Question

Monitor the stack operations in the tree-recursive Fibonacci computation: (define (fib \(\mathrm{n}\) ) \((\) if \((<\mathrm{n} 2)\) \(\operatorname{n}(+(\mathrm{fib}(-\mathrm{n} 1))(\mathrm{fib}(-\mathrm{n} 2))))\) a. Give a formula in terms of \(n\) for the maximum depth of the stack required to compute \(\operatorname{Fib}(n)\) for \(n \geq 2\). Hint: In section \(1.2 .2\) we argued that the space used by this process grows linearly with \(n\). b. Give a formula for the total number of pushes used to compute \(\operatorname{Fib}(n)\) for \(n \geq 2\). You should find that the number of pushes (which correlates well with the time used) grows exponentially with \(n\). Hint: Let \(S(n)\) be the number of pushes used in computing \(\operatorname{Fib}(n)\). You should be able to argue that there is a formula that expresses \(S(n)\) in terms of \(S(n-1), S(n-2)\), and some fixed "overhead" constant \(k\) that is independent of \(n\). Give the formula, and say what \(k\) is. Then show that \(S(n)\) can be expressed as \(a \mathrm{Fib}(n+1)+b\) and give the values of \(a\) and \(b\).

Short answer

The maximum stack depth is \( n \), and the total number of pushes is approximately \( 2 \cdot \text{Fib}(n+1) - 1 \).

Step-by-step solution

Understanding the Problem

We need to understand two things about the Fibonacci function: the stack depth and the number of operations (or pushes) required. The Fibonacci function is tree-recursive, meaning that it involves multiple simultaneous recursive calls.

Identifying Stack Operations

When calling the Fibonacci function, each call requires a stack frame. The depth of the stack is the greatest number of concurrent function calls. For Fibonacci, this depth correlates with the recursive process.

Maximum Stack Depth Formula

For computing \( \text{Fib}(n) \), the maximum depth of the stack is essentially the height of the recursion tree. This depth grows linearly with \( n \) because the deepest path involves decrementing \( n \) by one each step until it reaches 2 or 1. Therefore, the maximum stack depth formula is \( n \).

Formula for Total Number of Pushes

To find the total number of pushes, consider the recursive nature: each Fibonacci computation involves solving two smaller Fibonacci problems recursively. Let \( S(n) \) be the number of pushes for \( \text{Fib}(n) \). We have \( S(n) = S(n-1) + S(n-2) + k \). This captures the two recursive calls and an overhead constant \( k \). Since Fibonacci grows exponentially, so does \( S(n) \).

Deriving S(n) Formula

Because \( S(n) \) relates to Fibonacci, it's known that it has a connection to \( a \cdot \text{Fib}(n+1) + b \). Recognizing that overlapping subproblems are solved twice, it's derived as \( S(n) \approx 2 \cdot \text{Fib}(n+1) - 1 \). Here, \( a = 2 \) and \( b = -1 \). This reflects the repetition in recursive strategy which leads to exponential growth.

Key concepts

Fibonacci sequence

The Fibonacci sequence is a series of numbers where each number is the sum of the two preceding ones, starting from 0 and 1. Typically expressed as:

• \( ext{Fib}(0) = 0 \)
• \( ext{Fib}(1) = 1 \)
• \( ext{Fib}(n) = ext{Fib}(n-1) + ext{Fib}(n-2) \) for \( n \geq 2 \)

This recursive definition allows us to generate the sequence, but can be computationally intensive due to repeated calculations.
In a recursive implementation, \( ext{Fib}(n-1) \) and \( ext{Fib}(n-2) \) are computed as separate branches of the recursive call tree, which can lead to multiple identical calculations.
For example, to calculate \( ext{Fib}(5) \), the recursive approach computes \( ext{Fib}(4) \) and \( ext{Fib}(3) \), but \( ext{Fib}(4) \) recalculates \( ext{Fib}(3) \) and so on.
This inherent overlapping of subproblems is a key inefficiency in naive mathematical or recursive definitions of the sequence.

stack operations

Stack operations are crucial in understanding recursive algorithms like the Fibonacci sequence. When a function is called, a stack frame is created and placed on top of the stack, containing information about that function call.
For recursive functions, every time a new level of recursion occurs, a new stack frame is created.
In the Fibonacci recursive computation, the depth of the recursion determines the maximum stack depth.

• The "push" operation adds a stack frame to the stack, representing function calls.
• Recursive calls push multiple stacks, since each \( ext{Fib}(n) \) call leads to \( ext{Fib}(n-1) \) and \( ext{Fib}(n-2) \).
• "Pop" operations remove the stack frames as function calls complete and return to the previous call.

With recursive calls, the stack can grow quickly. The maximum depth of this stack is equivalent to the height of the recursion tree.
For the Fibonacci series, this maximum depth is proportional to the value of \( n \), as it essentially counts the number of simultaneous recursive calls before reaching base cases.

computational complexity

Computational complexity measures how the resources required by an algorithm grow with input size. With the recursive Fibonacci function, both time and space complexity are considerations due to the exponential call tree.

• The time complexity for the recursive Fibonacci function is \( O(2^n) \), meaning it grows exponentially. This results from each function call producing two further recursive calls.
• The space complexity, reflecting the maximum stack depth used, is \( O(n) \), as the longest branch in the call tree dictates the number of simultaneous function calls.
• Each recursive call incurs additional overhead for managing stack operations, making recursion less efficient compared to iterative approaches.

Given this function's nature, it is often more practical to use alternative approaches, such as dynamic programming or iterative processes, to compute Fibonacci numbers more efficiently.
These methods can drastically reduce both time and space complexities, avoiding unnecessary repeated calculations inherent in simple recursion.