Wikipedia says Camellia comes with a block size of 128 and a variable key size (128, 192, 256). Another site lists it as a 256 Bit cipher.
The OpenSSL API has a function named EVP_camellia_256_cbc. Does this refer to the key size or the block size? And does Camellia support 256 Bit block sizes at all?
The information on the Wikipedia page is correct: Camellia has a fixed block size of 128 bit and a variable key size of 128, 192 and 256 bit. You can compare that with other authoritative sources like its specification, e.g. found in RFC 3713.
The "256 bit" in "256 bit cipher" usually refers to its security level and that is determined by its key size (and potential attack vectors that might decrease it).
Therefore, EVP_camellia_256_cbc means Camellia with a 256 bit key size, so you should supply keys of that size. Supplying keys of the correct key size is important, because some implementations may behave differently than others and you will lose a lot of time debugging when trying to connect different implementations.
For example, if you define that you want to use Camellia-256, but you're passing a key of 192 bit, it may happen that
one implementation fills the passed key with 0x00 byte up to the specified key size,
another implementation doesn't care about the specification and only looks at the actual supplied key to then run Camellia-192 or
a broken implementation (for non-standard key sizes) that calculates the number of rounds (12 or 14 for Camellia) that need to be used and arrives at a non-standard number of rounds which makes the result non-compatible with all other implementations.
Related
To all homomorphic encryption experts out there:
I'm using the PALISADE library:
int plaintextModulus = 65537;
float sigma = 3.2;
SecurityLevel securityLevel = HEStd_128_classic;
uint32_t depth = 2;
//Instantiate the crypto context
CryptoContext<DCRTPoly> cc = CryptoContextFactory<DCRTPoly>::genCryptoContextBFVrns(
plaintextModulus, securityLevel, sigma, 0, depth, 0, OPTIMIZED);
could you please explain (all) the parameters especially intrested in ptm, depth and sigma.
Secondly I am trying to make a Packed Plaintext with the cc above.
cc->MakePackedPlaintext(array);
What is the maximum size of the array? On my local machine (8GB RAM) when the array is larger than ~8000 int64 I get an free(): invalid next size (normal) error
Thank you for asking the question.
Plaintext modulus t (denoted as t here) is a critical parameter for BFV as all operations are performed mod t. In other words, when you choose t, you have to make sure that all computations do not wrap around, i.e., do not exceed t. Otherwise you will get an incorrect answer unless your goal is to compute something mod t.
sigma is the distribution parameter (used for the underlying Learning with Errors problem). You can just set to 3.2. No need to change it.
Depth is the multiplicative depth of the circuit you are trying to compute. It has nothing to with the size of vectors. Basically, if you have AxBxCxD, you have a depth 3 with a naive approach. BFV also supports more efficient binary tree evaluation, i.e., (AxB)x(CxD) - this option will reduce the depth to 2.
BFV is a scheme that supports packing. By default, the size of packed ciphertext is equal to the ring dimension (something like 8192 for the example you mentioned). This means you can pack up to 8192 integers in your case. To support larger arrays/vectors, you would need to break them into batches of 8192 each and encrypt each one separately.
Regarding your application, the CKKS scheme would probably be a much better option (I will respond on the application in more detail in the other thread).
I have some experience with the SEAL library which also uses the BFV encryption scheme. The BFV scheme uses modular arithmetic and is able to encrypt integers (not real numbers).
For the parameters you're asking about:
The Plaintext Modulus is an upper bound for the input integers. If this parameter is too low, it might cause your integers to overflow (depending on how large they are of course)
The Sigma is the distribution parameter for Gaussian noise generation
The Depth is the circuit depth which is the maximum number of multiplications on a path
Also for the Packed Plaintext, you should use vectors not arrays. Maybe that will fix your problem. If not, try lowering the size and make several vectors if necessary.
You can determine the ring dimension (generated by the crypto context based on your parameter settings) by using cc->GetRingDimension() as shown in line 113 of https://gitlab.com/palisade/palisade-development/blob/master/src/pke/examples/simple-real-numbers.cpp
I'm using python and cryptography.io to sign and verify messages. I can get a DER-encoded bytes representation of a signature with:
cryptography_priv_key.sign(message, hash_function)
...per this document: https://cryptography.io/en/latest/hazmat/primitives/asymmetric/ec/
A DER-encoded ECDSA Signature from a 256-bit curve is, at most, 72 bytes; see: ECDSA signature length
However, depending on the values of r and s, it can also be 70 or 71 bytes. Indeed, if I examine length of the output of this function, it varies from 70-72. Do I have that right so far?
I can decode the signature to ints r and s. These are both apparently 32 bytes, but it's not clear to me whether that will always be so.
Is it safe to cast these two ints to bytes and send them over the wire, with the intention of encoding them again on the other side?
The simple answer is, yes, they will always be 32 bytes.
The more complete answer is that it depends on the curve. For example, a 256-bit curve has an order of 256-bits. Similarly, a 128-bit curve only has an order of 128-bits.
You can divide this number by eight to find the size of r and s.
It gets more complicated when curves aren't divisible by eight, like secp521r1 where the order is a 521-bit number.
In this case, we round up. 521 / 8 is 65.125, thus it requires that we free 66 bytes of memory to fit this number.
It is safe to send them over the wire and encode them again as long as you keep track of which is r and s.
I have a manufacturer unique number ID of 128 bits that I cannot change and it's size is just too long for our purpose (2^128). This is on some embedded micro controller.
One idea is to compute a (run time) CRC32 or hash for narrowing the results but I am not sure for unicity CRC32 as a example: this can be unique for 2^32
Or what king of cryptography function I can use for guarantee unicity of 32 bits output based on unique input?
Thanks for clarifications,
If you know all these ID values in advance, then you can check them using a hash table. You can save space by storing only as many bits of each hash value as are necessary to tell them apart if them happen to land in the same bucket.
If not, then you're going to have a hard time, I'm afraid.
Let's assume these 128-bit IDs are produced as the output of a cryptographic hash function (e.g., MD5), so each ID resembles 128 bits chosen uniformly at random.
If you reduce these to 32-bit values, then the best you can hope to achieve is a set of 32-bit numbers where each bit is 0 or 1 with uniform probability. You could do this by calculating the CRC32 checksum, or by simply discarding 96 bits — it makes no difference.
32 bits is not enough enough to avoid collisions. The collision probability exceeds 1 in a million after just 93 inputs, and 1 in a thousand after 2,900 inputs. After 77,000 inputs, the collision probability reaches 50%. (Source).
So instead, your only real options are to somehow reverse-engineer the ID values into something smaller, or implement some external means of replacing these IDs with sequential integers (e.g., using a hash table).
I'm a bit confused on the difference between SHA-2 and SHA-256 and often hear them used interchangeably. I think SHA-2 a "family" of hash algorithms and SHA-256 a specific algorithm in that family. Can anyone clear up the confusion.
The SHA-2 family consists of multiple closely related hash functions. It is essentially a single algorithm in which a few minor parameters are different among the variants.
The initial spec only covered 224, 256, 384 and 512 bit variants.
The most significant difference between the variants is that some are 32 bit variants and some are 64 bit variants. In terms of performance this is the only difference that matters.
On a 32 bit CPU SHA-224 and SHA-256 will be a lot faster than the other variants because they are the only 32 bit variants in the SHA-2 family. Executing the 64 bit variants on a 32 bit CPU will be slow due to the added complexity of performing 64 bit operations on a 32 bit CPU.
On a 64 bit CPU SHA-224 and SHA-256 will be a little slower than the other variants. This is because due to only processing 32 bits at a time, they will have to perform more operations in order to make it through the same number of bytes. You do not get quite a doubling in speed from switching to a 64 bit variant because the 64 bit variants do have a larger number of rounds than the 32 bit variants.
The internal state is 256 bits in size for the two 32 bit variants and 512 bits in size for all four 64 bit variants. So the number of possible sizes for the internal state is less than the number of possible sizes for the final output. Going from a large internal state to a smaller output can be good or bad depending on your point of view.
If you keep the output size fixed it can in general be expected that increasing the size of the internal state will improve security. If you keep the size of the internal state fixed and decrease the size of the output, collisions become more likely, but length extension attacks may become easier. Making the output size larger than the internal state would be pointless.
Due to the 64 bit variants being both faster (on 64 bit CPUs) and likely to be more secure (due to larger internal state), two new variants were introduced using 64 bit words but shorter outputs. Those are the ones known as 512/224 and 512/256.
The reasons for wanting variants with output that much shorter than the internal state is usually either that for some usages it is impractical to use such a long output or that the output need to be used as key for some algorithm that takes an input of a certain size.
Simply truncating the final output to your desired length is also possible. For example a HMAC construction specify truncating the final hash output to the desired MAC length. Due to HMAC feeding the output of one invocation of the hash as input to another invocation it means that using a hash with shorter output results in a HMAC with less internal state. For this reason it is likely to be slightly more secure to use HMAC-SHA-512 and truncate the output to 384 bits than to use HMAC-SHA-384.
The final output of SHA-2 is simply the internal state (after processing length extended input) truncated to the desired number of output bits. The reason SHA-384 and SHA-512 on the same input look so different is that a different IV is specified for each of the variants.
Wikipedia:
The SHA-2 family consists of six hash functions with digests (hash
values) that are 224, 256, 384 or 512 bits: SHA-224, SHA-256, SHA-384,
SHA-512, SHA-512/224, SHA-512/256.
Intel's official optimization guide has a chapter on converting from MMX commands to SSE where they state the fallowing statment:
Computation instructions which use a memory operand that may not be aligned to a 16-byte boundary must be replaced with an unaligned 128-bit load (MOVDQU) followed by the same computation operation that uses instead register operands.
(chapter 5.8 Converting from 64-bit to 128-bit SIMD Integers, pg. 5-43)
I can't understand what they mean by "may not be aligned to a 16-byte boundary", could you please clarify it and give some examples?
Certain SIMD instructions, which perform the same instruction on multiple data, require that the memory address of this data is aligned to a certain byte boundary. This effectively means that the address of the memory your data resides in needs to be divisible by the number of bytes required by the instruction.
So in your case the alignment is 16 bytes (128 bits), which means the memory address of your data needs to be a multiple of 16. E.g. 0x00010 would be 16 byte aligned, while 0x00011 would not be.
How to get your data to be aligned depends on the programming language (and sometimes compiler) you are using. Most languages that have the notion of a memory address will also provide you with means to specify the alignment.
I'm guessing here, but could it be that "may not be aligned to a 16-byte boundary" means that this memory location has been aligned to a smaller value (4 or 8 bytes) before for some other purposes and now to execute SSE instructions on this memory you need to load it into a register explicitly?
Data that's aligned on a 16 byte boundary will have a memory address that's an even number — strictly speaking, a multiple of two. Each byte is 8 bits, so to align on a 16 byte boundary, you need to align to each set of two bytes.
Similarly, memory aligned on a 32 bit (4 byte) boundary would have a memory address that's a multiple of four, because you group four bytes together to form a 32 bit word.