# Tag Info

17

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 ...

14

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 ...

13

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 ...

12

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 ...

12

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 ...

11

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 ...

11

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 ...

11

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 ...

11

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 ...

11

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 ...

11

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 ...

11

If you combine two affine ciphers, you obtain one affine cipher. Say the first cipher is $e_1(x) = a_1x+b_1$ and the second is $e_2(x) = a_2x+b_2$. Then $e_1(e_2(x)) = a_1(a_2x+b_2)+b_1 = (a_1a_2)x+(a_1b_2+b_1)$. Note that if $a_1$ and $a_2$ are both relatively prime with the modulus, then so is $a_1a_2$, so the new cipher can also be deciphered.

10

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 ...

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

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 ...

9

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 ...

9

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

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 ...

8

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 ...

8

First, you should start by guessing which symbols in the ciphertext are actually enciphered, and which are simply written in plain. (Don't worry if you guess wrong, you can always make several guesses.) For a Vigenère cipher, one also needs to guess whether any non-encrypted characters should advance the key position or not (usually they do not). For your ...

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 how ...

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 it'...

7

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 (...

7

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 ...

7

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 ...

7

One option would be to get them to select a one-time MAC of the form: $mac(m,k_0, k_1) = (k_0 \times m + k_1) \mod p$ You would select $p$ to be something like 29. $k_0$ and $k_1$ would be chosen at random from the values 0-29. $k_0$ has the additional restriction that it can't be 0. You can aid the computation by giving them a 29x29 matrix of all ...

7

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

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

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 \$f'...

6

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 ...

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