Full-adder sum implementation using XOR gates: In a standard 1-bit full adder, the SUM output (S) can be realized using two exclusive-OR (XOR) gates arranged so that S = A XOR B XOR Cin. Assess this claim.

Difficulty: Easy

Correct Answer: Correct

Explanation:


Introduction / Context:
The full adder is a fundamental combinational block that sums two bits and an incoming carry. Its outputs are SUM (S) and CARRY OUT (Cout). Understanding minimal gate implementations of S helps in efficient hardware design and HDL coding.



Given Data / Assumptions:

  • Inputs: A, B, Cin.
  • Outputs: S = A XOR B XOR Cin, and Cout = majority(A, B, Cin) = A*B + B*Cin + A*Cin.
  • Exclusive-OR gate property: X XOR Y = (X and not Y) + (not X and Y).


Concept / Approach:
The XOR function is the natural expression of bitwise addition without carry. For two inputs, S2 = A XOR B. Adding Cin requires XORing that intermediate sum with Cin: S = (A XOR B) XOR Cin. This can be implemented with exactly two XOR gates in cascade.



Step-by-Step Solution:

Compute partial sum: P = A XOR B using one XOR gate.Compute final sum: S = P XOR Cin using a second XOR gate.No additional gates are required for S; Cout is implemented separately.Therefore, two XORs suffice for the SUM path in a full adder.


Verification / Alternative check:
Build the truth table for A, B, Cin and compute S via XOR chaining. The resulting S matches binary addition modulo 2, confirming the implementation.



Why Other Options Are Wrong:
“Incorrect” ignores the standard identity. “Only correct for carry-lookahead adders” is irrelevant; the SUM path is the same across architectures. “Requires three XOR gates” overcounts the needed logic.



Common Pitfalls:
Mixing SUM and CARRY logic; assuming Cout also needs XOR chaining rather than majority logic; forgetting XOR associativity (A XOR B XOR Cin is well defined).



Final Answer:
Correct

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