Skip to content

Latest commit

 

History

History
142 lines (107 loc) · 2.65 KB

chapter1.md

File metadata and controls

142 lines (107 loc) · 2.65 KB

Exercise 1.1

> 10
10
> (+ 5 3 4)
12
> (- 9 1)
8
> (/ 6 2)
3
> (+ (* 2 4) (- 4 6))
6
> (define a 3)
> (define b (+ a 1))
> (+ 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 1.4

Given

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

The combination

(a-plus-abs-b 2 -3)

which evaluates to

((if (> -3 0) + -) 2 -3)

which evaluates to

((if #f + -) 2 -3)

which evaluates to

(- 2 -3)

which evaluates to

5

Exercise 1.5

Given

(define (p) (p))

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

Using applicative-order the combination (test 0 (p)) evaluates to (test 0 (p)) indefinately, resulting in an infinite loop.

Using normal-order evaluation, the combination (test 0 (p)) evaluates to

(if (= x 0) 0 (p))

which evaluates to

(if #t 0 (p))

which evaluates to

0

Exercise 1.6

The third operand of the new-if procedure will be evaluated before calling the procedure itself. This triggers another evalution of sqrt-iter, which in turn evaluates the third operand of new-if, etc. This causes unlimited recursion, Scheme's version of an endless loop. So the program will hang indefinately.

Exercise 1.7

The procedure good-enough? will accept the guess whenever the guess lies between (sqrt (- x 0.001)) and (sqrt (+ x 0.001)). When x is very small, the tolerance will dominate the equation of this accepted range and x is not part of the acceptance decision anymore. The guess will be accepted when it is close enough to (sqrt 0.001).

E.g. (sqrt 0) evaluates to 0.03125. Interestingly, evaluating (square (sqrt 0)) evaluates to 0.0009765625 which is indeed very close to the tolerance 0.001.

The procedure good-enough? will not accept the guess while its square is not closer than 0.001 from x. High numbers might not be representable with a preciesion of 0.001, so the guess will never be accepted.

E.g. evaluating (sqrt 1e13) hangs, while (sqrt 1e12) evaluates to 1e6.

Note that in the improved version (sqrt 0) will hang. This is because for 0 the next guess will always be half the previous guess and therefore the change will always be equal to the guess. This means that this fraction will always be 1, which is greater than the chosen tolerance of 0.001. Note that (sqrt 1e-323) can still be evaluated but (sqrt 1e-324) also hangs.