Arithmetic network to compute floor of binary logarithm

Question

I wonder how to build efficient arithmetic network (using logical gates only) to compute floor of binary logarithm of the given input number. I have read some articles on stackoverflow.com about this problem, respectively its implementation in C/C++ using bitwise operations only.

Stackoverflow: Computing the floor of log_2(x) using only bitwise operators in C
Stackoverflow: Computing the floor log of a binary number
Bit Twiddling Hacks

I am pretty sure there's a simpler way to get this done. Using De Bruijn sequences seems to be a magic for me.

Answer 1

It can be done with a fairly simple log-depth circuit without resorting to such hacks that only really make sense in software.

The "position of most significant set bit" function pmsb can be defined recursively:

$\begin{split}pmsb(``0") &= Nil \\ pmsb(``1") &= Nil \\ pmsb(a:b) &= ``0":pmsb(b) \text{ if a=0}\\ &= ``1":pmsb(a) \text{ otherwise} \end{split}$

Where : divides a power-of-two string down the middle or concatenates two strings, depending on its context.

This can already be implemented as a circuit, but it includes some annoying wide comparisons. They can be removed by returning a tuple with an extra bit that tracks the zero-ness of the whole string (it is applied to the halves so you get the result from the halves in the recursion):

$\begin{split}pmsb(``0") &= (Nil, T) \\ pmsb(``1") &= (Nil, F) \\ pmsb(a:b) &= \text{let } (pa, za) = pmsb(a) \\ &\phantom{=}\quad\quad (pb, zb) = pmsb(b) \\ &\phantom{=}\;\; \text{in if } za \text{ then } (``0":pb, zb)\\ &\phantom{=}\quad\quad\quad\quad\text{else } (``1":pa, F) \end{split}$

This can then be evaluated bottom-up by a circuit of depth O(log n) and size O(n) where n is the length of the bit-string at the top level, which must be a power of two and not one. The zeroness bit circuit has the structure of an AND-reduction, but it is used at the intermediate nodes instead of only at the end. The main result is computed by a similar tree of MUXes that take those bits as control input.

Answer 2

The floor of the binary logarithm of $n$ is the integer $i$ such that $2^i \leq x < 2^{i+1}$. This is also the leftmost (most significant) 1 in the binary expansion of $x$. If $x = x_{n-1} \ldots x_0$, then the answer is $i$ if $x_{n-1} = \cdots = x_{i+1} = 0$ and $x_i = 1$. This leads to a simple circuit that computes $i$.