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Suppose we define the procedure (define (f g) (g 2)) Then we have (f square) (f (lambda (z) (* z (+ z 1)))) 6 What happens if we (perversely) ask the interpreter to evaluate the combination (f \(f\) )? Explain.

Short Answer

Expert verified
Calling (f f) results in an error because f expects a function, but 2 is not callable.

Step by step solution

01

Understand the Function 'f'

The function \( f \) takes another function \( g \) as an argument and applies \( g \) to the number 2. So when we call \( f \) with any function \( g \), it produces the result of \( g(2) \).
02

Evaluate f(square)

The square function is defined as \( \text{square}(x) = x \times x \). Therefore, \( f(\text{square}) \) evaluates to \( \text{square}(2) = 2 \times 2 = 4 \).
03

Evaluate f on a Lambda Function

When we call \( f \) with the lambda function \( \lambda (z) \rightarrow (z \times (z + 1)) \), it evaluates \( \lambda(2) \). This results in \( 2 \times (2 + 1) = 2 \times 3 = 6 \).
04

Understand the Problematic Call f(f)

Now let's consider what happens when we call \( f(f) \). In this case, \( f \) is called with another function \( f \), so it attempts to apply \( f \) to the number 2—i.e., it tries to compute \( f(2) \).
05

Analyze the Call f(2)

The problem is that \( f \) expects its argument \( g \) to be a function that takes one argument (in this case, 2). However, 2 is a number, not a function, which leads to an error when \( f \) tries to call 2 as if it were a function.

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Key Concepts

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

Scheme Programming
Scheme is a minimalist dialect of Lisp, focusing on simplicity and flexibility. It's known for its elegant syntax and powerful features that are excellent for demonstrating and exploring concepts in functional programming.

One of the key aspects of Scheme is that it treats both code and data as lists. This makes it very versatile and allows procedures like the one in our example to be defined easily. In the procedure \( (\text{define} (f\ g) (g\ 2)) \), you can see the function "f" is defined to take another function "g" as an argument and apply it to the number 2.

Scheme's minimalist nature is beneficial for learning the core concepts of programming without the distractions of complex syntax. Its functionality bubbles up in higher-order functions, allowing exploration of different ways to manipulate data and procedures, making it an excellent introduction to languages that incorporate functional programming principles.
Higher-Order Functions
Higher-order functions are a crucial concept in functional programming. They are functions that can take other functions as arguments or return functions as results.

In our exercise, \( f \) is a higher-order function because it receives another function \( g \) and applies it. When you call \( f(\text{square}) \), \( \text{square} \) is applied to the number 2, producing 4.

An essential property of higher-order functions is their ability to make code more concise and expressive. By using functions as first-class citizens—able to pass functions as parameters and return them from other functions—you can develop more abstract and reusable code.
  • Example: Functions like map and filter are classical examples of higher-order functions.
  • Benefits: They allow for function composition and can lead to more elegant code.
Understanding these functions enhances your ability to craft solutions in functional languages, and it’s especially vital in languages like Scheme.
Lambda Calculus
Lambda calculus is a formal system for expressing computation based on function abstraction and application. It's the theoretical foundation of functional programming, offering a way to represent functions using simple expressions—lambdas.

In our example, we use the lambda expression \( \lambda (z) ightarrow (z \times (z + 1)) \). This defines an anonymous function that, when given a number "z", calculates \( z \times (z + 1) \).

Lambda calculus is not about numbers; it's about functions and the rules for manipulating them. It helps you think about computations abstractly, allowing the use of functional programming concepts effectively.

Using lambda expressions enables creating quick, unnamed functions, which can be passed around just like variables. For example, using \( f \) with this lambda function results in calculating \( 6 \). This aligns with Scheme's ability to handle functions as first-class citizens and makes it versatile for various computational problems.

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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

Modify fixed-point so that it prints the sequence of approximations it generates, using the newline and display primitives shown in exercise \(1.22\). Then find a solution to \(x^{x}=1000\) by finding a fixed point of \(x \mapsto \log (1000) / \log (x)\). (Use Scheme's primitive \(\log\) procedure, which computes natural logarithms.) Compare the number of steps this takes with and without average damping. (Note that you cannot start \(\mathrm{fixed}\)-point with a guess of 1 , as this would cause division by \(\log (1)=0\).)

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 $$ \frac{5+4+\left(2-\left(3-\left(6+\frac{4}{5}\right)\right)\right)}{3(6-2)(2-7)} $$

Define a procedure double that takes a procedure of one argument as argument and returns a procedure that applies the original procedure twice. For example, if inc is a procedure that adds 1 to its argument, then (double inc) should be a procedure that adds 2 . What value is returned by (((double (double double)) inc) 5)

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