3.1. Modified Syndrome-Based Method
We define the function of the indeterminate
z
and the syndrome function
, where
and
is given in Equation (
2). We can obtain in [Equation (18)] in [
17] that
Therefore, the can be regarded as the coefficient of of the polynomial . Then, the erased column is given by , where .
Note that the terms of set
are not involved in computing the coefficient of
of the polynomial
. Thus, we can just consider the first
terms (the
coefficients of degrees less than
) of
when computing these coefficients, but all the
r terms of
are calculated in [Step 1] in [
17]. This is one essential way our modified syndrome-based decoding method obtains a lower decoding complexity than the original method in [
17].
Moreover, the syndrome polynomials
satisfy
i.e., the
syndrome polynomials
either all have an even number of nonzero terms, or they all have an odd number of nonzero terms, from the definition of Equation (
2).
Let
and
. Then, we have
Thus,
is independent of the erasure index
i, and we only need to compute
once in the decoding procedure. Recall that
is the coefficient of
of the polynomial
; then, the
is also the coefficient of
of the polynomial
for all
. Suppose that
we can derive the recurrence formula
where
. Notice that
holds. Similar to
, we only compute the first
terms (the
coefficients of degrees less than
) of
, since the other coefficients of
are not needed, but all the
terms of
are calculated in [Step 2] in [
17]. This is another way our modified syndrome-based decoding method obtains a lower decoding complexity than the original method in [
17]. Algorithm 1 shows our modified syndrome-based decoding method over the ring
.
The following Lemma shows that we can always compute the divisions in steps 11–12 of Algorithm 2 by Lemmas 1 and 2 when .
Lemma 3. In steps 11–12 of Algorithm 2, the has an even number of nonzero terms for all , and we can employ Lemmas 1 and 2 to compute the divisions.
Proof. From Equation (
8) and steps 7–10 of Algorithm 2, we obtain
where
. If the number of polynomials in the set
, which has an odd number of nonzero terms, is an even number, then the
has an even number of nonzero terms for
. In the following, we will show this is true. According to Equation (
6) and step 3 of Algorithm 2,
holds.
Firstly, we consider . We denote the polynomials with as . Let be the initial for .
To prove that the number of polynomials with an odd number of nonzero terms in the set is even, it is equivalent to prove that .
Algorithm 2: Modified syndrome-based decoding method. |
|
According to Equation (
7) and steps 4–6 of Algorithm 2, we have
where
. The
holds for all
. We can obtain by induction
Note that
; we can suppose that there are an even number of polynomials in the set
, which has an odd number of nonzero terms, when
, i.e.,
first. We have
; so,
Equation (
11) comes from Equation (
10) with
. Therefore, there are an even number of polynomials in the set
, which has an odd number of nonzero terms.
Secondly, when , the argument is similar. This completes the proof. □
According to Lemma 3, we can use Lemmas 1 and 2 to compute the divisions in step 12. The number of divisions required in step 12 is recorded as
, which ranges from 1 to
for
. So, we can obtain
in step 12 by recursively computing the division
times, while the number of nonzero terms of the polynomial resulting from the first
divisions is even. Therefore, we can execute these divisions by Lemma 2 and execute the last division by Lemma 1. The computational complexity
in steps 11–12 of Algorithm 2 is
where
.
In steps 11–12 of Algorithm 2, we take the
division without Algorithm 1, in which
divisions are executed by Lemma 1 and
divisions are executed by Lemma 2; however, the number of the divisions can be reduced with Algorithm 1. In
Table 1, we show the average number of divisions in steps 11–12 of Algorithm 2 executed by Lemma 1 and Lemma 2 with Algorithm 1 for
.
We specify the computational complexity of Algorithm 2 as follows:
Steps 1–2 take XORs.
Steps 3–6 take XORs.
Steps 7–10 take XORs.
Steps 11–12 take
XORs by Equation (
12).
Then, the computational complexity
of Algorithm 2 is
where
Recall that the computational complexity of the decoding method in [
17] is
which is strictly larger than
.
Table 2 evaluates the computational complexity of the decoding method in [
17] and Algorithm 2 for some parameters. The results in
Table 2 demonstrate that Algorithm 2 has much lower decoding complexity, compared with the original decoding method in [
17]. For example, Algorithm 2 has 40.60% less decoding complexity than the decoding method in [
17] when
.
The reason why Algorithm 2 has lower decoding complexity than the decoding method in [
17] can be summarized as the following three points.
Firstly, we only consider the first
terms (the
coefficients of degrees less than
) for both
and
in computing the coefficients of
, while all
r terms of
and all
terms of
are calculated in the decoding method in [
17], where
.
Secondly, all the divisions in Algorithm 2 are executed over the ring
by Lemmas 1 and 2, which takes
XORs and
XORs for each division, respectively. In addition, the division in [
17] is executed over the ring
, which takes
XORs [
17] (Corollary 2).
Thirdly, we apply Algorithm 1 to steps 11–12 of Algorithm 2, which can significantly reduce the number of divisions, thus reducing the number of XORs required.
3.2. Modified Interpolation-Based Decoding Method
According to the decoding method in [
18], we can recover the erased column
with
by
where
and
. Let
where
. Then,
has an even number of nonzero terms, and we only need to compute once for
in the decoding procedure, since
is independent of the erasure index
i. Let
where
, and
. Algorithm 3 shows our modified interpolation-based method over the ring
.
After using Algorithm 1, the number of polynomial multiplications in step 2 ranges from 1 to
. Thus, the computational complexity
in steps 1–2 of Algorithm 3 is
Algorithm 3: Modified interpolation-based method. |
|
In steps 1–2, we need to take
multiplications without Algorithm 1, which takes
XORs; however, with Algorithm 1, the number of multiplications involved in steps 1–2 can be reduced. In
Table 3, we show the average number of XORs involved in steps 1–2 of Algorithm 3 with Algorithm 1 for
. The results in
Table 3 show that we can reduce the number of XORs with Algorithm 1, especially for a large value of
.
Only steps 4 and 6 of Algorithm 3 are needed to compute the division. We should employ Lemma 2 to execute the divisions in steps 3–4 in Algorithm 3, since in step 6 of Algorithm 3 should have an even number of nonzero terms. Notice that steps 5–6 of Algorithm 3 are exactly the same as steps 11–12 of Algorithm 2.
We specify the computational complexity of Algorithm 3 as follows:
Then, the computational complexity of Algorithm 3 is
where
Recall that the computational complexity of the decoding method in [
18] is
which is larger than that of our Algorithm 3.
Table 4 evaluates the computational complexity of the decoding method in [
18] and Algorithm 3 for some parameters. The results in
Table 4 demonstrate that our Algorithm 3 had much lower decoding complexity, compared with the original decoding method in [
18]. For example, Algorithm 3 had a 34.13% lower decoding complexity than the decoding method in [
18], when
.
The reason why Algorithm 3 has a lower decoding complexity than that of the decoding method in [
18] is summarized as follows.
Firstly, all the divisions in Algorithm 3 were executed over the ring
by Lemmas 1 and 2, which used
XORs and
XORs for each division, respectively. The division in the decoding method in [
18] was executed over the ring
, which used
XORs.
Secondly, we applied our Algorithm 1 to steps 1–2 and steps 5–6, which significantly reduced the number of multiplications, thus reducing the number of XORs required.