Double 3DES/TDES Encryption If 3DES/TDES (officially TDEA or Triple DEA - Triple Data Encryption Algorithm) encryption is used twice, using 6 different keys K1-K6 :

ciphertext = EK6(DK5(EK4(EK3(DK2(EK1(plaintext))))))
plaintext = DK1(EK2(DK3(DK4(EK5(DK6(ciphertext))))))

would be such encryption more strong and less vulnerable to attacks than a regular 3DES ?

Furthermore, instead of E-D-E-E-D-E encryption and D-E-D-D-E-D decryption, would such double 3DES work with D-E-D-E-D-E encryption and E-D-E-D-E-D decryption as well?
You'd have to move to 9DES rather than 6DES as undoubtedly 6DES would still fall to the meet in the middle attack that can be performed on 2DES, now with the two 3DES block ciphers as parts. That means that (given a large amount of memory), the attack only adds a single bit of security, 113 bits.

Even then, just like 3DES has only 112 bits security for 3 different keys, 9DES would seem to max out to 2 x 112 = 224 bits of security rather than 9 x 56 = 504 bits that you would expect. However, it might well be that the meet in the middle attack on 3DES now doesn't work anymore giving you 2 x 168 = 336 bits of security instead. You might want to prove that.

Of course all the other nasty properties of 3DES are kept. Weak keys, parity bits, small block size and horrible speed all have to be endured. However, now it adds huge keys and additional calculations for this ad hoc fix of an ancient, insecure cipher.



The 5DES scheme probably has 168 bits of security if you use a meet in the middle. DES and other modern ciphers cannot be split in two after all; they are protected against meet in the middle by design (it might be the first thing to fall when a cipher is designed by a well willing amateur that at least gets some things right).