Selenite Towers Move Faster Than Hanoï Towers, But Still Require Exponential Time

Jérémy Félix Barbay
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Abstract

The Hanoi Tower problem is a classic exercise in recursive programming: the solution has a simple recursive definition, and its complexity and the matching lower bound correspond to the solution of a simple recursive function (the solution is so simple that most students memorize it and regurgitate it at exams without truly understanding it). We describe how some minor change in the rules of the Hanoi Tower yields various increases of difficulty in the solution, so that to require a deeper mastery of recursion than the classical Hanoi Tower problem. In particular, we analyze the Selenite Tower problem, where just changing the insertion and extraction positions from the top to the middle of the tower results in a surprising increase in the intricacy of the solution: such a tower of n disks can be optimally moved in 3^(n/2) moves for n even (i.e. less than a Hanoi Tower of same height), via 5 recursive functions (or, equivalently, one recursion function with five states following three distinct patterns).
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亚硒酸盐塔比Hanoï塔移动得快,但仍然需要指数时间
河内塔问题是递归编程中的一个经典练习:解决方案有一个简单的递归定义,它的复杂度和匹配的下界对应于一个简单递归函数的解决方案(解决方案非常简单,以至于大多数学生在考试时背它,而没有真正理解它)。我们描述了河内塔规则的一些微小变化如何在解决方案中产生各种难度的增加,因此需要比经典的河内塔问题更深入地掌握递归。特别地,我们分析了Selenite Tower问题,其中只需将插入和提取位置从塔的顶部更改为塔的中间,就会导致解决方案复杂性的惊人增加:这样的n个磁盘的塔可以通过5个递归函数(或者,相当于一个递归函数,具有五个状态,以下三个不同的模式)以3^(n/2)次移动n偶数(即小于相同高度的河内塔)。
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