TAOCP 4.6.3 Exercise 9
We are asked to design an exponentiation procedure analogous to Algorithm A, but based on radix $m = 2^\nu$, such that the algorithm performs approximately \lg n + \nu + m multiplications, where $\nu$...
Section 4.6.3: Evaluation of Powers
Exercise 9. ▶ [25] Design an exponentiation procedure that is analogous to Algorithm A, but based on radix $m = 2^\nu$. Your method should perform approximately $\lg n + \nu + m$ multiplications, where $\nu$ is the number of nonzero digits in the $m$-ary representation of $n$.
Verified: yes
Solve time: 2m55s
Solution
We are asked to design an exponentiation procedure analogous to Algorithm A, but based on radix $m = 2^\nu$, such that the algorithm performs approximately
$$ \lg n + \nu + m $$
multiplications, where $\nu$ denotes the number of nonzero digits in the radix-$m$ representation of $n$.
The key insight is that a naive right-to-left table method, which updates all $m-1$ powers at each stage, costs too much. To achieve the required bound, we must avoid updating all table entries repeatedly. Instead, we adopt a precomputation and left-to-right scanning strategy, analogous to Algorithm A.
Step 1. Radix-$m$ representation
Write $n$ in radix $m$:
$$ n = d_0 + d_1 m + \cdots + d_t m^t, \qquad 0 \le d_j < m, $$
where $t = \lfloor \log_m n \rfloor$ and $\nu =$ the number of nonzero $d_j$.
Step 2. Precompute small powers
Define an array
$$ Z_0, Z_1, \dots, Z_{m-1}, $$
where
$$ Z_k = x^k, \quad 0 \le k \le m-1. $$
This requires $m-1$ multiplications.
Observation. Unlike the previous flawed solution, we will never update all $Z_k$ repeatedly. Each $Z_k$ is used only when the corresponding digit is nonzero.
Step 3. Left-to-right exponentiation
Let
$$ Y \leftarrow 1. $$
Scan the digits of $n$ from most significant digit $d_t$ to least significant $d_0$:
- Raise $Y$ to the $m$th power:
$$ Y \leftarrow Y^m. $$
Since $m = 2^\nu$, this requires exactly $\nu$ squarings. 2. Multiply by the precomputed power if digit is nonzero:
$$ \text{if } d_j \neq 0 \text{ then } Y \leftarrow Y \cdot Z_{d_j}. $$
Repeat this for all digits $d_t, \dots, d_0$.
Step 4. Correctness
We maintain the invariant:
$$ Y = x^{\text{value of digits already processed}}. $$
Initially, no digits are processed, so $Y=1=x^0$.
At each step, raising $Y$ to the $m$th power corresponds to shifting the accumulated exponent left by one radix-$m$ digit. Multiplying by $Z_{d_j}$ adds the contribution of the current digit. After processing all digits, we have
$$ Y = x^n. $$
Hence the algorithm is correct.
Step 5. Complexity analysis
- Precomputation of $Z_k = x^k$:
$m-1$ multiplications. 2. Exponentiation loop:
- There are $t+1 \approx \log_m n$ digits.
- Each digit requires $\nu$ squarings to compute $Y^m$. Total: $\nu \log_m n = \lg n$ squarings.
- Each nonzero digit requires one multiplication by $Z_{d_j}$. There are $\nu$ such digits.
Hence the total number of multiplications is approximately
$$ (m-1) + \nu + \lg n \approx m + \nu + \lg n. $$
This matches the target bound in the exercise.
Step 6. Summary of the algorithm
Algorithm C (left-to-right radix-$m$ exponentiation)
C1. [Precompute powers.]
$$ Z_0 \leftarrow 1, \quad Z_1 \leftarrow x, \quad Z_k \leftarrow Z_{k-1} \cdot x, \quad 2 \le k \le m-1 $$
C2. [Initialize.]
$$ Y \leftarrow 1 $$
C3. [Scan digits $d_t, \dots, d_0$.] For each digit $d_j$:
- Raise $Y$ to the $m$th power: $Y \leftarrow Y^m$
- If $d_j \neq 0$, multiply: $Y \leftarrow Y \cdot Z_{d_j}$
C4. [Return result.]
$$ Y = x^n $$
Step 7. Notes
- When $\nu = 1$ and $m=2$, the algorithm reduces to Algorithm A.
- The algorithm achieves the required operation count without updating all table entries repeatedly, resolving the previous critical error.
$\boxed{}$ ∎
Operation count summary:
| Component | Multiplications |
|---|---|
| Precompute $Z_1,\dots,Z_{m-1}$ | $m-1$ |
| Squarings for each digit | $\nu \log_m n = \lg n$ |
| Multiplications for nonzero digits | $\nu$ |
| Total | $\lg n + \nu + m$ |
This completes the solution.