Variable-Length Stop-Feedback Codes With Finite Optimal Decoding Times for BI-AWGN Channels

TL;DR

This paper investigates variable-length stop-feedback codes with finite decoding times for BI-AWGN channels, developing approximation methods and optimization algorithms to improve decoding efficiency and bounds.

Contribution

It introduces a two-step minimization approach and a gap-constrained SDO procedure to optimize decoding times, advancing the design of VLSF codes.

Findings

Finite decoding times can achieve Polyanskiy's bounds in certain error regimes.

The greedy algorithm provides suboptimal decoding times efficiently.

The gap-constrained SDO finds optimal real-valued decoding times.

Abstract

In this paper, we are interested in the performance of a variable-length stop-feedback (VLSF) code with $m$ optimal decoding times for the binary-input additive white Gaussian noise channel. We first develop tight approximations on the tail probability of length-$n$ cumulative information density. Building on the work of Yavas \emph{et al.}, for a given information density threshold, we formulate the integer program of minimizing the upper bound on average blocklength over all decoding times subject to the average error probability, minimum gap and integer constraints. Eventually, minimization of locally minimum upper bounds over all thresholds will yield the globally minimum upper bound and this is called the two-step minimization. For the integer program, we present a greedy algorithm that yields possibly suboptimal integer decoding times. By allowing a positive real-valued decoding time, we develop the gap-constrained sequential differential optimization (SDO) procedure that sequentially produces the optimal, real-valued decoding times. We identify the error regime in which Polyanskiy's scheme of stopping at zero does not improve the achievability bound. In this error regime, the two-step minimization with the gap-constrained SDO shows that a finite $m$ suffices to attain Polyanskiy's bound for VLSF codes with $m = \infty$.