程序代写代做 C database algorithm Chapter 12: Query Processing

Chapter 12: Query Processing
■ Overview
■ Measures of Query Cost ■ Selection Operation
■ Sorting
■ Join Operation
■ Other Operations
■ Evaluation of Expressions
Database System Concepts – 6th Edition
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Basic Steps in Query Processing
1. Parsing and translation 2. Optimization
3. Evaluation
query
parser and translator
evaluation engine
relational-algebra expression
query output
optimizer
execution plan
Database System Concepts – 6th Edition
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data
statistics about data

Basic Steps in Query Processing (Cont.)
■ Parsing and translation
● translate the query into its internal form. This is then
translated into relational algebra.
● Parser checks syntax, verifies relations
■ Evaluation
● The query-execution engine takes a query-evaluation plan,
executes that plan, and returns the answers to the query.
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Basic Steps in Query Processing : Optimization
■ Arelationalalgebraexpressionmayhavemanyequivalent expressions
● E.g., σsalary<75000(∏salary(instructor)) is equivalent to 
 ∏salary(σsalary<75000(instructor)) ■ Each relational algebra operation can be evaluated using one of several different algorithms ● Correspondingly, a relational-algebra expression can be evaluated in many ways. ■ Annotatedexpressionspecifyingdetailedevaluationstrategyis called an evaluation-plan. ● E.g., can use an index on salary to find instructors with salary < 75000, ● or can perform complete relation scan and discard instructors with salary ≥ 75000 Database System Concepts - 6th Edition Modified for CS6083 at NYU Tandon by T. Suel. Spring 2020 12.4 ©Silberschatz, Korth and Sudarshan 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 Database System Concepts - 6th Edition Modified for CS6083 at NYU Tandon by T. Suel. Spring 2020 12.5 ©Silberschatz, Korth and Sudarshan Measures 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. ● Number of seeks ● Number of blocks read ● Number of blocks written * average-block-write-cost Measured by taking into account * average-seek-cost * average-block-read-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 Database System Concepts - 6th Edition Modified for CS6083 at NYU Tandon by T. Suel. Spring 2020 12.6 ©Silberschatz, Korth and Sudarshan Measures of Query Cost (Cont.) ■ For simplicity we just use the number of block transfers from disk and the number of seeks as the cost measures ● tT – time to transfer one block ● tS – time for one seek ● Cost for b block transfers plus S seeks
 b*tT +S*tS ■ We ignore CPU costs for simplicity ● Real systems do take CPU cost into account ■ We do not include cost to writing output to disk in our cost formulae Database System Concepts - 6th Edition Modified for CS6083 at NYU Tandon by T. Suel. Spring 2020 12.7 ©Silberschatz, Korth and Sudarshan Measures of Query Cost (Cont.) ■ Several algorithms can reduce disk IO by using extra buffer space ● Amount of real memory available to buffer depends on other concurrent queries and OS processes, known only during execution ! We often use worst case estimates, assuming only the minimum amount of memory needed for the operation is available ■ Required data may be buffer resident already, avoiding disk I/O ● But hard to take into account for cost estimation Database System Concepts - 6th Edition Modified for CS6083 at NYU Tandon by T. Suel. Spring 2020 12.8 ©Silberschatz, Korth and Sudarshan Selection Operation ■ File scan ■ Algorithm A1 (linear search). Scan each file block and test all records to see whether they satisfy the selection condition. ● Cost estimate = br block transfers + 1 seek ! br denotes number of blocks containing records from relation r ● If selection is on a key attribute, can stop on finding record ! cost = (br /2) block transfers + 1 seek ● Linear search can be applied regardless of ! selection condition or ! ordering of records in the file, or ! availability of indices ■ Note: binary search generally does not make sense since data is not stored consecutively ● exceptwhenthereisanindexavailable, ● andbinarysearchrequiresmoreseeksthanindexsearch Database System Concepts - 6th Edition Modified for CS6083 at NYU Tandon by T. Suel. Spring 2020 12.9 ©Silberschatz, Korth and Sudarshan Selections Using Indices ■ Index scan – search algorithms that use an index ● selection condition must be on search-key of index. ■ A2 (primary index, equality on key). Retrieve a single record that satisfies the corresponding equality condition ● Cost=(hi +1)*(tT +tS) ■ A3(primaryindex,equalityonnonkey)Retrievemultiple records. ● Records will be on consecutive blocks ! Let b = number of blocks containing matching records ● Cost=hi *(tT +tS)+tS +tT *b Database System Concepts - 6th Edition Modified for CS6083 at NYU Tandon by T. Suel. Spring 2020 12.10 ©Silberschatz, Korth and Sudarshan Selections Using Indices ■ A4(secondaryindex,equalityonnonkey). ● Retrieve a single record if the search-key is a candidate key ! Cost=(hi +1)*(tT +tS) ● Retrieve multiple records if search-key is not a candidate key ! each of n matching records may be on a different block ! Cost= (hi +n)*(tT +tS) – Can be very expensive! Database System Concepts - 6th Edition Modified for CS6083 at NYU Tandon by T. Suel. Spring 2020 12.11 ©Silberschatz, Korth and Sudarshan Selections Involving Comparisons ■ Can implement selections of the form σA≤V (r) or σA ≥ V(r) by using ● a linear file scan, ● or by using indices in the following ways: ■ A5(primaryindex,comparison).(RelationissortedonA) ! For σA ≥ V(r) use index to find first tuple ≥ v and scan relation sequentially from there ! For σA≤V (r) just scan relation sequentially till first tuple > v; do not use index
■ A6(secondaryindex,comparison).
! For σA ≥ V(r) use index to find first index entry ≥ v and scan index
sequentially from there, to find pointers to records.
! For σA≤V (r) just scan leaf pages of index finding pointers to
records, till first entry > v
! In either case, retrieve records that are pointed to
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– requires an I/O for each record – Linear file scan may be cheaper
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Implementation of Complex Selections
■ Conjunction: σθ1∧ θ2∧. . . θn(r)
■ A7(conjunctiveselectionusingoneindex).
● Select a combination of θi and algorithms A1 through A7 that results in the least cost for σθi (r).
● Test other conditions on tuple after fetching it into memory buffer. ■ A8(conjunctiveselectionusingcompositeindex).
● Use appropriate composite (multiple-key) index if available. ■ A9(conjunctiveselectionbyintersectionofidentifiers).
● 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
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● If some conditions do not have appropriate indices, apply test in memory.
Database System Concepts – 6th Edition
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Algorithms for Complex Selections
■ Disjunction:σθ1∨ θ2 ∨. . . θn (r).
■ A10(disjunctiveselectionbyunionofidentifiers).
● 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: σ¬θ(r)
● Use linear scan on file
● If very few records satisfy ¬θ, and an index is applicable to θ
! Find satisfying records using index and fetch from file
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Join 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 of student: 5,000 ● Number of blocks of student: 100
takes: 10,000
takes:
400
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Nested-Loop Join
■ To compute the theta join r θ 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 θ 

if they do, add tr • ts to the result.
 end

end
■ 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.
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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 block transfers, plus
 nr + br seeks
■ If the smaller relation fits entirely in memory, use that as the inner relation.
● Reduces cost to br + bs block transfers and 2 seeks
■ Assuming worst case memory availability cost estimate is ● withstudentasouterrelation:
! 5000 * 400 + 100 = 2,000,100 block transfers,
! 5000 + 100 = 5100 seeks
● with takes as the outer relation
! 10000 * 100 + 400 = 1,000,400 block transfers and 10,400 seeks
■ If smaller relation (student) fits entirely in memory, the cost estimate will be
500 block transfers.
■ Block nested-loops algorithm (next slide) is preferable.
Database System Concepts – 6th Edition
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Block 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.
 end

end
 end

end
Database System Concepts – 6th Edition
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Block Nested-Loop Join (Cont.)
■ Worst case estimate: br * bs + br block transfers + 2 * br seeks ● Each block in the inner relation s is read once for each block
in the outer relation
■ Best case: br + bs block transfers + 2 seeks.
■ 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 block transfers +
 2 ⎡br / (M-2)⎤ seeks
● If equi-join attribute forms a key or 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) Database System Concepts – 6th Edition
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Indexed Nested-Loop Join
■ Index lookups can replace file scans if
● joinisanequi-joinornaturaljoinand
● anindexisavailableontheinnerrelation’sjoinattribute
! Can construct an index just to compute a join.
■ For each tuple tr in the outer relation r, use the index 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.
■ Costofthejoin: br (tT+tS)+nr *c
● Wherecisthecostoftraversingindexandfetchingallmatchings tuples for one tuple or r
● ccanbeestimatedascostofasingleselectiononsusingthejoin condition.
■ If indices are available on join attributes of both r and s,
 use the relation with fewer tuples as the outer relation.
Database System Concepts – 6th Edition
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12.20 ©Silberschatz, Korth and Sudarshan

Example of Nested-Loop Join Costs
■ Compute student takes, with student as the outer relation.
■ Let takes have a primary B+-tree index on the attribute ID, which
contains 20 entries in each index node.
■ Since takes has 10,000 tuples, the height of the tree is 4, and one more access is needed to find the actual data
■ student has 5000 tuples
■ Cost of block nested loops join
● 400*100 + 100 = 40,100 block transfers + 2 * 100 = 200 seeks ! assuming worst case memory
! may be significantly less with more memory
■ Cost of indexed nested loops join
● 100 + 5000 * 5 = 25,100 block transfers and seeks.
● CPUcostlikelytobelessthanthatforblocknestedloopsjoin
Database System Concepts – 6th Edition
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Merge-Join
1. Sortbothrelationsontheirjoinattribute(ifnotalreadysortedon the join attributes).
2. Mergethesortedrelationstojointhem
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
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Database System Concepts – 6th Edition
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r
12.22
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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 the cost of merge join is: 

br + bs block transfers + ⎡br / bb⎤ + ⎡bs / bb⎤ seeks ● + 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 .
● 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
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Hash-Join
■ Applicableforequi-joinsandnaturaljoins.
■ Ahashfunctionhisusedtopartitiontuplesofbothrelations
■ 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]).
● r0,, r1. . ., rn 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.
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Hash-Join (Cont.)
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partitions
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Hash-Join (Cont.)
■ 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.
● If that value is hashed to some value i, the r tuple has tobeinri andthestupleinsi.
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Hash-Join Algorithm
The hash-join of r and s is computed as follows.
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.
Relation s is called the build input and r is called the probe input.
Database System Concepts – 6th Edition
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Hash-Join algorithm (Cont.)
■ 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., with block size of 4 KB, recursive partitioning not needed for relations of < 1GB with memory size of 2MB, or relations of < 36 GB with memory of 12 MB Database System Concepts - 6th Edition Modified for CS6083 at NYU Tandon by T. Suel. Spring 2020 12.28 ©Silberschatz, Korth and Sudarshan Cost of Hash-Join ■ If recursive partitioning is not required: cost of hash join is
 3(br + bs) +4 * nh block transfers +
 2(⎡br/bb⎤+⎡bs/bb⎤) seeks ■ If recursive partitioning required: ● number of passes required for partitioning build relation 
 s is ⎡logM–1(bs) – 1⎤ ● best to choose the smaller relation as the build relation. ● Total cost estimate is: 
 2(br + bs) ⎡logM–1(bs) – 1⎤ + br + bs block transfers + 
 2(⎡br / bb⎤ + ⎡bs / bb⎤) ⎡logM–1(bs) – 1⎤ seeks ■ If the entire build input can be kept in main memory no partitioning is required ● Cost estimate goes down to br + bs. Database System Concepts - 6th Edition Modified for CS6083 at NYU Tandon by T. Suel. Spring 2020 12.29 ©Silberschatz, Korth and Sudarshan Example of Cost of Hash-Join instructor teaches ■ Assumethatmemorysizeis20blocks ■ binstructor= 100 and bteaches = 400. ■ instructor 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 teaches into five partitions,each of size 80. This is also done in one pass. ■ Therefore total cost, ignoring cost of writing partially filled blocks: ● 3(100 + 400) = 1500 block transfers +
 2( ⎡100/3⎤ + ⎡400/3⎤) = 336 seeks Database System Concepts - 6th Edition Modified for CS6083 at NYU Tandon by T. Suel. Spring 2020 12.30 ©Silberschatz, Korth and Sudarshan Other Operations ■ 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: ● perform projection on each tuple ● followed by duplicate elimination. Database System Concepts - 6th Edition Modified for CS6083 at NYU Tandon by T. Suel. Spring 2020 12.31 ©Silberschatz, Korth and Sudarshan Other Operations : Aggregation ■ Aggregationcanbeimplementedinamannersimilartoduplicate 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 Database System Concepts - 6th Edition Modified for CS6083 at NYU Tandon by T. Suel. Spring 2020 12.32 ©Silberschatz, Korth and Sudarshan Other 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 – ΠR(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. Database System Concepts - 6th Edition Modified for CS6083 at NYU Tandon by T. Suel. Spring 2020 12.33 ©Silberschatz, Korth and Sudarshan Evaluation of Expressions ■ So far: we have seen algorithms for individual operations ■ Alternativesforevaluatinganentireexpressiontree ● 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 ■ We study above alternatives in more detail Database System Concepts - 6th Edition Modified for CS6083 at NYU Tandon by T. Suel. Spring 2020 12.34 ©Silberschatz, Korth and Sudarshan Materialization ■ 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
 ■ 
 
 then compute the store its join with instructor, and finally compute the projection on name. Π name σ (department ) building ="Watson" σbuilding = “Watson” Database System Concepts - 6th Edition department Modified for CS6083 at NYU Tandon by T. Suel. Spring 2020 instructor 12.35 ©Silberschatz, Korth and Sudarshan 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 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 is full write it to disk while the other is getting filled ● Allows overlap of disk writes with computation and reduces execution time Database System Concepts - 6th Edition Modified for CS6083 at NYU Tandon by T. Suel. Spring 2020 12.36 ©Silberschatz, Korth and Sudarshan Pipelining ■ 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 result of
 
 ■ 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 σ building ="Watson" (department ) ● instead, pass tuples directly to the join.. Similarly, don’t store result of join, pass tuples directly to projection. Database System Concepts - 6th Edition Modified for CS6083 at NYU Tandon by T. Suel. Spring 2020 12.37 ©Silberschatz, Korth and Sudarshan