lecture8Alg.ppt
- 1. By relieving the brain of all unnecessary work, a good notation sets it free to concentrate on more advanced problems, and, in effect, increases the mental power of the race. -- Alfred North Whitehead (1861 - 1947) Relational Algebra CS 186 Spring 2006, Lecture 8 R & G, Chapter 4 p
- 2. Administrivia • Midterm 1 – Tues 2/21, in class – Covers all material up to and including Th 2/16 – Closed book – can bring 1 8.5x11” sheet of notes – No calculators – no cell phones • Midterm 2 – Th 3/21, in class – Focuses on material after MT 1 – Same rules as above. • Homework 1 – 2/14 • Homework 2 – Out after MT 1, Due 3/14
- 3. Relational Query Languages • Query languages: Allow manipulation and retrieval of data from a database. • Relational model supports simple, powerful QLs: – Strong formal foundation based on logic. – Allows for much optimization. • Query Languages != programming languages! – QLs not expected to be “Turing complete”. – QLs not intended to be used for complex calculations. – QLs support easy, efficient access to large data sets.
- 4. Formal Relational Query Languages Two mathematical Query Languages form the basis for “real” languages (e.g. SQL), and for implementation: Relational Algebra: More operational, very useful for representing execution plans. Relational Calculus: Lets users describe what they want, rather than how to compute it. (Non-procedural, declarative.) Understanding Algebra & Calculus is key to understanding SQL, query processing!
- 5. Preliminaries • A query is applied to relation instances, and the result of a query is also a relation instance. – Schemas of input relations for a query are fixed (but query will run over any legal instance) – The schema for the result of a given query is also fixed. It is determined by the definitions of the query language constructs. • Positional vs. named-field notation: – Positional notation easier for formal definitions, named-field notation more readable. – Both used in SQL
- 6. Relational Algebra: 5 Basic Operations • Selection ( ) Selects a subset of rows from relation (horizontal). • Projection ( ) Retains only wanted columns from relation (vertical). • Cross-product (x) Allows us to combine two relations. • Set-difference (–) Tuples in r1, but not in r2. • Union ( ) Tuples in r1 and/or in r2. Since each operation returns a relation, operations can be composed! (Algebra is “closed”.) p
- 8. Projection page S ( ) 2 • Examples: ; • Retains only attributes that are in the “projection list”. • Schema of result: – exactly the fields in the projection list, with the same names that they had in the input relation. • Projection operator has to eliminate duplicates (How do they arise? Why remove them?) – Note: real systems typically don’t do duplicate elimination unless the user explicitly asks for it. (Why not?) psname rating S , ( ) 2
- 10. Selection () rating S 8 2 ( ) p sname rating rating S , ( ( )) 8 2 • Selects rows that satisfy selection condition. • Result is a relation. Schema of result is same as that of the input relation. • Do we need to do duplicate elimination?
- 11. Union and Set-Difference • All of these operations take two input relations, which must be union-compatible: – Same number of fields. – `Corresponding’ fields have the same type. • For which, if any, is duplicate elimination required?
- 13. Set Difference S1 S2 S S 1 2 S2 – S1
- 14. Cross-Product • S1 x R1: Each row of S1 paired with each row of R1. • Q: How many rows in the result? • Result schema has one field per field of S1 and R1, with field names `inherited’ if possible. – May have a naming conflict: Both S1 and R1 have a field with the same name. – In this case, can use the renaming operator: ( ( , ), ) C sid sid S R 1 1 5 2 1 1
- 15. Cross Product Example R1 S1 R1 X S1 =
- 16. Compound Operator: Intersection • In addition to the 5 basic operators, there are several additional “Compound Operators” – These add no computational power to the language, but are useful shorthands. – Can be expressed solely with the basic ops. • Intersection takes two input relations, which must be union-compatible. • Q: How to express it using basic operators? R S = R (R S)
- 18. Compound Operator: Join • Joins are compound operators involving cross product, selection, and (sometimes) projection. • Most common type of join is a “natural join” (often just called “join”). R S conceptually is: – Compute R X S – Select rows where attributes that appear in both relations have equal values – Project all unique atttributes and one copy of each of the common ones. • Note: Usually done much more efficiently than this. • Useful for putting “normalized” relations back together.
- 19. Natural Join Example R1 S1 R1 S1 =
- 20. Other Types of Joins • Condition Join (or “theta-join”): • Result schema same as that of cross-product. • May have fewer tuples than cross-product. • Equi-Join: Special case: condition c contains only conjunction of equalities. R c S c R S ( )
- 22. Compound Operator: Division • Useful for expressing “for all” queries like: Find sids of sailors who have reserved all boats. • For A/B attributes of B are subset of attrs of A. – May need to “project” to make this happen. • E.g., let A have 2 fields, x and y; B have only field y: A/B contains all x tuples such that for every y tuple in B, there is an xy tuple in A.
- 23. Examples of Division A/B A B1 B2 B3 A/B1 A/B2 A/B3
- 24. Expressing A/B Using Basic Operators • Division is not essential op; just a useful shorthand. – (Also true of joins, but joins are so common that systems implement joins specially.) • Idea: For A/B, compute all x values that are not `disqualified’ by some y value in B. – x value is disqualified if by attaching y value from B, we obtain an xy tuple that is not in A. Disqualified x values: p p x x A B A (( ( ) ) ) A/B: p x A ( ) Disqualified x values
- 26. Find names of sailors who’ve reserved boat #103 • Solution 1: p sname bid serves Sailors (( Re ) ) 103 • Solution 2: p sname bid serves Sailors ( (Re )) 103
- 27. Find names of sailors who’ve reserved a red boat • Information about boat color only available in Boats; so need an extra join: p sname color red Boats serves Sailors (( ' ' ) Re ) A more efficient (???) solution: A query optimizer can find this given the first solution!
- 28. Find sailors who’ve reserved a red or a green boat • Can identify all red or green boats, then find sailors who’ve reserved one of these boats: ( ,( ' ' ' ' )) Tempboats color red color green Boats p sname Tempboats serves Sailors ( Re )
- 29. Find sailors who’ve reserved a red and a green boat • Previous approach won’t work! Must identify sailors who’ve reserved red boats, sailors who’ve reserved green boats, then find the intersection (note that sid is a key for Sailors): p ( , (( ' ' ) Re )) Tempred sid color red Boats serves p sname Tempred Tempgreen Sailors (( ) ) p ( , (( ' ' ) Re )) Tempgreen sid color green Boats serves
- 30. Find the names of sailors who’ve reserved all boats • Uses division; schemas of the input relations to / must be carefully chosen: p p ( , ( , Re ) / ( )) Tempsids sid bid serves bid Boats p sname Tempsids Sailors ( ) To find sailors who’ve reserved all ‘Interlake’ boats: / ( ' ' ) p bid bname Interlake Boats .....