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7

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


6

This is highly insecure, for the same reason that ECB mode and simple substitution ciphers are. Every time you use the word the in your message, it will be encrypted the same way. The same goes for other, lower-frequency (but still fairly common) words -- like as or with or will (or any of hundreds of other examples). This is a humongous clue to ...


5

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

One option would be to get them to select a one-time MAC of the form: $mac(m,k_0, k_1) = (k_0 \times m + k_1) \mod p$ You would select $p$ to be something like 29. $k_0$ and $k_1$ would be chosen at random from the values 0-29. $k_0$ has the additional restriction that it can't be 0. You can aid the computation by giving them a 29x29 matrix of all ...


5

"Frequency analysis of the output might help determine simple words in the ciphertext such as 'the' etc if that word is repeated and sent multiple times. This isn't necessarily a problem as it's only a simple word and doesn't convey much meaning to the message". If the word "the" doesn't convey much meaning, then why have you used that particular ...


4

This is a type of code book security. Code books can be very strong or very weak depending on operational security. If you never reuse a code book word even in a single message and the code words are genuinely random - this approach could work. Of course if you can't reuse code words and need perhaps 40 instances of THE and 30 instances BE to avoid ...


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

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

Since this is homework, I'm not going to give the answer, but will hopefully point you in the right direction. You are correct when you say for a perfect cipher, the probability should hold that $\mathbb{P}(P=m|C=c)=\mathbb{P}(P=m)$. In words, this means that given that you see the ciphertext is $c$ what is the probability of the plaintext being $m$ (note, ...


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 ...


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

Your scheme would make a nice puzzle for amateur codebreakers. That's about the best that can be said for it. It does not meet the generally accepted standards for a modern encryption scheme; in particular, it is not semantically secure. In fact, the security of your scheme would be seriously compromised if an attacker obtained even a small amount of ...


3

For chosen plaintext and a classic Vigenère cipher, you need only as many characters of plaintext as the length of the key to completely recover the key. It's a trivial reversal. (If the cipher uses 26 scrambled alphabets, it will take more.) if you don't know the length of the key, you will spot the repeating sequence after the second repeat. For an ...


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

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 ...


2

I think you're referring to Fleissner grilles, which are kind of a subset of Cardan grilles. According to wikipedia, to construct one you "make 16 perforations in an 8x8 grid – 4 holes in each quadrant. If the squares in each quadrant are [identically] numbered 1 to 16, all 16 numbers must be used once only. This allows many variations in placing the ...


2

I don't know where you get that the central element can not be $1$, because the Cardan grille does not have any special "requirements" as you're implying. In the original version, a piece of cardboard has several holes cut in it (= the grille) and when it is placed over an innocent looking message, the holes cover all but specific letters spelling out the ...


2

This scheme is very unsecure. In my humble opinion is like a complicated "translate your message into a unknown language". In my opinion it looks like an hashed version of the Windtalkers (WW2 native americans language used to encrypt messages). Your version add one more level: you have few languages (hash functions) to choose among.


2

Let's assume a ciphertext only attack, and compare it to Vigenere. Your scheme is on a quite similar level of security: It might be hard to break it without a computer, but with one it is probably done in seconds. Here's how: First, your permutation is static and only depends on the number of rounds and the length of the text, therefore you can just reorder ...


2

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 ...


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

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

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

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

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

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 ...


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. ...



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