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

204
appendix
wires must give rise to plain–cipher repeats, but the argument fails if the
wheel has an odd number of wires and then we can find wheels which
produce no repeats, thus:
Example (No repeats in any row of the encipherment matrix)
Consider the 7-point wheel having encipherment matrix with first column
(1, 3, 6, 2, 0, 5, 4). The first row is then
(1, 4�1, 5�2, 0�3, 2�4, 6�5, 3�6)
which is (mod 7)
(1, 5, 0, 3, 6, 4, 2),
a set which contains no repeats. Since the entire matrix is determined by
any row or column there can be no repeats in any row.
M15 Number of possible Enigma reflectors
In the reflector the 26 letters are joined in pairs. The first pair can be
chosen in
26�25
2
ways (we must divide by 2 because it doesn’t make any difference which of
the pair we choose first and which second). We can now choose the next
pair in
24�23
2
ways; and so on. Thus we can choose to join the 26 letters into pairs in
26!
2 13
ways. We would however get the same reflector if we chose the same 13
pairings in a different order and since we can re-order the 13 pairs in 13!
ways the total number of distinct reflectors is
26!
2 13 �13!
and this is more than 7�10 12 ; this is the same as the number of reciprocal
simple substitutions [M2].
The same calculation applies to the number of possible plugboards.