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<!DOCTYPE html>
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<title>Forbidden Salamanders · Key Commitment</title>
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<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"><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>
<span class="sep"> · </span>
<a href="/git/forbidden-salamanders">source code</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 AES-GCM 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
AES-GCM 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>8007941455b5af579bb12fff92ef31a3</code>
<br>
Key 2: <code>14ef746e8b1792e52b1d22ef124fae97</code>
<br>
Nonce: <code>4a4f5247454c424f52474553</code>
</p>
<br>
<div>
<button type="submit">Download 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,
then open <code>/tmp/polyglot-first.jpg</code> and <code>/tmp/polyglot-second.bmp</code>
in an image viewer. You may need to alter the path of <code>polyglot.enc</code> to reflect
your download directory.
</p>
<pre class="demo">
curl -L -o /tmp/decrypt-aes-gcm.go https://cyfraeviolae.org/forbidden-salamanders/static/decrypt-aes-gcm.go
go build -o /tmp/decrypt-aes-gcm /tmp/decrypt-aes-gcm.go
< polyglot.enc /tmp/decrypt-aes-gcm 8007941455b5af579bb12fff92ef31a3 4a4f5247454c424f52474553 > /tmp/polyglot-first.jpg
< polyglot.enc /tmp/decrypt-aes-gcm 14ef746e8b1792e52b1d22ef124fae97 4a4f5247454c424f52474553 > /tmp/polyglot-second.bmp
</pre>
<details>
<summary>
Attack outline.
</summary>
<p>
This attack was shown by Yevgeniy Dodis, Paul Grubbs, Thomas Ristenpart, and Joanne Woodage
in <a href="https://eprint.iacr.org/2019/016">Fast Message Franking: From Invisible Salamanders to Encryptment</a>.
</p>
<h4>Colliding MACs</h4>
<p>
First, we will describe a general strategy to create a ciphertext that yields the same MAC
with two different keys. Then we will show how to construct a ciphertext that yields
meaningful results when decrypted with those two keys.
</p>
<p>
Consider arbitrary keys \(k_1, k_2\), nonce \(n\), and ciphertext
\(c\) (additional associated data can be accounted for in a
straightforward manner). \(k_1, k_2\) are associated
with authentication keys \(h_1, h_2\) and blinds \(s_1, s_2\), respectively.
</p>
<p>
Given a ciphertext of three blocks, we will attempt to add a fourth ciphertext block \(c_3\), set the MACs equal to each other,
and solve for \(c_3\). Remember that in \(\mathbb{F}_{2^{128}}\), addition is the same as subtraction.
</p>
<p>
For the resulting ciphertext of four blocks, the MACs for each key are computed as
\[
mac_{1} = s_1 + \vert c\vert{}h_1 + c_3h_1^2 + c_2h_1^3 + c_1h_1^4 + c_0h_1^5
\]
\[
mac_{2} = s_2 + \vert c\vert{}h_2 + c_3h_2^2 + c_2h_2^3 + c_1h_2^4 + c_0h_2^5
\]
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. We can now set the MACs equal to each other and solve for \(c_3\).
\[
s_1 + \vert c\vert{}h_1 + c_3h_1^2 + c_2h_1^3 + c_1h_1^4 + c_0h_1^5 = s_2 + \vert c\vert{}h_2 + c_3h_2^2 + c_2h_2^3 + c_1h_2^4 + c_0h_2^5
\]
\[
c_3(h_1^2 + h_2^2) = (s_1 + s_2) + \vert c \vert (h_1 + h_2) + c_2(h_1^3+h_2^3) + c_1(h_1^4 + h_2^4) + c_0(h_1^5 + h_2^5)
\]
\[
c_3 = \frac{(s_1 + s_2) + \vert c \vert (h_1 + h_2) + c_2(h_1^3+h_2^3) + c_1(h_1^4 + h_2^4) + c_0(h_1^5 + h_2^5)}{h_1^2 + h_2^2}
\]
</p>
<p>
Note that the choice to place the extra block in the final position was arbitrary. For the attack below we will instead need
to change the penultimate block rather than adding a block; the computation is similar.
</p>
<h4>Magic Bytes</h4>
<p>
For the next phase, we construct a ciphertext that decrypts to a valid JPEG under one key and a valid BMP under another.
Recall that the ciphertext of AES-GCM, as in AES-CTR, is computed by taking the XOR of the keystream and the message. The keystream
is computed from the cipher key and the nonce.
</p>
<p>
The basic strategy is to place the JPEG bytes and BMP bytes at different locations, carefully arranging it so
each parser will ignore the other data for the file. JPEG files can include comments, in which we will include the
BMP data. The BMP parser will stop reading as soon as the indicated length of the BMP has been read, after which
we will include the JPEG data. In each decrypted file, the data for the other image will be scrambled as we are using
a different key, but it will not matter as the junk data will be in a location that is ignored by the image parser.
</p>
<p>
All JPEG files start with the magic bytes \(\mathtt{ffd8}\) and end
with \(\mathtt{ffd9}\). We will place a JPEG comment immediately
after the initial magic bytes, which is indicated by \(\mathtt{fffe}\) and is followed by a 2-byte big-endian encoding of the comment length \(J\).
Let \(J_i\) indicate the \(i\)th byte of \(J\); \(J_0\) being the least significant byte.
</p>
<p>
All BMP files start with the magic bytes \(\mathtt{424d}\) followed by a 4-byte little-endian encoding of the file length.
Because we need the BMP file to fit inside the JPEG comment, we set
\[
\begin{array}{|c|c|}\hline
& 0 & 1 & 2 & 3 & 4 & 5 & \ldots & -2 & -1 \\\hline
\mathsf{JPEG} & \mathtt{ff} & \mathtt{d8} & \mathtt{ff} & \mathtt{fe} & J_1 & J_0 & \ldots & \mathtt{ff} & \mathtt{d9} \\
\mathsf{BMP} & \mathtt{42} & \mathtt{4d} & J_0 & J_1 & \mathtt{00} & \mathtt{00} & \ldots & & \\\hline
\end{array}
\]
</p>
<p>
In addition to the file length at the beginning, BMP files also
include the size of the color array (the pixels of the image) in
the initial metadata. BMP parsers ignore any data after the color
array is supposed to be over, even if the file length has not been
exhausted yet. That means we can set \(J=\mathtt{ffff}=65536\), and the
resulting header will be valid for any BMP file less than \(J\) bytes.
</p>
<p>
Since these headers must be in the same location at the start of the file,
we search for two keys \(k_1, k_2\) and a nonce \(n\) such that
\[ \operatorname{AES-GCTR}(k_1, n, \mathtt{ffd8fffeffff}) = \operatorname{AES-GCTR}(k_2, n, \mathtt{424dffff0000}), \]
where \(\operatorname{AES-GCTR}\) returns the ciphertext portion of \(\operatorname{AES-GCM}\) but not the MAC.
</p>
<p>
The easiest way to do this is via a
birthday attack: fix an arbitrary nonce, then generate random keys
for both the JPEG header and the BMP header. Encrypt each and store
the ciphertext in a lookup table. Repeat until two keys are found that
encrypt their respective headers to the same ciphertext bytes. The search
takes less than a minute on a desktop computer.
</p>
<p>
We have now computed the ciphertext header \(C_{H}\) and two keys
which will decrypt it to the correct header bytes for both files.
Note that \(C_{H}\) only depends on the <em>maximum</em> size of
the BMP file, and thus can be precomputed. The remainder of the
attack that depends on the specific images is very fast.
</p>
<h4>Finishing the Polyglot</h4>
<p>
As explained before, we place the BMP bytes in the JPEG comment,
add padding to finish the comment, and add the JPEG bytes after the comment is over.
Below we show the structure of the ciphertext. For the blank cells in the ciphertext row,
use either the encryption of the JPEG cell under \(k_1\) or the BMP cell under \(k_2\) as indicated.
\[
\begin{array}{|c|c|}\hline
\mathsf{JPEG} && & & \textrm{JPEG} & \mathtt{ffd9} \\
C & C_H & & \mathtt{00}^{J-\vert \textrm{BMP}\vert}& & \\
\mathsf{BMP} && \textrm{BMP} & & & \\\hline
\end{array}
\]
</p>
<p>
Here, \(\textrm{JPEG}\) is the bytes of the JPEG file without the initial and final magic bytes,
and similarly \(\textrm{BMP}\) is the bytes of the BMP file without the initial magic bytes.
</p>
<p>
These ciphertexts do not have the same MAC yet. If we tried to use
the strategy outlined at the beginning where we add an extra block
at the end, the JPEG file would no longer end in \(\mathtt{ffd9}\)
and would be invalid. Instead, we modify it to change
the penultimate block. The collision algorithm will result in a
ciphertext block \(X\).
</p>
<p>
However, we don't want any data from the penultimate block to
corrupt our JPEG image. After \(\textrm{JPEG}\) ends, we start
another comment that will include the penultimate block, hiding it
from the parser. Care must be taken to ensure the penultimate block
is really on a block boundary. For AES-GCM, the block size is 16 bytes.
</p>
<p>
Below is the final structure of the polyglot ciphertext.
\[
J' = 16 - (6 + J + \vert \textrm{JPEG} \vert + 4) \pmod{16}
\]
\[
\begin{array}{|c|c|}\hline
\mathsf{JPEG} && & & \mathrm{JPEG} & \mathtt{fffe} & J' & & & & \mathtt{ffd9} \\
C & C_{H} & & \mathtt{00}^{J-\vert \textrm{BMP}\vert}& & & &\mathtt{00}^{J'-30} & X & \mathtt{00}^{14}& \\
\mathsf{BMP} && \textrm{BMP} & & \\\hline
\end{array}
\]
</p>
</details>
<details>
<summary>
Show me the code.
</summary>
<pre>
from <a href="/git/forbidden-salamanders">aesgcmanalysis</a> import att_merge_jpg_bmp
with open('first.jpg', 'rb') as h:
jpg = h.read()
with open('second.bmp', 'rb') as h:
bmp = h.read()
c, mac = att_merge_jpg_bmp(jpg, bmp, aad=b"")</pre></details>
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