Cryptography Reference
In-Depth Information
•
Case (b)
Δ
S
1
(
Z
k
Z
p
+
Z
k
Z
p
)+
S
2
(
Z
k
Z
p
+
Z
k
Z
p
)+
S
3
(
Z
k
Z
p
+
Z
k
Z
p
)
,
k
1
e
i
=
3
=
p
=
i,
(
i, k, p
)
∈{
1
,
2
,
3
}
Z
i
1
Z
i
2
Z
i
3
Δ=
1
3
i
1
+
i
2
+
i
3
=6
i
1
≤
i
1
,i
2
,i
3
≤
=
i
2
=
i
3
•
Case (c)
S
1
Z
p
+
S
2
Z
i
(
Z
1
+
Z
2
)
,p
2
e
i
=
=
i,
(
i, p
)
∈{
1
,
2
}
•
Case (d)
S
1
S
2
e
1
=
6. Correct the errors:
c
(
x
)=
r
(
x
)+
e
(
x
)
Example 4.14
To illustrate the decoding of an RS code using the direct method, we now
present an example considering an RS code correcting up to three errors (
t
=3
)
and having the following parameters:
m
=4
q
=16
n
=15
n
−
k
=6
Let us assume, for example, that the transmitted codeword is
c
(
x
)=0
and that
the received word has two errors.
r
(
x
)=
α
7
x
3
+
α
3
x
6
1. Calculate the components of the syndrome
S
1
=
α
10
+
α
9
=
α
13
S
4
=
α
19
+
α
27
=
α
6
S
2
=
α
13
+
α
15
=
α
6
S
5
=
α
22
+
α
33
=
α
4
S
3
=
α
16
+
α
21
=
α
11
S
6
=
α
25
+
α
39
=
α
13
2. Determine the number of errors
Δ
2
=
α
8
Δ
3
=0
Δ
3
being null and
Δ
2
=0
, we have two errors.
