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authorcyfraeviolae <cyfraeviolae>2022-08-25 02:16:03 -0400
committercyfraeviolae <cyfraeviolae>2022-08-26 04:18:14 -0400
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+<!DOCTYPE html>
+<html>
+ <head>
+ <title>Forbidden Salamanders &middot; 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>
+ <h4>Differences</h4>
+ <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 the two MACs, and \(d_i\) be
+ the difference between two blocks at position \(i\):
+ \[
+ e = (d_3)h^2 + (d_2)h^3 + (d_1)h^4 + (d_0)h^5,
+ \]
+ We want to achieve \(e=0\) but with at least one \(d_i \not= 0\) in order
+ to obtain a non-trivial forgery.
+ </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,
+ replacing \(h^{2^i}\) by \(S^ih\).
+ \[
+ e = M_{d_3}Sh + M_{d_2}h^3 + M_{d_1}S^2h + M_{d_0}h^5,
+ \]
+ </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
+ \[
+ e = M_{d_3}Sh + M_{d_1}S^2h = (M_{d_3}S + M_{d_1}S^2)h
+ \]
+ </p>
+ <p>
+ Let \(\hat{d}\) represent the concatenation of all the remaining \(d_i\)s.
+ The length of \(\hat{d}\) will be logarithmic in the size of the original
+ ciphertext as we only consider the power-of-2-indexed blocks.
+ </p>
+ <p>
+ \(A\) represents the action on \(h\) by \(\hat{d}\):
+ \[
+ A = (M_{d_3}S + M_{d_1}S^2)
+ \]
+ \[
+ e = Ah
+ \]
+ </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 \(A\) 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 \(A\)</h4>
+ <p>
+ Changing the bits of \(\hat{d}\) effects a linear change on the
+ bits of \(A\). Consider a set of \(8\vert\hat d\vert\) &ldquo;basis
+ vectors&rdquo; for \(\hat d\) (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 \(A\). Concatenate
+ the first \(M\) rows into a column vector of a dependency matrix \(T\).
+ Thus, \(T\) will have \(8\vert\hat{d}\vert\) columns and \(128M\) rows.
+ </p>
+ <p>
+ We want to compute some \(\hat d\) that result in the first
+ \(M\) rows of \(A\) equaling zero, which is equivalent to saying
+ \[ T\hat{d} = 0. \]
+ </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 \(\hat d\)
+ that sends the first \(M\) rows of \(A\) 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 \(\hat d\), since
+ \[ e[0:N-1] = 0 = A[0:N-1]h \]
+ we know that \(h\) is in the kernel of the first \(N\) rows of
+ \(A\). The first \(M\) rows of \(A\) are zero so this is trivial,
+ 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, 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 \(e\):
+ \[ e = Ah = A(Xh') = (AX)h', \]
+ where \(AX\) is a matrix of dimensions 128 by 112.
+ </p>
+ <p>
+ When we construct the corresponding dependency matrix \(T\),
+ we still have \(8\vert\hat d\vert\) 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 \(\hat d\) 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 \(A\) 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. Hint: in general the description has left out
+ whether variables are in \(\mathbb{F}_2\), \(\mathbb{F}_{2^{128}}\),
+ \(\mathbb{Z}\), etc. Consider this before you code each function.
+ </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"
+aad = b''
+mac_bytes = 4
+m = b'yellow_submarine'*(2**17)
+nonce = b'jorgelborges'
+c, mac = gcm_encrypt(k, nonce, aad, m, mac_bytes=MACBYTES)
+def oracle(base, aad, mac, nonce):
+ cipher = AES.new(k, mode=AES.MODE_GCM, nonce=nonce, mac_len=mac_bytes)
+ cipher.update(aad)
+ cipher.decrypt_and_verify(base, 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>