;; ====================================================================
;; SICP - 1.1: The Elements of Programming
;; ====================================================================

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


(define (sum-of-squares a b)
  (+ (square a) (square b)))

;; (define (abs x)
;;   (cond ((< x 0) (- x))
;;         ((= x 0) 0)
;;         ((> x 0) x)))

(define (abs x)
  (if (< x 0) (- x)
      x))

(define (average x y)
  (/ (+ x y) 2))

(define (sqrt x)
  (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 (good-enough? guess x)
    (< (abs (- (square guess) x)) 0.001))

  (sqrt-iter 1.0 x))


;; ====================================================================
;; *Exercise 1.1:* Below is a sequence of expressions.  What is the
;; result printed by the interpreter in response to each expression?
;; Assume that the sequence is to be evaluated in the order in which
;; it is presented.
;; -------------------------------------------------------------------

10
;; 10

(+ 5 3 4)
;; 12

(- 9 1)
;; 8

(/ 6 2)
;; 3

(+ (* 2 4) (- 4 6))
;; 10

(define a 3)
;; 3

(define b (+ a 1))
;; 4

(+ a b (* a b))
;; 19

(= a b)
;; #f

(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:* Translate the following expression into prefix
;; form.
;;
;;      5 + 4 + (2 - (3 - (6 + 4/5)))
;;      -----------------------------
;;             3(6 - 2)(2 - 7)
;; -------------------------------------------------------------------

(define ex1.2
  (/ (+ 5 4 (- 2 (- 3 (+ 6 (/ 4 5)))))
     (* 3 (- 6 2) (- 2 7))))

;; ===================================================================
;; *Exercise 1.3:* Define a procedure that takes three numbers as
;; arguments and returns the sum of the squares of the two larger
;; numbers.
;; -------------------------------------------------------------------

(define (ex1.3 a b c)
  (cond ((and (> b a) (> c a)) (sum-of-squares b c))
        ((and (> a b) (> c b)) (sum-of-squares a c))
        (#t (sum-of-squares a b))))

(define (ex1.3-fancy a b c)
  (let* ((sorted (sort (list a b c) >))
         (x (car sorted))
         (y (cadr sorted)))
    (+ (square x) (square y))))

;; ===================================================================
;; *Exercise 1.4:* Observe that our model of evaluation allows for
;; combinations whose operators are compound expressions.  Use this
;; observation to describe the behavior of the following procedure:
;;
     (define (a-plus-abs-b a b)
       ((if (> b 0) + -) a b))
;; -------------------------------------------------------------------

;; When the function is called, the following will happen:
;;
;;   * The first expression in the list, (if (> b 0) + -) will be
;;     evaluated. Within it, (> b 0) will be evaluated first, and
;;     based on the value of b, the result of the evaluation will be +
;;     or -.
;;   * The remaining expressions (a and b) will be evaluated to their
;;     passed-in values.
;;   * The resulting expression will be evaluated, e.g. (+ 3 2)
;;   * The final result will be the result of applying the + or -
;;     operator to the operands a and b

;; ===================================================================
;; *Exercise 1.5:* Ben Bitdiddle has invented a test to determine
;; whether the interpreter he is faced with is using
;; applicative-order evaluation or normal-order evaluation.  He
;; defines the following two procedures:
;;
     (define (p) (p))

     (define (test x y)
       (if (= x 0)
           0
           y))
;;
;; Then he evaluates the expression
;;
;;      (test 0 (p))
;;
;; What behavior will Ben observe with an interpreter that uses
;; applicative-order evaluation?  What behavior will he observe with
;; an interpreter that uses normal-order evaluation?  Explain your
;; answer.  (Assume that the evaluation rule for the special form
;; `if' is the same whether the interpreter is using normal or
;; applicative order: The predicate expression is evaluated first,
;; and the result determines whether to evaluate the consequent or
;; the alternative expression.)
;; -------------------------------------------------------------------

;; Applicative order evaluation will evaluate test, 0 and (p), then
;; evaluate the application of the operator test on its
;; operands. However, attempting to evaluate (p) will hang, as it is a
;; recursive function that never exits.
;;
;; Normal order evaluation will first apply the operator test on its
;; operands, which will then evaluate 0 in the if statment. The
;; conditional expression will succeed, and so the function will
;; return 0, never evaluating (p).

;; ===================================================================
;; *Exercise 1.6:* Alyssa P. Hacker doesn't see why `if' needs to be
;; provided as a special form.  "Why can't I just define it as an
;; ordinary procedure in terms of `cond'?" she asks.  Alyssa's friend
;; Eva Lu Ator claims this can indeed be done, and she defines a new
;; version of `if':
;; 
;;      (define (new-if predicate then-clause else-clause)
;;        (cond (predicate then-clause)
;;              (else else-clause)))
;; 
;; Eva demonstrates the program for Alyssa:
;; 
;;      (new-if (= 2 3) 0 5)
;;      5
;; 
;;      (new-if (= 1 1) 0 5)
;;      0
;; 
;; Delighted, Alyssa uses `new-if' to rewrite the square-root program:
;; 
;;      (define (sqrt-iter guess x)
;;        (new-if (good-enough? guess x)
;;                guess
;;                (sqrt-iter (improve guess x)
;;                           x)))
;; 
;; What happens when Alyssa attempts to use this to compute square
;; roots?  Explain.
;; -------------------------------------------------------------------

;; Calls to sqrt-iter will recurse indefinitely. This is because both
;; the then-clause and the else-clause passed to new-if will be
;; evaluated before the function new-if is applied.

;; ===================================================================
;; *Exercise 1.7:* The `good-enough?' test used in computing square
;; roots will not be very effective for finding the square roots of
;; very small numbers.  Also, in real computers, arithmetic operations
;; are almost always performed with limited precision.  This makes
;; our test inadequate for very large numbers.  Explain these
;; statements, with examples showing how the test fails for small and
;; large numbers.  An alternative strategy for implementing
;; `good-enough?' is to watch how `guess' changes from one iteration
;; to the next and to stop when the change is a very small fraction
;; of the guess.  Design a square-root procedure that uses this kind
;; of end test.  Does this work better for small and large numbers?
;; -------------------------------------------------------------------

;; ===================================================================
;; *Exercise 1.8:* Newton's method for cube roots is based on the
;; fact that if y is an approximation to the cube root of x, then a
;; better approximation is given by the value
;; 
;;      x/y^2 + 2y
;;      ----------
;;          3
;; 
;; Use this formula to implement a cube-root procedure analogous to
;; the square-root procedure.  (In section *Note 1-3-4:: we will see
;; how to implement Newton's method in general as an abstraction of
;; these square-root and cube-root procedures.)
;; -------------------------------------------------------------------