;; CS 3520
;; Fall 2001
;; HW 2 example solution

;; 2.1 (EoPL 1.15, page 24)

;; 1.

; <list-of-obj> = () | (<obj> . <list-of-obj>)

; duple : num object -> list-of-obj
(define (duple n x)
  (cond
   [(zero? n) '()]
   [else (cons x (duple (- n 1) x))]))

(equal? (duple 0 'a) '())
(equal? (duple 10 'a) '(a a a a a a a a a a))

;; 2.

; <2-list> = (<obj> <obj>)
; <list-of-2-list> = ()  | (<2-list> . <list-of-2-list>)

; invert : list-of-2-list -> list-of-2-list
(define (invert lst)
  (cond
   [(null? lst) '()]
   [else (cons (invert/2l (car lst))
	       (invert (cdr lst)))]))

; invert/2l : 2-list -> 2-list
;  Swaps the order of the elements in the given 2-list
;   (invert/2l '(a b)) = _(b a)_
(define (invert/2l l)
  (list (cadr l) (car l)))

(equal? (invert/2l '(a b)) '(b a))

(equal? (invert '()) '())
(equal? (invert '((a 1) (a 2) (b 1) (b 2)))
	'((1 a) (2 a) (1 b) (2 b)))

;; 3.

; filter-in : pred list-of-obj -> list-of-obj
(define (filter-in pred lst)
  (cond
   [(null? lst) '()]
   [else (cond
	  [(pred (car lst)) 
	   (cons (car lst) (filter-in pred (cdr lst)))]
	  [else
	   (filter-in pred (cdr lst))])]))

(equal? (filter-in number? '()) '())
(equal? (filter-in number? '(a 2 (1 3) b 7))
	'(2 7))

;; 4.

; every? : pred  list-of-obj -> bool
(define (every? pred lst)
  (cond
   [(null? lst) #t]
   [else (and (pred (car lst))
	      (every? pred (cdr lst)))]))

(equal? (every? number? '()) #t)
(equal? (every? number? '(a 2 (1 3) b 7)) #f)
(equal? (every? number? '(2 7)) #t)

;; 5.

; exists? : pred  list-of-obj -> bool
(define (exists? pred lst)
  (cond
   [(null? lst) #f]
   [else (or (pred (car lst))
	     (exists? pred (cdr lst)))]))

(equal? (exists? number? '()) #f)
(equal? (exists? number? '(a 2 (1 3) b 7)) #t)
(equal? (exists? symbol? '(2 7)) #f)

;; 7.

; list-set : list-of-obj num obj -> list-of-obj
;  where n must be less than (length lst)
(define (list-set lst n x)
  (cond
   [(zero? n) (cons x (cdr lst))]
   [else (cons (car lst)
	       (list-set (cdr lst) (- n 1) x))]))

(equal? (list-set '(1) 0 'a) '(a))
(equal? (list-set '(1 2 3) 1 'a) '(1 a 3))

;; 8.

; product : list-of-obj list-of-obj ->  list-of-2-list
(define (product lst1 lst2)
  (cond
   [(null? lst1) '()]
   [else (append
	  (product1 (car lst1) lst2)
	  (product (cdr lst1) lst2))]))

; product1 : obj list-of-obj ->  list-of-2-list
;  Pairs x with each element of lst, creating a
;  list of 2-lists.
;   (product1 'x '()) = _()_
;   (product1 'x '(a b)) = _((x a) (x b))_
(define (product1 x lst)
  (cond
   [(null? lst) '()]
   [else (cons (list x (car lst))
	       (product1 x (cdr lst)))]))

(equal? (product1 1 '()) '())
(equal? (product1 2 '(a b c)) '((2 a) (2 b) (2 c)))

(equal? (product '() '(a b c)) '())
(equal? (product '(1 2 3) '(a b c)) '((1 a) (1 b) (1 c)
					    (2 a) (2 b) (2 c)
					    (3 a) (3 b) (3 c)))

;; 9.

; down : list-of-obj -> list-of-obj
(define (down lst)
  (cond
   [(null? lst) '()]
   [else (cons (list (car lst))
	       (down (cdr lst)))]))

(equal? (down '()) '())
(equal? (down '(1 (2) 3)) '((1) ((2)) (3)))