Hacking
Wailin' Through The Nets
banshee at ucscb.UCSC.EDU
Tue Apr 2 14:44:43 AEST 1991
I seem to recall pkr at sgi.com (Phil Ronzone) saying:
>
>Well, assuming 100,000 words of 13 characters each, for each of 4096
>possibilities, that gives us 5,324,800,000 bytes. Now, with the
>750MB and 1.2G 5.25" disk drives around, I'd probably have to have
>several of the drives, OR, maybe use a smaller dictionary.
>
>100,000 words is a LOT of words .....
>
Don't be silly. Talk about 25000 words x 4096 permutations x 13 bytes.
That gives you 1.3 x 10^9 bytes. Now figure this is all ascii and you
can expect at least 50% compression ratios. So say 6.6 x 10^8 bytes.
That'll fit on my gigabyte drive with room to spare. Wanna tell me that
the NSA _doesn't_ have this kind of thing precomputed?
--
The Wailer at the Gates of Dawn | johnv at metaware.com |
#include <vomit.h> | uunet!metaware!johnv |
void puke(struct dinner *p) { free(p); } | banshee at ucsc{b,f}.UCSC.EDU |
2,3,5,7,13,17,19,31,61,89,107,127,521,607....| banshee at ucsc{b,f}.BITNET |
More information about the Comp.unix.wizards
mailing list