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?

Comments

Your keyspace wouldn't be that much bigger

[Profane MuthaFucka]

Seems like encrypting twice with a 64-bit key just means that instead of breaking a 64-bit key once, you have to do it twice. That's like using a 65-bit key, instead of a 64-bit key. Every bit doubles your keyspace.

Plus, you have to realize that the amount of time needed to brute force a single key of a certain size goes up non-linearly with each additional bit. If you just double the number of times you encrypt, you're pretty much just increasing the effort to brute force the thing linearly.

Re:Your keyspace wouldn't be that much bigger

[Kadin2048]

I think a big part of that comes down to the file headers and how you actually implement the cryptographic algorithms into a system.

If you take a plaintext file and encrypt it into a file which has headers ("BEGIN ENCRYPTED CONTENT---"), and then encrypt the result again, assuming the attacker knows how you did it and that the intermediate file has plaintext headers, then they'll know the moment they broke the first 64-bit encryption layer. So in this example, you're basically at 65 bits.

Now if you don't include any headers, so that there's no terribly good way to determine whether you've gotten the right key or not, as you're brute-forcing the first layer, then I think you're right -- the strength of the overall system is somewhere in a grey area between 65 and 128 bits.

If someone was just thinking that they could use a file-encryption utility twice (which produces output files that have plaintext headers) and double the keyspace, they are dead wrong.

Easy

[suricatta]

Each bit doubles the strength

Re:Bad math

[thefirelane]

In case anyone missed it though, there was a good insight in that humor: a clear way of expressing the answer to the posed question:

If two 64 bit encryptions equals one 128 bit, then 128 one-bit encryptions should also.

This means the file would have a password of either 1 or 0 ... you would be prompted for this password, and if you got it wrong, you could try again. Obviously you'd get it on the second try. You could repeat this 128 times, and that would be the total of all protection. Naturally, this means the most number of guesses you'd need to make is 256.

After thinking about it that way, it becomes quite clear: 128 bit encryption obviously requires more than 128 guesses to get the right key. As another poster pointed out, two 64 bit passes equals 65 bit encryption.

Re:Debunking this claim

[Coryoth]

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.

It isn't very common for the resulting encryption to be less secure, but "no appreciable improvement" is certainly very possible.

With block ciphers you can run into the issue that if the cipher scheme forms a group (as some do) then even if you use 2 different keys for each encryption round, there will be a single third key that provides identical encryption: an attacker never needs to break your two keys, they can simply find the single third key that is equivalent to your double encipherment. Depite encrypting twice with two different keys the result is no more secure than encrypting once. For a simple example of this, consider a ROT encryption scheme - if you encrypt a message with ROT9 then ROT12 that is just equivalent to ROT21, so instead of trying to find the two keys 9 and 12, the attacker can just solve the ROT21 problem (the worst case, of course, is if your combination of keys gives the identity element such as ROT9 and ROT17, or ROT13 twice, in which case you end up with an unencrypted document).

Even if that's not the case you can still get caught with a meet-in-the-middle approach if the attacker has some known plaintext (that is, they have some plaintext with the corresponding encrypted text and are attempting to extract your key) which does pretty much what it says, encrypting one way and decrypting the other to meet in the middle. Using such a scheme you can extract the keys with only marginally more effort than for single encryption.

Intuition doesn't work well in crypto

[billstewart]

Any sentence beginning with Seems like encrypting twice is likely to be doomed to bogosity, unless there's a later clause in it like but that's not what really happens. Crypto not only depends on a lot of deep math, it also depends on a lot of cryptographers spending time following bits around to see where they go and how to break them.

Sometimes things do what your mathematical intuition tells you, if you're mature enough in the field to have a solid intuition, but often they don't. Problems can be very hard if looked at from one direction and very easy (or at least less hard) when looked at from another direction, and a cryptographer's job is to make sure they've checked out _all_ the directions because it only takes one weak one to break something. NP-complete problems are especially that way - they're potentially useful for crypto because there's one easy direction if you know the key, but many problems can't be transformed in a way that you can use the easy path and the attacker's stuck on the hard paths.

But even the bit-twiddly algorithms, like most hash functions or the S-box building blocks inside Feistel-structured algorithms, can often be cracked by people examining them closely enough to find conditions under which they can deduce bits. For instance, MD5 is pretty much busted these days.

And both mathematical crypto and bit-twiddly crypto has to worry about Moore's Law and brute force - some algorithms scale well, so you can double the strength of an N-bit key by adding 1 bit or maybe logN bits, while others don't, or they form groups so that encrypting a message multiple times with an N-bit key still only gives you N bits of strength (leaving aside pathological cases like rot13.)

Re:Meet in the middle attack

[swillden]

The block size is still 64-bit, which means 2^64 possible combinations versus the 2^128 for ideal 128-bit.

When talking about block ciphers, block size and key size are separate things. Generally, if you say algorithm X is an n-bit cipher, you're talking about key size, not block size. Given use of good block chaining modes, ciphers with larger block sizes don't really offer significant additional security.

typically attackers find ways of reducing the effective number of possible combinations for an algorithm. Original DES has been reduced significantly, so triple-DES was designed to improve it

Original DES hasn't really been reduced significantly. There are attacks which reduce the computational complexity to, IIRC ~41 bits, but at the expense of requiring impractical space. All in all, DES as stood up extremely well and its practical strength hasn't been significantly reduced by 30+ years of scrutiny.

The problem 3DES was created to address was a deliberate limitation of the design: The small keyspace. The strongest variant of Lucifer (the original IBM cipher which became DES), used a 128-bit key, but the NSA deliberately had it reduced to a 56-bit key, presumably because they thought they could break the cipher with the smaller key size, but it was still secure enough against others. Advances in computation technology, of course, have put a brute force search of a 56-bit keyspace within reach of run-of-the-mill desktop computers. 3DES addresses this by increasing the effective keyspace to 112 bits.

Briefcases

[ajs318]

You can buy a briefcase with two separate, three-digit combination locks in any stationery store, and it won't cost you a lot of money. Invest in one as a "prop" for this demonstration. Have your co-worker set a combination for the left-hand lock and a different combination for the right-hand lock, lock the case and spin the wheels randomly. Now, there are 1000 possible combinations for the LH lock and 1000 possible combinations for the RH lock, giving you 1000000 possible combinations. If you can try one combination a second, it could take you up to 9 months to get into the case. Right? {If your co-worker actually believes this, perhaps it would be simplest just to have them put to S-L-E-E-P}.

Once you have cracked the left-hand lock {which will take at most 1000 tries - 17 minutes} you are free to concentrate on the RH lock {which also will take at most 1000 tries}. You don't have to try anymore combinations on the LH lock. So you can open the briefcase in a maximum of 2000 tries, or just over half an hour. The key fact that makes this possible is that the two locks are independent: you can undo one while the other remains fastened. The combined keyspace is then the sum, not the product,of the two keyspaces.

If the encryption operation adds a predictable header to the ciphertext {for example, BEGIN OPHIOL ENCRYPTED MESSAGE} {and most of them do} then you have something to look for that will let you know when you have cracked the first layer of encryption -- in effect, popped open the left-hand lock.