# Tag Info

14

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 ...

11

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 ...

10

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 ...

10

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 ...

9

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 ...

9

The real security of Vigenère 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 ...

9

DES actually demonstrated that a Feistel structure was not a guarantee against attacks. In "academic" terms, DES is broken by both differential and linear cryptanalysis, because they require, respectively, $2^{47}$ chosen plaintexts and $2^{43}$ known plaintexts, whereas the DES key is (effectively) 56 bits. Of course, for practical attacks, we would brute ...

8

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 ...

8

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 ...

7

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 ...

7

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 ...

6

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 ...

6

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 ...

5

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 ...

5

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 ...

5

Yes (guessing you are doing the cypher challenge?) The "Beaufort Decoder" is a really good decoding tool (saves you time), then trial and error keywords. Also, the "Vigenère cracking tool can be used to find the length of the keyword. Paste the texts you're decoding; the number of the column(s) with the most x's is the length of the keyword.

4

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.

4

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 ...

4

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 ...

4

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 ...

4

Given that the permutation is fixed and the key step is independent of the permutation you can reduce this to an ordinary text-substitution cipher. If the key is as long as the input you have a weak one-time pad, because the per letter change is limited to 10 instead of 26. However if the key is short then you have a Vigenère cipher (if you "decode" with ...

4

Yes, we can almost certainly break this, given enough ciphertext. One approach would be to use a dictionary and use word patterns. For instance, if the ciphertext word is qddxfozogf, then the plaintext word was probably ammunition. Notice how the 2nd and 3rd letters are the same; and the 5th and 10th letters, and the 6th and 8th letters? The word ...

4

A Michael Kjörling notes, what you basically have is a monographic substitution cipher. These are generally easy to crack using frequence analysis; for example, if we know that the underlying text is in English, we can be fairly sure that the two most frequently occurring symbols correspond to the letters T and E (and that any symbol that occurs unusually ...

3

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.

3

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 ...

3

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 ...

3

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 ...

3

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 ...

3

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 ...

3

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 ...

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