Random element from unordered_set in O(1)

问题

I've seen people mention that a random element can be grabbed from an unordered_set in O(1) time. I attempted to do so with this:

std::unordered_set<TestObject*> test_set;

//fill with data

size_t index = rand() % test_set.size();
const TestObject* test = *(test_set.begin() + index);

However, unordered_set iterators don't support + with an integer. begin can be given a size_t param, but it is the index of a bucket rather than an element. Randomly picking a bucket then randomly picking an element within it would result in a very unbalanced random distribution.

What's the secret to proper O(1) random access? If it matters, this is in VC++ 2010.

回答1:

I believe you have misinterpreted the meaning of "random access", as it was used in those cases you're referring to.

"Random access" doesn't have anything to do with randomness. It means to access an element "at random", i.e. access any element anywhere in the container. Accessing an element directly, such as with std::vector::operator[] is random access, but iterating over a container is not.

Compare this to RAM, which is short for "Random Access Memory".

回答2:

std::unordered_set don't provide a random access iterator. I guess it's a choice from the stl designers to give stl implementers more freedom...the underlying structure have to support O(1) insertion and deletion but don't have to support random access. For example you can code an stl-compliant unordered_set as a doubly linked list even though it's impossible to code a random access iterator for such an underlying container.

Getting a perfectly random element is then not possible even though the first element is random because the way the elements are sorted by hash in the underlying container is deterministic...And in the kind of algorithm that I am working on, using the first element would skew the result a lot.

I can think of a "hack", if you can build a random value_type element in O(1)... Here is the idea :

check the unordered set in not empty (if it is, there is no hope)
generate a random value_type element
if already in the unordered set return it else insert it
get an iterator it on this element
get the random element as *(it++) (and if *it is the last element the get the first element)
delete the element you inserted and return the value in (5)

All these operations are O(1). You can implement the pseudo-code I gave and templatize it quite easily.

N.B : The 5th step while very weird is also important...because for example if you get the random element as it++ (and it-- if it is the last iterator) then the first element would be twice less probable than the others (not trivial but think about it...). If you don't care about skewing your distribution that's okay you can just get the front element.

回答3:

std::unordered_set has no O(1) random access in the sense of an array. It is possible to access an element, based on key, in O(1) but it is impossible to find the k-th element.

Despite that, here is a way to get a random element with a uniform distribution from std::unordered_map (or with std::unordered_set if the key has a mutable field). I have layed out a similar technique in an answer to SO question Data Structure(s) Allowing For Alteration Through Iteration and Random Selection From Subset (C++).

The idea is to supplement each entry in std::unordred_set with a mutable index value into a vector of pointers into the unordered_set. The size of the vector is the size of the unordered_set. Every time a new element is inserted to the unordered_set, a pointer to that element is push_back-ed into the vector. Every time an element is erased from the unodrered_set, the corresponding entry in the vector is located in O(1), and is swapped with the back() element of the vector. The index of the previously back() element is amended, and now points to its new location in the vector. Finally the old entry is pop_back()-ed from the vector.

This vector points exactly to all elements in the unordered_set. It takes O(1) to pick a random element from the combined structure in uniform distribution. It takes O(1) to add or erase an element to the combined structure.

NOTE: Pointers to elements (unlike iterators) are guaranteed to stay valid as long as the element exists.

Here is how this should look:

For erasing element c:

swap element c_index and a_index and fix the pointers to them:
pop_back last element, which is element_c from the vector.
erase c from the unordred_set.

Randomization is trivial - simply select an element at random from the vector.

回答4:

I wrote a solution using buck_count() and cbegin(n) methods, to choose a bucket at random, and then choose an element at random in the bucket.

Two problems: - this is not constant time (worse case with a lot of empty buckets and many elements in one bucket) - the probability distribution is skewed

I think that the only way to peek an element at random is to maintain a separate container providing a random access iterator.

#include <random>
#include <iostream>
#include <unordered_set>
#include <unordered_map>
#include <cassert>

using namespace std;

ranlux24_base randomEngine(5);

int rand_int(int from, int to)
{
    assert(from <= to);

    return uniform_int_distribution<int>(from, to)(randomEngine);
}

int random_peek(const unordered_set<int> & container)
{
    assert(container.size() > 0);

    auto b_count = container.bucket_count();
    auto b_idx = rand_int(0, b_count - 1);
    size_t b_size = 0;

    for (int i = 0; i < b_count; ++i)
    {
        b_size = container.bucket_size(b_idx);
        if (b_size > 0)
            break;

        b_idx = (b_idx + 1) % b_count;
    }

    auto idx = rand_int(0, b_size - 1);

    auto it = container.cbegin(b_idx);

    for (int i = 0; i < idx; ++i)
    {
        it++;
    }

    return *it;
}

int main()
{
    unordered_set<int> set;

    for (int i = 0; i < 1000; ++i)
    {
        set.insert(rand_int(0, 100000));
    }

    unordered_map<int,int> distribution;

    const int N = 1000000;
    for (int i = 0; i < N; ++i)
    {
        int n = random_peek(set);
        distribution[n]++;
    }

    int min = N;
    int max = 0;

    for (auto & [n,count]: distribution)
    {
        if (count > max)
            max = count;
        if (count < min)
            min = count;
    }

    cout << "Max=" << max << ", Min=" << min << "\n";
    return 0;
}

来源：https://stackoverflow.com/questions/12761315/random-element-from-unordered-set-in-o1

标签

c++

stl

unordered-set