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Problem: Computing the nth Fibonacci number Input: n >= 1; o Output: the nth number in the Fibonacci sequence.

Problem: Computing the nth Fibonacci number Input: n >= 1; o Output: the nth number in the Fibonacci sequence.

Database System Concepts
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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### Problem: Computing the nth Fibonacci number **Input:** - \( n \geq 1 \) **Output:** - The nth number in the Fibonacci sequence. --- #### Recursive Approach **Algorithm:** `REC-FIBONACCI(n)` ```text if n = 1 or n = 2, return (1); else T1 = REC-FIBONACCI(n - 1); T2 = REC-FIBONACCI(n - 2); return (T1 + T2); ``` Explanation: - If \( n \) is 1 or 2, the function returns 1. - Otherwise, it recursively calculates the Fibonacci numbers for \( n - 1 \) and \( n - 2 \) and returns their sum. --- #### Iterative Approach **Algorithm:** `ITERATIVE-FIBONACCI(n)` ```text if n = 1 or n = 2 return (1); else M[1] = 1, M[2] = 1 for i = 3 to n do M[i] = M[i - 1] + M[i - 2] return (M[n]) ``` Explanation: - If \( n \) is 1 or 2, the function returns 1. - Otherwise, it initializes the first two Fibonacci numbers \( M[1] \) and \( M[2] \) to 1. - It then iteratively fills the array from \( M[3] \) to \( M[n] \) by summing the previous two values. - Finally, it returns \( M[n] \). --- ### Time Complexity Analysis - The **recursive approach** has exponential time complexity due to the repeated calculations, which results in \( O(2^n) \). - The **iterative approach** has linear time complexity \( O(n) \) as it calculates each Fibonacci number exactly once. **Discussion Point:** What are the upper bounds for the time complexity of the two solutions to the Fibonacci number?
Transcribed Image Text:### Problem: Computing the nth Fibonacci number **Input:** - \( n \geq 1 \) **Output:** - The nth number in the Fibonacci sequence. --- #### Recursive Approach **Algorithm:** `REC-FIBONACCI(n)` ```text if n = 1 or n = 2, return (1); else T1 = REC-FIBONACCI(n - 1); T2 = REC-FIBONACCI(n - 2); return (T1 + T2); ``` Explanation: - If \( n \) is 1 or 2, the function returns 1. - Otherwise, it recursively calculates the Fibonacci numbers for \( n - 1 \) and \( n - 2 \) and returns their sum. --- #### Iterative Approach **Algorithm:** `ITERATIVE-FIBONACCI(n)` ```text if n = 1 or n = 2 return (1); else M[1] = 1, M[2] = 1 for i = 3 to n do M[i] = M[i - 1] + M[i - 2] return (M[n]) ``` Explanation: - If \( n \) is 1 or 2, the function returns 1. - Otherwise, it initializes the first two Fibonacci numbers \( M[1] \) and \( M[2] \) to 1. - It then iteratively fills the array from \( M[3] \) to \( M[n] \) by summing the previous two values. - Finally, it returns \( M[n] \). --- ### Time Complexity Analysis - The **recursive approach** has exponential time complexity due to the repeated calculations, which results in \( O(2^n) \). - The **iterative approach** has linear time complexity \( O(n) \) as it calculates each Fibonacci number exactly once. **Discussion Point:** What are the upper bounds for the time complexity of the two solutions to the Fibonacci number?
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