1. Complete the definitions of move_botleft and move_right_or_up from the same-fringe solution in the week11 notes. Test your attempts against some example trees to see if the resulting make_fringe_enumerator and same_fringe functions work as expected. Show us some of your tests.

    type 'a tree = Leaf of 'a | Node of ('a tree * 'a tree)
    
    
    type 'a starred_level = Root | Starring_Left of 'a starred_nonroot | Starring_Right of 'a starred_nonroot
    and 'a starred_nonroot = { parent : 'a starred_level; sibling: 'a tree };;
    
    
    type 'a zipper = { level : 'a starred_level; filler: 'a tree };;
    
    
    let rec move_botleft (z : 'a zipper) : 'a zipper =
        (* returns z if the targetted node in z has no children *)
        (* else returns move_botleft (zipper which results from moving down from z to the leftmost child) *)
        _____
        (* YOU SUPPLY THE DEFINITION *)
    
    
    let rec move_right_or_up (z : 'a zipper) : 'a zipper option =
        (* if it's possible to move right in z, returns Some (the result of doing so) *)
        (* else if it's not possible to move any further up in z, returns None *)
        (* else returns move_right_or_up (result of moving up in z) *)
        _____
        (* YOU SUPPLY THE DEFINITION *)
    
    
    let new_zipper (t : 'a tree) : 'a zipper =
        {level = Root; filler = t}
        ;;
    

     

    let make_fringe_enumerator (t: 'a tree) =
        (* create a zipper targetting the botleft of t *)
        let zbotleft = move_botleft (new_zipper t)
        (* create a refcell initially pointing to zbotleft *)
        in let zcell = ref (Some zbotleft)
        (* construct the next_leaf function *)
        in let next_leaf () : 'a option =
            match !zcell with
            | Some z -> (
                (* extract label of currently-targetted leaf *)
                let Leaf current = z.filler
                (* update zcell to point to next leaf, if there is one *)
                in let () = zcell := match move_right_or_up z with
                    | None -> None
                    | Some z' -> Some (move_botleft z')
                (* return saved label *)
                in Some current
                )
            | None -> (* we've finished enumerating the fringe *)
                None
        (* return the next_leaf function *)
        in next_leaf
        ;;
    
    
    let same_fringe (t1 : 'a tree) (t2 : 'a tree) : bool =
        let next1 = make_fringe_enumerator t1
        in let next2 = make_fringe_enumerator t2
        in let rec loop () : bool =
            match next1 (), next2 () with
            | Some a, Some b when a = b -> loop ()
            | None, None -> true
            | _ -> false
        in loop ()
        ;;
    
  2. Here's another implementation of the same-fringe function, in Scheme. It's taken from http://c2.com/cgi/wiki?SameFringeProblem. It uses thunks to delay the evaluation of code that computes the tail of a list of a tree's fringe. It also involves passing "the rest of the enumeration of the fringe" as a thunk argument (tail-thunk below). Your assignment is to fill in the blanks in the code, and also to supply comments to the code, to explain what every significant piece is doing. Don't forget to supply the comments, this is an important part of the assignment.

    This code uses Scheme's cond construct. That works like this;

    (cond
        ((test1 argument argument) result1)
        ((test2 argument argument) result2)
        ((test3 argument argument) result3)
        (else result4))
    

    is equivalent to:

    (if (test1 argument argument)
        ; then
        result1
        ; else
        (if (test2 argument argument)
            ; then
            result2
            ; else
            (if (test3 argument argument)
                ; then
                result3
                ; else
                result4)))
    

    Some other Scheme details:

    • #t is true and #f is false
    • (lambda () ...) constructs a thunk
    • there is no difference in meaning between [...] and (...); we just sometimes use the square brackets for clarity
    • '(1 . 2) and (cons 1 2) are pairs (the same pair)
    • (list) and '() both evaluate to the empty list
    • (null? lst) tests whether lst is the empty list
    • non-empty lists are implemented as pairs whose second member is a list
    • '() '(1) '(1 2) '(1 2 3) are all lists
    • (list) (list 1) (list 1 2) (list 1 2 3) are the same lists as the preceding
    • '(1 2 3) and (cons 1 '(2 3)) are both pairs and lists (the same list)
    • (pair? lst) tests whether lst is a pair; if lst is a non-empty list, it will also pass this test; if lst fails this test, it may be because lst is the empty list, or because it's not a list or pair at all
    • (car lst) extracts the first member of a pair / head of a list
    • (cdr lst) extracts the second member of a pair / tail of a list

    Here is the implementation:

    (define (lazy-flatten tree)
      (letrec ([helper (lambda (tree tail-thunk)
                      (cond
                        [(pair? tree)
                          (helper (car tree) (lambda () (helper _____ tail-thunk)))]
                        [else (cons tree tail-thunk)]))])
        (helper tree (lambda () _____))))
    
    
    (define (stream-equal? stream1 stream2)
      (cond
        [(and (null? stream1) (null? stream2)) _____]
        [(and (pair? stream1) (pair? stream2))
         (and (equal? (car stream1) (car stream2))
              _____)]
        [else #f]))
    
    
    (define (same-fringe? tree1 tree2)
      (stream-equal? (lazy-flatten tree1) (lazy-flatten tree2)))
    
    
    (define tree1 '(((1 . 2) . (3 . 4)) . (5 . 6)))
    (define tree2 '(1 . (((2 . 3) . (4 . 5)) . 6)))
    
    
    (same-fringe? tree1 tree2)