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An Undecidable Nested Recurrence Relation

机译:一个不可思议的嵌套复发关系

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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))) + 4A (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,如果它包含表单的子表达式,则嵌套复发关系...... a(... a(...)......)。许多嵌套复发关系发生在文献中,并确定其行为似乎非常困难,高度依赖于他们的初始条件。如果未定定的问题,据说嵌套复发关系A(n)是不可识别的:给定A(n)的有限初始条件,是复发关系可计算的吗?这里可计算意味着对于每个n> 0,a(n)是初始条件,或者(n)的计算涉及仅在{0,1,...,n - 1}中的对参数的调用。我们表明复发关系a(n)= a(n - 4 - a(a(n - 4)))+ 4a(a(n - 4))+ a(2a(n - 4 - a(n - 2))+ a(n - 2))。通过展示如何使用它以及精心挑选的初始条件来模拟柱2标签系统,这是不可透明的,这是一个已知的图灵完整问题。

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