CPS216: Advanced Database Systems Notes 02:Query Processing (Overview)
CPS216: Advanced Database Systems
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1 CPS216: Advanced CPS216: Advanced Database Systems Database Systems Notes 05: Operators for Notes 05: Operators for Data Access Data Access Shivnath Babu Shivnath Babu
description
CPS216: Advanced Database Systems. Notes 05: Operators for Data Access Shivnath Babu. Problem. Relation: Employee (ID, Name, Dept, …) 10 M tuples (Filter) Query: SELECT * FROM Employee WHERE Name = “Bob”. Solution #1: Full Table Scan. Storage: - PowerPoint PPT Presentation
Transcript of CPS216: Advanced Database Systems
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CPS216: Advanced CPS216: Advanced Database SystemsDatabase Systems
Notes 05: Operators for Data Notes 05: Operators for Data AccessAccess
Shivnath BabuShivnath Babu
ProblemProblem
Relation: Employee (ID, Name, Dept, Relation: Employee (ID, Name, Dept, …)…)
10 M tuples10 M tuples (Filter) Query:(Filter) Query:
SELECT *SELECT * FROM Employee FROM Employee WHERE Name = “Bob” WHERE Name = “Bob”
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Solution #1: Full Table Solution #1: Full Table ScanScan
Storage:Storage: Employee relation stored in Employee relation stored in contiguouscontiguous
blocksblocks Query plan:Query plan:
Scan the entire relation, output tuples with Scan the entire relation, output tuples with Name = “Bob”Name = “Bob”
Cost:Cost: Size of each record = 100 bytesSize of each record = 100 bytes Size of relation = 10 M x 100 = 1 GBSize of relation = 10 M x 100 = 1 GB Time @ 20 MB/s ≈ 1 Minute Time @ 20 MB/s ≈ 1 Minute
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Solution #2Solution #2
Storage:Storage: Employee relation Employee relation sortedsorted on Name on Name
attributeattribute Query plan:Query plan:
Binary searchBinary search
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Solution #2Solution #2
Cost:Cost: Size of a block: 1024 bytesSize of a block: 1024 bytes Number of records per block: 1024 / Number of records per block: 1024 /
100 = 10100 = 10 Total number of blocks: 10 M / 10 = 1 Total number of blocks: 10 M / 10 = 1
MM Blocks accessed by binary search: 20Blocks accessed by binary search: 20 Total time: 20 ms x 20 = 400 msTotal time: 20 ms x 20 = 400 ms
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Solution #2: IssuesSolution #2: Issues
Filters on different attributes:Filters on different attributes:
SELECT * SELECT * FROM EmployeeFROM EmployeeWHERE Dept = “Sales”WHERE Dept = “Sales”
Inserts and DeletesInserts and Deletes
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IndexesIndexes
Data structures that efficiently evaluate Data structures that efficiently evaluate a class of filter predicates over a relationa class of filter predicates over a relation
Class of filter predicates:Class of filter predicates: Single or multi-attributes (Single or multi-attributes (index-key index-key
attributesattributes)) Range and/or equality predicatesRange and/or equality predicates
(Usually) independent of physical (Usually) independent of physical storage of relation:storage of relation: Multiple indexes per relationMultiple indexes per relation
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IndexesIndexes
Disk residentDisk resident Large to fit in memoryLarge to fit in memory PersistentPersistent
Updated when indexed relation Updated when indexed relation updatedupdated Relation updates costlierRelation updates costlier Query cheaperQuery cheaper
ProblemProblem
Relation: Employee (ID, Name, Dept, Relation: Employee (ID, Name, Dept, …)…)
(Filter) Query:(Filter) Query: SELECT *SELECT * FROM Employee FROM Employee WHERE Name = “Bob” WHERE Name = “Bob”
Single-Attribute Index on NameName that supports equality predicates
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RoadmapRoadmap
MotivationMotivation Single-Attribute Indexes: OverviewSingle-Attribute Indexes: Overview Order-based IndexesOrder-based Indexes
B-TreesB-Trees Hash-based Indexes (May cover in Hash-based Indexes (May cover in
future)future) Extensible HashingExtensible Hashing Linear HashingLinear Hashing
Multi-Attribute Indexes (Chapter 14 Multi-Attribute Indexes (Chapter 14 GMUW, May cover in future)GMUW, May cover in future)
Single Attribute Index: General Single Attribute Index: General ConstructionConstruction
b1
2b
ib
nb
a1
2a
ia
na
A B
Single Attribute Index: General Single Attribute Index: General ConstructionConstruction
b1
2b
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a1
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a1
2a
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A B
A = val
A > lowA < high
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ExceptionsExceptions
Sparse IndexesSparse Indexes Require specific physical layout of Require specific physical layout of
relationrelation Example: Relation sorted on indexed Example: Relation sorted on indexed
attributeattribute More efficientMore efficient
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Single Attribute Index: General Single Attribute Index: General ConstructionConstruction
b1
2b
ib
nb
a1
2a
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na
a1
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A = val
A > lowA < high
Textbook: Dense Index
Single Attribute Index: General Single Attribute Index: General ConstructionConstruction
a1
2a
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na
A = val
A > lowA < high
How do we organize(attribute, pointer) pairs?
Idea: Use dictionary data structures
Issue: Disk resident?
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RoadmapRoadmap
MotivationMotivation Single-Attribute Indexes: OverviewSingle-Attribute Indexes: Overview Order-based IndexesOrder-based Indexes
B-TreesB-Trees Hash-based IndexesHash-based Indexes
Extensible HashingExtensible Hashing Linear HashingLinear Hashing
Multi-Attribute IndexesMulti-Attribute Indexes
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B-TreesB-Trees
Adaptation of search tree data Adaptation of search tree data structurestructure 2-3 trees2-3 trees
Supports range predicates (and Supports range predicates (and equality)equality)
Use Binary Search Tree Use Binary Search Tree Directly?Directly?
16 32 54 71 74 83 92
16 74
71
54
32
92
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Use Binary Search Tree Use Binary Search Tree Directly?Directly?
Store records of type Store records of type <key, left-ptr, right-ptr, data-ptr><key, left-ptr, right-ptr, data-ptr>
Remember position of rootRemember position of root Question: will this work?Question: will this work?
YesYes But we can do better!But we can do better!
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Use Binary Search Tree Use Binary Search Tree Directly?Directly?
Number of keys: 1 MNumber of keys: 1 M Number of levels: log (2^20) = 20Number of levels: log (2^20) = 20 Total cost index lookup: 20 random Total cost index lookup: 20 random
disk I/Odisk I/O 20 x 20 ms = 400 ms20 x 20 ms = 400 ms
B-Tree: less than 3 random disk I/OB-Tree: less than 3 random disk I/O
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B-Tree vs. Binary Search B-Tree vs. Binary Search TreeTree
k k1 k2 k3 k40
1 Random I/O prunes tree by half
1 Random I/O prunes tree by 40
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B-Tree ExampleB-Tree Example
15 36 57 63 76 87 92 100
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B-Tree ExampleB-Tree Example
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15 36 57
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63 76 87
91
92 100
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Meaning of Internal Meaning of Internal NodeNode
84 91
key < 84 84 ≤ key < 91 91 ≤ key
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B-Tree ExampleB-Tree Example
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15 36 57
84
63 76 87
91
92 100
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Meaning of Leaf NodesMeaning of Leaf Nodes
63 76
pointer to record 63pointer to record 76
Next leaf
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Equality PredicatesEquality Predicates
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15 36 57
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63 76 87
91
92 100
key = 87
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Equality PredicatesEquality Predicates
null
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15 36 57
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63 76 87
91
92 100
key = 87
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Equality PredicatesEquality Predicates
null
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15 36 57
84
63 76 87
91
92 100
key = 87
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Equality PredicatesEquality Predicates
null
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15 36 57
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63 76 87
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92 100
key = 87
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Range PredicatesRange Predicates
null
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15 36 57
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63 76 87
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92 100
57 ≤ key < 95
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Range PredicatesRange Predicates
null
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15 36 57
84
63 76 87
91
92 100
57 ≤ key < 95
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Range PredicatesRange Predicates
null
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15 36 57
84
63 76 87
91
92 100
57 ≤ key < 95
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Range PredicatesRange Predicates
null
63
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15 36 57
84
63 76 87
91
92 100
57 ≤ key < 95
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Range PredicatesRange Predicates
null
63
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15 36 57
84
63 76 87
91
92 100
57 ≤ key < 95
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Range PredicatesRange Predicates
null
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15 36 57
84
63 76 87
91
92 100
57 ≤ key < 95
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General B-TreesGeneral B-Trees
Fixed parameter: nFixed parameter: n Number of keys: nNumber of keys: n Number of pointers: n + 1Number of pointers: n + 1
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B-Tree ExampleB-Tree Example
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15 36 57
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63 76 87
91
92 100
n = 2
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General B-TreesGeneral B-Trees
Fixed parameter: nFixed parameter: n Number of keys: nNumber of keys: n Number of pointers: n + 1Number of pointers: n + 1 All leaves at same depthAll leaves at same depth All (key, record pointer) in leavesAll (key, record pointer) in leaves
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B-Tree ExampleB-Tree Example
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15 36 57
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63 76 87
91
92 100
n = 2
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General B-Trees: General B-Trees: Space related constraintsSpace related constraints
Use at leastUse at least
Root: 2 pointersRoot: 2 pointers
Internal:Internal: (n+1)/2(n+1)/2pointerspointers
Leaf:Leaf: (n+1)/2(n+1)/2 pointers to pointers to datadata
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n=3
5 15 21 15
31 42 56 31 42
Internal
Leaf
Max Min
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Leaf NodesLeaf Nodes
n key slots
(n+1) pointer slots
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Leaf NodesLeaf Nodes
n key slots
(n+1) pointer slots
k kk1 2 mk3 … …
unused
record of k1record of k
2
… …
record of k m
…nextleaf
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Leaf NodesLeaf Nodes
n key slots
(n+1) pointer slots
k kk1 2 mk3 … …
unused
record of k1record of k
2
… …
record of k m
…
m ≥ (n+1)
2
nextleaf
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Internal NodesInternal Nodes
n key slots
(n+1) pointer slots
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Internal NodesInternal Nodes
n key slots
(n+1) pointer slots
k kk1 2 mk3
key < k 1k ≤ key < k1 2 k ≤ keym
unused
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Internal NodesInternal Nodes
n key slots
(n+1) pointer slots
k kk1 2 mk3
key < k 1k ≤ key < k1 2 k ≤ keym
unused(m+1) ≥
(n+1)
2
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Root NodeRoot Node
n key slots
(n+1) pointer slots
k kk1 2 mk3
key < k 1k ≤ key < k1 2 k ≤ keym
unused(m+1) ≥2
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LimitsLimits
Why the specific limitsWhy the specific limits (n+1)/2(n+1)/2 and and (n+1)/2(n+1)/2 ? ?
Why different limits for leaf and Why different limits for leaf and internal nodes?internal nodes?
Can we reduce each limit?Can we reduce each limit? Can we increase each limit?Can we increase each limit? What are the implications?What are the implications?
