By Valleylord

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SICP真是一本神书，看完第1章完全没有感觉到反感，虽然有难度，但是实在是很吸引人，马上开始第2章的征程。

2.1 代码如下

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(define (gcd a b)
  (if (= b 0)
      a
      (gcd b (remainder a b)) ))
(define (make-rat n d)
  (let ((g (gcd n d)))
    (cons (/ n g) (/ d g))))
(define (abs x)
  (if (< x 0) (- 0 x)
      x))
(define (make-rat-new n d) 
  (let ((g (gcd n d)))
    (cond ((= n 0)       (cons 0 1))
          ((= d 0)       (cons 1 0))
          ((< (* n d) 0) (cons (- 0 (abs (/ n g))) 
                               (abs (/ d g))))
          (else          (cons (abs (/ n g)) 
                               (abs(/ d g))))) ))
(define (numer x) (car x))
(define (denom x) (cdr x))

除去题目的要求，我还定义了0和无穷大2个值，可惜没有定义负无穷，偷懒了

2.2 这题并不是很麻烦

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(define (print-point p)
  (newline)
  (display "(")
  (display (x-point p))
  (display ",")
  (display (y-point p))
  (display ")"))
(define (make-point a b)
  (cons a b))
(define (x-point x) (car x))
(define (y-point x) (cdr x))
(define (make-segment a b)
  (cons a b))
(define (start-segment l) (car l))
(define (end-segment l) (cdr l))
(define (average x y)
  (/ (+ x y) 2))
(define (mid-segment l)
  (make-point (average (x-point (start-segment l)) 
                       (x-point (end-segment l)))
              (average (y-point (start-segment l)) 
                       (y-point (end-segment l))) ))

2.3 矩形的表示形式1，用左下角和右上角的点来表示矩形

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(define (make-point a b)
  (cons a b))
(define (x-point x) (car x))
(define (y-point x) (cdr x))
(define (make-square a b)
  (cons a b))
(define (leftdown-square s) (car s))
(define (rightup-square s) (cdr s))
(define (leftup-square s) (make-point (x-point (leftdown-square s))
                                      (y-point (rightup-square s))) )
(define (rightdown-square s) (make-point (x-point (rightup-square-square s))
                                      (y-point (leftdown-square-square s))) )
(define (length-square s)
  (- (x-point (rightup-square s)) (x-point (leftdown-square s))) )
(define (width-square s)
  (- (y-point (rightup-square s)) (y-point (leftdown-square s))) )
(define (area-square s)
  (* (length-square s) (width-square s)) )
(define (perimeter-square s)
  (* 2 (+ (length-square s) (width-square s))) )

如果要有另一种表示的话，也可以用左上角和右下角来表示，重新定义上面的函数就好了，主要是length-square和width-square。但是这个问题，似乎另一种定义更好，就是定义左下角点和两个边的边长。

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(define (make-square x a b)
  (cons x (cons a b)))
(define (leftdown-square s) (car s))
(define (rightup-square s) 
  (let ((x-leftdown (x-point (car s)))
        (y-leftdown (y-point (car s))))
    (make-point (+ x-leftdown (length-square s))
                (+ y-leftdown (width-square s))) ))
(define (leftup-square s) 
  (let ((x-leftdown (x-point (car s)))
        (y-leftdown (y-point (car s))))
    (make-point x-leftdown
                (+ y-leftdown (width-square s))) ))
(define (rightdown-square s) 
  (let ((x-leftdown (x-point (car s)))
        (y-leftdown (y-point (car s))))
    (make-point (+ x-leftdown (length-square s))
                y-leftdown) ))
(define (length-square s)
  (car (cdr s)) )
(define (width-square s)
  (cdr (cdr s)) )
(define (area-square s)
  (* (length-square s) (width-square s)) )
(define (perimeter-square s)
  (* 2 (+ (length-square s) (width-square s))) )

测试代码不贴了

2.4 代码如下，并不麻烦，关键是理解lambda

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(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)))

2.5 代码如下，借用1.16计算幂的函数

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(define (cons x y)
  (lambda (m) (m x y)))
(define (car z)
  (z (lambda (p q) (fast-expt1 2 p))))
(define (cdr z)
  (z (lambda (p q) (fast-expt1 3 q))))
; 1.16
(define (fast-expt1 x n)
  (fast-expt1-iter 1 x n))
(define (fast-expt1-iter a x n)
  (cond ((= n 0)   a)
        ((even? n) (fast-expt1-iter a (* x x) (/ n 2)))
        (else      (fast-expt1-iter (* a x) (* x x) (/ (- n 1) 2)))))

2.6 定义one和two，把zero带入到add-1中就好了

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; one = (add-1 zero)
(define one
  (lambda (f) (lambda (x) (f x))) )
; two = (add-1 one)
(define two
  (lambda (f) (lambda (x) (f (f x)))) )

可以猜测n=(lambda (f) (lambda (x) (f......<n个f> (f x))))，那么加法就可以定义成

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(define (add m n)
  (lambda (f) (lambda (x) (<m个f> (<n个f> x)))) )

更加好一点的表述是

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(define (add m n)
  (lambda (f) (lambda (x) (m ((n f) x)))))

这题主要思路还是lambda的展开

2.7 感觉跟前面的定义线段的那题很相似

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(define (make-interval a b) (cons a b))
(define (lower-bound x) (car x))
(define (upper-bound x) (cdr x))

2.8 区间相减的代码如下

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(define (sub-interval x y)
  (let ((p1 (- (lower-bound x) (lower-bound y)))
        (p2 (- (lower-bound x) (upper-bound y)))
        (p3 (- (upper-bound x) (lower-bound y)))
        (p4 (- (upper-bound x) (upper-bound y))))
    (make-interval (min p1 p2 p3 p4)
                   (max p1 p2 p3 p4))))

2.9 假设区间a和区间b的宽度是la和lb，那么a+b和a-b的宽度都是la+lb。可以证明，这里不证了。乘法和除法的长度都是非线性的，会跟区间的上下限有关。

2.10 这题浪费时间在找assert这个函数，其实错误处理还有error什么的一系列，没有去研究了

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(define (<= x y) (or (< x y) (= x y)))
(define (>= x y) (or (> x y) (= x y)))
(define (contain-0 y)
  (and (<= 0 (upper-bound y)) (>= 0 (lower-bound y))) )
(define (div-interval x y)
  (assert (not (contain-0 y)))
  (mul-interval x 
        (make-interval (/ 1.0 (upper-bound y))
                       (/ 1.0 (lower-bound y)))))

2.11 这题还比较有意思，区间a可以分3个情况：大于0、小于0、包含0；区间b也可以做这样的划分，这样就有3*3=9种不同的情况，代码如下

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(define (<= x y) (or (< x y) (= x y)))
(define (>= x y) (or (> x y) (= x y)))
(define (contain-0 y)
  (and (<= 0 (upper-bound y)) (>= 0 (lower-bound y))) )
(define (gt-0 y)
  (and (> (lower-bound y) 0) (> (upper-bound y) 0)) )
(define (lt-0 y)
  (and (< (lower-bound y) 0) (< (upper-bound y) 0)) )
(define (mul-interval-new x y)
  (let ((p1 (* (lower-bound x) (lower-bound y)))
        (p2 (* (lower-bound x) (upper-bound y)))
        (p3 (* (upper-bound x) (lower-bound y)))
        (p4 (* (upper-bound x) (upper-bound y))))
    (cond ((and (gt-0 x) (gt-0 y))
              (make-interval p1 p4))
          ((and (lt-0 x) (lt-0 y))
              (make-interval p1 p4))
          ((and (gt-0 x) (lt-0 y))
              (make-interval p3 p2))
          ((and (lt-0 x) (gt-0 y))
              (make-interval p2 p3)) 
          ((and (contain-0 x) (gt-0 y))
              (make-interval p2 p4))
          ((and (contain-0 x) (lt-0 y))
              (make-interval p4 p1))
          ((and (gt-0 x) (contain-0 y))
              (make-interval p3 p4)) 
          ((and (lt-0 x) (contain-0 y))
              (make-interval p4 p1))
          ((and (contain-0 x) (contain-0 y))
              (make-interval (min p2 p4) (max p1 p4)))
          )))

2.12 代码如下

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(define (make-center-percent c p)
  (let ((w (abs (* c p))))
    (make-interval (- c w) (+ c w)) ))
(define (center i)
  (/ (+ (lower-bound i) (upper-bound i)) 2))
(define (percent i)
  (/ (/ (- (upper-bound i) (lower-bound i)) 2) (center i)))

2.13

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