環狀網路的拓樸性質研究
In any , we prove that there exist cycles which have any length between 3 and 3n and paths which have any length between their smallest distance and longest Hamiltonian paths in any two different nodes; for any two nodes, there exist varied Hamiltonian cycles, making the two nodes locate on any possible counterpart position(only limited by the distance between the two nodes). In , there are 2n internally-disjoint spanning cycles, and 2n-1 internally-disjoint spanning paths. Besides, we also prove has no more than 2n disjoint spanning paths, and calculate its wide diameter.
本報告證明在環狀網路 中,存在有長度3到 3n 的所有迴圈;任何相異兩點都有各種不同的長度的路徑:從最短的距離到最長的漢米頓路徑;取定任意兩點,存在有各種不同的漢米爾頓迴圈,使得兩點位於所有可能的相對位置上(僅被兩點之間的距離限制)。在 中,也具有2n 個彼此不相交、經過所有點的迴圈,以及2n-1 個彼此不相交、經過所有點的路徑。除此之外,也證明了,在兩相異點間,具有個數不超過2n 且互斥的路徑,且這些路徑經過所有點。我們也估算了它的寬直徑。
「世紀難題-考拉茲猜想」 考拉茲猜想中循環的探討
自1930年代以來,考拉茲猜想(Collatz conjecture)一直是個未解之謎,其敘述如下:選定一個自然數,如果是偶數,則用2來除;如果是奇數,則乘以3再加1,經過有限次迭代,最後一定得到1。也就是說會得到1,4,2,1,4,2,…的數列,稱之為1-2-4循環。即使此猜想敘述簡單,卻是個橫跨世紀的難題,至近幾年才有一些證明方法出現。 其中一種證明考拉茲猜想的想法為證明所有不符合考拉茲猜想的狀況為假,而其中一種狀況為除了1-2-4循環還有其他組循環,即有些正整數在經過數次考拉茲猜想的計算後,會進入一組非1-2-4循環的循環。 因此,在此篇報告中我們透過討論每一個奇數在經由乘3再加1的計算後,所得到的偶數的2的冪次,再經由反證法證明除了1-2-4循環不會有其他組循環。
曲率的奧秘
我們研究的主題是曲率,且以高中所學的函數為主。雖然大學已有曲率公式,但我們將其表示成高中生較易了解的型式,並且以f(x) 的方式呈現。我們在函數曲線上取不共線三點,構成一個三角形,並求出此三角形的外接圓半徑。再將所取三點逼近,所求之半徑即為特定點的密切圓,也就是曲率半徑。而此曲率半徑的倒數,就是所求的曲率,同時我們將公式帶入高中各常見函數,以導出函數上各點曲率。;Our study is about curvature, especially about the fuctions we learn in senior high school. In university, there is a certain formula for curvature, but we hope to change it into a form that can be easily accepted by senior high school students, and express the formula with f(x), the symbol of functions. We pick three incollinear points from the curve of a function, making the three points into a triangle, and figure out the circumradius of this triangle. Then, we approximate the three points to one of them, and the circumradius will also be the radius of the osculating circle of the point. We define the radius as radius of curvature. The reciprocal of the radius of curvature will be the curvature. Then, we use the formula to figure out the curvature of the functions we learn in senior high school.
替機器人安排作業程序
編號1~mn 的mn 個物件已隨機置入m× n 階的矩陣中,另外有一行m 個空格的暫存區供物件暫存用。我們探討將這mn 個物件移至目標區並按照1,2,…, mn 的次序排列,所需的移動步數;每一步的移動中,只能移動每一行最頂層的物件到其他行(含暫存區)的最頂層或目標區。在這篇報告中,我們給出了一個適用於n ? m ?1時的移動方法,此方法在一般的情形下,所需的移動次數未必是最少;但是在最不利於移動的情形下,我們證明此方法所需的移動步數為最少。There are mn objects, numbered from 1 to mn, put on an m× n matrix randomly, and there is another column with m blank spaces for temporary storage purpose during moving. In each step of moving, we can only move the top object from one column to the top of another column or to the target pile. The total steps needed to move these mn objects to the target pile in increasing order from the bottom to the top is studied in this article. A general method for solving this problem when n ? m ?1 is given, and we prove that it provides an optimal solution in the worst cases. However, it may not always provide the minimal steps in all cases.