An algorithm thats gives all possible distinct arrays whose positive integer elements sum to a given number

An开发者_如何学JAVA algorithm which for any input positive integer, gives all possible distinct arrays of positive non-zero integers that can sum to it.

e.g Inputting 4 returns (1,1,1,1), (1,1,2), (1,2,1), (2,1,1), (1,3), (3,1), (2,2), (4)

Not homework, but "research". I just get lost trying.

Someone good at combinatorics would probably know this one.

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Here is some idea.

If I'm not mistaken the number of the arrays is 2N-1 and the arrays map to bit patterns codding integers form 0 to 2N-1-1 as follows:

I'll show an example for N = 4

The first array is all ones. Imagine every bit in the bit pattern corresponds to the boundary between two array cells

1 1 1 1 <- array elements
 | | |  <- sections
 0 0 0  <- bit pattern

Every 1 in the bit pattern means merging two neighbouring cells

1 (1+1) 1  <- array elements (1 2 1)
 |  |  |   <- sections
 0  1  0   <- bit pattern


1 (1+1+1)  <- array elements (1 3)
 |  | |    <- sections
 0  1 1    <- bit pattern


(1+1) (1+1)<- array elements (2 2)
  |  |  |  <- sections
  1  0  1  <- bit pattern


(1+1+1+1)  <- array elements (4)
  | | |    <- sections
  1 1 1    <- bit pattern

To enumerate all arrays you can generate integers from 0 to 2N-1-1 and for every bit pattern you get, generate the corresponding array. It might be helpful to convert the integer to the string of zeros and ones of length N-1. You decode the pattern as follows:

First cell contains 1 initially. Going through the pattern from left to right, for every bit, if it's 1 add 1 to the current cell, if it's 0 create new cell containing 1.

The pattern 1 1 0 0 1 0 1 for N = 8 would be decoded to an array

(3 1 2 2)

Here is some C++ code without argument validation and processing the pattern from right to left. It just changes the order of arrays produced and is simpler to code.

std::vector<std::vector<int> > generateArrays(unsigned int N)
{
    //validate the argument before processing
    // N > 0 and N <= numeric_limits<unsigned int>::digits

    unsigned int numOfArrays = (1U << (N-1));
    std::vector<std::vector<int> > result(numOfArrays);
    for(unsigned int i = 0; i < numOfArrays; ++i)
    {
        result[i].push_back(1);
        unsigned int mask = 1U;
        while(mask < numOfArrays)
        {
            if((i & mask) != 0)
            {
                result[i].back()++;
            }
            else
            {
               result[i].push_back(1);
            }
            mask <<= 1;
        }
    }
    return result;
}

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Recurse! The first entry (call it a[0]) could be any integer from 1 to N. Then you just need to find all the distinct arrays of positive nonzero integers that add up to N - a[0]...

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Recursive approach

# Pseudo code, not any real language
function all_arrays_summing_to(int N) {
    array_of_arrays All_solutions = ();
    if (N == 1) return { [[1]] }; # This is array of one array containing 1 element with value 1
    for each number x in (1 .. N-1)  {
        array_of_arrays AA = all_arrays_summing_to(N - x);
        for each array A in (AA) {
            push x onto array A;
            Add A to All_solutions;
        }
    }
    return All_solutions;
}

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Maybe this isn't the most elegant solution since I'm using "Distinct" to filter out duplicate results but here is one way in C#.

The general idea is that you split the number into an array of 1's then you just combine each node side-by-side together like a tree and select distinct combinations. I pictured it like this:

   [1,1,1]
   /     \
[2,1]   [1,2]
    \   /
     [3]


class Program
{
    static void Main(string[] args)
    {
        Console.Write("Enter an integer value: ");
        int num = int.Parse(Console.ReadLine());

        var y = new int[num];
        for (int x = 0; x < num; x++)
            y[x] = 1;

        var results = Combine(y, num)
            .Distinct(new ArrayComparer())
            .OrderByDescending(r => r.Length)
            .ToArray();

        foreach (var result in results)
        {
            Console.Write('[');
            for (int x = 0; x < result.Length; x++)
            {
                if (x > 0)
                    Console.Write(", ");
                Console.Write(result[x]);
            }
            Console.WriteLine(']');
        }

        Console.ReadKey(true);
    }

    public class ArrayComparer : IEqualityComparer<int[]>
    {
        bool IEqualityComparer<int[]>.Equals(int[] x, int[] y)
        {
            if (x.Length == y.Length)
            {
                for (int z = 0; z < x.Length; z++)
                    if (x[z] != y[z])
                        return false;

                return true;
            }

            return false;
        }

        int IEqualityComparer<int[]>.GetHashCode(int[] obj)
        {
            return 0;
        }
    }

    public static IEnumerable<int[]> Combine(int[] values, int num)
    {
        int val = 0;
        for (int x = 0; x < values.Length; x++)
            val += values[x];

        if (val == num)
        {
            yield return values;

            if (values.Length - 1 > 0)
            {
                for (int x = 0; x < values.Length; x++)
                {
                    int[] combined = new int[values.Length - 1];
                    for (int y = 0; y < x; y++)
                        combined[y] = values[x];

                    if (values.Length > x + 1)
                        combined[x] = values[x] + values[x + 1];

                    for (int y = x + 2; y < values.Length; y++)
                        combined[y - 1] = values[y];

                    foreach (var result in Combine(combined, num))
                        yield return result;
                }
            }
        }
    }
}

Outputs:

Enter an integer value: 4
[1, 1, 1, 1]
[2, 1, 1]
[1, 2, 1]
[1, 1, 2]
[3, 1]
[2, 2]
[1, 3]
[4]