SICP Chapter 1 Exercises https://repl.it/CKty/4

sicp-ch1.sc

;;--------------;;
;; Exercise 1.1 ;;
;;--------------;;
10
;=> 10
(+ 5 3 4)
;=> 12
(- 9 1)
;=> 8
(/ 6 2)
;=> 3
(- (* 2 4) (- 4 6))
;=> 10
(+ (* 2 4) (- 4 6))
;=> 6
(define a 3)
(define b (+ a 1))
(+ a b (* a b))
;=> 19
(if (and (> b a) (< b (* a b))) 
    b 
    a)
;=> 4
(cond ((= a 4) 6) 
      ((= b 4) (+ 6 7 a)) 
      (else 25))
;=> 16
(+ 2 (if (> b a) b a))
;=> 6
(* (cond ((> a b) a) 
         ((< a b) b) 
         (else -1)) 
   (+ a 1))
;=> 16

;;--------------;;
;; Exercise 1.2 ;;
;;--------------;;
(/ (+ 5 4 (- 2 (- 3 (+ 6 (/ 4 5))))) (* 3 (- 6 2) (- 2 7)))
;=> -0.24666666666666667

;;--------------;;
;; Exercise 1.3 ;;
;;--------------;;
(define (sq x) (* x x))
(define (sos x y) (+ (sq x) (sq y)))
(define (ex1.4 a b c) 
  (cond ((or (> a b c) (> b a c)) (sos a b)) 
        ((or (> a c b) (> c a b)) (sos a c)) 
        ((or (> b c a) (> c b a)) (sos b c))))

;; Test 
(ex1.4 1 2 3)
;=> 13
(ex1.4 3 2 1)
;=> 13
(ex1.4 2 3 1)
;=> 13

;;--------------;;
;; Exercise 1.4 ;;
;;--------------;;

; Given:
(define (a-plus-abs-b a b)
  ((if (> b 0) + -) a b))

; Explanation: operators are just functions that can be returned 
; from expressions. Thus, if b is positive, it will be added to a, 
; otherwise it will be substracted.

;;--------------;;
;; Exercise 1.5 ;;
;;--------------;;

; Given:
(define (p) (p))

(define (test x y)
  (if (= x 0)
      0
      y))
     

; Observation: (p) is self-referential, and will lead to an infinite loop 
; if evaluated. Therefore, the expression (test 0 (p)) will cause an infinite
; loop if the interpreter evaluates the expression starting from the innermost,
; but it will work fine if it is evaluated only as necessary.
; TODO: which of these is applicative order and which is normal order?


;;--------------;;
;; Exercise 1.6 ;;
;;--------------;;

;Given:
(define (sqrt-iter guess x)
  (if (good-enough? guess x)
      guess
      (sqrt-iter (improve guess x)
                 x)))
             
(define (improve guess x)
  (average guess (/ x guess)))
 
(define (average x y)
  (/ (+ x y) 2))
 
(define (good-enough? guess x)
  (< (abs (- (square guess) x)) 0.001))

(define (square x) (* x x))

; When redefining `sqrt-iter` using `new-if`, it causes an endless loop. Why?
(define (new-if predicate then-clause else-clause)
  (cond (predicate then-clause)
        (else else-clause)))
       
;;--------------;;
;; Exercise 1.7 ;;
;;--------------;;

;;--------------;;
;; Exercise 1.8 ;;
;;--------------;;

; Define a function compute a cube root using Newton's method
; analogous to the iterative square root function
(define (cbrt-iter guess x)
  (if (cb-good-enough? guess x)
      guess
      (cbrt-iter (cb-improve guess x)
                 x)))

(define (cb-improve guess x)
  (/ (+ (/ x (* guess guess)) (* 2 guess)) 3))
 
(define (cb-good-enough? guess x)
  (< (abs (- (* guess guess guess) x)) 0.001))