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Computational complexity of ecological and evolutionary spatial dynamics

机译:生态和进化空间动力学的计算复杂性

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摘要

There are deep, yet largely unexplored, connections between computer science and biology. Both disciplines examine how information proliferates in time and space. Central results in computer science describe the complexity of algorithms that solve certain classes of problems. An algorithm is deemed efficient if it can solve a problem in polynomial time, which means the running time of the algorithm is a polynomial function of the length of the input. There are classes of harder problems for which the fastest possible algorithm requires exponential time. Another criterion is the space requirement of the algorithm. There is a crucial distinction between algorithms that can find a solution, verify a solution, or list several distinct solutions in given time and space. The complexity hierarchy that is generated in this way is the foundation of theoretical computer science. Precise complexity results can be notoriously difficult. The famous question whether polynomial time equals nondeterministic polynomial time (i.e., P = NP) is one of the hardest open problems in computer science and all of mathematics. Here, we consider simple processes of ecological and evolutionary spatial dynamics. The basic question is: What is the probability that a new invader (or a new mutant) will take over a resident population? We derive precise complexity results for a variety of scenarios. We therefore show that some fundamental questions in this area cannot be answered by simple equations (assuming that P is not equal to NP).
机译:计算机科学与生物学之间存在着深远的联系,但很大程度上尚待探索。这两个学科都研究信息如何在时间和空间中扩散。计算机科学的主要结果描述了解决某些类型问题的算法的复杂性。如果算法可以解决多项式时间中的问题,则认为该算法有效,这意味着算法的运行时间是输入长度的多项式函数。对于某些较难的问题,最快的算法可能需要指数时间。另一个标准是算法的空间要求。可以在指定的时间和空间中找到解决方案,验证解决方案或列出几个不同解决方案的算法之间存在关键区别。以这种方式生成的复杂性层次结构是理论计算机科学的基础。精确的复杂度结果可能非常困难。多项式时间是否等于不确定性多项式时间(即P = NP)的著名问题是计算机科学和所有数学中最难解决的开放问题之一。在这里,我们考虑生态和演化空间动力学的简单过程。基本问题是:新的入侵者(或新的突变体)接管常住人口的概率是多少?我们针对各种情况得出精确的复杂度结果。因此,我们证明了该领域的一些基本问题不能通过简单的方程式来回答(假设P不等于NP)。

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