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

102
chapter 8
To take an easily verifiable case first: a sequence of order 4 might have a
period of 15, which is 2 4 �1. If we ignore the trivial case where all four
multipliers are 0 there are 15 possible binary linear sequences of order 4.
Do any of these produce a binary sequence of maximum period 15? Just 2
of them do; they are
and
U n �U (n�3) �U (n�4)
U n �U (n�1) �U (n�4) .
Example 8.1
Verify that the binary linear recurrence of the fourth order
U n �U (n�3) �U (n�4)
generates a sequence of maximum period 15.
Verification
We start with U 0 �U 1 �U 2 �U 3 �1 and generate each new term by adding
together the terms three and four places earlier in the sequence and
putting 0 or 1 according to whether the sum is even or odd. The sequence
is then found to be
1, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 1, 1, 1, 1...
and we see that the sequence begins to repeat from U 15 onwards, but not
before.
(Note that if a binary sequence generated by a linear recurrence of order k
has maximum period it must contain all possible 2 k binary sequences of
length k except the one consisting of all 0s. We may therefore take any
initial values except 00...00 and the period will be seen to be maximal in
every case. If the period is not maximal different starting values may
produce different sequences.)
Problem 8.1
Verify that the binary recurrence
U n �U (n�1) �U (n�4)
also generates a sequence of period 15 but that the recurrence
does not.
U n �U (n�1) �U (n�2) �U (n�3) �U (n�4)