Score:3

Why does CBC-MAC(M) = CFB-MAC(M)?

us flag

I don't understand why $\text{CBC-MAC}(M) = \text{CFB-MAC}(M)$. Has it something to do with $\text{CBC-MAC}(M) = C_L$ and $\text{CFB-MAC}(M) = E_K(C'_{L-1})$?

Maarten Bodewes avatar
in flag
You can use MathJax / $\LaTeX$ on our site, you just got to put it within dollar signs, e.g. '\$\square\$' gives you $\square$
Score:3
ru flag

For a block cipher $E_k()$, consider the following process applied to a message $M$ of length $\ell$ blocks $M_0,\ldots,M_{\ell-1}$ $$X_0=0$$ $$Y_i=X_i\oplus M_i$$ $$X_{i+1}=E_k(Y_i).$$ In CBC encryption with 0 IV the ciphertext is the sequence $X_1,\ldots,X_{\ell}$ and in CFB encryption with IV $E^{-1}_k(0)$ the ciphertext is the sequence $Y_0,\ldots, Y_{\ell-1}$.

CBC-MAC is simply the final block of ciphertext: in our notation $X_\ell$. CFB-MAC is the encryption of the final block of cipher (without this it is trivial to modify $M_{\ell-1}$ and forge a MAC): in our notation $E_k(Y_{\ell-1})=X_\ell$.

Note that CFB can be used to encrypt “segments” of data smaller than the block size whereas CBC cannot. In such cases, there is no equivalence.

mangohost

Post an answer

Most people don’t grasp that asking a lot of questions unlocks learning and improves interpersonal bonding. In Alison’s studies, for example, though people could accurately recall how many questions had been asked in their conversations, they didn’t intuit the link between questions and liking. Across four studies, in which participants were engaged in conversations themselves or read transcripts of others’ conversations, people tended not to realize that question asking would influence—or had influenced—the level of amity between the conversationalists.