# Practice Questions for Recursion | Set 6

Question 1
Consider the following recursive C function. Let len be the length of the string s and num be the number of characters printed on the screen, give the relation between num and len where len is always greater than 0.

 `void` `abc(``char` `*s) ` `{ ` `    ``if``(s == ``'\0'``) ` `        ``return``; ` ` `  `    ``abc(s + 1); ` `    ``abc(s + 1); ` `    ``printf``(``"%c"``, s);     ` `} `

Following is the relation between num and len.

` num = 2^len-1 `
```s is 1 time printed
s is 2 times printed
s is 4 times printed
s[i] is printed 2^i times
s[strlen(s)-1] is printed 2^(strlen(s)-1) times
total = 1+2+....+2^(strlen(s)-1)
= (2^strlen(s)) - 1
```

For example, the following program prints 7 characters.

 `#include ` ` `  `void` `abc(``char` `*s) ` `{ ` `    ``if``(s == ``'\0'``) ` `        ``return``; ` ` `  `    ``abc(s + 1); ` `    ``abc(s + 1); ` `    ``printf``(``"%c"``, s); ` `} ` ` `  `int` `main() ` `{ ` `    ``abc(``"xyz"``); ` `    ``return` `0; ` `} `

Thanks to bharat nag for suggesting this solution.

Question 2

 `#include ` `int` `fun(``int` `count) ` `{ ` `    ``printf``(``"%d\n"``, count); ` `    ``if``(count < 3) ` `    ``{ ` `      ``fun(fun(fun(++count))); ` `    ``} ` `    ``return` `count; ` `} ` ` `  `int` `main() ` `{ ` `    ``fun(1); ` `    ``return` `0; ` `} `

Output:

``` 1
2
3
3
3
3
3
```

The main() function calls fun(1). fun(1) prints “1” and calls fun(fun(fun(2))). fun(2) prints “2” and calls fun(fun(fun(3))). So the function call sequence becomes fun(fun(fun(fun(fun(3))))). fun(3) prints “3” and returns 3 (note that count is not incremented and no more functions are called as the if condition is not true for count 3). So the function call sequence reduces to fun(fun(fun(fun(3)))). fun(3) again prints “3” and returns 3. So the function call again reduces to fun(fun(fun(3))) which again prints “3” and reduces to fun(fun(3)). This continues and we get “3” printed 5 times on the screen.