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27

When trying to break an unknown cipher, one first needs to figure out what kind of cipher one it is. Generally, a good starting point would be to start with the most common and well known classical ciphers, eliminate those that obviously don't fit, and try the remaining ones to see if any of them might work. An obvious first step is to look at the ...


24

He is talking about the original version of the Caesar Cipher where the substitution was just a +3: A -> D B -> E C -> F D -> G E -> H F -> I G -> K H -> L ... X -> A Y -> B Z -> C Because the shift is fixed, it does not have a key (but you could say it is a substitution cipher with a key equal to +3). However it is ...


20

Using the book as a key is relatively similar to one-time pad, insofar as the book can be considered as a random stream of characters. But that's true only to some extent: a book consists of words, with meaning, which implies that characters which may appear at position 321:42:35 are not uncorrelated with characters which appear at positions 321:42:34 and ...


15

I think I understand what you're asking for. You're trying to learn how we know which algorithm was used, so we know how to attack it. That's a part of what is known as cryptanalysis, the task of breaking ciphers. If you are using a standard computer protocol, the encryption algorithm is defined as a part of the protocol. The computers can't talk unless ...


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

Solitaire by Bruce schneier is probably your best bet. It has a few issues but it will work well for most things. It ends up having a small bias, but it takes about 15 seconds per character after the initial key stream has been generated. It is not nearly as widely studied a field as most people are assumed to be able to get a computer or other ...


13

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.


12

Cipher details Cipher type The Felix cipher can be broken down into two algorithms: a substitution cipher and a permutation of the character pairs. We obtain the substitution if we read the number pairs in figure 3.3 vertically rather than horizontally. Since the permutation is fixed, it has no cryptographic value. Therefore, we'll only analyze the ...


12

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 ...


11

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 block ...


11

As the other poster rightly pointed out, it's a Playfair cipher. Even without the known plaintext, the program "playn" here will give the right text in less than a second. (you can compile it yourself, and it uses the bigram statistics of English) I ran it, and the result was the following: IT XT UR NS OU TX TH AT OR IG AM IX IS AB RI LX LI AN TW AY TO ...


11

I do not have a solution, but I pursued the cipher long enough to establish it wasn't one of the easy classical ciphers. This approach should get you started. The first thing you want to do is convert the text into numbers as many classic ciphers are mathematically-based (or at least easy represented mathematically). Using $A=0$, $B=1$, $\ldots$, the ...


11

Since this is an historical question, I am going to digress and make some historical corrections. In science, we give credit for important inventions to the people who published. If it turns out that someone else invented it earlier and didn't publish, they don't get credit. Obviously, they should be mentioned in passing or a footnote in the interests of ...


11

Wrapping up my comment as an answer: Imagine you’re a Japanese cryptanalyst in the year 1944. There is no such thing yet called “television”, and you’re still decades away from a wordwide network feeding you with all the knowledge you could wish for. In that case there’s only a minimal chance you’ve ever heard or seen a Navajo. So, you’ll be wondering ...


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

Some additions to the other answer: any given letter can only correspond to a fairly limited number of ciphertext letters: only the ones in the same column or row, and never to itself. So a highly frequent letter like E will still stick out in longer texts and then we will also find its row and column mates, which helps in reconstructing the square. There ...


9

I take your question to mean, how both historically and in the modern age one could construct a pen-and-paper cipher using the Chinese language. As pointed out in the question, Chinese is a logographic language and therefore has a far greater number of characters than Phonetic systems. Historically this has cause Chinese codes not to be based around the ...


9

The One Time Pad can be considered a secure hand executed cipher as long as you meet the security requirements of same. Yet, ask yourself why you are interested in such a method in this wonderfull age of high speed digital electronics.


9

Let's start by considering which cipher letters should correspond to the most common letters E and T. According to your frequency analysis, the most likely candidates are O, K, T and maybe D and N. Now, E is the fifth letter of the alphabet, so unless your keyword is very short, it's going to encrypt to some letter in the keyword (and if the keyword is ...


9

Kerckhoffs's principle states, that a cryptographic system shall be secure even if everything about the system, except the key, is known to the attacker. Typically an encryption algorithm has two inputs: a key the data In the case of Rot13, there is no key. So if you know the algorithm, there is nothing left to guess. Let's assume the algorithm was not ...


9

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 ...


9

Your interesting questions deserve to be answered more thoroughly, but here goes: According to a highly classified document that was written in 1947 and finally declassified in 2013, the Germans started using a one-time pad system for diplomatic traffic in 1925. This system (GEE) used a one-time pad of digits to encrypt codes by modulo addition. To be clear,...


8

If it is a simple substitution cipher, there are a few standard techniques: Frequency analysis. Count how many times each letter appears in the ciphertext. The most common ciphertext-letters probably correspond to the most-common letters in English. The most common letters in English are ETAOINSHRDLU... (in decreasing order of prevalence). Therefore, ...


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 ...


8

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.


7

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 ...


7

An obstacle to proving that a book cipher is secure is that the letters in (most) books are not chosen independently at random. Thus, in principle, if two indices are chosen too close to each other, an adversary could deduce some statistical information about how the corresponding plaintext letters may be correlated. As a toy example, suppose that an ...


7

I'll assume that the objective is to assert if the distribution of the $f'_i/n'$ is sufficiently similar with the distribution of the $f_i/n$ to support that a substitution cipher (including Caesar cipher) with the same permutation table and same frequency of plaintext characters could be used in both case. If $n \gg n'$, $f_i \gg 5$ and $f'...


7

This is essentially a Vigenère cipher; it's been known for centuries. As for how secure it is, well, it is actually fairly easy to break (unless the key is both as long as the ciphertext, and randomly chosen; however, at that point, if you could remember the key, you could have well just remembered the plaintext). As for your colleague, he's right, and it'...


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