Carry-Out of a 1-bit Full Adder (Gate-Level)

In any multi-bit adder, the carry-out signal determines whether the addition overflows into the next bit position. In a CPU's arithmetic logic unit, the carry chain is the critical timing path — carry-lookahead and prefix-tree adder architectures exist solely to speed up carry propagation. Understanding how carry-out is computed at the gate level is fundamental to arithmetic hardware design.

Your task is to implement only the carry-out of a 1-bit full adder. Given three 1-bit inputs A, B, and Cin, produce the carry-out Cout. The carry-out is 1 whenever two or more of the three inputs are 1 — this is the majority-of-three function. The complete truth table:

| A | B | Cin | Cout | |---|---|-----|------| | 0 | 0 | 0 | 0 | | 0 | 0 | 1 | 0 | | 0 | 1 | 0 | 0 | | 0 | 1 | 1 | 1 | | 1 | 0 | 0 | 0 | | 1 | 0 | 1 | 1 | | 1 | 1 | 0 | 1 | | 1 | 1 | 1 | 1 |

The algebraic expression is Cout = AB + BCin + ACin. However, there is a smarter decomposition: using one half-adder to first add A and B gives you a sum bit (A XOR B) and a partial carry (AB). The full carry-out is then AB OR (Cin AND (A XOR B)), which requires only 3 components total — one HA, one AND, and one OR. This is the optimal implementation.

| Signal | Direction | Width | Description | |--------|-----------|-------|-------------| | A | input | 1 | First addend bit | | B | input | 1 | Second addend bit | | Cin | input | 1 | Carry-in | | Cout | output | 1 | Carry-out: 1 when at least two inputs are 1 |

Constraints

• The circuit is purely combinational — no clock or state.
• Any gate types are allowed. The optimal solution uses 1 HA, 1 AND, and 1 OR (3 components total).
• Only Cout is required as an output. Do not produce a Sum output.
• All 8 input combinations must produce the correct Cout.

Topics

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