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Assignment 2: (Due at 10:30 a.m on Friday of Week 10)

Question 1 (Given in Tutorial 5) Question 2 (Given in Tutorial 7) If you do Question 1 only, you get 60 points. If you do Question 2 only, you get 90 points. If you correctly do both Question 1 and Question 2, you get 100 points.

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Assignment 2: (Due at 10:30 a.m on Friday of Week 10)

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  1. Question 1 (Given in Tutorial 5) • Question 2 (Given in Tutorial 7) • If you do Question 1 only, you get 60 points. • If you do Question 2 only, you get 90 points. • If you correctly do both Question 1 and Question 2, you get 100 points. • Bonus: 5 Points will be given to those who write a Java program for the Huffman code algorithm. Assignment 2: (Due at 10:30 a.m on Friday of Week 10)

  2. Lecture 1: Some concept: Pseudo code, Abstract Data Type. (Page 60 of text book.) Stack. Give the ADT of stack (slide 11 of lecture1) The interface is on slide 19. (Q: Is the interface equivalent to ADT? Not really. We need the method for insertion and deletion, i.e., first in last out. ) Applications: parentheses matching Review of Lecture 1 to Lecture 6

  3. Linked list Singly linked list Doubly linked list Just know how to setup a list. (Assignment 1) Lecture 3: Analysis of Algorithms (important) Primitive operations Count number of primitive operations for an algorithm big-O notation 2nO(n), 5n2+10n+11++>O(n2). Lecture 2:

  4. Definition of tree (slide 7) Tree terminology: root, internal node, external node (leaf), depth of a node, height of a node, height of a node. Inorder traversal of a binary tree Tree ADT, slide 11, Binary tree ADT, slide 17 In terms of programming, understand TreeInExample1.java. (If tested in exam, java codes will be given. I do not want to give long code.) Lecture 4: Tree

  5. Linked Structure for Binary Tree. Just understand the node: Preorder traversal for any tree Postorder traversal for any tree Array-Based representation of binary tree (slide 9) Algorithms for Depth(), Height() slide 12-15. Lecture 5: More on Trees 

  6. Priority Queue ADT (slide 2) • Heap: • definition of heap • What does “heap-order” mean? • Complete Binary tree (what is a complete binary?) • Height of a complete binary tree with n nodes is O(log n). • Insert a node into a heap runtimg time O(log n). • removeMin: remove a node with minimum key. Running time O(log n) • Array-based complete binary tree representation. • Show a sample exam paper. Lecture 6: Priority Queue (Heeps)

  7. Priority Queue ADT (slide 2) • Heap: • definition of heap • What does “heap-order” mean? • Complete Binary tree (what is a complete binary?) • Height of a complete binary tree with n nodes is O(log n). • Insert a node into a heap runtimg time O(log n). • removeMin: remove a node with minimum key. Running time O(log n) • Array-based complete binary tree representation. • Show a sample exam paper. Lecture 6: Priority Queue (Heeps)

  8. Exercise: Give some trees and ask students to give InOrder, PostOrder and PreOrder. Tutorial 6 of Question 2: Using PreOrder. Given a complete binary, write the array representation. Given an array, draw the complete binary tree. Given a heap, show the steps to removMin. Given a heap, show the steps to insert a node with key 3. (Do it for the tree version, do it for an array version.) Linear time construction of a heap.

  9. Huffman codes (Page 565 Chapter 12.4) • Binary character code: each character is represented by a unique binary string. • A data file can be coded in two ways: The first way needs 1003=300 bits. The second way needs 45 1+13 3+12 3+16 3+9 4+5 4=232 bits. Hash Tables

  10. Variable-length code • Need some care to read the code. • 001011101 (codeword: a=0, b=00, c=01, d=11.) • Where to cut? 00 can be explained as either aa or b. • Prefix of 0011: 0, 00, 001, and 0011. • Prefix codes: no codeword is a prefix of some other codeword. (prefix free) • Prefix codes are simple to encode and decode. Hash Tables

  11. Using codeword in Table to encode and decode • Encode: abc = 0.101.100 = 0101100 • (just concatenate the codewords.) • Decode: 001011101 = 0.0.101.1101 = aabe Hash Tables

  12. 100 0 100 0 1 1 86 a:45 14 0 1 0 0 1 0 1 1 58 28 14 0 0 1 0 1 0 1 c:12 b:13 d:16 14 30 0 1 55 25 a:45 b:13 c:12 d:16 e:9 f:5 f:5 e:9 • Encode: abc = 0.101.100 = 0101100 • (just concatenate the codewords.) • Decode: 001011101 = 0.0.101.1101 = aabe • (use the (right)binary tree below:) Tree for the fixed length codeword Tree for variable-length codeword Hash Tables

  13. Binary tree • Every nonleaf node has two children. • The fixed-length code in our example is not optimal. • The total number of bits required to encode a file is • f ( c ): the frequency (number of occurrences) of c in the file • dT(c): denote the depth of c’s leaf in the tree Hash Tables

  14. Constructing an optimal code • Formal definition of the problem: • Input:a set of characters C={c1, c2, …, cn}, each cC has frequency f[c]. • Output: a binary tree representing codewords so that the total number of bits required for the file is minimized. • Huffman proposed a greedy algorithm to solve the problem. Hash Tables

  15. c:12 b:13 a:45 d:16 0 1 f:5 e:9 14 (a) f:5 e:9 c:12 b:13 d:16 a:45 (b) Hash Tables

  16. a:45 0 1 c:12 b:13 d:16 0 1 a:45 f:5 e:9 0 1 1 0 c:12 b:13 d:16 0 1 f:5 e:9 14 14 30 25 25 (c) (d) Hash Tables

  17. a:45 0 1 0 100 1 0 1 1 0 a:45 c:12 b:13 d:16 0 1 0 1 f:5 e:9 0 1 1 0 c:12 b:13 d:16 14 14 30 30 0 1 55 55 25 25 f:5 e:9 (f) (e) Hash Tables

  18. HUFFMAN(C) 1 n:=|C| 2 Q:=C 3 for i:=1 to n-1 do 4 z:=ALLOCATE_NODE() 5 x:=left[z]:=EXTRACT_MIN(Q) 6 y:=right[z]:=EXTRACT_MIN(Q) 7 f[z]:=f[x]+f[y] 8 INSERT(Q,z) 9 return EXTRACT_MIN(Q) Hash Tables

  19. The Huffman Algorithm • This algorithm builds the tree T corresponding to the optimal code in a bottom-up manner. • C is a set of n characters, and each character c in C is a character with a defined frequency f[c]. • Q is a priority queue, keyed on f, used to identify the two least-frequent characters to merge together. • The result of the merger is a new object (internal node) whose frequency is the sum of the two objects. Hash Tables

  20. Time complexity • Lines 4-8 are executed n-1 times. • Each heap operation in Lines 4-8 takes O(lg n) time. • Total time required is O(n lg n). Note: The details of heap operation will not be tested. Time complexity O(n lg n) should be remembered. Hash Tables

  21. 10 0 1 c:6 b:9 d:11 e:4 a:6 Another example: e:4 a:6 c:6 b:9 d:11 Hash Tables

  22. 0 1 0 0 1 1 c:6 c:6 b:9 b:9 e:4 a:6 10 15 10 15 21 0 1 d:11 0 1 e:4 a:6 d:11 Hash Tables

  23. 0 1 36 10 15 21 0 1 0 1 d:11 c:6 b:9 0 1 e:4 a:6 Summary Huffman Code: Given a set of characters and frequency, you should be able to construct the binary tree for Huffman codes.Proofs for why this algorithm can give optimal solution are not required. Hash Tables

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