It’s been a while since I’ve blogged about my passion for crosswords, but this Sunday’s Azed puzzle in the Observer was in one of my favourite forms so I thought I’d mention it briefly here.
Azed is the pseudonym used by Jonathan Crowther who has been setting the Observer crossword since 1972; this week’s was number 1967. His puzzles are usually standard cryptic crosswords which, though quite difficult as such things go, are nevertheless set in a fairly straightforward style. Every now and again, however, he puts together a different type of puzzle that makes a different set of demands on the solver. To be honest, I don’t always like these “funny” ones, as they sometimes seem to me to be contrived and inelegant, but this last one was a type I really like as it combines the normal cryptic crossword style with another interest of mine, namely codes and codebreaking.
The interesting aspect of this particular puzzle, which is laid out on a normal crossword grid, is that it involves a type of code called a Playfair cipher. In fact, this particular scheme was invented by the scientist Charles Wheatstone whom most physicists will have heard of through “Wheatstone Bridge“. It was, however, subsequently popularized by Lord Playfair, whose name stuck rather than its real inventor’s. Stigler’s Law of Eponymy strikes again!
The Playfair scheme is built around the choice of a code word, which must have the special property that no letter occurs twice within it. Other than that, and the fact that the more letters in the codeword the better the code, there aren’t any real constraints on the choice. The particular example used by Azed to illustrate how it works is ORANGESTICK.
The codeword is used to construct a Playfair square which is a 5×5 arrangement of letters involving the codeword first and then afterwards the rest of the alphabet not used in the codeword, in alphabetical order. Obviously, there are 26 letters altogether and the square only holds 25 characters, so we need to ditch one: the usual choice is to make I stand for both I and J, doing double duty, which rarely causes ambiguity in the deciphering process. The Playfair square formed from ORANGESTICK is thus
This square is then used as the basis of a literal digraph substitution cipher, as follows. To encode a word it must first be split into pairs of letters e.g. CR IT IC AL. Each pair is then seen as forming the diagonally opposite corners of a rectangle within the word square, the other two corner letters being the encoded form. Thus, in the example shown, CR gives SG (not GS, which RC would give).
Where a pair of letters appears in the same row or column in the word square, its encoded form is produced from the letters immediately to the right of or below each respectively. For the last letters in a row or column the first letters in the same row or column become the encoded forms. Thus IC is encoded as CE. When all the pairs are encoded, the word is joined up again, thus CRITICAL is encoded as SGCICEOP.
Obviously, to decipher encrypted text into plain one simply inverts the process as it is completely reversible.
The advantage of this over simpler methods of encipherment is that a given letter in the plain text is not always rendered as the same letter in the encrypted form: that depends on what other letter is next to it in the digraph. That means it isn’t cracked so easily using letter frequencies, as simple subsitution ciphers are.
Now, what does this have to do with a crossword? Well, in a Playfair puzzle like the one I’m talking about a certain number of answers – in the case of the latest Azed puzzle, eight – have to be encrypted before they will fit in the diagram. These “special” clues, however, are to the unencrypted form of the answer words. The codeword is not given, but must be deduced.
What one has to do, therefore, is to solve the clues for the unencrypted words, then solve all the other clues that intersect with them on the grid. Given a sufficient number of digraphs in both plain text and encrypted form one can infer the codeword and hence encrypt the remaining (unchecked) letters for the special answers.
It probably sounds very convoluted, but in this puzzle it isn’t so bad because the “special” clues are relatively straightforward which generates enough “cribs” in the form of letter-pairs in both plain and encrypted form. There aren’t enough pairs to deduce the codeword exactly – as not every letter in the codeword appears among the digraphs – but it helps knowing that it is an actual word rather than a jumble of letters. That, together with the rules for encryption using the Playfair square, gives enough information to infer the codeword; the digraph PA which encrypts to AE is particularly useful in this case. In this case the codeword has 14 letters. I won’t tell you what it is because the competition is not yet closed!
What has to be done then is to use the codeword to complete the unchecked letters in the specials in their encrypted form. That bit is relatively straightforward but for the competition one also has to supply a “normal” cryptic clue for 9 down. That’s always the bit I find hardest!
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