; Oumar Ly
; Assignment 8.1
; 04/09/2019
; ------------------------------- Problem 1 ----------------------------
; append1 takes 2 lists as arguments, if first list is empty return the second list
; create another list with the first element of list 1
; then call append1 recursively with the tail of list1 as first argument along with list 2
; which will recursively repeat until both list are completed
(define (append1 ls1 ls2)
(if (null? ls1) ls2
(cons (car ls1) (append (cdr ls1) ls2))))
; testing append 1
(display (append1 '(1 2 3 4) '(a b c d e f)))
(newline)
(display (append1 '(1 2 3) '()))
(newline)
; ------------------------------- Problem 2 ----------------------------
; member? base case list is null/empty return false
; base2 x is the first number return true
; else calls member to recursive check the rest of the list for x
(define (member? x list)
(if (null? list) #f
(if (equal? x (car list)) #t
(member? x (cdr list)))))
; testing member? procedure
(display (member? 5 '(1 2 3 4 5 6 7 5)))
(newline)
(display (member? 10 '(1 2 3 4 5 6 7 5)))
(newline)
; ------------------------------- Problem 3 ----------------------------
; member1 base case list is null/empty return false
; case x is the first number return rest of the list
; else calls member1 to recursive check for x in the list and return rest of list
(define (member1 x list)
(cond ((null? list) #f)
((equal? x (car list)) cdr list)
(else (member1 x (cdr list)))))
; testing member1
(display (member1 2 '(a b c d e f)))
(newline)
(display (member1 3 '(1 2 3 4 5 3 34)))
(newline)
; ------------------------------- Problem 4 ----------------------------
; flatten checks if list is null return a literal empty list
; checks if not list, calls built-in list procedure to return it a list
; else use append1 to call the head of the list with flatten and the the rest of the list is flattened
(define (flatten lst)
(cond ((null? lst) `())
((list? lst) ; if list found append with arguments applied to flatten for the head and tail of list
(append1 (flatten (car lst)) (flatten (cdr lst))))
(else
(list lst)))) ; else use built in list to make into a list
; NOTE for some reason the cons of (car list) along with recursive call to flatten on the tail of list
; is throwing errors [*** ERROR: pair required, but got 1] (algorithm from hw tips)
; What is wrong with this ? (Note to Professor McGoff)
;(define (flatten lst)
; (cond ((null? lst) `())
; ((list? lst)
; (append1 (flatten (car lst)) (flatten (cdr lst))))
; (else (cons (car lst) (flatten (cdr lst))))))
; testing flatten with different values
(display (flatten '(1 (2) 3)))
(newline)
(display (flatten '((1) (2 3 4) (5 6))))
(newline)
(display (flatten '(a (b (c d (e) f (g h))) (i j))))
(newline)
(display (flatten '("nothing" "to" "flatten")))
(newline)
; ------------------------------- Problem 5 ----------------------------
; build a list of the first element of each list, call zip recursively going the car of the cdr of each list until both are empty
(define (zip ls1 ls2)
(cond [(null? ls1) '()]
[(null? ls2) '()]
(else (cons (cons (car ls1) (car ls2)) ; first cons creates the main list, then create the pairing of list and call with the cdr
(zip (cdr ls1) (cdr ls2))))))
; testing zip
(display (zip '(1 2 3) '(a b c)))
(newline)
(display (zip '(1 2 3) '()))
(newline)
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