In the Bouncy Castle libraries, the GCM cipher implementation has an interesting property that does not seem described in the GCM papers (neither the NIST or the original paper):

Some AAD was sent after the cipher started. We determine the difference b/w the hash value we actually used when the cipher started (S_atPre) and the final hash value calculated (S_at). Then we carry this difference forward by multiplying by H^c, where c is the number of (full or partial) cipher-text blocks produced, and adjust the current hash.

This scheme is present in the final calculations (in the doFinal() method).

Now what I understand is that addition of within the polynomial is equivalent to XOR. I can also see why exponentiation is required to carry the difference forward. What I don't see is how the complete scheme works, especially for partial 16 byte blocks.

Can somebody show how the adjustment in the doFinal method is defined mathematically?

Note that the $\operatorname{GHASH}$ function is performed over the follwing data within GCM:

$S = \operatorname{GHASH}_H (A || 0^v || C || 0^u || [len(A)]_{64} || [len(C)]_{64})$


  • $0^v$ and $0^u$ is padding (0..127 bits of zero's) up to the block size

$\operatorname{GHASH}$ itself is defined as follows:


  1. Let $X_1, X_2, ... , X_{m-1}, X_m$ denote the unique sequence of blocks such that $X = X_1 || X_2 || ... || X_{m-1} || X_m$.
  2. Let $Y_0$ be the “zero block,” $0^{128}$.
  3. For $i = 1, ..., m$, let $Y_i = (Y_{i-1} \oplus X_i) • H$.
  4. Return $Y_m$


  • $X•Y$ is the product of two blocks, $X$ and $Y$, regarded as elements within the binary Galois field

So in this scheme additional AAD ($A$) is send after ciphertext ($C$) was already put within the calculation.

  • $\begingroup$ I'm improving the GCM code of Bouncy Castle, making the nice properties of CTR return. I've already got everything running, but I still need to integrate this feature. I can do this better if I understand the involved mathematics. $\endgroup$
    – Maarten Bodewes
    Sep 16, 2014 at 13:16

1 Answer 1


Well, $\operatorname{GHASH}$ might be better understood as the polynomial:

$$\operatorname{GHASH}_H(X_1, X_2, ... , X_{m-1}, X_m) = X_1 H^{m} + X_2 H^{m-1} + ... + X_{m-1} H^2 + X_m H^1$$

where addition, multiplication and exponentiation are in the field $GF(2^{128})$. These addition, multiplication and exponentiation operations act algebraically quite a lot like the operators in a more traditional field (say, the real numbers); the usual tricks for rearranging polynomials work exactly the same.

What Bouncy Castle is doing is taking the vector $X_1, X_2, ... , X_{m-1}, X_m$ in two pieces, the AAD vector $A_1, A_2, ..., A_a = A || 0^v$ and the ciphertext $C_1, C_2, ..., C_c = C || 0^u$, and evaluating $\operatorname{GHASH}_H(A_1, A_2, ..., A_a, C_1, C_2, ..., C_c, X_{final})$ (where $X_{final}=[len(A)]_{64} || [len(C)]_{64}$ encodes the length of the AAD and the ciphertext). One way to rearrange this is:

$\operatorname{GHASH}_H(A_1, A_2, ..., A_a, C_1, C_2, ..., C_c, X_{final}) =$ $H^{c+1}\operatorname{GHASH}_H(A_1, A_2, ..., A_a) + H\operatorname{GHASH}_H(C_1, C_2, ..., C_c) + HX_{final}$

With this rearrangement, we can obviously evaluate $\operatorname{GHASH}_H(C_1, C_2, ..., C_c)$ before we see any AAD data (or even before we know how long it is).

Now, you ask about partial blocks. It turns out that we don't need to worry about partial blocks; GCM is defined so that both the ciphertext and AAD vectors are zero padded to the next full block; hence we never need to worry about partial blocks. If you are wondering a collision if we append a zero byte to the AAD, well, that's what $X_{final}$ is there for -- appending a 0 changes the AAD length, and hence the $X_{final}$ value.

  • $\begingroup$ So if there is already AAD data then we can just exponentiate the difference, where the difference is calculated with XOR, right? Bouncy also works for $A_{0..x} || C || A_{(x+1)..z}$... $\endgroup$
    – Maarten Bodewes
    Sep 17, 2014 at 23:13
  • $\begingroup$ @owlstead: I don't believe that quite works (unless you know the length of the AAD beforehand). The problem is that if you've already computed $\operatorname{GHASH}_H(A_{0..x}||C)$, there's no easy way to insert $A_{(x+1)...z}$ into the middle. If that's a requirement, I would suggest you keep running $\operatorname{GHASH}_H(A_{0..x})$ and $\operatorname{GHASH}_H(C_{0..y})$, and glue them together only when you've seen the entire $A$ $\endgroup$
    – poncho
    Sep 18, 2014 at 3:09
  • $\begingroup$ Well, it does work for sure. It seems that it first calculates $\operatorname{GHASH}_H(A_{0..x}||C)$ (including the zero padding), removes the padded $\operatorname{GHASH}_H(A_{0..x})$ and substitutes the new $A_{0...z}$, which was continued from the unpadded $\operatorname{GHASH}_H(A_{0..x})$... at least if I interpret the code correctly. $\endgroup$
    – Maarten Bodewes
    Sep 18, 2014 at 21:01
  • $\begingroup$ @owlstead: that'd work too... That's more work to handle the A||C||A case, however if you're optimizing for the A||C case (and just want to handle the A||C||A case correctly, and not dreadfully slowly), that's a reasonable solution. $\endgroup$
    – poncho
    Sep 18, 2014 at 21:09
  • $\begingroup$ What the value of $m$ correspond to? $\endgroup$
    – Melab
    Jan 2, 2018 at 18:52

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