<习题>第一章 构造过程抽象

Ch1.1 程序设计的基本元素

习题：

1.1

(+ 5 3 4）	12
(- 9 1)	8
(/ 6 2)	3
(+ (*2 4) (- 4 6 ))	-16
(define a 3)	a=3
(define b (+ a 1 ))	b=4
(if (and (>b a) (< b ( * a b )))      b      a)	4
(cond ((= a 4) 6)          ((= b 4) ( + 6 7 a))          (else 25))	16
(+ 2 ( if ( > b a ) b a ))	6
(* (cond (( > a b ) a)              (( < a b ) b)              (else -1))      (+ a 1))	16

 

1.2

(/ (+ 5 (+ 4 (- 2 (- 3 (+ 6 ( / 4 5)))))) (* 3 (- 6 2) (- 2 7))

 

1.3

(define (min x y)

           (if ( < x y ) x y ))

(define (larger-sum a b c)

           (- (+ a b c) (min (min a b) c ))

 

1.4

对a与b的绝对值求和，即a+|b|

 

1.5

采用应用序，会先对运算对象进行求值，在(test 0 (p))中，即先会对（p)进行求值，然而（define (p) (p))，所以对（p)进行求值会陷入死循环；而采用正则序，知道实际需要运算对象的值的时候才会进行求值，在题目的运算中，先进行if判断后，直接返回0，避免了对（p)进行求值。

 

1.6

使用自定义的new-if进行求平方根运算时，按照应用序会先对运算对象进行求值，在对第三个参数的求值过程中会出现对其本身的递归调用，同1.5相似，也会陷入死循环。

 

1.7

按照比值进行判断，如果两者之积为0，则无法进行判断

(define (good-enough? guess x)
(if (= 0 ( * guess x ))
    false
    (< (abs (- ( abs (/ guess x )) 1)) 0.001)))

 

1.8

(define (cubt-iter guess x)
  (define (cube guess) ( * guess guess guess))
  (define (good-enough? guess x)
    (if (= 0 ( * guess x ))
        false
        (< (abs (- ( abs (/ (cube guess) x )) 1)) 0.001)))
  (define (improve guess x)
    (/ (+ (/ x (* guess guess)) (* 2 guess)) 3))
  (if (good-enough? guess x)
      guess
      (cubt-iter (improve guess x) x)))

 

 

Ch1.2 过程与它们所产生的计算

习题：

1.9

(+ 4 5) (inc ( + 3 5 )) (inc ( inc ( + 2 5 ))) (inc ( inc ( inc ( + 1 5 )))) (inc ( inc ( inc ( inc ( + 0 5 ))))) (inc ( inc ( inc ( inc 5 )))) (inc ( inc ( inc 6 ))) (inc ( inc 7 )) (inc 8 ) 9	( + 4 5 ) ( + 3 6 ) ( + 2 7 ) ( + 1 8 ) ( + 0 9 ) 9
递归计算过程	迭代计算过程

 

1.10

( A 1 10 )

( A 0 ( A 1 9 ))

( A 0 ( A 0 ( A 1 8 )))

...

( A 0 ( A 0 ( A 0 ( A 0 ( A 0 ( A 0 ( A 0 ( A 0 ( A 0 ( A 1 1)))))))))))
( A 0 ( A 0 ( A 0 ( A 0 ( A 0 ( A 0 ( A 0 ( A 0 ( A 0 2))))))))))
( A 0 ( A 0 ( A 0 ( A 0 ( A 0 ( A 0 ( A 0 ( A 0 2^2)))))))))
( A 0 ( A 0 ( A 0 ( A 0 ( A 0 ( A 0 ( A 0 2^3))))))))

...

2^10

1024

 

( A 2 4 )

( A 1 ( A 2 3 ))

( A 1 ( A 1 ( A 2 2 )))

( A 1 ( A 1 ( A 1 ( A 2 1 ))))

( A 1 ( A 1 ( A 1 2 )))

( A 1 ( A 1 ( A 0 ( A 1 1))))

( A 1 ( A 1 ( A 0 2 )))

( A 1 ( A 1 4 ))

( A 1 2^4)

2^(2^4)

65536

 

( A 3 3 )

( A 2 ( A 3 2 ))

( A 2 ( A 2 ( A 3 1 )))

( A 2 ( A 2 2 ))

( A 2 ( A 1 ( A 2 1 )))

( A 2 ( A 1 2 ))

( A 2 ( A 0 ( A 1 1 )))

( A 2 ( A 0 2 )))

( A 2 4 )

65535

 

(define ( f n ) ( A 0 n ))

=>2*n

(define ( g n ) ( A 1 n ))

=>2^n

(define ( h n ) ( A 2 n ))

=>2^(2^(2^...(2^2)))  一共n个2