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It's a quite a weak cipher, being better than a simple substitution cipher by only using digraphs instead of monographs. An interesting weakness is the fact that a digraph in the ciphertext (AB) and it's reverse (BA) will have corresponding plaintexts like UR and RU (and also ciphertext UR and RU will correspond to plaintext AB and BA, i.e. the substitution ...

Yes, if you are willing to throw enough resources at it. Only the most fatally flawed schemes cannot be rescued (in practical terms) given enough additional computation. Since you are even willing to enhance the rotor complexity, you could actually use it to implement a modern algorithm exactly. The ability for a rotor to advance "forward a different number ...

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 ...

For designing a cipher, one first has to decide about the alphabet. This is a bit problematic for a language like Chinese, since it is not really clear how many (and which) characters should be used. The number of signs known by people differs greatly. You don't want that your encrypted message is un-decryptable just because you used some unknown character ...

Your recall isn't entirely inaccurate, although it's not completely right. The Allies were able to generate a given day's settings because they both knew the methods used to compose the messages had pitfalls and, generally, there were flaws in the composition of messages themselves; mistakes (known at Bletchley Park as Cillies) were pounced on and used as ...

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 langauge and therefore has a far greater number of characters than Phonetic systems. Historically this has cause chinese codes not to be based around the ...

This appears to be a simple substitution cipher. Substitution ciphers are vulnerable to frequency analysis, and would never be used in practice. It would be very easy to guess what character is being used to encrypt underscores (spaces), and from there an attacker could guess common one-, two-, and three-letter words and begin unraveling your messages from ...

non-phonetic language I think you want to use the term "non-alphabetic language". Chinese, like Japanese and Sanskrit is a syllabic language, where the tokens refer to syllables. Chinese, unlike most western languages is tonal. There is an example in the "mechanics" part of that wikipedia page that describes the syllable "ma" in Mandarin Chinese, and ...

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 ...

This is a lossy algorithm. You will lose information during the translation and reverse-translation steps. Introducing loss into any algorithm obviously increases the difficulty in pulling the clean plaintext out, since it's potentially impossible to pull the clean plaintext out even with the key. Even so, fairly normal cryptanalysis should apply here. You ...

This is a simple substitution cipher, specifically a mixed/deranged alphabet cipher. See wikipedia's description: Substitution of single letters separately—simple substitution—can be demonstrated by writing out the alphabet in some order to represent the substitution. This is termed a substitution alphabet. The cipher alphabet may be shifted or reversed ...

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 ...

It seems I can't comment on answers because the question is no longer on the Chinese Q&A, but I wanted to support fgrieu's suggestion. Certain web services don't have much care for security, but they want to avoid containing keywords that are blocked by the Chinese Firewall. One I'm familiar with is PIMCloud, a cloud-supported IME, which does exactly ...

The Vigenère cipher has many weaknesses, but perhaps the most obvious ones are: An attacker, who knows (or can guess) as many consecutive characters of any plaintext message as there are in the key, can trivially recover the key and thus decrypt all messages. (In fact, the characters need not even be consecutive, they just need to cover the entire key, or ...

As Paŭlo Ebermann says, this is (apparently) a homophonic cipher. Ciphers that obscure single-letter-frequencies, such as homophonic ciphers, the Alberti cipher, Vigenère cipher, the Playfair cipher, etc. are impossible to crack using single-letter frequency analysis, which is the only cryptanalysis technique published before 1863. However, other ...

If you consider your message as a series of Unicode code points (aka characters) then it does not matter that it is really pictograms you are encrypting but "letters" in a very large alphabet. Then you can use substitution or transposition or just even whatever modern cipher that work on bytes or characters.

A substitution cipher consist of a mapping from letters in the alphabet to letters in the alphabet (not necessarily the same alphabet, but probably is in this case). There are many forms that a key can take on. Ones I've seen in practice are: The key is the mapping (i.e. a->m, b->x, c->q,...). The key represents a shift. A key of 5 would mean the ...

yes (guessing ur doing the cypher challenge...?) This link has a really good decoding tool (saves you time), then trial and error keywords. also this tool; can be used to find the length of the keyword; paste the texts you're decoding then the number of the column(s) with the most x's is the length of the keyword

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 ...

If I understand your cipher idea right, you would have a larger ciphertext alphabet than the plaintext alphabet, where each plaintext symbol maps to multiple ciphertext symbols (and the number is dependent on the frequency of the plaintext symbol), one of which is used randomly. This is known as a Homophonic substitution, and with it the single-symbol ...

I can't tell you which method is the best, but I can point out some places to look. The longer the message the easier it is easier to identify. How long is the message? n-gram frequency: Look at the likelihood that groups of n-letters appear next to each other (often called n-grams). For example is does the n-gram AAA appear frequently (not many languages ...

First, linguistically this sounds like a stupid idea. Words in different languages don't correspond one-to-one to each other. Try to translate a text with Google translate between several languages and back to see how good this works. And good translation programs have the possibility to look at the context - your word-by-word translation doesn't have this ...

The real security of Vigenere is difficult to quantify. A million character plaintext with a 10 character password is easy to break. But a 10 character plaintext with a 10 character randomly chosen password is essentially a one-time-pad and theoretically unbreakable. Given the data you've told us (plaintext: 100 to 5000 characters; password: 30 to 100 ...

Eek! The Vigenere cipher is completely and totally insecure. You should never use it. Instead, use a modern authenticated encryption scheme. If you are protecting data in transit, I recommend using TLS (or SSL). If you are protecting data in storage, I recommend encrypting it with GPG (or PGP). This is the simplest, easiest way to get well-vetted ...

It's sometimes called a keyword cipher. As dr jimbob notes, it's a particular type of monoalphabetic substitution cipher. Ps. See also this recent question about breaking such ciphers.

If you save the current page and examine the file with an hex editor, you will likely find that your example Chinese string is represented by the bytes E6 88 91 E5 9C A8 E5 AD B8 E4 B8 AD E6 96 87 One option, very suitable for implementation by a machine, is to encipher the bytes representing the message in this (or other) format used by the machine. ...

hummm... some thoughts about it: I think that it could be secure depending on what you want to hide. The bigger and the more "real world words" you want to protect, the easier it gets to crack. Why? Because in books, in general, you'll only have letters, few numbers, and that's only. Ok, so I know your transmitting one or more words. By the size of the ...

One nice way to generate a word list is distilling it from a large amount of texts. There are a few interesting effects with this creation method: The word frequency depends a lot on the chosen texts. For example when feeding it wikipedia the language is rather formal, and common every-day words are pretty rare. It might be interesting to compare Simple ...

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 ...

The description isn't very clear but from what I understand it's a one time pad with a weird encoding of your input : If you take and encoded input you know that every even position is either $0$, $1$ or $2$ and if it is $2$ then the subsequent digit is between $0$ and $5$. That being said the whole security of your scheme relies on using a solid random ...