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<!DOCTYPE html>
<html>
<head>
<title>Forbidden Salamanders · Key Commitment</title>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
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<link rel="stylesheet" type="text/css" href="/forbidden-salamanders/static/styles.css">
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<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/key-commitment"><strong>key commitment</strong></a>
<span class="sep"> · </span>
<a href="/forbidden-salamanders/nonce-reuse">nonce reuse</a>
<span class="sep"> · </span>
<a href="/forbidden-salamanders/mac-truncation">mac truncation</a>
</div>
</div>
<p>
<strong>Key commitment.</strong> One of
our agents has infiltrated Roseacrucis’ inner circle, but all
secret keys are required to be surrendered to the
counterintelligence authority. Help her send ciphertexts back to
the Library that decrypt to confidential information under one key,
but innocuous banter under another.
</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/key-commitment" method="post" enctype="multipart/form-data">
<div><em>
The Library’s agent chooses two images: a JPEG file containing
confidential information, and a BMP file that looks innocuous.
</em></div><br>
<input type="radio" id="sample" name="mode" value="sample" required checked>
<label for="sample">Use sample JPEG and BMP files:</label><br>
<br>
<img class="responsive-img" src="/forbidden-salamanders/static/axolotl.jpg">
<img class="responsive-img" src="/forbidden-salamanders/static/kitten.bmp">
<br>
<br>
<input type="radio" id="custom" name="mode" value="custom" required>
<label for="custom">Select custom files:</label><br>
<div>
<label for="jpeg">JPEG file (<150KB)</label>
<input type="file" id="jpeg" name="jpeg" accept="image/jpeg">
</div>
<div>
<label for="bmp">BMP file (<50KB)</label>
<input type="file" id="bmp" name="bmp" accept="image/bmp">
</div>
<br><div><em>
The agent now computes two keys, a nonce and constructs a single
ciphertext. When decrypted under the first key, it will look
identical to the JPEG file; when decrypted under the second
key, it will look identical to the BMP file.
</em></div>
<p>
Key 1: <code>5c3cb198432b0903e58de9c9647bd241</code>
<br>
Key 2: <code>df923ae8976230008a081d23205d7a4f</code>
<br>
Nonce: <code>4a4f5247454c424f52474553</code>
</p>
<br>
<div>
<button type="submit">Download polyglot ciphertext</button>
</div>
</form>
<form action="/forbidden-salamanders/key-commitment" method="get">
<div>
<button type="submit">Reset</button>
</div>
</form>
<p>
You can test your ciphertext with Go. Run the following in a shell
and then try opening <code>first.jpg</code> and <code>second.bmp</code> in an image viewer.
</p>
<details>
<summary>
Show testing code.
</summary>
<pre style="font-size: small">
TEMP="$(mktemp).go"
cat > "$TEMP" <<EOF
package main
import ("crypto/aes"; "crypto/cipher"; "encoding/hex"; "os")
func main() {
var key, nonce, ciphertext, plaintext []byte; var block cipher.Block; var aesgcm cipher.AEAD; var err error
if len(os.Args) < 4 { panic("usage: go run salamander.go <key> <nonce> <ciphertext-filename>") }
if key, err = hex.DecodeString(os.Args[1]); err != nil { panic(err.Error()) }
if nonce, err = hex.DecodeString(os.Args[2]); err != nil { panic(err.Error()) }
if ciphertext, err = os.ReadFile(os.Args[3]); err != nil { panic(err.Error()) }
if block, err = aes.NewCipher(key); err != nil { panic(err.Error()) }
if aesgcm, err = cipher.NewGCM(block); err != nil { panic(err.Error()) }
if plaintext, err = aesgcm.Open(nil, nonce, ciphertext, nil); err != nil { panic(err.Error()) }
if _, err = os.Stdout.Write(plaintext); err != nil { panic(err.Error()) }
}
EOF
go run "$TEMP" 5c3cb198432b0903e58de9c9647bd241 4a4f5247454c424f52474553 polyglot.enc > first.jpg
go run "$TEMP" df923ae8976230008a081d23205d7a4f 4a4f5247454c424f52474553 polyglot.enc > second.bmp
</pre>
</details>
<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 monomial roots; in this
case, one can check each possibility against the enemy.
</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>
</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>
Show me the 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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