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FIT2004 Algorithms & Data Structures L9: Balanced Trees Prepared by: Bernd Meyer from lecture materials © 2004 Goodrich & Tamassia February 2007

3 prepared from lecture material © 2004 Goodrich & Tamassia ADT: Dictionary A dictionary is a searchable collection of key-value pairs ADT dictionary sorts dict, key, elem, bool; ops empty: -> dict ; insert: dict x key x elem -> dict; delete: dict x key -> dict; contains: dict x key -> bool; retrieve: dict x key -> elem; isempty: dict -> bool; etc… we want to implement a dictionary: –by lists --- bad idea: O(n) –by (binary) trees --- a good idea? O(log n)???

4 prepared from lecture material © 2004 Goodrich & Tamassia Search Tables A search table is a dictionary implemented by means of a sorted sequence –We store the items of the dictionary in an array-based sequence, sorted by key Performance: –find takes O(log n) time, using binary search –insert takes O(n) time since in the worst case we have to shift O(n) items to make room for the new item –remove take O(n) time since in the worst case we have to shift O(n) items to compact the items after the removal The lookup table is effective only for dictionaries of small size or for dictionaries on which searches are the most common operations, while insertions and removals are rarely performed (e.g., credit card authorizations)

5 prepared from lecture material © 2004 Goodrich & Tamassia Binary Search (Revision) Binary search can perform operation find(k) on a dictionary implemented by means of an array-based sequence, sorted by key –at each step, the number of candidate items is halved –terminates after O(log n) steps Example: find(7) m l h m l h m l h l  m  h

6 prepared from lecture material © 2004 Goodrich & Tamassia Binary Search Trees A binary search tree is a binary tree storing keys (or key-value entries) at its internal nodes and satisfying the following property: –Let u, v, and w be three nodes such that u is in the left subtree of v and w is in the right subtree of v. We have key(u)  key(v)  key(w) External nodes do not store items Which traversal of the tree enumerates the keys in increasing order?

7 prepared from lecture material © 2004 Goodrich & Tamassia Binary Search Trees   

8 prepared from lecture material © 2004 Goodrich & Tamassia Search To search for a key k, we trace a downward path starting at the root The next node visited depends on the outcome of the comparison of k with the key of the current node If we reach a leaf, the key is not found and we return notFound Example: find(4) Algorithm TreeSearch(k, v) if T.isExternal (v) return v if k  key(v) return TreeSearch(k, T.left(v)) else if k  key(v) return v else { k  key(v) } return TreeSearch(k, T.right(v))   

9 prepared from lecture material © 2004 Goodrich & Tamassia Insertion To perform operation insert(k, o), we search for key k (using TreeSearch) Assume k is not already in the tree, and let let w be the leaf reached by the search We insert k at node w and expand w into an internal node Example: insert    w w

10 prepared from lecture material © 2004 Goodrich & Tamassia Deletion To perform operation remove( k ), we search for key k Assume key k is in the tree, and let let v be the node storing k If node v has a leaf child w, we remove v and w from the tree with operation removeExternal( w ), which removes w and its parent Example: remove v w  

11 prepared from lecture material © 2004 Goodrich & Tamassia Deletion (cont.) We consider the case where the key k to be removed is stored at a node v whose children are both internal –we find the internal node w that follows v in an inorder traversal –we copy key(w) into node v –we remove node w and its left child z (which must be a leaf) by means of operation removeExternal( z ) Example: remove 3 Why is this correct? v w z v 2

12 prepared from lecture material © 2004 Goodrich & Tamassia Performance Consider a dictionary with n items implemented by means of a binary search tree of height h –the space used is O(n) –methods find, insert and remove take O(h) time The height h is O(n) in the worst case and O(log n) in the best case

13 prepared from lecture material © 2004 Goodrich & Tamassia Recall: Average Binary Tree Depth What is the depth of an average binary search tree? –generate by insertion only, all permutations equally likely: O(log n) --- we will show this –generated by insertion and deletion, very large sequences  (  n) --- very hard to show J. Culberson The Effects of Updates in Binary Search Trees. 17th annual ACM Symposium on Theory of Computing. The tree degenerates: This is not good as all operations become more expensive Solution: Self-adjusting trees that maintain their balance using specialize (more expensive) update operations: next lecture. randomly generated, 500 inserts after 250,000 insert/delete pairs

14 prepared from lecture material © 2004 Goodrich & Tamassia Dynamic Self-balancing Trees We cannot fully rebalance a binary tree after every operation. This is too costly. Thus, we cannot keep a binary tree perfectly balanced at all times. As an alternative we will try to relax the requirement in three different ways and hope that we still reach O(log n) access times. We will try to keep it –almost balanced at all times (AVL Trees) –almost balanced most of the time (Splay Trees) –perfectly balanced at all times but allow it to be it non-binary (2-4-Tree, …)

15 prepared from lecture material © 2004 Goodrich & Tamassia AVL Trees AVL trees are balanced. An AVL Tree is a binary search tree such that for every internal node v of T, the heights of the children of v can differ by at most 1. An example of an AVL tree with heights shown next to the nodes

16 prepared from lecture material © 2004 Goodrich & Tamassia Height of an AVL Tree Fact: The height of an AVL tree storing n keys is O(log n). Proof: Let us bound n(h): the minimum number of internal nodes of an AVL tree of height h. We easily see that n(1) = 1 and n(2) = 2 For h > 2, an AVL tree of height h contains the root node, one AVL subtree of height h-1 and another of height h-2. That is, n(h) = 1 + n(h-1) + n(h-2) Knowing n(h-1) > n(h-2), we get n(h) > 2n(h-2). So n(h) > 2n(h-2), n(h) > 4n(h-4), n(h) > 8n(h-6), … (by induction), n(h) > 2 i n(h-2i) Solving the base case we get: n(h) > 2 h/2-1 Taking logarithms: h < 2log n(h) +2 Thus the height of an AVL tree is O(log n) 3 4 n(1) n(2)

17 prepared from lecture material © 2004 Goodrich & Tamassia Insertion in an AVL Tree Insertion is as in a binary search tree Always done by expanding an external node. Example: w b=x a=y c=z before insertionafter insertion

18 prepared from lecture material © 2004 Goodrich & Tamassia Trinode Restructuring let (a,b,c) be an inorder listing of x, y, z perform the rotations needed to make b the topmost node of the three b=y a=z c=x T0T0 T1T1 T2T2 T3T3 b=y a=z c=x T0T0 T1T1 T2T2 T3T3 c=y b=x a=z T0T0 T1T1 T2T2 T3T3 b=x c=ya=z T0T0 T1T1 T2T2 T3T3 case 1: single rotation (a left rotation about a) case 2: double rotation (a right rotation about c, then a left rotation about a) (other two cases are symmetrical)

19 prepared from lecture material © 2004 Goodrich & Tamassia Trinode Restructuring Algorithm note that given the definitions above both types can be expressed let z be the first imbalanced node, y its higher child and x the higher child of y let a,b,c be an inorder enumeration of x,y,z let T0, …, T3 be the left-to-right listing of the subtrees rooted at x, y, z Alg trinode-restructure 1. replace subtree rooted at z with subtree rooted at b 2. let a be the left child of b, and let T0 (T1) be the left (right) child of a 3. let c be the right child of b and T2 (T3) be the left (right) child of c

20 prepared from lecture material © 2004 Goodrich & Tamassia Insertion Example, continued T 0 T 1 T 3 unbalanced......balanced

21 prepared from lecture material © 2004 Goodrich & Tamassia Restructuring (as Single Rotations) Single Rotations: T0T0 T1T1 T2T2 T3T3

22 prepared from lecture material © 2004 Goodrich & Tamassia Restructuring (as Double Rotations) double rotations: T0T0 T1T1 T2T2 T3T3 T0T0 T1T1 T2T2 T3T3

23 prepared from lecture material © 2004 Goodrich & Tamassia Removal in an AVL Tree Removal begins as in a binary search tree, which means the node removed will become an empty external node. Its parent, w, may cause an imbalance. Example: before deletion of 32after deletion

24 prepared from lecture material © 2004 Goodrich & Tamassia Rebalancing after a Removal Let z be the first unbalanced node encountered while travelling up the tree from w. Also, let y be the child of z with the larger height, and let x be the child of y with the larger height. If both subtrees have same height at y chose x to be the subtree on the same side of y as y’s side on z. We perform restructure(x) to restore balance at z. As this restructuring may upset the balance of another node higher in the tree, we must continue checking for balance until the root of T is reached w c=x b=y a=z

25 prepared from lecture material © 2004 Goodrich & Tamassia Running Times for AVL Trees a single restructure is O(1) –using a linked-structure binary tree find is O(log n) –height of tree is O(log n), no restructures needed insert is O(log n) –initial find is O(log n) –Restructuring at the node, restoring heights remove is O(log n) –initial find is O(log n) –Restructuring up the tree, maintaining heights is O(log n)

26 prepared from lecture material © 2004 Goodrich & Tamassia Splay Trees Splay Trees are Binary Search Trees BST Rules: –entries stored only at internal nodes –keys stored at nodes in the left subtree of v are less than or equal to the key stored at v –keys stored at nodes in the right subtree of v are greater than or equal to the key stored at v An inorder traversal will return the keys in order (20,Z) (37,P)(21,O) (14,J) (7,T) (35,R)(10,A) (1,C) (1,Q) (5,G) (2,R) (5,H) (6,Y) (5,I) (8,N) (7,P) (36,L) (10,U) (40,X)

27 prepared from lecture material © 2004 Goodrich & Tamassia Searching in a Splay Tree Start same as BST Search proceeds down the tree to found item or an external node. Example: Search for time with key 11. (20,Z) (37,P)(21,O) (14,J) (7,T) (35,R)(10,A) (1,C) (1,Q) (5,G) (2,R) (5,H) (6,Y) (5,I) (8,N) (7,P) (36,L) (10,U) (40,X)

28 prepared from lecture material © 2004 Goodrich & Tamassia Example Searching in a BST, continued search for key 8, ends at an internal node. (20,Z) (37,P)(21,O) (14,J) (7,T) (35,R)(10,A) (1,C) (1,Q) (5,G) (2,R) (5,H) (6,Y) (5,I) (8,N) (7,P) (36,L) (10,U) (40,X)

29 prepared from lecture material © 2004 Goodrich & Tamassia Rotations after Every Operation (Even Search) new operation splay: moves a node to the root using rotations right rotation makes the left child x of a node y into y’s parent; y becomes the right child of x left rotation makes the right child y of a node x into x’s parent; x becomes the left child of y y x T1T1 T2T2 T3T3 y x T1T1 T2T2 T3T3 y x T1T1 T2T2 T3T3 y x T1T1 T2T2 T3T3 (structure of tree above y is not modified) (structure of tree above x is not modified) a right rotation about ya left rotation about x

30 prepared from lecture material © 2004 Goodrich & Tamassia Splaying: “ x is a left-left grandchild” means x is a left child of its parent, which is itself a left child of its parent p is x ’s parent; g is p ’s parent is x the root? stop is x a child of the root? right-rotate about the root left-rotate about the root is x the left child of the root? is x a left-left grandchild? is x a left-right grandchild? is x a right-right grandchild? is x a right-left grandchild? right-rotate about g, right-rotate about p left-rotate about g, left-rotate about p left-rotate about p, right-rotate about g right-rotate about p, left-rotate about g start with node x no yes no yes zig-zig zig-zag zig-zig zig

31 prepared from lecture material © 2004 Goodrich & Tamassia Visualizing the Splaying Cases zig-zag y x T2T2 T3T3 T4T4 z T1T1 y x T2T2 T3T3 T4T4 z T1T1 y x T1T1 T2T2 T3T3 z T4T4 zig-zig y z T4T4 T3T3 T2T2 x T1T1 zig x w T1T1 T2T2 T3T3 y T4T4 y x T2T2 T3T3 T4T4 w T1T1

32 prepared from lecture material © 2004 Goodrich & Tamassia Splaying Example let x = (8,N) –x is the right child of its parent, which is the left child of the grandparent –left-rotate around p, then right- rotate around g (20,Z) (37,P)(21,O) (14,J) (7,T) (35,R)(10,A) (1,C) (1,Q) (5,G) (2,R) (5,H) (6,Y) (5,I) (8,N) (7,P) (36,L) (10,U) (40,X) x g p (10,A) (20,Z) (37,P)(21,O) (35,R) (36,L) (40,X) (7,T) (1,C) (1,Q) (5,G) (2,R) (5,H) (6,Y) (5,I) (14,J) (8,N) (7,P) (10,U) x g p (10,A) (20,Z) (37,P)(21,O) (35,R) (36,L) (40,X) (7,T) (1,C) (1,Q) (5,G) (2,R) (5,H) (6,Y) (5,I) (14,J) (8,N) (7,P) (10,U) x g p 1. (before rotating) 2. (after first rotation) 3. (after second rotation) x is not yet the root, so we splay again

33 prepared from lecture material © 2004 Goodrich & Tamassia Splaying Example, Continued now x is the left child of the root –right-rotate around root (10,A) (20,Z) (37,P)(21,O) (35,R) (36,L) (40,X) (7,T) (1,C) (1,Q) (5,G) (2,R) (5,H) (6,Y) (5,I) (14,J) (8,N) (7,P) (10,U) x (10,A) (20,Z) (37,P)(21,O) (35,R) (36,L) (40,X) (7,T) (1,C) (1,Q) (5,G) (2,R) (5,H) (6,Y) (5,I) (14,J) (8,N) (7,P) (10,U) x 1. (before applying rotation) 2. (after rotation) x is the root, so stop

34 prepared from lecture material © 2004 Goodrich & Tamassia Example Result of Splaying tree might not be more balanced e.g. splay (40,X) –before, the depth of the shallowest leaf is 3 and the deepest is 7 –after, the depth of shallowest leaf is 1 and deepest is 8 (20,Z) (37,P)(21,O) (14,J) (7,T) (35,R)(10,A) (1,C) (1,Q) (5,G) (2,R) (5,H) (6,Y) (5,I) (8,N) (7,P) (36,L) (10,U) (40,X) (20,Z) (37,P) (21,O) (14,J) (7,T) (35,R) (10,A) (1,C) (1,Q) (5,G) (2,R) (5,H) (6,Y) (5,I) (8,N) (7,P) (36,L) (10,U) (40,X) (20,Z) (37,P) (21,O) (14,J) (7,T) (35,R) (10,A) (1,C) (1,Q) (5,G) (2,R) (5,H) (6,Y) (5,I) (8,N) (7,P) (36,L) (10,U) (40,X) before after first splay after second splay

35 prepared from lecture material © 2004 Goodrich & Tamassia Splay Tree Definition a splay tree is a binary search tree where a node is splayed after it is accessed (for a search or update) –deepest internal node accessed is splayed –splaying costs O(h), where h is height of the tree – which is still O(n) worst-case >O(h) rotations, each of which is O(1)

36 prepared from lecture material © 2004 Goodrich & Tamassia Splay Trees & Ordered Dictionaries which nodes are splayed after each operation? use the parent of the internal node that was actually removed from the tree (the parent of the node that the removed item was swapped with) remove(k) use the new node containing the entry inserted insert(k,v) if key found, use that node if key not found, use parent of ending external node find(k) splay nodemethod

37 prepared from lecture material © 2004 Goodrich & Tamassia Amortized Analysis of Splay Trees “Amortized” means to balance the immediate cost of an operation with the future cost of further operations. –total cost = immediate cost + future cost –(note: future costs could be negative!) Cost (run time) of each operation is proportional to the cost for splaying. –splay at depth d is O(d), so is find, insert, delete >The immediate cost of an individual rotation are clear: –Costs: zig = $1, zig-zig = $2, zig-zag = $2. >splay at depth d performs at most d/2 zig-zig/zig-zag + 1 zig We only need to worry about splaying cost!

38 prepared from lecture material © 2004 Goodrich & Tamassia Amortized Analysis of Splay Trees To get a handle on the future cost we imagine that we maintain a virtual account at each node. We can pay into these accounts or withdraw from them. We will show that we can maintain a balance of $r(v) at each vertex v by paying a total of $O(log(n)) for each operation. Thus the amortized cost (run time) of each operation is O(log(n))

39 prepared from lecture material © 2004 Goodrich & Tamassia Invariant for the Analysis We define s(v) as the size of the subtree rooted in v (number of nodes) and r(v)=log 2 (s(v)) “rank of v” Invariant: the balance at each vertex is $r(v). –wisely chosen so that >the accounts are always positive >we don’t have to cheat by making an advance payment for an empty tree ie that we do not start the amortization with accounts that have money in them >we can show the amortized cost to be O(log n) The future cost of a rotation is the cost of maintaining this invariant.

40 prepared from lecture material © 2004 Goodrich & Tamassia Cost per zig Let r(x) be the rank of x before the rotation, Let r’(x) the rank after the rotation. Future cost of a zig at x is at most rank’(x) - rank(x): future cost = r’(x) + r’(y) - r (y) - r (x) < r’(x) - r (x). Total cost of zig at x is immediate cost + future cost: cost < 1 + (r’(x) - r (x)) < 1 + 3*(r’(x) - r (x)) zig x w T1T1 T2T2 T3T3 y T4T4 y x T2T2 T3T3 T4T4 w T1T1

41 prepared from lecture material © 2004 Goodrich & Tamassia Cost per zig-zig Future cost of a zig-zig at x is at most future cost < 3(r’(x) - r(x)) - 2. The proof uses the same thinking as the zig case, but is somewhat more complex (see Goodrich & Tamassia, proposition 9.2, page 440, or Weiss Sec or Kingston Sec 6.6) Total cost of zig-zig at x is immediate cost + future cost: cost < 2 + 3(r’(x) - r(x)) - 2 = 3(r’(x) - r(x)) y x T1T1 T2T2 T3T3 z T4T4 zig-zig y z T4T4 T3T3 T2T2 x T1T1

42 prepared from lecture material © 2004 Goodrich & Tamassia Cost per zig-zag Same as cost for a zig-zig: Future cost of a zig-zag or zig-zag at x is at most future cost < 3(r’(x) - r(x)) - 2. Total cost of zig-zig or zig-zag at x is immediate cost + future cost: cost < 2 + 3(r’(x) - r(x)) - 2 = 3(r’(x) - r(x)) zig-zag y x T2T2 T3T3 T4T4 z T1T1 y x T2T2 T3T3 T4T4 z T1T1

43 prepared from lecture material © 2004 Goodrich & Tamassia Cost of Splaying (= cost of find) Splaying a node x means to rotate it all the way up to the root (m rotations) The last operation is a zig, all others are zig-zig or zig-zag. total splay cost is the sum of (m-1) zig-zig/zig-zag plus a final zig Let r m (x) be the rank of x just after rotation step m

44 prepared from lecture material © 2004 Goodrich & Tamassia Cost of Deletion The tree shrinks (by one node) the total variation of all r(t) is negative we don’t have to worry about any extra payment for maintaining the invariant

45 prepared from lecture material © 2004 Goodrich & Tamassia Cost of Insertion inserting node v increases the ranks of all of v’s ancestors let v=v o, v 1 be its parent, and v i …v d all the ancestors on the way to the root. let n(v i ) be the size of the subtree rooted at v i and r(v i ) the corresponding rank (before insert) let n’(v i ) and r’(v i ) be the same values after the insert. –n’(v i ) = n(v i ) +1 –n(v i ) +1 ≤ n(v i+1 ) We have so the total variation is

46 prepared from lecture material © 2004 Goodrich & Tamassia Amortized Cost we now amortize all the costs for m operations (find, insert, delete) –start with empty tree –let n be the total number of insertions (maximum number of keys) –let n i be the number of keys in the tree after operation I the total cost for all these operation is thus the amortized cost of each operation is O(log n)

47 prepared from lecture material © 2004 Goodrich & Tamassia Performance of Splay Trees Thus, amortized cost of any splay operation is O(log n). Splay trees can adapt to perform searches on frequently-requested items much faster than O(log n) on average due to the “move- to-root” characteristics.