How to duplicate a list in Scheme x times? - scheme

Question

I am just starting to learn Scheme and using cons is a little confusing to me. I have a function duplicate (s number) where s is a list, and number is the number of times the list should be duplicated.
If I enter (duplicate '(1 2) 3), the output should be ((1 2) (1 2) (1 2))
My program looks like this, but when I run it, there is nothing in the output
(define (duplicate s number)
(cond [(null? s) '()]
[(> 0 number) (cons (list s) (duplicate s(- number 1)))]
))
What am I doing wrong here?

Answer 1

We want second input n become zero complete all list.
And we want output is a list so we use cons.
You can build your code by using minimum sample than add more complex data.
If input is (duplicate x 0) we want output is '().
If input is (duplicate x 1) we want output is '(x).
So your code should looks like this
(define (duplicate x n)
(cond
[(= n 0) '()]
[else
(cons x ...)]))
But we already know we want output is '(x) which is (cons x '()).
Obvious '() is (duplicate x 0)'s output. So we add (duplicate x (- n 1)) in second condition.
#lang racket
(define (duplicate x n)
(cond
[(= n 0) '()]
[else
(cons x (duplicate x (- n 1)))]))
;;; TEST
(duplicate '() 0)
(duplicate '() 3)
(duplicate '() 5) ; '(() () () () ())
(duplicate '(1 2) 5) ; '((1 2) (1 2) (1 2) (1 2) (1 2))
Or you can think like this way.
We have employee help us copy document.
employee-1 : We give him a number than he minus 1 than order employee-2 do his job
employee-2 : He copy a document than send message to employee-3
employee-3 : He supervision finish or not (number become zero). If not finish send message to employee-1.
So we want something like this
finish? -> no -> minus-1 -> copy -> finish? -> no -> minus-1 -> ...
#lang racket
(define x 1)
(define result '())
(define (employee-1 n)
(employee-2 (- n 1)))
(define (employee-2 n)
(begin
(set! result (cons x result))
(employee-3 n)))
(define (employee-3 n)
(if (= n 0)
result
(employee-1 n)))
;;; TEST
(employee-3 3) ; '(1 1 1)
Than we combine employee-1 to employee-3
(define x 1)
(define result '())
; (define (employee-1 n) (employee-2 (- n 1)))
(define (employee-2 n)
(begin
(set! result (cons x result))
(employee-3-v2 n)))
(define (employee-3-v2 n)
(if (= n 0)
result
(employee-2 (- n 1))))
;;; TEST
(employee-3-v2 3) ; '(1 1 1)
We use function input replace define global variable. So we have to remove set! and change input parameter.
; (define x 1)
; (define result '())
; (define (employee-1 n) (employee-2 (- n 1)))
(define (employee-2-v2 n x result)
(employee-3-v2 n x (cons x result)))
(define (employee-3-v2 n x result)
(if (= n 0)
result
(employee-2-v2 (- n 1) x result)))
;;; TEST
(employee-3-v2 3 1 '()) ; '(1 1 1)
Than we combine employee-2-v2 to employee-3-v2. Remember we have to change input parameter.
(define (employee-3-v3 n x result)
(if (= n 0)
result
(employee-3-v3 (- n 1) x (cons x result))))
;;; TEST
(employee-3-v3 3 'x '()) ; '(x x x)
Now we want remove not necessary input parameter result.
(define (employee-3-v4 n x)
(if (= n 0)
'()
(cons x (employee-3-v4 (- n 1) x))))
;;; TEST
(build-list 10 (λ (n) (employee-3-v4 n 'x)))
#|
output:
'(()
(x)
(x x)
(x x x)
(x x x x)
(x x x x x)
(x x x x x x)
(x x x x x x x)
(x x x x x x x x)
(x x x x x x x x x))
|#