According to this article, which looks at attacks on Caesar candidates beyond what they claim, they include an attack on page 6 for nonce-respecting adversaries for AES-GCM which allows IVs longer than 128-bits at the birthday bound (i.e., $2^{64}$).

My question is why is GCM not susceptible to this attack for 128-bit IVs? I could understand this if the GHASH calculation for a 128 bit IV acts as a permutation (and if so, could someone please prove this)?


1 Answer 1


Actually, it turns out that GHASH on 128 bit values is a permutation (unless $H=0$; this happens with probability $2^{-128})$

GHASH of a 128 bit value $X$ simplifies to $\text{GHASH}_H(X) = H \times X$, where $H$ is a function of the key (that is, will always be the same for a specific key), and $\times$ is multiplication in the field $GF(2^{128})$.

This is a permutation, as $GF(2^{128})$ is a field, and multiplication in a field by a fixed nonzero value is always a permutation.

As for the 'collision' you mention in your comments, that actually can't happen.

If we have $H \times I_1 = 128$, that means that $H \ne 0$ and $I_1 = 128 \times H^{-1}$

If we have $H \times ((H \times I_2) \oplus 128) = 0$, that can happen only if $H = 0$ (which can't happen if the first equation holds) or $(H \times I_2) \oplus 128 = 0$, which holds only if $H \times I_2 = 128$, that is, if $I_2 = 128 \times H^{-1}$

Hence, for both to hold, we must have $I_1 = I_2$, and thus this isn't a 'collision'

  • $\begingroup$ GHASH on 128-bit value is a permutation. However, the calculation of the initial block is a 256-bit value: $GHASH_H(IV||0^{64}||0^{56}10^7)$; i.e., the second block is 0 in 64 bits, then the length of the IV = 128 in 64 bits. $\endgroup$
    – MotiNK
    Aug 1, 2018 at 22:43
  • $\begingroup$ Ok, now that I think about it XOR-ing by a fixed second block still leaves a permutation (although at probability $2^{-128}$ the first permutation gave us the value of the second block, in which case after we're XOR we'll get a result of 0); and multiplying through again by $H$ is still a permutation. So there's some additional small probability of collision (i.e., for non-zero $H$ there exists an IV $I_1$ such that $H \times I_1 = 128$, and there exists another IV, $I_2$ such that $H \times (H \times I_2 \oplus 128) = 0$, however the probability of locating that pair is just over $2^{-256}$) $\endgroup$
    – MotiNK
    Aug 1, 2018 at 23:39
  • $\begingroup$ So please update your answer to include that reference to the second fixed block and I'll accept. $\endgroup$
    – MotiNK
    Aug 1, 2018 at 23:41
  • $\begingroup$ I am accepting since this was a good answer (even without the reference to XOR-ing by a fixed second block). $\endgroup$
    – MotiNK
    Sep 4, 2018 at 15:00

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