1、,B+-Trees and Hashing Techniques for Storage and Index Structures,Covers Chapters 8, 10, 11 Third EditionLast updated: January 30, 2003,Last changed: February 6, 03, 3p,Alternative File Organizations,Many alternatives exist, each ideal for some situation , and not so good in others: Heap files: Suit
2、able 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. Ha
3、shing function h: h(r) = bucket in which record r belongs. h looks at only some of the fields of r, called the search fields.,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. un
4、clustered: 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 i
5、ndex is clustered or not!,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
6、 for inserts. (Thus, order of data recs is close to, but not identical to, the sort order.),Index entries,Data entries,direct search for,(Index File),(Data file),Data Records,data entries,Data entries,Data Records,CLUSTERED,UNCLUSTERED,Index Classification (Contd.),Dense vs. Sparse: If there is at l
7、east 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,Smith, 44, 3000,Ashby,Cass,Smith,22,25,30
8、,40,44,44,50,Sparse Index,on,Name,Data File,Dense Index,on,Age,33,Bristow, 30, 2007,Basu, 33, 4003,Cass, 50, 5004,Tracy, 44, 5004,Daniels, 22, 6003,Jones, 40, 6003,Index Classification (Contd.),Composite Search Keys: Search on a combination of fields. Equality query: Every field value is equal to a
9、constant value. E.g. wrt 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.,sue,13,75,bob,cal,joe,12,10,20,80,11,12,name,age,sal,12,
10、20,12,10,11,80,13,75,20,12,10,12,75,13,80,11,11,12,12,13,10,20,75,80,Data records sorted by name,Data entries in index sorted by ,Data entries sorted by ,Examples of composite key indexes using lexicographic order.,Physical Database Design for Relational Databases,Select Storage Structures (determin
11、e 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),Introduction Indexing Techniques,As for any index, 3 alternatives for data entries k*:Data record with key valu
12、e kHash-based indexes are best for equality selections. Cannot support range searches. B+-trees are best for sorted access and range queries.,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
13、 belongs. (M = # of buckets),h(key) mod N,h,key,Primary bucket pages,Overflow pages,2,0,N-1,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
14、; 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.,Range Searches,Find all student
15、s 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.,Can do binary search on (smaller) index file!,Page 1,Page 2,Page N,Page 3,Data File,k2,kN,k1,Index File,B+ Tree
16、: 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.,Example B+ Tree (order p=5, m=4),Search begins at root, and key comparisons direct it to a
17、leaf (as in ISAM). Search for 5*, 15*, all data entries = 24* .,Based on the search for 15*, we know it is not in the tree!,Root,16,22,29,2*,3*,5*,7*,14*,16*,19*,20*,22*,24*,27*,29*,33*,34*,38*,39*,7,p=5 because tree can have at most 5 pointers in intermediate node; m=4 because at most 4 entries in
18、leaf node.,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 =
19、 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
20、 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.,Inserting 4* into Example B+ Tree,Observe how minimum occupancy is guaranteed in bo
21、th leaf and intermediate node splits. Note difference between copy-up and push-up; be sure you understand the reasons for this.,2*,3*,5*,7*,4,Entry to be inserted in parent node.,(Note that 4 is,continues to appear in the leaf.),s copied up and,appears once in the index. Contrast,4*,Example B+ Tree
22、After Inserting 4*,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.,2*,3*,Root,16,22,29,14*,16*,19*,20*,22*,24*,27*,29*,33*,34*,38*,39*,7,4,5*,4*,7*,Deleting a Data Entry from a B
23、+ 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 entr
24、y (pointing to L or sibling) from parent of L. Merge could propagate to root, decreasing height.,Example Tree (after inserting 4*, and deleting 19* and 20*) before deleting 24,Deleting 19* is easy. Deleting 20* is done with re-distribution. Notice that the intermediate node key had to be changed to
25、24.,2*,3*,Root,16,29,14*,16*,33*,34*,38*,39*,4,7*,5*,7,22*,24*,24,27*,29*,4*,. 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*,2*,3*,7*,14*,16*,22*,27*,29*,33*,34*,38*,39*,5*,4*,Root,29,7,4,16,Example of N
26、on-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,7,4,16,18,21,29,4*,After Re-distribution,Intuitively, entries are re-distributed by pushi
27、ng through the splitting entry in the parent node. It suffices to re-distribute index entry with key 20; weve re-distributed 17 as well for illustration.,14*,16*,33*,34*,38*,39*,22*,27*,29*,17*,18*,20*,21*,7*,5*,2*,3*,Root,7,4,16,29,18,21,4*,Clarifications B+ Tree,B+ trees can be used to store relat
28、ions 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 parameter
29、s 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 n
30、umber 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; th
31、en 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 poi
32、nters 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 f
33、or a tree of height 2 is: 2*(p+1)/2)*(m+1)/2) p*p*m of height 3 is: 2* (p+1)/2)* (p+1)/2)* (m+1/2) p*p*p*m of height n+1 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 shoul
34、d 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 blo
35、ck 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!,Block,B+-tree Block Meta Data,Storage for B+-tree node entries,B+-tree Block Meta Data: Neighbor pointers, #entri
36、es, Parent pointer, sibling bits,Choosing p and m (continued),Example2: Want to store an index for a relation E(ssn, name, salary) in a B+-tree using; storing ssns 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 f
37、or 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 cr
38、eate “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*) C
39、onsequences: 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 D
40、BMS. 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 structu
41、re Typically 67%-full pages at an average,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, splitt
42、ing 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!,Example,Directory is array of size 4. To find bucket for r, take last global depth # bits of h
43、(r); we denote r by h(r). If h(r) = 5 = binary 101, it is in bucket pointed to by 01.,Insert: If bucket is full, split it (allocate new page, re-distribute).,If necessary, double the directory. (As we will see, splitting abucket does not always require doubling; we can tell by comparing global depth
44、 with local depth for the split bucket.),13*,00,01,10,11,2,2,2,2,2,LOCAL DEPTH,GLOBAL DEPTH,DIRECTORY,Bucket A,Bucket B,Bucket C,Bucket D,DATA PAGES,10*,1*,21*,4*,12*,32*,16*,15*,7*,19*,5*,Insert h(r)=20 (Causes Doubling),20*,00,01,10,11,2,2,2,2,LOCAL DEPTH,2,2,DIRECTORY,GLOBAL DEPTH,Bucket A,Bucket
45、 B,Bucket C,Bucket D,Bucket A2,(split image,of Bucket A),1*,5*,21*,13*,32*,16*,10*,15*,7*,19*,4*,12*,19*,2,2,2,000,001,010,011,100,101,110,111,3,3,3,DIRECTORY,Bucket A,Bucket B,Bucket C,Bucket D,Bucket A2,(split image,of Bucket A),32*,1*,5*,21*,13*,16*,10*,15*,7*,4*,20*,12*,LOCAL DEPTH,GLOBAL DEPTH,
46、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 d
47、oes 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!),Directory Doubling,00,01,
