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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

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 ...


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

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 ...


10

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 ...


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 ...


8

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 and 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 ...


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 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 ...


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 ...


7

It's called a keyword cipher. See this question for some ways to break it.


6

When we consider that a Playfair key consists of the alphabet (reduced to 25 letters) spread on a 5x5 square, that's $25!$ keys (another formulation consider any string to be a key; then strings leading to the same square are equivalent keys). The rules of Playfair are such that any rotation of the lines in the square, and any rotation of its columns, lead ...


6

So the one's i'd bet on are either solitaire. 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 keystream has been generated. It is not nearly as widely studied a field since most people are assumed ...


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 ...


6

As the page explains, the cipher it describes is a simple variant of the bifid cipher, with the alphabet extended from the traditional 25 to 36 letters. As such, most techniques for breaking the bifid cipher ought to be more or less directly applicable to it. The bifid cipher is nowadays mainly used for crypto puzzles. Like most classical ciphers, it is ...


6

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 ...


6

This is highly insecure, for the same reason that ECB mode and simple substitution ciphers are. Every time you use the word the in your message, it will be encrypted the same way. The same goes for other, lower-frequency (but still fairly common) words -- like as or with or will (or any of hundreds of other examples). This is a humongous clue to ...


5

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 ...


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

Frequency analysis would work on average quite poorly on ciphertext of the proposed cipher. How exactly depends a lot on the value of the key: for some weak keys likes 1, 3, 30, 30000.. it works essentially as well as for any mono-alphabetic cipher. For 103, it still works well. For any key, given enough (lots of) ciphertext, it could still distinguish the ...


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.


5

The fact that a given cipher has a key length of 296 bits doesn't mean at all that it provides 296 bits of security or even that a brute force attack would take $2^{296}$ steps. The problem of mono-alphabetic substitution cipher is the ridiculously small block size (in this case, barely $\log 64 = 6$ bits). If absolutely nothing about the plaintext is ...


5

This looks like a sliding window approach to calculating the index of coincidence. So you would have something like: ABCDE FGHIJ KLMNO OACBD EFGHI JKLMN Given enough cyphertext, you'll discover a length at which the IC is high; this is a candidate keylength for the cyphertext, because you've shifted the two texts by one keylength. Multiples of this size ...


5

"Frequency analysis of the output might help determine simple words in the ciphertext such as 'the' etc if that word is repeated and sent multiple times. This isn't necessarily a problem as it's only a simple word and doesn't convey much meaning to the message". If the word "the" doesn't convey much meaning, then why have you used that particular ...


5

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 ...


4

I don't know the solution, but since you say you're only asking for hints, here's a few that occurred to me: If this is a Vigenère cipher, the missing character at the beginning should not matter (much): if you encrypt a message with the key FOOBAR and drop the first letter of the output, you can decrypt the resulting ciphertext with the key OOBARF. As ...


4

First guess the key length(Just try every plausible length, there aren't many). Then for each position where you know both plain- and ciphertext, calculate the key char. If you get a contradiction, the guessed key length was wrong. If the key length is short enough compared to the number of known pairs this will probably give you a large part of the key.


4

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


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

Your system is essentially: $c_0 = m_0 + k $ $c_i = m_i + c_{i-1} $ for $i>0$ It's absolutely trivial to break this, since the attacker knows the ciphertext, and thus knows both $c_i$ and $c_{i-1}$. Decrypting the first group is hard. But to decrypt any group but the first, simply subtract the previous group from the current group $ m_i = c_i - ...



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