TAOCP 7.1.3 Exercise 111
Let the input word be x = (\ldots x_2 x_1 x_0)_2, and fix $r \ge 1$.
Section 7.1.3: Bitwise Tricks and Techniques
Exercise 111. [23] [23] Explain the 01r pattern recognizer, (98).
Verified: yes
Solve time: 2m46s
Let the input word be
$$ x = (\ldots x_2 x_1 x_0)_2, $$
and fix $r \ge 1$. We want to mark exactly those positions $i$ such that the pattern $01^r$ begins at $i$, i.e.
$$ x_i = 0 \quad \text{and} \quad x_{i+1} = x_{i+2} = \cdots = x_{i+r} = 1. $$
Formula (98) in TAOCP does not use $r$ explicit shifts and intersections. Instead it first constructs, in a doubling scheme, the word that recognizes runs of $r$ consecutive 1s, and then combines this with the zeros of $x$.
1. Isolating the structure of the pattern
Define a word $b$ such that
$$ b_i = 1 \iff x_{i+1} x_{i+2} \cdots x_{i+r} = 1^r. $$
Then the desired output word is simply
$$ y = \bar{x} ,&, b. $$
Thus the problem reduces to computing $b$ efficiently.
2. Run-of-ones formulation
Shift $x$ right by one position:
$$ u = x \gg 1. $$
Then
$$ u_i = x_{i+1}. $$
So $b_i$ is equivalent to saying that $u$ has a run of at least $r$ consecutive 1s starting at position $i$.
We therefore define auxiliary words $b^{(k)}$ by
$$ b^{(k)}i = 1 \iff u_i u{i+1} \cdots u_{i+k-1} = 1^k. $$
We want $b^{(r)}$.
3. The doubling recurrence (core of formula (98))
The key observation used in (98) is:
If we already know the correctness of blocks of length $k$, then we can obtain correctness for blocks of length $2k$ by checking two adjacent blocks:
$$ b^{(2k)} = b^{(k)} ,&, (b^{(k)} \gg k). $$
Why this works
At position $i$, $b^{(k)}_i = 1$ means
$$ u_i, u_{i+1}, \ldots, u_{i+k-1} = 1. $$
Similarly,
$$ (b^{(k)} \gg k)i = b^{(k)}{i+k} = 1 $$
means
$$ u_{i+k}, \ldots, u_{i+2k-1} = 1. $$
Therefore both hold exactly when the full block
$$ u_i, \ldots, u_{i+2k-1} $$
consists of 1s. This proves the recurrence.
4. Connection with formula (98)
Formula (98) implements this recurrence in a SWAR/broadword manner:
- Start with
$$ b^{(1)} = u = x \gg 1. $$ 2. Repeatedly double the window size:
$$ b^{(2)} = b^{(1)} ,&, (b^{(1)} \gg 1), $$
$$ b^{(4)} = b^{(2)} ,&, (b^{(2)} \gg 2), $$
$$ b^{(8)} = b^{(4)} ,&, (b^{(4)} \gg 4), $$
and so on.
At stage $t$, the algorithm maintains a word $b^{(2^t)}$ such that:
$$ b^{(2^t)}i = 1 \iff u_i u{i+1} \cdots u_{i+2^t-1} = 1^{2^t}. $$
This is the invariant used implicitly in (98).
To handle arbitrary $r$, the construction combines appropriate stages (as in Knuth’s standard “binary decomposition of lengths”), yielding $b^{(r)}$ using $O(\log r)$ operations.
5. Final combination
Once $b$ is obtained, the pattern recognizer is
$$ y = \bar{x} ,&, b. $$
At position $i$, this equals 1 exactly when:
- $x_i = 0$,
- $x_{i+1}, \ldots, x_{i+r} = 1$,
which is exactly the occurrence of $01^r$ starting at $i$.
6. Conclusion
Formula (98) is an instance of a general broadword technique: it replaces a linear chain of $r$ shifted intersections by a logarithmic cascade based on the identity
$$ b^{(2k)} = b^{(k)} ,&, (b^{(k)} \gg k), $$
thereby computing runs of 1s of length $r$ efficiently. Combining this run detector with a single negation of $x$ yields the recognizer for the pattern $01^r$.