Exercise 1.3. Define a procedure that takes three numbers as arguments and returns the sum of the squares of the two larger numbers.
I used the definitions of
sum-of-squares that are in the text:
(define (square x) (* x x)) (define (sum-of-squares x y) (+ (square x) (square y)))
My First Solution
This procedure to find the sum of the squares of the two larger numbers in a triple uses
cond, which was introduced in this section of the book. The first line of the
cond checks if
a is both <=
b and <=
c. If so,
c will be the two larger numbers, so we call
sum-of-squares with them. If that first line isn’t true, then we know that
a is one of the two larger numbers so we can just check if
c. If it is, then we can pick
c for the two larger numbers to use in
sum-of-squares. If not, then the last case is that
b are the two larger numbers, and we use them in
sum-of-squares. Note that if two or three of the input numbers are the same, it doesn’t matter which we pick since the sum of the squares will be the same.
(define (sum-of-squares-of-two-larger-numbers a b c) (cond ((and (<= a b) (<= a c)) (sum-of-squares b c)) ((<= b c) (sum-of-squares a c)) (else (sum-of-squares a b))))
A More Readable Version
Someone pointed out to me that my first attempt isn’t very readable. That’s why I included a paragraph of explanation before it. It would be better to have the program clear enough that explanation isn’t necessary.
For my next version, I thought about what’s happening in the procedure. Two things:
- finding the larger two of the three given numbers
- finding the sum of the squares of those two numbers
So I wrote procedures to do those things separately, splitting up the first thing further into finding the largest number and finding the second largest number.
(define (sum-of-squares-of-two-largest-numbers x y z) (sum-of-squares (largest-of-three-numbers x y z) (second-largest-of-three-numbers x y z))) (define (largest-of-three-numbers x y z) (max (max x y) (max y z))) (define (second-largest-of-three-numbers x y z) (if (< x z) (min (max x y) z) (max (min x y) z))) (define (min x y) (if (< x y) x y)) (define (max x y) (if (> x y) x y))