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 Complexity Of Dictionary Operationsget(), put() and remove()

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

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

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

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

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

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

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

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

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.

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

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

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

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

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.

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).

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

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.

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. = +

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

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

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

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)

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

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.

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

-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

Dynamic Dictionaries