Size Balanced Tree. Chen Qifeng (Farmer John) Zhongshan Memorial Middle School, Guangdong, China. December 29, 2006.

Transcription

1 Size Balanced Tree Chen Qifeng (Farmer John) Zhonghan Memorial Middle School, Guangdong, China December 9, 006 Abtract Thi paper preent a unique trategy for maintaining balance in dynamically changing Binary Search Tree that ha optimal expected behavior at wort. Size Balanced Tree i, a the name ugget, a Binary Search Tree (abbr. BST) kept balanced by ize. It i imple, efficient and veratile in every apect. It i very eay to implement and ha a traightforward decription and a urpriingly imple proof of correctne and runtime. It runtime matche that of the fatet BST known o far. Furthermore, it work much fater than many other famou BST due to the tendency of a perfect BST in practice. It upport not only typical primary operation but alo Select and Rank. Key Word And Phrae Size Balanced Tree SBT Maintain Thi paper i dedicated to the memory of Heaven.

2 1 Introduction Before preenting Size Balanced Tree it i neceary to explicate Binary Search Tree and rotation on BST, Left-Rotate and Right-Rotate. 1.1 Binary Search Tree Binary Search Tree i a ignificant kind of advanced data tructure. It upport many dynamic-et operation, including Search, Minimum, Maximum, Predeceor, Succeor, Inert and Delete. It can be ued both a a dictionary and a a priority queue. A BST i an organized binary tree. Every node in a BST contain two children at mot. The key for compare in a BST are alway tored in uch a way a to atify the binary-earch-tree property: Let x be a node in a binary earch tree. Then the key of x i not le than that in left ubtree and not larger than that in right ubtree. For every node t we ue the field of leftt and rightt to tore two pointer to it children. And we define keyt to mean the value of the node t for compare. In addition we add t, the ize of ubtree rooted at t, to keep the number of the node in that tree. Particularly we call 0 the pointer to an empty tree and 0=0. 1. Rotation In order to keep a BST balanced (not degenerated to be a chain) we uually change the pointer tructure through rotation to change the configuration, which i a local operation in a earch tree that preerve the binary-earch-tree property. Figure1.1: The operation Left-Rotate(x) tranform the configuration of the two node on the right into the configuration on the left by changing a contant number of pointer. The configuration on the left can be tranformed into the configuration on the right by the invere operation, Right-Rotate(y).

3 1..1 Peudocode Of Right-Rotate The Right-Rotate aume that the left child exit. Right-Rotate (t) 1 k leftt leftt rightk rightk t 4 k t 5 t leftt+rightt+1 6 t k 1.. Peudocode Of Left-Rotate The Left-Rotate aume that the right child exit. Left-Rotate (t) 1 k rightt rightt leftk leftk t 4 k t 5 t leftt+rightt+1 6 t k Size Balanced Tree Size Balanced Tree (abbr. SBT) i a kind of Binary Search Tree kept balanced by ize. It upport many dynamic primary operation in the runtime of O(logn): Inert(t,v) Inert a node whoe key i v into the SBT rooted at t. Delete(t,v) Delete a node whoe key i v from the SBT rooted at t. In the cae that no uch a node in the tree, delete the node earched at lat. Find(t,v) Find the node whoe key i v and return it. Rank(t,v) Return the rank of v in the tree rooted at t. In another word, it i one plu the number of the key which are le than v in that tree. Select(t,k) Return the node which i ranked at the kth poition. Apparently it include operation of Get-max and Get-min becaue Get-min i equivalent to Select(t,1) and Get-max i equivalent to Select(t,t) Pred(t,v) Return the node with maximum key which i le than v. Succ(t,v) Return the node with minimum key which i larger than v.

4 Commonly every node of a SBT contain key, left, right and extra but ueful updated field: it ize, which ha been defined in the former introduction. The kernel of a SBT i divided into two retriction on ize: For every node pointed by t in a SBT, we guarantee that Property(a): Property(b): right t left left t, right left t left t right right t, left right t Figure.1:The node L and R are left and right children of the node T. The Subtree A and B, C and D are left and right ubtree of the node L and R repectively. Correpond to propertie (a) and (b), A, B R & C, D L Maintain Aume that we need to inert a node whoe key i v into a BST. Generally we ue the following procedure to accomplih the miion. Simple-Inert (t,v) 1 If t=0 then t NEW-NODE(v) Ele 4 t t+1 5 If v<keyt then 6 Simple-Inert(leftt,v) 7 Ele 8 Simple-Inert(rightt,v)

5 After the execution of Simple-Inert propertie (a) and (b) are probably not atified. In that cae we need to maintain the SBT. The vital operation on a SBT i a unique procedure, Maintain. Maintain(t) i ued to maintain the SBT rooted at t. The hypothei that the ubtree of t are SBT i applied before the performance. Since propertie (a) and (b) are ymmetrical we only dicu the cae that retriction (a) i violated in detail. Cae 1: leftleftt>rightt In thi cae that A>R correpond to Figure.1 after Inert(leftt,v), we can execute the following intruction to repair the SBT: (1) Firt perform Right-Rotate(T). Thi operation tranform Figure.1 into Figure.1; Figure.1: All the node are defined the ame a in Figure.1. () And then ometime it occur that the tree i till not a SBT becaue C>B or D>B by any poibility. So it i neceary to implement Maintain(T). () In ucceion the configuration of the right ubtree of the node L might be changed. Becaue of the poibility of violating the propertie it i needful to run Maintain(L) again. Cae : rightleftt>rightt In thi cae that B>R correpond to Figure. after Inert(leftt,v), the maintaining i kind of complicated than that in cae 1. We can carry out the following operation for repair.

6 Figure.: All the node are defined the ame a in Figure.1 except E, F and B. E and F are the ubtree of the node B. (1) Firt perform Left-Rotate(L). After that Figure. wa tranformed to Figure. hown below; Figure.: All the node are defined the ame a in Figure.. () And then perform Right-Rotate(T). Thi performance reult in the tranformation from Figure. to following Figure.4.

7 Figure.4: All the node are defined the ame a in Figure.. () After (1) and (), the tree become to be very unpredictable. But luckily, in Figure.4 ubtree A, E, F and R are till SBT. So we can perform Maintain(L) and Maintain(T) to repair the ubtree of the node B. (4) After tep (), thoe ubtree are already SBT. But propertie (a) and (b) may be till violated on the node B. Thu we need perform Maintain(B) again. Cae : rightrightt>leftt Thi cae i ymmetrical with cae 1. Cae 4: leftrightt>leftt Thi cae i ymmetrical with cae..1 Peudocode Of Standard Maintain According to the previou analyi it i eay to implement a normal Maintain. Maintain (t) 1 If leftleftt>rightt then Right-Rotate(t) Maintain(rightt) 4 Maintain(t) 5 Exit 6 If rightleftt>rightt then 7 Left-Rotate(leftt)

8 8 Right-Rotate(t) 9 Maintain(leftt) 10 Maintain(rightt) 11 Maintain(t) 1 Exit 1 If rightrightt>leftt then 14 Left-Rotate(t) 15 Maintain(leftt) 16 Maintain(t) 17 Exit 18 If leftrightt>leftt then 19 Right-Rotate(rightt) 0 Left-Rotate(t) 1 Maintain(leftt) Maintain(rightt) Maintain(t). Peudocode Of Fater And Simpler Maintain The tandard peudocode i a little complicated and low. Generally we can enure that property (a) or (b) ha been atified. Therefore we only need to check cae 1 and or cae and 4 to peed up. In that cae we can add a boolean variant, flag, to avoid meaningle checking. If flag i fale cae 1 and will be checked, otherwie cae and 4 will be checked. Maintain (t,flag) 1 If flag=fale then If leftleftt>rightt then Right-Rotate(t) 4 Eleif rightleftt>rightt then 5 Left-Rotate(leftt) 6 Right-Rotate(t) 7 Ele exit 8 Eleif rightrightt>leftt then 9 Left-Rotate(t) 10 Eleif leftrightt>leftt then 11 Right-Rotate(rightt) 1 Left-Rotate(t) 1 Ele exit 14 Maintain(leftt,fale) 15 Maintain(rightt,true) 16 Maintain(t,fale) 17 Maintain(t,true)

9 Why Maintain(leftt,true) and Maintain(rightt,fale) are expelled? What i the runtime of Maintain? You can find the anwer in part 6, analyi. 4 Inertion The inertion on a SBT i very imple. Here the peudocode of Inert on a SBT. 4.1 Peudocode Of Inert Inert (t,v) 1 If t=0 then t NEW-NODE(v) Ele 4 t t+1 5 If v<keyt then 6 Simple-Inert(leftt,v) 7 Ele 8 Simple-Inert(rightt,v) 9 Maintain(t,v keyt) 5 Deletion I augment the deletion for convenience. If no uch a value to delete in a SBT we delete the node earched at lat and record it. Here the tandard peudocode of Delete on a SBT. 5.1 Peudocode Of Standard Delete Delete (t,v) 1 If t then record keyt t leftt+rightt 4 Exit 5 t t-1 6 If v=keyt then 7 Delete(leftt,vt+1) 8 Keyt record 9 Maintain(t,true) 10 Ele

10 11 If v<keyt then 1 Delete(leftt,v) 1 Ele 14 Delete(rightt,v) 15 Maintain(t,v<keyt) 5. Peudocode Of Fater And Simpler Delete Actually thi i the implet deletion without other function. Delete(t,v) here i a function that return the value deleted. It can reult in a detroyed SBT. But with the inertion above, a BST i till kept at the height of O(logn*) where n* i the total number of inertion, not the current ize! Delete (t,v) 1 t t-1 If (v=keyt)or(v<keyt)and(leftt=0)or(v>keyt)and(rightt=0) then Delete keyt 4 If (leftt=0)or(rightt=0) then 5 t leftt+rightt 6 Ele 7 keyt Delete(leftt,vt+1) 8 Ele 9 If v<keyt then 10 Delete(leftt,v) 11 Ele 1 Delete(rightt,v) 6 Analyi Obviouly Maintain i a recurive procedure. Maybe the quetion about whether it can top or not ha appeared in your mind. Not to worry: it ha been proved that the amortized runtime for Maintain i only O(1). 6.1 Analyi Of Height Let fh be the minimum number of the node of a SBT whoe height i h. We have 1 ( h = 0) f h = ( h = 1) f h 1 + f h + 1 ( h > 1)

11 6.1.1 Proof (1) It i eay to work out that f 0 = 1and f 1 =. () Firt of all, f h f h 1 + f h + 1( h > 1). For each h>1, let aume that t point to a SBT whoe height i h. And then thi SBT contain a ubtree at the height of h-1. It doen t matter to uppoe it a the left ubtree. According to the previou definition of f h, we have that left t f h 1. And there i an h--tall ubtree of the left ubtree. In another word there i a ubtree whoe ize i at leat f h of the left ubtree. Owing to the property (b) we have that right t f h. Therefore we draw the concluion that t f h 1 + f h + 1. On the other hand, f h f h 1 + f h + 1( h > 1). We can contruct a SBT with exact f h node() and it height i h. We call thi kind of SBT treeh. a BST with one node tree h = any BST with two node a BST containing treeh -1 and treeh - a two ubtree ( h = 0) ( h = 1) ( h > 1) Hence that f h = f h 1 + f h + 1( h > 1) i obtained by umming up two upper apect The Wort Height In fact f h i exponential function. It precie value can be calculated from the recurion. h α f h = + β 5 h+ 1 = α h+ 5 1 = Fibonacci h + 1 = h i= 0 Fibonacci i where α = and β = 5

12 Some uual value of fh H Fh Lemma The wort height of an n-node SBT i the maximum h ubjected to f h n. Aume that maxh i the wort height of a SBT at the ize of n. According to the lemma above, we have α f maxh = α maxh+ 5 maxh log n maxh 1.44log maxh+ 5 5( n+ 1.5) α n n 1. Now it i clear that the height of a SBT i O(logn). 6. Analyi Of Maintain We can eaily prove that Maintain work quite efficiently uing the previou concluion. There i a very important value to etimate how well a BST i, the average of all node depth. It i the quotient obtained by SD, the um of the depth of all node, dividing n. Generally the le it i, the better a BST i. Becaue of the contant n, SD i expected to be a mall a poible. Now we need to concentrate on SD. It ignificance i the ability to retrict the time of Maintain. Recalling the condition to perform rotation in Maintain it i urpriing that SD alway decreae after rotation. In cae 1, for example, comparing Figure.1 with Figure.1, SD increae a negative value, rightt-leftleftt. And for intance comparing Figure. to Figure.4, SD increae a value le than -1, rightt-rightleftt-1. Due to the height of O(logn) SD i alway kept O(nlogn). And SD jut increae O(logn) after an inertion on a SBT. Therefore SD = n O(log n) T T = O( n log n) = O(log n) where T i the number of Maintain running rotation. The total number of Maintain i T plu the number of Maintain without rotation. Since the latter i O(nlogn)+O(T), the amortized runtime for Maintain i

13 O( T ) + O( nlog n) + O( T ) nlog n = O(1) 6. Analyi Of Each Operation Now that the height of SBT i O(logn) and Maintain i O(1), the runtime for all primary operation are O(logn). 6.4 Analyi Of Fater And Simpler Delete We call the tatement P(n*) that a BST with the fater and impler Delete and n* tandard Inert i at the height of O(logn*). We prove P(n*) i true for arbitrary poitive integer n* by mathematical induction Proof Here I jut give a rough proof. Aume that a node t i checked by Maintain(t,fale), we have left t 1 right t t 1 left t Therefore if all node on the path from a node to the root are checked by Maintain the depth of thi node i O(logn). (1) For n*=1, P(n*) i true apparently; () Aume that P(n*) i true for n*<k. For n*=k, after the lat conecutive inertion, the node checked by Maintain mut be connected together to form a tree. For every leaf of thi tree, the ubtree pointed by it doe not be changed by Maintain. So the depth of the node in thi ubtree i not larger than O(logn*)+O(logn)=O(logn*) () Therefore P(n*) i true for n*=1,, In thi way the amortized runtime of Maintain i till O(1). 6.5 Analyi Of Fater And Simpler Maintain Here the dicuion about why Maintain(leftt,true) and Maintain(rightt,fale) can be expelled. In cae 1 in Figure. we have

14 9 8, 9 8 1, R R F E R B F E R L B R L Thu Maintain(rightt,fale),correpond to Maintain(T,fale) in Figure.1, can be expelled. And Maintain(leftt,true) i unneceary apparently. In cae in Figure. we have R F E A Thee inequation alo mean that the ize of ubtree of E i le than A and the ize of ubtree of F i le than R. Thu Maintain(rightt,fale) and Maintain(leftt,true) can be expelled. 7 Advantage 7.1 How Fat A SBT Run Typical Problem Write a program to perform n operation given from the input. They are 1) Inert a given number into the et; ) Delete a given number from the et; ) Return if a given number i in the et; 4) Return the rank of a given number in the et; 5) Return the kth number in the et; 6) Return the previou number of a given number in the et; 7) Return the ucceeding number of a given number in the et.

15 7.1. Statitic Inert,000,000 node with random value econd SBT 7.1 AVL 8.4 Treap 11.6 Randomized BST Perform,000,000 operation of 66% inertion and % deletion with random value econd 6 4 SBT 5.59 AVL 7.4 Treap 8.86 Randomzied BST

16 Perform,000,000 operation of 0% inertion,10% deletion and 60% query with random value econd 4 1 SBT.9 AVL 4.1 Treap 5.67 Randomized BST In practice SBT work excellently. From the upper chart we ee that SBT run much fater than other balanced BST while the data are random. Moreover the more ordered the data are, the unexpectedly fater SBT run. It take only econd to inert,000,000 ordered node into a SBT. 7. How Efficiently A SBT Work The average of all node depth nonincreae while Maintain i running. For thi reaon a SBT alway tend to be a perfect balanced BST. Inert,000,000 node with random value Type SBT AVL Treap Randomized BST Splay Perfect BST Average Depth Height Time of Rotation ? Inert,000,000 node with ordered value Type SBT AVL Treap Randomized Perfect Splay BST BST Average Depth Height Time of Rotation ?

17 7. How Eay Debugging I At firt we can merely implement a imple BST to guarantee that the main of the code i correct. After that we add Maintain into the Inert and then debug. If an error i checked out we only need to debug Maintain. Moreover a SBT doen t bae on random o debugging i much tabler comparing to Treap, Skip lit, Randomized BST and o on. 7.4 How Simple A SBT I A SBT i almot the ame a a imple BST. Removing the only additional Maintain in Inert the former turn to be the latter. And Maintain i quite imple too. 7.5 How Compact A SBT I Lot of balanced BST uch a SBT, AVL, Treap, Red-Black BST and o on need extra field to be kept balanced. But many of them are uele like height, random factor and color. On the contrary a SBT contain a ueful extra field, ize. In poeion of it we can augment BST to upport election and ranking. 7.6 How Veratile A SBT I Now that the height of a SBT i O(logn), we can perform Select in O(logn) in the wort cae. But Splay can t upport it very well becaue the height can be degenerated to be O(n) eaily, which i expoed by the char above. Acknowledgment The author thank hi Englih teacher, Fiona, for enthuiatic help. Reference 1 G.M.Adelon-Velkii and E.M.Landi, An algorithm for the Organization of Information, Soviet.Mat.Doklady (196) L.J.Guiba and R.Sedgewick, A dichromatic Framework for Balanced Tree, Proceeding of the Nineteenth Annual IEEE Sympoium on Foundation of Computer Science (1978)

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