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Review of Chapter 5: Selection Trees

This review provides an overview of selection trees, including winner and loser trees, forest transformations, set representation, and binary tree operations. It also discusses the satisfiability problem, counting binary trees, and stack permutations.

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Review of Chapter 5: Selection Trees

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  1. Review of Chapter 5 張啟中

  2. Selection Trees • Selection trees can merge k ordered sequences (assume in non-decreasing order) into a single sequence easily. • Two kinds of selection trees • Winner trees • A winner tree is a complete binary tree • The root represents the smallest node in the tree. • Each leaf node represents the first record in the corresponding run. • Each non-leaf node in the tree represents the winner of its right and left subtrees. • Loser tress • A loser tree is a complete binary tree. • Each nonleaf node retains a pointer to the loser is called a loser tree. • Each leaf node represents the first record in the corresponding run. • An additional node, node 0, has been added to represent the overall winner of the tournament.

  3. Winner Tree (k=8) 1 6 2 3 6 8 4 5 7 6 9 6 8 17 8 9 10 11 12 13 14 15 10 9 20 6 8 9 90 17 15 20 20 15 15 11 95 18 16 38 30 25 50 16 99 20 28 run2 run1 run4 run8 run6 run5 run3 run7

  4. Winner Tree (k=8) O(nlogk) 1 8 2 3 9 8 4 5 7 6 9 15 8 17 8 9 10 11 12 13 14 15 10 9 20 15 8 9 90 17 15 20 20 25 15 11 95 18 16 38 30 28 50 16 99 20 run2 run1 run4 run8 run6 run5 run3 run7

  5. Loser Tree Overall winner 0 6 1 8 2 3 9 17 4 5 7 6 10 20 9 90 8 9 10 11 12 13 14 15 10 9 20 6 8 9 90 17 15 20 20 15 15 11 95 18 16 38 30 25 50 16 99 20 28 run2 run1 run4 run8 run6 run5 run3 run7

  6. Forests • Definition A forest is a set of n ≥ 0 disjoint trees. • When we remove the root of a tree, we’ll get a forest. A E G B C D F H I

  7. Transforming A Forest Into A Binary Tree • Definition If T1, …, Tn is a forest of trees, then the binary tree corresponding to this forest, denoted by B(T1, …, Tn) • is empty if n = 0 • has root equal to root (T1); has left subtree equal to B(T11, T12,…, T1m), where T11, T12,…, T1m are the subtrees of root (T1); and has right subtree B(T2, …, Tn).

  8. Transforming A Forest Into A Binary Tree A B E C F G D H I

  9. Forest Traversals • Preorder • Inorder • Postorder (not natural) • Level-order

  10. The Satisfiability Problem • Expression Rules • A variable is an expression • If x and y are expressions then are expressions • Parentheses can be used to alter the normal order of evaluation, which is not before and before or. • The satisfiablitity problem for formulas of proposition calculus asks if there is an assignment of values to the variables that causes the values of the expression to be true. • The satisfiablitity problem is NP-Complete problem.

  11. The Satisfiability Problem x3 x1 x3 O(2n) x2 x1

  12. Set Representation • Trees can be used to represent sets. • Pairwise Disjoint Sets If Si and Sj, i≠j, are two sets, then there is no element that is in both Si and Sj. • Set Operations • Disjoint set union If Si and Sj are two disjoint sets, then their union Si∪Sj = {all elements x such that x is in Si or Sj}. • Find(i) Find the set containing element i.

  13. Set Representation n=10 0 2 4 6 7 8 3 5 9 1 S1= {0,6,7,8} S2 = {1, 4, 9} S3={2,3,5}

  14. Disjoint Set Union S1 U S2 4 0 OR 9 4 0 1 6 7 8 9 6 7 8 1

  15. Data Representation for S1, S2, S3 Set Name Pointer S1 0 2 4 S2 6 7 8 3 5 1 9 S3

  16. Array Representation of S1, S2, S3 i [0] [1] [2] [3] [4] [5] [6] [7] [8] [9] parent -1 -1 0 0 0 4 -1 4 2 2

  17. Degenerate Tree union(0, 1), find(0) n-1 Union operation O(n) union(1, 2), find(1) n-2 Find operation O(n2) union(n-2, n-1), find(n-1) 0

  18. Improve the performance of Set Union and Find Algorithms • Weighting Rule[Weighting rule for union(I, j)] • If the number of nodes in the tree with root i is less than the number in the tree with root j, then make j the parent of i; otherwise make i the parent of j. • Collapsing Rule • If j is a node on the path from i to its root and parent[i]≠ root(i), then set parent[j] to root(i).

  19. Set Union with The Weighting Rule 0 1 n-1 0 2 n-1 0 3 n-1 1 1 2 0 0 4 n-1 1 2 3 1 n-1 2 3

  20. Set Find with Collapsing Rule [-8] [-8] 0 0 4 2 2 6 7 1 1 4 5 3 6 5 3 After collapsing 7 Before collapsing

  21. Equivalence Class [-1] [-1] [-1] [-1] [-1] [-1] [-1] [-1] [-1] [-1] [-1] [-1] 0 1 2 3 4 5 6 7 8 9 10 11 (a) Initial trees [-2] [-2] [-2] [-2] [-1] [-1] [-1] [-1] 0 3 6 8 2 5 7 11 4 1 10 9 (b) Height-2 trees following 0≡4, 3≡1, 6≡10, and 8≡9

  22. Equivalence Class (C) Trees following 7≡4, 6≡8, 3≡5, and 2≡11 [-2] [-4] [-3] [-3] 6 2 3 0 11 8 10 5 7 1 4 9 [-3] [-3] [-4] 3 0 6 5 1 8 10 4 7 2 11 (d) Thees following 11≡0 9

  23. Counting Binary Trees • Problem • Determine the number of distinct binary trees having n nodes. • Determine the number of distinct permutations of the numbers from 1 through n obtainable by a stack. • Determine the number of distinct ways of multiplying n+1 matrices.

  24. Distinct binary trees • n=0 or n=1  1 binary tree • n=2  2 distinct binary trees. • n=3  5 distinct binary trees (自己練習畫) bn bi bn-i-1

  25. Stack Permutations • How many permutations can we obtain by passing the numbers 1 through n through stack ? See chapter 3 about stack. • For example, the numbers 1, 2, 3 (1,2,3) (1,3,2) (2,1,3) (2,3,1) (3,2,1) but (3,1,2) • The recursive formula is bn = b0bn-1 + b1bn-2 + …… + bn-2b1+ bn-1b0

  26. Construct The Binary Tree from Preorder and Inorder Sequence • Give preorder and inorder sequence as follows. preorder sequence A B C D E F G H I Inorder sequence B C A E D G H F I • Problem • Does such a pair of sequences uniquely define a binary tree ? (Can this pair of sequences come from more than one binary tree?) • Conclusion • Every binary tree has a unique pair of preorder / inorder sequences.

  27. Construct The Binary Tree from Preorder and Inorder Sequence A A B, C D, E, F, G, H, I B D C F E A I G B H D, E, F, G, H, I C

  28. Inorder and Preorder Permutations 1 A 2 4 B D 3 6 5 C F E 9 7 I G 8 H Preorder: 1, 2, 3, 4, 5, 6, 7, 8, 9 Inorder: 2, 3, 1, 5, 4, 7, 8, 6, 9

  29. Stack Permutations Preorder permutation 1, 2, 3 1 1 1 1 1 2 3 2 2 2 2 3 3 3 3 (1, 2, 3) (1, 3, 2) (2, 1, 3) (2, 3, 1) (3, 2, 1) Inorder permutations

  30. Stack Permutations The number of distinct Binary Trees with n nodes ≡ Inorder permutations obtainable from binary trees having the preorder permutations, 1,2,3,….,n ≡ The number of distinct permutations by passing 1..n through stack

  31. Matrix Multiplication • Computing the product of n matrices M1 * M2 * … * Mn • By matrix multiplication associative law, we can perform these multiplications in any order. • For example, n=3 (n=4 自己練習) (M1 * M2) * M3 M1 * (M2 * M3) • The number of distinct ways to obtain M1 * M2 * … * Mn

  32. Number of Distinct Binary Trees • Let which is the generating function for the number of binary trees. • By the recurrence relation we get

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