Cryptography Reference
In-Depth Information
•
Step 2:
⎡
⎤
1101100
0101011
1011010
⎣
⎦
H
∗
=
After systematization on the last three columns, we obtain
⎡
⎤
1101100
1011010
1110001
⎣
⎦
H
=
•
Step 3: The hard decision on the
n
k
=4
highest received values gives
s
=(0
,
1
,
1
,
1)
.Matrix
H
without the
k
last columns makes it possible
to find the missing redundancy values:
−
⎡
⎤
⎡
⎤
⎡
⎤
0
1
1
1
1101
1011
1110
0
0
0
⎣
⎦
⎣
⎦
·
⎣
⎦
=
The initial decoded vector is therefore
(0
,
1
,
1
,
1
,
0
,
0
,
0)
which, after inverse
permutation, gives the vector
(1
,
1
,
0
,
0
,
0
,
0
,
1)
. The initial metric is
M
=
M
1
+
M
2
=
−
2
.
7
as
M
1
=(
−
0
.
9)
−
0
.
7
−
0
.
6
−
0
.
5=
−
2
.
7
and
M
2
=
(
−
0
.
3) + 0
.
2+0
,
1=0
.
0
.
•
Step 4: To list the concurrent words, we apply inversion masks to the
n
k
first bits (we can have from 1 to 4 inversions maximum). Each
inversion mask will increase the
M
1
part of the metric. For an inversion,
the bonus is at minimum 1.0 and at maximum 1.8. The minimum bonus
for two inversions is at minimum
2
−
(0
.
6+0
.
5) = 2
.
2
. The first concurrent
words to consider are therefore all those corresponding to a single inversion.
Moreover, modifications on the
M
2
part of the metric could decrease it.
However, the decrease cannot exceed
2
×
(0
.
2+0
.
1) = 0
.
6
compared to
the initial metric. We therefore already know that the found previously
word is the most likely word. However, if we decide to list them anyway,
we find the following concurrent words which, unlike the Chase-Pyndiah
×
