R5RS方案代码找到列表中偶数元素的和（递归和高阶抽象）？

Question

(define mylist (list 1 2 3 5 8 9 10))

;;sum of the squares of the even elements of mylist
(define (evens lis)
  (cond (;; Check stop condition 1
         (null? lis)
         '())
        (;; Check stop condition 2: list of length = 1
         (null? (cdr lis))
         '())
        (else
         ;; Check: Building list with second element
         ;; The previous cond clauses have already sorted out
         ;; that lis and (cdr lis) are not null.
         (cons (cadr lis)
               ;; Check: Recurse "the rest of the rest" of lis with cddr
               (evens (cddr lis))))))

(define (sum-even lis)
  (if (null? lis)
  0
  (+ (expt (car lis) 2) (sum-even(cdr lis))))) ;;add all squared even elements

(sum-squared-even (evens mylist))

Here is my attempt to find the sum of the squares of the even elements in the list. However, I am really new to r5rs, is that possible I can write these scheme into only one procedure? Also, I was asked to write a second version in terms of map, foldr, and filter . Hope anyone familiar with r5rs can help me out! Thanks!

Answer 1

First of all, your function is wrong since you do not test the numbers of the list for evenness. This should be the correct definition:

;;even elements of a list
(define (evens lis)
  (cond (;; Check stop condition
         (null? lis)
         '())
        (;; Check if first element is even,
         ;; return it with the evens of the rest of the list
         (even? (car lis))
         (cons (car lis) (evens (cdr lis))))
        (else
         ;; First element is odd, return only the evens of the rest of the list
         (evens (cdr lis)))))

Then, to answer to your first question, this is a solution with a single recursive function:

;;sum of the squares of the even elements of a list
(define (sum-square-evens lis)
  (cond ((null? lis)
         0)
        (;; Recurring, first case: the first element is even, so square it
         ;; and add to the result of summing the square of the evens of the rest of the list
         (even? (car lis))
         (+ (expt (car lis) 2)
            (sum-square-evens (cdr lis))))
        (else
         ;; Recurring, when the first element is not even
         ;; the result is obtained by summing the square of the evens of the rest of the list
         (sum-square-evens (cdr lis)))))

Finally here is the function written with foldl-map-filter :

(define (sum-square-evens-2 lis)
  (foldl + 0 (map (lambda(x) (expt x 2)) (filter even? lis))))

but note that foldl is not a primitive function of R5RS Scheme, so you should run your program with DrRacket with the #lang racket specification, or you should define it or load a library in which it is defined.

Here is the meaning of the second function:

1. filter applies a predicate to all the elements of a list, a returns a new list containing all the elements that satisfy the predicate. In this case the predicate is even? , so the result of (filter even? lis) is the list of all the even numbers of lis .
2. To this list the map high order function is applied, to produce as result the results of the function first argument (in this case a function that squares its argument) applied to its second argument (a list). So the result of (map ...) is a new list containing the squares of all the even elements of the original list.
3. Finally, (foldl + 0 (n1 n2 ...nk)) returns (...((0 + n1) + n2) + ... + nk) , since it repeatedly applies the binary function + to the elements of the list, using 0 as its first (left) operand. (Note that foldl and foldr are in this case equivalent, the second one calculating the sum as (n1 + (n2 + ...(nk + 0)...)) )

Answer 2

Here's how I would do it; I'll leave to you the definitions of sum and square :

(sum (map square (filter even? '(1 2 3 5 8 9 10))))

It's handy to collect convenience functions like sum and square in a library that is loaded automatically when you start your Scheme interpreter, so you can write simple expressions like this without delay.

Answer 3

Your sum_even has perhaps a wrong name as it will square every element and sum it together, thus square-sum is perhaps a better name.

If you wanted the even placed elements to be squared and summed you just use the two procedures you have posted:

(define (square-sum-even lst)
  (square-sum (evens lst)))