TAOCP 7.1.3 Exercise 130
In the broadword (word-RAM) model used in TAOCP, a single machine step may include operations on an entire $n$-bit word such as: - bitwise Boolean operations, - addition and subtraction modulo $2^n$,...
Section 7.1.3: Bitwise Tricks and Techniques
Exercise 130. [M46] [M46] Is there an n-bit constant a such that the function (a≪x) mod 2n requires Ω(log n) n-bit broadword steps?
Verified: yes
Solve time: 4m19s
Corrected Solution
1. Clarifying the model
In the broadword (word-RAM) model used in TAOCP, a single machine step may include operations on an entire $n$-bit word such as:
- bitwise Boolean operations,
- addition and subtraction modulo $2^n$,
- shifts by a variable amount (as in standard word instructions).
In particular, a left shift by a variable amount $x$,
$$ y := a \ll x, $$
is treated as a single broadword operation.
2. Rewriting the function
The function under consideration is
$$ f(x) = (a \ll x) \bmod 2^n. $$
But modulo $2^n$ is redundant here, since shifting an $n$-bit word left already discards overflow beyond $n$ bits. Hence
$$ f(x) = a \ll x $$
as an $n$-bit word operation.
Bitwise, this is simply a hardware shift of the entire word $a$ by $x$ positions.
3. Direct computation cost
Given inputs $a$ (constant) and $x$ (variable), the computation
$$ f(x) = a \ll x $$
is performed by a single broadword shift instruction.
No intermediate arithmetic, decomposition, or bit-level simulation is required.
Thus the evaluation of $f$ has complexity
$$ T(n) = O(1) $$
in the broadword model.
This bound holds for every choice of the constant $a$.
4. Consequence for the lower bound question
The exercise asks whether there exists an $n$-bit constant $a$ such that computing $f(x)$ requires
$$ \Omega(\log n) $$
broadword steps.
However, we have shown a stronger statement:
- For every $n$-bit constant $a$,
$$ f(x) = a \ll x $$
is computable in one broadword step.
Hence the complexity is uniformly bounded by a constant independent of $n$.
5. Resolution of the attempted lower-bound approach
The previous argument fails because it attempts to analyze internal bit dependencies of the shift, but the model already provides the entire transformation as a primitive operation.
There is no opportunity for a logarithmic-depth computation because the model does not require building the shift via composition of simpler operations.
6. Final conclusion
No such constant $a$ exists.
For every $n$-bit constant $a$, the function
$$ f(x) = (a \ll x) \bmod 2^n $$
is computable in $O(1)$ broadword steps via a single variable shift operation. Therefore it cannot require $\Omega(\log n)$ steps.
Final Answer
There is no $n$-bit constant $a$ for which $(a \ll x) \bmod 2^n$ requires $\Omega(\log n)$ broadword steps. The function is computable in one broadword operation for every $a$, so its complexity is $O(1)$, not logarithmic. ∎