Re: [Caml-list] Choosing a random element in a Map or a Set

[Ivan Gotovchits <ivg@ieee.org>]

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On Thu, Mar 18, 2021 at 5:41 AM <jean-marc.alliot@irit.fr> wrote:

> Dear list,
>
> I am trying to take a random element from a Map or a Set.
>
> Currently, I generate one random int between 1 et Card(map), and I iter
> until I reach this element. The solution is in O(n) which is not great...
>
> I was about to code the following solution: take the current Map code
> and add a function which follows the balanced binary tree from the root
> and takes randomly a left or right turn at each node until we reach a
> leaf (we only need to generate one random int and use its binary
> representation to choose the left or right direction at each node). This
> should be in O(log(n))
>
> Before I start coding like an idiot:
> 1) Is there another, more intelligent, solution?
>

I would create an array of keys and then randomly choose a key from that
array. If you want to be on the functional side, you can shuffle the array
and turn it into a list.
This solution involves some duplication but is easy to implement and is
especially useful if you do not what repetitions in your selection.

An alternative solution that doesn't suffer from an extra memory overhead
would be writing a function that generates a random key (instead of a
random integer). It is not always possible,
especially when the domain of keys is infinite. However, if your set/map is
total (i.e., maps all keys in the domain to values) then it is the perfect
solution.

Finally, as an amalgamation of the above approaches, you can hash-cons your
keys (essentially turning keys into integers). It usually involves an extra
data structure to maintain your index,
but gives you the additional benefits of hash-consing, e.g., faster and
tighter maps and sets.



> 2) Is my solution biased? I think it is not, as long as the tree is
> properly balanced but...
>

OCaml trees are relaxed AVL trees with the maximum difference between trees
height equal to 2 (not 1 as in classical AVL). For small trees, it is quite
a substantial difference, e.g., imagine a tree of 9 elements with the left
tree having a height of 3 and the right of height 1.
The right tree will have only one element and the left will have 7
elements. Therefore, the probability of selecting the right element will be
1/2 vs 1/9. Even for larger trees, it remains the problem as a large tree
is made of small subtrees :) Therefore your probability distribution
will be far from uniform.



>
> Thanks
>
> Jean-Marc
>
>
>

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