Skye asks (upon brute-force attacks):

If so, how about double-encrypting a file with two completely different and very complex programs? Then, even if it did get the first, it couldn't tell because the resulting data would still be largely gobbledegook.

Maybe. The question is the same as the mathematical question "does the encryption algorithm form a group?". "Groupness" refers to whether two applications of an encryption can be collapsed (by some arbitrary key) into a single application of the same encryption. [or, for two differing encryptions, a single application of some algorithm either less complex than the sum of the two original encryptions, or using a key shorter than the two original keys...] For example, consider Caesar rotations. Here, the key is just a number from 0 to 26 and rot13 (rotation by 13, a->n, being the USENET standard for encrypting dirty jokes). We can "collapse" any pair of Caesar rotations into a new single rotation; it's just rotate for the sum of the two keys. So, Caesar rotations form a group, and it does no good to encrypt twice, because brute force needs to solve only one problem, not two, as combinatorics would suggest. But what about something more... interesting? Say, a Caesar rotation followed by a N-skipped version of the alphabet (for N=1, this is the identity alphabet, for N=2, the alphabet is "a,c,e,g,i,k,m,o,q,s,u,w, y,b,d,f,h,j,l,n,p,r,t,v,x,z", for N=3, it's "a,d,g,...".) Now, there's no possibility of collapsing the two encryptions into one operation; no Caesar rotation can give any of the N-skip alphabets (except the trivial case of N=0), and most pairings of Caesar rotations followed by skipping alphabets cannot be faked by either a Caesar rotation or a skip-alphabet alone. Thus, we can say that Caesar followed by N-skip "does not form a group" and so is as hard to crack by brute force as combinatorics suggest. Back in the early days of DES, it was not known if DES encryption followed by another DES encryption formed a group. That's why triple DES encryption was designed to use an intermediate DEcryption (not encryption) stage, so that even if double-DES-encryption formed a group, encryption/decryption/encryption would not (since it's possible to DES-encrypt any possible message stream, therefore some set of cyphertext bits corresponds to some possible plaintext, and that plaintext can be reencoded) and so it would not be possible to collapse the first two operations into a single DES encode, collapse the and the third into yet another single encode and thereby save much time for the brute force attack. However, it's now been proven that DES encode followed by DES encode does NOT form a group, and so it doesn't really matter any more.

Probably a stupid question, but I was curious.

No, it's an *excellent* question. -Bill