How to find a maximal odd decomposition of integer M?

Question

Let M be an integer in range [1; 1,000,000,000].

A decomposition of M is a set of unique integers whose sum is equal to M.

A decomposition is odd

Answer 1

Here is a deterministic solution to the problem. Suppose M = {1, 3, 5, ..., 2*k-3, 2*k-1, r} where r <= 2*k + 1. It is 'obvious' that the maximal decomposition is not going to have more numbers than (k+1).

We have the following cases for k > 3 (the reasoning and handling of earlier cases is presented later):

Case 1. If r is odd and equal to 2*k+1: add r into the list thereby giving a decomposition of (k+1) elements.

Case 2. If r is even: replace {(2*k-1), r} by {2*k-1+r} giving a decomposition of k elements.

Case 3. If r is odd and not equal to 2*k+1: replace the first and the last two elements in the series {1, 2*k-1, r} by {2*k+r} giving a decomposition of (k-1) elements.

Note that the worst case of (k-1) elements will occur when the input is of the form n^2 + (odd number < 2*k+1).

Also note that (Case 3) will break in case the number of elements is less than 3. For example, the decomposition of 5 and 7. We will have to special-case these numbers. Likewise (Case 2) will break for 3 and will have to be special-cased. There is no solution for M=2. Hence the restriction k > 3 above. Everything else should work fine.

This takes O(sqrt(M)) steps.

Some C/C++ code:

#include 

int main(int argc, char *argv[])
{
    printf("Enter M:");
    int m = 0;
    scanf("%d", &m);

    int arr[100] = {0};
    printf("The array is:\n");
    switch(m) {
        case 2:
            printf("No solution\n");
            return 0;
        case 1:
        case 3:
        case 5:
        case 7:
            printf("%d\n", m);
            return 0;
    }

    int sum = 0;
    int count = 0;
    for (int i = 1; (sum + i) < m; i+= 2) {
        arr[count++] = i;
        sum += i;
    }
    int start = 0;
    int r = m - sum;
    if (r % 2 == 0) {
        arr[count - 1] += r;
    } else if (r > arr[count - 1]) {
        arr[count++] = r;
    } else {
        start = 1;
        arr[count - 1] += r + 1;
    }

    for (int i = start; i < count; i++) {
        printf("%d\n", arr[i]);
    }

    return 0;
}

Example:

Enter M:24
The array is:
1
3
5
15

Enter M:23
The array is:
3
5
15