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Query Processing Chapter 21 in Textbook

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Query Processing Chapter 21 in Textbook Slide 2 Overview What is Query Processing? What is Query Optimization? Example. Phases of Query Processing. 1. Decomposition. a. Analysis. b. Normalization. c. Semantic Analysis. d. Simplification. e. Restructuring. 2. Optimization. a. Heuristics. b. Comparing costs. 3. Code Generation. 4. Execution. 2 Slide 3 Query Processing 3 Activities involved in retrieving data from the database. Aims of QP: transform query from high-level language (SQL) into correct and efficient execution strategy in low-level language (Relational Algebra - RA); execute strategy to retrieve required data. Advantage of Declarative Languages (SQL): System performs optimization NOT user. Slide 4 Query Optimization 4 Activity of choosing an efficient execution strategy for processing query. As there are many equivalent transformations of same high-level query, choose one that minimizes resource usage. Generally, reduce total execution time of query. Disk access tends to be dominant cost in query processing for centralized DBMS. Slide 5 Example 21.1 - Different Strategies 5 Find all Managers who work at a London branch. SELECT * FROM Staff s, Branch b WHERE s.branchNo = b.branchNo AND (s.position = Manager AND b.city = London); Slide 6 Example 21.1 - Different Strategies 6 Three equivalent RA queries are: (1) (position='Manager') (city='London') (Staff.branchNo=Branch.branchNo) (Staff X Branch) (2) (position='Manager') (city='London') ( Staff Staff.branchNo=Branch.branchNo Branch) (3) ( position='Manager' (Staff)) Staff.branchNo=Branch.branchNo ( city='London' (Branch)) Slide 7 Example 21.1 - Different Strategies 7 Assume: 1000 tuples in Staff; 50 tuples in Branch; 50 Managers; 5 London branches; no indexes or sort keys; results of any intermediate operations stored on disk; cost of the final write is ignored; tuples are accessed one at a time. Slide 8 Example 21.1 - Cost Comparison 8 Cost (in disk accesses) are: (1) (1000 + 50) + 2*(1000 * 50) = 101 050 (2) 2*1000 + (1000 + 50) = 3 050 (3) 1000 + 2*50 + 5 + (50 + 5) = 1 160 Cartesian product and join operations much more expensive than selection, and third option significantly reduces size of relations being joined together. Slide 9 Phases of Query Processing 9 QP has four main phases: decomposition (consisting of parsing and validation); optimization; code generation; execution. Slide 10 Phases of Query Processing 10 Slide 11 1. Query Decomposition 11 Aims are: transform high-level query into RA query. check that query is syntactically and semantically correct. Typical stages are: a. analysis, b. normalization, c. semantic analysis, d. simplification, e. query restructuring. Slide 12 1.a. Analysis 12 Analyze query lexically and syntactically using compiler techniques. Verify relations and attributes exist. Verify operations are appropriate for object type. Slide 13 1.a. Analysis - Example 13 SELECT staff_no FROM Staff WHERE position > 10; This query would be rejected on two grounds: staff_no is not defined for Staff relation (should be staffNo). Comparison >10 is incompatible with type position, which is variable character string. Slide 14 1.a. Analysis 14 Finally, query transformed into a query tree constructed as follows: Leaf node for each base relation. Non-leaf node for each intermediate relation produced by RA operation. Root of tree represents query result. Sequence is directed from leaves to root. Slide 15 Example 21.1 Relational Algebra Tree (R.A.T.) 15 Slide 16 1.b. Normalization 16 Converts query into a normalized form for easier manipulation. Predicate can be converted into one of two forms: Conjunctive normal form: (position = 'Manager' salary > 20000) (branchNo = 'B003') Disjunctive normal form: (position = 'Manager' branchNo = 'B003' ) (salary > 20000 branchNo = 'B003') Slide 17 1.c. Semantic Analysis 17 Rejects normalized queries that are incorrectly formulated or contradictory. Query is incorrectly formulated if components do not contribute to generation of result. Query is contradictory if its predicate cannot be satisfied by any tuple. Algorithms to determine correctness exist only for queries that do not contain disjunction and negation. Slide 18 1.c. Semantic Analysis 18 For these queries (no disjunction and no negation) could construct: 1. A relation connection graph. 2. Normalized attribute connection graph. 1. Relation connection graph a. Create node for each relation and node for result. b. Create edges between two nodes that represent a join. c. Create edges between nodes that represent projection. If not connected, query is incorrectly formulated. Slide 19 Example 21.2 - Checking Semantic Correctness 19 SELECT p.propertyNo, p.street FROM Client c, Viewing v, PropertyForRent p WHERE c.clientNo = v.clientNo AND c.maxRent >= 500 AND c.prefType = Flat AND p.ownerNo = CO93; Slide 20 Example 21.2 - Checking Semantic Correctness 20 Relation Connection graph Relation connection graph not fully connected, so query is not correctly formulated. Have omitted the join condition (v.propertyNo = p.propertyNo). Slide 21 1.d. Simplification 21 Aims: 1. Detects redundant qualifications, 2. eliminates common sub-expressions, 3. transforms query to semantically equivalent but more easily and efficiently computed form. Apply well-known transformation rules of Boolean algebra. Slide 22 Transformation Rules for RA Operations 22 1. Conjunctive Selection operations can cascade into individual Selection operations (and vice versa). p q r (R) = p ( q ( r (R))) Sometimes referred to as cascade of Selection. branchNo='B003' salary>15000 (Staff) = branchNo='B003' ( salary>15000 (Staff)) Slide 23 Transformation Rules for RA Operations 23 2. Commutativity of Selection. p ( q (R)) = q ( p (R)) For example: branchNo='B003' ( salary>15000 (Staff)) = salary>15000 ( branchNo='B003' (Staff)) Slide 24 Transformation Rules for RA Operations 24 3. In a sequence of Projection operations, only the last in the sequence is required. L M N (R) = L (R) For example: lName branchNo, lName (Staff) = lName (Staff) Slide 25 Transformation Rules for RA Operations 25 4. Commutativity of Selection and Projection. If predicate p involves only attributes in projection list, Selection and Projection operations commute: Ai, , Am ( p (R)) = p ( Ai, , Am (R)) where p {A 1, A 2, , A m } For example: fName, lName ( lName='Beech' (Staff)) = lName='Beech' ( fName,lName (Staff)) Slide 26 Transformation Rules for RA Operations 26 5. Commutativity of Theta join (and Cartesian product). R p S = S p R R X S = S X R u Rule also applies to Equijoin and Natural join. For example: Staff staff.branchNo=branch.branchNo Branch = Branch staff.branchNo=branch.branchNo Staff Slide 27 Transformation Rules for RA Operations 27 6. Commutativity of Selection and Theta join (or Cartesian product). If selection predicate involves only attributes of one of join relations, Selection and Join (or Cartesian product) operations commute: p (R r S) = ( p (R)) r S p (R X S) = ( p (R)) X S where p {A 1, A 2, , A n } Slide 28 Transformation Rules for RA Operations 28 If selection predicate is conjunctive predicate having form (p q), where p only involves attributes of R, and q only attributes of S, Selection and Theta join operations commute as: p q (R r S) = ( p (R)) r ( q (S)) p q (R X S) = ( p (R)) X ( q (S)) Slide 29 Transformation Rules for RA Operations 29 For example: position='Manager' city='London' (Staff Staff.branchNo=Branch.branchNo Branch) = ( position='Manager' (Staff)) Staff.branchNo=Branch.branchNo ( city='London' (Branch)) Slide 30 Transformation Rules for RA Operations 30 7. Commutativity of Projection and Theta join (or Cartesian product). If projection list is of form L = L 1 L 2, where L 1 only has attributes of R, and L 2 only has attributes of S, provided join condition only contains attributes of L, Projection and Theta join commute: L1 L2 (R r S) = ( L1 (R)) r ( L2 (S)) Slide 31 Transformation Rules for RA Operations 31 If join condition contains additional attributes not in L (M = M 1 M 2 where M 1 only has attributes of R, and M 2 only has attributes of S), a final projection operation is required: L1 L2 (R r S) = L1 L2 ( ( L1 M1 (R)) r ( L2 M2 (S))) Slide 32 Transformation Rules for RA Operations 32 For example: position,city,branchNo (Staff Staff.branchNo=Branch.branchNo Branch) = ( position, branchNo (Staff)) Staff.branchNo=Branch.branchNo ( city, branchNo (Branch)) and using the latter rule: position, city (Staff Staff.branchNo=Branch.branchNo Branch) = position, city (( position, branchNo (Staff)) Staff.branchNo=Branch.branchNo ( city, branchNo (Branch))) Slide 33 Transformation Rules for RA Operations 33 8. Commutativity of Union and Intersection (but not set difference). R S = S R R S = S R Slide 34 Transformation Rules for RA Operations 34 9. Commutativity of Selection and set operations (Union, Intersection, and Set difference). p (R S) = p (S) p (R) p (R S) = p (S) p (R) p (R - S) = p (S) - p (R) Slide 35 Transformation Rules for RA Operations 35 10.Commutativity of Projection and Union. L (R S) = L (S) L (R) 11. Associativity of Union and Intersection (but not Set difference). (R S) T = S (R T) (R S) T = S (R T) Slide 36 Transformation Rules for RA Operations 36 12. Associativity of Theta join (and Cartesian product). Cartesian product and Natural join are always associative: (R S) T = R (S T) (R X S) X T = R X (S X T) If join condition q involves attributes only from S and T, then Theta join is associative: (R p S) q r T = R p r (S q T) Slide 37 Transformation Rules for RA Operations 37 For example: (Staff Staff.staffNo=PropertyForRent.staffNo PropertyForRent) ownerNo=Owner.ownerNo staff.lName=Owner.lName Owner = Staff staff.staffNo=PropertyForRent.staffNo staff.lName=lName (PropertyForRent ownerNo Owner) Slide 38 Example 21.3 Use of Transformation Rules 38 For prospective renters of flats, find properties that match requirements and owned by CO93. SELECT p.propertyNo, p.street FROM Client c, Viewing v, PropertyForRent p WHERE c.prefType = Flat AND c.clientNo = v.clientNo AND v.propertyNo = p.propertyNo AND c.maxRent >= p.rent AND c.prefType = p.type AND p.ownerNo = CO93; Slide 39 Example 21.3 Use of Transformation Rules 39 Slide 40 Example 21.3 Use of Transformation Rules 40 Slide 41 Example 21.3 Use of Transformation Rules 41 Slide 42 Overview Phases of Query Processing. 1. Decomposition. a. Analysis. b. Normalization. c. Semantic Analysis. d. Simplification. e. Restructuring. 2. Optimization. a. Heuristics. b. Comparing costs. 3. Code Generation. 4. Execution. 42 Slide 43 2. Query Optimization 2 kinds: a. Using heuristics on deciding the best plan. b. Comparing costs of different plans. 43 Slide 44 2.a. Heuristical Processing Strategies 44 Perform Selection operations as early as possible. Keep predicates on same relation together. Combine Cartesian product with subsequent Selection whose predicate represents join condition into a Join operation. Use associativity of binary operations to rearrange leaf nodes so leaf nodes with most restrictive Selection operations executed first. Slide 45 Heuristical Processing Strategies 45 Perform Projection as early as possible. Keep projection attributes on same relation together. Compute common expressions once. If common expression appears more than once, and result not too large, store result and reuse it when required. Useful when querying views, as same expression is used to construct view each time. Slide 46 2.b. Cost Estimation for RA Operations 46 Many different ways of implementing RA operations. Aim of QO is to choose most efficient one. Use formulae that estimate costs for a number of options, and select one with lowest cost. Consider only cost of disk access, which is usually dominant cost in QP. Many estimates are based on cardinality of the relation, so need to be able to estimate this. Slide 47 Database Statistics 47 Success of estimation depends on amount and currency of statistical information DBMS holds. Keeping statistics current can be problematic. If statistics updated every time tuple is changed, this would impact performance. DBMS could update statistics on a periodic basis, for example nightly, or whenever the system is idle. Slide 48 Selection Operation 48 Predicate may be simple or composite. Number of different implementations, depending on file structure, and whether attribute(s) involved are indexed/hashed. Main strategies are: Linear Search (Unordered file, no index). Binary Search (Ordered file, no index). Equality on hash key. Equality condition on primary key. Slide 49 Composite Predicates - Conjunction without Disjunction 49 May consider following approaches: - If one attribute has index or is ordered, can use one of above selection strategies. Can then check each retrieved record. - For equality on two or more attributes, with composite index (or hash key) on combined attributes, can search index directly. - With secondary indexes on one or more attributes (involved only in equality conditions in predicate), could use record pointers if exist. Slide 50 Composite Predicates - Selections with Disjunction 50 If one term contains an (OR), and term requires linear search, entire selection requires linear search. Only if index or sort order exists on every term can selection be optimized by retrieving records that satisfy each condition and applying union operator. Again, record pointers can be used if they exist. Slide 51 Join Operation 51 Main strategies for implementing join: Block Nested Loop Join. Indexed Nested Loop Join. Sort-Merge Join. Hash Join. Slide 52 Block Nested Loop Join 52 Simplest join algorithm is nested loop that joins two relations together a tuple at a time. Outer loop iterates over each tuple in R, and inner loop iterates over each tuple in S. As basic unit of reading/writing is a disk block, better to have two extra loops that process blocks. Slide 53 Indexed Nested Loop Join 53 If have index (or hash function) on join attributes of inner relation, can use index lookup. For each tuple in R, use index to retrieve matching tuples of S. Slide 54 Sort-Merge Join 54 For Equijoins, most efficient join is when both relations are sorted on join attributes. Can look for qualifying tuples merging relations. May need to sort relations first. Now tuples with same join value are in order. If assume join is *:* and each set of tuples with same join value can be held in database buffer at same time, then each block of each relation need only be read once. Slide 55 Hash Join 55 For Natural or Equijoin, hash join may be used. Idea is to partition relations according to some hash function that provides uniformity and randomness. Each equivalent partition should hold same value for join attributes, although it may hold more than one value. Slide 56 Projection Operation 56 To implement projection need to: remove attributes that are not required; eliminate any duplicate tuples produced from previous step. Only required if projection attributes do not include a key. Two main approaches to eliminating duplicates: sorting; hashing.