Cryptography Reference
In-Depth Information
3.31. Let
r
=6,
s
= 29, and
k
=(
k
1
k
2
k
3
k
4
k
5
k
6
) = (237591).
c
=
ARBHJOLSPGRORXJSIJGACFKWQAVZG
.
This is a famous quote attributed to the
Duchess of Windsor (Wallis Simp-
son) (1896-1986)
, who was the wife of the former King Edward VIII.
3.32. Let
r
=4,
s
= 16, and
k
=(
k
1
k
2
k
3
k
4
) = (7182).
c
=
SFITYXONEBHANMEP
.
This, and its conclusion in Exercise 3.33, is a quote from
John SheGeld
(First Duke of Buckingham and Normanby) (1648-1721)
, who was an
English poet and politician. The quote is taken from
“An Essay upon
Satire” (1689)
.
3.33. Let
r
=4,
s
= 17, and
k
=(
k
1
k
2
k
3
k
4
) = (1234).
c
=
PTQSHKBKKBHARBTPL
.
See Exercise 3.32 for the initial part of this quote.
In Exercises 3.34-3.44, use the information pertaining to LFSRs as de-
scribed on pages 156-158. For further reading on LFSRs, the reader may
consult
[111]or [283.
3.34. Given
=4,(
c
1
c
2
c
3
c
4
) = (1010) and initial state,
s
0
=(
k
(3
,
0)
k
(2
,
0)
k
(1
,
0)
k
(0
,
0)
) = (0111)
,
calculate the period length and each iteration for the LFSR.
3.35. Execute the calculations in Exercise 3.34 using matrix equations as de-
scribed on pages 157 and 158.
(
Hint: Your last calculation should be
CS
6
=
S
7
.
)
3.36. Show that the maximum number of possible internal states is 2
−
1.
3.37. Prove that an LFSR must be periodic with period length no larger than
2
1.
Note that if
c
=0
, then the LFSR is not periodic, but is
eventually
periodic since it must become periodic after ignoring a finite number of
initial terms. In other words,
s
L
=
s
N
for some
N
−
L>
0
. When
c
=0
, the LFSR is called
singular
, and is called
nonsingular
otherwise.
Hence, we are only considering nonsingular LFSRs since our assumption
on page 155 is that
c
=1
.
≥
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