B+-Trees and Hashing Techniques for Storage and Index Structures Covers Chapters [[8]], 10, 11 Third Edition Last updated: October 31, 2005 The slides for this text are organized into chapters. This lecture covers Chapter 10. Chapter 1: Introduction to Database Systems Chapter 2: The Entity-Relationship Model Chapter 3: The Relational Model Chapter 4 (Part A): Relational Algebra Chapter 4 (Part B): Relational Calculus Chapter 5: SQL: Queries, Programming, Triggers Chapter 6: Query-by-Example (QBE) Chapter 7: Storing Data: Disks and Files Chapter 8: File Organizations and Indexing Chapter 9: Tree-Structured Indexing Chapter 10: Hash-Based Indexing Chapter 11: External Sorting Chapter 12 (Part A): Evaluation of Relational Operators Chapter 12 (Part B): Evaluation of Relational Operators: Other Techniques Chapter 13: Introduction to Query Optimization Chapter 14: A Typical Relational Optimizer Chapter 15: Schema Refinement and Normal Forms Chapter 16 (Part A): Physical Database Design Chapter 16 (Part B): Database Tuning Chapter 17: Security Chapter 18: Transaction Management Overview Chapter 19: Concurrency Control Chapter 20: Crash Recovery Chapter 21: Parallel and Distributed Databases Chapter 22: Internet Databases Chapter 23: Decision Support Chapter 24: Data Mining Chapter 25: Object-Database Systems Chapter 26: Spatial Data Management Chapter 27: Deductive Databases Chapter 28: Additional Topics 1

Alternative File Organizations Many alternatives exist, each ideal for some situation , and not so good in others: Heap files: Suitable when typical access is a file scan retrieving all records. Sorted Files: Best if records must be retrieved in some order, or only a `range’ of records is needed. Hashed Files: Good for equality selections. File is a collection of buckets. Bucket = primary page plus zero or more overflow pages. Hashing function h: h(r) = bucket in which record r belongs. h looks at only some of the fields of r, called the search fields. 2

Physical Database Design for Relational Databases Select Storage Structures (determine how the particular relation is physically stored) Select Index Structures (to speed up certain queries) Select … … to minimize the runtime for a certain workload (e.g a given set of queries)

Focus of the Current Discussion Introduce popular storage structures / file organizations Introduce popular index structures Discuss advantages and disadvantages of the introduced storage and index structures Obtain a clear understanding how storage structure and index structures relate to query processing

Example of an Internal Schema for the Library Example INTERNAL Schema Library12 references Library. Book is stored sequentially, index on B# using hashing, index on Author using hashing. Person is stored using hashing on ssn. Check_out is stored sequentially, index on since using B+-tree.

Example: Stored Database Index on B# Block#= B# mod 10 index structure Index on Author Relation Book 1 11 51 20 30 W, … (1, C,W) (20, Y,W) (51, C, B) 1 (11, Y,W) (30, Z, B) Relation Checkout (101,…) storage structure (200,…) (500,…) Index on since Relation Person Block#= sss mod 10 1

Index Classification Primary vs. secondary: If search key contains primary key, then called primary index. Unique index: Search key contains a candidate key. Clustered vs. unclustered: If order of data records is the same as, or `close to’, order of data entries, then called clustered index. Alternative 1 implies clustered, but not vice-versa. A file can be clustered on at most one search key. Cost of retrieving data records through index varies greatly based on whether index is clustered or not! 11

Clustered vs. Unclustered Index Suppose that Alternative (2) is used for data entries, and that the data records are stored in a Heap file. To build clustered index, first sort the Heap file (with some free space on each page for future inserts). Overflow pages may be needed for inserts. (Thus, order of data recs is `close to’, but not identical to, the sort order.) Index entries UNCLUSTERED CLUSTERED direct search for data entries Data entries Data entries (Index File) (Data file) Data Records Data Records 12

Index Classification (Contd.) Dense vs. Sparse: If there is at least one data entry per search key value (in some data record), then dense. Alternative 1 always leads to dense index. Every sparse index is clustered! Sparse indexes are smaller; however, some useful optimizations are based on dense indexes. Ashby, 25, 3000 22 Basu, 33, 4003 25 Bristow, 30, 2007 30 Ashby 33 Cass Cass, 50, 5004 Smith Daniels, 22, 6003 40 Jones, 40, 6003 44 44 Smith, 44, 3000 50 Tracy, 44, 5004 Sparse Index Dense Index on on Data File Name Age 13

Index Classification (Contd.) Composite Search Keys: Search on a combination of fields. Equality query: Every field value is equal to a constant value. E.g. wrt <sal,age> index: age=20 and sal =75 Range query: Some field value is not a constant. E.g.: age =20; or age=20 and sal > 10 Data entries in index sorted by search key to support range queries. Lexicographic order, or Spatial order. Examples of composite key indexes using lexicographic order. 11,80 11 12,10 12 name age sal 12,20 12 13,75 bob 12 10 13 <age, sal> cal 11 80 <age> joe 12 20 10,12 sue 13 75 10 20,12 Data records sorted by name 20 75,13 75 80,11 80 <sal, age> <sal> Data entries in index sorted by <sal,age> Data entries sorted by <sal> 13

Introduction Indexing Techniques As for any index, 3 alternatives for data entries k*: Data record with key value k <k, rid of data record with search key value k> <k, list of rids of data records with search key k> Hash-based indexes are best for equality selections. Cannot support range searches. B+-trees are best for sorted access and range queries. 2

Static Hashing # primary pages fixed, allocated sequentially, never de-allocated; overflow pages if needed. h(k) mod M = bucket to which data entry with key k belongs. (M = # of buckets) h(key) mod N 2 key h N-1 Primary bucket pages Overflow pages 3

Static Hashing (Contd.) Buckets contain data entries. Hash fn works on search key field of record r. Must distribute values over range 0 ... M-1. h(key) = (a * key + b) usually works well. a and b are constants; lots known about how to tune h. Long overflow chains can develop and degrade performance. Two approaches: Global overflow area Individual overflow areas for each bucket (assumed in the following) Extendible and Linear Hashing: Dynamic techniques to fix this problem. 4

Range Searches ``Find all students with gpa > 3.0’’ If data is in sorted file, do binary search to find first such student, then scan to find others. Cost of binary search can be quite high. Simple idea: Create an `index’ file. Index File k1 k2 kN Data File Page 1 Page 2 Page 3 Page N Can do binary search on (smaller) index file! 3

B+ Tree: The Most Widely Used Index Insert/delete at log F N cost; keep tree height-balanced. (F = fanout, N = # leaf pages) Minimum 50% occupancy (except for root). Supports equality and range-searches efficiently. Index Entries Data Entries ("Sequence set") (Direct search) 9

Example B+ Tree (order p=5, m=4) Search begins at root, and key comparisons direct it to a leaf (as in ISAM). Search for 5*, 15*, all data entries >= 24* ... Root p=5 because tree can have at most 5 pointers in intermediate node; m=4 because at most 4 entries in leaf node. 7 16 22 29 2* 3* 5* 7* 14* 16* 19* 20* 22* 24* 27* 29* 33* 34* 38* 39* Based on the search for 15*, we know it is not in the tree! 10

B+ Trees in Practice Typical order: 200. Typical fill-factor: 67%. average fanout = 133 Typical capacities: Height 4: 1334 = 312,900,700 records Height 3: 1333 = 2,352,637 records Can often hold top levels in buffer pool: Level 1 = 1 page = 8 Kbytes Level 2 = 133 pages = 1 Mbyte Level 3 = 17,689 pages = 133 MBytes

Inserting a Data Entry into a B+ Tree Find correct leaf L. Put data entry onto L. If L has enough space, done! Else, must split L (into L and a new node L2) Redistribute entries evenly, copy up middle key. Insert index entry pointing to L2 into parent of L. This can happen recursively To split index node, redistribute entries evenly, but push up middle key. (Contrast with leaf splits.) Splits “grow” tree; root split increases height. Tree growth: gets wider or one level taller at top. 6

Inserting 4* into Example B+ Tree Entry to be inserted in parent node. Observe how minimum occupancy is guaranteed in both leaf and intermediate node splits. Note difference between copy-up and push-up; be sure you understand the reasons for this. 4 (Note that 4 is s copied up and continues to appear in the leaf.) 2* 3* 4* 5* 7* 4 22 29 16 7 Entry to be inserted in parent node. (Note that 16 is pushed up and only this with a leaf split.) appears once in the index. Contrast 12

Example B+ Tree After Inserting 4* Root 16 4 7 22 29 2* 3* 4* 5* 7* 14* 16* 19* 20* 22* 24* 27* 29* 33* 34* 38* 39* Notice that root was split, leading to increase in height. In this example, we can avoid split by re-distributing entries; however, this is usually not done in practice. 13

Deleting a Data Entry from a B+ Tree Start at root, find leaf L where entry belongs. Remove the entry. If L is at least half-full, done! If L has only d-1 entries, Try to re-distribute, borrowing from sibling (adjacent node with same parent as L). If re-distribution fails, merge L and sibling. If merge occurred, must delete entry (pointing to L or sibling) from parent of L. Merge could propagate to root, decreasing height. 14

Example Tree (after inserting 4. , and deleting 19. and 20 Example Tree (after inserting 4*, and deleting 19* and 20*) before deleting 24 Root 16 4 7 24 29 2* 3* 4* 5* 7* 14* 16* 22* 24* 27* 29* 33* 34* 38* 39* Deleting 19* is easy. Deleting 20* is done with re-distribution. Notice that the intermediate node key had to be changed to 24. 15

... And Then Deleting 24* Must merge. Observe `toss’ of index entry (on right), and `pull down’ of index entry (below). 29 22* 27* 29* 33* 34* 38* 39* Root 4 7 16 29 2* 3* 4* 5* 7* 14* 16* 22* 27* 29* 33* 34* 38* 39* 16

Example of Non-leaf Re-distribution Tree is shown below during deletion of 24*. (What could be a possible initial tree?) In contrast to previous example, can re-distribute entry from left child of root to right child. Root 21 4 7 16 18 29 14* 16* 17* 18* 20* 33* 34* 38* 39* 22* 27* 29* 21* 7* 5* 3* 2* 4* 17

After Re-distribution Intuitively, entries are re-distributed by `pushing through’ the splitting entry in the parent node. It suffices to re-distribute index entry with key 20; we’ve re-distributed 17 as well for illustration. Root 16 4 7 18 21 29 2* 3* 4* 5* 7* 14* 16* 17* 18* 20* 21* 22* 27* 29* 33* 34* 38* 39* 18

Clarifications B+ Tree B+ trees can be used to store relations as well as index structures In the drawn B+ trees we assume (this is not the only scheme) that an intermediate node with q pointers stores the maximum keys of each of the first q-1 subtrees it is pointing to; that is, it contains q-1 keys. Before B+-tree can be generated the following parameters have to be chosen (based on the available block size; it is assumed one node is stored in one block): the order p of the tree (p is the maximum number of pointers an intermediate node might have; if it is not a root it must have between ((p+1)/2) and p pointers; ‘/’ is integer division) the maximum number m of entries in the leaf node can hold (in general leaf nodes (except the root) must hold between (m+1)/2 and m entries) Intermediate nodes usually store more entries than leaf nodes

Why is the minimum number of pointers in an intermediate node (p+1)/2 and not p/2 + 1?? (p+1)/2: Assume p=10; then p is between 5 and 10; in the case of underflow without borrowing, 4 pointers have to be merged with 5 pointer yielding a node with 9 pointers. p/2 + 1: Assume p=10; then p is between 6 and 10; in the case of underflow without borrowing, 5 pointers have to be merged with 6 pointer yielding 11 pointers which is one too many. If p is odd: Assume p=11, then p is between 6 and 11; in the case of an underflow without borrowing a 5 pointer node has to be merged with a 6 pointer node yielding an 11 pointer node. Conclusion: We infer from the discussion that the minimum  maximum numbers of entries for a tree of height 3 is: 2*((p+1)/2)*((m+1)/2)  p*p*m of height 4 is: 2* ((p+1)/2)* ((p+1)/2)* ((m+1/2)  p*p*p*m of height n+2 is: 2*((p+1)/2)n *((m+1)/2)  pn+1*m Remark: Therefore the correct answer for the homework problem (p=10;m=100) should be: 2*5*50/10*10*100

What order p and leaf entry maximum m should I chose? Idea: One B+-tree node is stored in one block; choose maximal m and p without exceeding block size!! Example1: Want to store tuples of a relation E(ssn, name, salary) in a B+-tree using ssn as the search key; ssn, and salary take 4 Byte; name takes 12 byte. B+-pointers take 2 Byte; the block size is 2048 byte and the available space inside a block for B+-tree entries is 2000 byte. Choose p and m!! px2 + (p-1)x4 =<2000 p=<2004/6=334 m =< 2000/20 Answer: Choose p=334 and m=100! B+-tree Block Meta Data B+-tree Block Meta Data: Neighbor pointers, #entries, Parent pointer, sibling bits,… Block Storage for B+-tree node entries

Choosing p and m (continued) Example2: Want to store an index for a relation E(ssn, name, salary) in a B+-tree using; storing ssn’s take 4 Byte; index pointers take 4 Byte. B+-pointers take 4 Byte; the block size is 2048 byte and the available space inside the block for B+-tree entries is 2000 byte. Choose p and m!! px4 + (p-1)x4 =<2000 p=<2004/8=250 m =< 2000/8 = 250 Answer: Choose p=250 and m=250.

Coping with Duplicate Keys in B+ Trees Possible Approaches: Just allow duplicate keys. Consequences: Search is still efficient Insertion is still efficient (but could create “hot spots”) Deletion faces a lot of problems: We have to follow the leaf pointers to find the entry to be deleted, and then updating the intermediate nodes might get quite complicated (can partially be solved by creating two-way node pointers) Just create unique keys by using key+data (key*) Consequences: Deletion is no longer a problem p (because of the larger key size) is significantly lower, and therefore the height of the tree is likely higher.

Summary B+ Tree Most widely used index in database management systems because of its versatility. One of the most optimized components of a DBMS. Tree-structured indexes are ideal for range-searches, also good for equality searches (log F N cost). Inserts/deletes leave tree height-balanced; log F N cost. High fanout (F) means depth rarely more than 3 or 4. Almost always better than maintaining a sorted file Self reorganizing data structure Typically 67%-full pages at an average 23

Extendible Hashing Situation: Bucket (primary page) becomes full. Why not re-organize file by doubling # of buckets? Reading and writing all pages is expensive! Idea: Use directory of pointers to buckets, double # of buckets by doubling the directory, splitting just the bucket that overflowed! Directory much smaller than file, so doubling it is much cheaper. Only one page of data entries is split. No overflow page! Trick lies in how hash function is adjusted! 5

Example LOCAL DEPTH 2 Bucket A 4* 12* 32* 16* GLOBAL DEPTH 2 2 Bucket B 00 1* 5* 21* 13* Directory is array of size 4. To find bucket for r, take last `global depth’ # bits of h(r); we denote r by h(r). If h(r) = 5 = binary 101, it is in bucket pointed to by 01. 01 10 2 Bucket C 11 10* 2 DIRECTORY Bucket D 15* 7* 19* DATA PAGES Insert: If bucket is full, split it (allocate new page, re-distribute). If necessary, double the directory. (As we will see, splitting a bucket does not always require doubling; we can tell by comparing global depth with local depth for the split bucket.) 6

Insert h(r)=20 (Causes Doubling) LOCAL DEPTH 2 3 LOCAL DEPTH Bucket A 32* 16* GLOBAL DEPTH 32* 16* Bucket A GLOBAL DEPTH 2 2 3 2 Bucket B 00 1* 5* 21* 13* 000 1* 5* 21* 13* Bucket B 01 001 10 2 2 010 Bucket C 11 10* 011 10* Bucket C 100 2 DIRECTORY 101 2 Bucket D 15* 7* 19* 15* 7* 19* 110 Bucket D 111 2 3 Bucket A2 4* 12* 20* DIRECTORY (`split image' 4* 12* 20* Bucket A2 of Bucket A) (`split image' of Bucket A) 7

Points to Note 20 = binary 10100. Last 2 bits (00) tell us r belongs in A or A2. Last 3 bits needed to tell which. Global depth of directory: Max # of bits needed to tell which bucket an entry belongs to. Local depth of a bucket: # of bits used to determine if an entry belongs to this bucket. When does bucket split cause directory doubling? Before insert, local depth of bucket = global depth. Insert causes local depth to become > global depth; directory is doubled by copying it over and `fixing’ pointer to split image page. (Use of least significant bits enables efficient doubling via copying of directory!) 8

Directory Doubling Why use least significant bits in directory? Allows for doubling via copying! 6 = 110 3 6 = 110 3 000 000 001 100 2 2 010 010 1 00 1 00 011 110 6* 01 100 10 001 6* 10 6* 01 1 101 1 101 6* 11 6* 11 6* 110 011 111 111 Least Significant vs. Most Significant

Comments on Extendible Hashing If directory fits in memory, equality search answered with one disk access; else two. 100MB file, 100 bytes/rec, 4K pages contains 1,000,000 records (as data entries) and 25,000 directory elements; chances are high that directory will fit in memory. Directory grows in spurts, and, if the distribution of hash values is skewed, directory can grow large. Multiple entries with same hash value cause problems! Delete: If removal of data entry makes bucket empty, can be merged with `split image’. If each directory element points to same bucket as its split image, can halve directory. 9

Linear Hashing This is another dynamic hashing scheme, an alternative to Extendible Hashing. LH handles the problem of long overflow chains without using a directory, and handles duplicates. Idea: Use a family of hash functions h0, h1, h2, ... hi(key) = h(key) mod(2iN); N = initial # buckets h is some hash function (range is not 0 to N-1) If N = 2d0, for some d0, hi consists of applying h and looking at the last di bits, where di = d0 + i. hi+1 doubles the range of hi (similar to directory doubling) 10