[2021Fall] Lab10 of CS61A of UCB

Q2: Over or Under

Define a procedure over-or-under which takes in a number num1 and a number num2 and returns the following:

• -1 if num1 is less than num2
• 0 if num1 is equal to num2
• 1 if num1 is greater than num2

Challenge: Implement this in 2 different ways using if and cond!

(define (over-or-under num1 num2)
  'YOUR-CODE-HERE
)

代码其实本身不难, 主要是适应 scheme 语言的写法, 条件分支有两种写法:

1. (if <predicate> <consequent> <alternative>)
2. (cond (<condition> <consequent>) ...)

(define (over-or-under num1 num2) 
    (if (< num1 num2) 
            (print -1))
    (if (= num1 num2)
            (print 0))
    (if (> num1 num2)
            (print 1))
)

(define (over-or-under num1 num2)
  (cond ( (< num1 num2) (print -1) )
        ( (= num1 num2) (print 0)  )
        ( (> num1 num2) (print 1)  ))
)

Q3: Make Adder

Write the procedure make-adder which takes in an initial number, num, and then returns a procedure. This returned procedure takes in a number inc and returns the result of num + inc.

Hint: To return a procedure, you can either return a lambda expression or define another nested procedure. Remember that Scheme will automatically return the last clause in your procedure.

You can find documentation on the syntax of lambda expressions in the 61A scheme specification!

实现高阶函数的功能, 依旧是锻炼 scheme 语言的掌握程度的. 题目都是之前课上讲过的. 这里我用匿名函数来实现

(define (make-adder num)
  (lambda (inc) (+ num inc))
)

Q4: Compose

Write the procedure composed, which takes in procedures f and g and outputs a new procedure. This new procedure takes in a number x and outputs the result of calling f on g of x.

用 scheme 语言实现符合数学中的复合函数, 也就是高阶函数. 这里同样可以用 lambda 函数

(define (composed f g)
  (lambda (x) (f (g x) ) )
)

Q5: Make a List

In this problem you will create the list with the following box-and-pointer diagram:

Challenge: try to create this list in multiple ways, and using multiple list constructors!要求

题目要求我们按照给定的链表结构来生成对应的链表. 主要考察的是对 scheme 语言中 list 的掌握. 可以有多种实现方式

1. cons 的方式, 这个方式很容易眼花, 最好是写完之后在这里 验证一下. 这里我真的写得头有点晕
2. list 的方式, 这个代码会比较短, 注意我们每次在调用 (list ...) 相当于在链表中多创建了一个方向

(define lst 
  (cons (cons 1 nil)
        (cons 2 (cons (cons 3 (cons 4 nil))
                      (cons 5 nil))))
)

(define lst 
  (list (list 1)
        2 (list 3 4)
        5)
)

Q6: Remove

Implement a procedure remove that takes in a list and returns a new list with all instances of item removed from lst. You may assume the list will only consist of numbers and will not have nested lists.

Hint: You might find the built-in filter procedure useful (though it is definitely possible to complete this question without it).

You can find information about how to use filter in the 61A Scheme builtin specification!

这一题就是要我们在 scheme 的列表中移除掉值等于 item 的元素然后返回新的这个列表. 其实 scheme 的列表也就是链表. 所以这一题等效于我们要在链表中移除指定值的元素. 显然, 这可以用递归来解决! 而且这一道题说没有嵌套列表的情况存在, 这一道题就更简单了 !

显然 base case 就是链表为空的情况, 我们直接返回空. 否则:

1. 当前节点的值 = item, 我们删除它, 递归处理子链表
2. 当前节点的值 != item, 我们保留它, 递归处理子链表

(define (remove item lst)
  (cond ( (null? lst) '() )             ; base case
        ( (= item (car lst)) (remove item (cdr lst)))           ; exclude item
        ( else (cons (car lst) (remove item (cdr lst)))))       ; include item
)