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

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chapter 10
the numerical value of the slide being the usual one, given in Table 1.1
viz: A�0, B�1, . . . . , Z�25.
On machines that do not possess this feature the slide has a permanently
fixed value: on the M209, for example, it is Z, which is numerically
equivalent to 25, or �1 since all the arithmetic is (mod 26). So, on the
M209, a key value which appeared to be 25 would in reality have been
been generated by the six wheels as 0 or 26 and a key value which
appeared to be 0 would have been generated by the wheels as 1 or 27.
Identification of the slide would be a first step in key analysis.
Identifying the slide in a cipher message
The cryptanalyst may be able to identify the slide in a cipher message if he
knows the cage that is being used. To do this he needs to compute a ‘theoretical
cipher distribution’ and compare it statistically with the actual
cipher letter frequencies in the message. For further details see M20.
The existence of the slide doesn’t invalidate the differencing attack but
it makes the initial recognition of the 0 and 1 key values more difficult.
The slide would probably be changed for each message and the cipher
operator would have to have some means of communicating its identity.
Overlapping
This feature was available on all models of the Hagelin. Recall that there
are 27 bars and 54 lugs on the cage behind the wheels. On each bar the 2
lugs can be positioned opposite any of the six wheels or in one of the two
‘neutral’ positions. In an unoverlapped Hagelin one of the 2 lugs on each
bar would be in one of the neutral positions. In an overlapped Hagelin 1 or
more bars will have the 2 lugs opposite two of the wheels. This has the
effect that where two wheels have a lug on the same bar their contribution
to the overall key when both are active is 1, not 2. This is because the bars
only move the print wheel 1 position irrespective of whether 1 or 2 lugs
engage active pins. So, for example, if the 26-wheel has a kick of 5 and the
25-wheel has a kick of 6 and they share two bars where they both have an
active pin then their combined active contribution to the total key is not
11 but 11�2�9. So
26- and 25-wheels both inactive – contribution to the key is 0;
26-wheel active, 25 inactive – " " " " is 5;
26-wheel inactive, 25 active – " " " " is 6;
26- and 25-wheels both active – " " " " is 9.