SICP Exercise 1.3 request for comments

Question

I\'m trying to learn scheme via SICP. Exercise 1.3 reads as follow: Define a procedure that takes three numbers as arguments and returns the sum of the squares of the two la

Answer 1

I did it with the following code, which uses the built-in min, max, and square procedures. They're simple enough to implement using only what's been introduced in the text up to that point.

(define (sum-of-highest-squares x y z)
   (+ (square (max x y))
      (square (max (min x y) z))))

Answer 2

;exercise 1.3
(define (sum-square-of-max a b c)
  (+ (if (> a b) (* a a) (* b b))
     (if (> b c) (* b b) (* c c))))

Answer 3

I think this is the smallest and most efficient way:

(define (square-sum-larger a b c)
 (+ 
  (square (max a b))
  (square (max (min a b) c))))

Answer 4

What about something like this?

(define (p a b c)
  (if (> a b)
      (if (> b c)
          (+ (square a) (square b))
          (+ (square a) (square c)))
      (if (> a c)
          (+ (square a) (square b))
          (+ (square b) (square c)))))

Answer 5

Using only the concepts presented at that point of the book, I would do it:

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

(define (sum-of-squares x y) (+ (square x) (square y)))

(define (min x y) (if (< x y) x y))

(define (max x y) (if (> x y) x y))

(define (sum-squares-2-biggest x y z)
  (sum-of-squares (max x y) (max z (min x y))))

Answer 6

I've had a go:

(define (procedure a b c)
    (let ((y (sort (list a b c) >)) (square (lambda (x) (* x x))))
        (+ (square (first y)) (square(second y)))))