高阶函数

Question

given,

(define (reduce f id lis)
    (if (null? lis) id
      (f (car lis) (reduce f id (cdr lis)))))

The length of a list can be defined in terms of reduce (as opposed to using a recursive definition from scratch) as

(define (mylength lis) (reduce  (lambda (x n) (+ 1 n)) 0 lis)).

Define the list function mymap (similar to map ) that takes a unary function uf and a list lis in terms of reduce, that is, by determining the corresponding f and id in

(mymap uf lis) = (reduce f id lis),

Recall that mymap returns a list resulting from calling the function on every element in the input list such as (mymap (lambda(x) (* 2 x)) '(1 2 3)) = (2 4 6) .

A little Explanation to how this has been done would be helpful, rather than a blatant answer. Thank You.

Answer 1

we have

(mymap uf lis) = (reduce f id lis) = 
 = (if (null? lis) id
     (f (car lis) (reduce f id (cdr lis))))

which must be equal to

 = (if null? lis) '()
     (cons (uf (car lis)) (mymap uf (cdr lis))))

matching the corresponding entities, we get

id == '()

and, since (mymap uf lis) = (reduce f id lis) , it is also (mymap uf (cdr lis)) = (reduce f id (cdr lis)) , so we have

(f x y) = (cons (uf x) y)

Thus, we define

(define (mymap uf xs)   ; multiple `x`-es :)
  (reduce (lambda (x r)
             (cons .... r))
          '()
          xs ))

your reduce is a right fold : its combining function receives an element x of the argument list xs , and the recursive result of working on the rest of list, as r .

r has all the elements of the rest of xs , which were already mapped through uf . All that's left to do to combine the given element x with this recursive result r , is to cons the (uf x) onto r .

It should pose no problem now to complete the definition by writing the actual code in place of the dots .... there.