Half adder/subtractor; 1's and 2's complement subtraction

Half adder and half subtractor in one circuit

Inputs $A$, $B$. Outputs: sum $S$, carry $C$, difference $D$, borrow $B_o$ ($A - B$).

$$S = D = A \oplus B, \qquad C = AB, \qquad B_o = A'B$$

The sum and difference are the same function, so one XOR gate serves both. Add one AND gate for $C$ and one NOT plus one AND gate for $B_o$.

A --+---------------[XOR]------> S (= D)
B --|--+------------[   ]
    |  |
    +--|------------[AND]------> C
    |  +------------[   ]
    |  |
    +--|--[NOT]--A'-[AND]------> Bo
       +------------[   ]

Total: 1 XOR, 2 AND, 1 NOT.

(If one output pair with a mode input $M$ is preferred: $S/D = A \oplus B$, $C/B_o = (A \oplus M)B$, giving carry when $M=0$ and borrow when $M=1$.)

a) (47)₁₀ − (26)₁₀ by 2's complement (8-bit)

$$\begin{aligned} 47 &= 0010\,1111 \\ 26 &= 0001\,1010 \\ \text{2's complement of } 26 &= 1110\,0101 + 1 = 1110\,0110 \end{aligned}$$

    0010 1111   (+47)
  + 1110 0110   (-26)
  -----------
  1 0001 0101

An end carry occurs: discard it. The result is positive.

Answer: $0001\,0101_2 = 16 + 4 + 1 = 21$. Check: $47 - 26 = 21$.

b) (10010)₂ − (10011)₂ by 1's complement

$$\text{1's complement of } 10011 = 01100$$

   10010
 + 01100
 -------
   11110   (no end-around carry)

No end carry, so the result is negative; take its 1's complement:

$$\overline{11110} = 00001$$

Answer: $(10010)_2 - (10011)_2 = -(00001)_2 = -1$. Check: $18 - 19 = -1$.