As you know in the past few weeks it has emerged that NIST/NSA have been involved in weakening encryption standards over a long period of time so that they can retain the ability to break encryption used by corporations and the general public. We've already seen evidence of that with the Dual_EC_DRBG which has been recently disavowed by NIST after public outcry. That's probably a red herring, I think there's another smoking gun here.

I want to highlight a particular issue in the Advanced Encryption Standard process. The Rijndael algorithm had the option of block sizes of 128, 192 or 256 bits to match the key sizes. When that algorithm was decided as the winner, NIST went away and then defined a "restricted" block size of 128 bits for AES. Apparently that was what the spec called for in the first place.

However that doesn't make sense. If you were concerned about security, why would you have a 256 bit key size, but not have a matching block size? It would appear to be more secure to keep them at the same length. For example, a modern cipher like Threefish which has key sizes of 256, 512 or 1024 bits, also has matching block sizes of 256, 512 or 1024 bits.

  • Can the overall security of AES with a 256 bit key be reduced from 2256 to 2128 by attacking the lower block size?

  • Could that be combined that with various other attacks on AES which would reduce the 2128 bit security to even less e.g. 2100 bits?

  • Now that commercial quantum computers are viable (just one D-Wave Vesuvius is ranked in the top 10 of the TOP500 list), and assuming the NSA have taken advantage of that and constructed a quantum cluster in the basement of Fort Meade/Utah data center, if they employed Grover's algorithm which runs in O(N1/2) time, would that reduce the security of AES to 264 for all key sizes or worse 250 with attacks?

Extra links:

  • $\begingroup$ possible duplicate of Why most people use 256 bit encryption instead of 128 bit? $\endgroup$ Commented Sep 23, 2013 at 7:35
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    $\begingroup$ There is no reason for matching key and block size. You can't really compare Rijndael with ThreeFish, the latter was designed as a building block for a hashfunction (Skein) and its design choices reflect that goal. $\endgroup$ Commented Sep 23, 2013 at 9:49
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    $\begingroup$ Block size doesn't affect security much, for most modes of operations the only consequence is that you should encrypt significantly less than 2^(b/2) blocks of data using a single key. We currently encrypt nowhere near 2^68 bytes under one key, so 128 bit blocks are fine. $\endgroup$ Commented Sep 23, 2013 at 9:54
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    $\begingroup$ crypto.stackexchange.com/questions/6016/… $\endgroup$
    – hunter
    Commented Sep 23, 2013 at 12:24
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    $\begingroup$ The restriction to 128 bit block size happened either before the competition started, or at the start of the competition, not after the competition ended. Nobody has a quantum computer, but in the event, everyone will move to 256 bit keys, which is why 256 bit keys are there. The block size doesn't matter. $\endgroup$
    – K.G.
    Commented Sep 23, 2013 at 13:17

2 Answers 2


Security issues related to block size boil down to the following: a pseudorandom permutation is not a pseudorandom function, and the difference becomes visible when you query the function too many times. Imagine a function which accepts as inputs, and offers as outputs, elements from a set of size $N$. For instance, the inputs and outputs are blocks of $n$ bits, so $N = 2^n$. If you feed this function with different input values, then you will first get as many different output values. However, with a PRF, you expect to obtain a collision at some point: a new input which yields the same output as a previous distinct input. On average, this should occur after about $\sqrt{N}$ queries. With a PRP, you will never get a collision, because permutations are, by definition, injective.

So when encrypting with a block cipher, using blocks of $n$ bits, you begin to have problems when processing more than $2^{n/2}$ blocks with the same key. 3DES has 64-bit blocks, so encrypting more than $2^{32}$ 8-byte blocks with 3DES can be troublesome: that's 34 gigabytes or so, a value which is not so huge nowadays. To avoid this problem, NIST launched an open competition for an "Advanced Encryption Standard" (the AES) with the following rules:

  • 128-bit blocks.
  • Keys of 128, 192 and 256 bits.
  • Not slower than 3DES (many candidates turned out to be widely faster).

And that's what they defined. They did not really care that the selected candidate (Rijndael) had extended versions which could accept other block sizes; 128 bits were sufficient.

In all of the above, the important point is that the $2^{n/2}$ figure is related to the number of queries. It is generally estimated that, for an attacker, obtaining $x$ pairs plaintext/ciphertext is widely more difficult than trying out $x$ potential keys, since the former requires cooperation of the target system, while the latter requires only some computing power on the attacker's own machines. You'll have a hard time finding an "honest system" which needs to encrypt $2^{68}$ bytes with AES (that's close to 300 billions of gigabytes), and even more so an honest system where the attacker can get 300 billions of gigabytes of known plaintext while still being "locked out" (if the attacker can get the honest system to process that much data, then what left is there to attack ?).

Basically, the block size and the key size work on different scales, and it just happens, fortuitously, that "128 bits" is a nice level for keys (exhaustive search on $2^{128}$ keys is not feasible with current technology), and also a nice level for block size ($2^{64}$ blocks of 16 bytes of known plaintext is not a realistic attack scenario with current technology).

Notably, if a quantum computer ever sees the light, that computer would be able to break a 128-bit key in $2^{64}$ effort, but it would do absolutely nothing about the issues related to block size: 128-bit blocks would still be as good as they are today.

All of the above is for usage of a block cipher for encryption. When a block cipher is to be reused as a core element for something else, e.g. a hash function, then other constraints may apply. Typically, for the Skein hash function, the internal block cipher had to offer much larger blocks if the hash function was to live up to its promises (note that a hash function has no key, so we are far from the attack model of an encryption system). This is why Threefish, the block cipher in Skein, has big blocks. This is not because big blocks are good for encryption; this is because big blocks are necessary when a block cipher is turned into a hash function the way Skein does it.

A similar case is Whirlpool, which uses the "W" block cipher, which is a Rijndael derivative (with even larger blocks -- 512 bits -- and a revamped key schedule).

As for D-Wave systems, they are not "true" quantum computers; they compute a specific problem instance in a "somewhat quantumish way", but are not able to apply the nifty algorithms which eat crypto for breakfast. The difference is qualitative, not a question of mere tuning. See this for pointers.

  • $\begingroup$ Thanks Thomas. I see in DW's answer in 2011 he cites 7 sources to the same blog. Citing sources from a single guy is not evidence if the guy hasn't even had first hand experience with the system, just speculating about it. The links in my post show tests were done on the D-Wave Two and its raw speed ranks in the TOP10 supercomputer list. If it were not real, Lockheed Martin and Google would be asking for their $10 million back and that single system wouldn't be in the TOP10 beating out other supercomputers which are generally comprised of datacenters full of servers. $\endgroup$
    – Gabriel
    Commented Sep 26, 2013 at 3:10
  • $\begingroup$ If it's actually a working quantum computer it's now likely classified top secret need-to-know basis only. It's only a matter of time before NSA clusters 100 of them and figures out how to run quantum algorithms on them to crack most crypto. 100 D-Wave Twos = $1B = half the cost of the Utah datacenter. See NSA's black budget in the billions. Assume D-Wave Two is the real deal, does AES 128 have a sufficient "safety margin" against it? Going to AES 256 doesn't look much safer with the broken key schedule. $\endgroup$
    – Gabriel
    Commented Sep 26, 2013 at 3:18
  • $\begingroup$ D-Wave is not the "real deal" for breaking crypto. D-Waves are specific computers for finding approximate solutions to NP problems - and there is only "evidence for quantum entanglement"... which is not promising. For breaking crypto, the quantum entanglement is the only important aspect of qbits. But it would be interresting to know how many of their 512 qbits can actually be used reliably in this manner. Those quantum computers with full fledged qbits have much less qbits. I would be surprised if they would be ahead of academic research (the problems are mostly physical). $\endgroup$
    – tylo
    Commented Sep 26, 2013 at 14:42
  • $\begingroup$ About being in top10 supercomputers: Comparing a general purpose computer and a super-highly specialized computer in this specific field is not fair. Especially since they are not designed for crypto, this doesnt work. About "NSA clusters 100 of them"... exactly THIS is the main problem with real quantum computers: Creating a lot of qbits is easy, almost trivial. Quantum entanglement is what you need, and this is hard. You can't just "cluster a lot of them and be done". $\endgroup$
    – tylo
    Commented Sep 26, 2013 at 14:48
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    $\begingroup$ This sounds almost like "hey look, it's easy to factorize numbers under 1000. It has to be easy to factor numbers under $2^{1000}$". There's the fact that there aren't many facts researched about these systems. In my opinion this makes it more likely that their claim is bogus. Of course you can also interpret this swinging to the other side, saying it is more likely to work and be hidden from the public. This is a matter of paranoia. But I don't see the NSA as a gang with superpowers and that they are so far ahead dealing with physical problems. As I said, quantum entanglement is hard $\endgroup$
    – tylo
    Commented Sep 27, 2013 at 9:46

There is an issue with the key schedule needed to take take a 256 bit key and use it for a 128 bit block. It turns out that the key schedule for 256-bit AES keys was not as well worked out as for 128 bit keys. Whether this problem is a consequence of the mismatch between key size and block size isn't something I know.

Flaws in the 256-bit key schedule make it vulnerable to related key attacks (not something that should happen in well designed applications) and the "weakening" to date doesn't bring it below 128 bits. But this does have some people thinking that it is actually better to stay with 128-bit AES keys than move to larger ones:

for new applications I suggest that people don't use AES-256. AES-128 provides more than enough security margin for the forseeable future. But if you're already using AES-256, there's no reason to change.

I'm not sure I agree, but that is a bit of a digression from your principle question. To the extent that the problems with the 256 bit key schedule can be attributed to the 128 bit block size, then we have a problem because of that block size choice.

  • $\begingroup$ In that article you linked, due to the new attacks, Bruce Schneier recommended increasing the rounds to AES-128 at 16 rounds, AES-192 at 20 rounds and AES-256 at 28 rounds to retain the safety margin. The fact that there are published attacks on AES and no effort from NIST to strengthen the original algorithm means they aren't actually interested in improving it to keep up with future developments, therefore allowing NSA to retain the ability to crack it. Would you recommend a move to a different cipher or increasing the number of rounds (as per the article)? $\endgroup$
    – Gabriel
    Commented Sep 26, 2013 at 3:29
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    $\begingroup$ Changing a widely adopted standard is not something to do lightly. And there hasn't been a lot of demand to revise or develop a new standard, probably because under proper usage, related key attacks are irrelevant. $\endgroup$ Commented Sep 26, 2013 at 15:20

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