完全圖立方乘積之最小控制
完全圖Kn是指一個圖中有n個點，且任意一個點都跟其它的點有邊相連。兩個圖G和H的卡氏乘積G□H的點集V(G□H)＝{(g,h)| g∈V(G),h∈V(H)}，兩個點(g1,h1)和(g2,h2)有邊相連若且為若g1=g2 且h1~h2，或g1~g2且h1=h2。
三個完全圖Ka、Kb、Kc 的立方乘積是指Ka□Kb□Kc。一個圖G中的一點v所連的其它點稱為這個點v的鄰居，也就是N(v)={x | x~v}。一個點集S中的所有點的鄰居的聯集稱為這個點集的鄰居，也就是N(S)=∪v∈S N(v)。如果一個點集S和它的鄰居N(S)包含了一個圖G的所有的點，也就是S∪N(S)＝V(G)稱這個點集S是這個圖G的一個控制集。我們把圖G的所有控制集中點數最少的稱為最小控制集，並定最小控制集的點數為最小控制數γ(G)，也就是γ(G)＝min { | S |, S是G的控制}。
本文的目的在於研究完全圖立方乘積的最小控制，也就是要給γ(Ka□Kb□Kc)一個上界。特別當 a = b = c = n時，γ(Ka□Kb□Kc) = 。
A complete graph Kn is a graph with n vertices, which any vertex is adjacency to every other vertices. The Cartesian product of two graph G and H which is denoted G□H is define as follow: the vertex set V(G□H)＝{(g,h)| g∈V(G),h∈V(H)}，and two vertices (g1,h1) and (g2,h2) is adjacent if and only if g1=g2 and h1~h2 or g1~g2 and h1=h2. The Cartesian product of three complete graph Ka,Kb,Kc is Ka□Kb□Kc,which is the same with (Ka□Kb)□Kc.
In a graph G, the neighbor of a vertex v N(v) is the set of the vertices adjacent to the vertex v, that is N(v)={x | x~v}。 The neighbor of a vertex set S is N(S), which is the union of the neighbors of vertex v over S, that is N(S)=∪v∈SN(v). For a graph G, if a vertex set S unions its neighbor N(S) equal to the vertex set of G, that is S∪N(S)＝V(G), we say that S is a dominating set of G. The domination number of a graph G will be denoted as γ(G), which is the minimum size of all dominating set of G..
We give an upper bound to γ(Ka□Kb□Kc). And when a=b=c, γ(Ka□Kb□Kc) ≦
高中各學期成績與指定考科相關性
在升學主義越來越興盛的社會中，考試成績成為人人關心的重點，這\r 次研究就是藉由數理資優班同學的各學期在校成績和指定考科成\r 績，透過迴歸分析，找出各學期成績與指考成績之間的關係，並利用\r 圖表來解釋各種科目在各學期的課程，在高中三年所學的重要性，在\r 藉由此結果，希望能對目前老師的教育重點及學生學習方式能有所幫\r 助，亦可了解學生在高中求學過程中，哪些階段對指考成績較有正面\r 影響，進而強化該學習階段，以有助在指定考科時能充分發揮所學。\r \r In a society that emphasize on degrees, examination scores become the\r spotlight, and the ultimate goal for a high school student who had worked\r so hard for three years is to achieve high scores in the J.C.E.E. In the\r three years of high school, each subject has different topics each semester,\r but which semester has the most decisive effect on the J.C.E.E. score?\r This research is to study the effect of each semester on the J.C.E.E. by\r analyzing the grades of a science and math talented class in Senior High\r School using Regression analysis to find out the connections between\r term grades and the J.C.E.E. Then finding out which term grades had the\r most decisive effect in each subject. By using the result, we hope it can\r help teachers in their teaching and students in their learning. Also, it can\r provide the information about which stage in high school has positive\r effects on J.C.E.E. grades, therefore enabling students to emphasize on\r that stage in order to perform well on the J.C.E.E.
長方體中切割正立方體之研究
在1940 年代，Bouwkamp 提出一系列有關如何將矩形切割成若干個正方形的研究報告，但是如何找出正方形個數最少的方法仍是長久以來懸而未決的問題。在本研究報告中，首先引進「四角切割」的方法，並結合輾轉相除法的概念，來研究矩形的切割問題。我們的方法能大幅度降低正方形的個數，也適合做為此問題的上界函數。有關如何在長方體中切割出正立方體的組合，我們也將輾轉相除法的概念延伸到三維空間，進而建立所切割出最少個正立體數的一個上界模式。此外，藉由四角切割概念的延伸，我們也發現這個上界亦可再予修正。In 1940’s, Bouwkamp proposed the study of dissecting squares from rectangles. Among the study, the problem of the least number of dissected squares has been open for decades. In this project, we first propose a corner dissection method, associated with the famous Euclidean algorithm. By reducing nearly three fourths of the number dissected by the primitive Euclidian algorithm, our method indeed establish a suitable upper bound of the minimal number of dissected squares from the given rectangles Meanwhile, the Euclidean algorithm has also been considered to dissect the cubes from cuboids. We analyze the fundamental properties of the method and establish a prototype of upper bound function for the minimal number of dissected cubes. Moreover, the method of corner dissection has also been implemented for some cuboids, which also exhibits the acceptable improvement being a suitable upper bound.
棋子排列的平均值
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.
變形的橢圓—從距離及距離和談起
給定一平面E，A為平面上一點。取r>0，則我們知道到其距離為定值的點形成一圓，而A為此圓圓心。如果把A改成一平面圖形，則到其距離為定值的點形成的集合會是什麼樣子？類似地，給定平面上兩焦點F1及F2在平面上，則到其距離和為定值的點形成橢圓。同樣的，若把F1及F2改成平面圖形，其圖形會是什麼樣子？藉著GSP的輔助，到目前為止，我們得到了以下的結果： \r 1. 給定一平面E及此平面上的一個凸多邊形, 我們描繪出在此平面上到此凸多邊形之距離為定值的點所形成的圖形。\r 2. 設F1和F2分別為平面E上之點或線段或多邊形（未必是凸多邊形），我們利用包絡線描繪出所有滿足d(P,F1)+d(P,F2)=k（k夠大）的點所形成的圖形。 \r 3. 設C1,C2為平面E上之兩圓，我們討論所有滿足 d(P,C1)+d(P,C2)=k\r （k夠大）的點形成的圖形並討論其性質。 \r 4. 設L1和L2分別為平面E上之兩線段，我們討論所有滿足d(P,L1)+d(P,L2)=k（k夠大）的點形成的圖形並討論其性質。 \r 5. 設A為平面E上之一點，Γ為平面上一凸多邊形，我們討論所有滿足d(P,A)+D(P,Γ)=k（k夠大）的點形成的集合並討論其特性。 \r 6. 藉由和圓作比較，我們研究了變形圓的光學性質；而對變形橢圓也做類似的討論。\r Let E be a plane and A a fixed point on E. Given , it is known that all of the points on E with distance to 0r>rA form a circle and the point A is called the center of this circle. What is the corresponding graph if we replace the point A with a set (for example,a segament or a polygon) contained in FE? Similarly, what is the case when we modify the two focuses and in the definition of an ellcpse to sets and (or example,two segments or two polygons) contained in 1F2F1F2FE ? Taking advantages of GSP and analytic geomety, we research related situations and so far we have obtained the following results:\r 1. Let Γ?E be a segment, a convex polygon or a circle , etc. and r>0 be fixed. We sketch the graph of points on E with distance r to Γ and study properties of such graphs.\r 2. Let F1 and F2 be singletons, line segments , polygons(may not be convex), or circles,etc., on E Taking advantage of envelopes, we sketch the graph of those points P on E satisfying d(P,F1)=k(K>0 is large enough).\r 3. Let C1 and C2 be circles on 1C2CE. We sketch the graph of the points P on E that satisfiy d(P,C1)6d(P,C2)=k (k>0 is large enough) and study properties of this graph.\r 4. Let L1 and L2 be two line segments on E and be a large enough constant. We sketch the graph of points P on E that satisfy d(P,L1)+d(P,L2)=k(k >0is large enough) and research properties of this graph. 0k>\r 5. Let A?E and be a convex polygon on ΓE. We sketch the graph of points on E that satisfy d(P,L1)+d(P,L2)=k(k>0 is large enough) and research properties of this graph.\r 6.We compare the optical properties of metamorphic circles with circles and we deal with metamorphic ellipses similiarly.
重複圖形
「重複圖形」是本篇報告研究的問題，我們利用「方程式」建立一個尋找重複圖形，並証明其個數的方法。利用此方法得出下面的結論：1.會形成lap 2 的凸多邊形只有2 種，即三角形和四邊形。(1)「lap 2 三角形」只有1 種，即等腰直角三角形。(2)「lap 2 四邊形」只有1 種，即二邊之比為1： 且內角是45°、135°的平行四邊形。2.會形成lap 3 的凸多邊形只有2 種，即三角形和四邊形。(1)「lap 3 三角形」只有1 種，即內角為30°–60°–90°的直角三角形。3.其他的lap k 三角形：(1)任意內角為30°–60°–90°的直角三角形都是lap 3k²，其中k是正整數。(2)邊長比為1：m： 的直角三角形是lap (m²+1)k²三角形，其中m、k是正整數。
To find repeated figures, we construct a method to search them with the help of algebraic equations. Here we arrive at：1. There are only two kinds of lap 2 convex polygons, triangles and quadrilaterals. (1) The only lap 2 triangle is isogonal right-angled. (2) The only lap 2 quadrilateral is the one that contains angles 45°, 90° and two neighboring sides with the ratio 1: . 2. There are also two kinds of lap 3 convex polygons, triangles and quadrilaterals. (1) The only lap 3 triangle is the one with angles 30°, 60° and 90°. 3. Other kinds of lap k triangles are listed as following: (1) A triangle with angles 30o, 60°, 90° is a lap 3k², the k is a natural number. (2) A right-angled triangle whose ratio is 1 : m : is a lap (m2+1)k², the m and the k are natural numbers.
探討「避開矩形框」的配置方法與推廣
一、若Mn×n(s)表示在n×n 的正方形棋盤中，排列s 顆棋子在方格內，且每一方格最多只能排1子，其中s 顆棋子的配置需滿足兩個條件：1. 並無任意4 子可以形成矩形框的4 個頂點。（此矩形框的邊需與棋盤的邊平行）2. 在沒有棋子的方格中，無法再加入棋子。二、若Vn×n×n(a1,……,an) 表示在n×n×n 的正方體棋盤中，每層的棋子個數分別為a1,……,an，且s= a1+……+an，其中s 顆棋子的配置需滿足兩個條件：1. 並無任意8 子可以形成長方體的8 個頂點。（此長方體的邊需與立體棋盤的邊平行）2. 在沒有棋子的方格中，無法再加入棋子。本研究即在Mn×n(s)與Vn×n×n(a1,……,an) , s= a1+……+an 中探討s 的最小值、最大值及變化情形，並分析其配置方法。之後推廣至長方形Mn×m(s)及長方體Vn×m×k(a1,……,ak) , s= a1+……+ak。最後根據其研究結果設計一個「避開矩形框棋」，並加以分析出致勝的策略。一．If Mn×n(s) indicates in the n×n square chessboard, we put s chesses to line in the square and each square only can put one chess. Then the station of s chesses must satisfy the following two conditions：1. No any 4 chesses can form the tops of the rectangular frame ( The sides of rectangular frame must be parallel to the sides of chessboard )2. If there’s no chess in the square, we can’t add any chess. 二．If Mn×n×n(a1,……,an) indicates in the n×n×n square chessboard, the chess number in each layer are a1,……,an and s= a1+……+an. The station of s chesses must satisfy the following two conditions： 1. No any 8 chesses can form eight tops of the rectangular cube ( The sides of rectangular cube must be parallel to the sides of cubic chessboard ) 2. If there’s no chess in the square, we can’t add any chess. This research try to explore the minimum, maximum and variation of s which in Mn×n(s) and Mn×n×n(a1,……,an), s= a1+……+an, and analyze its station. Then we will extend the research to rectangle Mn×m(s) and rectangular cube Vn×m×k(a1,……,ak), s= a1+……+ak. Finally, according to the result of research we wish can design one “avert rectangular frame chess“ and analyze the strategies to triumph.
四面體體積平分面的包絡方程探討
剛開始考慮平分物件時，我們從二維的多邊形部分著手，後來發現已經有人做過相關研究，並且得到類似的結論。這個部份顯現出面積平分線與其包絡曲線間的密切關係。我們將其中的方法和結果加以歸納、改善，為了更全面地研究，我們推導出一般性的包絡方程。之後當我們推廣到三維領域時，發現四面體體積平分面與之前的結論有些相似之處，平分的情況卻也更複雜，我們將推導的結果用電腦軟體呈現出來，以便更深入地了解它。最後嘗試了相當抽象的高維積平分，結果仍具有工整的對稱性，讓我們充分領略了數學之美！When considering bisecting a subject, at first we focused our attention on 2-D case, polygons. But afterwards, we found there were already some similar studies conducted by other students, which indicated the close relation between the area-bisecting lines of a polygon and their envelope. We rearranged their methods and results, and then made further improvement. Moreover, in order to study the bisecting problem entirely, we derived the general envelope equation. Then when extending the generalization to the 3-D case, we came to the conclusion that tetrahedrons’ volume-bisecting planes is similar to that in 2-D, but the circumstances are more complex. We tried to show our result with the aid of software, hoping to understand it fully. Finally, we tried to do the case in higher dimension, which is very abstract, and the result was clear-cut symmetrical. During the studying process, we had seen “the beauty of mathematics.”