# Indexing schemes of binary sequences

I am looking for "low-complexity" indexing methods to enumerate binary sequences of a given length and a given weight.

Formally, let $T_k^n = \{x_1^n \in \{0,1\}^n: \sum_{i=1}^n x_i = k\}$. How to construct a bijective mapping $f: T_k^n \to \{1, 2, \ldots, \binom{n}{k}\}$ such that computing each $f(x_1^n)$ needs small number of operations?

For example, one could do lexicographical ordering, that is, e.g., $0110 < 1010$. Then this gives the following scheme:

$f(x_1^n) = \sum_{k=1}^n x_k \binom{n-k}{w_k}$

where $w_k=\sum_{i=k}^n x_i$. Computing $n$ binomial coefficients can be quite demanding. Any other ideas? Or is it impossible to avoid?

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thanks. will take a look! i heard before that this enumeration problem needs at least $O(n^2)$ binary operations. The above lexicographic enumeration needs to be computed using $n$ operations on a $n$-bit register, which is $O(n^2)$. Does DEK talk about this? –  mr.gondolier Jul 13 '10 at 23:29
I think there was some discussion about the complexity of generation, but I'm not sure if your specific issue was discussed - I'll check when I get home. Also, even if you have $O(n^2)$ worst-case performance, keep in mind that amortized performance may be much better (think of enumerating the numbers from $1$ to $2^n$ in a single register - you may have to change up to $n$ bits on any given operation, but the total number of bits changed is still just $2^{(n+1)}$, not $n2^n$) –  Steven Stadnicki Jul 14 '10 at 0:21
If you need frequently compute number of sequence you can pre-calculate all $\binom{i}{j}$. This would make you algorithm run in $O(n)$ time and $O(nk)$ memory. –  falagar Jul 14 '10 at 5:30