SICP – cap. 2 – sets come alberi binari – 85 – esercizio

Continuo da qui, copio qui.

Exercise 2.64: The following procedure list->tree converts an ordered list to a balanced binary tree. The helper procedure partial-tree takes as arguments an integer n and list of at least n elements and constructs a balanced tree containing the first n elements of the list. The result returned by partial-tree is a pair (formed with cons) whose car is the constructed tree and whose cdr is the list of elements not included in the tree.

(define (list->tree elements)
  (car (partial-tree 
        elements (length elements))))

(define (partial-tree elts n)
  (if (= n 0)
      (cons '() elts)
      (let ((left-size 
             (quotient (- n 1) 2)))
        (let ((left-result 
               (partial-tree 
                elts left-size)))
          (let ((left-tree 
                 (car left-result))
                (non-left-elts 
                 (cdr left-result))
                (right-size 
                 (- n (+ left-size 1))))
            (let ((this-entry 
                   (car non-left-elts))
                  (right-result 
                   (partial-tree 
                    (cdr non-left-elts)
                    right-size)))
              (let ((right-tree 
                     (car right-result))
                    (remaining-elts 
                     (cdr right-result)))
                (cons (make-tree this-entry 
                                 left-tree 
                                 right-tree)
                      remaining-elts))))))))

Write a short paragraph explaining as clearly as you can how partial-tree works. Draw the tree produced by list->tree for the list (1 3 5 7 9 11).

What is the order of growth in the number of steps required by list->tree to convert a list of n elements?

Qua non ci provo; chiamo Bill the Lizard 💥

The partial-tree procedure works by dividing the list into three parts, a center element (the root node of the tree), everything before the center element, and everything after the center element. All the elements before the center element are then passed to a recursive call to partial-tree to create the left branch of the tree, and all the elements after the center element are passed recursively to partial-tree to create the right branch. These recursive call continue until no elements are remaining, and the balanced binary tree is assembled.

The tree produced by list->tree for the list (1 3 5 7 9 11) is:

To verify this, we can simply call the procedure.

Next we’re asked what is the order of growth in the number of steps required by list->tree to convert a list of n elements? The procedure only needs to visit each element of the list once, and it only performs a cons for each element it visits, so the number of steps is proportional to the size of the list, or O(n).

sicp-ex risponde in modo simile, più conciso.
Risultato simile per Drewiki.
Insomma sono solo io che devo applicarmi di più 😐

🤢

Posta un commento o usa questo indirizzo per il trackback.

Trackback

Rispondi

Inserisci i tuoi dati qui sotto o clicca su un'icona per effettuare l'accesso:

Logo di WordPress.com

Stai commentando usando il tuo account WordPress.com. Chiudi sessione /  Modifica )

Google photo

Stai commentando usando il tuo account Google. Chiudi sessione /  Modifica )

Foto Twitter

Stai commentando usando il tuo account Twitter. Chiudi sessione /  Modifica )

Foto di Facebook

Stai commentando usando il tuo account Facebook. Chiudi sessione /  Modifica )

Connessione a %s...

Questo sito utilizza Akismet per ridurre lo spam. Scopri come vengono elaborati i dati derivati dai commenti.

%d blogger hanno fatto clic su Mi Piace per questo: