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5

I think the 10100 is a typo and should be $10^{100}$ as shown here The period would be something along the lines of how long until the byte stream repeats. For example if the byte stream were "ABCDABCDABCD" and so on, then the period would be 4. For security you want a large period so that you can encrypt large amounts of data.


5

A Diffie-Hellman exchange needs not be synchronous. In DH, each party has a private key (x) and a public key (gx mod p). If the sender knows the recipient's public key ga, then he can build his own key pair (b and gb), compute the shared secret (gab), and send both his public key (gb) and the encrypted message (symmetric key derived from gab) to the ...


5

There is a line between encryption and obfuscation, I would say this is on the latter side. If someone knows the method, and is able to correctly guess even 1 of the original values, simple math will reveal all the original values. Additionally, depending on the type of data in the field, guessing the floating point number may only take seconds even with no ...


3

In short: hybrid encryption. Encrypt the content with a random symmetric key. And encrypt this key with the public key of each recipient. Then all you need is a public key from all the people you want to share the content with. It doesn't even have to be the same public key cryptosystem for all the receivers. It might be they each come up with their ...


3

This is flawed: Cracking: The format leaks information on how often specific characters occur. For instance, if input message contains 6 o letters, there is likely much more 111 values than the most other values present. Such a small biases are sufficient for cryptoanalysis to break the message in many cases. Also, random.randint does not return random ...


2

This all depends on the IVs. If they are independently generated, the IVs will not only be different (so $IV_1 \neq IV_2$), but also their sequences will not overlap with overwhelming probability. In that case, then everything should be fine, so $C_2 = E(K,(nonce,IV_2))$, and $C_1 = M_1 \oplus E(K,(nonce,IV_1))$. However, if they are reused (so $IV_1 = ...


2

At its simplest, to encrypt a message for $n$ different recipients, you could just make $n$ copies of the message, encrypt each one with a different key, and join the encrypted messages together into a single long ciphertext. Of course, the disadvantage of this scheme is that the ciphertext length grows linearly with the number of recipients. To avoid ...


2

As Ricky Demer comments above, an obvious weakness of this scheme is that if you know (or can guess) the plaintext $M$ corresponding to any ciphertext $C$, then you can immediately recover the key by calculating $K = C/M$. Modern ciphers are generally expected to resist such known-plaintext attacks, making your cipher insecure by any current standards. In ...


2

In order to tell for certain which algorithm is used from the cipher text alone, you have to break the key as well. It is however possible to make some educated guesses about the algorithm. Your strings are only 88 bits long. That rules out every asymmetrical algorithm that I know of. If it is correct, that all of those strings encrypt the same message, ...


2

In addition to what Richie has noted, here are some problems with the encryption method that you normally don't like to see in encryption. It preserves order (if a > b, then E(a) > E(b)). So, say the cells encode salary information. Even if it doesn't reveal the salary, it still reveals a lot of information. It is not semantically secure. In other words, I ...


2

Here is the answer for why a deterministic public-key encryption scheme cannot be CPA secure. For CPA security it is sufficient if an adversary can distinguish between encryptions of two messages $m_0$ and $m_1$. That is, an adversary gets to see an encryption $c \gets \textsf{Enc}(pk,m_b)$ for a random bit $b$ together with the public key $pk$. Now in ...


2

None of Twofish, Serpent and AES are currently known as broken, so as far as security is concerned, you can use any of them. AES has a slight advantage because it's very widely used, so if it gets broken you're more likely to hear about it and get relevant software updates quickly. The Snowden postings haven't changed much as far as cryptography usage is ...


2

Yes, the same weaknesses apply. Text on computers is a bunch of numbers; a OTP encrypts a sequence of numbers modulo 2.


2

Could someone even recognize that the values are encrypted? Well, maybe, maybe not. You're correct that the values would all appear to be valid dates (this is known as format-preserving encryption, by the way), so they would not look obviously encrypted, the way, say, a random hex string would. If someone just saw a small number of such dates, with no ...


1

There is no out of the box tester which can tell you what you need, you need to do research. I'm having the same 'problem'. I'm doing an internship for a company which wants to protect their self-created protocol which works on top of TCP. So to tackle this I've create a plan of approach and defined my research parameters (quite broadly). Now i'm looking ...


1

Encryption & Decryption: It increases the messages secrecy. Sender encrypts the plain text using a secret Key and sends it(cipher text) to receiver. Then the receiver receives the cipher text and decrypt it using the same secret key. The secret key should be shared between the sender and receiver. Hashing: It ensures the integrity of a message (which ...


1

Is it theoretically possible? Yes. Is it practical? Absolutely not. In your scenario, there are two approaches, but both are pretty much theoretical only. First, there is fully homomorphic encryption (FHE). This means, you can do computations like addition and multiplication of ciphertexts and get another encrypted message. To actually have some of your ...


1

The fundamental difference between hash and encryption techniques is that hash is irreversible while encryption is reversible. Hash algorithms generate a digest of fixed length output cipher text for a given input plain text. The output text cannot be converted back to input text. The generated output will always be same for a given input plain text that ...


1

For a (fixed-length) cipher to meet your first condition, it needs to be the case that it's no easier to guess the plaintext if you have the ciphertext than it is to guess the plaintext without the ciphertext. Now, suppose I send a random 1024-bit string XORed with $G(k)$ for some 128-bit $k$ and computationally secure $G$. The probability that my message is ...


1

summmary In general, no. An attacker who has lots of ciphertext+plaintext pairs may never be able to reverse-engineer an algorithm from them. An attack may not even be able to distinguish which one of a large group of known encryption algorithms was used to generate those ciphertexts. However, various weaknesses in some algorithms and protocols are known. ...


1

Yes, key derivation with known parameters can add to security. For a start, in the scenario of the question (understood as a real-world scenario, with PDF encryption used in step 4), when key derivation is used, you can hand one encrypted PDF document and its derived key to a person, and that won't enable her to decipher other enciphered documents; this ...


1

To show that a family of functions is not a PRP, you have to either show that the functions are not permutations or that they do not behave pseudo-randomly. As it is already established that the functions are in fact permutation you need to show the latter. For a family of permutations to be a PRP means that it is computationally infeasible to distinguish a ...


1

If your nonce is 16 bytes, and your message pre-nonce is a multiple of 16 bytes (i.e. no padding is needed), sending the nonce in the clear opens you up to replay-ish attacks. Specifically, if an attacker captures one exchange with nonce $N$ and response $R$ (with $b$ blocks $R_1$ through $R_b$), and then impersonates the server and the client sends them ...


1

Malicious JavaScript could trigger calls against any SSL server. Although this JavaScript is not able to actually read the cookie for that server directly, the browser will send that cookie through the SSL tunnel. The JavaScript will then alter the URL, keeping the server but changing the path or request parameters. In that way it can inject data before ...


1

I generate a derived key for every document via KD=KDF∗KM∗filename where KDF and filename are known to the public. If I brute-force a document and get the correct KD, I know 3 out of 4 variables and therefore know KM. As a result, I’m able to derive every KD. The main problem here is, that you assume the existence of a function, which can calculate the ...


1

First of all the Additional Data (AD) is not a tag. It is data that is also authenticated by the authentication tag. This authentication tag is appended to the ciphertext by libsodium. The tag doesn't consist of separate portions for AD and the ciphertext (and IV), the AD is taken into account during calculation of the tag. The AD can be any data, including ...


1

Your scheme does mean that the server as a passive participant can't read the data, but if your threat model includes the server trying to get access to the data you'll need to do more. tylo mentions the server (or one of its agents) attempting to join a group, but it could probably also MITM the process of another user joining the group (by returning a ...


1

Your questions: if all the clients leave the group (at which moment they delete K from their local store), all the information on the server will be useless as nobody has K anymore if Client2 comes online for the first time and another group member is not online it will not be able to obtain K and will not be able to decrypt any of the data stored on ...


1

The entropy of a random password is given by the formula: $H = log_2 N^L = L\log_2 N$ where $N$ is the number of possible symbols, and $L$ is the length of the password. Since you want to know the minimum length to achieve a determined level of security, then the answer to your question is $L = \lceil \frac{H}{log_2 N} \rceil$. In your case, $N=20$, and ...



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