SICP 习题参考答案 2.3

win_hate

此帖为 SICP 章节 2.3 的参考答案。为保持可读性，请勿直接在本帖讨论。讨论请另开新帖。

• 本节电子版：[Symbolic Data]
  
• 答案目录：[点这里]
  
  

win_hate

Exercise 2.53.  What would the interpreter print in response to evaluating each of the following expressions?

(list 'a 'b 'c)

(list (list 'george))
(cdr '((x1 x2) (y1 y2)))

(cadr '((x1 x2) (y1 y2)))
(pair? (car '(a short list)))
(memq 'red '((red shoes) (blue socks)))

(memq 'red '(red shoes blue socks))

win_hate

1. 
   
2. > (list 'a 'b 'c)
   
3. (a b c)
   
4. 
   
5. > (list (list 'george))
   
6. ((george))
   
7. 
   
8. > (cdr '((x1 x2) (y1 y2)))
   
9. ((y1 y2))
   
10. 
    
11. > (cadr '((x1 x2) (y1 y2)))
    
12. (y1 y2)
    
13. 
    
14. > (pair? (car '(a short list)))
    
15. #f
    
16. 
    
17. > (memq 'red '((red shoes) (blue socks)))
    
18. #f
    
19. 
    
20. > (memq 'red '(red shoes blue socks))
    
21. (red shoes blue socks)
    

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win_hate

Exercise 2.54.  Two lists are said to be equal? if they contain equal elements arranged in the same order. For example,

(equal? '(this is a list) '(this is a list))

is true, but

(equal? '(this is a list) '(this (is a) list))

is false. To be more precise, we can define equal? recursively in terms of the basic eq? equality of symbols by saying that a and b are equal? if they are both symbols and the symbols are eq?, or if they are both lists such that (car a) is equal? to (car b) and (cdr a) is equal? to (cdr b). Using this idea, implement equal? as a procedure.

win_hate

1. 
   
2. (define (Equal? a b)
   
3.   (if (and (pair? a) (pair? b))
   
4.       (and (Equal? (car a) (car b))
   
5.            (Equal? (cdr a) (cdr b)))
   
6.       (eq? a b)))
   

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[ 本帖最后由 win_hate 于 2008-10-5 11:03 编辑 ]

win_hate

Exercise 2.55.  Eva Lu Ator types to the interpreter the expression

(car ''abracadabra)

To her surprise, the interpreter prints back quote. Explain.

[ 本帖最后由 win_hate 于 2008-10-5 11:08 编辑 ]

win_hate

这是因为 'x 是个缩写，等价于 (quote x)。所以 ''abracadabra 是 (quote (quote abracadabra)).

win_hate

Exercise 2.56.  Show how to extend the basic differentiator to handle more kinds of expressions. For instance, implement the differentiation rule



by adding a new clause to the deriv program and defining appropriate procedures exponentiation?, base, exponent, and make-exponentiation. (You may use the symbol ** to denote exponentiation.) Build in the rules that anything raised to the power 0 is 1 and anything raised to the power 1 is the thing itself.

win_hate

1. 
   
2. (define (exponentiation? exp)
   
3.   (and (pair? exp) (eq? (car exp) '**)))
   
4. 
   
5. (define base cadr)
   
6. (define exponent caddr)
   
7. 
   
8. (define (make-exponentiation b e)
   
9.     (cond ((=number? e 0) 1)
   
10.         ((=number? b 1) 1)
    
11.         ((and (number? b) (number? e)) (expt b e))
    
12.         (else (list '** b e))))
    

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1. 
   
2. (define (deriv exp var)
   
3.   (cond ((number? exp) 0)
   
4.         ((variable? exp)
   
5.          (if (same-variable? exp var) 1 0))
   
6.         ((sum? exp)
   
7.          (make-sum (deriv (addend exp) var)
   
8.                    (deriv (augend exp) var)))
   
9.         ((product? exp)
   
10.          (make-sum
    
11.           (make-product (multiplier exp)
    
12.                         (deriv (multiplicand exp) var))
    
13.           (make-product (deriv (multiplier exp) var)
    
14.                         (multiplicand exp))))
    
15.         ((exponentiation? exp)
    
16.          (make-product
    
17.           (make-product
    
18.            (exponent exp)
    
19.            (make-exponentiation (base exp)
    
20.                                 (- (exponent exp) 1)))
    
21.           (deriv (base exp) var)))
    
22.          (else
    
23.           (error "unknown expression type -- DERIV" exp))))
    

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1. 
   
2. > (deriv '(** x 1) 'x)
   
3. 1
   
4. > (deriv '(** x 10) 'x)
   
5. (* 10 (** x 9))
   
6. > (deriv (make-exponentiation 'x 0) 'x)
   
7. 0
   

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win_hate

Exercise 2.57.  Extend the differentiation program to handle sums and products of arbitrary numbers of (two or more) terms. Then the last example above could be expressed as

(deriv '(* x y (+ x 3)) 'x)

Try to do this by changing only the representation for sums and products, without changing the deriv procedure at all. For example, the addend of a sum would be the first term, and the augend would be the sum of the rest of the terms.