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 larger numbers. Please comment on how I can improve my solution.

(define (big x y)
    (if (> x y) x y))

(define (p a b c)
    (cond ((> a b) (+ (square a) (square (big b c))))
          (else (+ (square b) (square (big a c))))))

Answer 1

big is called max. Use standard library functionality when it's there.

My approach is different. Rather than lots of tests, I simply add the squares of all three, then subtract the square of the smallest one.

(define (exercise1.3 a b c)
  (let ((smallest (min a b c))
        (square (lambda (x) (* x x))))
    (+ (square a) (square b) (square c) (- (square smallest)))))

Whether you prefer this approach, or a bunch of if tests, is up to you, of course.

---

Alternative implementation using SRFI 95:

(define (exercise1.3 . args)
  (let ((sorted (sort! args >))
        (square (lambda (x) (* x x))))
    (+ (square (car sorted)) (square (cadr sorted)))))

As above, but as a one-liner (thanks synx @ freenode #scheme); also requires SRFI 1 and SRFI 26:

(define (exercise1.3 . args)
  (apply + (map! (cut expt <> 2) (take! (sort! args >) 2))))

Answer 2

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 3

(define (sum a b) (+ a b))
(define (square a) (* a a))
(define (greater a b ) 
  ( if (< a b) b a))
(define (smaller a b ) 
  ( if (< a b) a b))
(define (sumOfSquare a b)
    (sum (square a) (square b)))
(define (sumOfSquareOfGreaterNumbers a b c)
  (sumOfSquare (greater a b) (greater (smaller a b) c)))

Answer 4

;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 5

(define (sum-sqr x y)
(+ (square x) (square y)))

(define (sum-squares-2-of-3 x y z)
    (cond ((and (<= x y) (<= x z)) (sum-sqr y z))
             ((and (<= y x) (<= y z)) (sum-sqr x z))
             ((and (<= z x) (<= z y)) (sum-sqr x y))))

Answer 6

You can also sort the list and add the squares of the first and second element of the sorted list:

(require (lib "list.ss")) ;; I use PLT Scheme

(define (exercise-1-3 a b c)
  (let* [(sorted-list (sort (list a b c) >))
         (x (first sorted-list))
         (y (second sorted-list))]
    (+ (* x x) (* y y))))