Hairpin completion, derived from the hairpin formation observed in DNA biochemistry, is an operation applied to strings, particularly useful in DNA computing. Conceptually, a right hairpin completion operation transforms a string $S$ into $S\cdot S'$ where $S'$ is the reverse complement of a prefix of $S$. Similarly, a left hairpin completion operation transforms a string $S$ into $S'\cdot S$ where $S'$ is the reverse complement of a suffix of $S$. The hairpin completion distance from $S$ to $T$ is the minimum number of hairpin completion operations needed to transform $S$ into $T$. Recently Boneh et al. showed an $O(n^2)$ time algorithm for finding the hairpin completion distance between two strings of length at most $n$. In this paper we show that for any $\varepsilon>0$ there is no $O(n^{2-\varepsilon})$-time algorithm for the hairpin completion distance problem unless the Strong Exponential Time Hypothesis (SETH) is false. Thus, under SETH, the time complexity of the hairpin completion distance problem is quadratic, up to sub-polynomial factors.

翻译：发夹完成操作源于DNA生物化学中观察到的发夹结构形成，是应用于字符串的一种操作，尤其在DNA计算中具有重要价值。从概念上讲，右发夹完成操作将字符串$S$转换为$S\cdot S'$，其中$S'$是$S$前缀的反向互补序列；类似地，左发夹完成操作将字符串$S$转换为$S'\cdot S$，其中$S'$是$S$后缀的反向互补序列。从$S$到$T$的发夹完成距离是指将$S$转换为$T$所需的最少发夹完成操作次数。近期Boneh等人提出了一种$O(n^2)$时间算法，用于求解长度不超过$n$的两个字符串之间的发夹完成距离。本文证明：对于任意$\varepsilon>0$，不存在$O(n^{2-\varepsilon})$时间的发夹完成距离问题算法，除非强指数时间假说（SETH）不成立。因此，在SETH假设下，发夹完成距离问题的时间复杂度为二次阶，忽略次多项式因子。