AI 中文总结
该研究引入有界步递归与三参数层级细化Grzegorczyk层级，推导包含判据、坍缩性质等，证明H^m_{1,l}=E^m等结论，揭示H^2_{1,l}与多项式时间类的关系及P=NP相关推论。
AI 中文摘要
我们引入有界步递归和细化Grzegorczyk层级的三参数层级。对于严格递增函数φ:ℕ→ℕ且φ(x)≥x+1，其广义逆ρ_φ(y)=min{z:φ(z)≥y}替代普通前驱，生成下降序列y,ρ_φ(y),ρ_φ^[2](y),…,0。基于Grzegorczyk基B_m、复合及步长为g_n^[l]的有界步递归，我们定义类H^m_{n,l}，其中m衡量初始函数强度，n选择增长尺度，l固定通过其规范层的步幅。对于所有n,n'≥2，我们得到H^a_{n,l}⊆H^b_{n',l'}的精确判据。水平坍缩以下，固定步幅按反向可分性排序：同行包含由l'|l决定，而非l与l'的数值顺序。所有固定步幅从初始基m=n坍缩，且当m=n+1时，公共类恰好等于普通有界递归类E^m。正包含使用精确深度模拟；分离使用直接分段单调迹定理和选定依赖链的规范区域不变量。倍增行g_1(x)=2x+1在低基时特殊。我们证明对所有m≥3，H^m_{1,l}=E^m，在基2处构造首个垂直桥，并证明H^2_{1,l}中的每个固定元函数均为二元多项式时间可计算，且H^2_{1,l}⊂neqFP。等式H^2_{1,l}=E^2将意味着P=NP。
英文摘要
We ask whether asymptotic recursion depth determines the expressive strength of a bounded recursive algebra, and prove that it does not. We replace ordinary predecessor recursion by generalized-inverse descent along a fixed iterate $g_n^{[l]}$ and obtain classes $H^m_{n,l}$, where $m$ measures initial-function strength, $n$ the growth row, and $l$ the traversal stride. For all rows $n,n'\ge2$ we prove an exact inclusion criterion. At a fixed row $n\ge2$ three regimes occur: below the critical basis ($m<n$), equal-row inclusion is exactly reverse divisibility $l'\mid l$; at $m=n$ every stride collapses to one class; and from $m=n+1$ this class is ordinary bounded recursion $E^m$. Hence pairwise $Θ$-equivalent descent depths can induce infinite descending chains, infinite antichains, and copies of every finite partial order. The separation is therefore controlled by traversal alignment rather than by growth rate or recursion depth alone. The proof combines exact-depth simulation, trace sparsity, and selected dependency chains. The exceptional doubling row has the same reverse-divisibility order at basis zero, but all strides collapse from basis one onward; from basis three it equals ordinary bounded recursion, while at basis two $H^2_{1,l}\subsetneq FP$. At basis zero the doubling-row classes are proper subclasses of deterministic functional logspace, so the same dual-divisibility order already occurs inside $FL$.
