歐德斯-史特勞斯猜想之探討
歐德斯—史特勞斯猜想又稱為 4/n 問題，其內容為對於所有正整數n皆滿足 4/n=1/a+1/b+1/c ，其中a, b, c為正整數。於19世紀提出並在當代引起討論熱潮，至今此問題仍沒有完善的證明方法。經過查閱文獻資料後，我們發現他人研究重點著眼於如何將正整數n以同餘分類，且並未獲得一個系統性的研究結論。研究內容多執著於如何解決此猜想而非探討問題本身的規律性及各項性質。 此外，他人研究少有討論正整數n的解數者。因此本研究將方向設定在n, a, b, c的可行解數量。透過特例解切入n, a, b, c的表示方式，使問題簡化而較易於討論。以求對證明此猜想有所貢獻。
Reduction of traffic congestion in España Boulevard using graph theory
There have been numerous studies exploring the applications of graph theory in traffic management, often finding ways to reduce traffic congestion and make traveling more efficient. Such studies will be beneficial when applied to heavily congested areas such as España Boulevard, one of the busiest thoroughfares in Manila. This paper aimed tooptimize the road map of España Boulevard using graph theory. The current road map of España Boulevard was represented as a directed graphand subjected to the mutation method of edge removal, wherein an edge isremoved in each mutation based on a computed fitness function, F(G),which depicts better efficiency at lower values. Edges were removed until the graph got disconnected, which was tested using the Floyd-Warshall algorithm. The 28th mutation resulted in a minimum F(G) value of 144.4; this is a 50.18% decrease from the F(G) of the original graph, which is 290. After the 28th mutation, the removals resulted in an increase in the F(G). As a result, the final mutation resulted in an F(G) of 311.89, which characterized a less efficient graph. This study was able to apply graph theory concepts to optimize the España Boulevard road map using the mutation method, minimizing its F(G) by at most 50.18%. For future studies, the practicality of the alternate road map may be tested in simulations to examine its efficiency when other factors, such as traffic volume, are introduced.
距離便是美—多維空間的支配數之討論
本研究保持社交距離為發想，探討從一維到多維空間的支配數。我們從使得三個同色單位方格不相連的二維情況，拓展至m個同色單位方格不相連的一維、二維、三維情況。本研究從The Domination Number of Grids這篇論文中汲取靈感，其中”Domination Number”也是「支配數」此名詞的由來。我們定義L_nt={(x_1,x_2,…,x_n)|x_1+x_2+⋯+x_n≡t (mod m),x_1∈[1,l_1 ],x_2∈[1,l_2 ],……,x_n∈[1,l_n ]}，此處的l_n是邊長。對於一維情形的任意m，其支配數|A_1m |=⌊l_1/m⌋；對於二維情形且m=3時，我們經由列舉和畫圖證明其支配數|A_2 |=⌊(l_1 l_2)/3⌋。同樣的二維和三維情況在m=任意數時的支配數也可求得，不過在此我們改變了研究的方法，我們應用集合與同餘進行運算，除了減少窮舉將花費的時間，也可一次討論m=任意數的情況。
糖果傳遞問題之研究與推廣
n個人圍成一圈，面向圓心，且逆時針編號1,2,……,n。一開始每人手中有一個糖果，由1號開始，逆時針分別給右邊的人一個、兩個、一個、兩個……糖果，手上沒有糖果的人必須退出。我們將此傳遞規則定義為T_1,2，同理T_(1,2⋯,p)。這個傳遞遊戲，最終會有兩種情形，第一種是由一人獨得所有糖果(成功狀態)，第二種是數人間傳遞糖果且形成循環(循環狀態)。 研究後得知，在傳遞規則T_(1,2⋯,p) (p≥2)下，若p=〖p_1〗^(α_1 ) 〖p_2〗^(α_2 )⋯〖p_i〗^(α_i )⋯〖p_j〗^(α_j ) ( 為p的相異質因數)，任意的n值(n≥p+1)均可唯一表示成n=(p)^t×(〖p_1〗^(s_1 ) 〖p_2〗^(s_2 )⋯〖p_i〗^(s_i )⋅m)+q (t,m∈N, p ∤〖p_1〗^(s_1 ) 〖p_2〗^(s_2 )⋯〖p_i〗^(s_i ), (m,p)=1, q=1,2,⋯,p)，令S=(p^t (p-q)+(pq-1))/(p-1)+R⋅p^t，則當m=1時，最終為成功狀態，且獨得糖果者的初始編號為S；當m≥2時，最終為循環狀態，且由m人循環傳遞糖果，而此m人的初始編號是S, S+p^t 〖p_1〗^(s_1 ) 〖p_2〗^(s_2 )⋯〖p_i〗^(s_i ), ⋯⋯ , S+(m-1)⋅p^t 〖p_1〗^(s_1 ) 〖p_2〗^(s_2 )⋯〖p_i〗^(s_i )。上述公式中的R值，可透過我們研究出來的「R值迭代法」求得。更進一步，我們也找出達到成功狀態或循環狀態的最小傳遞數。