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7

It's called a keyword cipher. See this question for some ways to break it.


6

The VIC cipher and for something not as secure but easier for encryption and decryption, the double transposition cipher. VIC cipher: The VIC (short for VICTOR) was used by the Soviet spy Reino Häyhänen - a pencil paper cipher. To quote the wikipedia page: Although certainly not as complex or secure as modern computer operated stream ciphers or ...


5

Perhaps you could do something with Visual Cryptography. Maybe something like: Gather a few low-resolution images (symbols or short text phrases), perhaps a few more images than you have kids Use visual cryptography to split each image into 2 random-looking images, and print each random-looking image on its own piece of transparency paper Shuffle the pile ...


4

So if I understand correctly, the following describes your cipher: Take letters A-Z to be numbers 0-25. Let Mi represent letter i of the message Let Ci represent letter i of the ciphertext Let n be the ceasar shift number Let m be the skip number Let the key be described as the tuple (n,m) Then, Ci = (Mi + n + m*i) modulo 26 The system is a type of ...


4

The permutations in your question are given in Cauchy's two-line notation, where the upper line gives the input index to the permutation function, and the lower line gives the resulting permuted index. For example, the definition $$\sigma = {1\ 2\ 3 \choose 3\ 1\ 2}$$ means the same as $$\sigma(1) = 3,\quad \sigma(2) = 1,\quad \sigma(3) = 2.$$ Thus, if we ...


4

You are actually doing the right thing, however you do not include the "_" as part of the characterset. Assuming the alphabet consists of 27 letters (A-Z and "_"), and using the appended JavaScript script (you can simply paste it into the console), returns one sentence which actually does make sense: THE_FAILURE_MAY_BE_BOTH_DELIBERATE_AND_CLANDESTINE var ...


4

You could challenge them to devise low-tech, physical zero-knowledge proofs (of knowledge) for games like "Where's Waldo?" and Sudoku, then show them some methods that really work and why. I've done this before with high school CS students and they seemed to really like it. For "Where's Waldo?" one can prepare a large sheet of paper (at least twice as big ...


3

You can solve it at http://www.quipqiup.com/index.php in about 5 seconds. contrariwise continued tweedle dee if it was so it might be and if it were so it would be but as it isnt it aint thats logic It's an excerpt from Through the Looking-Glass by Lewis Carroll Information on how quipqiup works is available at http://www.quipqiup.com/howwork.php


3

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 ...


3

Since you encrypt just a single letter, there are $26^2$ combinations of $p$ and $c$ where $c=E(p)$. This is because there are $26$ possible shift keys in the key space, an therefore each $p$ can be mapped to one of $26$ letters in the code space. Now, assuming that the key is distributed uniformly in the key space, each of those combinations of $(p,c)$ has ...


3

Iterate through the alphabet (using a smaller alphabet is always a good way to start). So you take letter A and you can combine it with letter B..Z. Now if you take B, then you can combine it with letter A and C..Z. But the combination (A,B) is equivalent to (B,A). So you only have C..Z to consider. Thus you would get (n - 1) + (n - 2) + ... pairs, leaving ...


3

It took me a while, to actually understand the question. To your first question: No, this kind of scheme has no name, because it is actually worse than any of the classical ciphers. Well, if you consider any substitution table (over 26 letters) a variant of the Caesar cipher, then... you have such a variant. But it is much weaker than the random ...


3

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 ...


3

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 ...


3

How secure is this cipher? At first glance, not very. It would appear to be vulnerable to a ciphertext-only attack, for example, the attacker can recover the plaintext given a ciphertext of about 10k (actually, he probably can deal with less), even assuming that all the attacker initially knows is that the plaintext is "ASCII English", and he has no ...


2

It's not clear what they actually mean by half of their things, or if there is any benefit of using them together. That site doesn't want to create any meaningful security, it's just to play around with numbers and letters. They state to use Vigenere, that's okay, usual classic cipher, which can be broken if the text is long enough. "Caesar box code": I ...


2

Let $k=\left(\begin{array}{cc}k_0 & k_1\\k_2 & k_3\end{array}\right)$ be the key. And I'll assume the transformation $a=0$, $b=1$, and so on. So you know $k\left(\begin{array}{c}0\\1\end{array}\right)=\left(\begin{array}{c}c_0\\c_1\end{array}\right)$ where you know $c_0$ and $c_1$. Thus $c_0=0k_0+1k_1$ and $c_1=0k_2+1k_3$. So that is two equations, ...


2

The easiest way in this case to work out if it is vigenere is to brute force it. You know the key is only 6 characters. Take a vigenere decryption function and a dictionary file of 6 character words. Decrypt using the first word, then compute a histogram of the resulting plaintext. Compare with what you would expect to see given the distribution of ...


2

Known plaintext attacks and chosen plain-text attacks The attack here is straightforward. If you have any known plain-text at all, you can find codes for all the letters in the plain-text you have. You can then use this knowledge to decrypt other messages whose plain-text is unknown. A very small number of known plaintexts would completely compromise ...


2

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 ...


2

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 ...


2

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.


2

There are different approaches to crack a substitution cipher. A human would use a different strategy than a computer. But as the word boundaries are not preserved it will be rather challenging for a human solving this cipher. The quipqiuq tool mentioned by John is using word lists, but there are other methods as well. Resources: ...


1

The key to a mono-alphabetic substitution cipher is a substitution table. Thus you already have (most of) the key for that cipher, it is a b c d e f g h i j ? D E F G H I ? K L k l m n o p q r s t M N O P Q R S B T U u v w x y z V W X Y C ? where ...


1

There are two possible explanations. The literature you are reading swapped the positions of the plaintext and ciphertext in the second example. If you were to permute TEN, the result is ENT. There is a mistake in either the permutation value or in the resultant ciphertext. If the first example permutation was used, the ciphertext would be correct. ...


1

The answer to your question is easy: It depends. :-) First of all I think your two questions can be reduced to one. If you don't know the key length, then you can simply brute force over the different key lengths. For breaking the Vigenere cipher the proportion cipher_text_length/key_length is the most important number as it determines how many characters ...


1

As already stated it is a Vigenere cipher. Here even the length of the key is already known. A principle approach to break the cipher is as follows: Try different keys. For each key construct the corresponding clear text and check how similar the clear text is to the English language. How well this algorithm works depends on how good your check for the ...


1

For perfect secrecy: $$number\_ of\_keys >= number\_of\_cipher >= number\_of\_plaintext$$ According to Shannon's perfect secrecy theorem: let, $$number\_ of\_keys = number\_of\_cipher = number\_of\_plaintext$$ then we have perfect secrecy if and only if: each key is used with same probability, and for each (plain,cipher) pair there is unique ...


1

You have 3 equations and 2 unknowns, so it is solvable, assuming a solution exists. You can plug this into any linear equation solver. If you subtract equation 3 from equation 2, you get $a=-3$, and can solve for $b=75$. This fits equation 2 and 3, but not equation 1. So, no solution exists.



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