Relational Algebra IS264 Lecture PDF

Relational Algebra p 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 rac...

Relational Algebra p 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 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. Relational Algebra Overview Relational algebra is the basic set of operations for the relational model These operations enable a user to specify basic retrieval requests (or queries) The result of an operation is a new relation, which may have been formed from one or more input relations – This property makes the algebra closed (all objects in relational algebra are relations) Relational Algebra Overview (continued) The algebra operations thus produce new relations – These can be further manipulated using operations of the same algebra A sequence of relational algebra operations forms a relational algebra expression – The result of a relational algebra expression is also a relation that represents the result of a database query (or retrieval request) Relational Algebra Overview Relational Algebra consists of several groups of operations – Unary Relational Operations SELECT (symbol: s (sigma)) PROJECT (symbol: p (pi)) RENAME (symbol: r (rho)) – Relational Algebra Operations From Set Theory UNION ( È ), INTERSECTION ( Ç ), DIFFERENCE (or MINUS, – ) CARTESIAN PRODUCT ( x ) – Binary Relational Operations JOIN (several variations of JOIN exist) DIVISION – Additional Relational Operations OUTER JOINS, OUTER UNION AGGREGATE FUNCTIONS (These compute summary of information: for example, SUM, COUNT, AVG, MIN, MAX) 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!  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  Relational Algebra: 5 Basic Operations Selection ( σ ) Selects a subset of rows from relation (horizontal). Projection ( π ) 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 (È ) Tuples in r1 and/or in r2. Since each operation returns a relation, operations can be composed! (Algebra is closed.)  Example Instances R1 sid bid day 22 101 10/10/96 58 103 11/12/96 S1 sid sname rating age bid bname color 22 dustin 7 45.0 101 Interlake blue 31 lubber 8 55.5 102 Interlake red 58 rusty 10 35.0 103 Clipper green 104 Marine red S2 sid sname rating age Boats 28 yuppy 9 35.0 31 lubber 8 55.5 44 guppy 5 35.0 58 rusty 10 35.0  Projection Examples: π age(S2) ; π sname, rating(S2) 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?)  sname rating Projection yuppy 9 lubber 8 guppy 5 rusty 10 sid sname rating age π sname,rating (S 2) 28 yuppy 9 35.0 31 lubber 8 55.5 44 guppy 5 35.0 58 rusty 10 35.0 age S2 35.0 55.5 π age(S2)  Selection (s) 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? sid sname rating age 28 yuppy 9 35.0 sname rating 31 lubber 8 55.5 yuppy 9 44 guppy 5 35.0 58 rusty 10 35.0 rusty 10 σ rating >8(S2) π sname,rating(σ rating >8(S2))  Select Operator Produce table containing subset of rows of argument table satisfying condition s() Example: s Hobby=’stamps’ (Person) Id Name Address Hobby Id Name Address Hobby 1123 John 123 Main stamps 1123 John 123 Main stamps 1123 John 123 Main coins 9876 Bart 5 Pine St stamps 5556 Mary 7 Lake Dr hiking 9876 Bart 5 Pine St stamps Person Selection Condition Operators: , =, ¹ Simple selection condition: – operator – operator AND OR NOT NOTE: Each attribute in condition must be one of the attributes of the relation name Selection Condition - Examples s Id>3000 OR Hobby= hiking (Person) s Id>3000 AND Id 3000 AND Id < 3999) OR Hobby¹ hiking (Person)  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? Union Compatible Relations Two relations are union compatible if – Both have same number of columns – Names of attributes are the same in both – Attributes with the same name in both relations have the same domain Union compatible relations can be combined using union, intersection, and set difference Example Tables: Person (SSN, Name, Address, Hobby) Professor (Id, Name, Office, Phone) are not union compatible. However P Name (Person) and P Name (Professor) are union compatible and P Name (Person) - P Name (Professor) makes sense.  Union sid sname rating age sid sname rating age 22 dustin 7 45.0 22 dustin 7 45.0 31 lubber 8 55.5 31 lubber 8 55.5 58 rusty 10 35.0 58 rusty 10 35.0 44 guppy 5 35.0 S1 28 yuppy 9 35.0 sid sname rating age S1∪ S2 28 yuppy 9 35.0 31 lubber 8 55.5 44 guppy 5 35.0 58 rusty 10 35.0 S2  Set Difference sid sname rating age sid sname rating age 22 dustin 7 45.0 22 dustin 7 45.0 31 lubber 8 55.5 S1− S2 58 rusty 10 35.0 S1 sid sname rating age sid sname rating age 28 yuppy 9 35.0 28 yuppy 9 35.0 31 lubber 8 55.5 44 guppy 5 35.0 44 guppy 5 35.0 S2 – S1 58 rusty 10 35.0 S2  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: ρ (C(1→ sid1, 5 → sid2), S1× R1)  Cross Product Example sid sname rating age sid bid day 22 dustin 7 45.0 22 101 10/10/96 31 lubber 8 55.5 58 103 11/12/96 58 rusty 10 35.0 R1 S1 (sid) sname rating age (sid) bid day 22 dustin 7 45.0 22 101 10/10/9 S1XR1 = 22 dustin 7 45.0 58 103 11/12/9 31 lubber 8 55.5 22 101 10/10/9 31 lubber 8 55.5 58 103 11/12/9 58 rusty 10 35.0 22 101 10/10/9 58 rusty 10 35.0 58 103 11/12/9  Cartesian Product The Cartesian Product of two relations A (a tuples) and B (b tuples) , which have attributes A1... Am and B1... Bn, is the relation with m + n attributes containing row for every pair of rows one from A and one from B. The result has a x b tuples. EMPLOYEE DEPENDENT name NI dept ENI name sex ML 123 5 123 SM M JR 456 5 123 TC F 456 JA F EMPLOYEE x DEPENDENT name NI dept ENI name sex * ML 123 5 123 SM M * ML 123 5 123 TC F ML 123 5 456 JA F JR 456 5 123 SM M JR 456 5 123 TC F * JR 456 5 456 JA F  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 Ç S = R - (R - S)  Intersection sid sname rating age 22 dustin 7 45.0 31 lubber 8 55.5 sid sname rating age 58 rusty 10 35.0 31 lubber 8 55.5 S1 58 rusty 10 35.0 sid sname rating age 28 31 yuppy lubber 9 8 35.0 55.5 S1∩ S2 44 guppy 5 35.0 58 rusty 10 35.0 S2  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.  Natural Join Example sid sname rating age sid bid day 22 dustin 7 45.0 22 101 10/10/96 31 lubber 8 55.5 58 103 11/12/96 58 rusty 10 35.0 R1 S1 R1 S1 = sid sname rating age bid day 22 dustin 7 45.0 101 10/10/96 58 rusty 10 35.0 103 11/12/96  Other Types of Joins Condition Join (or theta-join ): R  c S = σ c ( R × S) 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.  Theta Join Example sid bid day sid sname rating age 22 101 10/10/96 22 dustin 7 45.0 58 103 11/12/96 31 lubber 8 55.5 58 rusty 10 35.0 R1 S1 S1 R1 = S1.sid< R1.sid € (sid) sname rating age (sid) bid day 22 dustin 7 45.0 58 103 11/12/96 31 lubber 8 55.5 58 103 11/12/96  sid bid day Reserves Examples 22 101 10/10/96 58 103 11/12/96 sid sname rating age Sailors 22 dustin 7 45.0 31 lubber 8 55.5 58 rusty 10 35.0 Boats bid bname color 101 Interlake Blue 102 Interlake Red 103 Clipper Green 104 Marine Red  Find names of sailors who ve reserved boat #103 Solution 1: π sname ((σ Re serves)  Sailors) bid =103 Solution 2: π sname (σ (Re serves  Sailors)) bid =103  Find names of sailors who ve reserved a red boat Information about boat color only available in Boats; so need an extra join: π sname ((σ Boats)  Re serves  Sailors) color =' red ' ❖ A more efficient (???) solution: π sname(π ((π (σ Boats)) Res) Sailors) sid bid color='red ' € ☛ A query optimizer can find this given the first solution!  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: ρ (Tempboats, (σ Boats)) color =' red ' ∨ color =' green ' π sname(Tempboats  Re serves  Sailors)  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): ρ (Tempred, π ((σ Boats)  Re serves)) sid color =' red ' ρ (Tempgreen, π ((σ Boats)  Re serves)) sid color =' green' π sname((Tempred ∩ Tempgreen)  Sailors)  Find the names of sailors who ve reserved all boats Uses division; schemas of the input relations to / must be carefully chosen: ρ (Tempsids, (π Re serves) / (π Boats)) sid, bid bid π sname (Tempsids  Sailors) ❖ To find sailors who ve reserved all Interlake boats:..... /π (σ Boats) bid bname =' Interlake'