Dictionaries A Dictionary stores keyelement pairs, called items . - PowerPoint PPT Presentation

Dictionaries A Dictionary stores key–element pairs, called items . Several elements might have the same key. Provides three methods: Inf 2B: AVL Trees I findElement ( k ) : If the dictionary contains an item with key k , then return its element; otherwise return the special Lecture 5 of ADS thread element NO_SUCH_KEY. I insertItem ( k , e ) : Insert an item with key k and element e . Kyriakos Kalorkoti I removeItem ( k ) : If the dictionary contains an item with key k , then delete it and return its element; otherwise return School of Informatics NO_SUCH_KEY. University of Edinburgh Assumption: we have a total order on keys (always the case in applications). Note: We are concerned entirely with fast access and storage so focus on keys. : : : : ADT Dictionary & its implementations Binary Search Trees Abstract definition: A binary tree is either empty or has a root vertex with a left and a right child each of which is a tree. List implementation: I Recursive datatype definition. Θ ( 1 ) time for InsertItem ( k , e ) but Θ ( n ) for findElement ( k ) and So every vertex v , either: removeItem ( k ) . (i) has two children ( v is an internal vertex ), or HashTable implementation (with Bucket Arrays): (ii) has no children ( v is a leaf ). Good average-case performance for n = Ω ( N ) . An internal vertex v has a left child and a right child which Worst-case running time: is InsertItem ( k , e ) Θ ( 1 ) , might be another internal vertex or a leaf. findElement ( k ) and removeItem ( k ) are both Θ ( n ) . A near leaf is an internal vertex with one or both children being Binary Search Tree implem. (without Balancing): leaves. Good in the average-case—about Θ ( lg n ) for all operations. Worst-case running time: Θ ( n ) for all operations. Definition A tree storing ( key , element ) pairs is a Binary Search Tree if for Balanced Binary search trees: every internal vertex v , the key k of v is: Worst-case is Θ ( lg n ) for all operations. I greater than or equal to every key in v ’s left subtree , and I less than or equal to every key in v ’s right subtree. : : : :

Key parameter for runtimes: height Binary Search Trees for Dictionary Leaves are kept empty. I Given any vertex v of a tree T and a leaf there is a unique path form the vertex to the leaf: I length of path defined as number of internal vertices. Algorithm findElement ( k ) I The height of a vertex is the maximum length over all 1. if isEmpty( T ) then return NO_SUCH_KEY paths from it to leaves. 2. else I The height of a tree is the height of the root. 3. u root 4. while (( u is not null) and u . key 6 = k ) do I Note that if v has left child l and right child r then 5. if ( k < u . key ) then u u . left height ( v ) = 1 + max { height ( l ) , height ( r ) } . 6. else u u . right 7. od I If we insert v r along the path v 1 , v 2 , . . . , v r then only the 8. if ( u is not null) and u . key = k then return u . elt heights of v 1 , v 2 , . . . , v r might be affected, all other vertices 9. else return NO_SUCH_KEY keep their previous height. findElement runs in O ( h ) time, where h is height. : : : : Binary Search Trees Binary Search Trees for Dictionary Algorithm insertItemBST ( k , e ) 1. Perform findElement ( k ) to find the “right" place for an item 4 24 with key k (if it finds k high in the tree, walk down to the “near-leaf” with largest key no greater than k ). 3 16 43 3 2. Neighbouring leaf vertex u becomes internal vertex, u . key k , u . elt e . 0 2 55 2 21 1 24 4 24 0 0 0 1 18 45 67 1 1 3 16 43 3 0 0 0 0 0 0 0 2 55 2 21 1 24 0 0 0 18 1 45 67 1 1 0 0 0 0 0 0 : : : :

Binary Search Trees for Dictionary Worst-case running time Algorithm removeItemBST ( k ) 1. Perform findElement ( k ) on the tree to get to vertex t . 2. if we find t with t . key = k , 3. then remove the item at t , set e = t . elt . Theorem: For the binary search tree implementation of 4. Let u be “near-leaf" closest to k . Move u ’s item up to t . Dictionary, all methods ( findElement , insertItemBST , 5. else return NO_SUCH_KEY removeItemBST ) have asymptotic worst-case running time 4 Θ ( h ) , where h is the height of the tree. (can be Θ ( n ) ). 24 16 43 3 3 0 2 21 1 24 55 2 0 0 0 1 18 45 67 1 1 0 0 0 0 0 0 : : : : AVL Trees (G.M. Adelson-Velsky & E.M. Landis, 1962) Not an AVL tree: 4 24 1. A vertex of a tree is balanced if the heights of its children 3 16 43 3 differ by at most 1. 0 2. An AVL tree is a binary search tree in which all vertices are 2 55 2 21 1 24 balanced. 0 0 0 1 18 45 67 1 1 0 0 0 0 0 0 : : : :

An AVL tree The height of AVL trees Theorem: The height of an AVL tree storing n items is O ( lg ( n )) . Corollary: The running time of the binary search tree methods findElement , insertItem , removeItem is O ( lg ( n )) on an AVL 24 tree. Let n ( h ) denote minimum number of items stored in an AVL 43 16 tree of height h . So n ( 1 ) = 1, n ( 2 ) = 2, n ( 3 ) = 4. 55 10 21 24 Claim: n ( h ) > 2 h / 2 � 1. n ( h ) 1 + n ( h � 1 ) + n ( h � 2 ) � 18 45 67 h − 1 h − 2 2 � 1 + 2 2 � 1 1 + 2 > ( 2 − 1 h 2 + 2 − 1 ) 2 2 � 1 = h 2 � 1 . 2 > Problem: After we apply insertItem or removeItem to an AVL tree, the resulting tree might no longer be an AVL tree. : : : : Example Example (cont’d) 24 24 43 16 16 43 10 21 24 55 10 21 24 55 18 45 67 18 45 67 60 AVL tree. INSERT 60 not AVL now . . . : : : :

Example (cont’d) Example (cont’d) 24 24 z 16 43 y 16 55 y 55 10 21 24 z x 10 21 43 67 Y x 18 45 67 18 24 45 60 X W 60 V Now is AVL tree. INSERT 44 We can rotate . . . : : : : Example (cont’d) Restructuring I z unbalanced vertex of minimal height 24 I y child of z of larger height I x child of y of larger height (exists because 1 ins/del 55 16 unbalanced the tree). I V , W subtrees rooted at children of x 10 21 43 67 I X subtree rooted at sibling of x I Y subtree rooted at sibling of y 18 24 45 60 44 Then height ( V ) � 1  height ( W )  height ( V ) + 1 max { height ( V ) , height ( W ) } = height ( X ) AVL tree. max { height ( V ) , height ( W ) } = height ( Y ) . : : : :

A clockwise single rotation An anti-clockwise single rotation z y z y y x z z x y x x Y Y V V Y W X X X W Y X V V W W (a) (b) (a) (b) : : : : An anti-clockwise clockwise double rotation A clockwise anti-clockwise double rotation x z z x y y z y z y rot 2 rot 2 x x Y Y rot 1 rot 1 X V V X V V Y W W Y X X W W (a) (b) (a) (b) : : : :

Rotations The insertion algorithm After an InsertItem () : We can always rebalance using just one single rotation or one double rotation (only 2x2 cases in total). Algorithm insertItem ( k , e ) single rotation: 1. Insert ( k , e ) into the tree with insertItemBST. We make y the new root (of rebalancing subtree), z moves Let u be the newly inserted vertex. down, and the X subtree crosses to become 2nd child of z 2. Find first unbalanced vertex z on the path from u to root. (with X as sibling). 3. if there is no such vertex, 4. then return double rotation: 5. else Let y and x be child, grandchild of z on z ! u path. We make x the new root, y and z become its children, and the 6. Apply the appropriate rotation to x , y , z . return two subtrees of x get split between each side. Θ ( 1 ) time for a single or double rotation. : : : : Example (cont’d) Example (cont’d) 24 24 z 55 16 55 16 y 10 21 43 67 21 43 67 x 18 24 45 60 18 24 45 60 44 44 AVL tree. REMOVE 10. Not AVL tree . . . We rotate : : : :

Example (cont’d) Example (cont’d) z 24 x 43 y 18 55 z y 24 55 16 21 x 43 67 18 24 45 67 44 24 45 60 60 16 21 44 AVL tree again. Still not AVL . . . We rotate again. : : : : Rotations The removal algorithm Algorithm removeItem ( k ) 1. Remove item ( k , e ) with key k from tree using removeItemBST. After a removeItem () : Let u be leaf replacing removed vertex. We may need to re-balance “up the tree”. 2. while u is not the root do This requires O ( lg n ) rotations at most, each takes O ( 1 ) time. 3. let z be the parent of u 4. if z is unbalanced then 5. do the appropriate rotation at z 6. let u be the parent of u 7. return e : : : :