Re: Word lists for passphrases

["Perry E. Metzger" <perry@piermont.com>]

AwakenToMe@aol.com, in a profound display of stubbornness, continues
to insist that his program to enumerate all possible words of length N
(that is, aaaaa, aaaab, aaaac, etc.) is somehow interesting. I am
therefore forced to drive in the nail with a sledgehammer. Forgive me.

He writes:
> > > > It's [...] trivial enough to be done by 99% of the people on
> > > > cypherpunks in their sleep.
> > >  
> > > Yes. But let me ask you this. Have you done it yet??
> >
> > Most of us don't bother writing up four line programs and shipping
> > them out, no.
>  
> really? Wow. four lines of code? You must be a really good programmer. duh.

Hardly. A ten year old could do it. I know, since I wrote substantially more
sophisticated stuff when I was ten.

Since you insist, here is less than a minute's work. Yes, I timed it.

------Cut Here------
/*
   This could be more elegant, but the point is obviousness.
*/
#include <stdio.h>

int main()
{
	char i[6];

	for (i[0] = 'a'; i[0] < 'z'; i[0]++)
	  for (i[1] = 'a'; i[1] < 'z'; i[1]++)
	    for (i[2] = 'a'; i[2] < 'z'; i[2]++)
	      for (i[3] = 'a'; i[3] < 'z'; i[3]++)
		for (i[4] = 'a'; i[4] < 'z'; i[4]++)
		  printf("%s\n", i);
}
------Cut Here------

The operative portion of the program is six lines ling, and five of
those lines are virtually identical.

You can write the thing much more elegantly, without redundant
code. However, I have elected to leave it as utterly brainless as
possible to demonstrate that ANYONE could write the thing.

> Youd be surprised at the # of requests from people who actually had a good
> use for it, and didnt have the time to spend writing it themselves.

Human stupidity is never a surprise.

Perry