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Dynamic Dictionaries

Dynamic Dictionaries. Primary Operations: get(key) => search put(key, element) => insert remove(key) => delete Additional operations: nearestGE(key), nearestLE(key) range(low,high) ascend() get(index) remove(index). n is number of elements in dictionary.

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Dynamic Dictionaries

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  1. Dynamic Dictionaries • Primary Operations: • get(key) => search • put(key, element) => insert • remove(key) => delete • Additional operations: • nearestGE(key), nearestLE(key) • range(low,high) • ascend() • get(index) • remove(index)

  2. n is number of elements in dictionary Complexity Of Dictionary Operationsget(), put() and remove()

  3. D is number of buckets Complexity Of Other Operationsascend(), get(index), remove(index)

  4. 20 10 40 6 15 30 25 2 8 The Operation put() 35 Put a pair whose key is 35.

  5. Three cases: • Element is in a leaf. • Element is in a degree 1 node. • Element is in a degree 2 node. The Operation remove()

  6. 20 10 40 6 15 30 18 25 35 2 8 7 Remove From A Leaf Remove a leaf element. key = 7

  7. 20 10 40 6 15 30 18 25 35 2 8 7 Remove From A Degree 1 Node Remove from a degree 1 node. key = 40

  8. 20 10 40 6 15 30 18 25 35 2 8 7 Remove From A Degree 1 Node (contd.) Remove from a degree 1 node. key = 15

  9. 20 10 40 6 15 30 18 25 35 2 8 7 Remove From A Degree 2 Node Remove from a degree 2 node. key = 10

  10. 20 Remove From A Degree 2 Node 10 40 6 15 30 18 25 35 2 8 Replace with largest key in left subtree (or smallest in right subtree). 7

  11. 20 Remove From A Degree 2 Node 10 40 6 15 30 18 25 35 2 8 Replace with largest key in left subtree (or smallest in right subtree). 7

  12. 20 Remove From A Degree 2 Node 8 40 6 15 30 18 25 35 2 8 Replace with largest key in left subtree (or smallest in right subtree). 7

  13. 20 Remove From A Degree 2 Node 8 40 6 15 30 18 25 35 2 8 7 Largest key must be in a leaf or degree 1 node.

  14. 20 10 40 6 15 30 18 25 35 2 8 7 Another Remove From A Degree 2 Node Remove from a degree 2 node. key = 20

  15. 20 Remove From A Degree 2 Node 10 40 6 15 30 18 25 35 2 8 7 Replace with largest in left subtree.

  16. 20 Remove From A Degree 2 Node 10 40 6 15 30 18 25 35 2 8 7 Replace with largest in left subtree.

  17. 18 Remove From A Degree 2 Node 10 40 6 15 30 18 25 35 2 8 7 Replace with largest in left subtree.

  18. 18 Remove From A Degree 2 Node 10 40 6 15 30 25 35 2 8 7 Complexity is O(height).

  19. Yet Other Operations • Priority Queue Motivated Operations: • find max and/or min • remove max and/or min • initialize • meld

  20. 20 10 40 6 15 30 25 2 8 Max And/Or Min • Follow rightmost path to max element. • Follow leftmost path to min element. • Search and/or remove => O(h) time.

  21. 20 10 40 6 15 30 25 2 8 Initialize • Sort n elements. • Initialize search tree. • Output in inorder (O(n)). • Initialize must take O(n log n) time, because it isn’t possible to sort faster than O(n log n).

  22. 5 10 1 12 6 15 10 17 7 2 8 6 17 8 2 15 7 12 1 5 Meld

  23. Meld And Merge • Worst-case number of comparisons to merge two sorted lists of size n each is 2n-1. • So, complexity of melding two binary search trees of size n each is W(n). • So, logarithmic time melding isn’t possible.

  24. 8 10 2 15 1 6 15 1 6 10 2 8 O(log n) Height Trees • Full binary trees. • Exist only when n = 2k –1. • Complete binary trees. • Exist for all n. • Cannot insert/delete in O(log n) time. = +

  25. Balanced Search Trees • Height balanced. • AVL (Adelson-Velsky and Landis) trees • Weight Balanced. • Degree Balanced. • 2-3 trees • 2-3-4 trees • red-black trees • B-trees

  26. AVL Tree • binary tree • for every node x, define its balance factor balance factor of x = height of left subtree of x – height of right subtree of x • balance factor of every node x is – 1, 0, or 1

  27. -1 1 1 Balance Factors -1 0 1 0 0 -1 0 This is an AVL tree. 0 0 0

  28. Height Of An AVL Tree The height of an AVL tree that has n nodes is at most 1.44 log2 (n+2). The height of every n node binary tree is at least log2 (n+1). log2 (n+1) <= height <=1.44 log2 (n+2)

  29. Proof Of Upper Bound On Height • Let Nh = min # of nodes in an AVL tree whose height is h. • N0 = 0. • N1 = 1.

  30. L R Nh, h > 1 • Both L and R are AVL trees. • The height of one is h-1. • The height of the other is h-2. • The subtree whose height is h-1 has Nh-1nodes. • The subtree whose height is h-2 has Nh-2nodes. • So, Nh =Nh-1 + Nh-2 + 1.

  31. Fibonacci Numbers • F0 = 0, F1 = 1. • Fi =Fi-1 + Fi-2 , i > 1. • N0 = 0, N1 = 1. • Nh =Nh-1 + Nh-2 + 1, i > 1. • Nh =Fh+2 – 1. • Fi ~ fi/sqrt(5). • f = (1 + sqrt(5))/2 • height <=1.44 log2 (n+2) follows

  32. -1 10 1 1 7 40 -1 0 1 0 45 3 8 30 0 -1 0 0 0 60 35 1 20 5 0 25 AVL Search Tree

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