Relational Algebra & Calculus

Transcription

1 Relational Algebra & Calculus Yanlei Diao UMass Amherst Slides Courtesy of R. Ramakrishnan and J. Gehrke 1

2 Outline v Conceptual Design: ER model v Logical Design: ER to relational model v Querying and manipulating data Practical language: SQL Declarative: say what you want, not how to get it Formal languages: Relational Algebra and Calculus Mathematical foundation for query processing 2

3 What is an Algebra? v A mathematical system consisting of: Operands: variables or values from which new values can be constructed. Operators: symbols denoting procedures that construct new values from given values. 3

4 What is Relational Algebra v Relational Algebra: Operands are relations. Each operator takes 1 or 2 relations and produces a relation. v Closure property: relational algebra is closed under the relational model. Relational operators can be arbitrarily composed! 4

5 Relational Algebra v Basic operations: Selection ( σ ) Selects a subset of rows from a relation. Projection ( π ) Deletes unwanted columns from a relation. Cross-product ( ) Allows us to combine two relations. Set-difference ( - ) Tuples in reln. 1, but not in reln. 2. Union ( ) Tuples in reln. 1 or in reln. 2. v Additional operations: Join ( ), Intersection ( ), Division ( / ), Renaming ( ρ ) Can be derived from basic operators. Not essential, but useful for writing simpler queries. 5

6 Example Instances Sailors S1 S2 22 dustin lubber yuppy lubber guppy R1 Reserves sid bid day /10/ /12/96 6

7 Projection v Retain only attributes in the projection list; delete others. v Schema of result contains exactly the fields in projection list. S2 28 yuppy lubber guppy sname yuppy 9 lubber 8 guppy 5 rusty 10 rating π ( S2) sname, rating 7

8 Projection (contd.) S2 28 yuppy lubber guppy age π age ( S2) v Projection operator has to eliminate duplicates! Real systems typically don t do duplicate elimination unless the user explicitly asks for it. (Why not?) 8

9 Selection v Select rows that satisfy the selection condition; discard others. v Schema of result identical to schema of input. S2 28 yuppy lubber guppy yuppy σ rating S >8 ( 2) 9

10 Selection (contd.) S2 28 yuppy lubber guppy σ (rating > 5 rating < 5) age < 50 (S2) 10

11 A Simple Example of Composition v Composition: result relation of an operator can be the input to another operator. 28 yuppy σ rating S >8 ( 2) sname rating yuppy 9 rusty 10 π σ ( ( S )) sname, rating rating>8 2 11

12 Union, Intersection, Set-Difference v v v Set operations: Union ( ) Intersection ( ) Set difference ( - ) Take two input relations, which must be union-compatible: Same number of fields. Corresponding fields have the the same type. What is the schema of result? S1 22 dustin lubber S2 28 yuppy lubber guppy

13 Example Set Operations S1 22 dustin lubber S2 28 yuppy lubber guppy dustin lubber guppy yuppy S1 S2 Duplicate elimination: remove tuples that have same values in all attributes. 13

14 Example Set Operations S1 22 dustin lubber S2 28 yuppy lubber guppy dustin S1 S2 31 lubber S1 S2 Duplicates? 14

15 Cross (Cartesian) Product v S1 R1: Each row of S1 is paired with each row of R1. S1 22 dustin lubber R1 sid bid day /10/ /12/96 S1 R1 (sid) sname rating age (sid) bid day 22 dustin /10/96 22 dustin /12/96 31 lubber /10/96 31 lubber /12/ /10/ /12/96 15

16 Cross-Product (contd.) (sid) sname rating age (sid) bid day 22 dustin /10/96 22 dustin /12/96 31 lubber /10/96 31 lubber /12/ /10/ /12/96 v Result schema inherits all fields of S1 and R1. Conflict: Both S1 and R1 have a field called sid. Renaming operator: ρ ( C( 1 sid1, 5 sid2), S1 R1) What if we want to find pairs s.t. S1.sid = S2.sid? 16

17 Join v Condition (theta) Join: R θ S = σ θ (R S) (sid) sname rating age (sid) bid day 22 dustin /12/96 31 lubber /12/96 S 1 R 1 S1. sid < R1. sid Result schema same as that of cross-product. But often fewer tuples, more efficient for computation. 17

18 Join v Equi-Join: A special case of condition join where the condition θ contains only equalities. bid day 22 dustin /10/ /12/96 S1 R1 sid Result schema contains only one copy of fields for which equality is specified. v Natural Join ( R w v S ): equi-join on all common fields. 18

19 Self-Join S1 22 dustin lubber S2 22 dustin lubber ρ(s1,s) ρ(s2,s) S1.rating>S2.rating S1.age<S2.age 19

20 Example Schema v Sailors(sid: integer, sname: string, rating: integer, age: integer) v Boats(bid: integer, color: string) v Reserves(sid: integer, bid: integer, day: date) 20

21 Find names of sailors who ve reserved boat #103 v Sailors(sid: integer, sname: string, rating: integer, age: integer) v Boats(bid: integer, color: string) v Reserves(sid: integer, bid: integer, day: date) v How many relations do we need to access? If more than 1, use cross-products or joins. v Filtering condition? Use selection. v Attributes to return? Use projection. 21

22 Find names of sailors who ve reserved boat #103 v Solution 1: π (( σ Re serves ) Sailors ) sname bid =103 v Solution 2: ρ ( Temp1, σ Re serves) bid =103 ρ ( Temp2, Temp1 Sailors) π sname ( Temp2) v Solution 3: π sname( σ (Re serves Sailors)) bid =103 Algebraic equivalence! 22

23 Find names of sailors who ve reserved a red boat v Sailors(sid: integer, sname: string, rating: integer, age: integer) v Boats(bid: integer, color: string) v Reserves(sid: integer, bid: integer, day: date) v Boat color is only available in Boats; so need an extra join: π sname (( σ color = ' red ' Boats) Re serves Sailors) 23

24 Find names of sailors who ve reserved a red or a green boat v Sailors(sid: integer, sname: string, rating: integer, age: integer) v Boats(bid: integer, color: string) v Reserves(sid: integer, bid: integer, day: date) 24

25 Find names of sailors who ve reserved a red or a green boat v Can identify all red or green boats, then find sailors who ve reserved one of these boats: ρ ( Tempboats, ( σ )) color = ' red ' color = ' green' Boats π sname ( Tempboats Re serves Sailors) v Can also define Tempboats using union. How? 25

26 Find names of sailors who ve reserved a red boat and a green boat v A single selection won t work. Why? v Instead, intersect sailors who ve reserved red boats and sailors who ve reserved green boats. sid is a key for Sailors ρ ( Tempred, π (( σ sid color = ' red ' ρ ( Tempgreen, π (( σ sid color = ' green' Boats) Re serves)) Boats) Re serves)) π sname (( Tempred Tempgreen) Sailors) Can we project onto name and then do the intersection? 26

27 Find the names of sailors who ve reserved all boats 27

28 Relational Calculus v Query has the form: ' ) ( ) *! # x1, x2,..., xn p x1, x2,..., xn # " $ + & ), & % ) - domain variables, or constants formula v Answer includes all tuples x1, x2,..., xn that make the formula! # p x1, x2,..., xn # " $ & & % be true. 28

29 Formulas v Formula is recursively defined: Atomic formulas: retrieving tuples from relations, or making comparisons of values Logical connectives:,, Quantifiers:, (optional) o x F(x) is equivalent to x F(x) 29

30 Find names of sailors rated > 7 who have reserved boat #103 Relational Algebra: π (( σ Reserves) ( σ sname bid = 103 rating> 7 Sailors)) Relational Calculus: {X sname X sid, X rating, X age Sailors(X sid, X sname, X rating, X age ) X rating > 7 X bid, X day Reserves(X sid, X bid, X day ) X bid =103 } v Where is the join? Use to find a tuple in Reserves that `joins with the Sailors tuple under consideration. 30

31 Find names of sailors who ve reserved all boats Sailors(sid: integer, sname: string, rating: integer, age: integer) Boats(bid: integer, color: string) Reserves(sid: integer, bid: integer, day: date) {X sname X sid, X rating, X age X sid, X sname, X rating, X age Sailors X bid, X color Boats ( X day X sid, X bid, X day Reserves) } v To find sailors who ve reserved all red boats: {X sname X sid, X rating, X age X sid, X sname, X rating, X age Sailors X bid, X color Boats (X color ʹ redʹ X day X sid, X bid, X day Reserves) } 31

32 Unsafe Queries, Expressive Power v Some syntactically correct calculus queries can have an infinite number of answers! Such queries are called unsafe. ( " %, * e.g., S $ S Sailors' * ) - * + $ # v Equivalence result: every query that can be expressed in relational algebra can be expressed as a safe query in relational calculus; the converse is also true. ' & *. v Query language (e.g., SQL) can express every query that is expressible in relational algebra/calculus. 32

33 Find names of sailors who ve reserved all boats {X sname X sid, X rating, X age X sid, X sname, X rating, X age Sailors X bid, X color Boats ( X day X sid, X bid, X day Reserves) } {X sname X sid, X rating, X age X sid, X sname, X rating, X age Sailors X bid, X color Boats ( X day X sid, X bid, X day Reserves) } 33

34 Find the names of sailors who ve reserved all boats v Step 1: for each sailor, check if there exists a boat that he has not reserved (called formula F). ρ(s _ neg, π sid ( (π sid Re serves) (π bid Boats) (π sid,bid Re serves))) v Step 2: find sailors for which F is not true and retrieve their names π sname ( (π sid Re serves S _ neg) Sailors ) - : the only way to express negation in relational algebra! 34