## Recent Content:

Steamed fish that is finished with a drizzle of hot oil is a classic Cantonese dish. The way I like to make it includes a few Japanese ingredients, and my method is a bit unorthodox.

### Hardware

- 12 inch frying pan with lid
- A small raised rack that will fit inside the pan that can provide a centimeter or so of clearance above the bottom of the pan; the removable rack that came with my toaster oven works perfectly
- A very small bowl
- A small saucepan (it can be tiny)
- A platter for serving that is large enough for the fish and deep enough to hold some sauce

### Software

- 2 tbsp. soy sauce
- 1 tbsp. mirin
- 1 tbsp. shaoxing cooking wine, or you can substitute an additional tbsp. of mirin
- 1 tbsp. sake
- 1/4 cup of either water, katsuo dashi stock, or water plus half of an instant dashi stock packet
- 1 fillet of a large flaky fish. Approximately 1 pound. I often use rockfish or wild striped bass. If you cannot find a large fillet, a smaller whole fish can be used.
- 1/4 cup coarsely chopped cilantro
- 2 in. knob of ginger, peeled and julienned
- 2 scallions, both whites and greens, julienned (or thinly sliced at an oblique angle, similar to the thicker Chinese “horse ear” cut)
- hot chili pepper, sliced thin at an oblique angle (optional)
- 1/8 tsp. five spice powder
- 1/4 tsp. toasted sesame oil
- 2 tbsp. vegetable oil

### Algorithm

- Mix the soy sauce, mirin, shaoxing wine, sake, and water/dashi in the frying pan
- Place the fish into the pan and marinade for 15 minutes, flipping once half way
- Meanwhile, combine the ginger, cilantro, scallion, optional chili pepper, five spice and toasted sesame oil in a small bowl and mix
- Remove the fish from the marinade, place the rack into the pan, and the fish onto the rack
- Bring to a boil
- Once boiling, cover the pan tightly, reduce the heat to simmer, and steam for 8 minutes
- Meanwhile, start heating the vegetable oil in the small saucepan
- When the fish is done steaming, remove it and the rack from the pan, placing the fish on the platter
- Increase heat to high and let the sauce reduce to the consistency of Grade A (runny) maple syrup
- Pour the sauce over the fish and put the herb mixture on top of the fish
- When the vegetable oil is very hot, spoon it over top of the herbs

## Lenticrypt: a Provably Plausibly Deniable Cryptosystem

### or, This Picture of Cats is Also a Picture of Dogs

Back in 2009, I wrote about a thought experiment on how to subvert copyright law via plausible deniability. A couple years ago I expanded on that thought experiment by proposing a seedling idea on how to accomplish it via cryptography. Since then, I’ve slowly been developing that seedling into a functioning proof-of-concept, which has culminated in the creation of the Lenticrypt project:

Lenticrypt can generate a single ciphertext file such that different plaintexts are generated depending on which key is used for decryption:

$ python lenticrypt.py -e key1 plaintext1 -e key2 plaintext2 -o output.enc $ python lenticrypt.py -d key1 output.enc | diff - plaintext1 -s Files - and plaintext1 are identical $ python lenticrypt.py -d key2 output.enc | diff - plaintext2 -s Files - and plaintext2 are identical

Unlike alternative plausibly deniable cryptosystems like the
recently discontinued TrueCrypt—whose ciphertext size grows in
proportion to the number of plaintexts (*i.e.*, hidden volumes) it
encrypts—Lenticrypt’s ciphertext size is proportional to the
*largest* plaintext it encrypts. This is because Lenticrypt
shares bytes in the cyphertext between each of the plaintexts it
encrypts; they are not stored in separate regions of the
ciphertext. Therefore, there is no straightforward way to estimate the
number of plaintexts that are “hidden” inside a single ciphertext.

In fact, Lenticrypt has the theoretical property that, under
reasonable assumptions, there is always a near 100% probability that
there exists an key in the public domain that will decrypt a given
ciphertext to *any* desired plaintext, even if that key is not
known. Therefore, even if an incriminating plaintext is revealed, the
author of the ciphertext can plausibly deny having created it because
there is a non-zero probability that the plaintext was legitimately
decrypted by random chance.

More technical details on the cryptosystem as well as additional use-cases are described in Issue 0x04 of The International Journal of PoC||GTFO.

Note that Issue 0x04 of PoC||GTFO is a polyglot: Among other
things, you can also treat the `.pdf` file as if it were a
`.zip`. If you extract it, there are some neat
Leticrypt-related Easter eggs inside the feelies.

## Random Solutions are Often Good Enough

### Sometimes taking the easy way out isn’t nearly as bad as it might seem!

### Hard Problems

Computer Scientists and Software Engineers deal with
computationally hard problems all the time. But those challenges
don’t solely apply to those people *creating* the software;
they are equally relevant to those people *analyzing* it. For
cybersecurity researchers, finding a hash collision…
determining the user inputs that can assign a certain value to a
tainted EIP… deciding whether a black-box binary is
malicious… they’re all really hard! As our work becomes more
and more automated with the help of program analysis techniques, these
computationally hard problems become more and more prevalent. But
what makes one problem harder than another? In the remainder of this
post I’ll answer that question, and introduce some surprising
new results of my research that demonstrate how good solutions to
inherently very hard problems can be quickly generated. Read on for more.

First of all, thanks for all of your positive feedback on our recent post on physical security. One of the comments we’ve received multiple times is that these types of locks and the practice of using mnemonics for their codes is primarily limited to government facilities. It turns out that’s not really a limitation. Like a muted post horn, now that these locks have piqued our curiosity we’re starting to see them everywhere we look. Check out this find we made today:

Don’t bother checking, we scrubbed the Exif. Bank of America: Feel free to contact us directly if you’d like to know the location ;-)

Another thing we neglected to emphasize in the original post was
that our time estimates are upper bounds based off of the assumption
that the lock will have a four minute timeout for successive failed
attempts. For locks that do not have such a timeout (*e.g.*,
basically all mechanical locks, and presumably the one on this ATM),
you can divide the brute-force cracking time by at least a factor of
four.

## Exploiting Password Weaknesses in Physical Security

### In which I channel the spirit of my eighth academic cousin thrice removed, Richard Feynman.

Digital spin locks like the Kaba Mas X-09 and X-10 are very common for high security applications like vault doors. US General Services Administration approval means that they are nearly ubiquitous in securing government filing cabinets that contain documents that are classified secret, as well as securing the doors to laboratories in which secret code is written and stored. Strolling through the hallways of any Department of Defense facility or government contractor’s office, one is likely to see many of these on the doors:

Kaba Mas X-09 |

These types of locks are very sophisticated: they are self-powered
by rotating the dial and they have a digital printout on the top of
the code as it is being entered. Security features to thwart automated
brute force attack include random jumps in the numbers as the dial is
being turned, algorithms for detecting “non-human” rotations of the
dial, and long timeouts for successive failed attempts. These types of
locks are usually configured to accept a six-digit code that is input
as a sequence of three numbers from 0 to 99. That means there are
$math$10^6 = 100^3 = 1,000,000$math$ possible combinations. With a
four minute timeout, brute forcing all of the combinations would take
almost *eight years!*

Humans are generally bad at remembering long sequences of numbers,
and in most facilities the combinations to these locks are changed
frequently. A single employee might have to memorize dozens of
different combinations (one for each safe, door,
*etc.*). Therefore, as a mnemonic, the security officers who set
the codes often use six-letter words which are translated into codes
using the mapping from a phone keypad. In fact, one will almost always
see a magnet or sticker like this right next to the lock:

For example, the word “HACKER” would equate to the code 42-25-37.

1 | 2abc | 3def |

4ghi | 5jkl | 6mno |

7pqrs | 8tuv | 9wxyz |

0 |

And therein lies a vulnerability. First of all, using a phone dial pad for the mapping from letters to numbers means that the numbers zero and one will never be used! This means the number of potential codes to test has been decreased from one million to $math$8^6 = 262,144$math$. That’s a decrease of almost 75%! But we can do even better.

Next, let’s assume that the code word is a six-letter English word
that occurs in the dictionary. Looking at
`/usr/share/dict/words`, there are only 17,706 six-letter
English words that have direct mappings using a phone keypad. That’s
*much* lower than the 262,144 possible codes without zero and
one, which means that we have reduced the number of possible codes to
brute force by *over *98% from the original! For example, the
code 55-57-59 doesn’t appear to correspond to any words in the
dictionary. That’s likely because almost all words need at least one
vowel, and the phone keys 5, 7, and 9 don’t provide any
vowels. Furthermore, many of those 17,706 words actually map to the
same code. For example, the words “DOSAGE”, “ENRAGE”, “FORAGE”, and
“FORBID” all map to the code 36-72-43. If we just count the number of
*distinct* codes generated by those six-letter English words,
we get only 14,684, a reduction of almost 99%! With a four minute
failure timeout, brute forcing those 14,684 codes would only take
about *forty days.*

But wait, there’s more!

Not all letters occur in English words with the same frequency; you’re much more likely to find the letter ‘e’ in a randomly chosen word than the letter ‘x’. For example, almost 10% of the six-letter English words in the dictionary end in the code “37”, just like in our example word “HACKER”. Therefore, if we cleverly order our brute force approach to try codes that occur most frequently in English words, then we can increase our chances of cracking the code quicker.

I sorted all of the 14,684 distinct codes in decreasing order
according to the number of words that map to that code. If a code word
is chosen uniformly from the set of all six-letter English words, then
this ordering will maximize our chances of cracking the code
quickest. In fact, after fewer than 250 attempts, we will have cracked
the code with about 5% probability. In other words, with only about
*half a day* of access to the lock, a brute force attempt using
this ordering will have a 5% chance of success. In contrast, if the
lock’s code were instead set using a randomly generated six-digit
number, it would require on average 50,000 attempts (~5 months)
to crack it with 5% probability. On the other hand, if the code was
set from a six-letter English word using a phone keypad mapping, the
lock can be cracked with 50% probability after about 16 days.

**Codes are not usually changed on a month-to-month
frequency, so an adversary with intermittent access to the lock
( e.g., after hours) could quite conceivably crack it before the
code were changed.**

If there exists an additional vulnerability that allows the
attacker to detect whether each successive two-digit subcode is
correct before entering the next two-digit subcode, then the expected
value for the length of time required to crack the code is on the
order of *minutes*.

So what can we do about it? The most obvious solution is to simply choose codes randomly from the set of all possible one million 6-digit numbers. If we really do want to have a mnemonic, though, we could do a lot better by using a letter-to-number mapping that makes use of zeros and ones, and also produces a relatively even distribution of codes from English words. Here is one such example of a mapping:

1a | 2e | 3br |

4co | 5dnq | 6ip |

7lmjz | 8hkt | 9su |

0fgvwxy |

It’s not necessarily the best one, though. Finding the optimal mapping—the one that ensures a uniform distribution across all six-digit code for six-letter English dictionary words—is actually an instance of the multiple knapsack problem (a variant known as the multiple subset sum problem) and is therefore NP-Complete, and likely intractable for problems even as “small” as this one. Let $math$L = \{a, b, \ldots, z\}$math$ be the set of letters in the alphabet and let $math$f : L \rightarrow \mathbb{N}$math$ be a function mapping letters to their frequency among six-letter words in the dictionary. Then we can relax the problem to the minimization of the mean squared error over the summed frequencies of the letters mapped to each number: Find a function $math$t : L \rightarrow \{0, \ldots, 9\}$math$ that maps letters to numbers such that $equation$\frac{1}{10}\sum_{i=0}^9 \left(\sum_{\ell \in t^{-1}(i)} \left(f(\ell) - \frac{\sum_{\ell \in L} f(\ell)}{10}\right)\right)^2$equation$ is minimized. The total number of legal mappings, $math$t$math$, is $math$10^{26} = 100$math$ million billion billion. With that said, the example mapping above will be at least a million billion billion times more secure than using a phone keypad mapping!

**The moral of the story here isn’t that these locks are
inherently vulnerable; if used properly, they are in fact incredibly
secure. The takeaway is that these locks are only as secure as the
codes humans choose to assign to them. Using a phone keypad mapping on
six-letter English dictionary words is the physical security
equivalent of a website arbitrarily limiting passwords to eight
characters.**

A couple days ago, Dominic Spill and Michael Ossman presented an interesting talk at Shmoocon on using specially crafted error correcting codes to have unambiguous encapsulation, preventing attacks like “Packet in Packet.” This appears to be the culmination of some work that was first proposed to the LANGSEC mailing list back in March. The underlying problem that they are trying to address is that the physical layer is almost always error-prone: bits often get flipped. The data link layer addresses this by using error-correcting codes: methods of encoding the bits to be transmitted such that certain types of errors can be caught and corrected, at the expense of some additional bandwidth overhead.

### Packet in Packet Attacks

If a sufficient number of bit-errors occur, however, error
correcting codes can still fail to correct or even *detect* an
error. Protocols usually address this by adding a checksum in the
packet header/preamble. To illustrate this, let’s consider a made-up
link layer protocol in which there is a header that starts with the
magic number `0x1337` followed by a checksum and the message
body. If a detectable but uncorrectable error occurs, then the packet
can be dropped. If an undetected bit-error occurs in the body, then
the checksum will fail and the packet will be dropped. If an
undetected error occurs in the header, then either the magic will be
broken and it will fail to be detected as a frame or the checksum will
fail, in both cases causing the packet to be dropped. The one
exception to this latter case is the basis for Packet in Packet
attacks: If we were to start the message body with `0x1337`
followed by a checksum and our desired body, then the link layer
protocol will skip over the corrupted header and think that the actual
frame starts at the beginning of our hand-crafted body! In effect, we
have embedded a complete packet inside the body of another. It turns
out that if we transmit this carefully crafted packet-in-a-packet a
bunch of times, then eventually (in what turns out to be a reasonable
amount of time) a bit-error will occur causing our payload to be
interpreted as the real link layer frame. Attacks like this have been
demonstrated on real data link layer protocols like IEEE 802.15.4.

### Isolated Complementary Binary Linear Block Codes

Spill and Ossman propose using a special error correcting code that has a property they call “isolation:” the code can be split into multiple subcodes that have additional error-correcting properties between them. One of the subcodes could be used strictly for bits of the header/preamble, while another subcode could be used strictly for the body. This provides additional error correcting capability specifically for cases like that which the attack above exploits, without increasing the bandwidth overhead of the code. They call this family of codes “Isolated Complementary Binary Linear Block Codes” (ICBLBC).

As an example, consider the following eight code words:

`00000` | `00011` | `00101` | `01001` |

`01110` | `10110` | `11010` | `11100` |

Note that each code word differs from the others by at least two
bits. This means that we can detect single-bit errors. For example, if
`01110` is corrupted in the physical layer to `01010`, then we can immediately know that
an error occurred because that code does not exist in our set of
eight. If two or more bit-errors occur, then an error may go
undetected, which is the case that Packet in Packet attacks
exploit. This is exactly like the types of error correcting codes used
in most data link layer protocols. The difference, however, is that
this particular set of eight code words has an additional property: it
is an isolated complementary binary linear block code. Let’s call the
first row of codes “Subcode 1” and the second row of codes
“Subcode 2.” Now, note that every code in Subcode 1 differs
from every other code in Subcode 2 by at least *three*
bits. That means that we can not only *detect* but also
*correct* single-bit errors *between* the two codes, giving
the two subcodes properties of a *Hamming Code.* If we encode the header/preamble
of the packet using one subcode and the body using the other, then we
can circumvent Packet in Packet attacks with little to no extra
bandwidth overhead!

### Generating the Codes

Spill and Ossman have provided some code that can enumerate these codes. Currently, they have reported discovery of 6-bit ICBLBCs with up to twenty-two codes in subcode 1 and two codes in subcode 2, or up to ten codes in each if they are divided evenly. While their implementation does work, it is limited by the fact that it essentially performs a brute-force search to discover them, causing them to investigate a hardware-based approach using FPGAs.

Out of curiosity, I wondered if this problem could be solved more
efficiently using constraint reasoning. Therefore, we implemented a
generator that encodes the ICBLBC constraints as a Satisfiability Modulo Theories problem and used
DOCO (Digital Operatives’ proprietary constraint optimizer) to
enumerate all feasible solution codes. Sure enough, our solver was
able to almost instantaneously reproduce Spill and Ossman’s results,
as well as enumerate codes of larger bit length that would likely be
intractable to discover using brute force search. For example, in less
than a second it was able to discover this 7-bit ICBLBC consisting of
*fifty* codes in subcode 1 and two codes in
subcode 2:

Subcode 1 | Subcode 2 |
---|---|

37, 118, 44, 21, 35, 28, 107, 61, 41, 25, 94, 13, 56, 84, 124, 31, 110, 117, 50, 76, 62, 49, 47, 81, 7, 19, 109, 87, 103, 22, 93, 82, 70, 26, 127, 115, 55, 52, 59, 73, 11, 91, 88, 67, 69, 42, 121, 14, 38, 122 | 0, 96 |

as well as this one with

*nineteen*codes in each subcode:

Subcode 1 | Subcode 2 |
---|---|

23, 89, 17, 27, 9, 65, 51, 5, 83, 85, 0, 18, 80, 54, 36, 33, 113, 3, 116 | 76, 122, 61, 127, 103, 14, 107, 104, 79, 56, 109, 98, 28, 70, 94, 47, 110, 74, 42 |

This 8-bit ICBLBC was discovered in a matter of seconds and consists of

*eighty*codes in one subcode:

Subcode 1 | Subcode 2 |
---|---|

1, 4, 7, 8, 13, 16, 19, 25, 30, 32, 35, 37, 41, 42, 44, 47, 49, 54, 56, 59, 61, 69, 76, 81, 84, 87, 88, 97, 100, 104, 109, 110, 112, 121, 122, 124, 127, 128, 131, 133, 137, 142, 145, 146, 148, 152, 155, 161, 162, 164, 168, 171, 173, 176, 179, 181, 185, 190, 193, 196, 199, 200, 205, 208, 211, 217, 218, 220, 223, 224, 227, 229, 233, 234, 236, 239, 241, 246, 248, 253 | 66, 75 |

I have reimplemented this approach using Microsoft’s “free” (as in beer) Z3 constraint solver and pushed it to a fork of Spill and Ossman’s project, available here. As always, feedback on this new approach is greatly appreciated.