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<!DOCTYPE html>
<html>
<head>
<title>Forbidden Salamanders · Nonce Reuse</title>
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<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="/forbidden-salamanders/key-commitment">key commitment</a>
<span class="sep"> · </span>
<a href="/forbidden-salamanders/nonce-reuse"><strong>nonce reuse</strong></a>
<span class="sep"> · </span>
<a href="/forbidden-salamanders/mac-truncation">mac truncation</a>
<span class="sep"> · </span>
<a href="/git/forbidden-salamanders">source code</a>
</div>
</div>
<p>
<strong>Nonce reuse.</strong> Due to rising entropy
prices, Roseacrucis has started to reuse AES-GCM nonces. You must perform the
Forbidden Attack in order to 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-reuse" method="post">
<div><em>
Roseacrucis chooses a key, a nonce, and encrypts two messages under the same nonce.
</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="m1">First message</label>
<input name="m1" id="m1" type="text" required maxlength=64 value="{{m1 if m1 else 'The universe (which others call the Library)'}}">
</div>
<div>
<label for="m2">Second message</label>
<input name="m2" id="m2" type="text" required maxlength=64 value="{{m2 if m2 else 'From any of the hexagons one can see, interminably'}}">
</div>
<br><div><em>
After intercepting the ciphertexts and recovering the
authentication key, you choose a new message to forge under the
same key and nonce.
</em></div><br>
<div>
<label for="mf">Forged message; shorter than the first 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-reuse" method="get">
<div>
<button type="submit">Reset</button>
</div>
</form>
{% if macs %}
<div class="solution">
<p>
Forged ciphertext: <code>{{ c_forged.hex() }}</code>
{% if macs|length == 1 %}
<br>
Forged MAC: <code>{{macs[0][2].hex()}}</code>
<br>
Authentication key: <code>{{macs[0][0].hex()}}</code></li>
{% endif %}
</p>
{% if macs|length != 1 %}
Forged MAC candidates:
<ul>
{% for h, _, mac in macs %}
<li>
MAC: <code>{{mac.hex()}}</code>
<ul class="inner-ul">
<li>Authentication key: <code>{{h.hex()}}</code></li>
</ul>
</li>
{% endfor %}
</ul>
{% endif %}
</div>
{% endif %}
<br>
<details>
<summary>
Attack outline.
</summary>
<p>
Recall that the AES-GCM ciphertext is computed as the XOR of the
keystream and the message. One can modify the bits of the
ciphertext arbitrarily to effect the same change in the decrypted
plaintext.
</p>
<p>
Where certain bits of the plaintext are already known, the attacker
can fully determine the same bits of the forged plaintext. If
nonces are reused, the keystream will be identical, allowing us to
recover plaintext via
<a href="https://samwho.dev/blog/toying-with-cryptography-crib-dragging/">
crib dragging</a>, which makes this attack particularly effective:
\[
c' = c \oplus m \oplus m'.
\]
</p>
<p>
However, we still need to compute a new MAC over the forged ciphertext.
Simplifying for a ciphertext \(c\) of two blocks and no additional
authenticated data, the GMAC MAC is computed as
\[
mac = s + \vert c\vert h + c_1h^2 + c_0h^3,
\]
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>
If we intercept a second ciphertext \(c'\) encrypted under the same key and nonce,
we can compute
\[
mac + mac' = (s + s') + (len + len')h + (c_1 + c'_1)h^2 + (c_0+c'_0)h^3,
\]
Since \(s = s'\) and \(x+x=0\) in \(\mathbb{F}_{2^{128}}\), we are
left with the polynomial equation
\[
0 = (mac + mac') + (len + len')h + (c_1 + c'_1)h^2 + (c_0+c'_0)h^3
\]
where all variables are known other than \(h\). Thus, recovering \(h\)
is a matter of finding the roots by <a href="https://en.wikipedia.org/wiki/Factorization_of_polynomials_over_finite_fields">factoring the
polynomial</a>.
</p>
<p>
We plug \(h\) back into the first equation to recover \(s\), and we
can forge the MAC for arbitary ciphertext under the same nonce.
Note that there may be multiple possible roots; in this
case, one can check each possibility against the enemy, or perform
the attack twice on two pairs of intercepted messages.
</p>
<p>
One can use SageMath to compute factors of a polynomial:
</p>
<pre>
K = GF(2**128, name='x', modulus=x^128+x^7+x^2+x+1)
x = K.gen()
S = PolynomialRing(K, 'y')
y = S.gen()
p = (1)*y^4 + (x^7)*y^3 + (x^9 + x^4 + 1)*y^2 + (x^12 + x^2)*y + (x^10 + x^5)
for factor, _ in p.factor():
if factor.degree() == 1:
print('Authentication key:', factor - y)</pre>
<p>
However, the library powering this demonstration implements <a href="https://en.wikipedia.org/wiki/Factorization_of_polynomials_over_finite_fields">polynomial factoring over finite fields</a> from scratch, which is an edifying exercise.
</p>
<p>
We present advice for those who wish to implement polynomial factorization as well:
</p>
<ul>
<li>The gcd of two polynomials is unique only up to multiplication by a non-zero constant because “greater” is defined for polynomials in terms of degree. When used in algorithms, gcd refers to the <em>monic</em> gcd, which is unique.</li>
<li>The <a href="https://math.stackexchange.com/a/943626/1084004">inverse Frobenius automorphism</a> (i.e., square root) in \(\mathbb{F}_{2^{128}}\) is given by \(\sqrt{x} = x^{2^{127}}\).</li>
<li>The authentication key <strong>must</strong> appear in one of the linear factors (those of the form \(y+h\)). This allows one to skip parts of the distinct-degree factorization and equal-degree factorization, making the algorithm much faster. Exercise: prove this claim.</li>
</ul>
<p>
Readers who wish to implement this attack themselves can try
<a href="https://cryptopals.com/">Cryptopals</a>; specifically
Set 8 Problem 62.
</p>
</details>
<details>
<summary>
Example with code.
</summary>
<pre>
from <a href="/git/forbidden-salamanders">aesgcmanalysis</a> import xor, gmac, gcm_encrypt, gcm_decrypt, nonce_reuse_recover_secrets
k = b"tlonorbistertius"
nonce = b"jorgelborges"
m1, aad1 = b"The universe (which others call the Library)", b""
m2, aad2 = b"From any of the hexagons one can see, interminably", b""
c1, mac1 = gcm_encrypt(k, nonce, aad1, m1)
c2, mac2 = gcm_encrypt(k, nonce, aad2, m2)
# Recover the authentication key and blind from public information
possible_secrets = nonce_reuse_recover_secrets(nonce, aad1, aad2, c1, c2, mac1, mac2)
# Forge the ciphertext
m_forged = b"As was natural, this inordinate hope"
c_forged, aad_forged = xor(c1, xor(m1, m_forged)), b""
for h, s in possible_secrets:
print("MAC candidate": gmac(h, s, aad_forged, c_forged))</pre></details>
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