Chapter 13: Query Processing

Chapter 13: Query Processing

اسلاید 1: Chapter 13: Query Processing

اسلاید 2: Chapter 13: Query ProcessingOverview Measures of Query CostSelection Operation Sorting Join Operation Other OperationsEvaluation of Expressions

اسلاید 3: Basic Steps in Query Processing1.Parsing and translation2.Optimization3.Evaluation

اسلاید 4: Basic Steps in Query Processing (Cont.)Parsing and translationtranslate the query into its internal form. This is then translated into relational algebra.Parser checks syntax, verifies relationsEvaluationThe query-execution engine takes a query-evaluation plan, executes that plan, and returns the answers to the query.

اسلاید 5: Basic Steps in Query Processing : OptimizationA relational algebra expression may have many equivalent expressionsE.g., balance2500(balance(account)) is equivalent to balance(balance2500(account))Each relational algebra operation can be evaluated using one of several different algorithmsCorrespondingly, a relational-algebra expression can be evaluated in many ways. Annotated expression specifying detailed evaluation strategy is called an evaluation-plan.E.g., can use an index on balance to find accounts with balance < 2500,or can perform complete relation scan and discard accounts with balance  2500

اسلاید 6: 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 cataloge.g. number of tuples in each relation, size of tuples, etc.In this chapter we studyHow to measure query costsAlgorithms for evaluating relational algebra operationsHow to combine algorithms for individual operations in order to evaluate a complete expressionIn Chapter 14We study how to optimize queries, that is, how to find an evaluation plan with lowest estimated cost

اسلاید 7: Measures of Query CostCost is generally measured as total elapsed time for answering queryMany factors contribute to time costdisk accesses, CPU, or even network communicationTypically disk access is the predominant cost, and is also relatively easy to estimate. Measured by taking into accountNumber of seeks * average-seek-costNumber of blocks read * average-block-read-costNumber of blocks written * average-block-write-costCost 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

اسلاید 8: 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 measurestT – time to transfer one blocktS – time for one seekCost for b block transfers plus S seeks b * tT + S * tS We ignore CPU costs for simplicityReal systems do take CPU cost into accountWe do not include cost to writing output to disk in our cost formulaeSeveral 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 executionWe often use worst case estimates, assuming only the minimum amount of memory needed for the operation is availableRequired data may be buffer resident already, avoiding disk I/OBut hard to take into account for cost estimation

اسلاید 9: Selection OperationFile 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 = br block transfers + 1 seekbr denotes number of blocks containing records from relation rIf selection is on a key attribute, can stop on finding recordcost = (br /2) block transfers + 1 seekLinear search can be applied regardless of selection condition orordering of records in the file, or availability of indices

اسلاید 10: Selection Operation (Cont.)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):cost of locating the first tuple by a binary search on the blockslog2(br) * (tT + tS)If there are multiple records satisfying selectionAdd transfer cost of the number of blocks containing records that satisfy selection condition Will see how to estimate this cost in Chapter 14

اسلاید 11: Selections Using IndicesIndex scan – search algorithms that use an indexselection condition must be on search-key of index.A3 (primary index on candidate key, equality). Retrieve a single record that satisfies the corresponding equality condition Cost = (hi + 1) * (tT + tS)A4 (primary index on nonkey, equality) Retrieve multiple records. Records will be on consecutive blocksLet b = number of blocks containing matching recordsCost = hi * (tT + tS) + tS + tT * bA5 (equality on search-key of secondary index).Retrieve a single record if the search-key is a candidate keyCost = (hi + 1) * (tT + tS)Retrieve multiple records if search-key is not a candidate keyeach of n matching records may be on a different block Cost = (hi + n) * (tT + tS) Can be very expensive!

اسلاید 12: Selections Involving ComparisonsCan implement selections of the form AV (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 thereFor AV (r) just scan relation sequentially till first tuple > v; do not use indexA7 (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 AV (r) just scan leaf pages of index finding pointers to records, till first entry > vIn either case, retrieve records that are pointed torequires an I/O for each record Linear file scan may be cheaper

اسلاید 13: Implementation of Complex SelectionsConjunction: 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 for i (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 fileIf some conditions do not have appropriate indices, apply test in memory.

اسلاید 14: Algorithms for Complex SelectionsDisjunction: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 fileNegation: (r)Use linear scan on fileIf very few records satisfy , and an index is applicable to  Find satisfying records using index and fetch from file

اسلاید 15: SortingWe may build an index on the relation, and then use the index to read the relation in sorted order. May lead to one disk block access for each tuple.For relations that fit in memory, techniques like quicksort can be used. For relations that don’t fit in memory, external sort-merge is a good choice.

اسلاید 16: External Sort-MergeCreate 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 NMerge the runs (next slide)…..Let M denote memory size (in pages).

اسلاید 17: External Sort-Merge (Cont.)Merge the runs (N-way merge). We assume (for now) that N < M. 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 pagerepeatSelect the first record (in sort order) among all buffer pagesWrite 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. until all input buffer pages are empty:

اسلاید 18: 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 runsRepeated passes are performed till all runs have been merged into one.

اسلاید 19: Example: External Sorting Using Sort-Merge

اسلاید 20: External Merge Sort (Cont.)Cost analysis:Total number of merge passes required: logM–1(br/M).Block transfers for initial run creation as well as in each pass is 2brfor 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 diskThus total number of block transfers for external sorting: br ( 2 logM–1(br / M) + 1)Seeks: next slide

اسلاید 21: External Merge Sort (Cont.)Cost of seeksDuring run generation: one seek to read each run and one seek to write each run 2 br / MDuring the merge phaseBuffer size: bb (read/write bb blocks at a time)Need 2 br / bb seeks for each merge pass except the final one which does not require a writeTotal number of seeks: 2 br / M + br / bb (2 logM–1(br / M) -1)

اسلاید 22: Join OperationSeveral different algorithms to implement joinsNested-loop joinBlock nested-loop joinIndexed nested-loop joinMerge-joinHash-joinChoice based on cost estimateExamples use the following informationNumber of records of customer: 10,000 depositor: 5000Number of blocks of customer: 400 depositor: 100

اسلاید 23: Nested-Loop JoinTo 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. end endr 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.

اسلاید 24: 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 seeksIf the smaller relation fits entirely in memory, use that as the inner relation. Reduces cost to br + bs block transfers and 2 seeksAssuming worst case memory availability cost estimate iswith depositor as outer relation:5000  400 + 100 = 2,000,100 block transfers,5000 + 100 = 5100 seeks with customer as the outer relation 10000  100 + 400 = 1,000,400 block transfers and 10,400 seeksIf smaller relation (depositor) fits entirely in memory, the cost estimate will be 500 block transfers.Block nested-loops algorithm (next slide) is preferable.

اسلاید 25: Block Nested-Loop JoinVariant 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

اسلاید 26: Block Nested-Loop Join (Cont.)Worst case estimate: br  bs + br block transfers + 2 * br seeksEach 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 relationBest 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) seeksIf equi-join attribute forms a key or inner relation, stop inner loop on first matchScan 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)

اسلاید 27: Indexed Nested-Loop JoinIndex lookups can replace file scans ifjoin is an equi-join or natural join andan index is available on the inner relation’s join attributeCan 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.Cost of the join: br (tT + tS) + nr  cWhere c is the cost of traversing index and fetching all matching s tuples for one tuple or rc 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.

اسلاید 28: Example of Nested-Loop Join CostsCompute 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.Since customer has 10,000 tuples, the height of the tree is 4, and one more access is needed to find the actual datadepositor has 5000 tuplesCost of block nested loops join400*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 join100 + 5000 * 5 = 25,100 block transfers and seeks.CPU cost likely to be less than that for block nested loops join

اسلاید 29: Merge-JoinSort both relations on their join attribute (if not already sorted on the join attributes).Merge the sorted relations to join themJoin step is similar to the merge stage of the sort-merge algorithm. Main difference is handling of duplicate values in join attribute — every pair with same value on join attribute must be matchedDetailed algorithm in book

اسلاید 30: Merge-Join (Cont.)Can be used only for equi-joins and natural joinsEach block needs to be read only once (assuming all tuples for any given value of the join attributes fit in memoryThus 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 attributeMerge the sorted relation with the leaf entries of the B+-tree . Sort the result on the addresses of the unsorted relation’s tuplesScan the unsorted relation in physical address order and merge with previous result, to replace addresses by the actual tuplesSequential scan more efficient than random lookup

اسلاید 31: Hash-JoinApplicable 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 tuplesEach tuple tr  r is put in partition ri where i = h(tr [JoinAttrs]).r0,, r1. . ., rn denotes partitions of s tuplesEach 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.

اسلاید 32: Hash-Join (Cont.)

اسلاید 33: 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 to be in ri and the s tuple in si.

اسلاید 34: Hash-Join Algorithm1.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.

اسلاید 35: 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 1.2The probe relation partitions si need not fit in memoryRecursive 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 sFurther partition the M – 1 partitions using a different hash functionUse same partitioning method on rRarely required: e.g., recursive partitioning not needed for relations of 1GB or less with memory size of 2MB, with block size of 4KB.

اسلاید 36: Handling of OverflowsPartitioning is said to be skewed if some partitions have significantly more tuples than some othersHash-table overflow occurs in partition si if si does not fit in memory. Reasons could beMany tuples in s with same value for join attributesBad hash functionOverflow resolution can be done in build phasePartition si is further partitioned using different hash function. Partition ri must be similarly partitioned.Overflow avoidance performs partitioning carefully to avoid overflows during build phaseE.g. partition build relation into many partitions, then combine themBoth approaches fail with large numbers of duplicatesFallback option: use block nested loops join on overflowed partitions

اسلاید 37: Cost of Hash-JoinIf recursive partitioning is not required: cost of hash join is 3(br + bs) +4  nh block transfers + 2( br / bb + bs / bb) seeksIf 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 seeksIf the entire build input can be kept in main memory no partitioning is requiredCost estimate goes down to br + bs.

اسلاید 38: Example of Cost of Hash-JoinAssume that memory size is 20 blocksbdepositor= 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, ignoring cost of writing partially filled blocks:3(100 + 400) = 1500 block transfers + 2( 100/3 + 400/3) = 336 seekscustomer depositor

اسلاید 39: Hybrid Hash–JoinUseful 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 memory1 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 80the first is used right away for probing, instead of being written outCost 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 >>

اسلاید 40: Complex JoinsJoin with a conjunctive condition:r 1  2...   n sEither use nested loops/block nested loops, orCompute the result of one of the simpler joins r i sfinal result comprises those tuples in the intermediate result that satisfy the remaining conditions 1  . . .  i –1  i +1  . . .  nJoin with a disjunctive condition r 1  2 ...  n s Either use nested loops/block nested loops, orCompute as the union of the records in individual joins r  i s:(r 1 s)  (r 2 s)  . . .  (r n s)

اسلاید 41: Other OperationsDuplicate 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.

اسلاید 42: Other Operations : AggregationAggregation 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 valuesFor count, min, max, sum: keep aggregate values on tuples found so far in the group. When combining partial aggregate for count, add up the aggregatesFor avg, keep sum and count, and divide sum by count at the end

اسلاید 43: Other Operations : Set OperationsSet 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 functionProcess each partition i as follows. Using a different hashing function, build an in-memory hash index on ri.Process si as followsr  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.

اسلاید 44: Other Operations : Outer JoinOuter 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 sIn 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 sIf r is probe relation, output non-matching r tuples padded with nullsIf 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

اسلاید 45: Evaluation of ExpressionsSo far: we have seen algorithms for individual operationsAlternatives for evaluating an entire expression treeMaterialization: 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 executedWe study above alternatives in more detail

اسلاید 46: MaterializationMaterialized 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 customer, and finally compute the projections on customer-name.

اسلاید 47: Materialization (Cont.)Materialized evaluation is always applicableCost of writing results to disk and reading them back can be quite highOur cost formulas for operations ignore cost of writing results to disk, soOverall cost = Sum of costs of individual operations + cost of writing intermediate results to diskDouble buffering: use two output buffers for each operation, when one is full write it to disk while the other is getting filledAllows overlap of disk writes with computation and reduces execution time

اسلاید 48: PipeliningPipelined 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 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

اسلاید 49: Pipelining (Cont.)In demand driven or lazy evaluationsystem repeatedly requests next tuple from top level operationEach operation requests next tuple from children operations as required, in order to output its next tupleIn between calls, operation has to maintain “state” so it knows what to return nextIn producer-driven or eager pipeliningOperators produce tuples eagerly and pass them up to their parentsBuffer maintained between operators, child puts tuples in buffer, parent removes tuples from bufferif buffer is full, child waits till there is space in the buffer, and then generates more tuplesSystem schedules operations that have space in output buffer and can process more input tuplesAlternative name: pull and push models of pipelining

اسلاید 50: Pipelining (Cont.)Implementation of demand-driven pipeliningEach operation is implemented as an iterator implementing the following operationsopen()E.g. file scan: initialize file scan state: pointer to beginning of fileE.g.merge join: sort relations; state: pointers to beginning of sorted relations next()E.g. for file scan: Output next tuple, and advance and store file pointerE.g. for merge join: continue with merge from earlier state till next output tuple is found. Save pointers as iterator state.close()

اسلاید 51: Evaluation Algorithms for PipeliningSome algorithms are not able to output results even as they get input tuplesE.g. merge join, or hash joinintermediate results written to disk and then read backAlgorithm variants to generate (at least some) results on the fly, as input tuples are read inE.g. hybrid hash join generates output tuples even as probe relation tuples in the in-memory partition (partition 0) are read inPipelined 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 tuplesWhen a new r0 tuple is found, match it with existing s0 tuples, output matches, and save it in r0Symmetrically for s0 tuples

اسلاید 52: End of Chapter

اسلاید 53: Figure 13.2

اسلاید 54: Complex JoinsJoin involving three relations: loan depositor customerStrategy 1. Compute depositor customer; 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 and 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 3 combines two operations into one special-purpose operation that is more efficient than implementing two joins of two relations.