如何从 SICP 调用方案编号函数

发布于 2024-07-18 11:03:25 字数 602 浏览 5 评论 0原文

在 SICP（例如 2.6）中，以下功能被描述为“无需数字即可度过难关”的方式。我正在努力理解这一点。首先，如何调用这些函数？我实际上可以以某种方式应用它们，输出将为 1 吗？（或任何其他数字？）

(define zero (lambda (f) (lambda (x) x)))

(define (add-1 n)
  (lambda (f) (lambda (x) (f ((n f) x)))))

我最初的尝试没有成功：

Welcome to DrScheme, version 4.1.5 [3m].
Language: Simply Scheme; memory limit: 128 megabytes.
> (add-1 (zero))
. . procedure zero: expects 1 argument, given 0
> (add-1 zero)
#<procedure>
> (add-1 1)
#<procedure>
> ((add-1 1))
. . #<procedure>: expects 1 argument, given 0
>

原文

In SICP, (ex 2.6) the following functions are described as ways of 'getting by without numbers'. I'm scratching trying to understand this. As a starting point, how do these functions get invoked? Can I actually apply them in some way where the output will be 1? (Or any other number?)

(define zero (lambda (f) (lambda (x) x)))

(define (add-1 n)
  (lambda (f) (lambda (x) (f ((n f) x)))))

My initial attempts haven't been successful:

Welcome to DrScheme, version 4.1.5 [3m].
Language: Simply Scheme; memory limit: 128 megabytes.
> (add-1 (zero))
. . procedure zero: expects 1 argument, given 0
> (add-1 zero)
#<procedure>
> (add-1 1)
#<procedure>
> ((add-1 1))
. . #<procedure>: expects 1 argument, given 0
>

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眼中杀气 2024-07-25 11:03:25

这些表示数字的函数称为“Church 数字”（如 SICP 规定）。它们的存在意味着您可以定义一个计算系统（例如 lambda 演算），而无需将数字作为第一类对象 - 您可以使用函数作为原始对象。这一事实主要具有理论意义；对于实际计算来说，教堂数字并不是一个好的选择。

通过将丘奇数字与其他对象作为参数应用，您可以看到丘奇数字定义的正确性。当您将代表 n 的丘奇数字应用于函数 f 时，您会得到另一个将 f 应用于其参数 n 次的函数，例如，对于 n=3，为 f(f(f(x)))。

> (define (double x) (* 2 x))
> (zero double)
#<procedure>
> ((zero double) 1)
1
> ((zero double) 100)
100
> (define one (add-1 zero))
> ((one double) 1)
2
> ((one double) 100)
200
> (define (cons-a x) (cons 'a x))
> ((zero cons-a) '())
()
> (((add-1 one) cons-a) '(1 2 3))
(a a 1 2 3)

These functions that represent numbers are called Church numerals (as SICP states). Their existence means that you can define a system of computation (such as the lambda calculus) without having numbers as first-class objects--you can use functions as the primitive objects instead. This fact is mainly of theoretical interest; Church numerals are not a good choice for practical computation.

You can see the correctness of your definitions of Church numerals by applying them with other objects as arguments. When you apply a Church numeral representing n to a function f, you get another function that applies f to its argument n times, e.g., f(f(f(x))) for n=3.

> (define (double x) (* 2 x))
> (zero double)
#<procedure>
> ((zero double) 1)
1
> ((zero double) 100)
100
> (define one (add-1 zero))
> ((one double) 1)
2
> ((one double) 100)
200
> (define (cons-a x) (cons 'a x))
> ((zero cons-a) '())
()
> (((add-1 one) cons-a) '(1 2 3))
(a a 1 2 3)

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八巷 2024-07-25 11:03:25

这是原始的 lambda 演算，它不产生数字，它完全用函数替换数字类型。

所以，你有一个“零”函数，如果你对它调用 add-1 ，你不会得到 1，你会得到另一个代表 1 的函数。关键是生成的函数符合与基本算术公理，所以它们等价于自然数

回复收藏 0 原文

旧时光的容颜 2024-07-25 11:03:25

(define (as-primitive-num church-num)
    (define (inc a) (+ a 1))
    ((church-num inc) 0))

;testing:
(define one (add-1 zero))
(define two (add-1 one))

(display (as-primitive-num one)) (newline)
(display (as-primitive-num two)) (newline)

和输出：

    1
    2

(define (as-primitive-num church-num)
    (define (inc a) (+ a 1))
    ((church-num inc) 0))

;testing:
(define one (add-1 zero))
(define two (add-1 one))

(display (as-primitive-num one)) (newline)
(display (as-primitive-num two)) (newline)

and the output: