1-27.rkt

#lang racket

;; CIS 352 -- Spring 2026
;; 1/27/26 (I believe)

;; quickanswer.harp-lab.com

;; the define form lets us define funcitons
;; (define (f x y z) e-body)
(define (both-even? n0 n1)
  'todo)

(define (f x)
  (if (< x 0)
      (* x x)
      (* x x x)))

;; the simplest list is
'()
'(x y z)

;; How do I set up conditional statements correctly?
;; Function returns true when l is equal to '()
;; and returns false otherwise
(define (is-the-empty-list? l)
  (equal? l '()))

(define (f+ x)
  (if (equal? x 0)
      5
      (if (equal? x 5)
          6
          (+ x 2))))

(define (f++ x)
  (cond [(equal? x 0) 5]
        [(equal? x 5) 6]
        [else (+ x 2)]))

;; Recursion

;; def sum_list(l):
;;   i = 0
;;   acc = 0
;;   while (i < len(l)):
;;     acc = acc + l[i]
;;     i = i + 1
;;   return acc

(define (sum-list l)
  (define (h i acc)
    (if (< i (length l))
        (h (+ i 1) (+ acc (list-ref l i)))
        acc))
  (h 0 0))

;; get the ith element from the list l
;; preconditions:
;; - l is not empty
;; - i < (length l) ∧ l >= 0
(define (list-ref l i)
  (if (equal? i 0)
      (first l)
      (list-ref (rest l) (- i 1))))


(define (sum-list+ l)
  (if (empty? l)
      0
      (+ (first l) (sum-list+ (rest l)))))

(define (recursive-funtcion l ...)
  (if (empty? l)
      'base-case
      ;; call (recursive-function (rest l)) and combine it
      ;; with (first-l)
      'recursive-case))

;; if l is '() then return #f
;; if l is a (cons x y), return (even (first l))
(define (is-first-list-element-even? l)
  (if (equal? '() l)
      #f
      (even? (first l))))

;; car is first 
;; cdr is rest

;; return all elements in l that satisfy the predicate f
(define (filter f l)
  (if (empty? l)
      '()
      (if (f (first l))
          (cons (first l) (filter f (rest l)))
          (filter f (rest l)))))

;; return the elements of l that are even?
(define (filter-evens l)
  (filter even? l))

;; return the sub-list of l whose elements are < 2
(define (filter-lt2 l)
  ;; a lambda says: "I am a function, without a name, defined
  ;; exactly here. I take the argument x, and I return my body expression
  #;(filter (lambda (x) (< x 2) l) l)
  'todo)