Comments on: How to be more confident about your own programs: an example using Perl

By: jff

jff — Thu, 26 Jul 2007 08:25:53 +0000

Thanks Rob! I’ve added the proof for your solution to the post. I don’t understand something in your argument, though. In equation 3, you say that n = y. But n is a natural and y is a real. And then you use this for equation 4. I believe the solution is correct, but I don’t understand your proof.

Regarding the less-than in comments, I’ll try to fix it. Sorry about that!

By: Rob Mayoff

Rob Mayoff — Thu, 26 Jul 2007 02:48:30 +0000

My previous comment got completely hosed because the blog software threw away every less-than-symbol and all following characters until the next greater-than symbol. Oh well. The point is that ceil(x) = -floor(-x).

By: Rob Mayoff

Rob Mayoff — Thu, 26 Jul 2007 02:44:12 +0000

There’s a simpler (IMHO) definition of ceil(x) in terms of floor(x):

For every integer n, n = -floor(-y). (eq. 2)
By algebra, n = y. (eq. 3)
Substitute eqs. 2 and 3 into eq. 1: for every integer n, -n >= -floor(-y) iff -n >= y. (eq. 4)
If predicate p(-n) is true for all integers n, then p(n) is also true for all integers n since both n and -n are integers. So replace -n with n in eq. 4.
For every integer n, n >= -floor(-y) iff n >= y.
For every integer n, n >= ceil(y) iff n >= y. (def’n of ceil)
Therefore ceil(y) = -floor(-y).

In perl:

sub ceil {
return ($_[0]