Consider the following recursive function F() in Java that takes an integer value and returns a string value :
public static String F(int N) {
if ( N <= 0) return "-";
return F(N - 3) + N + F(N - 2) + N;
}
The value of F(5) is :

Question

Consider the following recursive function F() in Java that takes an integer value and returns a string value :

public static String F(int N) {

if ( N <= 0) return "-";

return F(N - 3) + N + F(N - 2) + N;

}

The value of F(5) is :

Prepp Team · Accepted Answer

Understanding the Recursive Function F()

The problem asks us to find the output of a recursive Java function F(N) for a given integer input N=5. Let's first understand how the function works.

The function F(int N) is defined as follows:

public static String F(int N) {
    if ( N <= 0) return "-";
    return F(N - 3) + N + F(N - 2) + N;
}

This function has two main parts:

Base Case: If the input value N is less than or equal to 0 (N <= 0), the recursion stops for this branch, and the function simply returns the string "-".
Recursive Step: If N is greater than 0, the function calls itself twice: once with N - 3 and once with N - 2. It then concatenates the result of F(N - 3), the current value of N (converted to a string), the result of F(N - 2), and the current value of N (converted to a string) in that specific order.

Tracing the Execution of F(5)

We need to trace the calls starting from F(5) and see how the string is built through concatenation as the recursive calls return their values.

Step-by-Step Calculation of F(5)

Let's trace the execution:

1. F(5):

$N = 5$. Since $5 > 0$, we use the recursive step.
F(5) calls F(5 - 3) and F(5 - 2).
F(5) returns the result of F(2) + "5" + F(3) + "5".
We need to evaluate F(2) and F(3) first.

2. Evaluating F(2) (called by F(5)):

$N = 2$. Since $2 > 0$, we use the recursive step.
F(2) calls F(2 - 3) and F(2 - 2).
F(2) returns the result of F(-1) + "2" + F(0) + "2".
We need to evaluate F(-1) and F(0).

3. Evaluating F(-1) (called by F(2)):

$N = -1$. Since $-1 \le 0$, we hit the base case.
F(-1) returns the string "-".

4. Evaluating F(0) (called by F(2)):

$N = 0$. Since $0 \le 0$, we hit the base case.
F(0) returns the string "-".

5. Returning from F(2):

Now we substitute the results of F(-1) and F(0) back into the expression for F(2).
F(2) returns "-" + "2" + "-" + "2", which concatenates to the string "-2-2".

6. Evaluating F(3) (called by F(5)):

$N = 3$. Since $3 > 0$, we use the recursive step.
F(3) calls F(3 - 3) and F(3 - 2).
F(3) returns the result of F(0) + "3" + F(1) + "3".
We need to evaluate F(0) and F(1).

7. Evaluating F(0) (called by F(3)):

$N = 0$. Since $0 \le 0$, we hit the base case.
F(0) returns the string "-". (Note: This is a separate call from the previous F(0), but returns the same value).

8. Evaluating F(1) (called by F(3)):

$N = 1$. Since $1 > 0$, we use the recursive step.
F(1) calls F(1 - 3) and F(1 - 2).
F(1) returns the result of F(-2) + "1" + F(-1) + "1".
We need to evaluate F(-2) and F(-1).

9. Evaluating F(-2) (called by F(1)):

$N = -2$. Since $-2 \le 0$, we hit the base case.
F(-2) returns the string "-".

10. Evaluating F(-1) (called by F(1)):

$N = -1$. Since $-1 \le 0$, we hit the base case.
F(-1) returns the string "-". (Note: This is a separate call from previous F(-1) calls).

11. Returning from F(1):

Now we substitute the results of F(-2) and F(-1) back into the expression for F(1).
F(1) returns "-" + "1" + "-" + "1", which concatenates to the string "-1-1".

12. Returning from F(3):

Now we substitute the results of F(0) and F(1) back into the expression for F(3).
F(3) returns "-" + "3" + "-1-1" + "3", which concatenates to the string "-3-1-13".

13. Returning from F(5):

Finally, we substitute the results of F(2) and F(3) back into the expression for F(5).
F(5) returns F(2) + "5" + F(3) + "5".
Substitute the results: "-2-2" + "5" + "-3-1-13" + "5".
Concatenating these parts step-by-step:
"-2-2" + "5" results in "-2-25".
Then, "-2-25" + "-3-1-13" results in "-2-25-3-1-13".
Finally, "-2-25-3-1-13" + "5" results in "-2-25-3-1-135".

So, the value of F(5) is the string "-2-25-3-1-135".

Call	$N$ Value	Returns
F(-2)	-2	"-"
F(-1)	-1	"-"
F(0)	0	"-"
F(1)	1	F(-2) + "1" + F(-1) + "1" = "-" + "1" + "-" + "1" = "-1-1"
F(2)	2	F(-1) + "2" + F(0) + "2" = "-" + "2" + "-" + "2" = "-2-2"
F(3)	3	F(0) + "3" + F(1) + "3" = "-" + "3" + "-1-1" + "3" = "-3-1-13"
F(5)	5	F(2) + "5" + F(3) + "5" = "-2-2" + "5" + "-3-1-13" + "5" = "-2-25-3-1-135"

Conclusion

By tracing the recursive calls and concatenating the results according to the function definition, we find that the value of F(5) is the string "-2-25-3-1-135".

Java Recursive Function Revision Table

Concept	Description
Recursion	A function calling itself during its execution.
Base Case	The condition that stops the recursion, preventing infinite calls. Essential for correct recursive function design.
Recursive Step	The part of the function that makes one or more calls to itself, usually with smaller or simpler inputs.
Stack Overflow	An error that occurs when recursion goes too deep without reaching a base case, filling up the call stack memory.
String Concatenation	Joining two or more strings end-to-end to form a new string. In Java, the '+' operator can be used for this, including concatenating numbers (which are converted to strings).

Additional Information on Recursive Functions

Recursive functions are powerful tools in programming, often used to solve problems that can be broken down into smaller, self-similar subproblems. Here are some key points:

Problem Types: Recursion is well-suited for problems involving trees, graphs, fractals, and problems that can be defined by a recurrence relation (like the Fibonacci sequence).
Implementation: Every recursive function must have a base case to terminate the recursion. Without a base case, the function would call itself indefinitely. The recursive step must make progress towards the base case.
Recursion vs. Iteration: Many problems can be solved with either recursion or iteration (using loops). Recursive solutions can sometimes be more elegant and easier to understand for certain problems, while iterative solutions might be more efficient in terms of space and time complexity (due to function call overhead).
Tail Recursion: A special form of recursion where the recursive call is the last operation performed by the function. Some compilers can optimize tail recursion into iterative code, reducing stack usage. The function F(N) in this problem is not tail-recursive because there are operations (concatenations and the second recursive call) performed after the first recursive call returns.