plz help me....I can't understand the answer to the following program

[AndrewCool]

#include<stdio.h>

void crazy(int a,int b, int m, int n)
{
if(n==0)
return;
crazy(a+m,b,m+b,n-1);
printf("%d %d %d %d ",a,b,m,n);
printf("\n");
crazy(a,b+m,m+a,n-1);
}

int main()
{
crazy(2,5,1,3);
return (0);
}


what is the output?? i am confused about this

[weaknessforcats]

What don't you understand?

This is an example of recursion.

[donbock]

What output did you get?

[jaseel97]

recursion 1:
a=2 b=5 m=1 n=3

recursion1.1:
a=2+1 b=5 m=1+5 n=2

recursion 1.1.1:
a=3+6 b=5 m=6+5 n=1

recursion 1.1.1.1:
a=9+11 b=5 m=11+5 n=0

return
print-> 20 5 16 0

The second recursion call makes it very complex. you have to handle two recursions that are kinda nested.
Working out the second part....I guess ill leave it to you. :p

[weaknessforcats]

I am concerned that you have two calls to crazy() inside
crazy().

Usually there is only one recursive call:

void crazy(int a,int b, int m, int n)
{
   if(n==0)
     return;
   printf("%d %d %d %d ",a,b,m,n);
   printf("\n"); 
   crazy(a,b+m,m+a,n-1);
}
 

In this example you print if n is not zero then call crazy() with new argument values. Those value get printed in the next call, which changes the values and calls crazy()again.

Finally, n is zero and that remote crazy() returns causing the next remote to return, until the first call to crazy() returns.

The number of lines you print would be the number of calls in the recursion.

[donbock]

@jaseel97 I agree except for the last line. I think it will be
print-> 9 5 11 1
because crazy(a,b,m,0) doesn't print anything.

My pencil-and-paper analysis suggests a pattern -- for crazy(a,b,m,n), the number of lines printed will be 2^(0) + 2^(1) + ... + 2^(n-1) = 2^(n) - 1. The pattern looks like a family tree.

@AndrewCool Have you run this program yet? If so, what was the output?