CRYPTO NEWS

Ethereum Faces Resistance at the 100 Ema: ETH Consolidation Awaits!

Ethereum is not just a blockchain or cryptocurrency but an essential aspect of cryptography that showcases transformation and evolution along the way of its growth. Understanding the changing dynamics, ETH has even prepared the layout to transition towards Proof of Stake to address the energy consumption issue of its current operational Proof of Work. Of …

Equivalance of encryption and decryption in DES under bruteforce

I am studying the vulnerability of DES algorithm. Let there be a DES system with 5 keys, 56bits each. Encryption works as follows: (It’s in this order on purpose) $C=DEC_k1(ENC_k2(ENC_k3(DEC_k4(DEC_k5(M)))))$ With k2=k3 and k4=k5 Where C = Cipher, M = plain text DEC = Decryption ENC = Encryption -kn = Is the number of key::Listen

I am studying the vulnerability of DES algorithm.

Let there be a DES system with 5 keys, 56bits each. Encryption works as follows: (It’s in this order on purpose)

$C=DEC_k1(ENC_k2(ENC_k3(DEC_k4(DEC_k5(M)))))$

With k2=k3 and k4=k5

Where C = Cipher,
M = plain text
DEC = Decryption
ENC = Encryption
-kn = Is the number of key used, we have 5 keys so k1-k5.

So my question is, from what I understand, the difference between DES encryption and decryption is only the key used (or more like the order in which the keys are being used) so they just cancel out one another if the keys in two subsequent enc/dec operations are equal?

So, with bruteforce being applied, can C (the result) be figured in 2^(1×56) time complexity vulnerability or in a 2^(3×56) one?
This is unclear to me, because if DEC and ENC are equivalent, then $DEC_k4(Dec_k5(M))$ will infact cancel each other.
Same for $ENC_k2(ENC_k3(M))$.
Leaving us with $DEC_k1(M)$ which is, from what I understand, a single DES encryption on M. (Equal to $ENC_k1$).

Ethereum Faces Resistance at the 100 Ema: ETH Consolidation Awaits!

Shopping cart
There are no products in the cart!
Continue shopping
0