How do you efficiently generate a list of K non-repeating integers between 0 and an upper bound N

Question

The question gives all necessary data: what is an efficient algorithm to generate a sequence of K non-repeating integers within a given interval [0,N-1]. The trivial algorithm (generating random numbers and, before adding them to the sequence, looking them up to see if they were already there) is very expensive if K is large and near enough to N.

The algorithm provided in Efficiently selecting a set of random elements from a linked list seems more complicated than necessary, and requires some implementation. I've just found another algorithm that seems to do the job fine, as long as you know all the relevant parameters, in a single pass.

Answer 1

In The Art of Computer Programming, Volume 2: Seminumerical Algorithms, Third Edition, Knuth describes the following selection sampling algorithm:

Algorithm S (Selection sampling technique). To select n records at random from a set of N, where 0 < n ≤ N.

S1. [Initialize.] Set t ← 0, m ← 0. (During this algorithm, m represents the number of records selected so far, and t is the total number of input records that we have dealt with.)

S2. [Generate U.] Generate a random number U, uniformly distributed between zero and one.

S3. [Test.] If (N – t)U ≥ n – m, go to step S5.

S4. [Select.] Select the next record for the sample, and increase m and t by 1. If m < n, go to step S2; otherwise the sample is complete and the algorithm terminates.

S5. [Skip.] Skip the next record (do not include it in the sample), increase t by 1, and go back to step S2.

An implementation may be easier to follow than the description. Here is a Common Lisp implementation that select n random members from a list:

(defun sample-list (n list &optional (length (length list)) result)
  (cond ((= length 0) result)
        ((< (* length (random 1.0)) n)
         (sample-list (1- n) (cdr list) (1- length)
                      (cons (car list) result)))
        (t (sample-list n (cdr list) (1- length) result))))

And here is an implementation that does not use recursion, and which works with all kinds of sequences:

(defun sample (n sequence)
  (let ((length (length sequence))
        (result (subseq sequence 0 n)))
    (loop
       with m = 0
       for i from 0 and u = (random 1.0)
       do (when (< (* (- length i) u) 
                   (- n m))
            (setf (elt result m) (elt sequence i))
            (incf m))
       until (= m n))
    result))