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(load "displaylib.scm")
(title "Exercise 2.64")
(print "
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))))))))
a. 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).
b. What is the order of growth in the number of steps required by list->tree to convert a list of n elements?
")
; a.
;
; The procedure is recursive:
;
; it first constructs (recursively) the left-tree
; and using the remaining elements of the list
; construct the root of the tree and the right-tree (recursively)
;
; example: (1 3 5 7 9 11)
;
; => ( 1 3 5 7 9 11 ) 6
;
; make-left-tree => (1 3 5 7 9 11) 3 -> (3 (1) (5)) rest 7 9 11
; take-root => 7
; make-right-tree => (9 11) 2 -> (9 (11)), rest ()
;
; result => (7 (3 (1 5)) (9 (11)))
;
; b. Order of growth
;
; recursively: f(n) = c + 2f(n/2)
; f(0) = c
;
; f(0) = c
; f(1) = c + 2f(0) ~ 3c
; f(2) = c + 2f(1) ~ 7c
; f(4) = c + 2f(2) ~ 15c
; etc...
; f(2^n) = ϴ(2^n)
;
; generally => f(n) = ϴ(n)
;
; well played!