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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)

Short Answer

Expert verified
The final value is 13.

Step by step solution

01

Understand the Requirement

We need to define a procedure `double` that takes another procedure as input and returns a new procedure. This new procedure should apply the original one twice to any argument.
02

Define the `double` Procedure

The `double` procedure takes a function `f` as an argument and returns a new function that takes an argument `x` and applies `f` to `x` twice. In more formal writing: ```python def double(f): return lambda x: f(f(x)) ``` This means `double` will take a procedure `f`, and return a procedure that applies `f` two times to any argument.
03

Apply `double` to the `double` Function

Now we will apply `double` to itself twice as in \((double (double double))\). This results in a procedure that applies `double` four times to any function passed.
04

Apply the Resulting Function to `inc` and Evaluate

Now apply the resulting function to `inc`, a procedure that adds 1 to its argument, and finally apply it to 5. Each `double` doubles the application, so:\1st double: \(\text{inc}\) applied twice (result: \(5+2 = 7\))2nd double: double the number of applications to 4 times (result: \(5+4 = 9\))3rd double: double the number of applications to 8 times (result: \(5+8 = 13\)).
05

Evaluate the Final Expression

We have found that after applying all operations, the resulting evaluation is 13 when starting from 5 and applying `(double (double double))` to inc.

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Key Concepts

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

Procedures
Procedures in programming are essentially instructions that are executed in a sequence to perform a specific task. They are foundational elements in both functional and imperative programming paradigms. In functional programming, a procedure is often referred to as a function, and the focus is on applying these functions to transform data.

A procedure can take one or more inputs, which are called arguments, and produce an output. In the context of the `double` exercise, the procedure `double` is designed to accept another procedure as an argument and return a new procedure. This showcases how flexible and powerful procedures can be. They not only perform operations on data but also manipulate other procedures, allowing developers to create highly modular and reusable code components.

When defining procedures, it's crucial to clearly specify what inputs they accept, what they return, and any side-effects they might produce. Functional programming often emphasizes pure functions, which have no side effects and return the same output given the same input. This property is valuable for reasoning about code behavior and facilitates easier debugging and testing.
Higher-order Functions
Higher-order functions are a central concept in functional programming. A higher-order function is any function that can take one or more functions as arguments or return a function as its result. This paradigm allows for a great deal of flexibility in constructing programs and is key to writing concise and expressive code.

In the `double` exercise, `double` itself is a higher-order function. It takes another function as its argument and returns a new function. This capability allows developers to create functions that can manipulate other functions, thus creating more abstract and powerful operations that can be reused in different contexts.

Using higher-order functions simplifies patterns that would otherwise require more complex implementation. For example, operations like map, filter, and reduce, commonly used to process lists or collections, can be implemented effectively as higher-order functions, providing a streamlined approach to data transformation.
  • They promote function reuse and composition.
  • Enable more abstract operations.
  • Integral for defining operations like `map`, `filter`, and `reduce`.
Embracing higher-order functions leads to code that is not just more concise, but also more challenging to understand and modify, emphasizing the importance of clear documentation and naming conventions.
Lambda Calculus
Lambda calculus forms the theoretical foundation of functional programming. It was formulated by Alonzo Church in the 1930s as a formal system for expressing computation based on function abstraction and application.

In lambda calculus, functions are anonymous, and expressions are based on the idea of applying functions to arguments. A lambda expression in Python, for instance, provides a way to create small anonymous functions. The syntax `lambda x: f(f(x))` used in the `double` exercise is an example of how lambda calculus influences modern programming languages.

Lambda calculus introduces several essential concepts:
  • Function abstraction: The process of defining a function without giving it a name.
  • Application: Applying a function to its arguments.
  • Variables: Used as placeholders for actual data values in functions.
The simplicity of lambda calculus, with just a few elements at its core, enables complex computations through function composition and application. Understanding lambda calculus can illuminate many aspects of functional programming, including how recursion and iteration can be expressed solely through function calls and applications. This underlies many advanced concepts in modern programming languages, enhancing both theoretical and practical programming skills.

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Most popular questions from this chapter

Alyssa P. Hacker doesn't see why if needs to be provided as a special form. "Why can't I just define it as an ordinary procedure in terms of cond?" she asks. Alyssa's friend Eva Lu Ator claims this can indeed be done, and she defines a new version of if: (define (new-if predicate then-clause else-clause) (cond (predicate then-clause) (else else-clause))) Eva demonstrates the program for Alyssa: (new-if (=2 3) 0 5) 5 (new-if (= 1 1) 0 5) 0 Delighted, Alyssa uses new-if to rewrite the s (define (sqrt-iter guess \(x\) ) (new-if (good-enough? guess \(x\) ) $$ \begin{array}{l}\text { guess } \\ \text { (sqrt-iter (improve guess } x \text { ) } \\ x) \text { ) }\end{array} $$ What happens when Alyssa attempts to use this to compute square roots? Explain.

The sine of an angle (specified in radians) can be computed by making use of the approximation \(\sin x \approx x\) if \(x\) is sufficiently small, and the trigonometric identity \(\sin x=3 \sin \frac{x}{3}-4 \sin ^{3} \frac{x}{3}\) to reduce the size of the argument of \(\sin\). (For purposes of this exercise an angle is considered "sufficiently small" if its magnitude is not greater than \(0.1\) radians.) These ideas are incorporated in the following procedures: (define (cube \(x\) ) \((* x x x)\) ) (define (p \(x\) ) \((-(* 3 x)(* 4(\) cube \(x)))\) ) (define (sine angle) (if (not (> (abs angle) 0.1)) angle (p (sine (/ angle 3.0))))) a. How many times is the procedure \(\mathrm{p}\) applied when (sine 12.15) is evaluated? b. What is the order of growth in space and number of steps (as a function of \(a\) ) used by the process generated by the sine procedure when (sine a) is evaluated?

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.

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 \(\log (1)=0\).)

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.

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