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Debunking a Bogus Encryption Statement?
deviantphil asks: "Recently, a coworker tried to assert that encrypting a file twice with a 64 bit algorithm is equivalent to encrypting it once with a 128 bit algorithm. I know enough about encryption to know that isn't true, but I am having difficulties explaining why and how. Doesn't each pass of the encryption create a separate file header which makes this assertion untrue? Can anyone point me to references that would better help me explain this?" What other laughable claims have you heard attributed to encryption, and how were you able to properly lay them to rest?
http://en.wikipedia.org/wiki/Meet-in-the-middle_at tack
That's why we have triple-DES instead of double-DES.
The length of the key (64-bits, 128-bits) at a simple level describes the complexity or "hardness" of the encryption. Enrcypting with a 64-bit key will give you one level of "hardness" and 128-bit will give you another, greater, level of hardness.
If you encrypt twice with a 64-bit key you will just be taking a plain text file and encrypting it and then taking your encrypted file and encrypting it once again. To break encryption on this file, getting access to the plain text, you would need to break 64-bit encryption twice.
While this would be more of a challenge than breaking 64-bit standalone encryption, it wouldn't be equivalent to the security of a 128-bit key. The difficulty of breaking 128-bit encryption isn't 2 times 64-bit encryption. It is actually greater - I can't quantify it without additonal research, hopefully someone who can recall it off the top of their head can chime in with the details.
Now, breaking a doubly encrypted file would present some interesting challenges. The most obvious is that your plaintext for the first layer of encryption is anything but plaintext. Confirming that one set of encrypted text is right one vs. some other set of encrypted text would definitely be a challenge. It would make it difficult to determine if you had found the right key to decrypt your first layer of encryption.
As an aside, I do remember reading about code systems where double encryption acutally made the result encryption less secure. I don't remember the details, but now my brain is itching and I will have to do some research. Thanks!
Yours,
Jordan
I'm going to think of it as if you were trying to bruteforce it.
If you have 64 bits, that is 1.84467441 × 10^19 (2^64), meaning maximum that many tries to break the first layer of encryption. The second layer is the same number, meaning to break it would mean a maximum of 3.68934881 × 10^19 attempts.
With 128 bit encryption, there are 3.40282367 × 10^38 (2^128) different possibilities, MUCh more than the double 64 bit encryption.
Obviously people don't usually bruteforce encrypted files, but you can see there are much more possiblities for 128 bit encryption than with double 64 bit.
echo YOUR_OPINION >
- Applied Cryptography: Protocols, Algorithms, and Source Code in C by Bruce Schneier
- Handbook of Applied Cryptography by Alfred J. Menezes et al.
Either one will explain such things as why double-DES is not twice as strong as DES and common pitfalls in thought and design of cryptographic systems.Let me see if I can explain this in a few paragraphs....
A single encryption key defines a one-to-one mapping between each plaintext block and ciphertext block. (In ECB (electronic codebook) mode, but the chained block encoding follows a similar analysis.) It has to be bidirectional so that the decryption is unique.
This means that, AT A MAXIMUM, your meaningful keyspace can be no larger than the 2^(number of bits in block). In practice it takes a lot of hard work to ensure that you have 2^(number of bits in key) distinct and non-trivial mappings, and ideally for any arbitrary plaintext and ciphertext block you can guarantee that there is a unique key that will produce that result. However it's easy to screw up and only have, e.g., 2^40 actual encryptions even though you have a keysize of 64 or even 128 bits.
Double encryption also defines a one-to-one mapping between each plaintext block and ciphertext block... but you have the same blocksize. If the keys form a "group", then any double encryption will be equivalent to a single encryption with a different key. In fact any N-encryptions will be equavalent to encyrption with a single different key. Some combinations will leave you with no encryption. (XOR encryption is a trivial example of this -- you'll get no encryption whenever you have an even number of encryptions.)
Worse, your keys may not form a group because there's some property that gets cancelled out in a double encryption. E.g., maybe something drops out each time both keys have a '1' in the same spot. In this case double encryption would be significantly weaker than single encryption.
Triple-DES encryption is an oddball since it isn't (or wasn't) known whether DES keys form a group. Remember that a DES key is only 56 bits, but a DES cipherblock is 64 bits -- you know that can't get every possible mapping between plaintext and ciphertext by running through the keyspace.
Hence 3DES. However 3DES isn't simply encryption three times, e.g., the form we studied in my grad class used two (not three) 56-bit keys, in an EDE pattern. That is, you encrypt with the first key, then DECRYPT with the second key, then encrypt with the first key again. I didn't follow all of the reasoning but I seem to recall it was because of some property of the S and P boxes.
For every complex problem there is an answer that is clear, simple, and wrong. -- H L Mencken
This actually raises a question to which I don't know the answer: if you take a fairly standard symmetric cypher, say DES, and two keys K1 and K2, does there always exist a key K3 such that E_K1(E_K2(Message)) == E_K3(Message) ?
No, this is not always the case. For DES it is certainly NOT the case. The terminology you are looking for is "idempotent cipher." A cipher is idempotent if it obeys the equation you stated -- i.e., for any key pair K1,K2, there is a K3 such that E_K2(E_K1(x))=E_K3(x). DES is NOT an idempotent cipher.
These questions only seem deep to non-crypto-experts (no offense to you intended) -- and it was certainly accounted for in the design of DES (along with features which harden it against differential cryptanalysis, a technique that wasn't even publically known outside the NSA at the time -- the people who work on these things are far from stupid)
Wouldn't you have to do the "inner" 64bit bruteforce procedure for each possible key of the "outer" 64bit keyspace, thus making it 128bit again?
No, you do a meet-in-the-middle attack, which is basically 2^64 in complexity if you're using two 2^64 keys.
There are some optimizations that can be done, but the basic idea is this: You start with one ciphertext block and its corresponding known plaintext. Then you encrypt the plaintext with all 2^64 possible keys and store the results (with some indexes to make it easy to search for a particular block). Then you decrypt the ciphertext with all 2^64 possible keys, looking each result up. When you find a match, you have found the pair of 64-bit keys that turns that plaintext into that ciphertext. So to search the entire space, you have to do 2^64+2^64 = 2^65 trial operations. On average, you only have to search half of the second keyspace, so the complexity of the search is 2^64+2^63 = 2^64 (plus the huge storage cost).
Triple encryption is also weakened by a meet-in-the-middle attack. Using three 64-bit keys, it would be nice to think you have a 192-bit key. But a meet-in-the-middle requires encrypting with all 64-bit keys for the first step, and decryption with all 128-bit keys for the second step, giving an effective strenth of 2^64+2^127 = 2^127 (plus the huge storage cost).
Finally, keep in mind that DES keys aren't 64 bits. They're 56 bits. So 3DES has a brute force search computational complexity of 111 bits, on average.
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10x improvement over the last 10 years is about right (give or take 2 years).
A Celeron 566MHz has a memory bandwidth around 150-175MB/s. An Athlon64 3200+ has a memory bandwidth around 1600MB/s. (Approximate numbers, with all sorts of undocumented assumptions.) So it's a pretty good bet that modern single-core CPUs are at least 10x as fast as the ones from 8-10 years ago.
Of course, now we have things like dual/quad CPU machines with 2-4 cores per machine... so we could up that number to 150x as fast as a machine from 8-10 years ago. So instead of 3000 machines, we might only need 20-30.
Wolde you bothe eate your cake, and have your cake?