Database System Concepts
7th Edition
ISBN: 9780078022159
Author: Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher: McGraw-Hill Education
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Please answer the question in the screenshot. Please give full reasoning for the solution.
![### 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?](https://content.bartleby.com/qna-images/question/1725b614-c61a-470a-b05f-008b5304bcc7/25ad88a4-5ad2-47eb-92da-26ab6c240a29/yuidnl_processed.png)
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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