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arXiv 2609.29585cs.CCcs.LO

步进递归:混合步长谱、路径分解与同步几何

Step Recursion: Mixed Stride Spectra, Path Factorization, and Synchronization Geometry

Kirill Osipov

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中文总结 AI 辅助

本文证明混合步长递归不能产生新的规范步长,其可定义步长恰为原始步长生成的幺半群,并通过路径分解和同步几何刻画了计算过程与信息保留的边界。

中文摘要 AI 辅助

步进递归允许递归计算以跳跃或步长的形式在规范层级中移动。假设一个函数代数允许使用若干原始步长长度 $L$。复合操作立即产生这些长度的和,但尚不清楚任意嵌套的混合递归是否能创造任何真正新的规范步长。我们证明这是不可能的:在每一个固定的规范行 $n\ge2$ 和较低基 $m<n$ 上,由 $L$ 可定义的规范步长恰好是 $L$ 生成的加法幺半群 $\langle L\rangle$。证明提供的信息比成员资格本身更多。每个足够高的依赖路径计算步长 $p$ 的规范下降,其总标签权重恰好为 $p$,可能的路径签名正是 $p$ 由原始步长标签的加法分解。因此,计算既记住了哪些步长是可定义的,也记住了每个步长如何被组装。在饱和之后,混合步长类之间的包含关系恰好是其加法步长幺半群之间的包含关系,从而给出了具体的格描述和包含关系的有限证书。然后我们研究了几个同步递归时钟。联合下降映射可以编码 $\mathbb N^r$ 的任意加法子幺半群,对于 $r=2$ 已经给出连续序复杂度。然而,普通的标量观测会忘记时钟之间的相关性,仅保留独立的坐标步长。这精确地指出了同步信息在何处被保留,在何处被丢失。

英文摘要

Step recursion allows recursive computation to move through a canonical hierarchy in jumps, or strides. Suppose a function algebra is allowed to use several primitive stride lengths $L$. Composition immediately produces sums of these lengths, but it is not clear whether arbitrary nesting of mixed recursions can create any genuinely new canonical stride. We prove that it cannot: at every fixed canonical row $n\ge2$ and lower basis $m<n$, the canonical strides definable from $L$ are exactly the additive monoid $\langle L\rangle$ generated by $L$. The proof gives more information than membership alone. Every sufficiently high dependency path computing a canonical descent of stride $p$ carries total label weight exactly $p$, and the possible path signatures are precisely the additive factorizations of $p$ by the primitive stride labels. Thus the computation remembers both which strides are definable and how each one can be assembled. After saturation, inclusion between mixed-stride classes is exactly inclusion between their additive stride monoids, giving a concrete lattice description and finite certificates for inclusion. We then study several synchronized recursion clocks. Joint descent maps can encode arbitrary additive submonoids of $\mathbb N^r$, already giving continuum order complexity for $r=2$. Ordinary scalar observation, however, forgets correlations between the clocks and retains only the independent coordinate strides. This identifies precisely where synchronization information is preserved and where it collapses.

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