Code and ciphers: Julius Caesar, the Enigma and the internet

2, there are 26! possible simple substitution ciphers and the brute force
work factor is about 10 25 but such systems can be solved by hand in less
than an hour, given 200 letters of cipher. Many types of cipher, including
the Hagelin, are vulnerable to attacks involving mathematical or statistical
weaknesses in the cipher itself or in its mode of operation, as in the
Enigma. A cipher where there are no known weaknesses or ‘trapdoors’,
which enable the cryptanalyst to break the system by a special route,
would be formidable indeed. It is claimed that a modern system, the DES
(Data Encryption Standard), is such a system. A brief description of this is
given in Chapter 13.
Solving the Hagelin from a stretch of key
The Hagelin cipher machine 143
Recovery of the cage and pin patterns of a non-overlapped Hagelin is
straightforward if we have more than 131 consecutive key values, as is
shown below. ‘How, though,’ you might ask ‘do we obtain 131 consecutive
key values?’ The answer is: ‘By finding the plaintext of a message of
length 131 characters or more.’ This might be achieved in a number of
ways including:
(1) finding that the same message has been sent in another cipher that is
readable;
(2) finding that two or more Hagelin messages are in depth which may be
readable by the method of crib-dragging as explained in Chapter 7;
(3) by the cipher operator making a mistake and having to retransmit the
message on a slightly different set-up.
(4) by clandestine means, e.g. by an agent obtaining the plaintext of a
message.
To illustrate the method of solution when a sufficiently long stretch of
key has been obtained we look at the key that would be generated by a
‘mini-Hagelin’ with only three wheels. The corresponding attack for a
full scale Hagelin undoubtedly involves more work but the method is the
same and is based upon the fact that the key is the sum (mod 26) of the
contributions of six wheels which move regularly and so, by subtracting
the key stream from itself, suitably shifted, we can remove the contributions
of any five of these wheels and so discover the ‘kick’ on the sixth
wheel. Thus, if we generate a key by adding together the contributions of
two mini-wheels, one of length 7 and with a kick of 5 and the other of
length 9 and with a kick of 3 we might have