In the trace reconstruction problem, one seeks to reconstruct a binary string $s$ from a collection of traces, each of which is obtained by passing $s$ through a deletion channel. It is known that $\exp(\tilde O(n^{1/5}))$ traces suffice to reconstruct any length-$n$ string with high probability. We consider a variant of the trace reconstruction problem where the goal is to recover a "density map" that indicates the locations of each length-$k$ substring throughout $s$. We show that $\epsilon^{-2}\cdot \text{poly}(n)$ traces suffice to recover the density map with error at most $\epsilon$. As a result, when restricted to a set of source strings whose minimum "density map distance" is at least $1/\text{poly}(n)$, the trace reconstruction problem can be solved with polynomially many traces.
翻译:在追踪重建问题中, 一个人试图从一组痕迹中重建一个二进制字符串 $ $, 每一个痕迹都是通过一个删除频道传送 $ 。 已知 $\ exp(\ tilde O (n ⁇ 1/5}) ) 美元 的痕迹足以极有可能重建任何长度- $n 字符串 。 我们考虑一个追踪重建问题的变种, 其目标在于从“ 密度图” 中回收一个“ 密度图 ”, 该图显示每个长度- k$ 子字符串在美元中的位置 。 我们显示, $\ epsilon =2\ { ccdot\ text{poly} (n) 的痕迹足以以最多 $\ epslon$ 来恢复密度地图 。 结果, 当一个最小的“ 密度地图距离” 至少为 1/\ text{poly} (n) 的源字符串被限制时, 跟踪重建问题可以用多的痕迹解决 。