Forbidden Salamanders · Nonce Truncation

- Forbidden Salamanders - at cyfraeviolae.org -

- source code - · - nonce reuse - · - nonce truncation - -

- Nonce - truncation. The sorcerer aims to conserve - bandwidth by truncating nonces. Use the enemy as a decryption - oracle to once again, recover the authentication key and forge - arbitrary ciphertext. -

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- {% if form.errors %} -

- Errors: -

{{name}}: {{ error }}

- {% endif %} -

- Roseacrucis chooses a key, a nonce, and a truncated MAC length; - then encrypts an arbitrary 512-byte length message. -

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- Key (16 bytes in hex) - -

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- Nonce (12 bytes in hex) - -

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- MAC length - - -

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- After intercepting the ciphertext, you create new - specially-crafted messages and guess their corresponding MACs. - You send your guesses to Roseacrucis; whether he accepts gives - you enough information to recover the authentication key. -

- -

- Finally, you choose a new message to forge under the same key - and nonce. -

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- Forged message - -

- -

- - {% if h %} -

- Forged ciphertext: {{ c_forged.hex() }} -
- Forged MAC: {{mac.hex()}} -
- Authentication key: {{h.hex()}} -

- {% endif %} -
-

- Attack outline. -

- Review the nonce reuse attack - to learn why recovering the authentication key is enough to forge MACs over - arbitrary ciphertexts. -

- After intercepting a ciphertext and MAC, our initial goal is to compute - a variant of the ciphertext that has the same MAC. - Simplifying for a ciphertext \(c\) of four blocks and no additional - authenticated data, the GMAC MAC is computed as - \[ - mac = s + \vert c\vert{}h + c_3h^2 + c_2h^3 + c_1h^4 + c_0h^5, - \] - where \(s\) is a constant depending on the AES-GCM key and the nonce, and \(h\) - is the authentication key depending only on the AES-GCM key. -

- Given a different ciphertext \(c'\) of the same length encrypted - with the same key and nonce, the difference in their MACs can be computed - as - \[ - mac-mac' = (c_3-c_3')h^2 + (c_2-c_2')h^3 + (c_1-c_1')h^4 + (c_0-c_0')h^5, - \] -

Let \(e\) be the difference between two MACs, and \(d_i\) be - the difference between two blocks at position \(i\): - \[ - e(d_i, h) = (d_3)h^2 + (d_2)h^3 + (d_1)h^4 + (d_0)h^5, - \] - We want to achieve \(e=0\) but with at least one \(d_i \not= 0\) in order - to obtain a different ciphertext with the same MAC. -

Linear Operations in \(\mathbb{K}=\mathbb{F}_{2^{128}}\)

- As described in the mission home page, - each block of ciphertext, the authentication key \(h\), and the MAC are 16-byte - blocks that can be interpreted as elements of the finite field - \(\mathbb{K}=\mathbb{F}_{2^{128}}\). -

- The elements of \(\mathbb{K}\) are usually represented as - polynomials with coefficients in \(\mathbb{F}_2\) of degree less - than 128, where multiplication is performed modulo an irreducible - polynomial given by the AES-GCM specification. This gives us a way - to multiply, add, and even divide two blocks. -

- For this problem we will use an alternate representation: a - 128-length bit vector, where the \(i\)th bit of the vector - represents the coefficient of \(\alpha^i\) in the polynomial - representation. -

- The transformation \(f(a) = ca\) for \(c, a \in \mathbb{K}\) is - linear; thus, it can be represented as matrix \(M_c\). We set each - column to the transformation by \(f\) of the basis vectors \(1, \alpha, - \alpha^2, \ldots\): - \[ - M_c = \begin{bmatrix} - c & c\alpha & c\alpha^2 & \ldots & c\alpha^{127} - \end{bmatrix}. - \] -

- The squaring operation \(g(a) = a^2\) is also linear: since \(2 = 0 \in \mathbb{K}\), - \[(a+b)^2 = a^2 + 2ab + b^2 = a^2 + b^2,\] - and for \(k \in \mathbb{F}_2\), - \[(ka)^2 = k^2a^2 = ka^2. \] -

- Thus we can construct - \[ - S = \begin{bmatrix} - 1^2 & \alpha^2 & (\alpha^2)^2 & \ldots & (\alpha^{127})^2 - \end{bmatrix}, - \] - and \(g(a) = a^2\) can alternately be written \(g'(a) = Sa\), interpreting \(a\) - as a vector. -

Reframing the Problem

- We can now rewrite our equation for \(e\) in terms of matrices and vectors: - \[ - e(d_i, h) = M_{d_3}h^2 + M_{d_2}h^3 + M_{d_1}h^4 + M_{d_0}h^5, - \] - after which we replace \(h^{2^i}\) by \(S^ih\): - \[ - e(d_i, h) = M_{d_3}Sh + M_{d_2}h^3 + M_{d_1}S^2h + M_{d_0}h^5. - \] -

- To ensure that \(e\) will be a linear transformation on \(h\), we set - \(d_i = 0\) if the corresponding \(h^j\) term does not have \(j\) as - a power of 2, resulting in - \[ - e(d_i, h) = M_{d_3}Sh + M_{d_1}S^2h = (M_{d_3}S + M_{d_1}S^2)h - \] -

- Let \(d^*\) represent the concatenation of all the remaining \(d_i\)s. - The length of \(d^*\) will be logarithmic in the size of the original - ciphertext as we only consider the power-of-2-indexed blocks. -

- \(A_{d^*}\) represents the action on \(h\) by \(d^*\): - \[ - A_{d^*} = (M_{d_3}S + M_{d_1}S^2) - \] - \[ - e(d^*, h) = A_{d^*}h - \] -

- In order for a full forgery, we need \(e=0\), but if the MAC length - is reduced to \(N\) bits, we only need the first \(N\) bits of - \(e\) to be zero, rather than all 128 bits. This will be satisfied if the first \(N\) - rows of \(A_{d^*}\) are all zero, regardless of \(h\). In - practice, zeroing out all \(N\) rows will be too difficult, so we - settle for zeroing out \(M \lt N\) rows instead. -

- The remaining \(N-M\) relevant bits of \(e\) will be random, but we - shall see it will be small enough to brute force. -

Zeroing Out Rows of \(A_{d^*}\)

- Changing the bits of \(d^*\) effects a linear change on the - bits of \(A_{d^*}\). Consider a set of \(8\vert d^* \vert\) “basis - vectors” for \(d^*\) (one for each bit), the \(i\)th basis - vector having a 1 in the \(i\)th position and 0 everywhere else. -

- For each basis vector, compute the corresponding \(A_{d^*}\). Concatenate - the first \(M\) rows into a column vector of a dependency matrix \(T\). - Thus, \(T\) will have \(8\vert d^*\vert\) columns and \(128M\) rows. -

- We want to compute some \(d^*\) that result in the first - \(M\) rows of \(A_{d^*}\) equaling zero, which is equivalent to saying - \[ Td^* = 0. \] -

- The solution space is given by the null space (or kernel) of - \(T\). Note that we need \(T\) to have more columns than rows for - the matrix to be linearly dependent and thus have a non-trivial - kernel. -

- The vectors of \(\ker T\) each represents a potential \(d^*\) - that sends the first \(M\) rows of \(A_{d^*}\) to zero. Any linear - combination of the vectors of the kernel will have the same effect. -

Executing the Attack

- Consider a random linear combination of \(\ker T\). Since these - are difference vectors, they can be thought of as specifying - bit flips of the original ciphertext at the appropriate positions - (remember to leave the non-power-of-2 blocks alone). -

- By design, the first \(M\) bits of \(e\) will be zero, meaning that - the first \(M\) bits of the MAC of the modified ciphertext will - equal the original MAC. The remaining bits of the MAC will match - with \(\frac{1}{2^{N-M}}\) probability. -

- Say the MAC length is \(N=32\) bits, and we let \(M=16\) (this requires - intercepting a ciphertext of length \(2^{16+1}\)). We can compute - random linear combinations of \(\ker T\), sending the modified ciphertexts - and original MAC to Roseacrucis. If they accept (which they should after - roughly \(2^{16}=65536\) attempts), we've succeeded in a forgery. -

- In addition to the forgery, the acceptance gives us - information on \(h\): for the successful \(d^*\), since - \[ e(d^*, h)[0:N-1] = 0 = A_{d^*}[0:N-1]h \] - we know that \(h\) is in the kernel of the first \(N\) rows of - \(A_{d^*}\). The first \(M\) rows of \(A_{d^*}\) are zero, - but the next \(N-M\) rows are likely linearly independent rows. If - we put these rows into a new matrix \(K\), we have \[ 0 = Kh. \] - The dimensions of \(K\) are \((N-M, 128) = (16, 128)\). The kernel - of such a matrix is 112-dimensional by the rank-nullity theorem, - which is not enough to guess \(h\) yet. -

- However, each additional forgery (with good probability) gives us - more linearly independent vectors to add into \(K\). Once we - collect 127 linearly independent vectors (there can be no more - since we know the kernel is non-trivial), the kernel of \(K\) will be 1-dimensional, - and the only vector in the kernel will be \(h\). -

Speeding up the Attack

- Once we have our first successful forgery, we have \(K\) such that - \(h \in \ker K\), meaning that some linear combination - of basis vectors of \(\ker K\) equals \(h\). Let \(X\) be - a basis set for \(K\), and so write - \[ h = Xh'. \] - We don't know \(h'\), but it is a 112-dimensional vector. - Now rewrite our equation for \(e\): - \[ e(A_{d^*}, h) = A_{d^*}h = A_{d^*}(Xh') = (A_{d^*}X)h', \] - where \(A_{d^*}X\) is a matrix of dimensions 128 by 112. -

- When we construct the corresponding dependency matrix \(T\), - we still have \(8\vert d^*\vert\) columns, but each row - only takes 112 bits to zero out rather than 128. This lets us - set more rows to zero, which in turn gives us a better chance of - succeeding in the next \( d^*\) forgery. -

- We can continue in this fashion, each step getting more and more - efficient until we collect 127 vectors in \(K\). Remember to leave - at least one relevant row of \(A_{d^*}\) to be non-zero; otherwise, the - forgery will succeed but won't tell us any more information about - \(h\). -

- To complete the attack, one can recover the first \(N\) rows of \(s\) - and compute a forged MAC for arbitrary ciphertext under the same nonce - as in the nonce reuse attack. -

Addendum

- This attack was first shown by Dutch cryptographer Niels Ferguson - in his paper Authentication weaknesses in GCM. - He notes that a (then-)competing mode, CWC, avoids this attack by - encrypting the GMAC polynomial with the block cipher before adding - \(s\). This breaks the linear relationship between the ciphertext - and the MAC. -

- Readers who wish to implement this attack themselves can try - Cryptopals; specifically - Set 8 Problem 64. -

- Show me the code. -

-from aesgcmanalysis import xor, gmac, gcm_encrypt, nonce_truncation_recover_secrets
-from Crypto.Cipher import AES
-
-k = b"tlonorbistertius"
-mac_bytes = 4
-m, aad, = b"yellow_submarine"*(2**17) = b""
-nonce = b"jorgelborges"
-c, mac = gcm_encrypt(k, nonce, aad, m, mac_bytes=MACBYTES)
-def oracle(c, aad, mac, nonce):
-	cipher = AES.new(k, mode=AES.MODE_GCM, nonce=nonce, mac_len=mac_bytes)
-	cipher.update(aad)
-	cipher.decrypt_and_verify(c, mac)
-
-h, s = nonce_truncation_recover_secrets(c, mac, nonce, mac_bytes, aad, oracle)
-
-m_forged = b"As was natural, this inordinate hope"
-c_forged, aad_forged = xor(c, xor(m, m_forged)), b""
-mac_forged = gmac(h, s, aad_forged, c_forged)

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