I need some help to transform an iterative algorithm into a recursive one.

I need to generate all the 4 numbers that sum up to 11, where the first number is always 1 and the other 3 numbers at least 2.

  for(int i = 1 ; i < 7; i    ){
        for(int j = 2; j < 7; j    ){
            for(int k = 2; k < 7; k    ){
                for(int p = 2; p < 7; p    ){
                    if(i   j   k   p == 11 &&  i == 1){
                        printf(" %d %d %d %d \n", i, k, j, p);
                    }
                }
            }
        }
    }

Some output examples :

 1 2 2 6
 1 3 2 5
 1 4 2 4
 1 5 2 3

This would work but its designed horribly and I need to do it with recursion.

CodePudding user response：

The first number is always 1, you don't need a loop for that. For the rest you can create a function that gives all the possibilities summing to a specific number. Here is a program in Java syntax, but not using classes, so easily translatable to C:

public void printPossibilities(int rest, int[] buffer, int len, int total) {
    if (len   1 == total) {
        buffer[len] = rest;
        System.out.println(Arrays.toString(buffer));
    } else {
        // check if it is possible that the rest of the numbers are >= 2
        for (int i = 2; rest >= i   (total - len - 1) * 2; i  ) {
            buffer[len] = i;
            printPossibilities(rest - i, buffer, len   1, total);
        }
    }
}

// call with
int[] buffer = new int[4];
buffer[0] = 1;
printPossibilities(10, buffer, 1, 4);

CodePudding user response：

Firstly, a bit of math here: Say there are 4 numbers, num1, num2, num3 and num4, which need to sum up to 11.So,

num1 num2 num3 num4=11
num1=1 and num2>=2, num3>=2, num4>=2
num2 num3 num4=11-num1=10
(num2-2) (num3-2) (num4-2)=10-(2 2 2)=4
var2 var3 var4=4, where var2, var3 and var4>=0

Which boils down to finding 3 non negative numbers such that they sum to 4, in a recursive manner. Here's a general solution which can then be modified accordingly: Here's a code which can do that, in C :

#include <iostream>
#include <vector>

void get_target_sum(std::vector<std::vector<int>>& total_ans, std::vector<int>& curr_ans, int num_remaining, int targetnum){
// total_ans denotes all the tuples, curr_ans denotes the current 
// computation, num_remaining denotes the numbers remaining which 
// should sum up to targetnum.
// each time we find a valid number which can be a part of our answer, we 
// add it to curr_ans, decrement num_remaining since the numbers remaining 
// has decreased by 1, and decrease the target number to be found by the 
// value of the number
    if(num_remaining<1){return;}
    if(num_remaining==1){
        curr_ans.push_back(targetnum);
        total_ans.push_back(curr_ans);
        curr_ans.pop_back();
    }
    for(int curr_num=0;curr_num<targetnum;curr_num  ){
        curr_ans.push_back(curr_num);
        get_target_sum(total_ans,curr_ans,num_remaining-1,targetnum-curr_num);
        curr_ans.pop_back();
    }

}
int main()
{
    std::vector<std::vector<int>>final_ans;
    std::vector<int> curr_ans;
    get_target_sum(final_ans,curr_ans,3,4);
    for(int i=0;i<final_ans.size();i  ){
        std::cout<<1<<" ";
        // The first variable
        for(auto num:final_ans[i]){
            std::cout<<num 2<<" ";
            // the other 3 variables, incremented by 2.
        }
        std::cout<<"\n";
    }
    return 0;
}

The Output: https://onlinegdb.com/gdQdLgni8