Fast Addition of Integers (Carry-Lookahead)

In examining how incrementing can be made faster, we saw that the carry signal is propagated whenever there is a one in the input. If there is a one in the input, the carry bit is set if and only the previous carry bit is set. If there is a zero in the input, the carry bit is never set. We see similar behavior in general addition circuits.

Carry	$carry_{n-1}$	$carry_{n-2}$	$carry_{n-3}$	...	$carry(2)$	$carry(1)$	$carry(0)$
Input X		$x_{n-1}$	$x_{n-2}$	...	$x_3$	$x_2$	$x_1$	$x_0$
Input Y		$y_{n-1}$	$y_{n-2}$	...	$y_3$	$y_2$	$y_1$	$y_0$
Output	$out_n$	$out_{n-1}$	$out_{n-2}$	...	$out_3$	$out_2$	$out_1$	$out_0$

$$\begin{eqnarray*} carry_0 & = & x_0 \land y_0 \\ carry_i & = & \mathit{majority}(x_i, y_i, carry_{i-1}) \\ \end{eqnarray*}$$

If both $x_i$ and $y_i$ are set to one, then $carry_i$ is always set - this is called generating the carry bit. If one of these bits is one and the other is zero, then $carry_i = carry_{i-1}$ (with the exception of $carry_0$) - this is called propagating the carry bit. If both $x_i$ and $y_i$ are set to zero, then $carry_i$ is always zero. We abstract this into two bits $g_i$ and $p_i$ representing whether the carry value is generated and propagated respectively.

$$\begin{eqnarray*} g_i & = & (x_i = 1) \land (y_i = 1) \\ p_i & = & ((x_i = 1) \land (y_i = 0)) \lor ((x_i = 0) \land (y_i = 1)) \\ \end{eqnarray*}$$

We can reformulate the carry bits in terms of the $g$ and $p$ bits.

$$\begin{eqnarray*} carry_i & = & g_i \lor (p_i \land carry_{i-1}) \\ \end{eqnarray*}$$

If we unwind the recursion, we eventually get the following formula.

$$\begin{eqnarray*} carry_i & = & g_i \lor (p_i \land g_{i-1}) \lor \ldots \lor (p... ...p_2 \land g_1) \lor (p_i \land \ldots \land p_{1} \land g_0) \\ \end{eqnarray*}$$