In the beautifully simple-to-state problem of trace reconstruction, the goal is to reconstruct an unknown binary string $x$ given random "traces" of $x$ where each trace is generated by deleting each coordinate of $x$ independently with probability $p<1$. The problem is well studied both when the unknown string is arbitrary and when it is chosen uniformly at random. For both settings, there is still an exponential gap between upper and lower sample complexity bounds and our understanding of the problem is still surprisingly limited. In this paper, we consider natural parameterizations and generalizations of this problem in an effort to attain a deeper and more comprehensive understanding. We prove that $\exp(O(n^{1/4} \sqrt{\log n}))$ traces suffice for reconstructing arbitrary matrices. In the matrix version of the problem, each row and column of an unknown $\sqrt{n}\times \sqrt{n}$ matrix is deleted independently with probability $p$. Our results contrasts with the best known results for sequence reconstruction where the best known upper bound is $\exp(O(n^{1/3}))$. An optimal result for random matrix reconstruction: we show that $\Theta(\log n)$ traces are necessary and sufficient. This is in contrast to the problem for random sequences where there is a super-logarithmic lower bound and the best known upper bound is $\exp({O}(\log^{1/3} n))$. We show that $\exp(O(k^{1/3}\log^{2/3} n))$ traces suffice to reconstruct $k$-sparse strings, providing an improvement over the best known sequence reconstruction results when $k = o(n/\log^2 n)$. We show that $\textrm{poly}(n)$ traces suffice if $x$ is $k$-sparse and we additionally have a "separation" promise, specifically that the indices of 1's in $x$ all differ by $\Omega(k \log n)$.
翻译:在漂亮的简单到状态的追踪重建问题中,目标是重建一个未知的二进制字符串 $x$, 给它随机的“ 美元” $x$, 每一个痕迹都是通过删除每个坐标产生, 概率为 $p < 1 美元。 当未知字符串是任意的, 当它被任意选择时, 问题就得到了很好的研究。 对于这两个设置, 上下抽样复杂性的界限之间仍然存在着指数差距, 而我们对问题的理解仍然令人惊讶地有限 。 在本文中, 我们考虑这个问题的自然参数化和概括化, 以达成更深和更全面的理解。 我们证明$\ 美元( 1/ 美元 美元 美元 美元 美元 美元 美元 美元 。 在问题的矩阵版本中, 每行和列一个未知的 $qrqrtrequemation 之间, 当我们已知的上层重建结果是 $ $美元 ( =\\\\\ 3 3美元 美元 美元 美元 美元 ), 当我们知道的是, 我们最已知的上层的比结果要显示最低的为 美元 美元 美元。