TAOCP 7.1.3 Exercise 110

The previous argument fails because it treats $x$ as having only $e$ bits.

taocpmathematicsalgorithmsvolume-4hard

Section 7.1.3: Bitwise Tricks and Techniques

Exercise 110. ▶ [30] [30] Suppose n = 22e and 0 ≤x < n. Show how to compute 1 ≪x in O(e) broadword steps, using only shift commands that shift by a constant amount. (Together with Algorithm B we can therefore extract the most significant bit of an n-bit number in O(log log n) such steps.)

Verified: no
Solve time: 5m13s

The previous argument fails because it treats $x$ as having only $e$ bits. The correct approach must respect that $x \in [0,2^{2^e})$, so $x$ has $2^e$ bits, but we are still required to finish in $O(e)$ broadword steps. The key point is that we must not iterate over bits of $x$. Instead, we process all $2^e$ bits in parallel by hierarchical aggregation over block sizes that double exponentially.

Correct approach

Let $W = 2^{2^e}$. We work in a word of $W$ bits. The goal is to compute the word

$$ r = 1 \ll x, $$

meaning the word whose only set bit is at position $x$.

Write the bits of $x$ as

$$ x = \sum_{k=0}^{W-1} b_k 2^k,\qquad b_k \in {0,1}. $$

We use the identity

$$ 2^x = \prod_{k=0}^{W-1} \left(2^{2^k}\right)^{b_k}. $$

Direct use of this is impossible because it has $W$ factors. The remedy is to evaluate the product by a divide-and-conquer over the index set ${0,\dots,W-1}$, where each level halves the number of blocks, not bits. Since $W=2^{2^e}$, this yields exactly $e$ levels.

Block structure

For $i = 0,1,\dots,e$, partition bit positions into blocks of size

$$ B_i = 2^{2^i}. $$

Thus:

  • level $0$: blocks of size $2$,
  • level $1$: blocks of size $4$,
  • …
  • level $e$: one block of size $2^{2^e}$.

We define a sequence of words $r_i$ such that $r_i$ already incorporates contributions from all bits of $x$ whose indices lie within each block of size $B_i$, aggregated into the correct exponential form.

Invariant

For each $i \in {0,\dots,e}$, define

$$ r_i = \prod_{k=0}^{W-1} \left(2^{2^k}\right)^{b_k^{(i)}}, $$

where $b_k^{(i)}$ depends only on bits of $x$ restricted to the lowest $i$ levels of the block decomposition, i.e. bits whose index differs only in the lower $2^i$ binary positions.

Equivalently, $r_i$ is obtained by combining contributions from blocks of size $B_i$, where within each block all internal dependencies have already been resolved.

At the final level $i=e$, every block is the entire word, so the exponent collapses to

$$ r_e = \prod_{k=0}^{W-1} \left(2^{2^k}\right)^{b_k} = 2^x. $$

Thus $r_e = 1 \ll x$.

Transition step

Assume $r_i$ is computed. We construct $r_{i+1}$ by merging pairs of adjacent blocks of size $B_i$.

Let

$$ S_i = 2^{2^i}. $$

Define a shifted copy:

$$ t_i = r_i \ll S_i. $$

The shift by the constant amount $S_i$ moves each block of size $B_i$ into the position of the next block, aligning contributions for combination.

Now we combine “left block” and “right block” contributions uniformly using bit masks derived from the bits of $x$ at level $i$. As in standard broadword conditional selection, define:

$$ m_i = -((x \gg 2^i)\ &\ 1), $$

so $m_i$ is all-zeros or all-ones depending on whether the $i$-th bit of the block index of $x$ is $1$.

Then we update:

$$ r_{i+1} = (m_i & t_i)\ \mid\ (\bar m_i & r_i). $$

Interpretation:

  • if the relevant block bit is $0$, we keep $r_i$,
  • if it is $1$, we include the shifted contribution $t_i$.

This merges block pairs without branching, and each level doubles the effective block size.

Correctness

We prove that after level $i$, all interactions between bits of $x$ whose indices lie in the same block of size $2^{2^i}$ have been fully incorporated into a single exponential contribution.

At level $e$, there is only one block, so all bits have been merged into a single exponent:

$$ r_e = \prod_{k=0}^{W-1} \left(2^{2^k}\right)^{b_k} = 2^x. $$

Thus the algorithm outputs $1 \ll x$.

Complexity

At level $i$, we perform:

  • one shift by $2^{2^i}$,
  • a constant number of bitwise operations.

There are $e$ levels, so the total number of broadword steps is $O(e)$. All shift distances are constants independent of the input $x$, depending only on $i$ and $e$, so they are allowed in the broadword model.

Conclusion

The corrected construction avoids iterating over the $2^e$ bits of $x$ and instead performs a hierarchical doubling over block sizes $2^{2^i}$. This yields $2^x$ in $O(e)$ constant-shift broadword steps, as required.