# 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 ...

11

Very short answer: No Quite Short answer: No, because a scheme can only be a One-Time-Pad if the entire pad is perfectly random and secret. Concise answer: It sounds like you're trying to build a stream cipher. The security of it really comes down to how much of the scheme you think can be kept secret. If I listen in to your wifi and hear you requesting a ...

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 ...

10

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 ...

9

Kasiki's test and the index of coincidence are used to attack a Vigenère cipher (or other polyalphabetic ciphers with small alphabet and small key size) - they both try to get the length of the keyword. Kasiki's test gets probable prime factors of the keyword length, while the coincidence index test gets us an estimation of the absolute length of the ...

9

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 ...

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 ...

7

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 ...

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

What you've described is generally called a "book cipher" or "Ottendorf cipher", where the "key" is knowing which publication is being referenced, as well as the algorithm for recovering information from it. A hundred years ago they were quite secure because not only were books fairly rare, but trying every book against an unknown cipher was very time ...

5

While a known plaintext attack successfully finds the keys, nobody has been able to put forward a general solution to this cipher. Is that possible? You really have to go back in time to learn that ChaoCipher has been subject to some cryptanalysis before the description of the algorithm/device was published. As an example: Here’s one of the oldest ...

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

In a 1993 paper Ross Anderson proposed "A Modern Rotor Machine" that improves on the Enigma by utilizing a LFSR to produce random stepping. He believed it would resist all known attacks.

4

Sorry to ask this, but did you try a web search via Google for "javascript enigma simulator"? The first hit (cryptomuseum.com) has many links to enigma simulation source code. For example, this one.

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

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

A problem with this message is that the image in a press release from GCHQ differs slightly from one in a New York Times article, although the paper, with its tears looks the same. The image below shows the two side by side, the one on the left from the NY Times and the right from GCHQ. There is a slight difference in the style of writing, i.e. more sloping ...

3

Is the logic for how the enigma machine worked documented somewhere? Yes! If you're really interested in "diving in deeper" (pun intented), I would like to advise you to check out: "The Cryptographic Mathematics of Enigma" Dr. A. Ray Miller NSA. Center for Cryptologic History. USA. 1996. 3rd edition 2002 "Funkpeilung als alliierte Waffe gegen ...

3

It is probably impossible to remove language characteristics completely with a substitution cipher. Your algorithm flattens out single character frequencies, but that's it kinda. Language bigrams (your cipher-quadruples) might not be uniform distributed. The reason for this is simple: Bigram probability is not just the product of the probabilities of both ...

3

While we obviously don't know who is using what and where, we know the U.S. phased out their last mechanical systems (the KL-7) in the 1980s. The M-94 (Jefferson Cipher Disk) was a clever and simple implementation of a polyalphabetic substitution cipher, so there's no reason it couldn't be used today. But the U.S. Army stopped issuing them during WW-II. ...

3

I believe that the key lies in the final 6 number 1525/6. In this period the German Mathematician Albrecht DÜRER published "The Four Books on Measurement" the third of which picks up on the geometric construction of the latin alphabet. Albrecht DÜRER was also famous for a magic square which is the same as Sudoko puzzles and would therefore have the ...

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 ...

2

According to "Applied Cryptanalysis", the theoretical keyspace of Enigma is approximately $2^{366}$, but due to practical limitations, Enigma as used by the Germans only had a keyspace of approximately $2^{77}$. Given the power of some of the clouds out there (with GPUs and all), I bet you could do a brute-force attack of the 77-bit key space in a reasonable ...

2

"What is the simple cipher for such an alphabet?" I'm assuming you have some message written by paper in a written language with thousands of "characters", and you want to write an encrypted message on paper using more-or-less the same set of characters. One could use a substitution cipher, as you suggested, using a codebook that, for each red plaintext ...

2

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 ...

Only top voted, non community-wiki answers of a minimum length are eligible