关于广义Frame-Stewart数
On generalized Frame-Stewart numbers
摘要：
对于带$k \geqslant 4$桩的多桩塔Hanoi问题，迄今为止，最优解是用基于以下递推关系的Stewart算法得到的：$\mathrm{S}\_k(n)=\min\_{1 \leqslant t \leqslant n} \left\{2 \cdot \mathrm{S}\_k(n-t) + \mathrm{S}\_{k-1}(t)\right\}$，$\mathrm{S}\_3(n) = 2^n -- 1$ .在本文中，我们推广这种递归关系到$\mathrm{G}\_k(n) = \min\_{1\leqslant t\leqslant n}\left\{ p\_k\cdot \mathrm{G}\_k(n-t) + q\_k\cdot \mathrm{G}\_{k-1}(t) \right\}$，$\mathrm{G}\_3(n) = p\_3\cdot \mathrm{G}\_3(n-1) + q\_3$，为两个序列的任意正整数$\left(p\_i\right)\_{i \geqslant 3}$和$\left(q\_i\right)\_{i \geqslant 3}$，我们表明，序列的差异$\left(\mathrm{G}\_k(n)- \mathrm{G}\_k(n-1)\right)\_{n \geqslant 1}$由数的形式$\left(\prod\_{i=3}^{k}q\_i\right) \cdot \left(\prod\_{i=3}^{k}{p\_i}^{α\_i}\right)$，与$α\_i\geqslant 0$的所有$i$，安排在非递减顺序。我们也应用这个结果来分析几个图的河内塔问题的递归关系。
英文摘要：
For the multi-peg Tower of Hanoi problem with $k \geqslant 4$ pegs, so far the best solution is obtained by the Stewart's algorithm based on the the following recurrence relation: $\mathrm{S}\_k(n)=\min\_{1 \leqslant t \leqslant n} \left\{2 \cdot \mathrm{S}\_k(n-t) + \mathrm{S}\_{k-1}(t)\right\}$, $\mathrm{S}\_3(n) = 2^n -- 1$. In this paper, we generalize this recurrence relation to $\mathrm{G}\_k(n) = \min\_{1\leqslant t\leqslant n}\left\{ p\_k\cdot \mathrm{G}\_k(n-t) + q\_k\cdot \mathrm{G}\_{k-1}(t) \right\}$, $\mathrm{G}\_3(n) = p\_3\cdot \mathrm{G}\_3(n-1) + q\_3$, for two sequences of arbitrary positive integers $\left(p\_i\right)\_{i \geqslant 3}$ and $\left(q\_i\right)\_{i \geqslant 3}$ and we show that the sequence of differences $\left(\mathrm{G}\_k(n)- \mathrm{G}\_k(n-1)\right)\_{n \geqslant 1}$ consists of numbers of the form $\left(\prod\_{i=3}^{k}q\_i\right) \cdot \left(\prod\_{i=3}^{k}{p\_i}^{α\_i}\right)$, with $α\_i\geqslant 0$ for all $i$, arranged in nondecreasing order. We also apply this result to analyze recurrence relations for the Tower of Hanoi problems on several graphs.
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