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An Undecidable Nested Recurrence Relation

机译:无法确定的嵌套递归关系

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摘要

Roughly speaking, a recurrence relation is nested if it contains a subexpression of the form ... A(... A(...)...). Many nested recurrence relations occur in the literature, and determining their behavior seems to be quite difficult and highly dependent on their initial conditions. A nested recurrence relation A(n) is said to be undecidable if the following problem is undecidable: given a finite set of initial conditions for A(n), is the recurrence relation calculable? Here calculable means that for every n ≥ 0, either A(n) is an initial condition or the calculation of A(n) involves only invocations of A on arguments in {0,1,..., n - 1}. We show that the recurrence relation A (n) = A (n - 4 - A (A (n - 4))) + AA (A (n - 4)) + A (2A (n - 4 - A (n - 2)) + A (n - 2)) is undecidable by showing how it can be used, together with carefully chosen initial conditions, to simulate Post 2-tag systems, a known Turing complete problem.
机译:粗略地说,如果递归关系包含形式为... A(... A(...)...)的子表达式,则嵌套该递归关系。文献中出现了许多嵌套的递归关系,确定它们的行为似乎非常困难并且高度依赖于它们的初始条件。如果无法确定以下问题,则称为嵌套递归关系A(n)是不确定的:给定A(n)的一组初始条件有限,递归关系是否可计算?这里可计算的意味着,对于每n≥0,A(n)是一个初始条件,或者A(n)的计算仅涉及对{0,1,...,n-1}中的自变量的A调用。我们证明递归关系A(n)= A(n-4-A(A(n-4)))+ AA(A(n-4))+ A(2A(n-4-A(n- 2))+ A(n-2))无法确定,方法是展示如何将其与精心选择的初始条件一起使用,以模拟Post 2-tag系统,这是一个已知的图灵完全问题。

著录项

  • 来源
    《How the world computes》|2012年|107-117|共11页
  • 会议地点 Cambridge(GB)
  • 作者

    Marcel Celaya; Frank Ruskey;

  • 作者单位

    Department of Computer Science, University of Victoria,Victoria, BC, V8W 3P6, Canada;

    Department of Computer Science, University of Victoria,Victoria, BC, V8W 3P6, Canada;

  • 会议组织
  • 原文格式 PDF
  • 正文语种 eng
  • 中图分类
  • 关键词

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