Partition of an Integer + Number of partitions

Question

A partition of an integer n is a way of writing n as a sum of positive integers. For

example, for n=7, a partition is 1+1+5. I need a program that finds all the

partitions of an integer 'n' using 'r' integers. For example, all the partitions of n=7

using r=3 integers are 1+1+5, 1+2+4, 1+3+3, 2+2+3.

This is what I have so far:

#include <iostream>
#include <vector>

using namespace std;

void print (vector<int>& v, int level){
    for(int i=0;i<=level;i++)
        cout << v[i] << " ";
    cout << endl;
}

void part(int n, vector<int>& v, int level){
    int first; /* first is before last */

    if(n<1) return ;
    v[level]=n;
    print(v, level);

    first=(level==0) ? 1 : v[level-1];

    for(int i=first;i<=n/2;i++){
        v[level]=i; /* replace last */
        part(n-i, v, level+1);
    }
}

int main(){
    int num;
    cout << "Enter a number:";
    cin >> num;

    vector<int> v(num);

    part(num, v, 0);
}

The output of this program is:

Enter a number:5
5
1 4
1 1 3
1 1 1 2
1 1 1 1 1
1 2 2
2 3

Process returned 0 (0x0)   execution time : 1.837 s
Press any key to continue.

How can I change my code so I can have that 'r' variable?

EDIT:

In case it was not clear, the 'r' value represents the number of integers per partition. So in the case above, if r=2, then the partitions can only have two integers in them. The partitions would be 4+1, and 3+2. The 'r' value should be entered by the user.

Answer 1

How about this? Have an additional argument passed as reference for r, and increment r each time within the recursion block?

#include <iostream>
#include <vector>

using namespace std;

void print (vector<int>& v, int level){
    for(int i=0;i<=level;i++)
        cout << v[i] << " ";
    cout << endl;
}

void part(int n, vector<int>& v, int level, int &r){
    int first; /* first is before last */

    if(n<1) return ;
    v[level]=n;
    print(v, level);

    first=(level==0) ? 1 : v[level-1];

    for(int i=first;i<=n/2;i++){
        v[level]=i; /* replace last */
        r++;
        part(n-i, v, level+1, r);
    }
}

int main(){
    int num;
    cout << "Enter a number:";
    cin >> num;

    int r = 0;
    vector<int> v(num);

    part(num, v, 0, r);
    cout << "r = " << r << endl;
}

Output comes as:

Enter a number:5 
1 4 
1 1 3 
1 1 1 2 
1 1 1 1 1 
1 2 2 
2 3 
r = 6

Is this what you are looking for?

Answer 2

A sort of "hack" would be to make r an argument of part, pass it along recursively an just print the output if level equals r.

Answer 3

Essentially what Codor said, plus you don't need to recurse further into part() once you found a partition of the target length since they would be longer:

#include <iostream>
#include <vector>

using namespace std;

void print (vector<int>& v, int level){
    for(int i=0;i<=level;i++)
        cout << v[i] << " ";
    cout << endl;
}

void part(int n, vector<int>& v, int level, int r){
    int first; /* first is before last */

    if(n<1) return ;
    v[level]=n;
    if( level+1 == r ) {
        print(v, level);
        return;
    }

    first=(level==0) ? 1 : v[level-1];

    for(int i=first;i<=n/2;i++){
        v[level]=i; /* replace last */
        part(n-i, v, level+1, r);
    }
}

int main(){
    int num,r;
    cout << "Enter a number:";
    cin >> num;
    cout << "Enter size (r):";
    cin >> r;

    vector<int> v(num);

    part(num, v, 0, r);
}

Output:

Enter a number:5
Enter size (r):2
1 4
2 3