#lang racket

;; Mini Exercise 3 Solution

;; 1.  Version of my-or that avoids an extra "if"
 (define-syntax my-or
  (syntax-rules ()
    ((my-or) #f)
    ((my-or e) e)
    ((my-or e1 e2 ...)
     (let ((temp e1))
       (if temp
           temp
           (my-or e2 ...))))))

 
;; 2. my-if using delay and force
 (define (my-if a b c)
   (if a (force b) (force c)))
 
 ;; sample call:
(my-if (= 1 1) (delay (+ 2 4)) (delay (/ 10 0)))

;; 5. 
'( ( 1 . 2) (3 . 4))

;; 6. my-max.  We will use Racket's syntax for gathering an indefinite 
;; number of arguments - but we require at least 1
(define (my-max x . xs)
  (my-max-helper x xs))

;; my-max-helper takes a biggest element (so far) and a list, and finds the maximum
;; of biggest and the elements in rest
(define (my-max-helper biggest rest)
  (cond ((null? rest) biggest)
        ((> (car rest) biggest) (my-max-helper (car rest) (cdr rest)))
        (else (my-max-helper biggest (cdr rest)))))

;; alternative version of my-max that doesn't use a helper function.  We
;; use 'apply' to get around the fact that on the recursive call we've got
;; the arguments in a list already
(define (my-max2 x . xs)
  (cond ((null? xs) x)
        ((> (car xs) x) (apply my-max2 xs))
        (else (apply my-max2 (cons x (cdr xs))))))

;; yet another version, which checks that there is at least 1 argument in
;; the function rather than in the argument list
(define (my-max3 . xs)
  (cond ((null? xs) (error "my-max3 needs at least one argument"))
        ((null? (cdr xs)) (car xs))
        ((> (car xs) (cadr xs)) (apply my-max3 (cons (car xs) (cddr xs))))
        (else (apply my-max3 (cdr xs)))))