Circuit giving 2's complement of a 4-bit number

The circuit takes a 4-bit number $A_3A_2A_1A_0$ and gives $Y_3Y_2Y_1Y_0$ = its 2's complement ($\overline{A} + 1$, keeping 4 bits).

Truth table

Simplified expressions

From K-maps (or from the rule "copy bits up to and including the first 1 from the right, then invert the rest"):

• $Y_0 = 1$ for odd numbers only: $Y_0 = A_0$
• $Y_1 = 1$ for 01, 10 in $A_1A_0$: $Y_1 = A_1 \oplus A_0$
• $Y_2$ equals $A_2$ when $A_1A_0 = 00$, else $A_2'$: $Y_2 = A_2 \oplus (A_1 + A_0)$
• $Y_3$ equals $A_3$ when $A_2A_1A_0 = 000$, else $A_3'$: $Y_3 = A_3 \oplus (A_2 + A_1 + A_0)$

$$\begin{aligned} Y_0 &= A_0 \\ Y_1 &= A_1 \oplus A_0 \\ Y_2 &= A_2 \oplus (A_1 + A_0) \\ Y_3 &= A_3 \oplus (A_2 + A_1 + A_0) \end{aligned}$$

Circuit

A0 ------------------------------> Y0
A1, A0 ----------[XOR]-----------> Y1
A1, A0 --[OR]--> X1 = A1+A0
A2, X1 ----------[XOR]-----------> Y2
A2, X1 --[OR]--> X2 = A2+A1+A0
A3, X2 ----------[XOR]-----------> Y3

Gates: 3 XOR and 2 OR gates (the second OR adds $A_2$ to the first OR output).

Check: $A = 0110$: $Y_0 = 0$, $Y_1 = 1\oplus0 = 1$, $Y_2 = 1\oplus1 = 0$, $Y_3 = 0\oplus1 = 1$, so $Y = 1010$, and $16 - 6 = 10 = 1010$. Correct.

(Alternative: invert each bit with NOT gates and add 1 using a 4-bit parallel adder with $C_0 = 1$.)