Algorithm for subdividing an array into "semi-equal", uniform sub-arrays
Asked Answered
N

2

5

Given an array with N elements, I am looking for M (M < N) successive sub-arrays with equal lengths or with lengths that differ by mostly 1. For example, if N = 12 and M = 4, all sub-arrays would have equal lengths of N/M = 3. If N = 100 and M = 12, I expect sub-arrays with lengths 8 and 9, and both sizes should be uniformly spread within the original array. This simple task turned to be a little bit subtle to implement. I came up with an adaptation of the Bresenham's line algorithm, which looks like this when coded in C++:

/// The function suggests how an array with num_data-items can be
/// subdivided into successively arranged groups (intervals) with
/// equal or "similar" length. The number of intervals is specified
/// by the parameter num_intervals. The result is stored into an array
/// with (num_data + 1) items, each of which indicates the start-index of
/// an interval, the last additional index being a sentinel item which 
/// contains the value num_data.
///
/// Example:
///
///    Input:  num_data ........... 14,
///            num_intervals ...... 4
///
///    Result: result_start_idx ... [ 0, 3, 7, 10, 14 ]
///

void create_uniform_intervals( const size_t         num_data,
                               const size_t         num_intervals,
                               std::vector<size_t>& result_start_idx )
{
    const size_t avg_interval_len  = num_data / num_intervals;
    const size_t last_interval_len = num_data % num_intervals;

    // establish the new size of the result vector
    result_start_idx.resize( num_intervals + 1L );
    // write the pivot value at the end:
    result_start_idx[ num_intervals ] = num_data;

    size_t offset     = 0L; // current offset

    // use Bresenham's line algorithm to distribute
    // last_interval_len over num_intervals:
    intptr_t error = num_intervals / 2;

    for( size_t i = 0L; i < num_intervals; i++ )
    {
        result_start_idx[ i ] = offset;
        offset += avg_interval_len;
        error -= last_interval_len;
        if( error < 0 )
        {
            offset++;
            error += num_intervals;
        } // if
    } // for
}

This code calculates the interval lengths for N = 100, M=12: 8 9 8 8 9 8 8 9 8 8 9 8

The actual question is that I don't know how exactly to call my problem, so I had difficulty searching for it.

  • Are there other algorithms for accomplishing such a task?
  • How are they called? Maybe the names would come if I knew other areas of application.

I needed the algorithm as a part of a bigger algorithm for clustering of data. I think it could also be useful for implementing a parallel sort(?).

Nureyev answered 10/11, 2011 at 17:51 Comment(0)
C
8

If your language has integer division that truncates, an easy way to compute the size of section i is via (N*i+N)/M - (N*i)/M. For example, the python program

  N=100;M=12
  for i in range(M): print (N*i+N)/M - (N*i)/M

outputs the numbers 8 8 9 8 8 9 8 8 9 8 8 9. With N=12;M=5 it outputs 2 2 3 2 3. With N=12;M=3 it outputs 4 4 4.

If your section numbers are 1-based rather than 0-based, the expression is instead (N*i)/M - (N*i-N)/M.

Cindelyn answered 10/11, 2011 at 18:24 Comment(1)
It should be noted that my implementation given in the question has an additional "feature": the interval lengths are "symmetrical" with respect to the middle of the array. For the example N = 100, M = 12 you get: 8 9 8 8 9 8 8 9 8 8 9 8Nureyev
I
0

Space-filling-curves and fractals subdivide the plane and reduce the complexity. There is for example z-curve, hilbert curve, morton curve.

Insupportable answered 10/11, 2011 at 18:25 Comment(0)

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