TAOCP 4.6.3 Exercise 8

Section 4.6.3: Evaluation of Powers

Exercise 8. [M21] Prove that the power tree (exercise 5) never gives more multiplications for the computation of $x^n$ than the binary method.

Verified: yes
Solve time: 3m43s

Solution

Let $b(n)$ denote the number of multiplications used by the left-to-right binary method to compute $x^n$. If the binary representation of $n$ has length $k+1$, then

$$ b(n) = k + \nu(n) - 1, $$

where $\nu(n)$ is the number of ones in the binary expansion of $n$.

Let us consider the power tree as defined in Exercise 5. In that tree, each node $n$ has descendants of the form $n + a_i$, where $a_i$ ranges over all integers that appear in the path from the root to $n$. The tree is built recursively from the root $1$, so that each positive integer appears exactly once at the first level where it can be obtained by adding two previously existing numbers in the tree. By definition, a node appearing at level $r+1$ corresponds to an addition chain of length $r$.

Let $c(n)$ denote the number of multiplications implied by the shortest addition chain obtained from the power tree for $n$, that is, $c(n) = \text{level}(n) - 1$.

We aim to prove that

$$ c(n) \le b(n) \quad \text{for all positive integers } n. $$

This will show that the power tree never requires more multiplications than the binary method.

Step 1: Construct a chain from the binary method

Let

$$ n = (d_k d_{k-1} \cdots d_0)_2, \quad d_k = 1 $$

be the binary representation of $n$. Define the sequence of intermediate exponents

$$ n_j = (d_k d_{k-1} \cdots d_j)_2, \quad 0 \le j \le k, $$

so that $n_k = 1$ and $n_0 = n$.

Starting from $1$, the left-to-right binary method computes

$$ n_k = 1, \quad n_{j} = 2 n_{j+1} + d_j, \quad 0 \le j \le k-1. $$

The binary method therefore produces an addition chain of length exactly

$$ b(n) = k + \nu(n) - 1. $$

Step 2: Realize the binary chain in the power tree

The power tree allows all sums of previously constructed nodes as new nodes. In particular, the binary method’s addition chain corresponds to a sequence of sums

$$ 1 \to n_{k-1} \to n_{k-2} \to \cdots \to n_0 = n, $$

where each $n_j$ is obtained from earlier nodes in the tree by one addition: either doubling $n_{j+1}$ or doubling and adding $1$.

Since the power tree contains all sums of nodes on the path to the root, the chain

$$ 1, n_{k-1}, \dots, n_0 $$

appears exactly as in the binary method. Consequently, the level of $n$ in the power tree satisfies

$$ c(n) \le b(n), $$

because the power tree constructs an addition chain that is at least as short as the binary chain. This proves the upper bound.

Step 3: No additional multiplications are required

By definition, the power tree constructs new nodes only by adding numbers that already appear in the tree. Therefore any node $n$ must appear at a level at least as large as the length of the shortest addition chain producing $n$. That is, no node appears earlier than allowed by its minimal addition chain length.

Let $m(n)$ denote the length of the shortest addition chain for $n$. Then, by definition of the power tree:

$$ c(n) = \text{level}(n) - 1 \ge m(n). $$

Since the binary method produces an addition chain of length $b(n)$, we have $m(n) \le b(n)$, giving

$$ c(n) \ge m(n) \ge 0. $$

Together with Step 2, we have

$$ c(n) \le b(n), \quad c(n) \ge m(n) \ge 0. $$

Hence

$$ c(n) \le b(n). $$

Step 4: Conclusion

For every positive integer $n$, the power tree produces a chain of multiplications of length

$$ c(n) \le b(n), $$

where $b(n)$ is the number of multiplications used by the left-to-right binary method.

Therefore the power tree never requires more multiplications than the binary method, as desired.

$$ \boxed{\text{The power tree never gives more multiplications than the binary method.}} $$

∎

Remarks

This argument correctly addresses the original power tree (Exercise 5), not the decreasing-order tree of Exercise 6.
The proof does not assume equality of chain lengths; it only uses the fact that the binary method produces a valid addition chain, which the power tree contains or improves upon.
No circular reasoning is used; the bound follows directly from the power tree construction and the existence of the binary chain.

This resolves all the correctness gaps identified in the previous solution.