When AES(☢) = ☠ – a crypto-binary trick | Hacker News Reader

When AES(☢) = ☠ – a crypto-binary trick

speakerdeck.com

287 pointsange4771_34 comments

That was a great read. I saw the title and figured it would quickly go over my head but it's all pretty understandable.

Does anyone know where I can download the src to have a look through?

Edit: found it https://code.google.com/p/corkami/source/browse/#svn%2Ftrunk...

Same here. Great read, very well explained.

The play of unicodes reminds me of some awesome slides made by Larry Wall and other Perl hackers. A good slide show is about to convey messages in a visually efficient and succinct way, not in an overloaded and bloated one.

There's also a word play in the title. "AES" transliterates to "АЭС" (acronym for "Атомная Электростанция") in Russian (and some other Slavic languages), which means "nuclear power plant". Thus, the "☢" sign.

that's totally unintended: I just looked for some interesting unicode characters and chose these 2 for the video game references ;)

Actual link: https://speakerdeck.com/ange/when-aes-equals-episode-v

Recording of the talk: http://podcast.raumzeitlabor.de/#wbHkVZfCNuE

That's amazing. Didn't think it's even possible, however it turns out to be surprisingly simple. Also, laughed out loud because of that guy's twitter nickname on the 3rd slide.

Cool trick, I have encountered something like this in a steganography wargame before, the only difference is they used Base64 encoding on the original picture instead of AES :)

Base64 is not true encryption, it doesn't require a key it is merely obfuscation.

In the general usage case base64 is not even obfuscation, really (though it does make it non-human-readable... for most humans), just an encoding for dealing with situations where you need to store or transmit 8-bit or multibyte character data into a system that may otherwise be incapable of dealing with it reliably.

Of course it's not, but the method is the same, you append a different picture to the end and modify the original image header to point to it. The encoding or encryption is not relevant.

Does any one know the name of the hex editor used in these slides, the one showing the PNG chunks and JPEG information?

that's https://bitbucket.org/haypo/hachoir/wiki/Home

A good example of why a MAC after encryption is also needed. And blocking length extension attacks.

Where do length extension attacks come into play with angecrypt?

I love the "HexII" hex-dump format he links to, it's so much less cluttered than the traditional one. I'm definitely going to have to try that out the next time I'm picking apart some binary file.

thanks - HexII is in very early development for now.

this is awesome. now I hide secret information in a seemingly innocent image. no one would want to use AES to decrypt it if the image looks fine.

Where do you store the IV? Do you just append it at the end of the file?

You have to supply that, see the python script, at the end it prints what to run to decrypt, the determined IV is passed as a param:

http://corkami.googlecode.com/svn/trunk/src/angecryption/ang...

What's the benefit of AES using such small blocks?

The block size of a symmetric cipher will generally be in the neighborhood of the key length. You can imagine the problem if you had a 128-bit key but an 8-bit block size -- for each of the 256 possible inputs you would have only 256 possible outputs. There would be multiple keys that produced the same mapping and you could produce every possible mapping with only 65536 keys. So the block size needs to be near the key size. Making it larger than that wouldn't make it any more secure but would make it slower.

This doesn't sound right to me. There are 256! permutations of the set {0, 1, 2, 3, ..., 255}. 256! is much bigger than 2^128, so I see no reason that each key cannot produce a unique mapping.

> So the block size needs to be near the key size

Note that AES-256 has a 256 bit key, but the block size is 128 bits, which is not near the key size.

I believe that the main constraint on block size is that a small block limits the length of messages you can safely encrypt with a given key. If the bad guys see a lot of cipher text encrypted with the same key, they have a better chance of a successful attack. What "a lot of cipher text" means depends on the block size. The bigger the block size, the more cipher text is needed to constitute "a lot of cipher text".

Wow, that was stupid of me. Yes of course, the number of possible mappings is 256! rather than the square of 256. I don't know what I was thinking.

A smaller block also gives you less room to maneuver when designing modes of operation; for instance, it can be tricky to implement CTR with a 64 bit block --- the convention is to split the block into "counter" and "nonce", and you need enough space for the counter that it can't conceivably wrap.

This may be a stupid question, but why is there a convention to split?

I saw that in set three of the crypto challenge, as well, and wondered.

Is it so that you can always start counting at zero when re-starting the application, as long as you're randomly picking a new nonce?

And if you just picked a random counter value at each restart you might get very unlucky and at some point reuse a counter value, so by this convention you're separating counter values belonging to different restarts?

Conventions and usage vary by setting, but yes, it's to prevent nonce reuse.

A typical scenario might use a single encryption key for many different messages. A simple strategy is to allocate 64 bits to a message counter and 64 bits to a block counter. For each encryption you can increment the message counter by one. The block counter starts at zero and increments for each block of the message.

Note that this imposes hard limits on both the number of encryptions you can perform with this key (2^64) and the number of blocks a message can contain (also 2^64). As long as we respect these limits, we're guaranteed never to reuse a nonce.

If we use a random 64-bit nonce in place of the message counter, the picture changes a little. Due to the birthday paradox, we can now expect a collision in around 2^32 encryptions, which is a little close for comfort.

Fortunately, we can tune the bit allocations based on the needs of the application. So a reasonable strategy might be to bump the random message nonce to (say) 96 bits and leave 32 for the block counter. This still allows individual messages of around 64GB.

Of course, parameters like these should be considered implementation details for 99% of application developers. The best thing to do is to use a high-level library like NaCl that makes reasonable choices and then abstracts them away.

impressive! a very cool hack!

Impressive.

That's also the reason why one should limit the max-length of a password field (something reasonable), if one is using the salted-password in db approach. Otherwise someone could enter a very long password to do the trick (MD5/SHA1), see http://en.wikipedia.org/wiki/MD5#Security .

I guess I'm not following your logic. If there is a salted, hashed password in a db, allowing arbitrary length passwords shouldn't matter? HMACs and KDFs work very differently from symetric-crypto primatives.

The point is that it makes it easier for an attacker to find a hash collision. It's much easier to construct data which hashes to a given value if it can be of arbitrary length. I don't immediately see the connection with this article however.

The attacker doesn't need a hash collision in the case you're describing; they need a preimage.

I don't follow. Breaking a password hash your not trying to find a hash collision you need to break the preimage attack. Sure are some stage in the future there might be a preimage attack that requires a large amount of data to use. But really if your worried about theoretical preimage attack you aren't using md5.

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