instruction stringlengths 15 30k |
|---|
A sponge has by definition 'wasted' operations (the part of the state which always remains private but goes through all the ops of the permutation).
In return for that waste you get a MAC at the end - some claim that it's for free for some reason.
But the ops wasted on the hidden state cannot compete with somethi... |
[Full-State Keyed Sponge][1] (aka Donkey Sponge) is when a message is absorbed into the full state of the sponge.
Such an approach to MAC construction is considered secure. [1]
However for an AEAD or a stream cipher, exposing the state obviously breaks it.
Would utilizing a XEX-like approach be able to secu... |
Can a Full-State Keyed Sponge be used as AEAD by using XEX? |
It is well-known that, in the setup phase of Pedersen commitment, it is necessary to generate $g,h\in G$. Then a user can compute $c = g^vh^r$ to commit the value $v$.
Is a trusted center needed to generate $g$ and $h$?
In other words, can $g$ and $h$ be publicly generated? |
Is the NIZK constructed based on Pedersen commitment transparent? |
A sponge has by definition 'wasted' operations (the part of the state which always remains private but goes through all the ops of the permutation).
In return for that waste you get a MAC at the end - some claim that it's for free for some reason.
But the ops wasted on the hidden state cannot compete with somethi... |
[Full-State Keyed Sponge][1] (aka Donkey Sponge) is when a message is absorbed into the full state of the sponge.
Such an approach to MAC construction is considered secure. [1]
However for an AEAD or a stream cipher, exposing the state obviously breaks it.
Would utilizing a XEX-like approach be able to secu... |
[Full-State Keyed Sponge][1] (aka Donkey Sponge) is when a message is absorbed into the full state of the sponge.
Such an approach to MAC construction is considered secure. [1]
However for an AEAD or a stream cipher, exposing the state obviously breaks it.
Would utilizing a XEX-like approach be able to secu... |
Alice：has $x=(x_1，x_2,...x_m）$
Bob: has $(y_1,m_1),...,(y_n,m_n)$
For this, Alice wants to get some message from Bob, but does not want bob to know which one she gets
- Bob generate random $a \in Z_q$，$cid_i=H(y_i^b),c_i=AES_{enc}(y_i^b, m_i),\forall i \in n$
- Bob send all $(cid_i,c_i)$ to alice
- Alice ge... |
> I don't know if this is a point that $G$ could generate.
Yes it is.
As stated in the question, this point satisfies the equation of [secp256r1][1]. That defines a group of prime order $n$ (given in the linked reference). That is, there are $n$ elements in the group formed by the point at infinity $\mathcal O$ a... |
I am trying to implement the RSA cryptography algorithm using C language. I am not sure of the size that each operand should have. Let me explain:
RSA requires to generate two huge prime numbers **p** and **q**, and compute their product **n = pq**.
Let's suppose that we want to use a key size = 2048 bits.
So we... |
For the inverting the private key from public key, we have a well defined problem; **the discrete logarithm problem**; namely,
- given $P$ over a prime curve $E(\mathbb{F}_p)$ with the generator $G$, find $x$ such that $P= [x]G$ .
There are many works for speeding the discrete logarithm;
- [Baby-step-Giant-S... |
For the inverting the private key from public key, we have a well defined problem; **the discrete logarithm problem**; namely,
- given $P$ over a prime curve $E(\mathbb{F}_p)$ with the generator $G$, find $x$ such that $P= [x]G$ .
There are many works for speeding the discrete logarithm;
- [Baby-step-Giant-S... |
According to wikipedia(markdown is striped below) for [Decisional Diffie–Hellman assumption][1]:
> the DDH assumption does not hold in the multiplicative group Z(p),
> where p is prime. This is because if g is a generator of Z(p), then
> the Legendre symbol of g^a reveals if a is even or odd.
For example I hav... |
Alice：has $x=(x_1，x_2,...x_m)$
Bob: has $(y_1,m_1),...,(y_n,m_n)$
For this, Alice wants to get some message from Bob, but does not want bob to know which one she gets
- Bob generate random $a \in Z_q$，$cid_i=H(y_i^b),c_i=AES_{enc}(y_i^b, m_i),\forall i \in n$
- Bob send all $(cid_i,c_i)$ to alice
- Alice ge... |
[Since $g$ is generator then it is not square][1], otherwise cannot generate the group (See the bottom theorem). Therefore, its Legendre symbol is $−1$.
Now, consider $g^a \bmod p$ and the Legendre calculation
$$\left(\frac{g^a}{p}\right) \equiv (g^a)^{\frac{p-1}{2}} \pmod p \quad \text{ and } \quad\left(\frac{... |
According to wikipedia(markdown is striped below) for [Decisional Diffie–Hellman assumption][1]:
> the DDH assumption does not hold in the multiplicative group Z(p),
> where p is prime. This is because if g is a generator of Z(p), then
> the Legendre symbol of g^a reveals if a is even or odd.
For example I hav... |
I am struggling with using Microsoft's BCrypt library in AES-XTS mode.
I've confirmed that the Cipher Block Size is 16 bytes (by getting the BCRYPT_BLOCK_LENGTH property). I'm setting the Data Unit Size to 256 (by setting the key property BCRYPT_MESSAGE_BLOCK_LENGTH); I realize 256 is a bit unconventional, and 512 o... |
I am struggling with using Microsoft's BCrypt library in AES-XTS mode.
I've confirmed that the Cipher Block Size is 16 bytes (by getting the BCRYPT_BLOCK_LENGTH property). I'm setting the Data Unit Size to 256 (by setting the key property BCRYPT_MESSAGE_BLOCK_LENGTH); I realize 256 is a bit unconventional, and 512 o... |
[Full-State Keyed Sponge][1] (aka Donkey Sponge) is when a message is absorbed into the full state of the sponge.
Such an approach to MAC construction is considered secure. [1]
However for an AEAD or a stream cipher, exposing the state obviously breaks it.
Would utilizing a XEX-like approach be able to secu... |
[Since $g$ is generator then it is not square][1], otherwise cannot generate the group (See the bottom theorem). Therefore, as being a QNR, it's Legendre symbol is $−1$; $$\left(\frac{g}{p}\right) = -1 \tag{1}\label{r1}$$
Now, consider $g^a \bmod p$ and the [Legendre calculation][2]
$$\left(\frac{g^a}{p}\right)... |
[Since $g$ is generator then it is not square][1], otherwise cannot generate the group (See the bottom theorem). Therefore, as being a QNR, it's Legendre symbol is $−1$; $$\left(\frac{g}{p}\right) = -1 \tag{1}\label{r1}$$
Now, consider $g^a \bmod p$ and the [Legendre calculation][2]
$$\left(\frac{g^a}{p}\right)... |
A sponge has by definition 'wasted' operations (the part of the state which always remains private but goes through all the ops of the permutation).
In return for that waste you get a MAC at the end - some claim that it's for free for some reason.
But the ops wasted on the hidden state cannot compete with somethi... |
[Since $g$ is generator then it is not square][1], otherwise cannot generate the group (See the bottom theorem). Therefore, as being a QNR, it's Legendre symbol is $−1$; $$\left(\frac{g}{p}\right) = -1 \tag{1}\label{r1}$$
Now, consider $g^a \bmod p$ and the [Legendre calculation][2]
$$\left(\frac{g^a}{p}\right)... |
A sponge has by definition 'wasted' operations (the part of the state which always remains private but goes through all the ops of the permutation).
In return for that waste you get a MAC at the end - some claim that it's for free for some reason.
But the ops wasted on the hidden state cannot compete with somethi... |
> Are sponges inherently inefficient when compared to other
> constructions?
In a way, yes. The sponge/duplex is [serial][1] rather than [parallelisable][2]. However, there are ways of making permutation-based cryptography parallel, like [Farfalle][1], [NORX][3], and [AEGIS-128X/256X][4].
Furthermore, as mention... |
> Are sponges inherently inefficient when compared to other
> constructions?
In a way, yes. The sponge/duplex is [serial][1] rather than [parallelisable][2]. However, there are ways of making permutation-based cryptography parallel, like [Farfalle][1], [NORX][3], and [AEGIS-128X/256X][4].
Furthermore, as mention... |
So I've been trying to implement a Non-Deterministic RBG based on randomness polling from non-hardware sources (nothing novel, just a side-project). Since collecting entropy and further conditioning is a fairly expensive process, I am using these bits to seed a CSPRNG which finally provides the user with a stream of ra... |
So I've been trying to implement a Non-Deterministic RBG based on randomness polling from non-hardware sources (nothing novel, just a side-project). Since collecting entropy and further conditioning is a fairly expensive process, I am using these bits to seed a CSPRNG which finally provides the user with a stream of ra... |
So I've been trying to implement a Non-Deterministic RBG based on randomness polling from non-hardware sources (nothing novel, just a side-project). Since collecting entropy and further conditioning is a fairly expensive process, I am using these bits to seed a CSPRNG which finally provides the user with a stream of ra... |
So I’ve been reading about different ways to store passwords on servers (plain text, encrypted, hashed, hashed + salt, slow hash + salt) and I wanted to find a way where a leak of the stored passwords isn’t an issue without resorting to very slow hashes (since slow hashes means a lot more computer power as people sign ... |
Can you have a public password file that is still secure? |
So I’ve been reading about different ways to store passwords on servers (plain text, encrypted, hashed, hashed + salt, slow hash + salt) and I wanted to find a way where a leak of the stored passwords isn’t an issue without resorting to very slow hashes (since slow hashes means a lot more computer power as people sign ... |
The NTRU key generation involves polynomials and their arithmetic in polynomial rings, which is a bit different from arithmetic in modular integers.
In the NTRU cryptosystem, the public key $h$ is computed as:
$h=g⋅F_q mod q \quad$ (1)
where $g$ is a chosen polynomial, $F_q$ is the modular inverse of $f$ modu... |
in NTRU, can g be recovered given f and h? |
The NTRU key generation involves polynomials and their arithmetic in polynomial rings, which is a bit different from arithmetic in modular integers.
In the NTRU cryptosystem, the public key $h$ is computed as:
$h=g⋅F_q \bmod q \tag{1}\label{r1}$
where $g$ is a chosen polynomial, $F_q$ is the modular inverse ... |
> (In [SHA-512,][1]) what's the difference between message-length field and message length?
The message-length field is a 128-bit bitstring, when the message length is a (non-negative) integer $\ell$ in range $[0\ldots2^{128})$.
The integer $\ell$ is the size of the message in bits, thus 3967 for a 3967-bit messa... |
Are there a cryptographically strong hash function and encryption function(s) where `hash(encryptA(data)) == encryptB(hash(data))`. The functions `encryptA()` and `encryptB()` may be identical, or just related.
If no such functions currently exist, are there theoretical reasons (relating to the one-way nature of ha... |
Equality between encrypted hash of data and hash of encrypted data? |
[Since $g$ is generator then it is not square][1], otherwise cannot generate the group (See the bottom theorem). Therefore, as being a $\text{QNR}$, it's Legendre symbol is $−1$; $$\left(\frac{g}{p}\right) = -1 \tag{1}\label{r1}$$
Now, consider $g^a \bmod p$ and the [Legendre calculation][2]
$$\left(\frac{g^a}{... |
A simple way to consider the question is: Can we factor a given 2048-bit RSA modulus $N$, assumed to be the product of two 1024-bit primes $p$ and $q$, if also given the 256 high-order bits of $p$?
That class of problems is studied by Don Coppersmith: [_Small Solutions to Polynomial Equations, and Low Exponent RSA V... |
Is it possible to use a Distributed Key Generation protocol, like the Pedersen DKG, to generate multiple key sets (one secret key and the corresponding public key) instead of only one, but without modifying the exchanged messages? |
This might sound like a naive question (And maybe it is (It's an early morning Saturday thought, so there might be some clarity of thought missing)).
Suppose we havea protocol that sends messages as follows:
Encrypt(<Message><Padding*>Hash<Message>)
*Padding is optional, but if present pads o... |
This might sound like a naive question (And maybe it is (It's an early morning Saturday thought, so there might be some clarity of thought missing)).
Suppose we havea protocol that sends messages as follows:
Encrypt(<Message><Padding*>Hash(<Message>))
*Padding is optional, but if present pads... |
Is there an encryption scheme where it's hard to guess a valid ciphertext without knowing the key and the scheme is CPA secure? |
I have an implementation of PBKDF2, which I know
* Has two bytes of '=' at the end of the input
* Has an input length of 24 (which is a Base64 encoded character representation of 16 bytes of entropy)
* Uses one iteration
* Has a known 9 character salt.
It's using a SHA1 hasher.
Does knowing the two `=` byt... |
Does having a known suffix on the input to PBKDF2 make you more vulnerable? |
[Full-State Keyed Sponge][1] (aka Donkey Sponge) appears to cross over into block cipher mode territory such as [Full Block Cipher Feedback Mode][2]:
[![Full Block Cipher Feedback Mode][3]][3]
Full State Keyed Sponge (FKS) construction:
[![Full-State-Keyed-Sponge-FKS-construction][4]][4]
FKS has been shown ... |
Full-Block Cipher Feedback Mode as a complete AEAD with a free MAC? |
[Full-State Keyed Sponge][1] (aka Donkey Sponge) appears to cross over into block cipher mode territory such as [Full Block Cipher Feedback Mode][2]:
[![Full Block Cipher Feedback Mode][3]][3]
Full State Keyed Sponge (FKS) construction:
[![Full-State-Keyed-Sponge-FKS-construction][4]][4]
FKS has been shown ... |
[Full-State Keyed Sponge][1] (aka Donkey Sponge) appears to cross over into block cipher mode territory such as [Full Block Cipher Feedback Mode][2]:
[![Full Block Cipher Feedback Mode][3]][3]
Full State Keyed Sponge (FKS) construction:
[![Full-State-Keyed-Sponge-FKS-construction][4]][4]
FKS has been shown ... |
> Does knowing the two = bytes at the end of input (which represent the base64 encoding for padding) reduce the complexity of finding all possible collisions for a 24 characters input?
Since you processed the password with base64, then the base64 may need padding. Depending on the input, base64 may add none, one, or... |
> Does knowing the two = bytes at the end of input (which represent the base64 encoding for padding) reduce the complexity of finding all possible collisions for a 24 characters input?
Since you processed the password with base64, then the base64 may need padding. Depending on the input, base64 may add none, one, or... |
Are there any cryptographically strong hash function and encryption function(s) where `hash(encryptA(data)) == encryptB(hash(data))`. The functions `encryptA()` and `encryptB()` may be identical, or just related.
If no such functions currently exist, are there theoretical reasons (relating to the one-way nature of ... |
Each Enigma machine setting induces a certain encryption in the sense of a function from the space of plain texts to the set of cipher texts.
The number of different Enigma machine settings can be counted for a given Enigma version (and usage). This is answered here: <https://crypto.stackexchange.com/questions/33628... |
> Does knowing the two = bytes at the end of input (which represent the base64 encoding for padding) reduce the complexity of finding all possible collisions for a 24-character input?
Since you pre-processed the users' password with base64, then the base64 may need padding. Depending on the input, base64 may add non... |
[Since $g$ is generator then it is not square][1], otherwise cannot generate the group (See the bottom theorem). Therefore, as being a $\text{QNR}$, it's Legendre symbol is $−1$; $$\left(\frac{g}{p}\right) = -1 \tag{1}\label{r1}$$
Now, consider $g^a \bmod p$ and the [Legendre calculation][2]
$$\left(\frac{g^a}{... |
I am new to RLWE; I am learning by myself. I came across Prof. Bill Buchanan's video explaining the RLWE-KEX, I understood everything he explained until the last part, where he is talking about removing errors from the shared key using probabilistic encryption.
Alice has the shared key as: sh = **A.Sb.Sa**+*Eb.Sa*
... |
Single-key Even-Mansour is secure up to $2^{0.5 \times n}$. Where $n$ is permutation and key size.
Would using $n/2$-key retain same security as $n$-key?
Could the other half be used as a tweak? |
How secure is half-key Even-Mansour? |
> Does knowing the two = bytes at the end of input (which represent the base64 encoding for padding) reduce the complexity of finding all possible collisions for a 24-character input?
Since you pre-processed the users' password with base64, then the base64 may need padding. Depending on the input, base64 may add non... |
I came across Prof. Bill Buchanan's video "Lattice Crypto: Ring LWE with Key Exchange" explaining the RLWE-KEX. I understood everything he explained until [the last part](https://youtu.be/QZBkhmqyBTA?t=1398), where he is talking about removing errors from the shared key using probabilistic encryption.
- Alice has th... |
I have an implementation of PBKDF2, which I know
* Has two bytes of '=' at the end of the input
* Has an input length of 24 (which is a Base64 encoded character representation of 16 bytes of entropy)
* Uses one iteration
* Has a known 9-character salt.
It's using a SHA1 hasher.
Does knowing the two `=` byt... |
> Does knowing the two = bytes at the end of input (which represent the base64 encoding for padding) reduce the complexity of finding all possible collisions for a 24-character input?
Since you pre-processed the users' password with base64, then the base64 may need padding. Depending on the input, base64 may add non... |
According to wikipedia(markdown is striped below) for [Decisional Diffie–Hellman assumption][1]:
> the DDH assumption does not hold in the multiplicative group $Z(p)$,
> where $p$ is prime. This is because if $g$ is a generator of $Z(p)$, then
> the Legendre symbol of $g^a$ reveals if a is even or odd.
For exa... |
> Does the presence of the hash compromise the security of the exchange? I tend to think that sending any redundancy in a predictable format tends to open attack vectors - but am I overthinking this one?
If the IV may be repeated during encryption then the hash can be used to give more information to an adversary wh... |
A lot of cryptography is based on the assumption that ${\sf P} \neq {\sf NP}$.
Is it conceivable to construct a cryptography system based on a class of much harder problems than ${\sf NP}$-problems, namely [undecidable problems](https://en.wikipedia.org/wiki/List_of_undecidable_problems) such as the [mortal matrix ... |
Cryptography based on uncomputable problems? |
Alice can use DH key exchange to derive a shared secret from Alice's encryption key E<sub>A</sub> and Bob's decryption key D<sub>B</sub>. This requires a "standing" D<sub>B</sub> from Bob, i.e. Bob must have published it. If Bob has only published a verifying key V<sub>B</sub>, is there a scheme where the following can... |
Scheme to derive an encryption key from one's verification key, so they can reduce the decryption key from it and their signing key? |
What is the AEGIS design rationale for one way rounds and slow diffusion? |
A semantically secure encryption scheme can not preserve the length of an encrypted plaintext.
How to prove this? or are there some methods that can be used to obtain such information from a semantic secure ciphertext? |
I'm trying to understand which benefits can using of extension fields in elliptic curve cryptography bring over prime fields. Popular curves like secp256k1, curve25519, secp384r1 are defined over a prime field. Are there any useful elliptic curves over extension fields and what are their usecases? |
Encryption, signatures, etc., can always be broken in NP.
* *Any* encryption scheme can be broken by a polynomial-time algorithm for the following problem: "does there exist a secret key and encryption randomness that explains these ciphertexts as an encryption of these plaintexts?" This problem is in NP assuming t... |
I'm writing a function that will allow me to analyze the randomness of some input data and return some stats about the probability of this data to be random.
Obviously it should be based on two things:
* How often each byte value is found / not found (randomness / distribution)
* And how much data we have (cer... |
There is DH key exchange, but it requires an exchange before participants get to the shared secret. If Bob has only published a verifying key V<sub>B</sub>, is there a scheme where the following can be achieved?
1. Alice, from Bob's verification key V<sub>B</sub> and some nonce N, derives an ephemeral encryption key... |
Encryption, signatures, etc., can always be broken in NP.
* You can break *any* encryption scheme if you can solve the following decision problem: "does there exist a secret key and encryption randomness that explains these ciphertexts as an encryption of these plaintexts?" This problem is in NP assuming the encryp... |
I came across this thread accidentally, I am interested in this matter. Data with high entropy is hard to compress. Claud Shannon syas that it is impossible.
So, how can you tell how 'random' your test data is? There is a tool I tried, find it here: https://github.com/lsauer/entropy. It is called "ent", it will ana... |
I came across this thread accidentally, I am interested in this matter. Data with high entropy is hard to compress. Claud Shannon says that it is impossible.
So, how can you tell how 'random' your test data is? There is a tool I tried, find it here: https://github.com/lsauer/entropy. It is called "ent", it will ana... |
As stated for the question above here's an analogy:
*You are a robber looking for a house to rob with two different scenarios that might occur.*
*1. You have a key that you know belongs to a house and you can use it to let you inside. However, you have no clue for WHICH house it belongs to. At any neighborhood an... |
For Symmetric Cryptography, why is it considered more important to safeguard a key than the function/algorithm for encrypting/decrypting a message? |
As stated for the question above here's an analogy:
*You are a robber looking for a house to rob with two different scenarios that might occur.*
*1. You have a key that you know belongs to a house and you can use it to let you inside. However, you have no clue for WHICH house it belongs to. At any neighborhood an... |
As stated for the question above here's an analogy:
*You are a robber looking for a house to rob with two different scenarios that might occur.*
*1. You have a key that you know belongs to a house and you can use it to let you inside. However, you have no clue for WHICH house it belongs to. At any neighborhood an... |
Each Enigma machine setting induces a certain encryption in the sense of a function from the space of plain texts to the set of cipher texts.
The number of different Enigma machine settings can be counted for a given Enigma version (and usage). This is answered here: <https://crypto.stackexchange.com/questions/33628... |
Could anyone direct me to literature regarding privacy proofs in the MPC setting.
For example, how can one prove the following simple problem:
Suppose a setting with $n$ parties $S_1, \ldots, S_n$ wish to compute an additive sharing of $0$, so $r_i\in S_i$ such that $\sum r_i = 0$.
The parties can be assumed t... |
This won't be a full-fledged answer, since I don't have good pointers in mind (I read very few MPC textbooks). Still, a few comments on your proposal:
- the construction can be simplified a lot. First, you should view the $k_{ij} = k_{ji}$ as 2-party additive shares of zero. For that, simply have one of the parties... |
assuming you have encryptionvthat has a pass phrase then you can go no higher, but what if there is no pass phrase and everything is operated on and evaluated WHILE it is in it's encrypted state? An encryption where it is not reversible, like time where it is for almost all practical purposes irreversable? not sure how... |
I think if you consider how symmetric key cryptography is used in the real world, it becomes clear that the idea of keeping algorithms secret is not practical. A useful and ubiquitous example is [TLS][1]. TLS is typically thought of as an asymmetric system but one of the things that asymmetric interaction is used for ... |
Respected community,
I am using the OpenSSL API's, especially ```libcrypto``` and ```libssl``` and I am really depressed, thinking about side channel attacks on my implementation. My target platform is Windows, and I am programming in C and C++ and I am linking the libraries dynamically.
I am doing symmetric enc... |
Can attackers see the values I pass on to OpenSSL api functions? |
By Shannon's Theorem of perfect security if a give you a ciphertext 'LOUPL' you can do a bute forced attack you would found plaintexts like HELLO, 'APPLE', 'SPOON' but you can't distinguish the true plaintext from the fake one, unless with a probability in 1 of 23⁵. That's a theoretical information security warranty.
... |
Does Groover's algorithm really threaten symmetric cryptography? |
Does Grover's algorithm really threaten symmetric cryptography? |
I'm glad there are these types of questions that inspires the spirit of learning, eventhough this question is a bit too broad. Also, Chris Peikert is a member of this site, he may drop by if we're lucky.
----
First, key generation.
Generally, there are 2 major paradigms of key generation: NTRU, and learning-wi... |
The [MCOE][1] mode provides a really nice way to implement a nonce-misuse resistant AEAD and is based on an arbitrary tweak block cipher.
A specific requirement of the tweak block cipher is that the data block length is the same size as the tweak.
As I use Skein to perform key derivation and hashing, the Threefis... |
The existing answers are good. Here's a way of looking at it that might fit your intuition better:
> I just read for a key length being around 128 bits of length results in 340,282,366,920,938,463,463,374,607,431,768,211,456
And there are about 6 good algorithms to choose from.
Which is harder for the attacker -... |
The existing answers are good. Here's a way of looking at it that might fit your intuition better:
> I just read for a key length being around 128 bits of length results in 340,282,366,920,938,463,463,374,607,431,768,211,456
And there are about 6 good algorithms to choose from.
Which is harder for the attacker -... |
The existing answers are good. Here's a way of looking at it that might fit your intuition better:
> I just read for a key length being around 128 bits of length results in 340,282,366,920,938,463,463,374,607,431,768,211,456
And there are about 6 good algorithms to choose from.
Which is harder for the attacker -... |
The existing answers are good. Here's a way of looking at it that might fit your intuition better:
> I just read for a key length being around 128 bits of length results in 340,282,366,920,938,463,463,374,607,431,768,211,456
And there are about 6 good algorithms to choose from.
Which is harder for the attacker -... |
The existing answers are good. Here's a way of looking at it that might fit your intuition better:
> I just read for a key length being around 128 bits of length results in 340,282,366,920,938,463,463,374,607,431,768,211,456
And there are about 6 good algorithms to choose from.
Which is harder for the attacker -... |
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