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• 讲课笔记 (Lecture Notes)
• 线上教程 (Related Resources > Online Tutor for 6.001) ;需要注册，不过就算不是 MIT 的学生也可以注册的。

P.S.

线上教程的习题部分不错，难度低于 SICP 的习题，大部分可以自动批改（选择题除外）；个人感觉这种即时反馈很能增加信心和成就感。

如果先刷线上习题，之后再推 SICP，应该能缓和总体的难度曲线（；同步进行应该也不错）。

div-series应该求任何两个级数，您的代码只能求分母的常数项是1的情况。
我的代码：
; 假设X= S1 / S2
; X = S1 * (1 / (S2 / c)) * (1 / c) (c是S2的常数项)

(define (div-series s1 s2)
(let ((c (stream-car s2)))
(if (= c 0)
(error "The series in the denominator has 0! constant terms")
(scale-stream (mul-series s1
(reciprocal-series (scale-stream s2 (/ 1 c))))
(/ 1 c)))))

(define tan-series (div-series sine-series cosine-series))
;(display-top10 tan-series)
;0 1 0 1/3 0 2/15 0 17/315 0

你好，参考了你3.31习题的解答。这里想探讨一下。

我觉得你对half adder例子中inverter的分析是对的，不过加(proc)的原因我觉得措辞可以改一下。原因准确说不应该是让input不变的时候改变output，而是初始化的时候确保output会有正确的值。这里有一个初始化的限定，因为after-delay只是被调用了一次而已。

(define (eval-or exp env)
  (if (null? exp)
    true
    (or (eval (car exp) env)
        (eval-or (cdr exp) env))))

应该为

(define (eval-or exp env)
  (if (null? exp)
    false
    (or (eval (car exp) env)
        (eval-or (cdr exp) env))))

后面实现为派生表达式，我认为应该为

(define (eval-and exp env)
    (make-if (null? exp)
        true
        (make-if (car exp)
            (eval-add (cdr exp) env)
            false)))

但是，我感觉用派生表示式的话，就没有短路效果了！

可以看看我这里的解法4.19

我发现，如果将代码

(define (count-pairs x)
  (if (not (pair? x))
    0
    (if (seen? x) 
      0
      (begin
        (set! already-seen (cons x already-seen)) ;;;修改这里
        (+ (count-pairs (car x))
           (count-pairs (cdr x))
1)))))

修改为

(define (count-pairs x)
  (if (not (pair? x))
    0
    (if (seen? x) 
      0
      (begin
        (set! already-seen (cons already-seen x))
        (+ (count-pairs (car x))
           (count-pairs (cdr x))
1)))))

就得不到正确答案了，而且如果表成环的话也会无限递归
这是为什么

我的环境是win10 64位 DrRacket
在github上不知道如何将代码格式化，抱歉了

2.83
感觉, 像

(define (raise scheme-number)
    (make-rational x 1))

应该放在 install-rational-package 包中

sum-iter的body部分多了一层括号，应为：

(define (sum term a next b)
  (define (sum-iter a result)
    (if (> a b)
        result
        (sum-iter (next a) (+ result (term a)))))
  (sum-iter a 0))

您好
练习4.11 的add-binding-to-frame! 我觉得你实现错了
(define (add-binding-to-frame! var val frame) (cons (cons var val) frame))

这里不应该是简单的返回一个新的frame，而是要修改原来的frame

(define (add-binding-to-frame! var val frame)
  (let ((old-car (car frame)))
    (set-car! frame (cons var val))
    (set-cdr! frame (cons old-car (cdr frame)))))

;;test
(let ((frame (make-frame '(a b c) '(1 2 3))))
  (add-binding-to-frame! 'd 4 frame)
  (display frame))

2.51中关于paint-top的定义中第三个向量应该是(make-vect 0 1)吧，不应该是上半部分吗

我的不成熟的代码：
;重新定义make-sum过程,接受参数列表, 并整合成一个和式
(define (make-sum . items)
(define (iter result x)
(if (null? x)
result
(let ((first (car x)))
(if (=number? first 0)
(iter result (cdr x))
(iter (append result (list first))
(cdr x))))))
(let ((add (iter '() items)))
(if (null? (cdr add))
(car add)
(append '(+) add))))
;重新定义过程make-product, 接受参数列表, 并整合成一个乘式
(define (make-product . items)
(define (iter result x)
(if (null? x)
result
(let ((first (car x)))
(cond ((=number? first 0) (list 0))
((=number? first 1)
(iter result (cdr x)))
(else (iter (append result (list first))
(cdr x)))))))
(let ((mul (iter '() items)))
(if (null? (cdr mul))
(car mul)
(append '(*) mul))))

恩, 我看你 3.50 习题里面是直接(load "lib/stream.scm") 的
想问下, 怎么设置成相对路径加载
我目前每次都要(load "~/sicp/xxx/xxx.scm") 感觉比较繁琐--> 虽然就多几个字符
谢谢

试过scheme lib 么 https://github.com/scheme-lib/scheme-lib

sicp/exercises/01/1.28.scm

"1取模n的非平凡平方根"，也就是说，是不是存在不等于1或者n-1的数，其平方取模n等于1。

你的代码没判断不等于1和n-1。

习题2.11上，明确说了每种情况所需要的乘法不超过两次。但是答案中，当两个区间都横跨0轴的时候，计算了4次乘法。

(defn mul-interval-v1 [x y]
  (if (neg? (lower-bound x))
    (if (neg? (lower-bound y))
      (if (neg? (upper-bound x))
        (if (neg? (upper-bound y))
          ;;xl<0 xu<0 yl<0 yu<0
          (make-interval (* (upper-bound x) (upper-bound y))
                         (* (lower-bound x) (lower-bound y)))
          ;;xl<0 xu<0 yl<0 yu>0
          (make-interval (* (min (lower-bound x)
                                 (lower-bound y)) (upper-bound y))
                         (* (lower-bound x) (lower-bound y))))
        (if (neg? (upper-bound y))
          ;;xl<0 xu>0 yl<0 yu<0
          (make-interval (* (upper-bound x) (min (lower-bound y)
                                                 (lower-bound x)))
                         (* (lower-bound x) (lower-bound y)))
          ;;xl<0 xu>0 yl<0 yu>0
          (make-interval (* (min (lower-bound x)
                                 (lower-bound y))
                            (max (upper-bound x)
                                 (upper-bound y)))
                         (* (max (lower-bound x)
                                 (lower-bound y))
                            (min (upper-bound x)
                                 (upper-bound y))))))
      (if (neg? (upper-bound x))
        ;;xl<0 xu<0 yl>0 yu>0
        (make-interval (* (lower-bound x) (upper-bound y))
                       (* (upper-bound x) (lower-bound y)))
        ;;xl<0 xu>0 yl>0 yu>0
        (make-interval (* (lower-bound x) (upper-bound y))
                       (* (upper-bound x) (upper-bound y)))))
    (if (neg? (lower-bound y))
      (if (neg? (upper-bound y))
        (make-interval (* (lower-bound y) (upper-bound x))
                       (* (upper-bound y) (lower-bound x)))
        ;; xl>0 xu>0 yl<0 yu>0
        (make-interval (* (lower-bound y) (max (upper-bound x)
                                               (upper-bound y)))
                       (* (upper-bound x) (upper-bound y))))
      ;; xl>0 xu>0 yl>0 yu>0
      (make-interval (* (lower-bound x) (lower-bound y))
                     (* (upper-bound x) (upper-bound y))))))

由于没有 scheme 写所以就当一个补充了

while 的实现你没有写完，一次就停了：

(define (expand-predicate predicate i body)
  (if (predicate i)
    (cons 'begin body)))

(define (while->combination exp)
  (expand-predicate
    (while-predicate exp)
    (while-variable exp)
    (while-body exp)))

此外，for的实现我有疑问：

(define (expand-range start end proc)
  (cond
    ((< start end)
      (proc start)
      (expand-range (+ start 1) end proc))))

在这里面 (proc start)的求值需要环境吧？
是否需要写成这样？

(define (expand-range start end proc env)
  (cond
    ((< start end)
      (eval （lambda () (proc start)) env)
      (expand-range (+ start 1) end proc env)))

(define (make-zero-crossings input-stream smooth)
  (define (iter s last-value)
    (cons-stream (sign-change-detector (stream-car s) last-value)
                 (iter (stream-cdr s) (stream-car s))))
  (iter (smooth input-stream)))

iter 是双参吧

;(b)
(define (product2 a term b next)
(define (iter a result)
(if (> a b)
result
((iter (next a) (* (term a) result))))
(iter a 1))

https://github.com/jiacai2050/sicp/blob/master/exercises/02/lib/complex_number.scm#L63

should be (put 'magnitude '(rectangular) magnitude)

(define (install-polynomial-package) 
  ;; ... 
  (put '=zero? 'polynomial 
    (lambda(poly) 
      (or (empty-termlist? (term-list poly))
          (= 0 (first-term (coeff (term-list poly)))))))
  'done)

应该不能只判断最高项的系数，而且如果想获得更好的抽象能力的话使用=zero?代替= 0应该会更好，(coeff (term-list poly))应该也有错

下面是我自己写的

(define (install-polynomial-package) 
...
 (define (=zero? poly) 
   (define (zero-terms? termlist) 
     (if (empty-termlist? termlist)
         true 
         (and (=zero? (coeff (first-term termlist))) 
              (zero-terms? (rest-terms termlist))))) 
   (zero-terms? (term-list poly))) 

(put '=zero '(polynomial) =zero?)
...
)

2.87----------

如果有错的话，欢迎大家纠正

for the first pair procedure, you have a clock tick in normal speed
for the second pair procedure, 2 times slower
3: 4 times slower per tick
n: 2^(n-1) n s p t

then to get to the next pair procedure,
you have to take two step, per step is going to be a tick of a clock in that pair pro's world.
so you get total time of getting to (n, 1)'s location.
(n, 1):2^n - 1

then you consider walk to the other direction,
walk to (n, m) from (n, 1):
you get that it take one step to (n, 2)
and two step each to the next.
then you have every thing you get.