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Solution to Exercise 1.3:
(define (square x)
(* x x))
(define (sum-of-squares x y)
(+ (square x) (square y)))
(define (sum-of-squares-of-two-larger x y z)
(cond ((and (<= x y) (<= x z)) (sum-of-squares y z))
((and (<= y x) (<= y z)) (sum-of-squares x z))
(else (sum-of-squares x y))))
Posted by Ken Dyck in Programming
Alternatively:
(define (sum-of-squares-of-two-larger a b c)
(define (lt a b) (if (> a b) b a))
(define (least a b c) (lt a (lt b c)))
(define (square x) (* x x))
(+ (square a)
(square b)
(square c)
(- (square (least a b c)))))Another, though bad, alternative:
(define (test x y z)
(cond ((> x y) (if (> y z)
(+(* x x) (* y y))
(+(* x x) (* z z)))
(> x z) (+(* y y) (* x x))
(else (+(* z z) (* y y)))—-Thanks for posting your answers to SICP. I’m new to programming and am looking for other, likely better, solutions for this book.
Another alternative:
(define (sum-of-squares-two-largest a b c)
(define (square x) (* x x))
(define (sum-of-squares x y) (+ (square x) (square y)))
(define (greater x y)
(if (> x y) x y))
(if (> a b)
(sum-of-squares a (greater b c))
(sum-of-squares b (greater a c))))
my silly newbie alternative:
(define (square x) (* x x))
(define (trio a b c) (cond (
(and (
ups… whats wrong with my post ?? it get truncated..
Yeah, newbie, it looks that way. Would you mind emailing me your code? I’ll update your comment with your entire solution.
ken when you input the same number into your code it might affect the result. for example when i put this into your code:
(sum-of-squares-of-two-larger 2 2 3)
it respondes “9″.
By the way really great page. I’m so glad to see that so many people out there are working on SICP as well. It’s a really great book.
All so here is my code. For some reason i really feel like this is a round about way of getting things done but it seems to work so…
(define (sum-of-square-of-2-largest-of-3 a b c)
(define (square x)
(* x x))
(define (sum-of-squares t y)
(+ (square t) (square y)))
(define (all-equal?)
(if (and (equal? a b) (equal? a c))
#t
#f))
(define (any-equal?)
(if (or (equal? a b) (equal? b c) (equal? a c))
#t
#f))
(define (largest)
(cond ((and (>= a b) (>= a c)) a)
((and (>= b c) (>= b a)) b)
(else c)))
(define (equal)
(cond ((equal? a b) a)
((equal? b c) b)
(else c)))
(define (2nd-largest)
(cond ((or (and (> a b) ( a c))) a)
((or (and (> b a) ( b c))) b)
(else c)))
(cond ((all-equal?) (sum-of-squares a a))
((any-equal?) (sum-of-squares (largest) (equal)))
(else (sum-of-squares (largest) (2nd-largest)))))
Well isn’t that embarassing? Thanks for pointing that out, b. I’ll make a fix at my next available opportunity.
Great Idea to post some of these Ken. Hope there’s no flack from profs using the book. I too have decided once again to learn this since I’ve read and been told soooooo many times it makes one a much better programmer…which is my job too.
Any way here is my example using the regular sum-of-squares everone else has:
(define (biggest x y) (if (> x y) x y))
(define (smallest x y) (if (Oops seems maybe I need to escape the less than character for this to post without truncating(?).. anyway the body of the procedure is:
(define (sumSq2Largest a b c)
(sum-of-squares (biggest a b) (biggest (smallest a b) c)))Rob, I hope you don’t mind that I formatted your code… but it seems as though it got cut off along the way.
I presume your implementation of smallest is :
(define (smallest x y) (if (< x y) x y))
I really like your solution, btw. Very elegant.
[...] Еще один очень элегантный вариант решения я обнаружил в комментариях к блогу Кена Дика: [...]
(define (maxm a b)
(cond ((> a b) a) (else b)))
(define (sq x) (* x x))
(define (proc a b c)
(if (and (> c a)
(> c b))
(+ (sq c)
(sq (maxm a b)))
(if (and (
This is one more solution to the sum of squares problem:-
(define (sq x)
(* x x))
(define (sq-lrg a b c)
(cond ((and ( a c))(= a b c)) (+ (sq a)))
(else(sq-lrg b c a))
))
Here is the solution the previous one was a little messed up .(forgot the tags !!)
(define (sq x)
(* x x))
(define (sq-lrg a b c)
(cond ((and (<a> a c)) (= a b c)) (+(sq a)(sq a)))
(else (sq-lrg b c a))
))
This is my solution
(define (sum-squares-of-largest-2 x y z)
(cond ((and (
here is my solution:
;procedure square
;given a number, returns the square
(define square
(lambda (x)
(* x x)
))
;procedure biggest-2-out-of-3
;given 3 numbers, returns an unordered vector of the two
;largest
(define biggest-2-out-of-3
(lambda (x y z)
(cond
((>= x y z) (vector x y))
((>= x z y) (vector x z))
((>= z y x) (vector z y))
)))
;procedure sum-squares
;given 2 numbers, returns the sum of their squares
(define sum-squares
(lambda (x y)
(+ (square x) (square y))
))
;procedure sum-squares-of-3
;given 3 numbers, returns the sum of the squares of the
;two largest
(define sum-squares-of-3
(lambda (x y z)
(let*
(
(top-2 (biggest-2-out-of-3 x y z))
(first (vector-ref top-2 0))
(second (vector-ref top-2 1))
)
(sum-squares first second)
)))
;main
(display(sum-squares-of-3 2 3 3))
Argh, the formatting got all messed up. How do you post so that whitespace is preserved?
here is my solution:
;procedure square
;given a number, returns the square
(define square
(lambda (x)
(* x x)
))
;procedure biggest-2-out-of-3
;given 3 numbers, returns an unordered vector of the two
;largest
(define biggest-2-out-of-3
(lambda (x y z)
(cond
((>= x y z) (vector x y))
((>= x z y) (vector x z))
((>= z y x) (vector z y))
)))
;procedure sum-squares
;given 2 numbers, returns the sum of their squares
(define sum-squares
(lambda (x y)
(+ (square x) (square y))
))
;procedure sum-squares-of-3
;given 3 numbers, returns the sum of the squares of the
;two largest
(define sum-squares-of-3
(lambda (x y z)
(let*
(
(top-2 (biggest-2-out-of-3 x y z))
(first (vector-ref top-2 0))
(second (vector-ref top-2 1))
)
(sum-squares first second)
)))
;main
(display(sum-squares-of-3 2 3 3))
(define (exercise a b c)
(define (^2 x) (* x x))
(if (> a b)
(+ (^2 (if (> b c) b c)) (^2 a))
(+ (^2 (if (> a c) a c)) (^2 b)))
This only uses what has been shown up to the moment in the book.
Just using what I learned of lisp from this book so far. And keeping it short as possible:
(define (max_square a b c)
(cond
((>= a b c) (+ (* a a) (* b b)))
((>= a c b) (+ (* a a) (* c c)))
(else (+ (* b b) (* c c)))
))
Here’s my solution:
(define (square x)
(* x x))
(define (f x y z)
(- (+ (square x)(square y)(square z)) (square (min x y z))))
Well, I think that perhaps it’s not pretty good to use approaches that was not explained in the book before this exercise was suggested.
Here is my solution:
(define (sum-max2-squares x y z)
(cond ((>= x y) (+ (* x x) (if (>= y z) (* y y) (* z z))))
(else (+ (* y y) (if (>= x z) (* x x) (* z z))))))