ivanjovanovic / sicp Goto Github PK

It's hard to read for me. Many thanks !

; excercise 1.3
; Define procedure that takes three numbers as arguments and returns
; the sum of the squares of the two larger numbers
(define (sum-larger-square a b c)
(cond ((and (< a b) (< a c)) (+ (* b b) (* c c)))
((and (< b a) (< b c)) (+ (* a a) (* c c)))
(else (+ (* a a) (* b b)))))

(sum-larger-square 5 5 6)
50

so, the right answer is :
(define (sum-larger-square a b c)
(cond ((and (<= a b) (<= a c)) (+ (* b b) (* c c)))
((and (<= b a) (<= b c)) (+ (* a a) (* c c)))
(else (+ (* a a) (* b b)))))
(sum-larger-square 5 5 6)
61

I don't know if you are looking for feedback or not. If not, ignore this. I have used your answers when I get stuck or am looking for alternative implementations.

What happens when the duplicates are in list b instead of list a? Example;
(intersection-set (list 1 2 3 1 1 7) (list 2 3 4 1 1 7 7))