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XEX Mode (with two tweaks) is the following scheme:

$X = E_{k}(I) \otimes \alpha^j \\ C = E_{K}(P \oplus X) \oplus X$

where $I$ and $j$ are both tweaks (sector number and block number respectively). If I wanted to use only one tweak, $I$, is the following correct and safe?

$X = E_{k}(I) \\ C = E_{K}(P \oplus X) \oplus X$

Essentially $j$ is treated as $0$, which I believe eliminates it from the equation.

To clarify what I mean by "safe"; does the second scheme I show have the same security guarantees that the first scheme has, as outlined in Rogaway's paper? I skimmed the paper, but remain unclear as to whether the second scheme is included in the XEX family as Rogaway defined it.

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Word safe is ambiguous. Please, define in more detail what is your intended security model. –  user4982 Apr 30 at 14:40
    
@user4982: Updated to clarify. Thank you. –  Xor Apr 30 at 19:02
    
It would also be quite slower than original XEX, because you need two encryptions for each block, instead of one per block + one per sector. –  Paŭlo Ebermann May 1 at 9:36
    
@PaŭloEbermann: Correct, but not in all cases. If the application performs many random, block sized reads/writes then the simplified scheme (as corrected in the answer below) would be faster (requiring 2 encryptions, instead of 2 encryptions + GF(2^128) multiplication). But really, I was just curious how to do it; less so actually implementing XEX/XTS like that. –  Xor May 1 at 20:24

1 Answer 1

up vote 1 down vote accepted

No, using XEX mode with $j = 0$ is not entirely safe. As noted in section 6 of the Rogaway (2004) paper (emphasis mine):

"Some added care is needed to address the security of XEX. Suppose, to be concrete, that we are looking at $\mathsf{XEX}[E,2^{\mathbf I}]$ and $\mathbf I=[0\mathop{..}2^{n-2}]$. Let the adversary ask a deciphering query with ciphertext $C=0^n$ and tweak $(0^n,0)$. If the adversary has a construction-based deciphering oracle then it will get a response of $M=\tilde D_K^{0^n}\,^0(0^n)=D_K(Δ)⊕Δ=D_K(\mathsf N)⊕\mathsf N=0^n⊕\mathsf N=\mathsf N$, where $\mathsf N = E_K(0^n) = Δ$. This allows the adversary to defeat the CCA-security. For example, enciphering $2M=2\mathsf N$ with a tweak of $(0^n,1)$ and enciphering $4M=4\mathsf N$ with a tweak of $(0^n,2)$ will give identical results (if the adversary has the construction-based enciphering oracle). Corresponding to this attack we exclude any tweak $(N,i_1,\dotsc,i_k)$ for which $(i_1,\dotsc,i_k)$ is a representative of $1$—that is, any tweak $(N,i_1,\dotsc,i_k)$ for which $α_1^{i_1} \dotsb α_k^{i_k} = 1$. In particular, this condition excludes any tweak $(N,0,\dotsc,0)$."

However, using $j = 1$ is safe, and, indeed, is explicitly used as an example of a valid tweak by Rogaway e.g. in section 3:

"Tweaks that are not the increment of a prior tweak will also arise, and they will typically look like $(N,1,0\dotsc,0)$. Constructions should be reasonably efficient in dealing with such tweaks."


Also, in their comments on the XTS mode, Liskov and Minematsu confirm that, indeed, $j=0$ is a bad choice for XEX (emphasis original):

"Note that $j = 0$ must be excluded, as $f(0, v) = v$ for any $v$, which implies $ρ = 1$. Moreover, if $j = 0$ was allowed, a simple attack based on this fact existed, as pointed out by [6] and [3]. Hence if XEX is used, one must be careful to avoid $j$ being $0$."

However (while generally criticizing the choice to use two keys for XTS), they also note that this requirement to avoid $j = 0$ only applies to the single-key XEX mode, and not to XTS, which uses separate keys for the tweak and encryption stages:

"Note that XTS does not require to avoid $j = 0$, as the offset function of XTS needs not be $(ϵ,γ,ρ)$-uniform, but only be $ϵ$-AXU if it is combined with URP. This difference is significant in security, but has no impact on effectiveness for practical applications."

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