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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 ...

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 ...

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 ...

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 ...

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 ...

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 ...

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 ...

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 ...

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 ...

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 ...

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 ...

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 ...

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 ...

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 ...

It was never determined how to derive the keyword PALIMPSEST from the sculpture or its text. It was, as you stated, found by brute-force hill-climbing algorithms. One theory is that crib-dragging the morse phrase SHADOW FORCES reveals IMPS at SHAD, and working form there retrieving the entire keyword, but I find this far-fetched and implausible. The ...

In principle even a one time pad isn't unbreakable if the pad is a short sequence that is repeated. If that were the case then a brute force search comparing each output message with the statistical likelihood of that message existing in English should yield an answer. Seems that we really need to know more about the encryption methods that may have been ...

If TVSHOW would be the keyword, you would have this substitution table (written horizontally for space reasons): ABCDEFGHIJKLMNOPQRSTUVWXYZ TVSHOWABCDEFGIJKLMNPQRUXYZ This is not quite what you have in your table. The frequency distribution only gives you an approximate solution, and will normally not give the exact one for the less often letters. You'll ...

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 ...