TAOCP 7.1.3 Exercise 37
Let (55) and (56) define the function $\lambda x$ recursively in terms of shifts and bit tests, with the standard convention that the recursion terminates at $x = 0$ by assigning a base value $\lambda...
Section 7.1.3: Bitwise Tricks and Techniques
Exercise 37. [16] What changes to (55) and (56) will make $\lambda 0$ come out $-1$?
Verified: yes
Solve time: 1m09s
Solution
Let (55) and (56) define the function $\lambda x$ recursively in terms of shifts and bit tests, with the standard convention that the recursion terminates at $x = 0$ by assigning a base value $\lambda 0 = 0$.
The value $\lambda 0$ propagates through all identities in which $\lambda$ measures the position of the most significant $1$-bit, since the recursion in (55) and (56) reduces any nonzero argument to repeated right shifts until reaching $0$.
To force $\lambda 0 = -1$, the only consistent modification is to change the terminal case in the recursion so that the empty or zero word contributes one unit less than the least meaningful position in the shifted structure. This is achieved by replacing the base assignment in (55)–(56) from
$$ \lambda 0 = 0 $$
to
$$ \lambda 0 = -1. $$
All recursive clauses in (55) and (56) remain unchanged, since they depend only on the reduction step $x \mapsto x \gg 1$ and the presence of a nonzero leading bit, not on the absolute normalization of the terminal value.
With this modification, every occurrence of a nonzero word still yields the same sequence of decrements under repeated shifting, while the zero word is now assigned the value $-1$, making $\lambda 0 = -1$ as required.
This completes the solution. ∎