chapter 13: query processing
DESCRIPTION
Chapter 13: Query Processing. Database System Concepts. Chapter 1: Introduction Part 1: Relational databases Chapter 2: Relational Model Chapter 3: SQL Chapter 4: Advanced SQL Chapter 5: Other Relational Languages Part 2: Database Design - PowerPoint PPT PresentationTRANSCRIPT
Database System Concepts, 5th Ed.
©Silberschatz, Korth and SudarshanSee www.db-book.com for conditions on re-use
Chapter 13: Query ProcessingChapter 13: Query Processing
©Silberschatz, Korth and Sudarshan13.2Database System Concepts - 5th Edition, Aug 27, 2005.
Chapter 1: Introduction Part 1: Relational databases
Chapter 2: Relational Model Chapter 3: SQL Chapter 4: Advanced SQL Chapter 5: Other Relational Languages
Part 2: Database Design Chapter 6: Database Design and the E-R Model Chapter 7: Relational Database Design Chapter 8: Application Design and Development
Part 3: Object-based databases and XML Chapter 9: Object-Based Databases Chapter 10: XML
Part 4: Data storage and querying Chapter 11: Storage and File Structure Chapter 12: Indexing and Hashing Chapter 13: Query Processing Chapter 14: Query Optimization
Part 5: Transaction management Chapter 15: Transactions Chapter 16: Concurrency control Chapter 17: Recovery System
Database System ConceptsDatabase System Concepts
Part 6: Data Mining and Information Retrieval Chapter 18: Data Analysis and Mining Chapter 19: Information Retreival
Part 7: Database system architecture Chapter 20: Database-System Architecture Chapter 21: Parallel Databases Chapter 22: Distributed Databases
Part 8: Other topics Chapter 23: Advanced Application Development Chapter 24: Advanced Data Types and New Applications Chapter 25: Advanced Transaction Processing
Part 9: Case studies Chapter 26: PostgreSQL Chapter 27: Oracle Chapter 28: IBM DB2 Chapter 29: Microsoft SQL Server
Online Appendices Appendix A: Network Model Appendix B: Hierarchical Model Appendix C: Advanced Relational Database Model
©Silberschatz, Korth and Sudarshan13.3Database System Concepts - 5th Edition, Aug 27, 2005.
Part 4: Data storage and querying Part 4: Data storage and querying (Chapters 11 through 14).(Chapters 11 through 14).
Chapter 11: Storage and File Structure
deals with disk, file, and file-system structure.
Chapter 12: Indexing and Hashing
A variety of data-access techniques are presented including hashing and B+ tree indices.
Chapters 13: Query Processing and Chapter 14: Query Optimization
address query-evaluation algorithms, and query optimization techniques. These chapters provide an understanding of the internals of the storage and retrieval components of a database which are necessary for query processing and optimization
©Silberschatz, Korth and Sudarshan13.4Database System Concepts - 5th Edition, Aug 27, 2005.
Chapter 13: Query ProcessingChapter 13: Query Processing
13.1 Overview
13.2 Measures of Query Cost
13.3 Selection Operation
13.4 Sorting
13.5 Join Operation
13.6 Other Operations
13.7 Evaluation of Expressions
13.8 Summary
©Silberschatz, Korth and Sudarshan13.5Database System Concepts - 5th Edition, Aug 27, 2005.
Basic Steps in Query ProcessingBasic Steps in Query Processing
1. Parsing and translation
2. Optimization
3. Evaluation
©Silberschatz, Korth and Sudarshan13.6Database System Concepts - 5th Edition, Aug 27, 2005.
Basic Steps in Query Processing (Cont.)Basic Steps in Query Processing (Cont.)
Parsing and translation First, translate the query into its internal form. This is then translated into relational algebra. Parser checks syntax, verifies relations
Optimization Enumerate all possible query-evaluation plans Compute the cost for the plans Pick up the plan having the minimum cost
Evaluation The query-execution engine takes a query-evaluation plan, executes that plan, and returns the answers to the query.
©Silberschatz, Korth and Sudarshan13.7Database System Concepts - 5th Edition, Aug 27, 2005.
Basic Steps in QP: OptimizationBasic Steps in QP: Optimization
A relational algebra expression may have many equivalent expressions
E.g., balance2500(balance(account)) is equivalent to balance(balance2500(account))
Each relational algebra operation can be evaluated using one of several different algorithms
Correspondingly, a relational-algebra expression can be evaluated in many ways.
Query evaluation-plan.
annotated expression specifying detailed evaluation strategy
Ex1: can use an index on balance to find accounts with balance < 2500,
Ex2: can perform complete relation scan and discard accounts with balance 2500
©Silberschatz, Korth and Sudarshan13.8Database System Concepts - 5th Edition, Aug 27, 2005.
Basic Steps: Optimization (Cont.)Basic Steps: Optimization (Cont.)
Query Optimization
Amongst all equivalent evaluation plans, choose the one with lowest cost.
Cost is estimated using statistical information from the database catalog
e.g. number of tuples in each relation, size of tuples, etc.
In this chapter we study
How to measure query costs
Algorithms for evaluating relational algebra operations
How to combine algorithms for individual operations in order to evaluate a complete expression
In Chapter 14
We study how to optimize queries, that is, how to find an evaluation plan with lowest estimated cost
©Silberschatz, Korth and Sudarshan13.9Database System Concepts - 5th Edition, Aug 27, 2005.
Chapter 13: Query ProcessingChapter 13: Query Processing
13.1 Overview
13.2 Measures of Query Cost
13.3 Selection Operation
13.4 Sorting
13.5 Join Operation
13.6 Other Operations
13.7 Evaluation of Expressions
13.8 Summary
©Silberschatz, Korth and Sudarshan13.10Database System Concepts - 5th Edition, Aug 27, 2005.
Measures of Query CostMeasures of Query Cost
Cost is generally measured as total elapsed time for answering query
Many factors contribute to time cost
disk accesses, CPU, or even network communication
Typically disk access is the predominant cost, and is also relatively easy to estimate. Measured by taking into account
Number of seeks X average-seek-cost
Number of blocks read X average-block-read-cost
Number of blocks written X average-block-write-cost
Cost to write a block is greater than cost to read a block
– data is read back after being written to ensure that the write was successful
©Silberschatz, Korth and Sudarshan13.11Database System Concepts - 5th Edition, Aug 27, 2005.
Measures of Query Cost (Cont.)Measures of Query Cost (Cont.) For simplicity we just use number of block transfers from disk as the cost
measure
We ignore the difference in cost between sequential and random I/O for simplicity
We also ignore CPU costs for simplicity
Costs depends on the size of the buffer in main memory
Having more memory reduces need for disk access
Amount of real memory available to buffer depends on other concurrent OS processes, and hard to determine ahead of actual execution
We often use worst case estimates, assuming only the minimum amount of memory needed for the operation is available
Real systems take CPU cost into account, differentiate between sequential and random I/O, and take buffer size into account
We do not include cost to writing output (query result) to disk in our cost formulae
©Silberschatz, Korth and Sudarshan13.12Database System Concepts - 5th Edition, Aug 27, 2005.
Chapter 13: Query ProcessingChapter 13: Query Processing
13.1 Overview
13.2 Measures of Query Cost
13.3 Selection Operation
13.4 Sorting
13.5 Join Operation
13.6 Other Operations
13.7 Evaluation of Expressions
13.8 Summary
©Silberschatz, Korth and Sudarshan13.13Database System Concepts - 5th Edition, Aug 27, 2005.
Selection OperationSelection Operation
File scan
search algorithms that locate and retrieve records that fulfill a selection condition.
Algorithm A1 (linear search):
Scan each file block and test all records to see whether they satisfy the selection condition.
Cost estimate (number of disk blocks scanned) = br
br denotes number of blocks containing records from relation r
If selection is on a key attribute, cost = (br /2)
stop on finding record
Linear search can be applied regardless of
selection condition or
ordering of records in the file, or
availability of indices
©Silberschatz, Korth and Sudarshan13.14Database System Concepts - 5th Edition, Aug 27, 2005.
Selection Operation (Cont.)Selection Operation (Cont.)
Algorithm A2 (binary search):
Applicable if selection is an equality comparison on the attribute on which file is ordered.
Assume that the blocks of a relation are stored contiguously
Cost estimate (number of disk blocks to be scanned):
log2(br) — cost of locating the first tuple by a binary search on the blocks
Plus number of blocks containing records that satisfy selection condition
– Will see how to estimate this cost in Chapter 14
©Silberschatz, Korth and Sudarshan13.15Database System Concepts - 5th Edition, Aug 27, 2005.
Selections Using IndicesSelections Using Indices Index scan – search algorithms that use an index
selection condition must be on search-key of index. HT: height of index
A3 (primary index on candidate key, equality). Retrieve a single record that satisfies the corresponding equality condition Cost = HTi + 1
A4 (primary index on nonkey, equality) Retrieve multiple records. Records will be on consecutive blocks Cost = HTi + number of blocks containing retrieved records
A5 (equality on search-key of secondary index). Retrieve a single record if the search-key is a candidate key
Cost = HTi + 1 Retrieve multiple records if search-key is not a candidate key
Cost = HTi + number of records retrieved
– can be very expensive! each record may be on a different block
– one block access for each retrieved record
©Silberschatz, Korth and Sudarshan13.16Database System Concepts - 5th Edition, Aug 27, 2005.
Selections Involving ComparisonsSelections Involving Comparisons
Can implement selections of the form AV (r) or A V(r) by using
a linear file scan or binary search, or by using indices in the following ways:
A6 (primary index, comparison). (Relation is sorted on A) For A V(r) use index to find first tuple v and scan relation
sequentially from there For AV (r) just scan relation sequentially till first tuple > v; do not use
index A7 (secondary index, comparison).
For A V(r) use index to find first index entry v and scan index sequentially from there, to find pointers to records.
For AV (r) just scan leaf pages of index finding pointers to records, till first entry > v
In either case, retrieve records that are pointed to
– requires an I/O for each record
– Linear file scan may be cheaper if many records are to be fetched!
©Silberschatz, Korth and Sudarshan13.17Database System Concepts - 5th Edition, Aug 27, 2005.
Complex SelectionsComplex Selections
Conjunction: 1 2. . . n(r)
A8 (conjunctive selection using one index).
Select a combination of i and algorithms A1 through A7 that results in the
least cost fori (r).
Test other conditions on tuple after fetching it into memory buffer.
A9 (conjunctive selection using multiple-key index).
Use appropriate composite (multiple-key) index if available.
A10 (conjunctive selection by intersection of identifiers).
Requires indices with record pointers.
Use corresponding index for each condition, and take intersection of all the obtained sets of record pointers.
Then fetch records from file
If some conditions do not have appropriate indices, apply test in memory.
©Silberschatz, Korth and Sudarshan13.18Database System Concepts - 5th Edition, Aug 27, 2005.
Complex Selections (Cont’)Complex Selections (Cont’)
Disjunction:1 2 . . . n (r).
A11 (disjunctive selection by union of identifiers).
Applicable if all conditions have available indices.
Otherwise use linear scan.
Use corresponding index for each condition, and take union of all the obtained sets of record pointers.
Then fetch records from file
Negation: (r)
Use linear scan on file
If very few records satisfy , and an index is applicable to Find satisfying records using index and fetch from file
©Silberschatz, Korth and Sudarshan13.19Database System Concepts - 5th Edition, Aug 27, 2005.
Chapter 13: Query ProcessingChapter 13: Query Processing
13.1 Overview
13.2 Measures of Query Cost
13.3 Selection Operation
13.4 Sorting
13.5 Join Operation
13.6 Other Operations
13.7 Evaluation of Expressions
13.8 Summary
©Silberschatz, Korth and Sudarshan13.20Database System Concepts - 5th Edition, Aug 27, 2005.
SortingSorting
We may build an index on the relation, and then use the index to read the relation in sorted order.
This may lead to one disk block access for each tuple.
For relations that fit in memory, techniques like quicksort can be used.
Pick a middle element P in an array A
Push the elements having less value than P to the left array LA
Push the elements having bigger value than P to the right array RA
Apply the same idea recursively on LA and RA
For relations that don’t fit in memory, external sort-merge is a good choice.
©Silberschatz, Korth and Sudarshan13.21Database System Concepts - 5th Edition, Aug 27, 2005.
External Sort-MergeExternal Sort-Merge
1. Create sorted runs. Let i be 0 initially. Repeatedly do the following till the end of the relation: (a) Read M blocks of relation into memory (b) Sort the in-memory blocks (c) Write sorted data to run Ri; increment i.Let the final value of I be N
2. Merge the runs (N-way merge). We assume (for now) that N < M. 1. Use N blocks of memory to buffer input runs, and 1 block to buffer
output. Read the first block of each run into its buffer page
2. repeat Select the first record (in sort order) among all buffer pages Write the record to the output buffer. If the output buffer is full
write it to disk. Delete the record from its input buffer page.
If the buffer page becomes empty then read the next block (if any) of the run into the buffer.
3. until all input buffer pages are empty:
Let M denote memory size (in pages).
©Silberschatz, Korth and Sudarshan13.22Database System Concepts - 5th Edition, Aug 27, 2005.
External Sort-Merge (Cont.)External Sort-Merge (Cont.) If N M, several merge passes are required.
In each pass, contiguous groups of M - 1 runs are merged.
A pass reduces the number of runs by a factor of M -1, and creates runs longer by the same factor.
E.g. If M=11, and there are 90 runs, one pass reduces the number of runs to 9, each 10 times the size of the initial runs
Repeated passes are performed till all runs have been merged into one.
©Silberschatz, Korth and Sudarshan13.23Database System Concepts - 5th Edition, Aug 27, 2005.
External Merge Sort (Cont.)External Merge Sort (Cont.)
Cost analysis:
br : the number of blocks in R
M : the number of blocks in each run
br / M : the initial number of runs
Total number of merge passes required: logM–1(br/M)
Disk accesses for initial run creation as well as in each pass is 2br
for final pass, we don’t count write cost
– we ignore final write cost for all operations since the output of an operation may be sent to the parent operation without being written to disk
Thus total number of disk accesses for external sorting:
br ( 2 logM–1(br / M) + 1)
©Silberschatz, Korth and Sudarshan13.24Database System Concepts - 5th Edition, Aug 27, 2005.
Chapter 13: Query ProcessingChapter 13: Query Processing
13.1 Overview
13.2 Measures of Query Cost
13.3 Selection Operation
13.4 Sorting
13.5 Join Operation
13.6 Other Operations
13.7 Evaluation of Expressions
13.8 Summary
©Silberschatz, Korth and Sudarshan13.25Database System Concepts - 5th Edition, Aug 27, 2005.
Join OperationJoin Operation
Several different algorithms to implement joins
Nested-loop join
Block nested-loop join
Indexed nested-loop join
Merge-join
Hash-join
Choice based on cost estimate
Examples use the following information
Number of records
customer: 10,000 depositor: 5000
Number of blocks
customer: 400 depositor: 100
©Silberschatz, Korth and Sudarshan13.26Database System Concepts - 5th Edition, Aug 27, 2005.
Nested-Loop JoinNested-Loop Join
To compute the theta join r s
for each tuple tr in r do begin
for each tuple ts in s do begin
test pair (tr,ts) to see if they satisfy the join condition
if they do, add tr • ts to the result.
endend
r is called the outer relation and s the inner relation of the join.
Requires no indices and can be used with any kind of join condition.
Expensive since it examines every pair of tuples in the two relations.
©Silberschatz, Korth and Sudarshan13.27Database System Concepts - 5th Edition, Aug 27, 2005.
Loan number Branch name amount
L- 170 Downtown 300
L – 42 Redwood 400
L – 48 Redwood 1500
L – 112 Perryridge 2300
L – 321 Redwood 3100
L – 90 Downtown 800
L – 112 Perryridge 2300
L – 31 Redwood 200
L - 70 Perryridge 600
L - 221 Downtown 1000
L – 155 Redwood 800
L - 320 Downtown 2500
Customer name
Loan number
Jones L - 170
Bahn L – 82
Kim L – 42
Lee L – 48
Jane L – 112
Smith L – 34
Hwang L – 321
Choi L – 109
Pedro L – 90
Sammy L – 112
Jun L – 31
Jung L – 62
Shin L – 99
Koh L – 70
Mark L – 221
Harry L - 116
Loan relation : N = 12, B = 3
Borrower relation: N = 16, B = 4
Nested-loop joinNested-loop join보조자료
©Silberschatz, Korth and Sudarshan13.28Database System Concepts - 5th Edition, Aug 27, 2005.
Nested-loop joinNested-loop join
Loan number Branch name amount
L- 170 Downtown 300
L – 42 Redwood 400
L – 48 Redwood 1500
L – 112 Perryridge 2300
L – 321 Redwood 3100
L – 90 Downtown 800
L – 112 Perryridge 2300
L – 31 Redwood 200
L - 70 Perryridge 600
L - 221 Downtown 1000
L – 155 Redwood 800
L - 320 Downtown 2500
Customer name Loan number
Jones L - 170
Bahn L – 82
Kim L – 42
Lee L – 48
Jane L – 112
Smith L – 34
Hwang L – 321
Choi L – 109
Pedro L – 90
Sammy L – 112
Jun L – 31
Jung L – 62
Shin L – 99
Koh L – 70
Mark L – 221
Harry L - 116
Loan relationBorrower relation
보조자료
©Silberschatz, Korth and Sudarshan13.29Database System Concepts - 5th Edition, Aug 27, 2005.
Nested-loop joinNested-loop join
Loan number Branch name amount
L- 170 Downtown 300
L – 42 Redwood 400
L – 48 Redwood 1500
L – 112 Perryridge 2300
L – 321 Redwood 3100
L – 90 Downtown 800
L – 112 Perryridge 2300
L – 31 Redwood 200
L - 70 Perryridge 600
L - 221 Downtown 1000
L – 155 Redwood 800
L - 320 Downtown 2500
Customer name Loan number
Jones L - 170
Bahn L – 82
Kim L – 42
Lee L – 48
Jane L – 112
Smith L – 34
Hwang L – 321
Choi L – 109
Pedro L – 90
Sammy L – 112
Jun L – 31
Jung L – 62
Shin L – 99
Koh L – 70
Mark L – 221
Harry L - 116
Loan relationBorrower relation
보조자료
©Silberschatz, Korth and Sudarshan13.30Database System Concepts - 5th Edition, Aug 27, 2005.
Nested-loop joinNested-loop join
Loan number Branch name amount
L- 170 Downtown 300
L – 42 Redwood 400
L – 48 Redwood 1500
L – 112 Perryridge 2300
L – 321 Redwood 3100
L – 90 Downtown 800
L – 112 Perryridge 2300
L – 31 Redwood 200
L - 70 Perryridge 600
L - 221 Downtown 1000
L – 155 Redwood 800
L - 320 Downtown 2500
Customer name Loan number
Jones L - 170
Bahn L – 82
Kim L – 42
Lee L – 48
Jane L – 112
Smith L – 34
Hwang L – 321
Choi L – 109
Pedro L – 90
Sammy L – 112
Jun L – 31
Jung L – 62
Shin L – 99
Koh L – 70
Mark L – 221
Harry L - 116
Loan relation
Borrower relation
보조자료
©Silberschatz, Korth and Sudarshan13.31Database System Concepts - 5th Edition, Aug 27, 2005.
Nested-loop joinNested-loop join
Loan number Branch name amount
L- 170 Downtown 300
L – 42 Redwood 400
L – 48 Redwood 1500
L – 112 Perryridge 2300
L – 321 Redwood 3100
L – 90 Downtown 800
L – 112 Perryridge 2300
L – 31 Redwood 200
L - 70 Perryridge 600
L - 221 Downtown 1000
L – 155 Redwood 800
L - 320 Downtown 2500
Customer name Loan number
Jones L - 170
Bahn L – 82
Kim L – 42
Lee L – 48
Jane L – 112
Smith L – 34
Hwang L – 321
Choi L – 109
Pedro L – 90
Sammy L – 112
Jun L – 31
Jung L – 62
Shin L – 99
Koh L – 70
Mark L – 221
Harry L - 116
Loan relationBorrower relation
borrower relation 을 outer relation 으로 하면 16*3 + 4 = 52 회의 disk access 가 필요
Loan relation 을 outer relation 으로 했을 경우 12*4 + 3 = 51 회의 disk access 가 필요
보조자료
©Silberschatz, Korth and Sudarshan13.32Database System Concepts - 5th Edition, Aug 27, 2005.
Nested-Loop Join (Cont.)Nested-Loop Join (Cont.)
In the worst case, if there is enough memory only to hold one block of each relation, the estimated cost is nr bs + br disk accesses
where nr is number of record in R and bs and br are number of disk blocks in S and R
If the smaller relation fits entirely in memory, use that as the inner relation.
Reduces cost to br + bs disk accesses.
Assuming worst case memory availability cost estimate is
5000 400 + 100 = 2,000,100 disk accesses with depositor as outer relation, and
1000 100 + 400 = 1,000,400 disk accesses with customer as the outer relation.
If smaller relation (depositor) fits entirely in memory, the cost estimate will be 500 disk accesses.
Block nested-loops algorithm (next slide) is preferable.
©Silberschatz, Korth and Sudarshan13.33Database System Concepts - 5th Edition, Aug 27, 2005.
Block Nested-Loop JoinBlock Nested-Loop Join
Variant of nested-loop join in which every block of inner relation is paired with every block of outer relation.
for each block Br of r do begin
for each block Bs of s do begin
for each tuple tr in Br do begin
for each tuple ts in Bs do begin
Check if (tr,ts) satisfy the join condition
if they do, add tr • ts to the result.
endend
endend
Won Kim’s Join Method
One chapter of ’80 PhD Thesis at Univ of Illinois at Urbana-Champaign
©Silberschatz, Korth and Sudarshan13.34Database System Concepts - 5th Edition, Aug 27, 2005.
Block nested-loop joinBlock nested-loop join
Loan number Branch name amount
L- 170 Downtown 300
L – 42 Redwood 400
L – 48 Redwood 1500
L – 112 Perryridge 2300
L – 321 Redwood 3100
L – 90 Downtown 800
L – 112 Perryridge 2300
L – 31 Redwood 200
L - 70 Perryridge 600
L - 221 Downtown 1000
L – 155 Redwood 800
L - 320 Downtown 2500
Customer name Loan number
Jones L - 170
Bahn L – 82
Kim L – 42
Lee L – 48
Jane L – 112
Smith L – 34
Hwang L – 321
Choi L – 109
Pedro L – 90
Sammy L – 112
Jun L – 31
Jung L – 62
Shin L – 99
Koh L – 70
Mark L – 221
Harry L - 116
Loan relationBorrower relation
보조자료
©Silberschatz, Korth and Sudarshan13.35Database System Concepts - 5th Edition, Aug 27, 2005.
Block nested-loop joinBlock nested-loop join
Loan number Branch name amount
L- 170 Downtown 300
L – 42 Redwood 400
L – 48 Redwood 1500
L – 112 Perryridge 2300
L – 321 Redwood 3100
L – 90 Downtown 800
L – 112 Perryridge 2300
L – 31 Redwood 200
L - 70 Perryridge 600
L - 221 Downtown 1000
L – 155 Redwood 800
L - 320 Downtown 2500
Customer name Loan number
Jones L - 170
Bahn L – 82
Kim L – 42
Lee L – 48
Jane L – 112
Smith L – 34
Hwang L – 321
Choi L – 109
Pedro L – 90
Sammy L – 112
Jun L – 31
Jung L – 62
Shin L – 99
Koh L – 70
Mark L – 221
Harry L - 116
Loan relationBorrower relation
보조자료
©Silberschatz, Korth and Sudarshan13.36Database System Concepts - 5th Edition, Aug 27, 2005.
Block nested-loop joinBlock nested-loop join
Loan number
Branch name
amount
L- 170 Downtown 300
L – 42 Redwood 400
L – 48 Redwood 1500
L – 112 Perryridge 2300
L – 321 Redwood 3100
L – 90 Downtown 800
L – 112 Perryridge 2300
L – 31 Redwood 200
L - 70 Perryridge 600
L - 221 Downtown 1000
L – 155 Redwood 800
L - 320 Downtown 2500
Customer name Loan number
Jones L - 170
Bahn L – 82
Kim L – 42
Lee L – 48
Jane L – 112
Smith L – 34
Hwang L – 321
Choi L – 109
Pedro L – 90
Sammy L – 112
Jun L – 31
Jung L – 62
Shin L – 99
Koh L – 70
Mark L – 221
Harry L - 116
Loan relationBorrower relation
보조자료
©Silberschatz, Korth and Sudarshan13.37Database System Concepts - 5th Edition, Aug 27, 2005.
Block nested-loop joinBlock nested-loop join
Loan number Branch name amount
L- 170 Downtown 300
L – 42 Redwood 400
L – 48 Redwood 1500
L – 112 Perryridge 2300
L – 321 Redwood 3100
L – 90 Downtown 800
L – 112 Perryridge 2300
L – 31 Redwood 200
L - 70 Perryridge 600
L - 221 Downtown 1000
L – 155 Redwood 800
L - 320 Downtown 2500
Customer name Loan number
Jones L - 170
Bahn L – 82
Kim L – 42
Lee L – 48
Jane L – 112
Smith L – 34
Hwang L – 321
Choi L – 109
Pedro L – 90
Sammy L – 112
Jun L – 31
Jung L – 62
Shin L – 99
Koh L – 70
Mark L – 221
Harry L - 116
Loan relationBorrower relation
borrower relation 을 outer relation 으로 하면 4*3 + 4 = 16 회의 disk access 가 필요
Loan relation 을 outer relation 으로 했을 경우 3*4 + 3 = 15 회의 disk access 가 필요
보조자료
©Silberschatz, Korth and Sudarshan13.38Database System Concepts - 5th Edition, Aug 27, 2005.
Block Nested-Loop Join (Cont.)Block Nested-Loop Join (Cont.)
Worst case estimate: br bs + br block accesses. Each block in the inner relation s is read once for each block in the outer
relation (instead of once for each tuple in the outer relation
Best case: br + bs block accesses.
Improvements to nested loop and block nested loop algorithms: In block nested-loop, use M - 2 disk blocks as blocking unit for outer
relations, where M = memory size in blocks; use remaining two blocks to buffer inner relation and output
Cost = br / (M-2) bs + br If equi-join attribute forms a key of inner relation, stop inner loop on first
match Scan inner loop forward and backward alternately, to make use of the
blocks remaining in buffer (with LRU replacement) Use index on inner relation if available (next slide)
©Silberschatz, Korth and Sudarshan13.39Database System Concepts - 5th Edition, Aug 27, 2005.
Indexed Nested-Loop JoinIndexed Nested-Loop Join
Index lookups can replace file scans if join is an equi-join or natural join and an index is available on the inner relation’s join attribute
Can construct an index just to compute a join.
For each tuple tr in the outer relation r, use the index (B+ tree) to look up tuples in s that satisfy the join condition with tuple tr.
Worst case: buffer has space for only one page of r, and for each tuple in r, we perform an index lookup on s.
Cost of the join: br + nr c Where c is the cost of traversing index and fetching all matching s tuples for
one tuple or r c can be estimated as cost of a single selection on s using the join condition.
If indices are available on join attributes of both r and s, use the relation with fewer tuples as the outer relation.
©Silberschatz, Korth and Sudarshan13.40Database System Concepts - 5th Edition, Aug 27, 2005.
Indexed nested-loop joinIndexed nested-loop join
Borrower relation 의 16 개의 tuple 에 대해 각각 그 tuple 과 join 할 수 있는 loan relation 의 tuple 을 B+tree로 검색 (3 회의 disk access)
Tree 의 height 가 2, 그리고 , 실제의 data access 를 위한 1 회 , 따라서 , 3 회의 disk access 필요 따라서 , 총 cost 는 4 block access +16*3 block access = 43 회의 disk access 필요
Loan number Branch name amount
L- 170 Downtown 300
L – 42 Redwood 400
L – 48 Redwood 1500
L – 112 Perryridge 2300
L – 321 Redwood 3100
L – 90 Downtown 800
L – 112 Perryridge 2300
L – 31 Redwood 200
L - 70 Perryridge 600
L - 221 Downtown 1000
L – 155 Redwood 800
L - 320 Downtown 2500
Customer name Loan number
Jones L - 170
Bahn L – 82
Kim L – 42
Lee L – 48
Jane L – 112
Smith L – 34
Hwang L – 321
Choi L – 109
Pedro L – 90
Sammy L – 112
Jun L – 31
Jung L – 62
Shin L – 99
Koh L – 70
Mark L – 221
Harry L - 116
B+ treeIndex
보조자료
©Silberschatz, Korth and Sudarshan13.41Database System Concepts - 5th Edition, Aug 27, 2005.
Example of Nested-Loop Join CostsExample of Nested-Loop Join Costs
Compute depositor customer, with depositor as the outer relation.
Let customer have a primary B+-tree index on the join attribute customer-name, which contains 20 entries in each index node.
Then, since customer has 10,000 tuples, the height of the tree is 4, and one more access is needed to find the actual data
depositor has 5000 tuples
Cost of block nested-loop join
400*100 + 100 = 40,100 disk accesses assuming worst case memory (may be significantly less with more memory)
Cost of indexed nested-loop join
100 + 5000 * 5 = 25,100 disk accesses
For each depositor record, search B+ tree of customer
CPU cost likely to be less than that for block nested loops join
©Silberschatz, Korth and Sudarshan13.42Database System Concepts - 5th Edition, Aug 27, 2005.
Merge-JoinMerge-Join
1. Sort both relations on their join attribute (if not already sorted on the join attributes).
2. Merge the sorted relations to join them
1. Join step is similar to the merge stage of the sort-merge algorithm.
2. Main difference is handling of duplicate values in join attribute — every pair with same value on join attribute must be matched
3. Detailed algorithm in book
©Silberschatz, Korth and Sudarshan13.43Database System Concepts - 5th Edition, Aug 27, 2005.
Merge Join Merge Join on The on The borrower and loan borrower and loan RelationRelation
보조자료
Custom-name
Loan -number
Loan-number
Branch-name
amount
Smith L-11 L-11 Round Hill 900
Jackson L-14 L-14 Downtown 1500
Hayes L-15 L-15 Perryridge 1500
Adams L-16 L-16 Perryridge 1300
Jones L-17 L-17 Downtown 1000
Williams L-17
Smith L-23 L-23 Redwood 2000
Curry L-93 L-93 Mianus 500
©Silberschatz, Korth and Sudarshan13.44Database System Concepts - 5th Edition, Aug 27, 2005.
Merge-Join (Cont.)Merge-Join (Cont.)
Can be used only for equi-joins and natural joins
Each block needs to be read only once (assuming all tuples for any given value of the join attributes fit in memory
Thus number of block accesses for merge-join is br + bs + the cost of sorting if relations are unsorted.
hybrid merge-join: If one relation is sorted, and the other has a secondary B+-tree index on the join attribute
Merge the sorted relation with the leaf entries of the B+- tree
Using the order property of the leaf entries of the B+- tree
Sort the result on the addresses of the unsorted relation’s tuples
Scan the unsorted relation in physical address order and merge with previous result, to replace addresses by the actual tuples
Sequential scan more efficient than random lookup
©Silberschatz, Korth and Sudarshan13.45Database System Concepts - 5th Edition, Aug 27, 2005.
Hash-JoinHash-Join
Applicable for equi-joins and natural joins. A hash function h is used to partition tuples of both relations h maps JoinAttrs values to {0, 1, ..., n}, where JoinAttrs denotes the common
attributes of r and s used in the natural join.
r0, r1, . . ., rn denote partitions of r tuples
Each tuple tr r is put in partition ri where i = h (tr [JoinAttrs]).
s0,, s1. . ., sn denotes partitions of s tuples
Each tuple ts s is put in partition si, where i = h (ts [JoinAttrs]).
Note: In book, ri is denoted as Hri, si is denoted as Hsi and n is denoted as nh
r tuples in ri need only to be compared with s tuples in si
Need not be compared with s tuples in any other partition, since: an r tuple and an s tuple that satisfy the join condition will have the same
value for the join attributes..
©Silberschatz, Korth and Sudarshan13.46Database System Concepts - 5th Edition, Aug 27, 2005.
Hash-Join (Cont.)Hash-Join (Cont.)
JOIN
JOIN
JOIN
JOIN
JOIN
©Silberschatz, Korth and Sudarshan13.47Database System Concepts - 5th Edition, Aug 27, 2005.
Hash Join Hash Join on The on The borrower and loan borrower and loan RelationRelation
보조자료
Custom-name
Loan -number
해시집합
(R)
해시집합
(S)
Loan-number
Branch-name amount
Adams L-16 3 0 L-11 Round Hill 900
Curry L-93 6 1 L-14 Downtown 1500
Hayes L-15 2 2 L-15 Perryridge 1500
Jackson L-14 1 3 L-16 Perryridge 1300
Jones L-17 4 4 L-17 Downtown 1000
Smith L-11 0 5 L-23 Redwood 2000
Smith L-23 5 6 L-93 Mianus 500
Williams L-17 4
Custom-name
Loan -number
해시집합
(R)
해시집합
(S)
Loan-number
Branch-name amount
Smith L-11 0 0 L-11 Round Hill 900
Jackson L-14 1 1 L-14 Downtown 1500
Hayes L-15 2 2 L-15 Perryridge 1500
Adams L-16 3 3 L-16 Perryridge 1300
Jones L-17 4 4 L-17 Downtown 1000
Williams L-17 4
Smith L-23 5 5 L-23 Redwood 2000
Curry L-93 6 6 L-93 Mianus 500
©Silberschatz, Korth and Sudarshan13.48Database System Concepts - 5th Edition, Aug 27, 2005.
Hash-Join AlgorithmHash-Join Algorithm
1. Partition the relation s using hashing function h.
When partitioning a relation, one block of memory is reserved as the output buffer for each partition.
2. Partition r similarly.
3. For each i:
(a) Load si into memory and build an in-memory hash index on it using the join attribute.
This hash index uses a different hash function than the earlier one h.
(b) Read the tuples in ri from the disk one by one.
For each tuple tr locate each matching tuple ts in si using the in-memory hash index.
Output the concatenation of their attributes.
The hash-join of r and s is computed as follows.
Relation s is called the build input and r is called the probe input.
©Silberschatz, Korth and Sudarshan13.49Database System Concepts - 5th Edition, Aug 27, 2005.
Hash-Join Algorithm AnalysisHash-Join Algorithm Analysis
The value n and the hash function h is chosen such that each si should fit in
memory.
Typically n is chosen as bs/M * f where f is a “fudge factor”, typically
around 1.2
The probe relation partitions si need not fit in memory
Recursive partitioning required if number of partitions n is greater than number of pages M of memory.
instead of partitioning n ways, use M – 1 partitions for s
Further partition the M – 1 partitions using a different hash function
Use same partitioning method on r
Rarely required: e.g., recursive partitioning not needed for relations of 1GB or less with memory size of 2MB, with block size of 4KB.
©Silberschatz, Korth and Sudarshan13.50Database System Concepts - 5th Edition, Aug 27, 2005.
Handling of Overflows in Hash-JoinHandling of Overflows in Hash-Join
Hash-table overflow occurs in partition si if si does not fit in memory.
Reasons could be
Many tuples in s with same value for join attributes
Bad hash function
Partitioning is said to be skewed if some partitions have significantly more tuples than some others
Overflow resolution can be done in build phase
Partition si is further partitioned using different hash function.
Partition ri must be similarly partitioned.
Overflow avoidance performs partitioning carefully to avoid overflows during build phase
E.g. partition build relation into many partitions, then combine them
Both approaches fail with large numbers of duplicates
Fallback option: use block nested loops join on overflowed partitions
©Silberschatz, Korth and Sudarshan13.51Database System Concepts - 5th Edition, Aug 27, 2005.
Cost of Hash-JoinCost of Hash-Join
If recursive partitioning is not required: cost of hash join is 3(br + bs) +2 nh
If recursive partitioning required, number of passes required for partitioning s is logM–1(bs) – 1. This is because each final partition of s should fit in memory.
The number of partitions of probe relation r is the same as that for build relation s; the number of passes for partitioning of r is also the same as for s.
Therefore it is best to choose the smaller relation as the build relation.
Total cost estimate with recursive partitioning is:
2(br + bs logM–1(bs) – 1 + br + bs
If the entire build input can be kept in main memory, n can be set to 0 and the algorithm does not partition the relations into temporary files.
Cost estimate goes down to br + bs.
©Silberschatz, Korth and Sudarshan13.52Database System Concepts - 5th Edition, Aug 27, 2005.
Example of Cost of Hash-JoinExample of Cost of Hash-Join
Assume that memory size is 20 blocks
bdepositor= 100 and bcustomer = 400.
depositor is to be used as build input.
Partition it into five partitions, each of size 20 blocks.
This partitioning can be done in one pass.
Similarly, partition customer into five partitions, each of size 80.
This is also done in one pass.
Therefore total cost: 3(100 + 400) = 1500 block transfers
ignores cost of writing partially filled blocks
customer depositor
©Silberschatz, Korth and Sudarshan13.53Database System Concepts - 5th Edition, Aug 27, 2005.
Hybrid Hash–JoinHybrid Hash–Join Useful when memory sized are relatively large, and the build input is bigger than
memory. Main feature of hybrid hash join:
Keep the first partition of the build relation in memory. E.g. With memory size of 25 blocks, depositor can be partitioned into five
partitions, each of size 20 blocks. Division of memory:
The first partition occupies 20 blocks of memory 1 block is used for input, and 1 block each for buffering the other 4 partitions.
customer is similarly partitioned into five partitions each of size 80; the first is used right away for probing, instead of being written out and read
back. Cost of 3(80 + 320) + 20 +80 = 1300 block transfers for
hybrid hash join, instead of 1500 with plain hash-join.
Hybrid hash-join most useful if M >> sb
©Silberschatz, Korth and Sudarshan13.54Database System Concepts - 5th Edition, Aug 27, 2005.
Complex JoinsComplex Joins
Join with a conjunctive condition:
r 1 2... n s
Either use nested loops/block nested loops, or
Compute the result of one of the simpler joins r i s
final result comprises those tuples in the intermediate result that satisfy the remaining conditions
1 . . . i –1 i +1 . . . n
Join with a disjunctive condition
r 1 2 ... n s
Either use nested loops/block nested loops, or
Compute as the union of the records in individual joins r i s:
(r 1 s) (r 2 s) . . . (r n s)
©Silberschatz, Korth and Sudarshan13.55Database System Concepts - 5th Edition, Aug 27, 2005.
Chapter 13: Query ProcessingChapter 13: Query Processing
13.1 Overview
13.2 Measures of Query Cost
13.3 Selection Operation
13.4 Sorting
13.5 Join Operation
13.6 Other Operations
13.7 Evaluation of Expressions
13.8 Summary
©Silberschatz, Korth and Sudarshan13.56Database System Concepts - 5th Edition, Aug 27, 2005.
Other Operations: Duplicate eliminationOther Operations: Duplicate elimination
Duplicate elimination can be implemented via hashing or sorting.
On sorting duplicates will come adjacent to each other, and all but one set of duplicates can be deleted.
Optimization: duplicates can be deleted during run generation as well as at intermediate merge steps in external sort-merge.
Hashing is similar – duplicates will come into the same bucket.
Projection is implemented by performing projection on each tuple followed by duplicate elimination.
©Silberschatz, Korth and Sudarshan13.57Database System Concepts - 5th Edition, Aug 27, 2005.
Other Operations : AggregationOther Operations : Aggregation
Aggregation can be implemented in a manner similar to duplicate elimination.
Sorting or hashing can be used to bring tuples in the same group together, and then the aggregate functions can be applied on each group.
Optimization: combine tuples in the same group during run generation and intermediate merges, by computing partial aggregate values
For count, min, max, sum:
– keep aggregate values on tuples found so far in the group.
– When combining partial aggregate for count, add up the aggregates
For avg:
– keep sum and count,
– and divide sum by count at the end
©Silberschatz, Korth and Sudarshan13.58Database System Concepts - 5th Edition, Aug 27, 2005.
Other Operations : Set OperationsOther Operations : Set Operations
Set operations (, and ):
can either use variant of merge-join after sorting, or variant of hash-join.
E.g., Set operations using hashing:
Partition both relations using the same hash function, thereby creating, r0,, r1, ..,
rn and s0 , s1, .., sn
Process each partition i as follows.
Using a different hashing function, build an in-memory hash index on ri after it is brought into memory.
r s: Add tuples in si to the hash index if they are not already in it.
At end of si add the tuples in the hash index to the result.
r s: Output tuples in si to the result if they are already there in the hash index.
r – s: For each tuple in si, if it is there in the hash index, delete it from the index.
At end of si add remaining tuples in the hash index to the result.
©Silberschatz, Korth and Sudarshan13.59Database System Concepts - 5th Edition, Aug 27, 2005.
Other Operations : Outer JoinOther Operations : Outer Join
Outer join can be computed either as
A join followed by addition of null-padded non-participating tuples.
by modifying the join algorithms.
Modifying merge join to compute r s
In r s, non participating tuples are those in r – R(r s)
Modify merge-join to compute r s: During merging, for every tuple tr from r that do not match any tuple in s, output tr padded with nulls.
Right outer-join and full outer-join can be computed similarly.
Modifying hash join to compute r s
If r is probe relation, output non-matching r tuples padded with nulls
If r is build relation, when probing keep track of which r tuples matched s tuples. At end of si output non-matched r tuples padded with nulls
©Silberschatz, Korth and Sudarshan13.60Database System Concepts - 5th Edition, Aug 27, 2005.
Chapter 13: Query ProcessingChapter 13: Query Processing
13.1 Overview
13.2 Measures of Query Cost
13.3 Selection Operation
13.4 Sorting
13.5 Join Operation
13.6 Other Operations
13.7 Evaluation of Expressions
13.8 Summary
©Silberschatz, Korth and Sudarshan13.61Database System Concepts - 5th Edition, Aug 27, 2005.
Evaluation of ExpressionsEvaluation of Expressions
So far, we have seen algorithms for individual operations
Need to consider alternatives for evaluating an entire expression tree
Materialization:
generate results of an expression whose inputs are relations or are already computed,
materialize (store) it on disk. Repeat.
Pipelining:
pass on tuples to parent operations even as an operation is being executed
Alternatives for evaluating an entire expression tree may have different costs
We study above alternatives in more detail
©Silberschatz, Korth and Sudarshan13.62Database System Concepts - 5th Edition, Aug 27, 2005.
MaterializationMaterialization Materialized evaluation:
evaluate one operation at a time, starting at the lowest-level. Use intermediate results materialized into temporary relations to evaluate next-
level operations. E.g., in figure below,
compute and store (materialize) then compute the store its join with customer, and finally compute the projections on customer-name.
)(2500 accountbalance
©Silberschatz, Korth and Sudarshan13.63Database System Concepts - 5th Edition, Aug 27, 2005.
Materialization (Cont.)Materialization (Cont.)
Materialized evaluation is always applicable
Cost of writing results to disk and reading them back can be quite high
Our cost formulas for operations ignore cost of writing final results to disk, so
Overall cost = sum of costs of individual operations + cost of writing intermediate results to disk
Double buffering: (use two output buffers for each operation)
When one buffer is full, write it to disk while the other buffer is getting filled
Allows overlap of disk writes with computation
Reduces the total execution time
©Silberschatz, Korth and Sudarshan13.64Database System Concepts - 5th Edition, Aug 27, 2005.
PipeliningPipelining
Pipelined evaluation :
evaluate several operations simultaneously,
passing the results of one operation on to the next.
E.g., in previous expression tree, don’t store the result of
instead, pass tuples directly to the join.
Similarly, don’t store result of join, pass tuples directly to projection.
Much cheaper than materialization: no need to store a temporary relation to disk.
Pipelining may not always be possible – e.g., sort, hash-join.
For pipelining to be effective, use evaluation algorithms that generate output tuples even as tuples are received for inputs to the operation.
Pipelines can be executed in two ways
demand driven and producer driven
)(2500 accountbalance
©Silberschatz, Korth and Sudarshan13.65Database System Concepts - 5th Edition, Aug 27, 2005.
Pipelining (Cont.)Pipelining (Cont.) In demand driven or lazy evaluation
System repeatedly requests next tuple from top level operation
Each operation requests next tuple from children operations as required, in order to output its next tuple
In between calls, operation has to maintain “state” so it knows what to return next
Each operation is implemented as an iterator implementing the following operations
open()
– E.g. file scan: initialize file scan, store pointer to beginning of file as state
– E.g.merge join: sort relations and store pointers to beginning of sorted relations as state
next()
– E.g. for file scan: Output next tuple, and advance and store file pointer
– E.g. for merge join: continue with merge from earlier state till next output tuple is found. Save pointers as iterator state.
close()
©Silberschatz, Korth and Sudarshan13.66Database System Concepts - 5th Edition, Aug 27, 2005.
Pipelining (Cont.)Pipelining (Cont.)
In produce-driven or eager pipelining
Operators produce tuples eagerly and pass them up to their parents
Buffer maintained between operators, child puts tuples in buffer, parent removes tuples from buffer
if buffer is full, child waits till there is space in the buffer, and then generates more tuples
System schedules operations that have space in output buffer and can process more input tuples
©Silberschatz, Korth and Sudarshan13.67Database System Concepts - 5th Edition, Aug 27, 2005.
Evaluation Algorithms for PipeliningEvaluation Algorithms for Pipelining
Some algorithms are not able to output results even as they get input tuples
E.g. merge join, or hash join
These result in intermediate results being written to disk and then read back always
Algorithm variants are possible to generate (at least some) results on the fly, as input tuples are read in
E.g. hybrid hash join generates output tuples even as probe relation tuples in the in-memory partition (partition 0) are read in
Pipelined join technique: Hybrid hash join, modified to buffer partition 0 tuples of both relations in-memory, reading them as they become available, and output results of any matches between partition 0 tuples
When a new r0 tuple is found, match it with existing s0 tuples, output
matches, and save it in r0
Symmetrically for s0 tuples
©Silberschatz, Korth and Sudarshan13.68Database System Concepts - 5th Edition, Aug 27, 2005.
Multiple JoinsMultiple Joins Join involving three relations: loan depositor customer
Strategy 1.
Compute depositor customer first
use result to compute loan (depositor customer)
Strategy 2.
Computer loan depositor first,
and then join the result with customer.
Strategy 3.
Perform the pair of joins at once.
Build an index on loan for loan-number, and on customer for customer-name.
For each tuple t in depositor, look up the corresponding tuples in customer and the corresponding tuples in loan.
Each tuple of deposit is examined exactly once.
Strategy 1 & 2 may use materialization or pipelining
Strategy 3 combines two operations into one special-purpose operation that is more efficient than implementing two joins of two relations.
©Silberschatz, Korth and Sudarshan13.69Database System Concepts - 5th Edition, Aug 27, 2005.
Chapter 13: Query ProcessingChapter 13: Query Processing
13.1 Overview
13.2 Measures of Query Cost
13.3 Selection Operation
13.4 Sorting
13.5 Join Operation
13.6 Other Operations
13.7 Evaluation of Expressions
13.8 Summary
©Silberschatz, Korth and Sudarshan13.70Database System Concepts - 5th Edition, Aug 27, 2005.
Chapter 13. Summary (1)Chapter 13. Summary (1)
The first action that the system must perform on a query is to translate the query into its internal form, which (for relational database systems) is usually based on the relational algebra.
In the process of generating the internal form of the query, the parser checks the syntax of the user’s query, verifies that the relation names appearing in the query are names of relations in the database, and so on.
If the query was expressed in terms of a view, the parser replaces all references to the view name with the relational-algebra expression to compute the view.
Given a query, there are generally a variety of methods for computing the answer.
It is the responsibility of the query optimizer to transform the query as entered by the user into an equivalent query that can be computed more efficiently.
Chapter 14 covers query optimization.
©Silberschatz, Korth and Sudarshan13.71Database System Concepts - 5th Edition, Aug 27, 2005.
Chapter 13. Summary (2)Chapter 13. Summary (2) We can process simple selection operations by performing a linear scan, by doing
a binary search, or by making use of indices.
We can handle complex selections by computing unions and intersections of the results of simple selections.
We can sort relations larger than memory by the external merge-sort algorithm.
Queries involving a natural join may be processed in several ways, depending on the availability of indices and the form of physical storage for the relations.
If the join result is almost as large as the Cartesian product of the two relations, a block nested-loop join strategy may be advantageous.
If indices are available, the indexed nested-loop join can be used.
If the relations are sorted, a merge join may be desirable. It may be advantageous to sort a relation prior to join computation (so as to allow use of the merge join strategy).
The hash join algorithm partitions the relations into several pieces, such that each piece of one of the relations fits in memory. The partitioning is carried out with a hash function on the join attributes, so that corresponding pairs of partitions can be joined independently.
©Silberschatz, Korth and Sudarshan13.72Database System Concepts - 5th Edition, Aug 27, 2005.
Chapter 13. Summary (3)Chapter 13. Summary (3)
Duplicate elimination, projection, set operations (union, intersection and difference), and aggregation can be done by sorting or by hashing.
Outer join operations can be implemented by simple extensions of join algorithms.
Hashing and sorting are dual, in the sense that any operation such as duplicate elimination, projection, aggregation, join, and outer join that can be implemented by hashing can also be implemented by sorting, and vice versa
that is, any operation that can be implemented by sorting can also be implemented by hashing.
An expression can be evaluated by means of materialization, where the system computes the result of each subexpression and stores it on disk, and then uses it to compute the result of the parent expression.
Pipelining helps to avoid writing the results of many subexpressions to disk, by using the results in the parent expression even as they are being generated.
©Silberschatz, Korth and Sudarshan13.73Database System Concepts - 5th Edition, Aug 27, 2005.
Ch13. Bibliographical Notes (1)Ch13. Bibliographical Notes (1) A query processor must parse statements in the query language, and must
translate them into an internal form. Parsing of query languages differs little from parsing of traditional programming
languages.
Most compiler texts, such as Aho et al. [1986], cover the main parsing techniques,
and present optimization from a programming-language point of view.
Knuth [1973] presents an excellent description of external sorting algorithms, including an optimization that can create initial runs that are (on the average) twice the size of memory. Based on performance studies conducted in the mid-1970s, database systems
of that period used only nested-loop join and merge join. These studies, which were related to the development of System R, determined
that either the nested-loop join or merge join nearly always provided the optimal join method(Blasgen and Eswaran [1976]); hence, these two wer the only join algorithms implemented in System R.
The System R study, however, did not include an analysis of hash join algorithms. Today, hash joins are considered to be highly efficient.
©Silberschatz, Korth and Sudarshan13.74Database System Concepts - 5th Edition, Aug 27, 2005.
Ch13. Bibliographical Notes (2)Ch13. Bibliographical Notes (2)
Hash join algorithms were initially developed for parallel database systems. Hash join techniques are described in Kitsuregawa et al. [1983], and
extensions including hybrid hash join are described in Shapiro [1986]. Zeller and Gray [1990] and Davison and Graefe [1994] describe hash join
techniques that can adapt to the available memory, which is important in systems where multiple queries may be running at the same time.
Graefe et al. [1998] describes the use of hash joins and hash teams, which allow pipelining of hash-joins by using the same partitioning for all hash-joins in a pipeline sequence, in the Microsoft SQL Server.
Graefe [1993] presents an excellent survey of query-evaluation techniques. An earlier survey of query-processing techniques appears in Jarke and Koch
[1984]. Query processing in main memory database is covered by DeWitt et al. [1984]
and Whang and Krishnamurthy [1990]. Kim [1982] and Kim [1984] describe join strategies and the optimal use of
available main memory.
©Silberschatz, Korth and Sudarshan13.75Database System Concepts - 5th Edition, Aug 27, 2005.
Chapter 13: Query ProcessingChapter 13: Query Processing
13.1 Overview
13.2 Measures of Query Cost
13.3 Selection Operation
13.4 Sorting
13.5 Join Operation
13.6 Other Operations
13.7 Evaluation of Expressions
13.8 Summary
Database System Concepts, 5th Ed.
©Silberschatz, Korth and SudarshanSee www.db-book.com for conditions on re-use
End of ChapterEnd of Chapter