Cryptography/Meet In The Middle Attack

An extremely specialized attack, meet in the middle is a known plaintext attack that only affects a specific class of encryption methods - those which achieve increased security by using one or more "rounds" of an otherwise normal symmetrical encryption algorithm. An example of such a compound system is 3DES.

However, to explain this attack let us begin with a simpler system defined as follows: Two cryptographic systems denoted $$encrypt_\alpha$$ and $$encrypt_\beta$$ (with inverse functions $$decrypt_\alpha$$ and $$decrypt_\beta$$ respectively) are combined simply (by applying one then the other) to give a composite cryptosystem. each accepts a 64 bit key (for values from 0 to 18446744073709551615) which we can call $$key_\alpha$$ or $$key_\beta$$ as appropriate.

So for a given plaintext, we can calculate a cryptotext as

$$cryptotext=encrypt_\beta( key_\beta, encrypt_\alpha(key_\alpha,plaintext))$$

and correspondingly

$$plaintext=decrypt_\alpha( key_\alpha, decrypt_\beta(key_\beta,cryptotext))$$

Now, given that each has a 64 bit key, the amount of key needed to encrypt or decrypt is 128 bits, so a simple analysis would assume this is the same as a 128 bit cypher.

However, given sufficient storage, you can reduce the effective key strength of this to a few bits larger than the largest of the two keys employed, as follows.


 * 1) Given a plaintext/cyphertext pair, apply $$encrypt_\alpha$$ to the plaintext with each possible key in turn, generating $$2^{64}$$ intermediate cryptotexts $$cryptotext_1$$$$\rightarrow$$$$cryptotext_n$$ where $$n=2^{64}$$
 * 2) Store each of the $$n$$ cryptotexts in a hash table so that each can be referenced by its cryptotext, and give the key used to generate that cryptotext
 * 3) Apply $$decrypt_\beta$$ to the ciphertext for each possible key in turn, comparing the intermediate plaintext to the hash table calculated earlier. this gives a pair of keys (one for each of the two algorithms employed, $$\alpha$$ and $$\beta$$)
 * 4) Taking the two keys from stage 3, test each against a second plaintext/cryptotext pair. if this also matches, odds are extremely high you have a valid keypair for the message - not in $$2^{128}$$ operations, but a "mere" $$2x2^{64}$$ operations (which nonetheless are significantly longer due to the hash table operations, but not so much as to add more than a couple of extra bits worth of time to the complexity of the task)

The downside to this approach is storage. Assuming you have a 64 bit key, then you will need at least $$2^{64}$$ units of storage - where each unit is the amount of space used by a single hash record. Even given a minimal implementation (say, 64 bits for the key plus four bits hash collision overhead), if you implemented such a system using 160GB hard drives, you would need close to one billion of them to store the hash table alone.