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It's the difference between an active and a passive attacker: Known plaintext attack: The attacker knows at least one sample of both the plaintext and the ciphertext. In most cases, this is recorded real communication. If the XOR cipher is used for example, this will reveal the key as plaintext xor ciphertext. Chosen plaintext attack: The attacker can ...

A known-plaintext attack (i.e. knowing a pair of corresponding plaintext and ciphertext) always allows a brute-force attack on a cipher: Simply try all keys, decrypt the ciphertext and see if it matches the plaintext. This always works for every cipher, and will give you the matching key. (For very short plaintext-ciphertext pairs, you might get multiple ...

I suspect that KCV's are in general not used because they don't add enough to be worth the small overhead. There are a number of cryptographical attacks on encrypted methods that involve the attacker modifying a valid ciphertext, and then having the receiver decrypt the modified message (and watch how the receiver reacts). Because of these attacks, it is ...

I know no published attack against the XTEA algorithm, much less one working with a few plaintext/ciphertext pairs and a random key. There are published attacks against reduced variants of XTEA, AFAIK either with much less than half of the 64 rounds, or up to 36 rounds but assuming related or/and weak keys in addition of known plaintext. Often, especially ...

No matter how bad a protocol built on top of RSA is, there is no known risk that a private key leaks from valid plaintext/ciphertext pairs, even if the adversary chooses plaintext or ciphertext; that's one of RSA's virtues. Thus to the question is it possible with this information for Cindy to find $pri_A$ we can answer: as far as we know, no; and more, ...

Well, no, modern ciphers (such as AES) do not share that weakness; they are designed specifically to be secure without any assumption of what the plaintext looks like. You can encrypt the same message with multiple keys; in fact, you can encrypt the same message a bunch of times with the same key, and the attacker still cannot deduce any information for ...

As far as is publicly known, no, you can't. If you could, that would constitute a practical known-plaintext key recovery attack on AES, and the existence of such an attack would mean that AES would be considered totally insecure by modern cryptographic standards. If you do figure out how to do that, publish it and you'll be famous. (Or, if you'd prefer ...

Simply put no. As per the abstract, those attacks take at most 4 bits off the key space, this still results minimally in 124 bits of security. Put another way, to use these attacks you would need to expend effort roughly proportional to brute forcing an AES key where you already knew four bits of the key and this would take approximately $2^{124}$ ...

Well, the answer to 'why is AES resistant to known-plaintext attacks' is that, well, lots of really bright people have thought hard about how to break AES, and no one has come up with a practical way, either assuming known plaintext or chosen plaintext. See how-much-would-to-cost-to-brute-force-AES for a discussion on what it would take, given the current ...

You got three equations with two unknowns ($k$ and $x$). You only need two signatures to solve the private key $x$: $s_1k \equiv h_1 + xr_1 \pmod q$ $s_2k + s_2 \equiv h_2 + xr_2 \pmod q$ This might be solved using Gaussian elimination. Step 1: $s_1k/r_1 \equiv h_1/r_1 + x \pmod q$ - Divide 0.1 by $r_1$ $s_2k + s_2 - s_1kr_2/r_1 \equiv h_2 - ...

You have to define precisely what Eve can and cannot do. For instance, has Eve occasional physical access to the Arduino-based device ? If yes, then she can (at least conceptually) grab the device, "open" it, extract the shared secret, and replace the device with one of here own which does the same job, except that it also sends a copy of the data to another ...

Eli Biham examined this, along with a long series of other "internal chaining" modes in the paper "Cryptanalysis of Multiple Modes of Operation" in AsiaCrypt 1994 (alas, there doesn't appear to be a free copy on the Web). His conclusion was that there were ways of attacking such modes that were strictly easier than the standard "external chaining" mode that ...

Well, I went and solved the puzzle using brute force and Maple. I won't spoil the actual answer, but here are some tips that ought to make the process a bit more quicker. Solving the linear system modulo 2 gives you the parity of the second and third letters of the unknown plaintext. Note that all vowels in the English alphabet map to even numbers using ...

There are 18 plaintext and ciphertext letters $p_j$ and $c_j$, $0\le j<18$ (with $j<6$ for the "first plaintext"), all of which are known except $p_7..p_{17}$. Let $M=\pmatrix{m_{0,0}&m_{0,1}&m_{0,2}\\m_{1,0}&m_{1,1}&m_{1,2}\\m_{2,0}&m_{2,1}&m_{2,2}}$ be the key matrix (unknown, except that it is invertible). We have 18 linear ...

Your reasoning is correct. However, there is still some information to be exploited: as you know already, the matrix used to encrypt (the secret key) need to be invertible in order to allow decryption. Since you basically know (say) the first column and the second column of the secret key from your plaintext/ciphertext, you derive from the invertibility of ...

No for practical definitions of possible, assuming the key was chosen truly randomly, and no side-channel information is available (such as the power-consumption traces of the encrypting device, or the time it took, for many encryptions). The design of AES strives to be such that the best way to find the key from plaintext-ciphertext examples is to try keys ...

Known-plaintext attacks are key-recovery attacks. By design, an asymmetric cipher cannot be susceptible to known-plaintext attacks; it would be completely worthless. The property you describe that asymmetric ciphers such as textbook RSA lack is called semantic security. Some considerations: In practice, RSA might be used only to encrypt a randomly ...

If the encryption is any good, no. What you're describing is a known-plaintext (or possibly chosen-plaintext) key recovery attack, and any encryption system that was even suspected of being vulnerable to such attacks would be considered hopelessly broken by modern standards. The "gold standard" that modern encryption methods aim for is generally taken to ...

What you are trying to do is called a known-plaintext attack. A proper cipher will not be susceptible to a known-plaintext attack, so if the encryption method used on the strings in your database is any good, it can't be done. Imagine that you were in fact using a (128 bit block) cipher that permits deducting the password given the plaintext and the ...

Probably not in this case, although not for the reason you gave. As @Sadeq points out in the comments, AES is resistant to known-plaintext attacks. This means that by simply knowing M+P or P (the entire plaintext you've encrypted) it should not be possible to recover the AES key. However, key recovery isn't the only attack and it raises a deeper question: ...

Both the Vigenère and autokey ciphers are classified as polyalphabetic substitution ciphers, so the cipher in your exam is not likely to be either of those. Rather, the phrasing of the question suggests that it belongs to the other branch of classical ciphers, transposition ciphers. Indeed, looking at the letter frequencies of the ciphertext strongly ...

Using your recap, your problem is the same as the person who wants to securely send their credit card number to a remote server. If you get rid of the need to securely agree on a one-time symmetric key, you get rid of the majority of the complications. In your case, the symmetric key is hard-coded and the attacks from the adversary are confined to ...

As others have pointed out, there are some ciphers that can be broken if all you have is a known plaintext and the ciphertext. In general, because of this, those ciphers are considered very vulnerable and are not used anywhere. Or I should say, where they are used, the keys are generated (pseudo-) randomly and only used once. However, if the attacker can ...

Defining security against known-plaintext attacks does make sense in PKE. As it happens, any scheme that is secure against an eavesdropper is automatically secure against known plaintext attack as well as chosen plaintext attack. Nevertheless, it does make sense to define these classes of attacks and then show formally that security for an eavesdropper ...

Well, I'll assume that we'll use the same mapping between letters and integers both to translate the plaintext into integers (to be matrix multipled), and the integers (after the matrix multiply) back into ciphertext. And, we don't know that mapping, the key matrix $K$, and possibly the value of $n$. If so, the obvious place to start is to attempt to solve ...

For a key recovery attack, you'd basically need to break AES itself. There are no known practical key recovery attacks on AES (and if there were, it would not be considered safe to use), so your pretty much only hope would be to find some kind of side-channel attack on the AES implementation, or on the overall crypto framework it is part of. Alternatively, ...

According to Schneier, there is an attack on 42 rounds of ThreeFish-512, but if you check out the analysis, it relies on multiple plaintext values. Wikipedia lists no singular known plaintext attack, only a boomerang attack, which also depends on multiple known plaintexts. So given these sources, it would seem there is not a well known and effective ...

Typically the cost of brute force so greatly out weighs the cost of decrypting a single block, that the fact that an adversary can check for proper decryption after decrypting a single block is not a serious concern. Otherwise protocols like SSL would not be secure. That said, as you point out, this relates to the key space. If the key space is too small, ...

A known plaintext attack is that if you know any of the plaintext that has been encrypted and have the resulting encrypted file, with a flawed encryption algorithm you can use that to break the rest of the encryption. Example: We saw this with the old pkzip encryption method. In this case if you had any of the unencrypted files in the archive, you could ...

The classic attack using known plaintext essentially runs the encryption backwards and constructs the key. No brute force is needed, you just need enough matching plaintext and cyphertext, where "enough" can be as little as the key length (for sufficiently vulnerable cyphers). Resistant cyphers do internal mixing of the cypher state so this is not ...