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| courses:cs211:winter2011:journals:chen:chapter_4 [2011/02/16 11:55] – [4.2 Scheduling to Minimize Lateness: An Exchange Argument] zhongc | courses:cs211:winter2011:journals:chen:chapter_4 [2011/03/02 03:13] (current) – zhongc | ||
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| + | Readable/ | ||
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| di < dj but j scheduled before i. | di < dj but j scheduled before i. | ||
| + | Greedy Exchange Proofs on slide 4.2 p7. | ||
| + | |||
| + | Running time: nlongn --- need to sort the deadlines first. | ||
| + | |||
| + | |||
| + | Readable/ | ||
| + | ===== 4.4 Shortest Paths in a Graph ===== | ||
| + | problem: | ||
| + | |||
| + | Given a directed weighted graph, what is the shortest path? | ||
| + | |||
| + | Dijkstra' | ||
| + | The algorithm maintains a set S of vertices u for which we have determined a shortest-path distance d(u) | ||
| + | from s; this is the " | ||
| + | |||
| + | Now, for each node v ~ V-S, we determine the shortest path that can be constructed by traveling along a path through the explored part S to some u ~ $, followed by the single edge (u, v). That is, we consider the quantity d’(v) = mine=(a, | ||
| + | |||
| + | |||
| + | We always form shortest new s-v path from a path in S followed by a single edge | ||
| + | • Proof of optimality: Stays ahead of all other | ||
| + | solutions | ||
| + | Each time selects a path to a node v, that path is | ||
| + | shorter than every other possible path to v (stay ahead proof) | ||
| + | |||
| + | proof of correctness: | ||
| + | **inductive proof** | ||
| + | Inductive hypothesis: Assume true for |S| = k, k ≥ 1 | ||
| + | Let v be the next node added to S by Greedy, and let uv be | ||
| + | the chosen edge. | ||
| + | The shortest s->u path plus u->v is an s->v path of length π(v) | ||
| + | Consider any s->v path P. It's no shorter than π(v). | ||
| + | Let x->y be the first edge in P that leaves S,and let P' be the subpath to x. | ||
| + | P is already too long as soon as it leaves S. | ||
| + | |||
| + | |||
| + | Question: Still cannot see why the order matter. If we do a BFS and visit all vertices and update all distance labels, wouldn' | ||
| + | |||
| + | |||
| + | |||
| + | Readable/ | ||
| + | |||
| + | ===== 4.5 The Minimum Spanning Tree Proble ===== | ||
| + | Section Summary | ||
| + | |||
| + | **Motivation of the problem/ | ||
| + | How to minimize cost of laying cables? | ||
| + | How to find the minimum spanning tree for a connected weighed graph? | ||
| + | |||
| + | **Definition: | ||
| + | Min Spanning tree: | ||
| + | spanning: spans all nodes in graph | ||
| + | Tree: | ||
| + | Minimum: | ||
| + | |||
| + | Claim: MST contains no cycle | ||
| + | Proof: Contradiction, | ||
| + | Thus the tree is not minimum. | ||
| + | |||
| + | **Two Properties: | ||
| + | When is it safe to invclude an edge in the MST? | ||
| + | Cut property. Let S be any subset of nodes, and let e | ||
| + | be the min cost edge with exactly one endpoint in S. | ||
| + | Then MST contains e. | ||
| + | proof on P170 | ||
| + | |||
| + | When is it safe to remove an edge from the MST? | ||
| + | Cycle property. Let C be any cycle, and let f be the | ||
| + | max cost edge belonging to C. Then MST does not | ||
| + | contain f. | ||
| + | proof on P170 | ||
| + | |||
| + | cycle-cut intersection proof | ||
| + | |||
| + | **Three algorithms** | ||
| + | |||
| + | Prim: | ||
| + | Start with an arbitrary node s and add it to the tree T. Grow T outward by choosing the cheapest edge e with on endpoint in T and the other out of T. | ||
| + | Proof on P172 | ||
| + | Cut property | ||
| + | |||
| + | Kruskal: | ||
| + | Sort the edges in G in ascending orders. pick the smallest edge that does not creat a cycle. | ||
| + | |||
| + | Reverse-delete: | ||
| + | It is the rever version of Kruskal | ||
| + | |||
| + | **Implementations of the two algorithms** | ||
| + | |||
| + | |||
| + | |||
| + | |||
| + | |||
| + | Questions: | ||
| + | I had some trouble understanding the proof for Cayley' | ||
| + | |||
| + | |||
| + | ===== 4.6 Implementing Kruskal’s Algorithm: The Union-Find Data Structure ===== | ||
| + | |||
| + | Section Summary: | ||
| + | |||
| + | Problem Motivation: | ||
| + | Try to implement the Kruskal' | ||
| + | |||
| + | pointer based implementation | ||
| + | |||
| + | Find Operations | ||
| + | Log(n) | ||
| + | |||
| + | Union Operations | ||
| + | Log(n) | ||
| + | |||
| + | |||
| + | The Kruskal Algorithm in detail | ||
| + | |||
| + | |||
| + | Intereting/ | ||
| + | |||
| + | |||
| + | |||
| + | ===== 4.7 Clustering ===== | ||
| + | |||
| + | Section Summary | ||
| + | definition: | ||
| + | Divide objects into clusters so that points in | ||
| + | different clusters are far apart. | ||
| + | |||
| + | Motivation/ | ||
| + | Identify patterns in gene expression | ||
| + | |||
| + | K-Clustering: | ||
| + | farthest two elements in one cluster are not farther apart than any | ||
| + | elements outside of the cluster. | ||
| + | |||
| + | It is basically Kruskal' | ||
| + | |||
| + | proof on P184 | ||
| + | |||
| + | Interesting/ | ||
| + | |||
| + | |||
| + | ===== 4.8 Huffman Codes and Data Compression ===== | ||
| + | |||
| + | summary | ||
| + | |||
| + | Motivation: | ||
| + | How to encode data so that transmission could be more efficient? | ||
| + | Answer: use less bits on the data without much differentiation!(Low entropy data?) | ||
| + | |||
| + | We use Huffman codes. | ||
| + | |||
| + | If we use all letters in the same frequency, then there is nothing to encode or compress, but when we do not, which is often the case, | ||
| + | we can always represent the more frequently used words with less bits. | ||
| + | |||
| + | Watch out for prefix ambiguity! | ||
| + | |||
| + | **variable-length encoding** | ||
| + | |||
| + | Goal: minimize Average number of Bits per Letter (ABL): | ||
| + | calculate abl using the expected value. | ||
| + | |||
| + | Algorithm | ||
| + | |||
| + | implemenation | ||
| + | |||
| + | * Binary tree for the prefix codes | ||
| + | * Priority queue (with heap) choosing the node with lowest frequency | ||
| + | Cost (nlogn) Why? logn inserting and dequeuing. do it n times. | ||
| + | Interesting/ | ||
