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

Thanks.Code:`Fr(n):`

if n=1

print n;

else

for i=1 to n-1 do

Fr(i)

Printable View

- 10-23-2004smitskyRecursive 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)

- 10-23-2004louis_mine
if ypou are going to play n-1 as an ending parameter you must decrease n somewhere,

- 10-24-2004smitskySeems to Work
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);

}

}

- 10-24-2004Prelude
>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