Thread: Recursion

Hello, my name is Igor and I'm a student. In the near future, we will have our exams, and I'm preparing for them right now. The issue I have is some of the problems that we have to practice are hard to solve, so I'm asking you to help me solve it.



This is the problem that I have to solve.

This is my code:

Code:

double recursion(int n, int k)
{
    if(n == 0)
        return ;
    if(k == 1)
        return sqrt(k/n+recursion(n--, k++));
    else
        return sqrt(k/n+recursion(n--, k--));
}




int main()
{
    int n;


    scanf("%d", &n);


    printf("%f", recursion(n, 1));


    return 0;
}

I think I'm having a problem with the first part of my recursion function. I think I'm very close to solving it. Thanks for reading this and please help if you can!

Originally Posted by Salem

> return sqrt(k/n+recursion(n--, k++));
Yeah, write it as
return sqrt(k/n+recursion(n-1, k+1));

Relying on the post-increment to actually pass a changed value to the recursive function will lead to weirdness.

Nope, still not working.

Code:

double rekurzija(int n, int k){
    if(n == 0)
        return 0;
    if(k == 1)
        return sqrt(k/n+rekurzija(n-1, k+1));
    else
        return sqrt(k/n+rekurzija(n-1, k-1));
}

for n = 6, i get 1.010889, instead of 1.212149

@Igor, I know that is not the right solution, and it is iterativ, but maybe a way there . . .
It is an interesting task. - At until the solution: Take it easy!

Code:

#include <stdio.h>
#include <stdlib.h>
#include <math.h>


int main(void)
{
    double zaehler = 1.0, nenner = 6.0;
    double ergebnis, endsum = 0.0;
    
    for (int i = 1; i <= 6; i++)
    {
        ergebnis = sqrt(zaehler++ / nenner--);
        printf("\nErgebnis = %2.6lf\n", ergebnis);
        if (i % 2 == 0)
        {
            zaehler = 1.0;
        }    
        endsum += ergebnis;        
    }    
    printf("\nErgebnis = %2.6lf\n", endsum);
    
    return(0);
}

YouTube

Originally Posted by laserlight

What do you mean?

See my correction - he have to start from the end.

Look that:



He have to start from the end.
That's a first try:

Code:

#include <stdio.h>
#include <stdlib.h>
#include <math.h>


int main(void)
{
    double zaehler = 2.0, nenner = 1.0;
    double ergebnis, endsum = 0.0;
    double zwischen_sum;
    
    for (int i = 1; i <= 6; i++)
    {
        ergebnis = sqrt(zaehler-- / nenner++);
        printf("\nErgebnis = %2.6lf\n", ergebnis);
        if (i % 2 == 0)
        {
            zaehler = 1.0;
        }    
        zwischen_sum += ergebnis;
        //printf("\n%2d -- %2.6lf\n", i, ergebnis);
    }    
    printf("\nErgebnis = %2.6lf\n", endsum);
    
    return(0);
}

You understand?

I don't know whether the recursion ist part of the exerice but whether the solution is recursiv or iterativ the solution have to start from the last calculation.



I dont no whether this problem is resolvable iterativ. With "printf...":

Code:

#include <stdio.h>
#include <stdlib.h>
#include <math.h>


int main(void)
{
    double zaehler = 2.0, nenner = 1.0;
    double tmp, tmp2, tmp3, zsum;
    double endsum = 0.0;
    
    for (int i = 6; i >= 1; i--)
    {
        zsum = (zaehler-- / nenner++);
        
        if (i % 2 == 0)
        { zaehler = 1.0; }
        
        printf("\nzsum %2d -- %2.6lf", i, zsum);
        printf("\ntmp %2d -- %2.6lf", i, tmp);
        
        tmp = (zsum + tmp3);
        printf("\n%2c -- %2.6lf", 64 + i, tmp);
        
        tmp2 = sqrt(tmp);                
        printf("\ntmp2 %2d -- %2.6lf", i, tmp2);
        
        tmp3 = tmp2;                
        printf("\ntmp3 %2d -- %2.6lf\n", i, tmp3);
        zsum = 0;        
    }    
    printf("\nEndsumme = %2.6lf\n", endsum);
    
    return(0);
}

Without "printf()":

Code:

#include <stdio.h>
#include <stdlib.h>
#include <math.h>


int main(void)
{
    double zaehler = 2.0, nenner = 1.0;
    double tmp, tmp2, tmp3, zsum;
    double endsum = 0.0;
    
    for (int i = 6; i >= 1; i--)
    {
        zsum = (zaehler-- / nenner++);
        
        if (i % 2 == 0)
        { zaehler = 1.0; }
        
        tmp = (zsum + tmp3);        
        
        tmp2 = sqrt(tmp);                        
        
        tmp3 = tmp2;                        
        zsum = 0;        
    }    
    printf("\nEndsumme = %2.6lf\n", endsum);
    
    return(0);
}

Last output:

@laserlight - elegant solution! Clear and straight. Unfortunately not from me.
The search for an iterative solution was nonsense.

Thank you!