SICP Exercises 1.1 - 1.5
1.1
10, 12, 8, 3, 6, a, b, 19, #f, 4, 16, 6, 16
1.2
(/ (+ 5 4 (- 2 (- 3 (+ 6 (/ 4 5)))))
(* 3 (- 6 2) (- 2 7)))
1.3
$ cat ex1-5.ss
(define (square x)
(* x x))
(define (sum-of-squares x y)
(+ (square x) (square y)))
(define (sum-of-squares-larger-nums x y z)
(cond ((and (< x y) (< x z)) (sum-of-squares y z))
((and (< y z) (< y x)) (sum-of-squares x z))
((and (< z x) (< z y)) (sum-of-squares x y))))
1.4
When b is greater than 0 the + operator is used, otherwise the - operator is used.
1.5
Because applicative-order evaluation evaluates the arguments and then applies them, the procedure p defined as (define (p) (p)) would call itself over and over when called with (test 0 (p)) where the procedure test is defined as (define (test x y) (if (= x 0) 0 y)). Normal-order evaluation however will not have this problem because it does not evaluate the arguments, but rather expands them and reduces them. (test 0 (p)) would simply return 0.
Notes: I see now that I could have simplified 1.3's sum-of-squares-larger-num to use min. It would have made what was going on clearer, but I wrote it using and's first and tested it out. It works so I just left it.
