# 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 ...

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 ...

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 ...

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 ...

4

There are several algorithms available which can attack a playfair cipher. Hill climbing might be one option. Basically it starts with a random key (assuming it's the best one) and decrypts the cipher. The resulting clear text is scored using a fitness function. Then small changes are applied to the key and if the resulting clear text of the modified key ...

3

Counting duplicate keys, it would be $25!$ (remember i=j in playfair). Basically the key is the 5x5 square. There are $25!$ permutations of the $25$ characters which populate that square.

3

It's not elegant, but the brute force method is to write a program that creates a table of 25x25 digraphs (assuming i=j), yielding 625 rows. I'd also add a column that lists the relative frequency of each digraph (given enough ciphertext you can use that to identify frequent substitutions, as you already have done). You start off with 625! possible ...

3

Your math is off. The key space of "four squares of random letters" is not four times as big as the one provided by one square, but the fourth power of that size. To see this easier, consider the case of one $2×2$-square (or four ones), where you can still calculate everything by hand. There are $4! = 24$ possible ways to fill a $2×2$-square. If you have ...

2

The square used is P L A Y F I R B C D E G H K M N O Q S T U V W X Z Dataconfidentiality becomes bf qf rs tp ri nu nd ya dn cy (assuming we pad with x) Done using this simulation

2

Let's start by explaining how Playfair works normally to encrypt a message. First, you create a 5x5 table by writing the keyword letter-by-letter across the top of the table, from left to right, skipping duplicate letters; you then fill in the remaining characters in alphabetical order after the keyword (combining i&j or j&k into a single box). ...

2

The unicity distance is defined to be the minimum number of ciphertext characters needed to have a unique significant decryption. It answers the question 'if we try all the keys, how much ciphertext would we need to be sure our solution was the true solution?'. The answer depends on the redundancy of the language in which the plaintext was written. To find ...

1

The short answer is yes. It would be considerably more secure. But nowadays, classical encryption methods like Playfair and Vigenère are so easily broken by computer analysis that they offer next to no security whatsoever. Aiming for something "considerably more secure" than either of these is really setting the bar very low. Specifically, although the ...

1

use letters to represent digits "a captured German revealed under interrogation that Enigma operators had been instructed to encode numbers by spelling them out ... Alan Turing reviewed decrypted messages and determined that the number "eins" ("one") was the most common string in the plaintext." -- Wikipedia: known-plaintext_attack "The Enigma machine ...

1

The Playfair cipher has a key consisting of a square of $5 \times 5$ letters (usually the J is not used, or I/J are considered one letter). Filling the square can be done in $25!$ ways (pick a letter for left upper corner, a new one for the place next to it, and so on), but then every square has equivalent forms, formed by rotating the columns and/or ...

1

There are diagnostic programs that will tell you the cipher type from a statistical analysis. For example: http://bionsgadgets.appspot.com/gadget_forms/refscore.html tells you immediately that this cipher is a Playfair.

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