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6

You'd be trying each possible displacement (offset). Suppose the ciphertext is CXEKCWCOZKUCAYZEKW. Here's displacement 1: CXEKCWCOZKUCAYZEKW CXEKCWCOZKUCAYZEKW At displacement 1, there are no matches (nothing where the a letter in the top line is equal to the letter immediately below it). Here's displacement 2: CXEKCWCOZKUCAYZEKW CXEKCWCOZKUCAYZEKW ...


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

This looks like a sliding window approach to calculating the index of coincidence. So you would have something like: ABCDE FGHIJ KLMNO OACBD EFGHI JKLMN Given enough cyphertext, you'll discover a length at which the IC is high; this is a candidate keylength for the cyphertext, because you've shifted the two texts by one keylength. Multiples of this size ...


5

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 ...


4

"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 special case of the affine cipher where $m=26$. Let's encrypt a single letter using your $E$. Let it be m, say, which is at index 12. So, $$E(12) = (7 \cdot 12 + 10) \mod{26} = 16$$ Now if we try to use the $D$ in your question, we decrypt this as: $$D(16) = (7 \cdot 16 - 10) \mod{26} = 24$$ which is obviously not right. The issue is that your ...


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

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

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 ...


3

This can be broken. The exact nature of the attack will depend what modulus you use for the Hill cipher: are you working modulo a prime number, or working modulo 26? Working modulo a prime $p$ A simple attack, with no fancy mathematics needed. One simple attack is to start by requesting the encryption of the 26 messages AAAA, BBBB, CCCC, DDDD, ..., ZZZZ. ...


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

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

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

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 ...


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

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 ...


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

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

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 ...


1

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, ...


1

Identification There isn't a specific name that describes this as an individual cipher. It is indeed a "Caesar cipher" that merely differs from the cipher Julius Caesar himself used as it uses two (or more) numbers to shift the alphabet. Looking at your description a bit closer, one could maybe even think it is somewhere half-way between a Caesar cipher and ...


1

LFSR theory, for an $n$-bit register uses an isomorphism between the shift operation, and the multiplication of a polynomial in $\mathbb{F}_2[X]$ by $X$; the multiplication is taken modulo a given polynomial $P$ of degree $n$; that modulus polynomial is the one defined by the position of the "feedback" bits and we usually want it to be primitive (i.e. ...


1

Generally, that is possible. crypto-analytics Including solutions that collect statistical information, there are several solutions which are used during cryptographic analytics that can cope with shifting (read: changing) values, frequencies, and pattern detections. Yet, merely having knowledge of the cyphertext will not be enough most of the time. For ...


1

Your math is off. The key space of "four squares of random letters" is not four times as big as the one provided by one square, but the fourth power of that size. To see this easier, consider the case of one $2×2$-square (or four ones), where you can still calculate everything by hand. There are $4! = 24$ possible ways to fill a $2×2$-square. If you have ...



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