TAOCP 7.1.2 Exercise 41

Let $A(n)$ denote the number of binary gates in an $n$-bit conditional-sum adder constructed recursively as described, and let $D(n)$ denote its depth.

taocpmathematicsalgorithmsvolume-4math-medium

Section 7.1.2: Boolean Evaluation

Exercise 41. [M22] (Conditional-sum adders.) One way to accomplish binary addition (25) with depth $O(\log n)$ is based on the multiplexer trick of exercise 4: If $(xx')_2 + (yy')_2 = (zz')_2$, where $|x'| = |y'| = |z'|$, we have either $(x)_2 + (y)_2 = (z)_2$ and $(x')_2 + (y')_2 = (z')_2$, or $(x)_2 + (y)_2 + 1 = (z)_2$ and $(x')_2 + (y')_2 = (1x')_2$. To save time, we can compute both $(x)_2 + (y)_2$ and $(x)_2 + (y)_2 + 1$ simultaneously as we compute $(x')_2 + (y')_2$. Afterwards, when we know whether or not the less significant part $(x')_2 + (y')_2$ produces a carry, we can use multiplexers to select the correct bits for the most significant part.

If this method is used recursively to build $2n$-bit adders from $n$-bit adders, how many gates are needed when $n = 2^m$? What is the corresponding depth?

Verified: no
Solve time: 2m58s

Let $A(n)$ denote the number of binary gates in an $n$-bit conditional-sum adder constructed recursively as described, and let $D(n)$ denote its depth. Assume $n = 2^m$.

1. Correct size recurrence

The construction of a $2n$-bit adder proceeds as follows.

Split the inputs into low and high halves:

  • The low $n$-bit adder computes the lower sum and produces a carry bit.

  • The high $n$-bit portion is computed twice in parallel:

  • once assuming carry-in $0$,

  • once assuming carry-in $1$.

  • A selection network (multiplexers) chooses the correct high-half result based on the carry from the low half.

Thus the total work for size $2n$ consists of three full $n$-bit adder constructions plus selection overhead:

$$ A(2n) = A(n) + 2A(n) + \alpha n = 3A(n) + \alpha n, $$

where $\alpha n$ accounts for the $n$ multiplexers, each of constant size.

2. Solving the recurrence

Let $n = 2^m$ and define $a_m = A(2^m)$. Then

$$ a_m = 3a_{m-1} + \alpha 2^{m-1}. $$

Divide by $3^m$. Define $b_m = \frac{a_m}{3^m}$. Then

$$ b_m = \frac{a_m}{3^m} = \frac{3a_{m-1}}{3^m} + \frac{\alpha 2^{m-1}}{3^m} = b_{m-1} + \alpha \cdot \frac{2^{m-1}}{3^m}. $$

Rewrite the additive term:

$$ \frac{2^{m-1}}{3^m} = \frac{1}{3}\left(\frac{2}{3}\right)^{m-1}. $$

Hence

$$ b_m = b_{m-1} + \frac{\alpha}{3}\left(\frac{2}{3}\right)^{m-1}. $$

Unrolling,

$$ b_m = b_0 + \frac{\alpha}{3} \sum_{i=0}^{m-1} \left(\frac{2}{3}\right)^i. $$

The geometric sum gives

$$ \sum_{i=0}^{m-1} \left(\frac{2}{3}\right)^i = \frac{1 - (2/3)^m}{1 - 2/3} = 3\left(1 - (2/3)^m\right). $$

Therefore

$$ b_m = b_0 + \alpha \left(1 - (2/3)^m\right). $$

Multiplying back by $3^m$,

$$ a_m = 3^m(b_0 + \alpha) - \alpha 3^m (2/3)^m. $$

Since $3^m (2/3)^m = 2^m$, this becomes

$$ a_m = (b_0 + \alpha)3^m - \alpha 2^m. $$

Returning to $n = 2^m$,

$$ A(n) = \Theta(3^m) = \Theta!\left(n^{\log_2 3}\right). $$

The subtraction term $\alpha n$ is lower order and does not affect the asymptotic growth.

3. Depth recurrence

The depth structure is unchanged from the original argument.

  • The low-half adder has depth $D(n)$.
  • The two high-half adders run in parallel, also contributing depth $D(n)$.
  • The multiplexing stage has constant depth $\beta$.

Thus

$$ D(2n) = D(n) + \beta. $$

Letting $d_m = D(2^m)$,

$$ d_m = d_{m-1} + \beta, \quad\text{so}\quad d_m = d_0 + \beta m. $$

Since $m = \log_2 n$,

$$ D(n) = \Theta(\log n). $$

Final result

The conditional-sum adder constructed recursively satisfies

$$ A(n) = \Theta!\left(n^{\log_2 3}\right), \qquad D(n) = \Theta(\log n). $$

This completes the corrected solution. ∎