DNA链的数据驱动蜜蜂识别
Data-Driven Bee Identification for DNA Strands
AI总结：
本文提出一种数据驱动的DNA链识别方法，在二进制擦除信道下几乎确定正确识别，并通过剪枝程序将平均数据比较次数降至M^2的一小部分，具体界为κM^2。
摘要：
我们研究了DNA链蜜蜂识别问题的数据驱动方法。蜜蜂识别问题由Tandon等人（2019）提出，要求通过一组M个噪声测量来识别M只蜜蜂，每只蜜蜂都标记有独特的条形码。后来，Chrisnata等人（2022）将模型扩展到观察到每只蜜蜂的N个噪声测量的情况，并将该模型应用于解决DNA存储系统的无序性。在此类系统中，通常在每个DNA数据块前添加一个唯一地址以形成DNA链，但地址可能被损坏。虽然聚类通常用于识别DNA链的地址，但这需要M^2次数据比较（其中M是读取次数）。相比之下，Chrisnata等人（2022）的方法完全避免了数据比较。在这项工作中，我们研究了这种识别任务的中间数据驱动方法。对于二进制擦除信道，我们首先证明在某些温和假设下，我们可以几乎确定地正确识别所有DNA链。然后，我们提出了一种数据驱动的剪枝程序，并证明平均而言，该程序仅使用M^2次数据比较的一小部分。具体来说，对于M=2^n和擦除概率p，该程序执行的预期数据比较次数为κM^2，其中((1+2p-p^2)/2)^n ≤ κ ≤ ((1+p)/2)^n。
英文摘要：
We study a data-driven approach to the bee identification problem for DNA strands. The bee-identification problem, introduced by Tandon et al. (2019), requires one to identify $M$ bees, each tagged by a unique barcode, via a set of $M$ noisy measurements. Later, Chrisnata et al. (2022) extended the model to case where one observes $N$ noisy measurements of each bee, and applied the model to address the unordered nature of DNA storage systems. In such systems, a unique address is typically prepended to each DNA data block to form a DNA strand, but the address may possibly be corrupted. While clustering is usually used to identify the address of a DNA strand, this requires $\mathcal{M}^2$ data comparisons (when $\mathcal{M}$ is the number of reads). In contrast, the approach of Chrisnata et al. (2022) avoids data comparisons completely. In this work, we study an intermediate, data-driven approach to this identification task. For the binary erasure channel, we first show that we can almost surely correctly identify all DNA strands under certain mild assumptions. Then we propose a data-driven pruning procedure and demonstrate that on average the procedure uses only a fraction of $\mathcal{M}^2$ data comparisons. Specifically, for $\mathcal{M}= 2^n$ and erasure probability $p$, the expected number of data comparisons performed by the procedure is $κ\mathcal{M}^2$, where $\left(\frac{1+2p-p^2}{2}\right)^n \leq κ\leq \left(\frac{1+p}{2}\right)^n $.
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