Improved decoding and error floor analysis of staircase codes

Abstract

Staircase codes play an important role as error-correcting codes in optical communications. In this paper, a low-complexity method for resolving stall patterns when decoding staircase codes is described. Stall patterns are the dominating contributor to the error floor in the original decoding method. Our improvement is based on locating stall patterns by intersecting non-zero syndromes and flipping the corresponding bits. The approach effectively lowers the error floor and allows for a new range of block sizes to be considered for optical communications at a certain code rate or, alternatively, a significantly decreased error floor for the same block size. Further, an improved error floor analysis is introduced which provides a more accurate estimation of the contributions to the error floor.

Appendix

Lemma 2

Denote by $r_1,\ldots ,r_K$ the weight of the rows and by $s_1,\ldots ,s_L$ the weight of the columns of a matrix $\mathbf {A} \in \mathbb {F}_{2}^{K\times L}$. The cardinality of the set

$$\begin{aligned} \mathscr {Z}_{K,L}^{\varepsilon } = \left\{ \mathbf {A} | {{\mathrm{wt_H}}}(\mathbf {A}) = \varepsilon , r_i \ge t+1 \; \forall \; i \in [1,K], s_i \ge t+1 \; \forall \; i \in [1,L] \right\} \end{aligned}$$

(14)

is given by

$$\begin{aligned} |\mathscr {Z}_{K,L}^\varepsilon | = D(\varepsilon ,K,L,0) \end{aligned}$$

(15)

with

$$\begin{aligned} D(a,b,c,d) = \sum _{w_{b+d}=\max \left\{ t+1,a-c(b-1)\right\} }^{\min \left\{ a-(b-1)(t+1),c\right\} } D(a-w_{b+d},b-1,c,d) \end{aligned}$$

(16)

and

$$\begin{aligned} D(w_{d+1},1,c,0)&= D(\varepsilon ,L,K,K), \end{aligned}$$

(17)

$$\begin{aligned} D(w_{d+1},1,c,K)&= \mathscr {A}([w_1,\ldots ,w_K],[w_{K+1},\ldots ,w_{K+L}]), \end{aligned}$$

(18)

where $\mathscr {A}(\cdot ,\cdot )$ is given in [12].

Proof

The inputs of D(a, b, c, d) give the number of non-zero positions a that are to be distributed within the rows or columns (determined by an index offset d) of a matrix of size $b\times c$, such that the conditions of (14) are fulfilled. Each recursion step of (15) determines the number of non-zero positions in one row, i.e., its Hamming weight. Trivially, the weight $r_i$ of a row of an $b \times c$ matrix is upper bounded by $r_i \le c$. It follows that $r_b \ge a-c(b-1)$, because otherwise the resulting matrix cannot be of weight a, even if all remaining rows are of maximal weight. By definition $r_b \ge t+1$ and as this condition also has to hold for the other $b-1$ rows, the Hamming weight of the b-th row can be at most $a-(b-1)(t+1)$. Combining these conditions gives the limits of the sum in (16). The problem is reduced to distributing the remaining $a-r_b$ non-zero positions among the rows of a $b-1 \times c$ matrix, where the sum limits guarantee that the conditions on the weight of the rows can be met. When the case (17) is reached, the remaining non-zero positions have to be in the last row and its weight $r_1$ is therefore given by the first function input. The recursion giving the weights $w_{K+1},\ldots , w_{K+L}$ of the L columns is initiated, where $d=K$ gives the offset in the indices of the weights. The same arguments as above hold for the limits of the sum. When the last column is reached , i.e., the case (18), the function $\mathscr {A}([w_1,\ldots ,w_K],[w_{K+1},\ldots ,w_{K+L}])$ from [12] (see also [10]) gives the number of matrices corresponding to the unique weight vector $\mathbf {w}$ of size $K+L$. $\square$

Note that while this notation offers a mathematically correct way to calculate the cardinality, when implemented it would be beneficial to call the function $\mathscr {A}(\mathbf {r}_{},\mathbf {b}^{})$ as few times as possible, as it is the computationally most complex part. As permutations of the input vectors lead to the same result, many calls of this function can be avoided by, e.g., maintaining a look-up table.

Lemma 3

Denote by $r_1,\ldots ,r_K$ the weight of the rows of a matrix $\mathbf {A} \in \mathbb {F}_{2}^{K\times L}$. The cardinality of the set

$$\begin{aligned} \mathscr {Z}_{K,L}^{\varepsilon } = \left\{ \mathbf {A}_{K \times L} | {{\mathrm{wt_H}}}(\mathbf {A}) = \varepsilon , r_i \ge t+1 \;\; \forall \; i \in [1,K] \right\} \end{aligned}$$

is given by

$$\begin{aligned} | \mathscr {Z}_{K,L}^\varepsilon | = F(\varepsilon ,K) \end{aligned}$$

(19)

with

and $F(a,1) = \left( {\begin{array}{c}L\\ a\end{array}}\right)$.

Proof

There are $\left( {\begin{array}{c}L\\ h\end{array}}\right)$ unique binary vectors $\mathbf {r}_{}$ of length L with ${{\mathrm{wt_H}}}(\mathbf {r}_{}) = h$. Summing over all $t+1 \le h \le L$ gives the number of unique binary vectors $\mathbf {r}_{}$ of length L, such that $t+1 \le {{\mathrm{wt_H}}}(\mathbf {r}_{}) \le L$, as

(20)

Let there be K vectors of weight

$$\begin{aligned} t+1 \le r_i&\le L \; \forall \; i \in [1,K] \end{aligned}$$

(21)

and let $\sum _{i=1}^K r_i = \varepsilon$. For every distribution with $r_1 = \delta$, it holds that $\sum _{i=2}^K = \varepsilon -\delta$, and by (21) we obtain

$$\begin{aligned} \max \{t+1,\varepsilon -L(K-1)\} \le \delta \le \min \{\varepsilon -(K-1)(t+1),L\} \end{aligned}$$

(22)

as the region for $\delta$ for which the constraints on the other rows are still satisfiable.

By setting the weight $r_1 = \delta$, the problem is reduced to finding the number of unique distributions of weight $\sum _{i=2}^K = \varepsilon -\delta$ over $K-1$ rows, such that (21) holds. This results in the recursive equation

and $F(a,1) = \left( {\begin{array}{c}L\\ a\end{array}}\right)$. The partial term $\left( {\begin{array}{c}L\\ j\end{array}}\right)$ is given in (20), the bounds of the sum in (22). The partial term $F(a-j,b-1)$ results in the number of unique distributions of the reduced problem. When $b=1$ the last row is reached, which has to take the remaining difference between $\varepsilon$ and the combined weight of the previous rows. The bounds on the sums assure, that the constraints for the last row are always met. $\square$