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

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appendix
Solution
A linear recurrence of order 5 takes the form
U n �aU (n�1) �bU (n�2) �cU (n�3) �dU (n�4) �eU (n�5)
where a, b, c, d and e are unknown constants whose values are either 0 or 1,
since all the arithmetic is (mod 2).
We number the bits 1 to 15 from left to right and then put n�6, 7, 8, 9
and 10 in the recurrence to obtain five linear equations in the five
unknowns a, b, c, d and e:
a(1)�b(0)�c(1)�d(0)�e(1)�0, (A.1)
a(0)�b(1)�c(0)�d(1)�e(0)�0, (A.2)
a(0)�b(0)�c(1)�d(0)�e(1)�1, (A.3)
a(1)�b(0)�c(0)�d(1)�e(0)�1, (A.4)
a(1)�b(1)�c(0)�d(0)�e(1)�0. (A.5)
From equations (A.1) and (A.3) we find that a�1 and then from equation
(A.4) it follows that d�0. From equation (A.2) we then find that b�0 and
from equation (A.5) that e�1 and, finally, from equation (A.1) that c�0.
The solution to these five equations is therefore
U n �U (n�1) �U (n�5) .
We now need to confirm that this gives the correct values when n�11, 12,
13, 14 and 15 and it will be seen that this is the case. We have therefore
verified that the recurrence of order 5 that we have just found does generate
the given stretch of key.
As mentioned in Chapter 8 it can happen that there are no solutions of
order k or that there is more than one solution. In the former case the
equations are inconsistent and in the latter the ambiguities can usually be
resolved if extra key digits are available. The following examples illustrate
these situations.
Example (More than one solution)
Verify that the 10-bit binary key
0110110110
can be generated by either of two binary linear recurrences of order 5.