（n选择k）和长度为n的位串之间的双射，设置k位

Question

While I know how to generate all ( n choose k ) bitstrings of size n with exactly k bits set to one, I'm struggling finding a bijection, that gets as input a number i between 1 and ( n choose k ) and outputs the i -th vector of that kind in an arbitrary ordering.

Obviously, one could simply enumerate all of that vectors in a list and then output the i -th entry of the list, but unfortunately that approach has to high memory requirements for my setting.

Edit: also it should be an efficient computation, computing the list of all vectors for each call to the bijection is also not an option.

Answer 1

The straightforward way:

If i < (n-1 choose k) , then the leftmost bit is 0 and i determines the remainder of the bits recursively. Otherwise, the leftmost bit is 1 , and i - (n-1 choose k) determines the remainder of the bits recursively. In the second case there are at most (n-1 choose k-1) possible values of i - (n-1 choose k) .