Tricky Problem: Evaluating a Full Adder for A=1, B=0, Cin=1

Correctly apply the full adder's SUM (three-input XOR) and Cout (majority) equations to concrete input values, without substituting OR-style reasoning for the required XOR parity logic.

A full adder receives inputs A=1, B=0, Cin=1. Find SUM and Cout, applying SUM = A XOR B XOR Cin and Cout = AB + ACin + BCin.

Why This Problem Is Tricky

Two of the three inputs (A and Cin) are 1. It is tempting to reason informally — 'two inputs are high, so the sum should be high too' — which is exactly the kind of OR-flavored intuition that gives the wrong answer for SUM. SUM is not 'is anything 1'; it is a three-input XOR, which reports the parity (odd/even count) of the 1s among A, B, and Cin, not their mere presence.

Cout carries its own, different trap: it is easy to forget one of its three product terms (AB, ACin, BCin) when substituting values, especially the BCin term when B=0 makes it look 'obviously zero' and tempts a solver to drop it from the written-out formula entirely rather than substituting and confirming it evaluates to zero.

Step-by-Step Worked Solution

SUM = A XOR B XOR Cin. Substitute A=1, B=0, Cin=1. Evaluate left to right: A XOR B = 1 XOR 0 = 1. Then (A XOR B) XOR Cin = 1 XOR 1 = 0. So SUM = 0.

Cout = AB + ACin + BCin. Substitute the same values into each term: AB = 1·0 = 0. ACin = 1·1 = 1. BCin = 0·1 = 0. Sum these with OR: Cout = 0 + 1 + 0 = 1.

So the full adder outputs SUM = 0 and Cout = 1 for this input combination. Sanity check by binary addition: A+B+Cin = 1+0+1 = 2 in decimal, and 2 in 2-bit binary is '10' — the low bit (SUM) is 0 and the high bit (Cout) is 1, exactly matching the equation-based result.

Why the Obvious Wrong Answer Is Wrong

A common wrong answer is SUM = 1, reached by informally OR-ing the three bits (1 OR 0 OR 1 = 1) instead of XOR-ing them. XOR is a parity function: it outputs 1 only when an odd number of its inputs are 1, and 0 when an even number are 1 (including zero). Here, exactly two of the three inputs (A and Cin) are 1 — an even count — so the correct XOR result is 0, the opposite of what OR-style reasoning suggests.

The binary-addition sanity check above is the fastest way to catch this mistake: since A+B+Cin totals 2, and 2 is an even number represented as '10' in binary, the SUM bit (the units place) must be 0. Any answer of SUM=1 is immediately inconsistent with that arithmetic check.

Full Adder Equations

SUM reports the parity of the three inputs; Cout reports whether at least two of the three inputs are 1. Together, (Cout, SUM) form the 2-bit binary representation of the count of 1s among A, B, and Cin (0, 1, 2, or 3).

SUM  = A xor B xor Cin
Cout = AB + ACin + BCin   (the 'majority function': 1 when at least two of A,B,Cin are 1)

Code Examples

Each equation is evaluated term by term with the given values, then the binary-addition identity A+B+Cin = 2*Cout + SUM is used as an independent cross-check, confirming both results without relying on the equations alone. Expected result: SUM = 0, Cout = 1 SUM = 0, Cout = 1

SUM = A xor B xor Cin
    = 1 xor 0 xor 1
    = (1 xor 0) xor 1
    = 1 xor 1
    = 0

Cout = AB + ACin + BCin
     = (1*0) + (1*1) + (0*1)
     = 0 + 1 + 0
     = 1

Check: A+B+Cin = 1+0+1 = 2 (decimal) = '10' (binary) -> Cout=1, SUM=0  (matches)

Full Adder Block with This Problem's Values

+-------------+
      A=1 --->|             |
      B=0 --->| FULL ADDER  |---> SUM  = 0
    Cin=1 --->|             |---> Cout = 1
              +-------------+

  Internally: SUM  = A xor B xor Cin = 1 xor 0 xor 1 = 0
              Cout = AB + ACin + BCin = 0 + 1 + 0 = 1

Why Precise Bit-Level Reasoning Matters

Full adders chain together to build ripple-carry adders inside every ALU. A single incorrectly reasoned bit — mistaking XOR parity logic for OR logic while hand-tracing or debugging a design — produces an adder that is wrong only for specific input patterns, which is exactly the kind of bug that slips past a quick test and surfaces only in specific, hard-to-reproduce numeric edge cases in the field.

SUM = 1 (from treating A, B, Cin with OR-style 'any bit is 1' reasoning)

SUM is a three-input XOR, a parity function, not an OR. With A=1, B=0, Cin=1, exactly two inputs are 1 — an even count — so XOR correctly outputs 0. Confirm with binary addition: 1+0+1=2='10', so the SUM (low) bit is 0.

For a full adder with A=1, B=0, Cin=1, what are SUM and Cout? A. SUM=1, Cout=1 B. SUM=0, Cout=1 C. SUM=1, Cout=0 D. SUM=0, Cout=0

B. SUM=0, Cout=1 — SUM = A xor B xor Cin = 1 xor 0 xor 1 = 0 (two of three inputs are 1, an even count, so XOR parity is 0). Cout = AB+ACin+BCin = 0+1+0 = 1. Binary check: 1+0+1=2='10', so Cout=1 and SUM=0, confirming the result.

Cout of a full adder equals 1 whenever which condition holds? A. Any one of the three inputs is 1 B. At least two of the three inputs (A, B, Cin) are 1 C. All three inputs are 1 D. A and B are both 1, regardless of Cin

B. At least two of the three inputs (A, B, Cin) are 1 — Cout = AB + ACin + BCin is a majority function on three inputs: each product term is 1 exactly when a particular pair of the three inputs is simultaneously 1, so the OR of all three terms is 1 whenever at least two of the three original inputs are 1 — covering the two-high and all-three-high cases alike.

Evaluate a full adder for A=1, B=1, Cin=1.

SUM = 1 xor 1 xor 1 = 0 xor 1 = 1. Cout = AB+ACin+BCin = 1+1+1 = 1 (OR of any 1s is 1). Check: A+B+Cin=3='11', so Cout=1, SUM=1 — matches.

Use the same two equations; check your SUM against the binary value of A+B+Cin.