Recursive Function in Pseudo Code

Hi. Can someone help me analyze this code. I think it runs into an infinite loop at n=3.

Code:

---

Fr(n):
 if n=1
  print n;
 else
  for i=1 to n-1 do
    Fr(i)

---

Thanks.

if ypou are going to play n-1 as an ending parameter you must decrease n somewhere,

Look at this. It seems to run as I was told it should. For example when n=6 the function outputs "1" 16 times. Why?!

Code:

---

#include<iostream>
#include<cstdlib>
using namespace std;

void Fr(int);

int main()
{
  int n;
  n=0;
  cout<<"Please enter a value for n: "<<endl;
  cin>>n;
  Fr(n);
  system("pause");
  return 0;
}

void Fr(int m)
{
  int i;
  if(m==1)
  {
    cout<<m<<endl;
  }
  else
  {
    for(i=1; i<=m-1; i++)
    Fr(i);
  }
}

---

I can't understand it.

>I can't understand it.
That's no surprise. You're mixing recursion with iteration and loop indices with the original argument that was not a loop index to begin with. :rolleyes:

Start at the top with, say, m = 3. Going from the outside in, we realize that the outer most call will loop twice because i starts at 1 and goes to m - 1. Then we break away from m and call Fr with the loop index 1 that starts from 1 every time.

The first iteration of the loop is 1, so m is printed in the first recursive call. In the second recursive call, m is 2 so m is not printed and the first recursive loop begins. Because m is 2, the loop only executes once with the argument of 1, thus printing the value of m.

After this all recursive paths return and the output is:

Code:

---

1
1

---

Use the same logical progression for larger values of m and you'll have a good understanding of how this function works.