Tail-Recursive Power

Problem Statement

Write a Java program that implements a tail-recursive method to compute (x^n) (x raised to the power n) using an accumulator to track the running product. The program should also include an iterative method for comparison. Test both methods with various inputs, including edge cases like (n = 0) and (n = 1), and compare their performance and readability. You can visualize this as calculating the total growth of an investment ((x)) compounded (n) times, either by iteratively multiplying or by recursively passing the accumulated product to the next step.

Input: A double (x) and a non-negative integer (n) (e.g., (x = 2.0, n = 3)). Output: A double representing (x^n) (e.g., 8.0 for (2.0^3)). Constraints:

• (-100.0 \leq x \leq 100.0).
• (0 \leq n \leq 10^3).
• The result should fit within a double’s precision. Example:
• Input: (x = 2.0, n = 3)
• Output: 8.0
• Explanation: (2.0^3 = 2.0 * 2.0 * 2.0 = 8.0).
• Input: (x = 5.0, n = 0)
• Output: 1.0
• Explanation: Any non-zero number raised to the power 0 is 1.

Pseudocode

FUNCTION tailRecursivePower(x, n, accumulator)
    IF n equals 0 THEN
        RETURN accumulator
    ENDIF
    SET newAccumulator to accumulator * x
    RETURN tailRecursivePower(x, n - 1, newAccumulator)
ENDFUNCTION

FUNCTION iterativePower(x, n)
    IF n equals 0 THEN
        RETURN 1.0
    ENDIF
    SET result to 1.0
    FOR i from 1 to n
        SET result to result * x
    ENDFOR
    RETURN result
ENDFUNCTION

FUNCTION mainPower(x, n)
    CALL tailRecursivePower(x, n, 1.0)
    CALL iterativePower(x, n)
    RETURN results
ENDFUNCTION

Algorithm Steps

1. For the tail-recursive approach: a. Define a tail-recursive helper function that takes (x), (n), and an accumulator as parameters. b. Implement the base case: if (n = 0), return the accumulator. c. For the recursive case, multiply the accumulator by (x) and recursively call the function with (n - 1) and the updated accumulator. d. Call the helper function with the initial accumulator set to 1.0.
2. For the iterative approach: a. Check if (n = 0). If so, return 1.0. b. Initialize a variable result to 1.0. c. Iterate (n) times, multiplying result by (x) in each iteration. d. Return the final result.
3. Test both methods with various inputs (e.g., (n = 0, 1, 5)) and compare performance and readability.

Java Implementation

public class TailRecursivePower {
    // Computes x^n using tail recursion
    public double tailRecursivePower(double x, int n) {
        // Call tail-recursive helper with initial accumulator
        return tailRecursivePowerHelper(x, n, 1.0);
    }

    // Helper function for tail-recursive power
    private double tailRecursivePowerHelper(double x, int n, double accumulator) {
        // Base case: x^0 = accumulator
        if (n == 0) {
            return accumulator;
        }
        // Recursive case: multiply accumulator by x and recurse
        return tailRecursivePowerHelper(x, n - 1, accumulator * x);
    }

    // Computes x^n using iteration
    public double iterativePower(double x, int n) {
        // Base case: x^0 = 1
        if (n == 0) {
            return 1.0;
        }
        // Initialize result
        double result = 1.0;
        // Multiply x by itself n times
        for (int i = 1; i <= n; i++) {
            result *= x;
        }
        // Return final result
        return result;
    }
}

Output

For the input (x = 2.0, n = 3), both methods output:

8.0

Explanation: (2.0^3 = 2.0 * 2.0 * 2.0 = 8.0).

For the input (x = 5.0, n = 0), both methods output:

1.0

Explanation: Any non-zero number raised to the power 0 is 1.

For the input (x = 3.0, n = 2), both methods output:

9.0

Explanation: (3.0^2 = 3.0 * 3.0 = 9.0).

How It Works

• Tail-Recursive Approach:
  • Step 1: The tailRecursivePower method calls the helper with (x = 2.0, n = 3, accumulator = 1.0).
  • Step 2: In tailRecursivePowerHelper:
    • For (n = 3): accumulator = 1.0 * 2.0 = 2.0, recurse with (n = 2, accumulator = 2.0).
    • For (n = 2): accumulator = 2.0 * 2.0 = 4.0, recurse with (n = 1, accumulator = 4.0).
    • For (n = 1): accumulator = 4.0 * 2.0 = 8.0, recurse with (n = 0, accumulator = 8.0).
    • For (n = 0): Base case, return accumulator = 8.0.
  • Example Trace: For (x = 2.0, n = 3), the accumulator builds: (1.0 \rightarrow 2.0 \rightarrow 4.0 \rightarrow 8.0), returning 8.0.
  • Tail Recursion Note: The function is tail-recursive because the recursive call is the last operation, though Java does not optimize tail recursion.
• Iterative Approach:
  • Step 1: The iterativePower method checks if (n = 0). For (x = 2.0, n = 3), it initializes result = 1.0.
  • Step 2: Iterates 3 times:
    • Iteration 1: result = 1.0 * 2.0 = 2.0.
    • Iteration 2: result = 2.0 * 2.0 = 4.0.
    • Iteration 3: result = 4.0 * 2.0 = 8.0.
  • Example Trace: For (x = 2.0, n = 3), multiplies (2.0 * 2.0 * 2.0 = 8.0).
• Comparison:
  • Readability: Tail-recursive is concise and intuitive for recursive thinking but abstract; iterative is straightforward and explicit.
  • Performance: Both are O(n) time, but iterative uses O(1) space, while tail-recursive uses O(n) space due to the call stack in Java.

Complexity Analysis Table

Note:

• n is the exponent.
• Time complexity: O(n) for both, as tail-recursive makes n calls, and iterative performs n multiplications.
• Space complexity: O(n) for tail-recursive due to the call stack (Java does not optimize tail recursion); O(1) for iterative, using only a single variable.
• Best, average, and worst cases are O(n) for both.

✅ Tip: Use tail recursion for elegant recursive solutions, but prefer the iterative approach for large (n) to minimize memory usage. Test with edge cases like (n = 0, 1) and large (n) (e.g., 1000) to verify correctness and compare performance.

⚠ Warning: Java does not optimize tail recursion, so the O(n) space complexity may cause stack overflow for very large (n). Be cautious with floating-point precision for large (x) or (n), as results may exceed double’s capacity.