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Last changed: February 6, 03, 3p. B+-Trees and Hashing Techniques for Storage and Index Structures. Covers Chapters [[8]], 10, 11 Third Edition Last updated: January 30, 2003. Alternative File Organizations. Many alternatives exist, each ideal for some situation , and not so good in others:

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b trees and hashing techniques for storage and index structures

Last changed: February 6, 03, 3p

B+-Trees and Hashing Techniques for Storage and Index Structures

Covers Chapters [[8]], 10, 11 Third Edition

Last updated: January 30, 2003

alternative file organizations
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 moreoverflow pages.
    • Hashing functionh: h(r) = bucket in which record r belongs. h looks at only some of the fields of r, called the search fields.
index classification
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!
clustered vs unclustered index
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

index classification contd
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, 50, 5004

Cass

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

index classification contd6
Index Classification (Contd.)

Examples of composite key

indexes using lexicographic order.

  • 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.

11,80

11

12

12,10

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

20,12

Data records

sorted by name

75,13

75

80,11

80

<sal, age>

<sal>

Data entries in index

sorted by <sal,age>

Data entries

sorted by <sal>

physical database design for relational databases
Physical Database Designfor 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)

introduction indexing techniques
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 equalityselections. Cannot support range searches.
  • B+-trees are best for sorted access and range queries.
static hashing
Static Hashing
  • # primary pages fixed, allocated sequentially, never de-allocated; overflow pages if needed.
  • h(k) mod M = bucket to which data entry withkey k belongs. (M = # of buckets)

0

h(key) mod N

2

key

h

N-1

Primary bucket pages

Overflow pages

static hashing contd
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 LinearHashing: Dynamic techniques to fix this problem.
range searches
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

kN

k2

k1

Data File

Page N

Page 3

Page 1

Page 2

  • Can do binary search on (smaller) index file!
b tree the most widely used index
B+ Tree: The Most Widely Used Index

Index Entries

(Direct search)

Data Entries

("Sequence set")

  • 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
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* ...

p=5 because tree can have at most 5 pointers in intermediate node; m=4 because at most 4 entries in leaf node.

Root

29

7

16

22

39*

3*

5*

19*

20*

22*

24*

27*

38*

2*

7*

14*

16*

29*

33*

34*

  • Based on the search for 15*, we know it is not in the tree!
b trees in practice
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
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 splitL (into L and a new node L2)
      • Redistribute entries evenly, copy upmiddle key.
      • Insert index entry pointing to L2 into parent of L.
  • This can happen recursively
    • To split index node, redistribute entries evenly, but push upmiddle 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
Inserting 4* into Example B+ Tree

Entry to be inserted in parent node.

(Note that 16 is pushed up and only

16

this with a leaf split.)

4

7

22

29

Entry to be inserted in parent node.

(Note that 4 is

s copied up and

  • Observe how minimum occupancy is guaranteed in both leaf and intermediate node splits.
  • Note difference between copy-upand push-up; be sure you understand the reasons for this.

4

continues to appear in the leaf.)

3*

5*

4*

2*

7*

appears once in the index. Contrast

example b tree after inserting 4
Example B+ Tree After Inserting 4*

Root

16

22

4

7

29

39*

2*

3*

19*

20*

22*

24*

27*

38*

29*

33*

34*

14*

16*

4*

5*

7*

  • 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.
deleting a data entry from a b tree
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, mergeL and sibling.
  • If merge occurred, must delete entry (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
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 24.

Root

16

24

4

29

7

39*

2*

3*

5*

7*

22*

24*

27*

29*

38*

33*

34*

14*

16*

4*

and then deleting 24
... And Then Deleting 24*
  • Must merge.
  • Observe `toss’ of index entry (on right), and `pull down’ of index entry (below).

29

39*

22*

27*

38*

29*

33*

34*

Root

4

7

16

29

3*

39*

2*

5*

7*

22*

38*

27*

33*

34*

14*

16*

29*

4*

example of non leaf re distribution
Example of Non-leaf Re-distribution

2*

3*

5*

7*

39*

17*

18*

38*

20*

21*

22*

27*

29*

33*

34*

14*

16*

  • 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

29

16

18

4

7

4*

after re distribution
After Re-distribution
  • Intuitively, entries are re-distributed by `pushingthrough’ 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*

5*

7*

39*

4*

17*

18*

38*

20*

21*

22*

27*

29*

33*

34*

14*

16*

clarifications b tree
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
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 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 should i chose
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
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
Coping with Duplicate Keysin 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
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
extendible hashing
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. Nooverflowpage!
    • Trick lies in how hash function is adjusted!
example
Example

LOCAL DEPTH

2

Bucket A

16*

4*

12*

32*

GLOBAL DEPTH

2

2

Bucket B

00

5*

1*

21*

13*

01

  • 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.

2

10

Bucket C

10*

11

2

DIRECTORY

Bucket D

15*

7*

19*

DATA PAGES

  • Insert: If bucket is full, splitit (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.)
insert h r 20 causes doubling
Insert h(r)=20 (Causes Doubling)

2

LOCAL DEPTH

3

LOCAL DEPTH

Bucket A

16*

32*

32*

16*

GLOBAL DEPTH

Bucket A

GLOBAL DEPTH

2

2

2

3

Bucket B

5*

21*

13*

1*

00

1*

5*

21*

13*

000

Bucket B

01

001

2

10

2

010

Bucket C

10*

11

10*

Bucket C

011

100

2

2

DIRECTORY

101

Bucket D

15*

7*

19*

15*

7*

19*

Bucket D

110

111

2

3

Bucket A2

4*

12*

20*

DIRECTORY

12*

20*

Bucket A2

4*

(`split image'

of Bucket A)

(`split image'

of Bucket A)

points to note
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!)
directory doubling
Directory Doubling
  • Why use least significant bits in directory?
    • Allows for doubling via copying!

6 = 110

6 = 110

3

3

000

000

001

100

2

2

010

010

00

00

1

1

011

110

6*

01

10

0

0

100

001

6*

6*

10

01

1

1

101

101

6*

11

11

6*

6*

110

011

111

111

vs.

Least Significant

Most Significant

comments on extendible hashing
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.
linear hashing
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)