Hi,
I want to calculate the storage requirements for a possible attack on a cryptographic algorithm known as 2DES. The algorithm is given below:
The algorithm as described in (a Cryptography book) . It is based on the observation that, if we have
C = E(K2, E(K1, P))
then (see Figure 7.1a)
X = E(K1, P) = D(K2, C)
Given a known pair, (P, C), the attack proceeds as follows. First, encrypt P for all
2^56 possible values of K1. Store these results in a table and then sort the table by the
values of X. Next, decrypt C using all 256 possible values of K2. As each decryption
is produced, check the result against the table for a match. If a match occurs, then
test the two resulting keys against a new known plaintext–ciphertext pair. If the two
keys produce the correct ciphertext, accept them as the correct keys.


Storage for:
Given a known pair, (P, C), the attack proceeds as follows. First, encrypt P for all
2^56 possible values of K1- 2^56 units of storage
If each hash key takes n bytes of data then: n * 2^56
For decryption, we are not storing anything in the hard disk.

Is the above result correct? Some body please guide me.

Zulfi.

 

Where is this hash coming into play?

How long is X?

How long is each key?

What needs to be stored for each entry?

 

Hi,
Thanks a lot. There are 2^56 key, and each key requires 7 bytes (b/c each key is of 56 bits). For each key we have to create cipher text so there are 2^56 copies of cipher text and & if each cipher text is of 8 bytes then:

8* 7 * 2^56 * 2^56 bytes of storage

2^40 = 1 Tera Byte

Around 816 Tera bytes.

Zulfi.