Coursera - Algorithmic Thinking - Module 2

# Coursera - Algorithmic Thinking - Module 2 課程綱要 Terminology worst-case running time best-case running time tightest upper bound BFS = Breadth-first Search The running time is O(n^2) or O(m+n) or O(m). m is the number of edge. n is the number of node. O(n^2) when the graph is closed to complete graph. O(m+n) when the graph is sparse. O(m) when the graph is sparse and m > n. For the tree search algorithm, you also need to consider how tree is stored. Adjacency matrix and adjacency list might have different complexity. 有沒有按讚的地方？CZChen++ The worse case is O(n^2). Homework Question 8 n0=1 的意思是從 n 等於 1 開始都要成立 Project Application Question 1 Question 2 兩千次 computer network graph 的結果，大家可以參考一下 Question 3 Question 4 Question 5 Bonus Question 6 Challenge - Disjoint set algorithm What is this? Project 中有說過： Challenge problem: In the Application, you will compute the resilience of graphs with several thousand nodes and edges. In desktop Python, this computation will take on the order of a few seconds. In CodeSkulptor, this computation will take 3-5 minutes per graph. As a challenge, investigate other asymptotically faster approaches to implementing compute_resilence. We have implemented one approach based on a simple disjoint set algorithm that has a running time of O(nlog(n)+m). This method can perform all of the analysis in the Application part of this Module in CodeSkulptor in a few seconds. Progress czchen Homework Project Application FourDollars Homework Project Application Kommodore Kezza Homework Project Application Peer Assessment Q6 要等 rubric 改動過後再看看 allenwangs Homework Project Application Peer Assessment