Re: Please Cure My Ignorance (I know that may be difficult :-)) onHash Functions

From: Benjamin Goldberg (goldbb2@earthlink.net)
Date: 02/20/03 

From: Benjamin Goldberg <goldbb2@earthlink.net>
Date: Thu, 20 Feb 2003 00:51:05 -0500

"Douglas A. Gwyn" wrote:
>
> TC wrote:
> > I used that fact to question the icy certaintly of the statement:
> > "if there are more possible inputs than outputs, multiple inputs
> > must map to the same output". The joke is, there is *no* output!
>
> No joke, you're just wrong. For that particular function,
> *all* inputs map to the same output. To disprove that,
> exhibit two inputs that have different outputs; can you?

That's too difficult a proof -- how about I turn it around, ok?

To demonstrate that it's *possible* for all inputs to map to the same
output, you would have to exhibit two inputs that have the same output.

Considering that the body of the so-called "function" was nothing more
than while(true){no-op}, a person would be hard-pressed to produce even
one output, let alone two. 

-- 
$;=qq qJ,krleahciPhueerarsintoitq;sub __{0 &&
my$__;s ee substr$;,$,&&++$__%$,--,1,qq;;;ee;
$__>2&&&__}$,=22+$;=~y yiy y;__ while$;;print