Divide an integer range into nearly equal integer ranges

Question

I am writing an algorithm that involves taking a set of numbers and putting them into buckets. Can you help me implement these two simple methods? Let me know if I need to explain more.

// return a vector where each element represents the
// size of the range of numbers in the corresponding bucket
// buckets should be equal in size +/- 1
// doesn't matter where the bigger/smaller buckets are
vector<int> makeBuckets(int max, int numberOfBuckets);

// return which bucket n belongs in
int whichBucket(int max, int numberOfBuckets, int n); 

Example output

makeBuckets(10, 3) == { 3, 3, 4 }; // bucket ranges: (0, 2), (3, 5), (6, 9)
whichBucket(10, 3, 0) == 0;
whichBucket(10, 3, 1) == 0;
whichBucket(10, 3, 2) == 0;
whichBucket(10, 3, 3) == 1;
whichBucket(10, 3, 4) == 1;
whichBucket(10, 3, 5) == 1;
whichBucket(10, 3, 6) == 2;
whichBucket(10, 3, 7) == 2;
whichBucket(10, 3, 8) == 2;
whichBucket(10, 3, 9) == 2;

Answer 1

Given an integral range [B, E), the most evenly (The size differences never exceed 1.) split N_bin bins are:

[ ⌊B＋E×N_index÷N_bin⌋, ⌊B＋E×(N_index＋1)÷N_bin⌋ ) _{where Nindex＝0 to (Nbin－1)}.

---

Example ― Getting pixel ranges to feed multiple threads:

for(unsigned int worker_index = 0; worker_index < n_workers; ++worker_index) {
    std::cout
        << "[" << n_pixels * worker_index / n_workers << ", "
        << n_pixels * (worker_index + 1) / n_workers << ")"
        << std::endl;
}

---

[0, 5) into 16 bins

[0, 0)
[0, 0)
[0, 0)
[0, 1)
[1, 1)
[1, 1)
[1, 2)
[2, 2)
[2, 2)
[2, 3)
[3, 3)
[3, 3)
[3, 4)
[4, 4)
[4, 4)
[4, 5)

[0, 18) into 16 bins

[0, 1)
[1, 2)
[2, 3)
[3, 4)
[4, 5)
[5, 6)
[6, 7)
[7, 9)
[9, 10)
[10, 11)
[11, 12)
[12, 13)
[13, 14)
[14, 15)
[15, 16)
[16, 18)

[0, 36152320) into 16 bins

[0, 2259520)
[2259520, 4519040)
[4519040, 6778560)
[6778560, 9038080)
[9038080, 11297600)
[11297600, 13557120)
[13557120, 15816640)
[15816640, 18076160)
[18076160, 20335680)
[20335680, 22595200)
[22595200, 24854720)
[24854720, 27114240)
[27114240, 29373760)
[29373760, 31633280)
[31633280, 33892800)
[33892800, 36152320)

Answer 2

Let's say you need to divide a range [0,n] into k buckets.

1. e = n / k (integer divide!) will tell you the minimal size of each bucket.
2. o = n % k will tell you how many buckets need to grow in size.
3. Now loop over k:
   • If o > 0, create a bucket of size e+1, decrease o.
   • If o == 0, create a bucket of size e.

How to best create buckets depends on the size of n. For example, if you have a small n, you could just have an array of size n that stores the bucket index for each number. In the loop above, you would fill up that array. Then the query whichBucket() would run in O(1).

If n is large, however, this is impractical. In this case, you would do your bucket sorting completely implicitely. That means, for each incoming query, you can directly compute the corresponding bucket index using e and o.

Answer 3

Thanks ypnos for the answer. Here's my implementation in C++.

vector<int> makeBuckets(int max, int numberOfBuckets) {
    int e = max / numberOfBuckets;
    vector<int> b(numberOfBuckets, e);
    fill(b.begin(), b.begin() + (max % numberOfBuckets), e + 1);
    return b;
}

int whichBucket(int max, int numberOfBuckets, int n) {
    return n * numberOfBuckets / max;
}

Answer 4

You may use the naive implementation:

std::vector<std::size_t> parts(std::size_t m, std::size_t size)
{
    std::vector<std::size_t> res(size);

    for (std::size_t i = 0; i != m; ++i) {
        ++res[i % size];
    }
    return res;
}

std::size_t whichPart(std::size_t m, std::size_t size, std::size_t n)
{
    std::size_t index = 0;
    for (auto i : parts(m, size)) {
        if (n < i) {
            return index;
        }
        ++index;
        n -= i;
    }
    throw std::runtime_error("invalid argument");
}