Question

Consider the following function: int func (int x)\\{\\[\text { if } \quad(x=0)\\]return 2 else if \((x=1)\) return 3 else return \((\text { func }(x-1)+\text { func }(x-2))\)} What is the output of the following statements? a. \(\quad\) cout \(\langle\langle\text { func }(0)<<\text { end } 1\) b. \(\quad\) cout \(\langle\langle\text { func }(1)<<\text { end } 1\) c. cout \(<<\) func \((2)<<\) endl d. \(\quad\) cout \(<\langle\text { func }(5)<<\text { end } 1\)

Short answer

a. 2, b. 3, c. 5, d. 21

Step-by-step solution

Understand the Function

The function `func` is defined recursively with the following rules:1. If \(x = 0\), return 2.2. If \(x = 1\), return 3.3. Otherwise, return the sum of `func(x-1)` and `func(x-2)`.

Calculate func(0)

By applying the rule for \(x = 0\) directly from the function:\[\text{func(0)} = 2\]

Evaluate func(1)

Apply the function's rule for \(x = 1\):\[\text{func(1)} = 3\]

Solve for func(2)

Using the recursive rule for \(x > 1\):\[\text{func(2)} = ext{func}(1) + ext{func}(0) = 3 + 2 = 5\]

Compute func(5)

We need `func` values for 4, 3 before calculating:- \( ext{func}(3) = ext{func}(2) + ext{func}(1) = 5 + 3 = 8\)- \( ext{func}(4) = ext{func}(3) + ext{func}(2) = 8 + 5 = 13\)- \( ext{func}(5) = ext{func}(4) + ext{func}(3) = 13 + 8 = 21\)

Key concepts

Recursion

Recursion is a method where a function calls itself in order to solve a problem. In the given exercise, the function `func` uses recursion to calculate its result for any integer input. When `func` is called with a number greater than 1, it returns the sum of `func(x-1)` and `func(x-2)`. This means that the function will keep calling itself with decreasing arguments until it reaches the base cases for which the results are already known, specifically when the argument is either 0 or 1.
This approach breaks down complex problems into simpler sub-problems, making it a powerful concept in programming. However, it's important to understand the base case, which prevents the function from calling itself indefinitely. For `func`, the base cases are defined at `func(0)=2` and `func(1)=3`.

Function Evaluation

Function evaluation involves determining the result of a function given an input. For the `func` function, evaluating it means applying the specific rules defined by its recursive relations.
With `func(0)`, evaluation is straightforward because it matches the base case directly and results in 2. The same goes for `func(1)`, yielding 3. For `func(2)`, the function evaluates to \ \[\text{func}(2) = \text{func}(1) + \text{func}(0) = 3 + 2 = 5\]
Each evaluation requires substitution of the previously evaluated values unless you are working with the base cases, which simplify the process.

Programming Logic

Programming logic involves the reasoning and thinking that structures the code. For recursive functions like `func`, understanding the flow and how each recursive call is made is crucial.
In the `func` exercise, the logic is premised on two key points: base cases and the recursive step. The base cases provide a stopping point for recursion, ensuring that the function does not end in an infinite loop. The recursive step provides a way to break down the problem into smaller pieces by expressing the function in terms of itself, but with smaller indices. This logic is essentially the pattern or algorithm that the code follows to arrive at a solution.

• Base Cases: Direct evaluations return a constant (2 for `func(0)`, 3 for `func(1)`).
• Recursive Cases: Involve breaking the problem down further (`func(x)` uses `func(x-1)` and `func(x-2)`).

Step-by-Step Solution

A step-by-step solution is a methodic approach to solving problems, breaking them into their constituent parts for enhanced clarity. In solving the `func` problem, the exercise is addressed by evaluating the function at specific points using recursion.
The steps involve:

• Recognizing the base conditions, which provide direct outcomes for `func(0)` and `func(1)`.
• Using these outcomes to compute higher values recursively, like `func(2)`=5, `func(3)`=8, `func(4)`=13, and `func(5)`=21.

Each step builds on the last, illustrating how recursion unfolds and leads to the final results by systematically substituting and calculating.