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A function \(f\) is defined by the rule that \(f(n)=n\) if \(n<3\) and \(f(n)=\) \(f(n-1)+2 f(n-2)+3 f(n-3)\) if \(n \geq 3\). Write a procedure that computes \(f\) by means of a recursive process. Write a procedure that computes \(f\) by means of an iterative process.

Short Answer

Expert verified
Two procedures to compute \( f(n) \): 1. Recursive `f_recursive(n)`: Directly implements the given recursive relation. 2. Iterative `f_iterative(n)`: Uses a loop to efficiently compute \( f(n) \).

Step by step solution

01

Understand the Recursive Definition

The function \( f(n) \) is defined in two parts. For \( n < 3 \), \( f(n) = n \). For \( n \geq 3 \), it is defined as \( f(n) = f(n-1) + 2f(n-2) + 3f(n-3) \). Our task is to implement a recursive process to compute this function.
02

Implement Recursive Procedure

To create the recursive procedure named `f_recursive(n)`, check if \( n < 3 \) first (base case). Return \( n \) if true. Otherwise, use the recursive rule: \( f(n-1) + 2f(n-2) + 3f(n-3) \).
03

Recursive Procedure Code

```python def f_recursive(n): if n < 3: return n else: return f_recursive(n - 1) + 2 * f_recursive(n - 2) + 3 * f_recursive(n - 3) ```
04

Understand the Iterative Process

The goal is to use an iterative approach to compute \( f(n) \). We need to keep track of the previous three computed values (let's call them \( a \), \( b \), and \( c \)) because the formula depends on them. Begin with \( f(0) = 0 \), \( f(1) = 1 \), and \( f(2) = 2 \). Then, iterate up to \( n \) updating these values at each step.
05

Implement Iterative Procedure

Create an iterative procedure named `f_iterative(n)`. Use a loop to calculate values iteratively. Initialize \( a = 0 \), \( b = 1 \), and \( c = 2 \). For each \( i \geq 3 \) up to \( n \), calculate \( f(i) = c + 2b + 3a \), then update \( a \), \( b \), and \( c \) to progress through the recursion iteratively.
06

Iterative Procedure Code

```python def f_iterative(n): if n < 3: return n a, b, c = 0, 1, 2 for i in range(3, n+1): result = c + 2*b + 3*a a, b, c = b, c, result return c ```

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Iterative Process
An Iterative Process in programming is a method of solving problems where a sequence of steps is repeated in a loop to achieve a result. In this context, instead of calling a function repeatedly (as in recursion), the process simply updates variables to compute values. This can be more efficient in terms of memory usage because it avoids the overhead of repeated function calls.
  • Starts with initial values: In our example, the initial values are set as \(f(0) = 0\), \(f(1) = 1\), and \(f(2) = 2\).
  • Updates variables: Instead of solving new instances of the problem with a function call, the process simply updates the values of \(a, b,\) and \(c\) in each iteration.
  • Less Memory Usage: Iterative processes typically use fewer resources since they maintain state using variables rather than creating new function calls.
Understanding and leveraging iterative processes can greatly enhance how efficiently an algorithm runs, especially for large values of \(n\).
Recursive Function
A Recursive Function is defined as a function that calls itself in order to solve a problem. It breaks down a problem into smaller, more manageable chunks, often leading to elegant and easy-to-read code. However, recursive functions can be resource-intensive compared to iterative solutions, particularly when calculating deep recursive calls.
  • Base Case: This is the condition that ends the recursion. For the function \(f(n)\), if \(n < 3\), then \(f(n) = n\) is the base case, which stops further recursive calls.
  • Recursive Step: This is where the function calls itself with new parameters, aiming to reach the base case. Here, it is \(f(n) = f(n-1) + 2f(n-2) + 3f(n-3)\).
  • Stack Space: Recursive functions use the call stack to keep track of function calls, which might lead to stack overflow for large inputs.
Recursive functions are powerful tools in algorithm design, often providing clear, simple solutions to complex problems.
Algorithm Design
Algorithm Design is the process of creating a step-by-step solution to solve a problem efficiently. Good design principles balance between correctness and optimization of resources like time and space. Understanding the nature of your problem is crucial in deciding whether to use recursion, iteration, or another approach.
  • Define the Problem: Clearly understand what the function \(f(n)\) is intended to accomplish, such as how it differs in definition for \(n < 3\) and \(n \geq 3\).
  • Choose the Right Approach: Determine whether an iterative or recursive solution is more optimal based on constraints on space and time. In our case, choosing between \(f\_iterative(n)\) and \(f\_recursive(n)\).
  • Efficiency and Optimization: Consider trade-offs between different design choices like ease of understanding versus running efficiency, particularly for large \(n\).
Effective algorithm design requires a deep understanding of your problem domain, the environment in which the solution operates, and potential constraints like memory and execution time.

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Most popular questions from this chapter

Let \(f\) and \(g\) be two one-argument functions. The composition \(f\) after \(g\) is defined to be the function \(x \mapsto f(g(x))\). Define a procedure compose that implements composition. For example, if inc is a procedure that adds 1 to its argument, ((compose square inc) 6) 49

Design a procedure that evolves an iterative exponentiation process that uses successive squaring and uses a logarithmic number of steps, as does \(f\) ast- expt. (Hint: Using the observation that \(\left(b^{n / 2}\right)^{2}=\left(b^{2}\right)^{n / 2}\), keep, along with the exponent \(n\) and the base \(b\), an additional state variable \(a\), and define the state transformation in such a way that the product \(a b^{n}\) is unchanged from state to state. At the beginning of the process \(a\) is taken to be 1 , and the answer is given by the value of \(a\) at the end of the process. In general, the technique of defining an invariant quantity that remains unchanged from state to state is a powerful way to think about the design of iterative algorithms.)

a. An infinite continued fraction is an expression of the form $$ f=\frac{N_{1}}{D_{1}+\frac{N_{2}}{D_{2}+\frac{N_{3}}{D_{3}+\cdots}}} $$ As an example, one can show that the infinite continued fraction expansion with the \(N_{i}\) and the \(D_{i}\) all equal to 1 produces \(1 / \phi\), where \(\phi\) is the golden ratio (described in section 1.2.2). One way to approximate an infinite continued fraction is to truncate the expansion after a given number of terms. Such a truncation-a so-called \(k\)-term finite continued fraction - has the form $$ \frac{N_{1}}{D_{1}+\frac{N_{2}}{\ddots+\frac{N_{K}}{D_{K}}}} $$ Suppose that \(\mathrm{n}\) and \(\mathrm{d}\) are procedures of one argument (the term index \(i\) ) that return the \(N_{i}\) and \(D_{i}\) of the terms of the continued fraction. Define a procedure cont-frac such that evaluating (cont- frac \(\mathrm{n} \mathrm{d} \mathrm{k}\) ) computes the value of the \(k\)-term finite continued fraction. Check your procedure by approximating \(1 / \phi\) using \(\begin{aligned} \text { (cont-frac (lambda (i) } & 1.0 \text { ) } \\ &\text { (lambda (i) } 1.0) \\ & \text { k) } \end{aligned}\) for successive values of \(\mathrm{k}\). How large must you make \(\mathrm{k}\) in order to get an approximation that is accurate to 4 decimal places? b. If your cont-frac procedure generates a recursive process, write one that generates an iterative process. If it generates an iterative process, write one that generates a recursive process.

Louis Reasoner is having great difficulty doing exercise \(1.24\). His fast- prime? test seems to run more slowly than his prime? test. Louis calls his friend Eva Lu Ator over to help. When they examine Louis's code, they find that he has rewritten the expmod procedure to use an explicit multiplication, rather than calling square: (define (expmod base exp m) (cond \(((=\exp 0) 1)\) \(((\) even? exp) \((\) remainder \((*(\) expmod base \((/ \exp 2) \mathrm{m})\) \((\) expmod base \((/ \exp 2) \mathrm{m}))\) \((\) else \(\quad\) m)) \((\) remainder \((*\) base \((\) expmod base \((-\exp 1) \mathrm{m}))\) \(\quad\) m) \())\) "I don't see what difference that could make," says Louis. "I do." says Eva. "By writing the procedure like that, you have transformed the \(\Theta(\log n)\) process into a \(\Theta(n)\) process." Explain.

Translate the following expression into prefix form $$ \frac{5+4+\left(2-\left(3-\left(6+\frac{4}{5}\right)\right)\right)}{3(6-2)(2-7)} $$

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