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19

Yes, this is a widely-used cryptographic construction called a stream cipher. For more information about this and other encryption schemes, Coursera's cryptography class is a good resource.


17

It's a good question. As pg1989 said, this is the basis behind stream ciphers, which are very fast in practice. I thought I'd quickly expand upon your statement that "the one-time pad is the perfect cipher and impossible to crack." This is true, in a sense, but it's worth pointing out that sometimes an attacker wants to do something simpler than "cracking" ...


12

First, I'll assume we're talking about encrypting/decrypting exactly 128 bits of data, i.e. the block size of AES. Otherwise, you'll need to specify a mode of operation — and if your data's length isn't a multiple of the block size, well, that'll be more difficult to deal with. So, I'll assume we're working with a single block. (If you are using a mode ...


8

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 $n$ ...


6

The answer is yes, non-US ciphers exist and are in fact very popular. Actually, some who are looking for alternatives, opt for non-NSA/NIST ciphers, for instance Salsa/ChaCha from DJB (who is US citizen). A lot of ciphers have been developed in EU and Japan. China definitely has developed ciphers for its own use, just like many other countries. But long ...


6

A stream cipher where you can calculate the stream at any offset without deriving the prior stream bytes is probably the simplest option. AES-CTR is a mode that uses AES like a stream cipher. To decrypt at a random spot, you need only know the offset from the beginning and you can perform a single AES encryption call. AES-CTR overview: Generate a unique ...


6

As noted in this answer and this answer to another question, permutation is just a mathematical term for a function $\sigma:X{\rightarrow}X$ that maps a finite set $X$ onto itself, in such way that for each $y \in X$ there exists exactly one $x \in X$ such that $\sigma(x) = y$. This is also equivalent to how the term substitution is used in cryptography, so ...


6

Well, it turns out that depends on what you mean by "the AES cipher". If you are talking about the block cipher primitive, that is, if you define an alternate block cipher by taking AES, and swapping the 'encrypt' and 'decrypt' directions, well, that alternative block cipher is precisely as strong as AES. It can be used in any mode of operation we would ...


6

You should think of Rijndael's S-box as a function that maps bytes to bytes, where a byte (octet) is considered to be a member of a finite field of size $2^8$ (with xor as addition). It's not seen as a 16x16 octet array, really. The substitution is then just done byte-wise: every octet in the 4x4 block is replaced by its function value under the S-box ...


6

Some brief thoughts: Shared secret Generation: $$s=E_a(B)=E_b(A)$$ The shared secret is generated by encrypting the other users public key with your private key. This is effectively an ECDH step, which is very reasonable, and one of the key aims of C25519$^{[1]}$. Key Generation: $$s_0=\mathrm{SHA256}(s); s_i=\mathrm{SHA256}(s_{i-1})$$ First, using the ...


5

Congratulations you just reinvented the stream cipher. The main strength of the one-time pad is that the key space is as large as the message space. This means that any cipher-text only attacks always fail because all plaintexts are valid. This automatically means that any construct that decreases the key space (like using a seed for a PRNG) severely ...


5

If using a cryptographically-secure random number generator then the result is a stream cipher. If using actual random numbers, then it's a one-time pad. Any output you get from a random source needs to be run through a randomness extractor anyway in a 2:1 ratio (2 bits in, 1 bit out). Don't forget to provide a MAC along with the ciphertext to prevent an ...


5

As an Iranian Cryptology student in one of the most well-known Iranian Universities called Sharif University of Technology, I want to add this to the answers. There doesn't seem to be any National Standard Cipher here in Iran. But It doesn't mean that there shouldn't be any classified cipher being used by the military or the revolutionary guards. As I am ...


5

Any block cipher in CTR mode can be used to encrypt and decrypt data in arbitrary order. Basically, to encrypt something in CTR mode, you use the block cipher to encrypt a simple sequence of values, like (1, 2, 3, 4, 5, etc.), and concatenate the results to produce a pseudorandom bitstream, and then XOR this "keystream" with the data you want to encrypt. ...


4

Salsa20 has strong rotational symmetry. The main point of these constant is that they're not invariant under rotations, introducing an asymmetry. The precise value isn't very important, as long as it's sufficiently asymmetric. Bernstein - Salsa20 security says: Notes on the diagonal constants Each Salsa20 column round affects each column in the ...


4

From what I understand from your question, you are describing a stream cipher. If the one-time pad is the perfect cipher and impossible to crack, why would the following algorithm not be one of the strongest ... You're on the right track; a one-time pad is essentially a perfect (unbreakable) stream cipher. Without going into (any) mathematical ...


4

This is a special case of the affine cipher where $m=26$. Let's encrypt a single letter using your $E$. Let it be m, say, which is at index 12. So, $$E(12) = (7 \cdot 12 + 10) \mod{26} = 16$$ Now if we try to use the $D$ in your question, we decrypt this as: $$D(16) = (7 \cdot 16 - 10) \mod{26} = 24$$ which is obviously not right. The issue is that your ...


4

Since this is homework, I'm not going to give the answer, but will hopefully point you in the right direction. You are correct when you say for a perfect cipher, the probability should hold that $\mathbb{P}(P=m|C=c)=\mathbb{P}(P=m)$. In words, this means that given that you see the ciphertext is $c$ what is the probability of the plaintext being $m$ (note, ...


4

Well, for one thing, you are not using a "One Time Pad". A "One Time Pad" means, by definition, that someone generates a pad of numbers using true randomness (and not algorithmicly), and that no potential adversary has any information on what that pad may contain. Then, that pad is given to both the sender and the receiver, and then the sender uses it to ...


3

This is indeed a good question; let me try to make it a bit more precise. Suppose: Alice has a plaintext message of some number of bits, call it p. Alice and Bob share a crypto-strength random number generator that generates n truly random bits. Alice and Bob share a pseudo-random number generator that can take a seed of size n and produce one of 2n ...


3

That sort of thing is known as multi-party computation, and you should use a Socialist Millionaire Protocol for your particular instance.


3

standard AES disclaimer: Given the questions you've asked, you should not implement AES yourself in a real-world system because there are lots of security considerations when implementing ciphers. Think of the S-box as a function from byte $ \to $ byte. So, to look up the image of $x$ under the s-box transformation, you simply use $S_\text{box}(x)$, which ...


3

Item 2 has been answered satisfactorily, so this will focus on point 1: the s-box. The size of the s-box is not a 16x16 array unless it is viewed as such. The s-box is actually an 8-bit non linear transformation of the input, and is only viewed as a 16x16 array if you arrange it as a table of such dimensions. This array would then be a 1 to 1 representation ...


3

What is the significance of the repetition? Does it mean that a 6 character key was used and repeated across the same characters within P1 and P2??? I'm assuming your assumption about this being an xor cipher with the pad used twice for two ciphertexts. It does not mean a 6 character key was used. A one time pad used twice could result in this ...


2

If it is a long text and you are sure it is a substitution cipher there are a lot of statistical methods to find the key. Statistical methods rely on the frequency at which alphabets occur in a language. Not just single alphabets but even pairs. For eg in the english alphabet 'E' is the most frequent alphabet followed by 'T' and 'A'. Also some characters ...


2

One time pad is definitely both easy to do and has perfect secrecy, but key management is a pain and can compromise security. Basically a Vigenère cipher with a key as long as the the message should be secure, because different keys can create ALL possible messages with equal probabilities. Again, it's a one time pad, so no KPA, CPA, or CCA security. ...


2

Let's assume a ciphertext only attack, and compare it to Vigenere. Your scheme is on a quite similar level of security: It might be hard to break it without a computer, but with one it is probably done in seconds. Here's how: First, your permutation is static and only depends on the number of rounds and the length of the text, therefore you can just reorder ...


2

In the asymmetric encryption context I think something can be done in this direction with a double trapdoor function. I studied few examples of them in the past and briefly you can build up an encryption scheme with a "local" trapdoor and a "global" one. If you keep the secret for the global trapdoor for you you'll have a sort of escrow key allowing you to ...


2

Here is your algorithm in formula form, which might be easier to understand for non-objective-C-experts: Encryption: $P_1 || P_2 || \dots || P_n := P$ (split plaintext in blocks, same size as hash output, except the last one) $K := SHA256(key)$ $O_0 := IV (= K)$ $O_i := HMAC_{SHA256}(K, O_{i-1})$ for all $i \in \{ 1, \dots, n\}$ (key stream) $C_i := P_i ...


2

Keeping the public key private: Does not help in protecting data. Does not work: depending on the algorithm, the public key could be recomputed from the exchanged messages. Does not work: something which has been communicated to more than two people cannot be considered secret anymore. Especially since public keys are commonly written everywhere without ...



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