數學

高斯曾經提出八皇后問題：八個皇后在8 × 8 的棋盤上有幾種放法可以使任意兩皇后不會互相攻擊？我們在原來的問題上加上一些條件，改變皇后攻擊規則，使得皇后失去一條對角線的攻擊方向，稱之為「跛腳皇后」。我們稱一個在棋盤上放置最多跛腳皇后使其不互相攻擊的放法為好放法；研究跛腳皇后放置在各類棋盤上其好放法的個數和性質。我們分別在六種棋盤上做討論：（1） 在平面n x n 棋盤上，我們證明了其好放法與完美極致史考倫型數列之間的對應關係，並歸納出相關的性質和定理。（2） 在平面m x n 棋盤上，我們固定一邊長度n，做出n 較小時好放法數的通式；我們也將其好放法對應至廣義史考倫。（3） 在環面n x n 棋盤上，我們說明了其好放法與完全剩餘系排列之間的對應關係，並歸納出相關的性質和定理。（4） 在環面m x n 棋盤上，我們固定gcd（m,n），做出gcd（m,n）較小時好放法數的通式。（5） 在柱面n x n 棋盤上，我們證明其與環面n x n 棋盤等價，說明其好放法具有和環面n x n 棋盤好放法相同的性質和定理。（6） 在柱面m x n 棋盤上，分成左右柱面以及上下柱面來做討論。我們歸納出相關性質和定理；並固定一邊長度n，做出n 較小時好放法數的通式。Gauss had researched about putting eight queens on the chessboard on the way that doesn’t make any queen attacks another one. Thus, we added some rules on the question: the queen loses one diagonal attacking-way and become the “lame queen”. We call a way that doesn’t make any lame queen attacks another one “a good way”. We have been investigating the amount and properties of good ways based on six kinds of chessboard: （1）We found the correspondences between the “good way” on n × n plane chessboard and the Perfect extremal Skolem-type sequence, and concluded some associated properties and theorems. （2） On m× n plane chessboard, we fixed the length n of one side of the chessboard, and accomplished the amount of good ways when n is small. We also correspond the good way to the Generalized Skolem.（3）We found the correspondences between the “good way” on n × n torus chessboard and the arrayal of complete residue system, and concluded some associated properties and theorems.（4）On m× n torus chessboard, we fixed the gcd(m,n) （greatest common divisor of m and n）, and accomplished the amount of good ways when gcd(m,n) is small.（5）On n × n cylinder chessboard, we proved that this kind of chessboard is equal to torus chessboard. So the good ways, characters, and theorems on cylinder chessboard are the same as on the torus one.（6）On m× n cylinder chessboard, we separate it into two cases: left-right cylinder chessboard and up-down cylinder chessboard. We concluded some associated properties and theorems, and we also fixed the length n of one side of the chessboard and accomplished the amount of\r the good ways when n is small.

本文我們研究下列問題： 1. 何時心律會成穩定的狀態呢？ 2. 是否能建立出心律穩定的數學模型？ 3. 什麼樣的函數會使得心律趨於穩定？我們以xmp+i 表示第p+1 個訊息傳到第i 個細胞之前，第i 個細胞的舒張時間，而且xk 和 xk−m, xk−m+1,..., xk−1的關係為 In this paper we study the problems as follows: When will the rhythm of heart beats approach to a steady state? Can we set up the mathematics model with steady rhythm of heart beats? What kind of function will make the rhythm of the heart beating tend to be stable? The result of our study is as follows:

At the website “MathLinks EveryOne,” we found a problem “Snakes on a chessboard,” which was raised by Prof. Richard Stanley. The following is the problem. A snake on the m n chessboard is a nonempty subset S of the squares of the board with the following property: Start at one of the squares and continue walking one step up or to the right, stopping at any time. The squares visited are the squares of the snake. Prove that the total number of ways to cover an m × n chessboard with disjoint snakes is a product of Fibonacci numbers. We call the total number of ways to cover a chessboard with disjoint snakes “the snake-covering number.” This problem hasn’t been solved since it was posted on September 18, 2004, so it aroused our interest to study it. First, we used the way in which we added each block to the chessboard, and therefore we discovered some regulations about the snake-covering number of the1 × n , 2 × n and 3 × n chessboard. Through “recursive relation” and “mathematical induction”, we proved the general term of the snake-covering number of the1 × n , 2 × n and 3 × n chessboard. In the following study, we found a key method in which we added a group of blocks to the chessboard. Finally, we proved the general term of the snake-covering number of the m × n chessboard. Also, we discovered the way to figure out the snake-covering number of the nonrectangular chessboard.在網站“ MathLinks EveryOne ”中，我們找到了一個有趣的問題“棋然上的蛇” ( Snakes on a chessboard ) ，這個問題是由教授 Richard Stanley 所提出。問題如下：在m x n棋盤形格子上，蛇由任意一格出發，但蛇的走法只能往右 → ，往上↑，或停住 ‧ 若此蛇已停住，將由另一條蛇來走，且不同蛇走過的格子不可重疊”證明：將 m × n 棋盤形格子完全覆蓋的總方法數為費氐（ Fibonacci ）數列某些項的乘積。我們將把棋盤形格子完全覆蓋的所有方法數稱之為“蛇填充數” 由於這個問題自從 2004年 9 月 18 日被登在網站上後，還沒有人提出解答，於是引發了我們研究的興趣。首先，我們使用了將一個一個格子加到棋盤上的方法，並發現了 l × n 、 2 x n、 3 × n 棋盤形格子蛇填充數的一些規律。我們使用遞迴關係及數學歸納法來證明 l x n 、 2 x n ， 3 × n 棋盤形格子蛇填充數的一般項。在接下來的研究中我們發現一個特別的方法，一次增加數個方塊 ‧ 最後我們證明了，m x n, ，棋然形格子的蛇填充數的一般項 ‧ 而且，我們也找到如何求出不規則棋盤形格子的蛇填充數。

正方形兩條對角線的交點（即中心點）距四頂點等長，也與四邊等距。如果將正方形的頂點比擬成它的「手」，兩對角線的交點當成它的「心」，則兩個正方形頂點間、中心點間、或頂點與中心點間的線段相連（或重合），就如同「手」或「心」彼此相連。本文即探索當多個正方形間「心手相連」時，衍生圖形間的面積關係。而四個正方形中某幾個頂點相接（邊未重疊），恰圍出兩個三角形的圖形則是本內容討論圖形的主體架構，我們以此架構向外作出「層出不窮」的正方形，再配合中心點連接成四邊形，將推導出這些四邊形與基準正方形（Reference Square）間的面積關係。In a square, the lengths from the intersection point （center point） of two diagonal lines to the four apexes are the same, and so are they from that point to the four sides. If the apexes are “hands” and the intersection point of two diagonal lines is the “heart” of a square, the connection or overlap of two squares’ apexes and apexes, center point and center point, or apexes and center points is just like the connection of hands with hearts. In this article, hence, we are to explore the relation in area of derivative graphs formed by several squares connected “heart in hand.” When some apexes of four squares are overlain without sides overlapped, two triangles are created. And that’s the theme we are going to discuss. Furthermore, we extend the operation to infinitely overlain squares and frame out quadrangles referring to the center points of some squares. Then, the relation in areas of these overlapped squares and the Reference Square would be deduced.

令Sn 為{1,2,…,n}任意排列所成的集合，π ? Sn 為其中的一個元素，我們記π = (π(1), π(2),…, π(n))。今給定π ? Sn ，若對所有i,1? i ? n，都有π (π (i)) = i 時，我們稱π 為involution。假設π ? Sn ，並給定σ ? Sm (m ? n)，當π 中任取m 項，其大小關係的順序都和σ 不同，我們稱π 避開σ，或稱π 是一個σ-avoiding 排列。在這篇報告中，我們主要分析了2143-avoiding involution，1234-avoiding involution，和3412-avoiding involution 中的一些統計量，給出了十數個結果與幾個猜想。Let Sn be the set of permutations on {1,2,…,n} and π ? Sn be an element in Sn. Denote π as π = (π(1), π(2),…, π(n)). We say that π is an involution if π(π(i)) = i for every i, 1? i ? n. Given π ? Sn and σ ? Sm (m ? n) , we say that π avoids σ (or π is an σ-avoiding permutation) if π does not contain any m-term subsequence in the order of σ. In this paper, we discuss some classic statistics on 2143-avoiding involutions, 1234-avoiding involutions and 3412-avoiding involutions. We get many new results in this field and give some interesting conjectures.

如果n-1 個正四面體在n 個面的多面體棋盤中翻轉(如：正八面體、正二十面體、五邊雙菱錐及三角錐台棋盤)，則正四面體相互之間會受到很多限制。本研究為探討n-1 個正四面體在n 個面的多面體棋盤中，利用其翻轉的特性，翻轉至相鄰空格，進而完成：1. 每個正四面體的底面及側面均能「同色共面」。2. 在滿足以上的兩個條件下做「數字排序」。3. 探討每個遊戲在各階段是否有解的情形。最後，將遊戲推廣至正八面體在多面體棋盤中的翻轉。If n-1 pyramids can turn in the chessboard of n planes pyramids (ex: 8 planes pyramid, 20 planes pyramid, 5 sides bi rhombus pyramid and triangular pyramid chessboard) there are lost of restrictions between these pyramids. This research is discussing about the n-1 pyramids in the chessboard of n planes multi pyramid, which have the peculiarity of turn that can turn to nearby space. We found that: 1. Each pyramid’s button plane and aide plane can be “same plane same color”. 2. Matching about 2 conditions then can do “order by numbers”. 3. Search for answers of each levels in each game. Finally, the game will be propagated to eight planes pyramid can turn on the chessboard of multi pyramid.

給定下面範例：\r 058823529411764705882 +235294117647058823532\r =0588235294117647058823529411764705882353，\r 其等式結果與質數17 的倒數結果(1/17)有某種關聯(卻沒有一個決定性的證據)，意即\r 1/17=0.0588235294117647=\r 0.058823529411764705882352941176470588235294117647...... ( Len(17) =16 )\r \r 曾經在下列網站上發現過幾組數字(挑戰試題)，引起我們極大的興趣。\r http://www.domino.research.ibm.com/Comm/wwwr_pondernsf/challenges/March2000.html\r http://www.math.smsu.edu/~les/POW08_96.html\r \r \r The two examples that I have are 0588 2+23532=05882353 and 058823529411764705882+23529411764705882353 2=0588235294117647058823529411764705882353 These were found by the Canadian professor Alf van der Poorten, and he gave a talk on these identities in December at the west coast number theory conference. He was unspecific as to exactly where these identities were coming from, but they are connected with reciprocals of primes:1/17 = 0.0588235294117647= 0.058823529411764705882352941176470588235294117647 ΛΛ ( Len(17) = 16 ) Though not mentioning how to obtain these equations, Prof. Poorten demonstrated the relationship between the above examples and the reciprocal of the prime numbers 17 (1/17 ) without a definitive proof.

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。

本研究是藉由數學手法探討；如何由一疊 36 張四種花色的撲克牌中，尋找出保證可猜中最多張花色的方法。研究過程是以在適當的猜牌時機，以邏輯推理、二進位、分析與歸納 … … 等數學原理與方法，搭配巧妙的策略運用而達到目的。 猜牌方法：先約定好猜牌規則，助手將 36 張牌背圖樣相同但非對稱的撲克牌，以旋轉牌背的方向傳達訊息。在本研究中得出「經由巧妙的猜牌方法保證可以猜中不少於 26 張花色」，並得出「當總張數趨近於無窮大時，保證可以猜中不少於 81 . 07 ％的牌，並且證出若僅使用獨立的訊息猜牌，無論任何猜牌方法皆無法猜中多於 87 . 37 ％的牌」 · 其中一個猜中多於 80 ％的例子是‘「當總張數等於 23006 張時，保證可以猜中不少於 1 8405 張牌（18405/23006 > 4/5）」 ·The study is mathematically based with reasonable explanations behind it. We are to correctly guess as many cards as possible from a deck of 36 cards, with random numbers and four different suits. We will apply mathematical methods, such as logic inference, binary system, and analytical reduction, upon right timing. Using careful arrangement of the principles and reasoning, we can reach our ultimate goal. To state guessing: Conference between the guesser and the assistant about the guessing rules, the assistant will have 36 cards with the same exact pattern on the back but not symmetrical. The pattern of the cards will be different when rotated 180o. The only communication between the two is by rotating cards. In this study, we can prove that through mathematical method, we can assure 26 or more cards can be correctly guessed. Furthermore, when the total amount of cards is close to infinity, we can assure 81.07% or more of the cards can be correctly guessed, and prove that if the cards are guessed from independent information, no more than 87.37% of the cards will be correctly guessed by any guessing methods. One of the examples, which 80% of the cards are correctly guessed, is that when the amount of the cards is 23006, 18405 or more of the cards can be correctly guessed. （18405/23006 > 4/5）