# Tag Info

59

Encrypt your document, and embed a web address (and login details) in the packaging from which a reader can get the decryption key. The website must be trusted. The website logs will tell you when software has requested the key to decrypt it. If you also want to protect confidentiality, encrypt with two keys. One is the usual private key used to protect ...

22

With classical information, there is no way as you correctly surmise: someone could always duplicate the data. However with quantum information there is a no-cloning theorem. With quantum information it is possible to bound the amount of information that has been extracted from a system based on the fidelity of the system. This gives the concept of tamper ...

21

No no no! You don't get to pick which key is private and which public. That false sense of freedom is due to people not understanding that public-key cryptography is conceptually different from ciphers, and the most popular public-key algorithm RSA being a bijective permutation. With discrete logarithm for example, your private key is always a scalar ...

20

This question can be summarized: the attacker found a $d$ that did not satisfy $e \cdot d \equiv 1 \pmod{ \phi(n) }$, but it works; what's going on. It turns out that $e \cdot d \equiv 1 \pmod{ \phi(n) }$ is not necessary (it is sufficient). The necessary and sufficient conditions are: $$e \cdot d \equiv 1 \pmod{p-1}$$ $$e \cdot d \equiv 1 \pmod{q-1}$$ If ...

20

Strictly speaking, all hash functions are compressing since the output can be smaller than the input, but I imagine you're asking about compressing data that can later be losslessly decompressed. This is impossible due to the pigeonhole principle. The fact that the fixed output space of a hash algorithm is smaller than the input space means that there will ...

18

Why don't we use Blowfish if it hasn't been cracked? The reason is well-known, it has 64-bit block size and therefore it is vulnerable to birthday attacks. This is done for HTTPS and for more information see sweet32; $$\text{Sweet32: Birthday attacks on 64-bit block ciphers in TLS and OpenVPN}$$ Is it safe to use Blowfish to encrypt strings of less than ...

17

TLDR: None worth academic interest. All modern encryption techniques intended for digital computers are applicable, and secure, for all kinds of digitized data, including text and image. There is no need for such encryption specialized to image or text. Argument: the modern baseline for encryption security is resisting Chosen Plaintext Attack, and that ...

16

Encrypting $M$ using $H(M)$ as the key is a natural and well-studied approach to deduplication. It is known in the literature as convergent encryption or message-locked encryption. The natural problem with this approach is that it cannot achieve the standard notions of security for encryption (IND-CPA, IND-CCA, etc). Indeed, anyone who knows $M$ will be able ...

14

Kindly, let me know what was the actual problem which leads us to use groups in cyptogrpahy? Well, we use groups and other similar mathematical constructs because: We found there are problems that appeared to be difficult to solve with those groups We found ways to translate the difficulty of solving those problems into the cryptographical strength of ...

14

I believe a potential application can be found in the so-called "bounded storage model" introduced by Maurer here: https://crypto.ethz.ch/publications/Maurer92b.html In summary, the bounded storage model is a theoretical research direction that studies cryptographic constructions where one assumes that only the adversary's storage is limited. This ...

13

Not with a file, as you say in your question You can always take a bitwise copy of a file. Always. Even if some specific OS makes it inconvenient, you can change to an OS which does let you. This leaves you with two possibilities for confirming opening. The file is encrypted in some way which requires you to access an external website to get the key, and ...

13

TLS 1.3 has huge clean up after failures. We have only 5 cipher suites in TLS 1.3, with their IDs: {0x13,0x01} - TLS_AES_256_GCM_SHA384 {0x13,0x02} - TLS_CHACHA20_POLY1305_SHA256 {0x13,0x03} - TLS_AES_128_GCM_SHA256 {0x13,0x04} - TLS_AES_128_CCM_8_SHA256 {0x13,0x05} - TLS_AES_128_CCM_SHA256 As of current RFC 8446: A TLS-compliant application MUST ...

12

If the characters are from $[A..Z]$ and $[0..9]$ then that makes 39 possibilities per character (byte). Then the keyspace will be $39^{16}$ and that is approximately 83-bit. This is space is achievable with supercomputers but not for ordinary people. Therefore, there must be something else around that cracks your system. Used key & IV which was only ...

12

Grover's algorithm on AES-128 If all of the problems of Grover's algorithm are solved then yes, otherwise no! Grover's algorithm provides quadratic speed up so AES-128 has 64-bit security in the terms of searching complexity of Grover's method. Grover's attack for AES-128 requires approximately $2^{64}$ successive AES evaluations. That is one has to set up ...

11

Can I modify encrypted data without accessing it? if there is an example appreciate it If access means altering the data without decryption as an attacker then the answer is yes for messages without integrity. Take OTP, execute bit flipping attack, done. Take CTR mode, execute bit flipping attack, done. Take CBC, execute bit flipping attack, done. ...

10

RSA private key can be found in two ways with $n = p\cdot q$, $p = 11$ and $q = 13$ if Euler's totient function is used as in RSA paper: $$\varphi(n)= (p-1)(q-1) = 120$$ is used then $d = 67 = e^{-1} \bmod 120$ If Carmichael Function used as requried in FIPS 180.4 and allowed in PKCS#1 v2.2 standards: $$\lambda(n) = \text{LCM}(p-1,q-1) = 60$$ is used ...

10

If the question was about (current form) Reversible cryptographic hash functions Then No! One-wayness property of the cryptographic secure hash functions will prevent that. Hash functions don't use keys. So if you can reverse, everybody will reverse and there will be no secure hash function at all. Besides, mathematically impossible, too; hash functions use ...

10

There's simply no way to know. I find it very unlikely that AES will be vulnerable to a ciphertext-only attack in the next 20 years (remember, it has been around for over 20 years already and attacks haven't gotten very far). There's no reason to believe the same won't be true for ChaCha20. If you use a good cipher with 256-bit keys (to avoid potential ...

10

If (and ONLY if) it was encrypted with a special "homomorphic" encryption scheme, then you can do the operations allowed by that scheme. Those are almost never used in practice currently (they're very slow, and an area of active research). This demo is quite good. More commonly data is encrypted using an "Authenticated Encryption with ...

10

A standard candidate that I mentioned recently here is to setup a common reference string, assuming (1) the string received from the celestial object contains randomness, and (2) the observation of this string can be reliably made by distant parties. This gives an untamperable common random string, a very useful object for cryptographic protocols (in ...

9

$y^2 = x^3 + ax + b\bmod p$ is the Short Weierstrass equation. The theory behind it is here Using Bezout’s Theorem, it can be shown that every irreducible cubic has a flex (a point where the tangent intersects the curve with multiplicity three) or a singular point (a point where there is no tangent because both partial derivatives are zero). [Reducible ...

9

After generating a key-pair can we pick which key will be private or public? No, in general we cannot. For most asymmetric cryptosystems the private and public keys are completely different kinds of objects (e.g. one may be a number and the other a point on an elliptic curve), and there's no way to use one in place of the other. There is, however, one ...

9

Is there another way to do this or a way to solve this problem? We hope not; otherwise, you can break RSA. Suppose you did have a way that, given $m^{e_1} \bmod n$ and $m^{e_2} \bmod n$ (and $e_1$ and $e_2$), you could recover $m$ (even if $e_1, e_2$ were not relatively prime). Then, given $m^e \bmod n$ and $e$ (which is standard RSA), here is what you can ...

9

It seems like what you are actually describing is a way to encode data with Lego bricks, rather than encrypting with them. But, maybe the way that you encode data is hard to invert without knowing a secret key, for example. I don't think the lego adds security to this encoding, though. Suppose you design a method $f$ to encode your message space $\mathcal{M}$...

8

Since when decrypting we always want to get the correct message back, there's no reason why we would want to make this ambiguous. It would have no security advantage (if the adversary can guess with any non-negligible probability, you have already lost, so ambiguous decryption can't make that harder). Thus, unlike probabilistic encryption, which is needed ...

8

80-bit First, note that the Bitcoin miner's hash rate reached $> 2^{93}$ SHA-256 hashes per year in February 2021. This makes the 80-bit security absolute for known entities. They even passed that years ago (that is around 2012 per year). And even around 2011, it was absolute and NIST increased security strength from 80 to 122 for Federal Government in ...

8

Slightly tongue-in-cheek answer, but why not put the document on a usb key, then put the usb key in a box and wrap it with the tamper evident physical seals you pictured. That way the document is secure (inside the box) and people will know if anyone has attempted to read the data (because of the physical seals you have to break to get at the usb key).

8

Groups have properties which are useful for many cryptographic operations When you multiply 2 numbers in a cryptographic operation you want the result of the multiplication also to be in the same set. For e.g. if you are multiplying something which fits in a byte (or n bytes) by something similar, you also want the result also to fit in a byte (or n bytes). ...

8

TL;DR There was a customized encryption protocol built into an existing app popular with criminal figures, based on information given in a plea bargain with a criminal. Edit:9/6/2021 More background: https://en.m.wikipedia.org/wiki/Phantom_Secure This company as well as an RCMP insider were involved in an earlier app. The archived page below seems to be some ...

8

In RSA encryption and decryption are similar. If you chose e randomly and calculate matching d you could then chose to swap their roles and pick either as the public key. Usually we don't do this, we pick a small public exponent with few set bits. This makes public key operations much faster. We can not swap the roles and make the private key operations ...

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