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DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan and Korth
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DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

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Page 1: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

DATABASE MANAGEMENT SYSTEMS

TERM 2008-09

B. Tech II/IT II Semester

UNIT-VI PPT SLIDES

Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan and Korth

Page 2: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

INDEXUNIT-6 PPT SLIDES

S.NO Module as per Lecture PPT Session planner No Slide NO-----------------------------------------------------------------------------------------

--------1. Data on external storage & File organization and indexing L1 L1- 1 to L1- 4

2. Index data structures L2 L2- 1 to L2- 7

3. Comparison of file organizations L3 L3- 1 to L3- 5

4. Comparison of file organizations L4 L4- 1 to L4- 2

5. Indexes and performance tuning L5 L5- 1 to L5- 4

6. Indexes and performance tuning L6 L6- 1 to L6 -5

7. Intuition for tree indexes & ISAM L7 L7- 1 to L7- 7

8. B+ tree L8 L8- 1 to L8- 9

Page 3: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L1-1

Data on External StorageData on External Storage• Disks: Can retrieve random page at fixed cost

– But reading several consecutive pages is much cheaper than reading them in random order

• Tapes: Can only read pages in sequence– Cheaper than disks; used for archival storage

• File organization: Method of arranging a file of records on external storage.– Record id (rid) is sufficient to physically locate record– Indexes are data structures that allow us to find the record ids of

records with given values in index search key fields• Architecture: Buffer manager stages pages from

external storage to main memory buffer pool. File and index layers make calls to the buffer manager.

Page 4: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L1-2

Alternative File OrganizationsAlternative File Organizations

Many alternatives exist, each ideal for some situations, and not so good in others:– Heap (random order) 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.– Indexes: Data structures to organize records via trees or

hashing. • Like sorted files, they speed up searches for a subset

of records, based on values in certain (“search key”) fields

• Updates are much faster than in sorted files.

Page 5: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L1-3

Index ClassificationIndex 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; in practice, clustered also implies

Alternative 1 (since sorted files are rare).– 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!

Page 6: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L1-4

Clustered vs. Unclustered IndexClustered 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

Data entries

direct search for

(Index File)

(Data file)

Data Records

data entries

Data entries

Data Records

CLUSTERED UNCLUSTERED

Page 7: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L2-1

IndexesIndexes• An index on a file speeds up selections on

the search key fields for the index.– Any subset of the fields of a relation can be the

search key for an index on the relation.– Search key is not the same as key (minimal set of

fields that uniquely identify a record in a relation).• An index contains a collection of data

entries, and supports efficient retrieval of all data entries k* with a given key value k.– Given data entry k*, we can find record with key k

in at most one disk I/O. (Details soon …)

Page 8: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L2-2

B+ Tree IndexesB+ Tree Indexes

Leaf pages contain data entries, and are chained (prev & next) Non-leaf pages have index entries; only used to direct searches:

P0 K 1 P 1 K 2 P 2 K m P m

index entry

Non-leaf

Pages

Pages (Sorted by search key)

Leaf

Page 9: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L2-3

Example B+ TreeExample B+ Tree

• Find 28*? 29*? All > 15* and < 30*• Insert/delete: Find data entry in leaf, then

change it. Need to adjust parent sometimes.

– And change sometimes bubbles up the tree

2* 3*

Root

17

30

14* 16* 33* 34* 38* 39*

135

7*5* 8* 22* 24*

27

27* 29*

Entries <= 17 Entries > 17

Note how data entriesin leaf level are sorted

Page 10: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L2-4

Hash-Based IndexesHash-Based Indexes

• Good for equality selections. • Index is a collection of buckets.

– Bucket = primary page plus zero or more overflow pages.

– Buckets contain data entries. • Hashing function h: h(r) = bucket in

which (data entry for) record r belongs. h looks at the search key fields of r.

– No need for “index entries” in this scheme.

Page 11: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L2-5

Alternatives for Data Entry Alternatives for Data Entry k*k* in Indexin Index• In a data entry k* we can store:

– Data record with key value k, or

– <k, rid of data record with search key value k>, or

– <k, list of rids of data records with search key k>

• Choice of alternative for data entries is orthogonal to the indexing technique used to locate data entries with a given key value k.– Examples of indexing techniques: B+ trees, hash-based

structures

– Typically, index contains auxiliary information that directs searches to the desired data entries

Page 12: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L2-6

Alternatives for Data Entries (Contd.)Alternatives for Data Entries (Contd.)• Alternative 1:

– If this is used, index structure is a file organization for data records (instead of a Heap file or sorted file).

– At most one index on a given collection of data records can use Alternative 1. (Otherwise, data records are duplicated, leading to redundant storage and potential inconsistency.)

– If data records are very large, # of pages containing data entries is high. Implies size of auxiliary information in the index is also large, typically.

Page 13: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L2-7

Alternatives for Data Entries (Contd.)Alternatives for Data Entries (Contd.)• Alternatives 2 and 3:

– Data entries typically much smaller than data records. So, better than Alternative 1 with large data records, especially if search keys are small. (Portion of index structure used to direct search, which depends on size of data entries, is much smaller than with Alternative 1.)

– Alternative 3 more compact than Alternative 2, but leads to variable sized data entries even if search keys are of fixed length.

Page 14: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L3-1

Cost Model for Our AnalysisCost Model for Our AnalysisWe ignore CPU costs, for simplicity:

– B: The number of data pages– R: Number of records per page– D: (Average) time to read or write disk

page– Measuring number of page I/O’s ignores

gains of pre-fetching a sequence of pages; thus, even I/O cost is only approximated.

– Average-case analysis; based on several simplistic assumptions.

Page 15: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L3-2

Comparing File OrganizationsComparing File Organizations

• Heap files (random order; insert at eof)• Sorted files, sorted on <age, sal> • Clustered B+ tree file, Alternative (1),

search key <age, sal>• Heap file with unclustered B + tree index

on search key <age, sal>• Heap file with unclustered hash index on

search key <age, sal>

Page 16: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L3-3

Operations to CompareOperations to Compare

• Scan: Fetch all records from disk• Equality search• Range selection• Insert a record• Delete a record

Page 17: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L3-4

Assumptions in Our AnalysisAssumptions in Our Analysis• Heap Files:

– Equality selection on key; exactly one match.• Sorted Files:

– Files compacted after deletions.• Indexes:

– Alt (2), (3): data entry size = 10% size of record – Hash: No overflow buckets.

• 80% page occupancy => File size = 1.25 data size

– Tree: 67% occupancy (this is typical).• Implies file size = 1.5 data size

Page 18: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L3-5

Assumptions (contd.)Assumptions (contd.)

• Scans:

– Leaf levels of a tree-index are chained.– Index data-entries plus actual file

scanned for unclustered indexes.• Range searches:

– We use tree indexes to restrict the set of data records fetched, but ignore hash indexes.

Page 19: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L4-1

Cost of Operations Cost of Operations

(a) Scan (b) Equality (c ) Range (d) Insert (e) Delete

(1) Heap BD 0.5BD BD 2D Search +D

(2) Sorted BD Dlog 2B D(log 2 B + # pgs with match recs)

Search + BD

Search +BD

(3) Clustered

1.5BD Dlog F 1.5B D(log F 1.5B + # pgs w. match recs)

Search + D

Search +D

(4) Unclust. Tree index

BD(R+0.15) D(1 + log F 0.15B)

D(log F 0.15B + # pgs w. match recs)

Search + 2D

Search + 2D

(5) Unclust. Hash index

BD(R+0.125) 2D BD Search + 2D

Search + 2D

Page 20: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L4-2

Understanding the WorkloadUnderstanding the Workload

• For each query in the workload:– Which relations does it access?– Which attributes are retrieved?– Which attributes are involved in selection/join

conditions? How selective are these conditions likely to be?

• For each update in the workload:– Which attributes are involved in selection/join

conditions? How selective are these conditions likely to be?

– The type of update (INSERT/DELETE/UPDATE), and the attributes that are affected.

Page 21: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L5-1

Choice of IndexesChoice of Indexes

• What indexes should we create?

– Which relations should have indexes? What field(s) should be the search key? Should we build several indexes?

• For each index, what kind of an index should it be?

– Clustered? Hash/tree?

Page 22: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L5-2

Choice of Indexes (Contd.)Choice of Indexes (Contd.)• One approach: Consider the most important

queries in turn. Consider the best plan using the current indexes, and see if a better plan is possible with an additional index. If so, create it.– Obviously, this implies that we must understand how

a DBMS evaluates queries and creates query evaluation plans!

– For now, we discuss simple 1-table queries.• Before creating an index, must also consider the

impact on updates in the workload!– Trade-off: Indexes can make queries go faster,

updates slower. Require disk space, too.

Page 23: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L5-3

Index Selection GuidelinesIndex Selection Guidelines• Attributes in WHERE clause are candidates for

index keys.– Exact match condition suggests hash index.– Range query suggests tree index.

• Clustering is especially useful for range queries; can also help on equality queries if there are many duplicates.

• Multi-attribute search keys should be considered when a WHERE clause contains several conditions.– Order of attributes is important for range queries.– Such indexes can sometimes enable index-only

strategies for important queries.• For index-only strategies, clustering is not

important!

Page 24: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L5-4

Examples of Clustered IndexesExamples of Clustered Indexes• B+ tree index on E.age can be

used to get qualifying tuples.– How selective is the condition?– Is the index clustered?

• Consider the GROUP BY query.– If many tuples have E.age > 10,

using E.age index and sorting the retrieved tuples may be costly.

– Clustered E.dno index may be better!

• Equality queries and duplicates:– Clustering on E.hobby helps!

SELECT E.dnoFROM Emp EWHERE E.age>40

SELECT E.dno, COUNT (*)FROM Emp EWHERE E.age>10GROUP BY E.dno

SELECT E.dnoFROM Emp EWHERE E.hobby=Stamps

Page 25: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L6-1

Indexes with Composite Search Keys Indexes with Composite Search Keys • 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.

sue 13 75

bob

cal

joe 12

10

20

8011

12

name age sal

<sal, age>

<age, sal> <age>

<sal>

12,20

12,10

11,80

13,75

20,12

10,12

75,13

80,11

11

12

12

13

10

20

75

80

Data recordssorted by name

Data entries in indexsorted by <sal,age>

Data entriessorted by <sal>

Examples of composite keyindexes using lexicographic order.

Page 26: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L6-2

Composite Search KeysComposite Search Keys• To retrieve Emp records with age=30 AND

sal=4000, an index on <age,sal> would be better than an index on age or an index on sal.– Choice of index key orthogonal to clustering etc.

• If condition is: 20<age<30 AND 3000<sal<5000: – Clustered tree index on <age,sal> or <sal,age> is

best.• If condition is: age=30 AND 3000<sal<5000:

– Clustered <age,sal> index much better than <sal,age> index!

• Composite indexes are larger, updated more often.

Page 27: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L6-3

Index-Only PlansIndex-Only Plans

• A number of queries can be answered without retrieving any tuples from one or more of the relations involved if a suitable index is available.

SELECT E.dno, COUNT(*)FROM Emp EGROUP BY E.dno

SELECT E.dno, MIN(E.sal)FROM Emp EGROUP BY E.dno

SELECT AVG(E.sal)FROM Emp EWHERE E.age=25 AND E.sal BETWEEN 3000 AND 5000

<E.dno>

<E.dno,E.sal>

Tree index!

<E. age,E.sal> or<E.sal, E.age>Tree index!

Page 28: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L6-4

SummarySummary

• Many alternative file organizations exist, each appropriate in some situation.

• If selection queries are frequent, sorting the file or building an index is important.

– Hash-based indexes only good for equality search.

– Sorted files and tree-based indexes best for range search; also good for equality search. (Files rarely kept sorted in practice; B+ tree index is better.)

• Index is a collection of data entries plus a way to quickly find entries with given key values.

Page 29: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L6-5

Summary (Contd.)Summary (Contd.)

• Data entries can be actual data records, <key, rid> pairs, or <key, rid-list> pairs.

– Choice orthogonal to indexing technique used to locate data entries with a given key value.

• Can have several indexes on a given file of data records, each with a different search key.

• Indexes can be classified as clustered vs. unclustered, primary vs. secondary, and dense vs. sparse. Differences have important consequences for utility/performance.

Page 30: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L7-1

IntroductionIntroduction• 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>• Choice is orthogonal to the indexing technique

used to locate data entries k*.• Tree-structured indexing techniques support

both range searches and equality searches.• ISAM: static structure; B+ tree: dynamic,

adjusts gracefully under inserts and deletes.

Page 31: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L7-2

Range SearchesRange 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.

Page 1 Page 2 Page NPage 3 Data File

k2 kNk1 Index File

Page 32: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L7-3

ISAMISAM

• Index file may still be quite large. But we can apply the idea repeatedly!

P0

K1 P

1K 2 P

2K m

P m

index entry

Non-leaf

Pages

Pages

Overflow page

Primary pages

Leaf

Page 33: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L7-4

Comments on ISAMComments on ISAM

• File creation: Leaf (data) pages allocated sequentially, sorted by search key; then index pages allocated, then space for overflow pages.

• Index entries: <search key value, page id>; they `direct’ search for data entries, which are in leaf pages.

• Search: Start at root; use key comparisons to go to leaf. Cost log F N ; F = # entries/index pg, N = # leaf pgs

• Insert: Find leaf data entry belongs to, and put it there.

• Delete: Find and remove from leaf; if empty overflow page, de-allocate.

Data Pages

Index Pages

Overflow pages

Page 34: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L7-5

Example ISAM TreeExample ISAM Tree

• Each node can hold 2 entries; no need for `next-leaf-page’ pointers. (Why?)

10* 15* 20* 27* 33* 37* 40* 46* 51* 55* 63* 97*

20 33 51 63

40

Root

Page 35: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L7-6

After Inserting 23*, 48*, 41*, 42* ...After Inserting 23*, 48*, 41*, 42* ...

10* 15* 20* 27* 33* 37* 40* 46* 51* 55* 63* 97*

20 33 51 63

40

Root

23* 48* 41*

42*

Overflow

Pages

Leaf

Index

Pages

Pages

Primary

Page 36: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L7-7

... Then Deleting 42*, 51*, 97*... Then Deleting 42*, 51*, 97*

10* 15* 20* 27* 33* 37* 40* 46* 55* 63*

20 33 51 63

40

Root

23* 48* 41*

Page 37: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L8-1

B+ Tree: Most Widely Used IndexB+ Tree: 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). Each

node contains d <= m <= 2d entries. The parameter d is called the order of the tree.

• Supports equality and range-searches efficiently.

Index Entries

Data Entries("Sequence set")

(Direct search)

Page 38: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L8-2

Example B+ TreeExample B+ Tree

• Search begins at root, and key comparisons direct it to a leaf (as in ISAM).

• Search for 5*, 15*, all data entries >= 24* ...

Root

17 24 30

2* 3* 5* 7* 14* 16* 19* 20* 22* 24* 27* 29* 33* 34* 38* 39*

13

Page 39: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L8-3

B+ Trees in PracticeB+ Trees in Practice

• Typical order: 100. 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

Page 40: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L8-4

Inserting a Data Entry into a B+ TreeInserting 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.

Page 41: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

Slide No:L8-5

Inserting 8* into Example B+ TreeInserting 8* into Example B+ Tree

• Observe how minimum occupancy is guaranteed in both leaf and index pg splits.

• Note difference between copy-up and push-up; be sure you understand the reasons for this.

2* 3* 5* 7* 8*

5

Entry to be inserted in parent node.(Note that 5 iscontinues to appear in the leaf.)

s copied up and

appears once in the index. Contrast

5 24 30

17

13

Entry to be inserted in parent node.(Note that 17 is pushed up and only

this with a leaf split.)

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Slide No:L8-6

Example B+ Tree After Inserting 8*Example B+ Tree After Inserting 8*

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

17

24 30

14* 16* 19* 20* 22* 24* 27* 29* 33* 34* 38* 39*

135

7*5* 8*

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Slide No:L8-7

Deleting a Data Entry from a B+ TreeDeleting 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.

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Slide No:L8-8

Example Tree After (Inserting 8*, Then) Example Tree After (Inserting 8*, Then) Deleting 19* and 20* ...Deleting 19* and 20* ...

• Deleting 19* is easy.• Deleting 20* is done with re-distribution.

Notice how middle key is copied up.

2* 3*

Root

17

30

14* 16* 33* 34* 38* 39*

135

7*5* 8* 22* 24*

27

27* 29*

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Slide No:L8-9

... And Then Deleting 24*... And Then Deleting 24*

• Must merge.• Observe `toss’ of

index entry (on right), and `pull down’ of index entry (below).

30

22* 27* 29* 33* 34* 38* 39*

2* 3* 7* 14* 16* 22* 27* 29* 33* 34* 38* 39*5* 8*

Root30135 17

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Chapter 6: Indexing and HashingChapter 6: Indexing and Hashing

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Chapter 6: Indexing and HashingChapter 6: Indexing and Hashing

Basic Concepts Ordered Indices B+-Tree Index Files B-Tree Index Files Static Hashing Dynamic Hashing Comparison of Ordered Indexing and

Hashing Index Definition in SQL Multiple-Key Access

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Basic ConceptsBasic Concepts

Indexing mechanisms used to speed up access to desired data. E.g., author catalog in library

Search Key - attribute to set of attributes used to look up records in a file.

An index file consists of records (called index entries) of the form

Index files are typically much smaller than the original file

Two basic kinds of indices: Ordered indices: search keys are stored in

sorted order Hash indices: search keys are distributed

uniformly across “buckets” using a “hash function”.

search-key pointer

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Index Evaluation MetricsIndex Evaluation Metrics

Access types supported efficiently. E.g., records with a specified value in the

attribute or records with an attribute value

falling in a specified range of values. Access time Insertion time Deletion time Space overhead

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Ordered IndicesOrdered Indices

In an ordered index, index entries are stored sorted on the search key value. E.g., author catalog in library.

Primary index: in a sequentially ordered file, the index whose search key specifies the sequential order of the file. Also called clustering index The search key of a primary index is usually but

not necessarily the primary key. Secondary index: an index whose search key

specifies an order different from the sequential order of the file. Also called non-clustering index.

Index-sequential file: ordered sequential file with a primary index.

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Dense Index FilesDense Index Files

Dense index — Index record appears for every search-key value in the file.

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Sparse Index FilesSparse Index Files

Sparse Index: contains index records for only some search-key values. Applicable when records are sequentially ordered on

search-key To locate a record with search-key value K we:

Find index record with largest search-key value < K Search file sequentially starting at the record to which

the index record points

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Sparse Index Files (Cont.)Sparse Index Files (Cont.)

Compared to dense indices: Less space and less maintenance

overhead for insertions and deletions.

Generally slower than dense index for locating records.

Good tradeoff: sparse index with an index entry for every block in file, corresponding to least search-key value in the block.

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Multilevel IndexMultilevel Index If primary index does not fit in memory, access

becomes expensive. Solution: treat primary index kept on disk as a

sequential file and construct a sparse index on it. outer index – a sparse index of primary index inner index – the primary index file

If even outer index is too large to fit in main memory, yet another level of index can be created, and so on.

Indices at all levels must be updated on insertion or deletion from the file.

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Multilevel Index (Cont.)Multilevel Index (Cont.)

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Index Update: DeletionIndex Update: Deletion

If deleted record was the only record in the file with its particular search-key value, the search-key is deleted from the index also.

Single-level index deletion: Dense indices – deletion of search-key:similar to file

record deletion. Sparse indices –

if an entry for the search key exists in the index, it is deleted by replacing the entry in the index with the next search-key value in the file (in search-key order).

If the next search-key value already has an index entry, the entry is deleted instead of being replaced.

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Index Update: InsertionIndex Update: Insertion

Single-level index insertion: Perform a lookup using the search-key value

appearing in the record to be inserted. Dense indices – if the search-key value does

not appear in the index, insert it. Sparse indices – if index stores an entry for

each block of the file, no change needs to be made to the index unless a new block is created. If a new block is created, the first search-key

value appearing in the new block is inserted into the index.

Multilevel insertion (as well as deletion) algorithms are simple extensions of the single-level algorithms

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Secondary IndicesSecondary Indices

Frequently, one wants to find all the records whose values in a certain field (which is not the search-key of the primary index) satisfy some condition. Example 1: In the account relation stored

sequentially by account number, we may want to find all accounts in a particular branch

Example 2: as above, but where we want to find all accounts with a specified balance or range of balances

We can have a secondary index with an index record for each search-key value

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Secondary Indices ExampleSecondary Indices Example

Index record points to a bucket that contains pointers to all the actual records with that particular search-key value.

Secondary indices have to be dense

Secondary index on balance field of account

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Primary and Secondary IndicesPrimary and Secondary Indices

Indices offer substantial benefits when searching for records.

BUT: Updating indices imposes overhead on database modification --when a file is modified, every index on the file must be updated,

Sequential scan using primary index is efficient, but a sequential scan using a secondary index is expensive Each record access may fetch a new block

from disk Block fetch requires about 5 to 10

milliseconds versus about 100 nanoseconds for memory

access

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BB++-Tree Index Files-Tree Index Files

Disadvantage of indexed-sequential files performance degrades as file grows, since many

overflow blocks get created. Periodic reorganization of entire file is required.

Advantage of B+-tree index files: automatically reorganizes itself with small, local,

changes, in the face of insertions and deletions. Reorganization of entire file is not required to

maintain performance. (Minor) disadvantage of B+-trees:

extra insertion and deletion overhead, space overhead.

Advantages of B+-trees outweigh disadvantages B+-trees are used extensively

B+-tree indices are an alternative to indexed-sequential files.

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BB++-Tree Index Files (Cont.)-Tree Index Files (Cont.)

All paths from root to leaf are of the same length

Each node that is not a root or a leaf has between n/2 and n children.

A leaf node has between (n–1)/2 and n–1 values

Special cases: If the root is not a leaf, it has at least 2

children. If the root is a leaf (that is, there are no

other nodes in the tree), it can have between 0 and (n–1) values.

A B+-tree is a rooted tree satisfying the following properties:

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BB++-Tree Node Structure-Tree Node Structure

Typical node

Ki are the search-key values

Pi are pointers to children (for non-leaf nodes) or pointers to records or buckets of records (for leaf nodes).

The search-keys in a node are ordered

K1 < K2 < K3 < . . . < Kn–1

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Leaf Nodes in BLeaf Nodes in B++-Trees-Trees

For i = 1, 2, . . ., n–1, pointer Pi either points to a file record with search-key value Ki, or to a bucket of pointers to file records, each record having search-key value Ki. Only need bucket structure if search-key does not form a primary key.

If Li, Lj are leaf nodes and i < j, Li’s search-key values are less than Lj’s search-key values

Pn points to next leaf node in search-key order

Properties of a leaf node:

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Non-Leaf Nodes in BNon-Leaf Nodes in B++-Trees-Trees

Non leaf nodes form a multi-level sparse index on the leaf nodes. For a non-leaf node with m pointers:

All the search-keys in the subtree to which P1 points are less than K1

For 2 i n – 1, all the search-keys in the subtree to which Pi points have values greater than or equal to Ki–1 and less than Ki

All the search-keys in the subtree to which Pn points have values greater than or equal to Kn–1

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Example of a BExample of a B++-tree-tree

B+-tree for account file (n = 3)

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Example of BExample of B++-tree-tree

Leaf nodes must have between 2 and 4 values ((n–1)/2 and n –1, with n = 5).

Non-leaf nodes other than root must have between 3 and 5 children ((n/2 and n with n =5).

Root must have at least 2 children.

B+-tree for account file (n = 5)

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Observations about BObservations about B++-trees-trees

Since the inter-node connections are done by pointers, “logically” close blocks need not be “physically” close.

The non-leaf levels of the B+-tree form a hierarchy of sparse indices.

The B+-tree contains a relatively small number of levels

Level below root has at least 2* n/2 valuesNext level has at least 2* n/2 * n/2 values.. etc.

If there are K search-key values in the file, the tree height is no more than logn/2(K)

thus searches can be conducted efficiently. Insertions and deletions to the main file can be

handled efficiently, as the index can be restructured in logarithmic time (as we shall see).

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Queries on BQueries on B++-Trees-Trees

Find all records with a search-key value of k.

1. N=root2. Repeat

1. Examine N for the smallest search-key value > k.2. If such a value exists, assume it is Ki. Then set N = Pi

3. Otherwise k Kn–1. Set N = Pn

Until N is a leaf node3. If for some i, key Ki = k follow pointer Pi to the desired record

or bucket. 4. Else no record with search-key value k exists.

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Queries on BQueries on B+-+-Trees (Cont.)Trees (Cont.)

If there are K search-key values in the file, the height of the tree is no more than logn/2(K).

A node is generally the same size as a disk block, typically 4 kilobytes and n is typically around 100 (40 bytes per

index entry). With 1 million search key values and n = 100

at most log50(1,000,000) = 4 nodes are accessed in a lookup.

Contrast this with a balanced binary tree with 1 million search key values — around 20 nodes are accessed in a lookup above difference is significant since every node

access may need a disk I/O, costing around 20 milliseconds

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Updates on BUpdates on B++-Trees: Insertion-Trees: Insertion

1. Find the leaf node in which the search-key value would appear

2. If the search-key value is already present in the leaf node

1. Add record to the file

2. If necessary add a pointer to the bucket.

3. If the search-key value is not present, then

1. add the record to the main file (and create a bucket if necessary)

2. If there is room in the leaf node, insert (key-value, pointer) pair in the leaf node

3. Otherwise, split the node (along with the new (key-value, pointer) entry) as discussed in the next slide.

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Updates on BUpdates on B++-Trees: Insertion (Cont.)-Trees: Insertion (Cont.)

Splitting a leaf node: take the n (search-key value, pointer) pairs

(including the one being inserted) in sorted order. Place the first n/2 in the original node, and the rest in a new node.

let the new node be p, and let k be the least key value in p. Insert (k,p) in the parent of the node being split.

If the parent is full, split it and propagate the split further up.

Splitting of nodes proceeds upwards till a node that is not full is found. In the worst case the root node may be split

increasing the height of the tree by 1. Result of splitting node containing Brighton and Downtown on inserting ClearviewNext step: insert entry with (Downtown,pointer-to-new-node) into parent

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Updates on BUpdates on B++-Trees: Insertion (Cont.)-Trees: Insertion (Cont.)

B+-Tree before and after insertion of “Clearview”

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Redwood

Insertion in BInsertion in B++-Trees (Cont.)-Trees (Cont.)

Splitting a non-leaf node: when inserting (k,p) into an already full internal node N Copy N to an in-memory area M with

space for n+1 pointers and n keys Insert (k,p) into M Copy P1,K1, …, K n/2-1,P n/2 from M back

into node N Copy Pn/2+1,K n/2+1,…,Kn,Pn+1 from M into

newly allocated node N’ Insert (K n/2,N’) into parent N

Read pseudocode in book!

Downtown Mianus Perryridge Downtown

Mianus

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Updates on BUpdates on B++-Trees: Deletion-Trees: Deletion

Find the record to be deleted, and remove it from the main file and from the bucket (if present)

Remove (search-key value, pointer) from the leaf node if there is no bucket or if the bucket has become empty

If the node has too few entries due to the removal, and the entries in the node and a sibling fit into a single node, then merge siblings: Insert all the search-key values in the two nodes into

a single node (the one on the left), and delete the other node.

Delete the pair (Ki–1, Pi), where Pi is the pointer to the deleted node, from its parent, recursively using the above procedure.

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Updates on BUpdates on B++-Trees: Deletion-Trees: Deletion

Otherwise, if the node has too few entries due to the removal, but the entries in the node and a sibling do not fit into a single node, then redistribute pointers: Redistribute the pointers between the node

and a sibling such that both have more than the minimum number of entries.

Update the corresponding search-key value in the parent of the node.

The node deletions may cascade upwards till a node which has n/2 or more pointers is found.

If the root node has only one pointer after deletion, it is deleted and the sole child becomes the root.

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Examples of BExamples of B++-Tree Deletion-Tree Deletion

Deleting “Downtown” causes merging of under-full leaves leaf node can become empty only for n=3!

Before and after deleting “Downtown”

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Examples of BExamples of B++-Tree Deletion (Cont.)-Tree Deletion (Cont.)

Leaf with “Perryridge” becomes underfull (actually empty, in this special case) and merged with its sibling.

As a result “Perryridge” node’s parent became underfull, and was merged with its sibling

Value separating two nodes (at parent) moves into merged node Entry deleted from parent

Root node then has only one child, and is deleted

Deletion of “Perryridge” from result of previous example

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Example of BExample of B++-tree Deletion (Cont.)-tree Deletion (Cont.)

Parent of leaf containing Perryridge became underfull, and borrowed a pointer from its left sibling

Search-key value in the parent’s parent changes as a result

Before and after deletion of “Perryridge” from earlier example

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BB++-Tree File Organization-Tree File Organization

Index file degradation problem is solved by using B+-Tree indices.

Data file degradation problem is solved by using B+-Tree File Organization.

The leaf nodes in a B+-tree file organization store records, instead of pointers.

Leaf nodes are still required to be half full Since records are larger than pointers, the

maximum number of records that can be stored in a leaf node is less than the number of pointers in a nonleaf node.

Insertion and deletion are handled in the same way as insertion and deletion of entries in a B+-tree index.

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BB++-Tree File Organization (Cont.)-Tree File Organization (Cont.)

Good space utilization important since records use more space than pointers.

To improve space utilization, involve more sibling nodes in redistribution during splits and merges Involving 2 siblings in redistribution (to avoid split / merge

where possible) results in each node having at least entries

Example of B+-tree File Organization

3/2n

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Indexing StringsIndexing Strings

Variable length strings as keys Variable fanout Use space utilization as criterion for splitting,

not number of pointers Prefix compression

Key values at internal nodes can be prefixes of full keyKeep enough characters to distinguish entries

in the subtrees separated by the key value–E.g. “Silas” and “Silberschatz” can be

separated by “Silb” Keys in leaf node can be compressed by

sharing common prefixes

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B-Tree Index FilesB-Tree Index Files

Similar to B+-tree, but B-tree allows search-key values to appear only once; eliminates redundant storage of search keys.

Search keys in nonleaf nodes appear nowhere else in the B-tree; an additional pointer field for each search key in a nonleaf node must be included.

Generalized B-tree leaf node

Nonleaf node – pointers Bi are the bucket or file record pointers.

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B-Tree Index File ExampleB-Tree Index File Example

B-tree (above) and B+-tree (below) on same data

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B-Tree Index Files (Cont.)B-Tree Index Files (Cont.)

Advantages of B-Tree indices: May use less tree nodes than a corresponding B+-Tree. Sometimes possible to find search-key value before

reaching leaf node. Disadvantages of B-Tree indices:

Only small fraction of all search-key values are found early

Non-leaf nodes are larger, so fan-out is reduced. Thus, B-Trees typically have greater depth than corresponding B+-Tree

Insertion and deletion more complicated than in B+-Trees Implementation is harder than B+-Trees.

Typically, advantages of B-Trees do not out weigh disadvantages.

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Multiple-Key AccessMultiple-Key Access

Use multiple indices for certain types of queries. Example:

select account_numberfrom accountwhere branch_name = “Perryridge” and balance =

1000 Possible strategies for processing query using indices on

single attributes:

1. Use index on branch_name to find accounts with branch name Perryridge; test balance = 1000

2. Use index on balance to find accounts with balances of $1000; test branch_name = “Perryridge”.

3. Use branch_name index to find pointers to all records pertaining to the Perryridge branch. Similarly use index on balance. Take intersection of both sets of pointers obtained.

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Indices on Multiple KeysIndices on Multiple Keys

Composite search keys are search keys containing more than one attribute E.g. (branch_name, balance)

Lexicographic ordering: (a1, a2) < (b1, b2) if either

a1 < b1, or

a1=b1 and a2 < b2

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Indices on Multiple AttributesIndices on Multiple Attributes

With the where clause where branch_name = “Perryridge” and balance = 1000the index on (branch_name, balance) can be used to fetch only records that satisfy both conditions. Using separate indices in less efficient — we may fetch

many records (or pointers) that satisfy only one of the conditions.

Can also efficiently handle where branch_name = “Perryridge” and balance < 1000

But cannot efficiently handle where branch_name < “Perryridge” and balance = 1000 May fetch many records that satisfy the first but not the

second condition

Suppose we have an index on combined search-key(branch_name, balance).

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Non-Unique Search KeysNon-Unique Search Keys

Alternatives: Buckets on separate block (bad idea) List of tuple pointers with each key

Extra code to handle long listsDeletion of a tuple can be expensive if there

are many duplicates on search key (why?)Low space overhead, no extra cost for queries

Make search key unique by adding a record-identifierExtra storage overhead for keysSimpler code for insertion/deletionWidely used

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Other Issues in IndexingOther Issues in Indexing

Covering indices Add extra attributes to index so (some) queries can

avoid fetching the actual recordsParticularly useful for secondary indices

–Why? Can store extra attributes only at leaf

Record relocation and secondary indices If a record moves, all secondary indices that store

record pointers have to be updated Node splits in B+-tree file organizations become very

expensive Solution: use primary-index search key instead of

record pointer in secondary indexExtra traversal of primary index to locate record

–Higher cost for queries, but node splits are cheapAdd record-id if primary-index search key is non-

unique

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HashingHashing

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Static HashingStatic Hashing

A bucket is a unit of storage containing one or more records (a bucket is typically a disk block).

In a hash file organization we obtain the bucket of a record directly from its search-key value using a hash function.

Hash function h is a function from the set of all search-key values K to the set of all bucket addresses B.

Hash function is used to locate records for access, insertion as well as deletion.

Records with different search-key values may be mapped to the same bucket; thus entire bucket has to be searched sequentially to locate a record.

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Example of Hash File OrganizationExample of Hash File Organization

There are 10 buckets, The binary representation of the ith

character is assumed to be the integer i.

The hash function returns the sum of the binary representations of the characters modulo 10 E.g. h(Perryridge) = 5 h(Round

Hill) = 3 h(Brighton) = 3

Hash file organization of account file, using branch_name as key (See figure in next slide.)

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Example of Hash File Organization Example of Hash File Organization

Hash file organization of account file, using branch_name as key(see previous slide for details).

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Hash FunctionsHash Functions

Worst hash function maps all search-key values to the same bucket; this makes access time proportional to the number of search-key values in the file.

An ideal hash function is uniform, i.e., each bucket is assigned the same number of search-key values from the set of all possible values.

Ideal hash function is random, so each bucket will have the same number of records assigned to it irrespective of the actual distribution of search-key values in the file.

Typical hash functions perform computation on the internal binary representation of the search-key. For example, for a string search-key, the binary

representations of all the characters in the string could be added and the sum modulo the number of buckets could be returned. .

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Handling of Bucket OverflowsHandling of Bucket Overflows

Bucket overflow can occur because of Insufficient buckets Skew in distribution of records. This

can occur due to two reasons:multiple records have same search-

key valuechosen hash function produces non-

uniform distribution of key values Although the probability of bucket

overflow can be reduced, it cannot be eliminated; it is handled by using overflow buckets.

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Handling of Bucket Overflows (Cont.)Handling of Bucket Overflows (Cont.)

Overflow chaining – the overflow buckets of a given bucket are chained together in a linked list.

Above scheme is called closed hashing. An alternative, called open hashing,

which does not use overflow buckets, is not suitable for database applications.

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Hash IndicesHash Indices

Hashing can be used not only for file organization, but also for index-structure creation.

A hash index organizes the search keys, with their associated record pointers, into a hash file structure.

Strictly speaking, hash indices are always secondary indices if the file itself is organized using hashing, a

separate primary hash index on it using the same search-key is unnecessary.

However, we use the term hash index to refer to both secondary index structures and hash organized files.

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Example of Hash IndexExample of Hash Index

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Deficiencies of Static HashingDeficiencies of Static Hashing

In static hashing, function h maps search-key values to a fixed set of B of bucket addresses. Databases grow or shrink with time. If initial number of buckets is too small, and file grows,

performance will degrade due to too much overflows. If space is allocated for anticipated growth, a significant

amount of space will be wasted initially (and buckets will be underfull).

If database shrinks, again space will be wasted. One solution: periodic re-organization of the file with a new

hash function Expensive, disrupts normal operations

Better solution: allow the number of buckets to be modified dynamically.

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Dynamic HashingDynamic Hashing

Good for database that grows and shrinks in size Allows the hash function to be modified dynamically Extendable hashing – one form of dynamic hashing

Hash function generates values over a large range — typically b-bit integers, with b = 32.

At any time use only a prefix of the hash function to index into a table of bucket addresses.

Let the length of the prefix be i bits, 0 i 32. Bucket address table size = 2i. Initially i = 0Value of i grows and shrinks as the size of the

database grows and shrinks. Multiple entries in the bucket address table may

point to a bucket (why?) Thus, actual number of buckets is < 2i

The number of buckets also changes dynamically due to coalescing and splitting of buckets.

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General Extendable Hash Structure General Extendable Hash Structure

In this structure, i2 = i3 = i, whereas i1 = i – 1 (see next slide for details)

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Use of Extendable Hash StructureUse of Extendable Hash Structure

Each bucket j stores a value ij

All the entries that point to the same bucket have the same values on the first ij bits.

To locate the bucket containing search-key Kj:

1. Compute h(Kj) = X

2. Use the first i high order bits of X as a displacement into bucket address table, and follow the pointer to appropriate bucket

To insert a record with search-key value Kj

follow same procedure as look-up and locate the bucket, say j. If there is room in the bucket j insert record in the bucket. Else the bucket must be split and insertion re-attempted (next

slide.)Overflow buckets used instead in some cases (will see shortly)

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Insertion in Extendable Hash Structure (Cont) Insertion in Extendable Hash Structure (Cont)

If i > ij (more than one pointer to bucket j) allocate a new bucket z, and set ij = iz = (ij + 1) Update the second half of the bucket address table entries

originally pointing to j, to point to z remove each record in bucket j and reinsert (in j or z) recompute new bucket for Kj and insert record in the bucket

(further splitting is required if the bucket is still full) If i = ij (only one pointer to bucket j)

If i reaches some limit b, or too many splits have happened in this insertion, create an overflow bucket

Elseincrement i and double the size of the bucket address

table.replace each entry in the table by two entries that point to

the same bucket.recompute new bucket address table entry for Kj

Now i > ij so use the first case above.

To split a bucket j when inserting record with search-key value Kj:

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Deletion in Extendable Hash StructureDeletion in Extendable Hash Structure To delete a key value,

locate it in its bucket and remove it. The bucket itself can be removed if it becomes empty

(with appropriate updates to the bucket address table).

Coalescing of buckets can be done (can coalesce only with a “buddy” bucket having same value of ij and same ij –1 prefix, if it is present)

Decreasing bucket address table size is also possibleNote: decreasing bucket address table size is an

expensive operation and should be done only if number of buckets becomes much smaller than the size of the table

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Use of Extendable Hash Structure: Example Use of Extendable Hash Structure: Example

Initial Hash structure, bucket size = 2

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Example (Cont.)Example (Cont.)

Hash structure after insertion of one Brighton and two Downtown records

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Example (Cont.)Example (Cont.)

Hash structure after insertion of Mianus record

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Example (Cont.)Example (Cont.)

Hash structure after insertion of three Perryridge records

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Example (Cont.)Example (Cont.)

Hash structure after insertion of Redwood and Round Hill records

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Extendable Hashing vs. Other SchemesExtendable Hashing vs. Other Schemes

Benefits of extendable hashing: Hash performance does not degrade with growth of file Minimal space overhead

Disadvantages of extendable hashing Extra level of indirection to find desired record Bucket address table may itself become very big (larger

than memory)Cannot allocate very large contiguous areas on disk

eitherSolution: B+-tree structure to locate desired record in

bucket address table Changing size of bucket address table is an expensive

operation Linear hashing is an alternative mechanism

Allows incremental growth of its directory (equivalent to bucket address table)

At the cost of more bucket overflows

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Comparison of Ordered Indexing and HashingComparison of Ordered Indexing and Hashing

Cost of periodic re-organization Relative frequency of insertions and deletions Is it desirable to optimize average access time at the

expense of worst-case access time? Expected type of queries:

Hashing is generally better at retrieving records having a specified value of the key.

If range queries are common, ordered indices are to be preferred

In practice: PostgreSQL supports hash indices, but discourages

use due to poor performance Oracle supports static hash organization, but not

hash indices SQLServer supports only B+-trees

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Bitmap IndicesBitmap Indices

Bitmap indices are a special type of index designed for efficient querying on multiple keys

Records in a relation are assumed to be numbered sequentially from, say, 0 Given a number n it must be easy to retrieve

record nParticularly easy if records are of fixed size

Applicable on attributes that take on a relatively small number of distinct values E.g. gender, country, state, … E.g. income-level (income broken up into a small

number of levels such as 0-9999, 10000-19999, 20000-50000, 50000- infinity)

A bitmap is simply an array of bits

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Bitmap Indices (Cont.)Bitmap Indices (Cont.)

In its simplest form a bitmap index on an attribute has a bitmap for each value of the attribute Bitmap has as many bits as records In a bitmap for value v, the bit for a

record is 1 if the record has the value v for the attribute, and is 0 otherwise

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Bitmap Indices (Cont.)Bitmap Indices (Cont.)

Bitmap indices are useful for queries on multiple attributes not particularly useful for single attribute queries

Queries are answered using bitmap operations Intersection (and) Union (or) Complementation (not)

Each operation takes two bitmaps of the same size and applies the operation on corresponding bits to get the result bitmap E.g. 100110 AND 110011 = 100010 100110 OR 110011 = 110111

NOT 100110 = 011001 Males with income level L1: 10010 AND 10100 =

10000 Can then retrieve required tuples. Counting number of matching tuples is even faster

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Bitmap Indices (Cont.)Bitmap Indices (Cont.)

Bitmap indices generally very small compared with relation size E.g. if record is 100 bytes, space for a single bitmap is

1/800 of space used by relation. If number of distinct attribute values is 8, bitmap is

only 1% of relation size Deletion needs to be handled properly

Existence bitmap to note if there is a valid record at a record location

Needed for complementationnot(A=v): (NOT bitmap-A-v) AND ExistenceBitmap

Should keep bitmaps for all values, even null value To correctly handle SQL null semantics for NOT(A=v):

intersect above result with (NOT bitmap-A-Null)

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Efficient Implementation of Bitmap OperationsEfficient Implementation of Bitmap Operations

Bitmaps are packed into words; a single word and (a basic CPU instruction) computes and of 32 or 64 bits at once E.g. 1-million-bit maps can be and-ed with just 31,250

instruction Counting number of 1s can be done fast by a trick:

Use each byte to index into a precomputed array of 256 elements each storing the count of 1s in the binary representationCan use pairs of bytes to speed up further at a higher

memory cost Add up the retrieved counts

Bitmaps can be used instead of Tuple-ID lists at leaf levels of B+-trees, for values that have a large number of matching records Worthwhile if > 1/64 of the records have that value, assuming

a tuple-id is 64 bits Above technique merges benefits of bitmap and B+-tree

indices

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Index Definition in SQLIndex Definition in SQL

Create an index

create index <index-name> on <relation-name>

(<attribute-list>)E.g.: create index b-index on

branch(branch_name) Use create unique index to indirectly specify and

enforce the condition that the search key is a candidate key is a candidate key. Not really required if SQL unique integrity

constraint is supported To drop an index

drop index <index-name> Most database systems allow specification of type of

index, and clustering.

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End of ChapterEnd of Chapter

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Partitioned HashingPartitioned Hashing

Hash values are split into segments that depend on each attribute of the search-key.

(A1, A2, . . . , An) for n attribute search-key

Example: n = 2, for customer, search-key being (customer-street, customer-city)

search-key value hash value(Main, Harrison) 101 111(Main, Brooklyn) 101 001(Park, Palo Alto) 010 010(Spring, Brooklyn) 001 001(Alma, Palo Alto) 110 010

To answer equality query on single attribute, need to look up multiple buckets. Similar in effect to grid files.

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Sequential File For Sequential File For account account RecordsRecords

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Sample Sample accountaccount File File

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Figure 12.2Figure 12.2

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Figure 12.14Figure 12.14

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Figure 12.25Figure 12.25

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Grid FilesGrid Files

Structure used to speed the processing of general multiple search-key queries involving one or more comparison operators.

The grid file has a single grid array and one linear scale for each search-key attribute. The grid array has number of dimensions equal to number of search-key attributes.

Multiple cells of grid array can point to same bucket

To find the bucket for a search-key value, locate the row and column of its cell using the linear scales and follow pointer

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Example Grid File for Example Grid File for accountaccount

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Queries on a Grid FileQueries on a Grid File

A grid file on two attributes A and B can handle queries of all following forms with reasonable efficiency

(a1 A a2)

(b1 B b2)

(a1 A a2 b1 B b2),. E.g., to answer (a1 A a2 b1 B

b2), use linear scales to find corresponding candidate grid array cells, and look up all the buckets pointed to from those cells.

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Grid Files (Cont.)Grid Files (Cont.)

During insertion, if a bucket becomes full, new bucket can be created if more than one cell points to it. Idea similar to extendable hashing, but on multiple

dimensions If only one cell points to it, either an overflow

bucket must be created or the grid size must be increased

Linear scales must be chosen to uniformly distribute records across cells. Otherwise there will be too many overflow buckets.

Periodic re-organization to increase grid size will help. But reorganization can be very expensive.

Space overhead of grid array can be high. R-trees (Chapter 23) are an alternative

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Indexing Structures for FilesIndexing Structures for Files

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Slide 14-132

FIGURE FIGURE 14.114.1

Primary Primary index on index on

the ordering the ordering key field of key field of

the file the file shown in shown in

Figure 13.7.Figure 13.7.

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Slide 14-133

FIGURE 14.2FIGURE 14.2A clustering index on A clustering index on the DEPTNUMBER the DEPTNUMBER ordering nonkey field ordering nonkey field

of an EMPLOYEE file.of an EMPLOYEE file.

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Slide 14-134

FIGURE 14.3FIGURE 14.3Clustering index Clustering index with a separate with a separate

block cluster for block cluster for each group of each group of

records that share records that share the same value for the same value for

the clustering the clustering field.field.

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Slide 14-135

FIGURE 14.4FIGURE 14.4A dense A dense

secondary secondary index (with index (with

block pointers) block pointers) on a on a

nonordering nonordering key field of a key field of a

file.file.

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Slide 14-136

FIGURE 14.5FIGURE 14.5A secondary index (with recored pointers) on a A secondary index (with recored pointers) on a nonkey field implemented using one level of nonkey field implemented using one level of

indirection so that index entries are of fixed length indirection so that index entries are of fixed length and have unique field values.and have unique field values.

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Slide 14-137

FIGURE FIGURE 14.614.6

A two-level A two-level primary primary index index

resembling resembling ISAM ISAM

(Indexed (Indexed Sequential Sequential

Access Access Method) Method)

organization.organization.

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Slide 14-138

FIGURE 14.7FIGURE 14.7A tree data structure that shows an A tree data structure that shows an

unbalanced tree.unbalanced tree.

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Slide 14-139

FIGURE 14.8FIGURE 14.8A node in a search tree with pointers to A node in a search tree with pointers to

subtrees subtrees below it.below it.

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Slide 14-140

FIGURE 14.9FIGURE 14.9A search tree of order p = 3.A search tree of order p = 3.

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Slide 14-141

FIGURE 14.10FIGURE 14.10B-tree structures. (a) A node in a B-tree with B-tree structures. (a) A node in a B-tree with q – 1 search values. (b) A B-tree of order p = q – 1 search values. (b) A B-tree of order p = 3. The values were inserted in the order 8, 5, 3. The values were inserted in the order 8, 5,

1, 7, 3, 12, 9, 6.1, 7, 3, 12, 9, 6.

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Slide 14-142

FIGURE 14.11FIGURE 14.11The nodes of a B+-tree. (a) Internal node of a The nodes of a B+-tree. (a) Internal node of a

B+-tree with q –1 search values. (b) Leaf B+-tree with q –1 search values. (b) Leaf node of a B+-tree with q – 1 search values node of a B+-tree with q – 1 search values

and q – 1 data pointers.and q – 1 data pointers.

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Slide 14-143

FIGURE 14.12FIGURE 14.12An example of An example of

insertion in a B+-insertion in a B+-tree with q = 3 and tree with q = 3 and

ppleafleaf = 2. = 2.

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Slide 14-144

FIGURE 14.13FIGURE 14.13An example of An example of deletion from a deletion from a

B+-tree.B+-tree.

Page 145: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

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FIGURE 14.14FIGURE 14.14Example of a grid array on DNO and AGE Example of a grid array on DNO and AGE

attributes.attributes.

Page 146: DATABASE MANAGEMENT SYSTEMS TERM 2008-09 B. Tech II/IT II Semester UNIT-VI PPT SLIDES Text Books: (1) DBMS by Raghu Ramakrishnan (2) DBMS by Sudarshan.

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FIGURE 14.15FIGURE 14.15B+-tree B+-tree

insertion with insertion with left left

redistribution.redistribution.