Cryptography Reference
In-Depth Information
3. Calculate coecients
σ
1
and
σ
2
of the error locator polynomial.
[
S
1
S
4
+
S
2
S
3
]=
α
19
+
α
17
α
8
1
Δ
2
=
α
11
+
α
9
=
α
2
σ
1
=
S
2
S
4
+
S
3
=
α
12
+
α
22
α
8
1
Δ
2
=
α
4
+
α
14
=
α
9
σ
2
=
The error locator polynomial is therefore equal to:
σ
d
(
x
)=
x
2
+
α
2
x
+
α
9
4. Look for the two roots of the error locator polynomial.
Looking through the elements of field
F
16
we find that
α
3
and
α
6
cancel
the polynomial
Λ(
x
)
. The errors therefore concern the terms in
x
3
and in
x
6
of word
r
(
x
)
.
5. Calculate the error coecients
e
1
and
e
2
.
=
α
19
+
α
6
α
9
+
α
6
=
α
12
α
5
S
1
Z
2
+
S
2
Z
1
Z
2
+
Z
1
=
α
7
e
1
=
=
α
16
+
α
6
α
9
+
α
12
=
α
11
α
8
S
1
Z
1
+
S
2
Z
1
Z
2
+
Z
2
=
α
3
e
2
=
6. Correct the errors
c
(
x
)=(
α
7
x
3
+
α
3
x
6
)+(
α
7
x
3
+
α
3
x
6
)=0
The transmitted codeword is the null word; the two errors have therefore
been corrected.
Simplification of Peterson's algorithm for binary codes
For BCH codes with binary symbols it is not necessary to calculate the coe
-
cients
e
j
. Indeed, as these coe
cients are binary, they are necessarily equal to 1
in the presence of an error in position
j
. The computation of coecients
σ
j
can
also be simplified by taking into account the fact that for a code with binary
symbols we have:
S
2
j
=
e
(
α
2
j
)=
e
(
α
j
)
2
=
S
j
For a BCH code with binary symbols correcting up to
t
=3
errors, taking into
account the previous remark and using the expressions of the three coecients
σ
j
of the error locator polynomial, we obtain:
σ
1
=
S
1
S
1
S
3
+
S
5
S
1
σ
2
=
(4.40)
+
S
3
=
S
1
+
S
3
)+
S
1
S
1
S
3
+
S
5
σ
3
S
1
+
S
3
