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We know that DES has key length that is too short to defend brute-force attack. One idea to improve DES was to encrypt a message using DES twice with two different keys. Let E be the encryption algorithm of DES and k1, k2 be two DES keys. Such a double DES encryption performs like c = E(k2; E(k1; m)) We might hope now the total key length is 112 bits, enough for defending brute-force attack which should take 2112 steps. It turns out, such a system doesn’t provide that level of brute-force security. Given one pair of plaintext-ciphertext pair, show how to brute-force the double encryption with complexity 257 rather than 2112. This attack demonstrates that double symmetric ciphering doesn’t increase the security much. In fact, To get 112 bits security, Triple-DES (a.k.a 3-DES) was introduced and standarlised. You will see that in many security protocols, 3-DES remains an option.

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    $\begingroup$ Attacking 2DES efficiently $\endgroup$ – kelalaka Jun 10 at 10:07
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    $\begingroup$ As a new contributor, a little advice; first search this site. There are tons of questions that contain the answer that you were looking for. You can see some of them on the right in the related part. $\endgroup$ – kelalaka Jun 10 at 10:25

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