數學

單淘汰賽是一種失敗一次即遭淘汰的賽制；在此假定每位選手都有一相對應的能力數值，本文主要探討在均高的單淘汰賽程表之下，若賽程安排完全依照種子安排原則(亦即最強的選手對最弱的選手、次強隊次弱….)，則對於能力越強的選手越有保障，直觀上而言能力最強的選手應有最大的奪冠機率，探討此種賽程安排是否滿足能力較強的選手有較大的勝率？因發現在某些特殊的選手能力數值分佈之下會發生次強選手勝率大於最強選手的情況，令A、B代表最強與次強選手，P(A)、P(B)代表A、B奪冠的機率，故擬定P(B)/ P(A)為參考依據，尋求P(B)/ P(A)的最大值發生處作為最極端的狀況。發現四位選手的情況下，P(B) / P(A)最大值 = 1；八位選手的情況下，P(B) / P(A)最大值=(196+98) / 343=1.0938，當選手數為2n時，P(B)/ P(A)最大值隨n的增加而遞增。 Knockout Tournament is a highly competitive system in which any player losing a game can no longer play in the tournament. Here we suppose that every player has a numerical value that corresponds to his ability. We consider a totally-seeded knockout tournament with 2n players where in the first round, the strongest player matches the weakest player, the second strongest player matches the second weakest player, and so on. We examine whether a stronger player always has a greater probability of winning the tournament. The answer is in the affirmative for n = 2. For a tournament with eight players(n = 3), the situation is much more complicated. In certain cases, the second strongest player has the greatest probability of winning the tournament. Specifically, let A and B denote the strongest and second strongest players, P(A) and P(B) their respective probability of winning the tournament. We find that the maximum value of P(B)/P(A)equals (196+98) / 343 = 1.0938. For n > 3, we have not obtained the maximum value of P(B) / P(A) . However, it can be readily seen that the maximum value of P(B) / P(A) is non-decreasing as n increases.

此份報告主要是探討一個最優化問題，即是給定特定的移動方式，求出至少要多少操作，才能夠把棋子從某一種分布狀態調整成另外一種。而經過許多例子的嘗試，發現因為原本的問題不帶有數學的式子於其中，故沒辦法用上任何已知的工具進行解題，所以引入座標系以及排列矩陣，讓其含有許多數學式，能夠幫助問題具體化。接著給出好的上界以及得知最大值達成時滿足的條件後，便是開始進行一連串的構造，而這我採取「先猜後調」的策略，簡言之，就是先放置一種看似不錯的分布狀態，再利用討論出來的一些工具，對此狀態進行調整，同時用簡單之圖論方法觀察內部的數學結構，以獲得最優的構造之一。

本文主要研究以數學方法探究物理上的平衡問題，將物理與數學的代數、數論、組合及幾何做了緊密的結合，並導出幾個有趣的結果。文中得出n的因數結構與平衡多項式的關係，以及平衡多項式與平衡問題一般解的關係，並進一步探討其平衡的排序策略及個數。再則將平衡多項式引入複數平面，可以看到許多有趣的幾何性質，並進而回歸至物理的應用。

This research mainly talks about how someone, by observing the non-congruent patterns on the backs of the playing cards and by working with the dealer on a pre-arranged lay out, can call out the cards as if he possessed the magic power to see through them. During the card-predicting game, one can use the patterns on the backs of the cards as visual clues （Observing whether it was places upside down or not）to help figure out the probability of where the card is going to show up. Suck a mathematical formula is known as the Pigeonhole Principle. Upon an analysis of the formula, we find that when given that the value of n is greater than 24, we can successfully call out a number of cards that is greater than 2n/3 . The possibility of such mathematical studies in other directions is endless. 中文摘要： 本研究主要探討利用橋牌非對稱的牌背，猜牌者經由和傳遞牌的人的一種事先約定的方 式（排法），彷彿（魔術）透視般的將一疊牌的花色逐步猜出。其猜牌過程是利用牌背 圖案的朝前朝後的指示，配合適當的猜牌張的分配，而運用到的數學法則包含鴿籠原 理，分析與討論歸納。最後我們得到一疊由四種花色張數相等所混合的n 張牌，可猜出 的張數恆大於 2n/3 （n>24 時）。後續可研究的方向仍然甚廣。

我們的題目是”座標平面上任意三點與x軸的最短距離和”。我們一開始直接由尺規作圖著手，把距離和最小時的那點設為p，但始終無法找到一個方法精確畫到p點。後來變轉入公式的研究，我們設第一象限中A B 和C三點，固定A B C的y座標為d，並討論不同d時的情況，求出B點的範圍。另外我們也從橢圓方面著手，但此部份目前還待努力。;A B and C are three points in the plane, and L is a line in the same plane. Our target is to find a point P in L make the sum of PA+PB+PC minimum. At first we undertake by drawing pictures but we can’t find the solution we want in any case. We make L to be X-axis make A B C’s Y-coordinate to be d, and then we discuss various situations of d, and figure out the range of B. We also analyze the situation in oval, but still have no conclusion.

原題目是環球城市盃中，一個圖論的問題。而題目提供了一個證明，是證 明此種連線都是偶數的圖形，一定會在三的倍數邊形成立。在經過一番思考過 後，我們希望能將原本的偶數連線性質加以驗證，並確定奇存在性。此時，我們 也不禁聯想到：奇數是否也有所特別的性質。因此，我們也向奇數連線做研究。 就在平面得到了部分結論的同時，我們想到這個問題是否可以推廣至三維 空間。然而在推至三維空間的過程中，我們又聯想到，另一種平面：球面。在球 面上放點，能否也找到一些不同的性質。因此，我們分別從平面、球面、立體圖 下手。 基本上，探討平面和立體問題的方法，是以土法煉鋼的方式來求出結果。 然而這種圖論的問題，不可能嘗試到無限多點的情形。因此，我們是著找出一個 關鍵的key，那就是結合性質和外接合性質。以這兩種方法，我們可以將一個簡 單的基本圖形，推向無限多點和無限多邊的情況。 接下來，還有討論一些特殊狀況，例如： deg v＝3n+1，探討其結果。 最後得到的結論是： 1、平面偶圖成立的條件為：此多邊形為三倍數邊形， 而且除了內 部一、二、四點以外， 其他點數都可以成為偶圖。 2、平面奇圖成立的條件為：奇數邊形的情形下，除了三點以外，其 他的內部奇數點的都可以成為奇圖。偶數情形下， 除了四 點以外， 其他的內部奇數點的都可以成為偶圖。 3、三角形平面圖，d eg n 皆為m 成立的條件：2＜ m＜ 6( m? N ) 4、三角形內外任意點d eg 皆為3n ( n ? N )的成立條件： 三角形內部4 x+1 個點( x ? N )。 5、三角形內外任意點d eg 皆為3n+1 ( n ? N )的成立條件： 三角形內部3 x 個點( x ? N )。 6、立體偶圖n 頂點(n＞4)面體的成立條件為： 內部點數為5m+ n- 3、5m+ n- 1、5m+ n、5m+ n +1、5m+ n +3。(m 為大於或等於零的整數) 7、立體奇圖四面體的成立條件為： 內部點數為偶數皆存在。 The original problem is a question of Graph Theory in IMTOT ,which provides\r a proof that proving the figure which its linking-line number is even ,should also be\r contented in the triple-sides figure. After profound consideration ,we try to make sure\r the existence of the properties the we mentioned above. Meanwhile ,it also occurs to\r us that whether the properties would be contented ,in the figure which its linking-line\r number is odd. So we make our way to it. Additionally ,three-dimensional and\r spherical figures are part of our research as well.\r Basically ,we discuss the problem in two-dimensional and three-dimensional\r aspects with the simplest method .However ,it is impossible to discuss the problem in\r unlimited dots .Hence , we are going to find a “key” to solve this problem .As a\r result ,we can find a simple basic-picture , and expand to infinite-multiple lateral\r pictures.\r Next step ,we also discussed some special situations , for example: for each\r point v , deg v=3n+1.\r At last the conclusion is following:\r 1、The conditions of linking-line number is even: triple-sides. And the amount of\r points inside the figure is without 1,2,and 3.\r 2、The conditions of linking-line number is odd: In the odd-sides figure , all number\r of the points inside the figure can be content without 3 point. In the even-sides\r figure , all number of the points inside the figure can be content without 4 point.\r 3、In a triangle , each point’s deg is the same number m: 2

數字波是探討在直線上的起始點、位移速度、總數相互變化的節點關係。在直線上，將全部格子數做為總數(m)，開始彈跳的點為起始點(i)，每次彈跳的格子數為位移速度(s)，被踩到的格子就是節點。節點是由位移速度和起始點決定，起始點本身可視為節點之一，之後的節點是由起始點加n 個位移速度產生。我們分別以三種型式討論：起始點等於位移速度，總數增加：使起始點和位移速度所代表的數字相同的彈跳。節點呈2、s、s+2…起始點固定，位移速度與總數增加：觀察位移速度和總數的關係。兩節點的和=s+2位移速度固定，起始點與總數增加：探討起始點和總數的關係。發現節點隨起始點有規律的變化在上述討論的型式中，我們再進一步將位移速度分為質數和合數，進而依其因數變化，可觀測到很多特殊的節點變化。The number wave is to discuss the relationship of the starting point, the moving speed, and the variations of total amount. In straight lines, let all the trellises be total amount (m), and let the starting jumping point be the starting point (i). The trellis number of each jump is the moving speed(s). The trodden trellises are knots. And knots are decided by the moving speed and the starting point. The starting point itself can be viewed as a knot. The following knots produce with the starting point and n moving speeds. We respectively discuss them in three types: When the starting point equals the moving speed, the total amount increases. The number of the starting point is the same with the jumping moving point; the knots are 2, s, s+2…. When the starting point is fixed, the moving speed and the total number increase. From observing the relationship between the moving speed and the total number, the sum of two knots is s+2. When the moving speed is fixed, the starting point and the total number increase. After our research into the relationship between the starting point and the total amount, we find the knots have regular variations with the starting point. From the types discussed above, we further divide the moving speed into prime numbers and non-prime numbers. Furthermore, according to the factor variations, we can see a lot of specific knot variations.

This is a study about the solution to a chess flipping game. The game is based on a 4*4 game grid. First, place some chesses on the grid randomly to start a game. Move any chess by jumping over one or two neighboring chesses in the same row (left or right), same column (up or down), or on the same diagonal. Chesses which get jumped over should be flipped. The ultimate aim is to make all the chesses upside down. In this study, I try to find the rules of the beginning arrangement that ensure solution. Here are the steps I take: First, break the restriction of the 4*4 grid, and set the coordinate system. Second, find out a few “basic illustrations” that can be solved and moved in order to cope with certain complicated problems. Third, with “basic illustrations”, find the rules applying to games on n*n grid. 這是關於翻棋遊戲的可行解之探討。棋盤是一個4*4 的方格，遊戲開始時在棋盤上任意擺上一些棋子，均是正面朝上，利用相鄰棋子的水平、垂直、斜向跳躍，棋子被跳躍過一次則翻面一次，遊戲目的在於使所有的棋子都翻為反面。我要探討的是關於棋局可解不可解的問題，找出棋盤上可解棋局的規則。研究步驟大致如下：一、打破棋盤4*4 的限制，將棋盤座標化。二、找出若干個可解並可移動的「基本圖」。三、利用基本圖，找出n 列棋盤可解的規律。

After reading “CKSH Communication 20 “, we are interested in Question 5 . We try to explain that in all the No. Crunches, whether we can put the numbers in the No. Crunches any certain position. We specify the question, according to the “Point Symmitry Homing” in mxn No. Crunches, to find the correlated characters of the rules. Finally, we find a “switch” – we can get the better way to rotate the numbers quickly by some programs.從「建中通訊解題」第20 期第5 題出發，本研究嘗試去解釋對所有的數字轉盤而言，是否能將其中的數字歸位到任意指定的位置？接下來將題目特殊化，藉由m× n 數字轉盤的「點對稱歸位」，尋找遊戲規則衍生出的相關性質。最後，利用研究出的性質找到一個「判斷式」可藉由程式設計，快速的找到較佳轉法。