數學

這個題目是源自2003年的TRML思考賽的題目，原題目並不難，它只有用到簡單的排列方法，主要是討論 an 、bn 兩種數字的排列，其中 an 為滿足下列所有條件之N位數A的個數。 I. A中每一個數字為1或2 II. A中至少有相鄰的兩數字是1 而 bn 表示滿足下列所有條件的N位數B的個數 I. B中每一個數字為0或1 II. B中至少有相鄰的兩數字是1 以及探討an 、bn 與費氏數列cn之關係，其中 cn = cn-1 + cn-2 ，n≧3 ,c1=1, c2=2 。 其中 an 如果改成考慮為一數列，其值不變；而 bn 如果改為數列，那麼就不需要考慮0不能為首位數字的情況。如此，讓人聯想到一個用生成函數解的題目「一個N項數列，其中每一項只能是0或1或2，其中0和2永不能相鄰，求這個數列個數的一般式。」，因此，我們嘗試將這個題目改變它的要求繼續做下去，發現其中有某些規則，例如：不只是原來的11相鄰，甚至是排列其它種方式，都可能從其遞迴式看出它排列的意義，甚至這種排列數是可以用遞迴式求出來的。這提供了我們另一種求數字排列的方法，也是我們覺得有趣的地方。 在過程中我們初步得到以下結論： This solution is according to power contest of 2003 TRML. It is composed of two number arrangements, an , bn . First, suppose an is the total number conforming to the following rules. I. Each number is 1 or 2 in A. II. There is a couple of (11) in A at least. Then, suppose bn is the total number conforming to the following conditions. I. Each number is 0 or 1 in B. II. There is a couple of (11) in B at least. Furthermore , we give the thought to the relation among an ， bn ，and cn (Fibonacci Sequence). By the way, if an is changed to a sequence, and the result is the same. But if bn is to arrange number, we have to give thought to the fact that the first number can’t be zero. If it is a sequence, we don’t have to consider it. The problem belongs to combinatorics. After we do this problem, we find not only original question but also other permutation can be understood by its formula. The problem provides us with other means to solve permutation and combination question. Then, we get the conclusion as follows：

本研究是[對於正n 邊形A1A2…An邊上一點P(含頂點)，想像自定點P 朝鄰邊發出一條光線，若依逆(順)時針方向依序與每邊皆碰撞一次，經一圈而可回到P 點，則此路徑稱為「光圈」。過程試著追蹤在正n 邊形內能形成光圈的光線行進路徑及其相關問題。 本研究令，且以逆時針得光圈來討論： 1.根據[光的反射原理]，探討光圈之存在性，發現除定點P 在正2m 邊形或正三角形的頂點外，其餘皆有光圈。 2.將可形成光圈的路徑圖展開成[直線路徑圖]來探討。 3.由[直線路徑圖]，觀察到形成光圈的光線行進路徑，可能存在下列情況： (1)不通過正n 邊形的頂點，且產生路徑循環與不循環問題。 (2)通過正n 邊形的頂點。 4.發現正2m 邊形光圈皆為[完美光圈]。 5.發現正2m+1 邊形光圈之路徑與有理數、無理數之特質有關。即當s 值為有理數時，路徑會循環；當s 值為無理數時，路徑不循環。 The research is about [on Point P (including the angles) on the side of regular polygons A1、A2…An , imagine the light goes from Point P to the closest side, then bumps each side sequentially counterclockwise. After going a circle, it’s back to Point P. The track is called “the circle of light.” I try to trace the light track of the circle of light and other correlative questions.] In this research, we suppose,and we discuss the circle of light according counterclockwise direction:1.According to the light reflective principles, we discuss whether the circle of light exists or not. And then we discover that the circle of light really exists except when Point P is on the angles of regular triangle or regular 2m polygons. 2.Spread out the circle of light’s track to [rectilinear track.] 3.By [the picture of rectilinear track], observing there are two kinds of the circle of light’s track: (1)If the light doesn’t go through the angles of regular polygons, it can be a circulative track or a non-circulative track. (2)When the light goes through the angles, it stops. 4.We discover that all the circles of light in regular 2m polygons are [the perfect circles of light.] 5.We discover the circle of light’s track is correlative with rational numbers and irrantional numbers. When s is a rational number, the track is circulative, if s is a irrantional number, the track is not circulative.

思考公車車門開關時在地面上掃過的區域形狀與面積時，發現其中變動的直線為過定點與座標軸圍成三角形面積最小的直線，我們很好奇，這類圍成最小面積的直線更進一步的包絡現象，所以就動手去嘗試做研究。\r \r The research is done out of the curiosity we had when we pondered over the area and the shape a bus door sweeps on the floor when it opens or closes, and thus to discover the graph of the varying line which forms the smallest area so that we attempt to do research on such envelopment phenomena.

現在的市面上，四處充斥著各式各樣的電玩或是３Ｄ立體動畫，但是在呈現動畫的\r 時候，依然存在著很多地方的不足，與狀況表現上的矛盾情形！於是令我興起：一個普\r 通的高中學生，是否也有機會運用所學的知識，創造出自己的虛擬實境？\r 我嘗試地寫出各種物體架構的函數，再對這函數圖形所呈現出的立體加工處理，使\r 它能自由的運動，甚至使它能多采多姿就有如我們在現實生活中所看見的一般，有著自\r 己的花紋與圖案！\r 期望在架構完整之後，有一天，我們這些學生可以不再被那些軟體公司牽著走，花\r 大把的錢買電腦動畫中種種的不合理，而可以自己擁有自己的3D 世界！！

我們試著尋找所需最小的城堡個數以看守整個a × b × c (a,b,c ? N) 的長方體。所謂城堡是一種棋子，當放置城堡的位置是(x, y, z) ，則(x, y,t)、(x,t, z)、(t, y, z) (t 是任何不超出邊界的正整數)是這個城堡可以看守的格子。我們用這些城堡來完全看守長方體，試著找出其最小值。在2005 年我們猜測了a = b = c 、a = b c 、a > b > c 的上界，而在2006 年時完成了a = b = c 、a = b c 的大部分情況的證明，少數不能解決的部份也提供了不錯的上界。目前我們在a = b = c 、a = b c 的情況幾乎完全解決，目前正在向a > b > c 的部份發展。A generalized searching method of finding the minimum number of castle which can oversee all over the rectangular box, defined as a × b× c (a,b,c ? N) , is presented. The castle here is defined as one kind of chess. The castle positioned as (x, y, z) can direct the lattice points of (x, y,t) 、(x,t, z) 、(t, y, z) (t is the positive integer and smaller than the box size). These castles we use here is to oversee the rectangular box and to help us to find the minimum number. In 2005, we got the upper bound of overseeing the rectangular box in the conditions of a = b = c、a = b c、a > b > c , while in 2006 we complete the proofs of the minimum number of castles based on the conditions of a = b = c 、a = b c . The further work we want to attain is to complete the case of a > b > c.

Quantum Rings are defined to be polygons with sides all of the same unit length that are connected with a fixed positive or negative angle. In the research, the number of Quantum Rings corresponding to a given number of sides and a fixed angle will be discussed. Quantum Rings could be expressed by many sequences which would involve the theory of partitions and ways to eliminate the many to one nature of the sequences in order to evaluate the upper and lower bound. Besides estimating the upper and lower bound, a lot of the qualities of the Quantum Rings under certain circumstances will be mentioned.「能量環」為許多單位長度的線段以定角首尾相接，並且最後接回原點的多邊形。本研究將要探討對於給定邊長個數與相接角度的「能量環」的個數。「能量環」可以被表示成許多種數列的形式。在數列的運算中會牽涉到許多數字分割的理論與排列組合的排除重複以求得能量環個數的上下界。除了定量的求算出上下界以外，報告中也定性的歸納出許多給予特殊條件的能量環的性質。

在這個科展中我們要研究兩個非常有趣的問題：\r 停車場問題 與 彈硬幣遊戲.\r 停車場問題是這樣的：在一條單行道上有n個車位，編號從1到n。現在有n個司機排成一排要進入停車。但是每個司機都有怪癖，各自有最想要停的位子。他們依序將車子開進單行道，如果想要停的位子是空的，當然停在這個位子。但是如果不巧那個位子已經被停了，不得已只好找下一個空位，姑且停之。但是如果往下找都沒有空位，由於是單行道，司機就只好開走不停了。\r 比如說，如果現在有五輛車，司機的喜好分別是(3,1,2,5,2)。則五輛車都可以順利停車。但是司機的喜好如果是(3,1,4,5,4)，有些車就無法停車了。\r 彈硬幣遊戲是這樣的：考慮圓內接正n+1邊形，任意兩點都連線。這正n+1邊形中有一個頂點P是特殊的，每個頂點上一開始都放有一些硬幣(各點硬幣數可以不同)。如果P以外的某個頂點上的硬幣數n個，我們可以對這個頂點進行操作：一次操作是指將這個頂點上的硬幣各分一個給每個其他頂點。點P只在其他點都無法操作時操作。我們不理會頂點P上的錢數，因此這個遊戲可以無限地玩下去。

本研究由下述問題開始：將n1 個黑色棋子和n2 個白色棋子排成一列，規定第一個棋子必為黑棋；對於每一種排列方法中，同色棋相鄰處記為1，異色棋相鄰處記為-1，所有1 和-1 的總和記為 t (n1,n2 )。對所有可能的排列方法所算出來的t( n1,n2 ) 值求其平均值，記為a (n1,n2 ) 。我們先由觀察各種n1 和n2 值，得到這平均值的可能公式，隨後並嚴格證明其正確性，證明方法也經過多次精鍊到十分簡潔的方式。以此為基礎，我們並做了各方向的推廣，研究涉及下列各點：(一) 利用組合數探討原來的問題。(二) 在第一個棋子不限定為黑棋的假設下，求平均值a( n1,n2 ) 。(三) 將棋子由兩種增加到多種。(四) 改變棋子排列以及相鄰的方式。經由研究，我們發現，每一次愈將問題推廣時，愈能找出清晰的概念涵蓋並印證先前的想法。Our study starts with the following problem. Suppose n1 black chesses and n2 white chesses are arranged in a line under the condition that the first chess is black. For any arrangement of these chesses, an adjacent pair of chesses having the same (respectively, different) colors is associated with a value of 1 (respectively, -1). Let t(n1,n2 ) denote the sum of these values. The purpose of this problem is to calculate the average value a (n1,n2 ) of these t (n1,n2 )which runs over all possible arrangements of the chesses described above. We begin from observing various values of n1 and n2 and find a possible formula for the solution. We then give a rigorous proof for the formula. After some refinements, simple proofs are also established. Based on this, we also make some generalizations. In summary, the research includes the following: 1. Study the problem by using binomial coefficients. 2. Calculate a(n1,n2 ) when t( n1,n2 ) runs over all possible arrangements in which the first chess can be black or white. 3. Increase the types of chesses from two to many. 4. Variant the arrangement method of the chesses from a line to other configurations. During the study, we find that whenever we extend the problem to a more general case, we make the ideas for the original problem clearer.

Generalized Petersen graph P(n,k)，定義為n 為不小於2 的整數以及1≤ k ≤ n−1，有頂點{ u0, u1, . . . , un−1, v0 , v1 , . . . , vn−1 }，及路徑{ uiui+1 , uivi , vivi+k：1≤ i ≤ n−1 }。在 [2] 中，我們可以知道P(n,5) 是Hamiltonian 等價於當n≠11。 在這一篇報告中，我們證明當generalized Petersen graph P(n,5) 是hyper Hamiltonian（一種Hamiltonian graph 再去掉任何一點後，仍然是Hamiltonian graph）的充要條件是n 為不等於11 的奇數且n ≥ 7。 The generalized Petersen graph P(n,k), n ≥ 2 and 1≤ k ≤ n−1, has vertex-set { u0, u1, . . . , un−1, v0 , v1 , . . . , vn−1 } and edge-set { uiui+1 , uivi , vivi+k：1≤ i ≤ n−1 with subscripts reduced modulo n}. And we can know that P(n,5) is Hamiltonian if and only if n≠11 from [2].In this paper it is proved that generalized Petersen graph P(n,5) is Hyper Hamiltonian (A Hamiltonian graph can still be a Hamiltonian graph when any one of the nodes fault) if and only if n is odd and n≠11.