SICP-2.1.2节练习
SICP-2.1.4节练习

SICP-2.1.3节练习

lispor posted @ Feb 25, 2011 03:02:40 PM in Scheme with tags SICP , 1457 阅读

练习 2.4 - 2.6

 
练习 2.4:
Here is an alternative procedural representation of pairs. For this representation, verify that (car
(cons x y)) yields x for any objects x and y.
(define (cons x y)
  (lambda (m) (m x y)))
(define (car z)
  (z (lambda (p q) p)))
What is the corresponding definition of cdr? (Hint: To verify that this works, make use of the
substitution model of section 1-1-5.)
我的解答:
程序:
(define (cons x y)
  (lambda (m)
    (m x y)))

(define (car z)
  (z (lambda (p q) p)))

(define (cdr z)
  (z (lambda (p q) q)))

运行结果:
scheme@(guile-user)> (define a (cons 1 2))
scheme@(guile-user)> (car a)
1
scheme@(guile-user)> (cdr a)
2
 
练习 2.5:
Show that we can represent pairs of nonnegative integers using only numbers and arithmetic
operations if we represent the pair a and b as the integer that is the product 2^a 3^b. Give the
corresponding definitions of the procedures cons, car, and cdr.
我的解答:
程序:
(define (cons a b)
  (* (expt 2 a)
     (expt 3 b)))

(define (car z)
  (define (iter a z)
    (if (not (= 0 (remainder z 2)))
        a
        (iter (+ a 1) (/ z 2))))
  (iter 0 z))

(define (cdr z)
  (define (iter b z)
    (if (not (= 0 (remainder z 3)))
        b
        (iter (+ b 1) (/ z 3))))
  (iter 0 z))

运行结果:
scheme@(guile-user)> (define a (cons 3 5))
scheme@(guile-user)> (car a)
3
scheme@(guile-user)> (cdr a)
5
 
练习 2.6:
In case representing pairs as procedures wasn't mind-boggling enough, consider that, in a language
that can manipulate procedures, we can get by without numbers (at least insofar as nonnegative
integers are concerned) by implementing 0 and the operation of adding 1 as
(define zero (lambda (f) (lambda (x) x)))
(define (add-1 n)
  (lambda (f) (lambda (x) (f ((n f) x)))))
This representation is known as Church numerals, after its inventor, Alonzo Church, the logician who
invented the [lambda] calculus.
Define one and two directly (not in terms of zero and add-1). (Hint: Use substitution to evaluate
(add-1 zero)). Give a direct definition of the addition procedure + (not in terms of repeated
application of add-1).
我的解答:
(define zero
  (lambda (f)
    (lambda (x) x)))

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

(define one
  (lambda (f)
    (lambda (x)
      (f x))))

(define two
  (lambda (f)
    (lambda (x)
      (f (f x)))))

(define (+ a b)
  (lambda (f)
    (lambda (x)
      ((a f) ((b f) x)))))

 

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