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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 ...


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

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 ...


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 ...


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 ...


2

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.


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

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 ...



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