Many applications require the selection of a subset of objects from
a larger set. For example, suppose you are down-sampling a large data
set by choosing the indices that you want to keep at random. If you are
satisfied with obtaining duplicate indices (in essence, if you are
sampling *with* replacement), this is a trivial matter in most
programming language. Python certainly makes this absurdly easy:

```
import random
def sample(k,n):
return [random.choice( [x for x in range(n)] ) for _ in range(k)]
```

But what about sampling *without* replacement? More precisely, what if
you require the selection of `k`

indices from an array of `n`

values
without choosing duplicate values but still want the probability of
choosing a certain element to be uniform? Well, if you use Python, you
are in luck again:

```
import random
def sample_without_replacement(k,n):
return random.sample([x for x in range(n)], k)
```

But what about other programming languages that may not have similar
builtin functionality, such as C++? In this case, a very simple and
elegant technique is given by none other than the great Donald E.
Knuth. The following
algorithm originally was given in *The Art of Computer Programming*,
Volume 2, Section 3.4.2 on p. 137. Briefly put, it works like this:

Set

`t = 0`

and`m = 0`

.`m`

represents the number of items selected so far, while`t`

is the total number of items that have already been traversed.Generate a random number

`U`

that is uniformly distributed between zero and one.If

`(N-t)*U >= n-m`

,*skip*the current item by increasing`t`

by 1 and going back to step 2. Else,*select*the current item by increasing both`t`

and`m`

by 1. Afterwards, either go back to step 2 or stop the algorithm (if sufficiently many items have been sampled already).

A very basic implementation of this algorithm (in C++) is given in the gist below:

I do not know about you, but my gut reaction upon seeing this algorithm
for the first time was *How can this possibly work*? So, let us briefly
prove the algorithm to be correct. Basically, we have to show that every
item has the same probability of being chosen. The key insight is to
realize that the probability has to change depending on the number of
elements that have already been selected. More precisely, we need to
determine the probability of choosing item `t+1`

if `m`

items have
already been selected. This requires some combinatorics. There are
ways of choosing `n`

items from `N`

such that `m`

items are picked as
the first `t`

items. Or, equivalently, these are the number of potential
permutations of the *remaining* `n-m`

elements. Out of these, we are
interested in all the ones that contain item `t+1`

—but this is
easy, because we can just take that item as granted and count the
*remaining* combinations as
, i.e. the number of ways to choose `n-m-1`

items out of `N-t-1`

ones. The quotient of these two numbers is exactly the probability with
which we must choose item `t+1`

if we want to have a uniform probability
for choosing a certain item. This quotient turns out to be
, which looks familiar. Finally, note that since `U`

was chosen to
be uniformly distributed between zero and one, the condition ```
(N-t)*U >=
n-m
```

in the algorithm is satisfied with the required probability.
Consequently, this method samples without replacement in the required
manner!

If you are interested or require a more efficient algorithm, please
refer to the paper *An Efficient Algorithm for Sequential Random
Sampling*
by Jeffrey Scott Vitter in *ACM Transactions on Mathematical Software*,
Volume 13, Issue 1, pp. 58–67, for more details. The paper
*should* be available free-of-charge from the link provided above, but
there is also
a mirror
available, thanks to the good folks at INRIA,
who require their publications to be kept in an open archive. Time
permitting, I might provide a second blog post and implementation about
the more efficient algorithm.

That is all for now, until next time!