Design and Analysis of Algorithms Lecture- 9: B- Trees
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1 Design and Analysis of Algorithms Lecture- 9: B- Trees Dr. Chung- Wen Albert Tsao 1/12/16 Slide Source: SVU CS
2 Motivation for B-Trees Index structures for large datasets cannot be stored in main memory Storing it on disk requires different approach to efficiency Crudely speaking, one disk access takes about the same time as 200,000 instructions B-Trees 2
3 Motivation (cont.) Assume that we use an Red-Black tree to store about 20 million records We end up with a very deep binary tree with lots of different disk accesses; log 2 20,000,000 is about 24 We can t improve on the log n lower bound on search for a binary tree But, we can reduce slow disk accesses with more branches and thus shorter tree! As branching increases, depth decreases B-Trees 3
4 1) All leaves are at same level. 2) B-Tree is defined by the term minimum degree t. That is, a node can have between t and 2t child nodes. Exception: Root may contain as few as 1 node. The value of t depends upon disk block size. The "fill factor" of (50%) is used to control the growth and the shrinkage. 3) Number of keys per-node = Number of children per-node -1 Every node must contain between t-1 and 2t 1 keys All keys of a node are sorted in increasing order. The child between keys k1 and k2 contains all keys between k1 and k2 B-Trees 4
5 An example B-Tree B-Tree of minimum degree t=3 Note that all the leaves are at the same level
6 Constructing a B-tree B-Tree of minimum degree t=3 The splitchild operation moves a key up, so B-Trees grow up Binary search trees like red-black tree grow down. B-Trees 6
7 Inserting k into a B-Tree A proactive insertion algorithm 1) Initialize x as root. 2) While x is not leaf, do following..a) Find the child of x that is going to to be traversed next...b) If child node y is not full, x=y...c) If y is full, split it and move a key from y to its parent x. If k < mid-key in y, x = first part of y. Else x = second part of y. 3) The loop in step 2 stops when x is leaf. Insert k to x. (x must have space for 1 extra key as we have been splitting all nodes in advance.) B-Trees 6
8 Inserting into a B-Tree A re-active insertion algorithm 1)Attempt to insert the new key k into a leaf 2)If leaf is full, split it into two, promoting the middle key to the leaf s parent 3)If parent becomes full, split it into two, promoting the middle key 4)Repeat the above steps all the way to the top 5)If necessary, the root is split in two and the middle key is promoted to a new root, making the tree one level higher B-Trees 11
9 B-Tree of minimum degree t=3 Insert Insert 8: splitchild splitchild Insert 6: Insert 9: Insert 7: Insert 10: B-Trees 6
10 B-Tree of minimum degree t=3 Insert 11: Insert 12: Insert 13: splitchild Insert 14: Insert 15: Insert 16: splitchild B-Trees 6
11 B-Tree of minimum degree t=3 Insert 17: splitchild Insert 18: Insert 19: Insert 20: B-Trees 6
12 Insert 21: Insert 22: Insert 23:
13 Insert 24: Insert 25: Insert 26:
14 Exercise in Inserting a B-Tree Insert the following keys to a B-Tree of minimum degree t=3: 3, 7, 9, 23, 45, 1, 5, 14, 25, 24, 13, 11, 8, 19, 4, 31, 35, B-Trees 12
15 Exercise in Inserting a B-Tree Insert the following keys to a B-Tree of minimum degree t=2: 3, 7, 9, 23, 45, 1, 5, 14, 25, 24, 13, 11, 8, 19, 4, 31, 35, 56 B-Trees 12
16 Removal from a B-tree Case 1. If key k a leaf node x, delete k from x. (Simple leaf deletion) Case 2. If key k an internal node x, do the following. Let y/z = x s child node that precedes/follows k. Let k0/k1 = x s predecessor/successor. (Simple non-leaf deletion) 2.a) If y has t keys, k = k0. Recursively delete k0. 2.b) If y has < t keys, but z has t keys, k = k1. Recursively delete k1. (Too few keys in child node y and y s siblings z) 2.c) If both y, z have only t-1 keys, delete k from x, and merge y with z. Case 3. If key k x, find child node y who is the ancestor of k. Let y = x s child node that is the ancestor of k. Let z = y s immediate sibling. 3.a) If y has t keys, continue to 3.d) 3.b) If y has only t-1 keys but z has t keys, move key: z x y, (Too few keys in child node y and y s siblings z) 3.c) If both y and z have t-1 keys, merge y with z, and move key: x y. 3.d) Set x = y; Recursively delete k. 16
17 Delete F: Case 1 Removal from a B-tree: Case 1 1. If key k a leaf node x, delete k from x. (Simple leaf deletion) x x 17
18 Removal from a B-tree: Case 2a 2. If key k an internal node x, do the following. a) If y has t keys, replace k with k s predecessor. Recursively delete k s predecessor. (Simple non-leaf deletion) Delete M: Case 2a x child y x child y 18
19 Case 2b) If y has < t keys, but z has t keys, k = k1. Recursively delete k1. (Simple non-leaf deletion) Remove
20 Removal from a B-tree: Case 2c Case 2c (k x: Too few keys in child node y and y s sibling z) delete k & merge y with z. x y z y y merge z y y merge z 20
21 Case 2c: (k x: Too few keys in child node y and y s siblings z) Delete Delete
22 Removal from a B-tree: Case 3b Case 3b (k x: Too few keys in child node y and y s sibling z) Delete D: x y y y merge z z node x x y merge z D 22
23 Removal from a B-tree: 3a Case 3a (k x: Too few keys in child node y but enough in y s sibling z) Delete B x y z x y B y z 23
24 Exercise in Removal from a B-Tree Given B-tree created by these data: Minimum degree t=2 3, 7, 9, 23, 45, 1, 5, 14, 25, 24, 13, 11, 8, 19, 4, 31, 35, 56 Add these further keys: 2, 6,12 Delete these keys: 4, 5, 7, 3, 14 B-Trees 21
25 Analysis of B-Trees The maximum number of items in a B-tree of order m=2t and height h: root m 1 level 1 m(m 1) level 2 m 2 (m 1)... level h m h (m 1) So, the total number of items is (1 + m + m 2 + m m h )(m 1) = [(m h+1 1)/ (m 1)] (m 1) = m h+1 1 When m = 5 and h = 2 this gives = 124 B-Trees 22
26 Reasons for using B-Trees When searching tables held on disc, the cost of each disc transfer is high but doesn't depend much on the amount of data transferred, especially if consecutive items are transferred If we use a B-tree of order 101, say, we can transfer each node in one disc read operation B-tree of order 101 and height 3 can hold items (approximately 100 million) and any item can be accessed with 3 disc reads (assuming we hold the root in memory) If we take m = 3, we get a 2-3 tree, in which non-leaf nodes have two or three children (i.e., one or two keys) B-Trees are always balanced (since the leaves are all at the same level), so 2-3 trees make a good type of balanced tree B-Trees 23
27 Comparing Trees Binary trees Can become unbalanced and lose their good time complexity (big O) AVL trees are strict binary trees that overcome the balance problem Heaps remain balanced but only prioritise (not order) the keys Multi-way trees B-Trees can be m-way, they can have any (odd) number of children One B-Tree, the 2-3 (or 3-way) B-Tree, approximates a permanently balanced binary tree, exchanging the AVL tree s balancing operations for insertion and (more complex) deletion operations B-Trees 24
28 B + TREE Slides source:
29 INTRODUCTION OF B+ TREE The properties of a B+ tree, Similar to the B-Tree, All the leaf nodes are at the same bottom level. Each internal node of the tree has between [t] and [2*t] children Number of keys per-node = Number of keys per-node -1 Use a "fill factor" of (50%) to control the growth and the shrinkage. In contrast to a B-tree, All records are stored at the leaf level of the tree; only keys are stored in internal nodes. All the leaf nodes are linked as a list for faster (sequential) disk access.
30 B+ TREE Internal (Index) Node Leaf (data) nodes are linked for a rapid sequential file read.
31 OPERATIONS IN B+ TREE SEARCH INSERTION DELETION
32 B+ TREE- SEARCH OPERATION TWO CASES: Successful Search Unsuccessful Search
33 SEARCHING Compare the key value with the data in the tree, then give the result back. For example: find the value 45, and 15 in below tree.
34 B+ TREE- INSERTION OPERATION Inserting a record when Case 1: leaf node: not full, index node: not full. Case 2: leaf node: full, index node: not full. Case 3: leaf node: full, index node: full.
35 INSERTION CASE 1 LEAF NODE: NOT FULL, INDEX NODE: NOT FULL Add Record with Key 28
36 INSERTION CASE 2 LEAF NODE: FULL, INDEX NODE: NOT FULL 70 Add Record with Key 70
37 This record should go in the leaf node containing 50, 55, 60, and 65. Left Leaf node Right Leaf node AFTER INSERTING A Record With Key 70.
38 INSERTION CASE 3 LEAF NODE: FULL, INDEX NODE: FULL Add a record containing a key value of 95 to the following tree. 95
39 This record belongs in the node containing 75, 80, 85, and 90. Since this node is full we split it into two nodes: Left Leaf Node Right Leaf node The middle key, 85, rises to the index node. But the index node is also full, so we split the index node: Left Index node Right Index node New Index node
40 split add 95 split Leaf nodes are at same level only.
41 B+ TREE DELETION OPERATION Deleting a record from B+ tree may result in Case 1: both leaf/index nodes above the fill factor. Case 2: leaf/index node below/above fill factor Case 3: Both leaf/index nodes below the fill factor.
42 Case 1: both leaf/index nodes above the fill factor. Delete 70 from the following B+ Tree This node will contain 2 records after the deletion. So, simply delete 70 from the leaf node. 70
43 Delete 25 from the B+ tree when we delete 25 we must replace it with 28 in the index node.
44 DELETE 60 FROM THE B+ TREE The leaf node containing 60 will be below the fill factor after the deletion. Thus, we must combine leaf nodes. With recombined nodes, the index node will be reduced by one key. Hence, it will also fall below the fill factor. Thus, we must combine index nodes. 60 appears as the only key in the root index node.
45
46 B+ TREES AS FILE INDEXES B+ Trees are descendants of B Trees. Retrieval of records from large files or databases stored in external memory is time consuming. To promote Efficient Retrievals, file indexes are used. An index is a <Key, Address> pair.
47 The records of the file are sequentially stored and for each block of records, the largest key and the block address is stored in an index. In B+ Tree to retrieve a record given its key, it is essential that the search traverses down to a leaf node to retrieve its address. The non leaf nodes only serve to help the process traverse downwards towards the appropriate leaf node.
48 ADVANTAGE OF B+ TREE B+ Trees good for a full scan the leaf nodes form a linked list. B tree will need a complete in-order traversal DISADVANTAGE OF B+ TREE Any search will end at leaf node only. Time complexity for every search results in O(h), where h is the height of the B+ tree. Waste of Memory. In comparing to B+ trees, B trees are efficient.
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