Latest Crypto related questions

Score: 0
salt avatar
Is there anything stopping me from making my own NFT of an existing NFT?
cn flag

I feel like I some understanding of cryptocurrencies, but what never made sense to me are NFTs. This is my current understanding of NFT's

  • NFT's are a type of smart contract
  • Smart contracts aren't necessarily open source, but you do have access to the bytecode
  • The actual NFT isn't stored on the blockchain, but rather on a file hosting server

So if I have access to the smart contract's bytecode and t ...

Score: 1
BoostMatch avatar
Cracking secret key, n, and m by hand with the Diffie-Hellman key exchange brute force
us flag

Problem: You see Michael and Nikita agree on a secret key using the Diffie-Hellman key exchange. Michael and Nikita choose $p = 97$ and $g = 5$. Nikita chooses a random number n and tells Michael that $g^n \equiv 3\pmod{97}$, and Michael chooses a random number $m$ and tells Nikita that $g^m ≡ 7 \pmod{97}$. Brute force crack their code: What is the secret key that Nikita and Michael agree upon? What i ...

Score: 1
user77340 avatar
Is there a blind signature that can sign on a commitment?
ie flag

For the blind signature that can sign on a commitment, I mean finally, the user can get a signed commitment, rather than a signed message inside the commitment. It means the verification also takes a commitment as input. Is there such a blind signature?

Score: 1
Fabian Schmitthenner avatar
Why does HMAC use the hash twice?
cn flag

According to the HMAC specification in RFC2104, an HMAC is computed in the following way:

HMAC(K, text) = H(K XOR opad, H(K XOR ipad, text))

where H is the underlying hash function, , is concatenation and K has the length of one block.

Now I wonder, what is the benefit of applying the hash function twice here, that is, why wouldn't it be defined like this:

HMAC'(K, text) = H(K XOR ipad, text)

Are ther ...

Score: 0
Hunger Learn avatar
Secure protocols by implementing cheap talk instead on a centralized mediator to compute any function $f(s_1,...,s_n) = (y_1,...,y_n)$?
ua flag

Based on this paper a protocol is secure if and only if it satisfies secrecy and resiliency. Most of the papers in ecnomic and computer since deal with the following problem. They consider the case where $n$ parties with private information $s1,...,s_n$ wish to compute any function $f(s_1,...,s_n) = (y_1,...,y_n)$ in such a way that no party $i=1,2,...,n$ learns more than their input $s_1$ and output

Score: 2
Ozzie Nano avatar
Does encryption also hide file hash?
jp flag

I have files in my computer that are being uploaded to a cloud storage provider (Sync) that specifies that files are encrypted on the client's side, then uploaded to their servers, and that only has the key to decrypt the files. So files can only be decrypted on my computer.

Now my question is:

Do encryption hide file name and hash?
Can such cloud providers identify the files by their names or hash if the ...

Score: 2
Generic avatar
Existence of algorithm predicting next bit in output sequence
lu flag

Let $X = [0, 1]\cap \mathbb{Q}$, and let $f:X \rightarrow X$ be a chaotic map (i.e. the logistic map with rational parameter). My question is as follows, and is purely theoretical in nature. Pick some value $x_0$ from $X$ (note that $X$ is infinite here, so pick a value using the axiom of choice), and then consider sequences of bits generated by iterating $f$ over $x_0$, returning a $0$ if $f^n ...

Score: 9
BoostMatch avatar
Is encrypting every number separately using RSA secure?
us flag

Suppose RSA is considered a "secure" method for encryption. RSA is meant to encode a sequence of integers base $27$. If we use an $n=pq$ that is hard to factor, Is it still secure if we encode every integer (letter) separately rather than the whole phrase once?

Edit: I didn't expect to get such great answers. Thanks everyone!

Score: 0
Why is it hard to compute $g^{xy}\bmod p$ from $g^x\bmod p$ and $g^y\bmod p$?
sa flag

Why is it hard to compute $$(g^x\bmod p, g^y\bmod p) \longmapsto g^{xy}\bmod p $$ when can we quickly compute $$x \longmapsto g^x\bmod p$$ ?

Score: 7
matthias_buehlmann avatar
signatures that are verifiably generated by one of several private keys without revealing which one
in flag

is there an encryption algorithm that allows to:

  • generate a set of different private keys
  • sign data with those private keys
  • allow to publicly verify that such a signature has been created by ONE of keys in the set without revealing which one it was
  • give the person that did create the signature a way to create a proof that the signature was created by him specifically

If so, which one has these prope ...

Score: 1
Building an Adversary for a PRF game
sa flag

enter image description here

enter image description here

Here is the game:

enter image description here

How can I make an $\mathcal{O}(k^2)$-time adversary making only one query to its Fn oracle and achieving advantage $= 1 - 1/(p-1)$

Here is my idea so far: query $2^{-1}$, which when it goes through the Encryption algorithm, will return 1. So,

Adversary A:
C <- Fn(2^{-1})
if C == 1 return 1
else return 0

When we query 2^{-1}: \begin{align} Y_1 &= (2^{-1})^e &\bmod p \\  ...

Score: 1
Which of the following is considered cryptographically hard/easy?
sa flag

Which of the following are easy, if any? Which are hard? and why.

Case 1) Given $x^3 \bmod N$, where $N$ is a composite number and we don't know any of the factors of $N$, find $x$.

Case 2) Given $x^3 \bmod p$, where $p$ is prime, find $x$.

Here is what I think but I don't fully understand it.

For case 1, this is assumed to be hard? In the RSA assumption, where $e = 3$, imagine $N$ being a large  ...

Score: 0
Building an adversary for a OW-CCA game
sa flag

Let K_rsa be a RSA genertor with associated security parameter k >= 1024. Let game OW-CCA_Krsa be as follows:

enter image description here

How can I build a O(k^3)-time adversary A making at most 2 queries to Invert and achieving advantage = 1.

Here is the idea that I have: if we query Invert(ya^e) then multiply that with a^{-1}, then in the end we get x:

$$\text{Adversary A:}\\z \leftarrow ya^e\\C \leftarrow Invert(z)\\return (a ...

Score: 0
How can we prove that the advantage for this hide game for any adversary is equal 0?
sa flag

Here is the Scheme:

enter image description here

enter image description here

Here is the HIDE game:

enter image description here

Here is my idea but I am not quite sure. I would appreciate some input.

We want to bring advantage = 0 for all adversaries. We can show that advantage = 0 if we can prove that all of the C values are uniformly random and independent of the Message we give. If we prove that then we can argue that the adversary won't be able to tell which game its in.

So L is ...

Score: 2
Bharat Malhotra avatar
Distinguishing the correct IV from incorrect IV in AES CBC when key is known
us flag

Currently, I'm using a static IV value for all encryption and decryption but I would like it to be dynamic for each encyption/decryption request so I started using new byte[16] and it works. The problem is how to detect and decrypt old data. Below is my code to decrypt and I'm passing static IV stored on a secret in keyvault.

private static byte[] DecryptFromBytes_Aes(byte[] dataBytes, byte[] key,  ...

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