# Tag Info

6

First of all, you cannot uniquely determine the keyword of a Playfair cipher, or even the key table constructed from it, simply because there are multiple equivalent key tables that will produce the same ciphertext (and multiple keywords that will produce each table). In particular, the following key tables are all equivalent: Original: Row shift: ...

5

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 ...

4

This looks like a Playfair cipher with MAN as the key. For characters not in the same row / column, you make a rectangle with the encoded characters at the corners, and the characters in the same row on the opposing sides of the rectangle form the decoded letters. For NP we have - [N]--B-->C F G H L<--O--[P] Which yields CL. So for ET we have ...

4

As it turns out, the encryption method used was "ADFGX", an earlier version of the proposed "ADFGVX" method suggested in a comment by fgrieu: You want to look at the ADFGVX cipher, a German, 1st World War cipher. This made me search and find ADFGX - a German cipher based on a combination of the Polybius checkerboard and ciphers using key words. ...

3

Playfair key grids are often generated by filling in the letters of a key phrase, then completing the grid with the remaining unused letters of the alphabet. It looks to me like the partial grid you provided has been created in this way. If so, you can fill in several more letters straight away (assuming there is no J): ? A ? D ? ? A ? D ? ? ? ? ...

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 ...

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 ...

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 ...

2

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 ...

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

The Playfair key matrix contains a permutation of the 25-letter cipher alphabet. The total number of such permutations is \begin{aligned} 25! &= 1 \times 2 \times 3 \times \dots \times 25 \\ &= 15{,}511{,}210{,}043{,}330{,}985{,}984{,}000{,}000. \end{aligned} However, as I noted in a previous answer, the Playfair cipher has equivalent keys: moving ...

1

As Maarten notes in the comments, you're indeed expecting rather too much from an old classical cipher designed to be used by hand. Ciphers from that era were simply not designed to be able to mechanically and unambiguously encode all possible messages. Basically, as you've noted, the Playfair cipher as you've described it cannot encrypt all messages; ...

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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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 ...

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