Comments on: SICP section 2.1.4

By: eliben

eliben — Tue, 30 Sep 2008 12:31:47 +0000

@gorilych,
I’ve fixed the typo, thanks.

By: gorilych

gorilych — Tue, 30 Sep 2008 09:44:46 +0000

Ops. Sorry. It proves. Only one typo:

“if we multiply it by another interval with width 3, say 2,5″

should be

“if we multiply it by another interval with width 1.5, say 2,5″

By: gorilych

gorilych — Tue, 30 Sep 2008 09:42:39 +0000

“Now, to prove that this isn’t true for multiplication, consider the interval 2,4 (width = 1). Multiplied by 7,10 (width = 1.5) we get 14,40 (width = 13). Now, if we multiply it by another interval with width 3, say 2,5 we get 4,20 (width = 8). Therefore, the width of the multiplication doesn’t depend only on the widths of the multiplicands.”

Well. It proves nothing. Think better.

By: eliben

eliben — Fri, 25 Jul 2008 12:12:03 +0000

Wang, you’re right - posting fixed code.

By: Wang Shengyi

Wang Shengyi — Fri, 25 Jul 2008 08:39:23 +0000

I don’t think Exercise 2.12 is correct. When center is negative and p is positive, the make-center-percent will construct an error interval. Then the result of percent is negative too.

By: eliben

eliben — Sun, 13 Jul 2008 16:00:49 +0000

Peter,
Thanks for the tip - I added 0 into the range check.

By: Peter Michaux

Peter Michaux — Fri, 11 Jul 2008 05:44:24 +0000

I think that Exercise 2.10 is still not correct. For example, the following passes the current conditional check suggested by David Clark, yet this causes a divide by zero error.

(div-interval (make-interval 1 2) (make-interval 0 1))

I used the following helper predicate in my code

(define (contains-zero? interval)
(and (<= (lower-bound interval) 0)
(>= (upper-bound interval) 0)))

By: eliben

eliben — Sun, 06 Jul 2008 16:11:53 +0000

David,

You’re right, and I fixed the relevant code snippet. Thanks !

By: David Clark

David Clark — Sat, 05 Jul 2008 21:22:07 +0000

I don’t think that Exercise 2.10 is correct. They want you to check if the y interval spans zero, not if the upper and lower bounds are zero. The correct test should be

(and (> (upper-bound y) 0) (< (lower-bound y) 0))

The reason for this is because the make-interval constructor would then make an illegal interval, the lower bound would be greater than the upper bound, which does not make sense.

Your solution catches a divide by zero error, but since any Lisp will already error on that condition, that check is already implicit.