SUB
IC
=
1 0 1 1
1 0 0 1
0 1 0 1
0 1 1 1
0 0 0 1
we can launch a cryptanalytic attack against the shrinking generator in order to obtain
the initial states of both LFSRs. Table 1 shows the calculations carried out for cryptan-
alyzing the above described generator. The most left column represents the indices n
i
numbered (0,··· , 2
A
− 2 = 30). Next column shows from Theorem 3.1 the position of
the terms (a
0
,a
1
,··· ,a
4
) of the sequence {a
i
} in the first column {d
o
} of the matrix IC.
The following columns of the Table 1 represent the matrix IC: in boldface the (5 × 4)
sub-matrix with the known bits, the remaining bits of {d
0
} are the bits computed to
determine the initial states of SRA and SRS, and the symbol - corresponds to unknown
bits of the shrunken sequence.
Computation of the SRA Initial State: According to Theorem 4.2, we compute the
positions of the (n
i
+ 1)-th elements of the first column of IC by solving the equation
system
n
i
· 15 ≡ mod 31 (i = 0,1, ..., 4).
That is to say, n
0
= 0,n
1
= 29,n
2
= 27,n
3
= 25,n
4
= 23. Then, by means of the char-
acteristic polynomial P
D
(x) we determine the values of the (n
i
+ 1)-th (i = 0,1,...,4)
elements of the first column {d
0
} of IC. Consequently, a
0
= 1,a
1
= 0,a
2
= 0,a
3
=
1,a
4
= 1 (see Table 1). Therefore, the initial state of the register SRA (1, 0, 0, 1, 1) has
been determined.
Computation of the SRS initial state: According to Theorem 4.3, we compute the
relative shifts between consecutive columns in the matrix IC:
- Computation of o1: We know a
1
at the (29+1)-th position of the first column fd0g
and compute its S−1= 4 successive bits. We compare these 5 bits (0, 0, 1, 1, 0) with the
first 5 bits (0, 0, 1, 1, 0) of the second column {d
1
} (see Table 1). There is coincidence,
thus o1 = 1.
- Computation of o2: We know a
2
at the (27+1)-th position of the {d
0
} and compute
its 4 successive bits. We compare these 5 bits (0, 1, 0, 0, 1) with the first 5 bits (1, 0, 0, 1,
0) of the third column {d
2
} . There is no coincidence, thus we analyze the following bit
a
3
.We know a
3
at the (25+1)-th position of {d
0
} and compute its 4 successive bits.We
compare these 5 bits (1, 0, 0, 1, 0) with the first 5 bits (1, 0, 0, 1, 0) of { d
2
} (see Table
1). There is coincidence, thus o
2
= 3. Since o
2
= 3 ≥ S − 1, we have determined the
initial state of SRS. In fact, s
0
= 1, o
1
= 1 implies s
1
= 1, o
2
= 3 implies s
2
= 0 and
s
3
= 1. Therefore, the SRS initial state is (s
0
,s
1
,s
2
,s
3
) = (1,1,0,1). Remark that only
the knowledge of three columns of the sub-matrix has been necessary to identify the
initial state of SRS. Indeed, this number equals the number of bits 1 in the initial state
of the selector register. The maximum number of known bits corresponds to SRS initial
state with all bits 1. In the remaining cases, less bits are sufficient.
Once the initial states of both register are determined, the whole shrunken sequence that
is the keystream sequence can be computed.
64