# Tag Info

15

Edit: I think the edit to the question makes it as vigenere cipher; which invalidates my answer below. @galvatron answer gives the suitable answer why vigenere is not secure. The old answer below ( applies only to substitution) Baiscally this is a simple substitution cipher, where each letter is mapped to another letter (i.e. the shift). The answers for ...

14

If you combine two affine ciphers, you obtain one affine cipher. Say the first cipher is $e_1(x) = a_1x+b_1$ and the second is $e_2(x) = a_2x+b_2$. Then $e_1(e_2(x)) = a_1(a_2x+b_2)+b_1 = (a_1a_2)x+(a_1b_2+b_1)$. Note that if $a_1$ and $a_2$ are both relatively prime with the modulus, then so is $a_1a_2$, so the new cipher can also be deciphered.

11

Encryption is naïvely viewed as a way to send messages from A to B that cannot be deciphered (at least in practice) by an adversary. Sure, encryption does do that, but modern ciphers do so much more... A common attack scenario is the known plaintext attack (KPA). Of course, if the adversary already knows the entire plaintext, there's not much to be gained ...

9

If the message is shorter than the key, then the Vigenere cipher is essentially the one-time pad, which is unbreakable for a random key. If the key is not random, then you may get some information on the plaintext.

9

Yes. Remember that, in a Vigenère cipher, the $n$-th ciphertext letter is calculated by adding the $n$-th plaintext letter and the $n$-th key letter (where the key is repeated as many times as necessary to make it as long as the plaintext) modulo 26 (for the standard English alphabet), i.e.: $$c_n \equiv p_n + k_n \mod 26 \tag1$$ (Here, I'll assume the ...

8

First, you should start by guessing which symbols in the ciphertext are actually enciphered, and which are simply written in plain. (Don't worry if you guess wrong, you can always make several guesses.) For a Vigenère cipher, one also needs to guess whether any non-encrypted characters should advance the key position or not (usually they do not). For your ...

7

Applying another round of vigenere would make the ciphtertext (in nearly every case) harder to break, yes. The problem is: This "new" algorithm is just a normal vigenere algorithm with a longer key (if the key lengths of both keys are not equal and not 1). You don't need to apply vigenere a second time, you can just calculate the new key in advance. Example:...

6

Well, if we have plaintext^shiftedPlaintext, what we have is the values $P_i \oplus P_{i+k}$, where $k$ is the length of the shift. What does this imply? Well, if we consider the values $P_i \oplus P_{i+k}$, $P_{i+k} \oplus P_{i+2k}$, $P_{i+2k} \oplus P_{i+3k}$, ... we get a chain where, if we guess one of the values $P_{i+nk}$ for some $n$, we can ...

6

There is imprecision in what is stated in your notes. The Kasisky test only works if the corresponding letters in the two segments are separated by a distance that is a multiple of the key length (in other words they are encrypted by the same letter in the key). For example (source Wikipedia): [abcdea]bcdeabcdeabcde[abcdea]bcdeabc [crypto] is short for [...

6

Is this a real one-time pad? No, it's not a one time pad; as the keystream is not generated from a truly random process, hence you don't get the informational security guarrantees that you'd get from a true OTP. Instead, it's a stream cipher; in particular, counter mode (with SHA512 is the "block cipher"; in counter mode, the block cipher needn't be ...

6

The Vigenère cipher is essentially a series of interwoven Caesar ciphers, in which the size of the shift is different for each character and is defined by the keyword. The one-time pad (OTP) is a very similar technique in which the only difference is that key is generally: truly random, at least as long as the message, and never reused (thus the name), ...

5

For breaking a Vigenere cipher by frequency analysis the length of the cipher text alone is not the crucial part. What really matters is the proportion cipher_text_len/key_len, as this indicates how many characters of the clear text are encoded by the same character of the key. For the example you provided this proportion is below 3. Frequency analysis ...

5

This is, in fact, not a Vigenère cipher. One clue to this is the fact that the ciphertext (which, conveniently, includes unenciphered word breaks) contains lots of repeated words like UTL, VCI, V, UB and QVRY that would be very unlikely to occur by chance in the output of a polyalphabetic cipher like Vigenère. Another clue can be obtained by examining the ...

5

To be honest, I think all of them do NOT make much difference, since Caesar Cipher is a weak cipher. You can see Breaking the cipher on Wikipedia. It uses a method called frequency analysis, which ignores key strength. Also, the three methods you mentioned all have their weaknesses. You should use a secure random generator.

5

As far as I can tell, your original reasoning is correct. To obtain perfect secrecy with a Vigenère cipher, the key must not repeat. The key does not repeat whenever it is at least as long as the message, so option 2 is clearly a sufficient (and indeed necessary) condition for perfect secrecy, at least as far as only key length is considered. Options 3 ...

5

ADLY is the order those letters appear in the alphabet, so A=1, D=2, L=3, Y=4. But I've never seen such a substitution in a Vigenere cipher, and it's dumb, since it makes a weak algorithm even weaker. Normally you just identify A with 0 or 1 and then assign increasing numbers to subsequent letters. A=0, B=1, C=2, ... Z=25, so LADY would map to 11,0,3,24 (or ...

5

Let's consider claim 1 first. If the ciphertext alphabet has, say, 26 letters, then a Vigenère cipher with a 100-character key length has $26^{100}$ possible keys. You may wish to do your own back-of-the envelope calculation of how long it would take to exhaustively search that keyspace, assuming e.g. that you have access to a billion computers that can ...

4

I believe you need a few clarifications to answer this question yourself. The first is the one time pad (OTP). This is the only truly unbreakable system if it's used correctly. Using correctly means that for every symbol of the message there is exactly one truly random symbol in the key. Specifically, this means that there is no chosen symbol of the key ...

4

Vigenere Cryptosystem is as follow: You chose a key $(K_0,...,K_{m-1})$ consisting of elements in $Z_{26}$. Then a ciphertext for the message $(M_0,...,M_{n-1})$ is $$(M_i+K_{i\mod m}\mod 26)_{i \in [0..n-1]}$$ It is easy to see that you can generate a ciphertext for the message $(M_0+1,...,M_{n-1}+1)$ by adding 1 to each letter. It is therefore by ...

4

NO, the question does not contain the mathematical equation for the Vigenere Cipher with plaintext t and ciphertext c in thet set $\{1\dots 26\}$ (with the letter a as 1 and the letter z as 26), and displacement (or key) n, for two reasons: the best interpretation I can make of the expression given is by considering -> to be the $\implies$ mathematical ...

4

A Vigenère cipher is easily breakable, when the ciphertext is reasonably larger than the key size. This is due to the fact that its cipher-texts leak statistical information about the pair plaintext - key. Applying a Vigenère cipher several times, with different keys and key-sizes, would not make your scheme much more secure. It might, at most, requires a ...

4

A Vigenère cypher can be unbreakable if, and only if: the key is random, the key is at least as long as the message being encrypted the key is used only once. In that case a Vigenère cypher is mathematically equivalent to a One Time Pad, and the same mathematical proof of unbreakability applies. You proposal is closer to being unbreakable than standard ...

4

Forming my comment into an answer: If the key has the same length as the message and is used only once, it is basically a One-Time-Pad. This means, that in theory you can match any ciphertext to any plaintext with $a key$. If this key has to match certain criteria (e.g. be a word of a certain language), the information theoretic aspect will be lost. It ...

4

All the Quagmire ciphers (see e.g. here for definitions) are combinations of a Vigenère shift cipher and a keyword-based simple substitution cipher, where the substitution cipher is used to scramble the alphabet before and/or after the Vigenère encryption. In particular, the Quagmire III cipher in effect scrambles the plaintext alphabet before doing a ...

4

The first thing you should do is estimate the key length. While you may be able to do this using Kasiski examination, in this modern era of computers I very much recommend instead using the index of coindicence, which you can compute very easily, e.g. with this Python program: import sys msg = sys.stdin.read() # read ciphertext from standard input for ...

4

2 can't be correct and it's not a trick, message space should be equal or smaller than key space, in 2 message space is much larger, its : $\sum_{i=1}^{6} 26^{i}$

4

Sari's answer is very good at explaining why this method isn't particularly good. However I'd like to add that if your array is chosen from a very good random number source (think radioactive decay, the LSB of radio noise etc.), (at least) as long as your plaintext, and you do not use it to encrypt more than a single message, then what you have is a one time ...

4

The project Cryptool does some if not all of what you want. I have not used it extensively, but it seems quite well documented. Below from the webpage: CrypTool 1 (CT1) was the first version of CrypTool. It was released in 1998 and allows to experiment with different cryptographic algorithms. CT1 runs under Windows and has two successors: CT2 and JCT. ...

3

This is a bit contrived, because you have given the correct answer to your test: WHATANICEDAYTODAY was the plain text and the key is crypto. However, it shows one way to attack a short Vigenère cipher, where you have a message only a few times longer than the key. I made the following assumptions: Plain text was a short English text The key was a 6-...

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