Implementation of a reduce procedure in racket with three arguments: function, base and a list

Question

My task is to implement a reduce procedure that takes a function that adds, or multiplies, or divides, each number in a list. I am stuck with the argument base. I have been able to check through the list of numbers and apply the function passed as argument. However, in the recursive call, I don't know how to handle the base number. I don't want the recursion to take into account the argument base in every call (base could be any number).

Here is what I have been able to do:

(define (reduce f base lst)
  (cond ((empty? lst) base)
        (else (f (f base (first lst)) (reduce f base (rest lst))))))

When base is 0 it passes the test for addition:

> (test (reduce + 0 '(1 2 3)) 6)
good (reduce + 0 '(1 2 3)) at line 3
  expected: 6
  given: 6

However, when base is 1, the test fails because base is added in the recursion call every time:

> (test (reduce + 1 '(1 2 3)) 7)
bad (reduce + 1 '(1 2 3)) at line 13
  expected: 7
  given: 10

ad absurdum · Accepted Answer · 2021-02-07 04:24:02Z

1

If there are arguments in lst to process, then you want to combine the first argument with the result of processing the rest of lst. But, with (f (f base (first lst)) (reduce f base (rest lst))) the posted code is combining the result of (f base (first lst)) with the result of processing the rest of lst. Since the test is using +, this means that at every step base is added to the next item of lst, and that sum is combined with the remainder of the computation.

Instead, just use f to combine the next element of lst with the result of reducing the rest of lst:

(define (reduce f base lst)
  (cond ((empty? lst) base)
        (else (f (first lst) (reduce f base (rest lst))))))

scratch.rkt> (reduce + 1 '(1 2 3))
7

answered Feb 7, 2021 at 4:24

ad absurdum

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