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4杆汉诺塔的最优移动次数
引用本文:许道云.4杆汉诺塔的最优移动次数[J].贵州大学学报(自然科学版),2012,29(5):49-52,62.
作者姓名:许道云
作者单位:贵州大学计算机科学与信息学院,贵州 贵阳,550025
基金项目:基金项目:国家自然科学基金(No.6126006)
摘    要:通常汉诺塔问题只带三根杆,当圆盘数为n时,最优移动次数为T3(n)=2n-1.对于带4杆的汉诺塔问题,最优移动次数满足关系T4(n)=2T4(m)+T3(n-m),其中m=arglmin{2T4(l)+T3(n-l)}依赖于n.对于正数整k,当k(k-1)/2+1≤n≤k(k+1)/2,n=k(k-1)/2+l时,T4(n)=(l+k-2)2k-1+1.特别,T4(sk)=2T4(sk-1)+T3(k),其中s0=0,sk=sk-1+k(k≥1).

关 键 词:4杆汉诺塔  最优移动方案  移动次数

Optimal Number of Moving Discs in the Tower of Hanoi with Four Pegs
XU Dao-yun.Optimal Number of Moving Discs in the Tower of Hanoi with Four Pegs[J].Journal of Guizhou University(Natural Science),2012,29(5):49-52,62.
Authors:XU Dao-yun
Institution:XU Dao-yun ( College of Computer Science and Information, Guizhou University, Guiyang 550025, China)
Abstract:The optimal number of moving discs is T3 (n) = 2n - 1 in the tower of Hanoi with three pegs and n discs. For the tower of Hanoi with four pegs, the optimal number satisfies the equation T4 (n) = 2 T4 (m) + T3 ( n-m) ,where m = arg, mint2T4(l) + T3(n - l) } depending on n. Given k(k-1)/2+1≤n≤k(k+1)/2 and n=k(k - 1)/2 + l for some integers k and 1, T4(n) = (l + k - 2)2^k-1 + 1. Specially, T4(sk) = 2T4(sk-1) +T3(k) ,whereso = O,sk = sk-1 + k(k ≥ 1)..
Keywords:Tower of Hanoi with four pegs  optimal frame of moving  number of moving
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