is 1 but output stays 0.

When you translate your truth table into a Karnaugh map, you get the checker-board pattern typical for XORs. In the end you are right: It does in fact help to "see it".

(Karnaugh map image copied from here)