Take the 2-minute tour ×
Cryptography Stack Exchange is a question and answer site for software developers, mathematicians and others interested in cryptography. It's 100% free, no registration required.

I am studying for an exam right now. And I wanted to make sure I got this point correct.

AES is not a Feistel cipher because the operations in AES are not invertible.

Is the above statement correct? If not, why isn't it a Feistel cipher?

share|improve this question
2  
If the operations were not invertible we wouldn't be able to decrypt anything. Read the links in Reid's answer and try to memorize the schematics and what a Feistel network looks like. That's very different from a simple sub-perm network (link also in Reid's answer). A Feistel cipher is a cipher that uses a Feistel structure in its design - AES does not. –  rath Sep 28 '13 at 6:36
1  
@rath you would not be able to decrypt anything if the cipher was not augmented with a mode that only requires one way encryption such as CTR mode –  owlstead Sep 28 '13 at 12:51
    
@owlstead Good point. [+1] –  e-sushi Sep 29 '13 at 7:10

2 Answers 2

Well, AES is not a Feistel cipher because it's a substitution-permutation network instead. If I were taking a test that asked me why AES was not a Feistel cipher, this would be my argument: namely, that the structure of substitution-permutation networks is fundamentally different from that of Feistel networks. (Here one could elaborate on invertibility and other differences.)

That said, your statement is not correct. In a Feistel cipher, the round function is not necessarily invertible (DES's round function is not), but in AES, like any substitution-permutation network, the rounds are invertible. This is a property of the construction itself.

share|improve this answer

By definition, a Feistel network uses a series of rounds that split the input block into two sides, uses one side to permute the other side, then swaps the sides. As always, Wikipedia has a nice diagram.

AES doesn't do this. Performing a round necessarily permutes the entire state. Each round consists of the SubBytes, ShiftRows, MixColumns, and AddRoundKey steps, none of which behave in a Feistel network-like manner:

  • SubBytes performs byte-wise substitution from a constant table, no byte's value influences another byte's permuted value.
  • ShiftRows permutes 4-byte words at a time using only those 4 bytes, no byte from another word influences their permuted output.
  • MixColumns permutes 4-byte words at a time using only those 4 bytes, no byte from another word influences their permuted output.
  • AddRoundKey is a permutation using the derived round key, no byte's value influences another byte's permuted value.

So only the ShiftRows and MixColumns steps even allow a byte to influence the permutation of any other bytes in the state, and in both of those steps a given byte only influences the permutation of other bytes when it itself is also being permuted.

None of that matches the "split the block into A and B and use A to permute B" style of a Feistel network.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.