Improved syndrome decoding of lifted $L$-interleaved Gabidulin codes

Abstract

A syndrome decoding algorithm for lifted interleaved Gabidulin codes of order L is proposed. The algorithm corrects L times more deviations (packet insertions) than known syndrome decoding methods with probability at least $1-8q^{-n}$, where n is the length of the (interleaved) Gabidulin code. For $n<L$, the proposed scheme has L times less computational complexity than known interpolation-factorization based decoders which attain the same decoding region. Upper bounds on the decoding failure probability are derived. Up to our knowledge this is the first syndrome-based scheme for interleaved subspace codes that can correct deviations beyond the unique decoding radius.

Appendix

1.1 Proof of Lemma 1

Proof

Let us first consider

$$\begin{aligned} \tilde{s}^{(j)}_\text {D}(x)&=\sum _{l=1}^{L}\overline{s}^{(l)}(x)\otimes \overline{\varGamma }_{l}^{(j)}\left( x\right) \end{aligned}$$

(59)

$$\begin{aligned}&= \overline{s}^{(1)}(x)\otimes \overline{\varGamma }_{1}^{(j)}\left( x\right) +\overline{s}^{(2)}(x)\otimes \overline{\varGamma }_{2}^{(j)}\left( x\right) +\dots + \overline{s}^{(L)}(x)\otimes \overline{\varGamma }_{L}^{(j)}\left( x\right) \end{aligned}$$

(60)

for all $j=1,\dots ,L$. Define the vectors

$$\begin{aligned} \mathbf {s}_{j}^{[i]}\overset{{{\mathrm{def}}}}{=}\left( \overline{s}_{j}^{(1)[i]} \ \overline{s}_{j}^{(2)[i]} \ \dots \ \overline{s}_{j}^{(L)[i]}\right) \end{aligned}$$

(61)

and

$$\begin{aligned} \mathbf {z}_{\ell }^{(l)[i]}\overset{{{\mathrm{def}}}}{=}\left( \overline{\varGamma }_{1,\ell }^{(l)[i]} \ \overline{\varGamma }_{2,\ell }^{(l)[i]} \ \dots \ \overline{\varGamma }_{L,\ell }^{(l)[i]} \right) ^T =\left( \varGamma _{1,D-\ell }^{(l)[i-D]} \ \varGamma _{2,D-\ell }^{(l)[i-D]} \ \dots \ \varGamma _{L,D-\ell }^{(l)[i-D]} \right) ^T \end{aligned}$$

(62)

for all $\ell \in [0,D]$. The equality in (62) follows from the definition of the q-reverse (see (4)). Defining

$$\begin{aligned} \tilde{\mathbf {S}}_\text {D}= \begin{pmatrix} \tilde{s}_{\text {D},0}^{(1)} &{}\quad \tilde{s}_{\text {D},0}^{(2)} &{}\quad \dots &{}\quad \tilde{s}_{\text {D},0}^{(L)} \\ \tilde{s}_{\text {D},1}^{(1)[-1]} &{}\quad \tilde{s}_{\text {D},1}^{(2)[-1]} &{}\quad \dots &{}\quad \tilde{s}_{\text {D},1}^{(L)[-1]} \\ \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots \\ \tilde{s}_{\text {D},D+d-2}^{(1)[-(D+d-2)]} &{}\quad \tilde{s}_{\text {D},D+d-2}^{(2)[-(D+d-2)]} &{}\quad \dots &{}\quad \tilde{s}_{\text {D},D+d-2}^{(L)[-(D+d-2)]} \end{pmatrix} \end{aligned}$$

(63)

we can write (59) as

$$\begin{aligned} \tilde{\mathbf {S}}_\text {D}= \underbrace{ \begin{pmatrix} \mathbf {s}_0^{[0]}\phantom {^{-}} &{} &{} &{} \\ \mathbf {s}_1^{[-1]} &{} \quad \mathbf {s}_0^{[-1]} &{} &{} \\ \vdots &{} \quad \mathbf {s}_1^{[-2]} &{} \quad \ddots &{} \\ \mathbf {s}_{d-2}^{[-(d-2)]} &{}\quad \vdots &{}\quad \ddots &{}\quad \mathbf {s}_{0}^{[-D]} \\ &{}\quad \ddots &{}\quad \ddots &{}\quad \mathbf {s}_{1}^{[-(D+1)]\phantom {(-1)}} \\ &{} &{}\quad \ddots &{}\quad \vdots \\ &{} &{} &{}\quad \mathbf {s}_{d-2}^{[-(D+d-2)]} \end{pmatrix} }_{\mathbf {S}} \cdot \underbrace{ \begin{pmatrix} \mathbf {z}_{0}^{(1)} &{}\quad \mathbf {z}_{0}^{(2)} &{}\quad \dots &{}\quad \mathbf {z}_{0}^{(L)} \\ \mathbf {z}_{1}^{(1)[-1]} &{}\quad \mathbf {z}_{1}^{(2)[-1]} &{}\quad \dots &{}\quad \mathbf {z}_{1}^{(L)[-1]} \\ \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots \\ \mathbf {z}_{D}^{(1)[-D]} &{}\quad \mathbf {z}_{D}^{(2)[-D]} &{}\quad \dots &{}\quad \mathbf {z}_{D}^{(L)[-D]} \end{pmatrix}. }_{\mathbf {Z}} \end{aligned}$$

(64)

We now show that the rank of $\mathbf {Z}$ in (64) is full, i.e. ${{\mathrm{rk}}}_{q^m}(\mathbf {Z})=L$. The matrix

$$\begin{aligned} \tilde{\mathbf {Z}}\overset{{{\mathrm{def}}}}{=}\begin{pmatrix} \mathbf {z}_{0}^{(1)} &{}\quad \mathbf {z}_{0}^{(2)} &{}\quad \dots &{}\quad \mathbf {z}_{0}^{(L)} \\ \mathbf {z}_{1}^{(1)} &{}\quad \mathbf {z}_{1}^{(2)} &{}\quad \dots &{}\quad \mathbf {z}_{1}^{(L)} \\ \vdots &{}\quad \vdots &{}\quad \ddots &{}\quad \vdots \\ \mathbf {z}_{D}^{(1)} &{} \mathbf {z}_{D}^{(2)} &{} \dots &{} \mathbf {z}_{D}^{(L)} \end{pmatrix} \end{aligned}$$

(65)

contains the coefficients of $\varGamma ^{(1)},\dots ,\varGamma ^{(L)}$ as columns. Since $\varGamma ^{(1)},\dots ,\varGamma ^{(L)}$ are $\mathbb F_{q^m}$-linearly independent, we have ${{\mathrm{rk}}}_{q^m}(\tilde{\mathbf {Z}})=L$. Thus, using elementary column operations, we can transform $\tilde{\mathbf {Z}}$ into a matrix that contains $L$ unit row vectors. These unit rows are invariant under the row-wise Frobenius automorphisms applied in (64) and thus we have ${{\mathrm{rk}}}_{q^m}(\mathbf {Z})=L$. Hence, the mapping

$$\begin{aligned} \left( s^{(1)}(x) \ s^{(2)}(x) \ \dots \ s^{(L)}(x)\right) \mapsto \left( \tilde{s}^{(1)}_\text {D}(x) \ \tilde{s}^{(1)}_\text {D}(x) \ \dots \ \tilde{s}^{(L)}_\text {D}(x)\right) \end{aligned}$$

is bijective. The Frobenius powers $\overline{\varGamma }_{l}^{(j)}\left( x^{[d-2]}\right) ^{[-(d-2)]}$ in (43) correspond to raising each coefficient of $\overline{\varGamma }_{l}^{(j)}(x)$ to the $[-(d-1)]$ which does not affect the rank of $\mathbf {Z}$ in (64). Since $\varLambda _\text {U}(x)$ is nonzero and common for all modified syndromes it follows that the mapping $\left( s^{(1)}(x) \ s^{(2)}(x) \ \dots \ s^{(L)}(x)\right) \mapsto \left( \tilde{s}^{(1)}_\text {UD}(x) \ \tilde{s}^{(1)}_\text {UD}(x) \ \dots \ \tilde{s}^{(L)}_\text {UD}(x)\right)$ is bijective. $\square$