diff options
| author | cyfraeviolae <cyfraeviolae> | 2022-08-27 02:55:45 -0400 |
|---|---|---|
| committer | cyfraeviolae <cyfraeviolae> | 2022-08-27 02:55:45 -0400 |
| commit | 00e2704d0039ed9dbbfec54b2643da395f642f66 (patch) | |
| tree | e950941b75e49d8b65fac23c1a656bfb767f1f51 /templates/nonce-truncation.html | |
| parent | 0812e77c8d980773806bd7810a5e0992c180b589 (diff) | |
key commitment
Diffstat (limited to 'templates/nonce-truncation.html')
| -rw-r--r-- | templates/nonce-truncation.html | 386 |
1 files changed, 0 insertions, 386 deletions
diff --git a/templates/nonce-truncation.html b/templates/nonce-truncation.html deleted file mode 100644 index 0095bb1..0000000 --- a/templates/nonce-truncation.html +++ /dev/null @@ -1,386 +0,0 @@ -<!DOCTYPE html> -<html> - <head> - <title>Forbidden Salamanders · Nonce Truncation</title> - <meta charset="utf-8"> - <meta name="viewport" content="width=device-width, initial-scale=1.0"> - <link rel="stylesheet" type="text/css" href="/static/styles.css"> - <link rel="stylesheet" type="text/css" href="/forbidden-salamanders/static/styles.css"> - <link rel="shortcut icon" type="image/x-icon" href="/forbidden-salamanders/static/favicon.ico"> - </head> - <body> - <div class="container"> - <div> - <div class="home"> - <a href="/forbidden-salamanders" class="home-title">Forbidden Salamanders</a> - <span> at </span><a href="/">cyfraeviolae.org</a> - </div> - <div class="crumbs"> - <a href="/git/forbidden-salamanders">source code</a> - <span class="sep"> · </span> - <a href="/forbidden-salamanders/nonce-reuse">nonce reuse</a> - <span class="sep"> · </span> - <a href="/forbidden-salamanders/nonce-truncation"><strong>nonce truncation</strong></a> - <!-- - <span class="sep"> · </span> - <a href="/forbidden-salamanders/key-commitment">key commitment</a> - --> - </div> - </div> - <p> - <strong><a href="/forbidden-salamanders/nonce-truncation">Nonce - truncation</a>.</strong> 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. - </p> - <br> - {% if form.errors %} - <div class="errors"> - Errors: - <ul> - {% for name, errors in form.errors.items() %} - {% for error in errors %} - <li> {{name}}: {{ error }} </li> - {% endfor %} - {% endfor %} - </ul> - </div> - {% endif %} - <form action="/forbidden-salamanders/nonce-truncation" method="post"> - <div><em> - Roseacrucis chooses a key, a nonce, and a truncated MAC length; - then encrypts an arbitrary 512-byte length message. - </em></div><br> - - <div> - <label for="key">Key (16 bytes in hex)</label> - <input name="key" id="key" type="text" value="{{ key if key else '746c6f6e6f7262697374657274697573' }}" minlength=32 maxlength=32 required> - </div> - - <div> - <label for="nonce">Nonce (12 bytes in hex)</label> - <input name="nonce" id="nonce" type="text" value="{{ nonce if nonce else '4a4f5247454c424f52474553' }}" minlength=24 maxlength=24 required> - </div> - - <div> - <label for="mac_len">MAC length</label> - - <select name="mac_len" id="mac_len"> - <option value="1">1 byte</option> - </select> - </div> - - <br><div><em> - 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. - </em></div> - - <br><div><em> - Finally, you choose a new message to forge under the same key - and nonce. - </em></div><br> - - <div> - <label for="mf">Forged message</label> - <input name="mf" id="mf" type="text" required maxlength=64 value="{{mf}}"> - </div> - <div> - <button type="submit">Recover authentication key and forge MAC</button> - </div> - </form> - <form action="/forbidden-salamanders/nonce-truncation" method="get"> - <div> - <button type="submit">Reset</button> - </div> - </form> - {% if h %} - <div class="solution"> - <p> - Forged ciphertext: <code>{{ c_forged.hex() }}</code> - <br> - Forged MAC: <code>{{mac.hex()}}</code> - <br> - Authentication key: <code>{{h.hex()}}</code> - </p> - </div> - {% endif %} - <br> - <details> - <summary> - Attack outline. - </summary> - <p> - Review the <a href="/forbidden-salamanders/nonce-reuse">nonce reuse attack</a> - to learn why recovering the authentication key is enough to forge MACs over - arbitrary ciphertexts. - </p> - <p> - 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. - </p> - <p> - 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 - \[ - mac-mac' = (c_3-c_3')h^2 + (c_2-c_2')h^3 + (c_1-c_1')h^4 + (c_0-c_0')h^5, - \] - </p> - <p> 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. - </p> - <h4>Linear Operations in \(\mathbb{K}=\mathbb{F}_{2^{128}}\)</h4> - <p> - As described in the <a href="/forbidden-salamanders">mission home page</a>, - 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 - \(\mathbb{K}=\mathbb{F}_{2^{128}}\). - </p> - <p> - 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. - </p> - <p> - 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. - </p> - <p> - 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 \(1, \alpha, - \alpha^2, \ldots\): - \[ - M_c = \begin{bmatrix} - c & c\alpha & c\alpha^2 & \ldots & c\alpha^{127} - \end{bmatrix}. - \] - </p> - <p> - 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 \(k \in \mathbb{F}_2\), - \[(ka)^2 = k^2a^2 = ka^2. \] - </p> - <p> - 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. - </p> - <h4>Reframing the Problem</h4> - <p> - 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 \(S^ih\): - \[ - e(d_i, h) = M_{d_3}Sh + M_{d_2}h^3 + M_{d_1}S^2h + M_{d_0}h^5. - \] - </p> - <p> - 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 - \[ - e(d_i, h) = M_{d_3}Sh + M_{d_1}S^2h = (M_{d_3}S + M_{d_1}S^2)h - \] - </p> - <p> - 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. - </p> - <p> - \(A_{d^*}\) represents the action on \(h\) by \(d^*\): - \[ - A_{d^*} = (M_{d_3}S + M_{d_1}S^2) - \] - \[ - e(d^*, h) = A_{d^*}h - \] - </p> - <p> - 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 \(N\) - <em>rows</em> 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. - <p> - The remaining \(N-M\) relevant bits of \(e\) will be random, but we - shall see it will be small enough to brute force. - </p> - <h4>Zeroing Out Rows of \(A_{d^*}\)</h4> - <p> - 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. - </p> - <p> - 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. - </p> - <p> - We want to compute some \(d^*\) that result in the first - \(M\) rows of \(A_{d^*}\) equaling zero, which is equivalent to saying - \[ Td^* = 0. \] - </p> - <p> - 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. - </p> - <p> - 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. - </p> - <h4>Executing the Attack</h4> - <p> - 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). - </p> - <p> - 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. - </p> - <p> - 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. - </p> - <p> - 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. - </p> - <p> - 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 \(h\). - </p> - <h4>Speeding up the Attack</h4> - <p> - 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. - </p> - <p> - 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. - </p> - <p> - 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 - \(h\). - </p> - <p> - 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 <a href="/forbidden-salamanders/nonce-reuse">nonce reuse attack</a>. - </p> - <h4>Addendum</h4> - <p> - This attack was first shown by Dutch cryptographer Niels Ferguson - in his paper <a href="https://csrc.nist.gov/csrc/media/projects/block-cipher-techniques/documents/bcm/comments/cwc-gcm/ferguson2.pdf">Authentication weaknesses in GCM</a>. - 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. - </p> - <p> - Readers who wish to implement this attack themselves can try - <a href="https://cryptopals.com/">Cryptopals</a>; specifically - Set 8 Problem 64. - </p> - </details> - <details> - <summary> - Show me the code. - </summary> - <pre> -from <a href="/git/forbidden-salamanders">aesgcmanalysis</a> 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)</pre></details> - -<script> -MathJax = { - tex: { - extensions: ["AMSmath.js", "AMSsymbols.js"] - } -}; -</script> -<script id="MathJax-script" async - src="https://cdn.jsdelivr.net/npm/mathjax@3/es5/tex-chtml.js"> -</script> - </body> -</html> |