#lang racket

; find position of max in a list
; helper function simply passes along all the state that is needed for
; the computation.

; note the tail recursion, so the stack is never more than constant size

; Pre: L is a non-empty list of numbers
; Post: returns position of first maximum in list, 0-origin indexing
(define lmax-index
  (lambda (L)
    (lmax-index-r (first L) 0 1 (rest L) )
    )
  )

; lmax-index recursor
; cur-max - value of the current maximum
; cur-index - position of the current maximum
; L-index - position of the first element of the remaining list
; L - remaining list of numbers to be processed

(define lmax-index-r 
  ( lambda (cur-max cur-index L-index L)
     (if (null? L) 
         cur-index
         (if (>= cur-max (first L))
             ; max still good (it is at least as big as the first element)
             (lmax-index-r cur-max cur-index (+ 1 L-index) (rest L))
             ; first is new max
             (lmax-index-r (first L) L-index (+ 1 L-index) (rest L))
             )
         )
     )
  )

(lmax-index '(1 2 3 4))
; NOTE: what happens if the >= test above is changed to a > test?
  
; Now create a function that returns a list of the positions of 
; all the maximums in L.
; This would work for an empty list, since the list of positions
; would be empty