A Linear-Complexity Channel-Independent Code Construction Method for List Sphere Polar Decoder

Abstract

Besides the main decoding methods for polar codes such as successive cancellation (SC) decoding, SC list decoding (SCL), and belief propagation (BP) decoding, list sphere decoding (LSD) is an alternative for short codes regarding its lower complexity. Though the improved LSD (ILSD) can achieve better performance than LSD by a synchronous determination scheme, its performance is still not satisfactory with an inappropriate code construction. Good polar code construction is to pick the K most reliable bits from N bits. However, existing polar construction methods are designed for SC or BP decoders. These construction methods might not be suitable for ILSD, which searches based on the Euclidean distance. In this paper, a linear-complexity channel-independent polar code construction method for ILSD is proposed to improve the performance. Considering the Hamming distance as the reliability for ILSD, the proposed method arranges more frozen bits in the early synchro sets and improves the evaluation of the paths at each decoding level. With linear complexity, this method is channel-independent, which is efficient for offline constructions. Numerical results show the proposed construction with low code rates outperforms the state-of-the-art by 2 dB with ILSD.

Appendices

Appendix A: Proof of Lemma 1

Proof by mathematical induction:

• (i) Consider $\mathbf {G}_{2^{1}}= \mathbf {G}_{2} = \left [\begin {array}{ccc} 1 & 0 \\ 1 & 1 \end {array}\right ]$, g_(2,1) = 1 and H₂ > H₁. Then Lemma 1 is true for $n=1$.

• (ii) Assume Lemma 1 is true for n. Consider $\mathbf {G}_{2^{n+1}}=\left [\begin {array}{ccc} \mathbf {G}_{2^{n}} & \mathbf {0} \\ \mathbf {G}_{2^{n}} & \mathbf {G}_{2^{n}} \end {array}\right ]$, for any i, j satisfying $1 \le j<i \le 2^{n+1}$ and $g_{(i,j)}=1$, there are 3 cases:

  • 1) $0<j<i \le 2^{n}$, then $H_{i}$ and $H_{j}$ are the same as that in the $\mathbf {G}_{2^{n}}$, so $H_{i}>H_{j}$;

  • 2) $0<j\le 2^{n}<i \le 2^{n+1}$, then $H_{i}=2H_{i-2^{n}}$ and $g_{(i-2^{n},j)}=1$. If $i-2^{n} = j$, then $H_{i}=2H_{j}$; If $i-2^{n} > j$, then $H_{i}=2H_{i-2^{n}}>2H_{j}$. So $H_{i}>H_{j}$;

  • 3) $2^{n}<j<i\le 2^{n+1}$, then $H_{i}=2H_{i-2^{n}}$, $H_{j}=2H_{j-2^{n}}$ and $g_{(i-2^{n},j-2^{n})}=1$. Because $i-2^{n}>j-2^{n}$, then $H_{i-2^{n}}>H_{j-2^{n}}$, so $H_{i}>H_{j}$.

Lemma 1 is true for n implies that Lemma 1 is true for $n+1$. Then Lemma 1 is true for all positive integers n.

Appendix B: Proof of Lemma 2

Proof by mathematical induction:

• (i) Consider $\mathbf {G}_{2^{1}}= \mathbf {G}_{2} = \left [\begin {array}{ccc} 1 & 0 \\ 1 & 1 \end {array}\right ]$. Lemma 2 is clearly true for $n=1$.

• (ii) Assume Lemma 2 is true for n. Consider $\mathbf {G}_{2^{n+1}}=\left [\begin {array}{ccc} \mathbf {G}_{2^{n}} & \mathbf {0} \\ \mathbf {G}_{2^{n}} & \mathbf {G}_{2^{n}} \end {array}\right ]$, for any i, j satisfying $1 \le i<j \le 2^{n+1}$ and $g_{(i+1,i)}=g_{(i+2,i)}=...=g_{(j,i)}=0$, there are 3 cases:

  • 1) $0<i<j \le 2^{n}$, then $H_{i}$ and $H_{j}$ are the same as that in the $\mathbf {G}_{2^{n}}$, so $H_{i} \ge H_{j}$;

  • 2) $0<i\le 2^{n}<j\le 2^{n+1}$, notice that $g_{(2^{n},1)}=g_{(2^{n},2)}=...=g_{(2^{n},2^{n})}=1$, to satisfy $g_{(i+1,i)}=g_{(i+2,i)}=...=g_{(j,i)}=0$, then i must be $2^{n}$. Notice $H_{2^{n}}=2^{n}$ is the largest value in {$H_{1}$, $H_{2}$, ...,$H_{2^{n+1}-1}$}, $g_{(2^{n+1},2^{n})}=1$, so $H_{i} \ge H_{j}$;

  • 3) $2^{n}<i<j\le 2^{n+1}$, then $H_{i}=2H_{i-2^{n}}$, $H_{j}=2H_{j-2^{n}}$ and $g_{(i-2^{n}+1,i-2^{n})}=g_{(i-2^{n}+2,i-2^{n})}=...=g_{(j-2^{n},i-2^{n})}=0$. Lemma 2 is true for n, so $H_{i-2^{n}}>= H_{j-2^{n}}$, $H_{i} \ge H_{j}$.

Lemma 2 is true for n implies that Lemma 2 is true for $n+1$. Then Lemma 2 is true for all positive integers n.