Relational Algebra Chapter 4 - part I

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. 2

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-operational, rather declarative.)

Preliminaries A query is applied to relation instances. The result of a query is also a relation instance. Schemas of input relations for a query are fixed ; (but query will run regardless of instance!) The schema for result of given query is also fixed! Determined by definition of query language . 4

Preliminaries Positional vs. named-field notation: Pros/Cons: Positional field notation e.g., S.1 Named field notation e.g., S.sid Pros/Cons: Positional notation easier for formal definitions, named-field notation more readable. Both used in SQL Assume that names of fields in query results are `inherited’ from names of fields in query input relations. 4

Example Instances Sailors Reserves R1 S1 Sailors S2 5

Relational Algebra Basic operations: Additional operations: Selection ( ) Selects a subset of rows from relation. Projection ( ) Deletes unwanted columns from relation. Cartesian-product ( ) Allows us to combine two relations. Set-difference ( ) Tuples in reln. 1, but not in reln. 2. Union ( ) Tuples in reln. 1 and in reln. 2. Additional operations: Intersection, join, division, renaming: Not essential, but (very!) useful. Since each operation returns a relation, operations can be composed! (Algebra is “closed”.) 6

Selection Sailors S2 5

Selection condition (R) Selects rows that satisfy selection condition. attribute op constant attribute op attribute Op is {<,>,<=,>=, =, =} No duplicates in result! Schema of result identical to schema of (only) input relation. 8

Selection Result relation can be input for another relational algebra operation! (Operator composition.) 8

Projection Sailors S2 5

Projection  projectlist (R) Deletes attributes that are not in projection list. Schema of result = contains fields in projection list Projection operator has to eliminate duplicates! (Why??) Note: real systems typically don’t do duplicate elimination unless the user explicitly asks for it. (Why not?) 7

Union, Intersection, Set-Difference All of these operations take two input relations, which must be union-compatible: Same number of fields. `Corresponding’ fields have same type. What is the schema of result? 9

Example Instances : Union 5

Difference Operation S1 S2 5

Intersection Operation 5

Cross-Product (Cartesian Product) S1 R1 : Each row of S1 is paired with each row of R1. Reserves R1 Sailors S1 10

Cross-Product (Cartesian Product) S1 R1 : Result schema has one field per field of S1 and R1, with field names `inherited’ if possible. Conflict: Both S1 and R1 have a field called sid. Renaming operator: 10

Why we need a Join Operator ? In many cases, Join = Cross-Product + Select + Project However : Cross-product is too large to materialize Apply Select and Project "On-the-fly" 11

Condition Join / Theta Join Result schema same as that of cross-product. Fewer tuples than cross-product, more efficient. 11

EquiJoin Equi-Join: A special case of condition join where the condition c contains only equalities. Result schema similar to cross-product, but only one copy of fields for which equality is specified. An extra project: PROJECT ( THETA-JOIN) 12

Natural Join Natural Join: Equijoin on all common fields.

Division Not supported as a primitive operator, but useful for expressing queries like: Find sailors who have reserved all boats. 13

Division Let A have 2 fields x and y; B have only field y: A/B = A/B contains all x tuples (sailors) such that for every y tuple (boat) in B, there is an xy tuple in A. If set of y values (boats) associated with an x value (sailor) in A contains all y values in B, then x value is in A/B. A/B is the largest relation instance Q such that QB A. 13

Division Example B= relation parts e.g., A = all parts supplied by suppliers, B= relation parts A/B = suppliers who supply all parts listed in B 13

Examples of Division A/B 14

Expressing A/B Using Basic Operators 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: A/B: 15

A few example queries

Find names of sailors who’ve reserved boat #103 Solution 1 Solution 2 Solution 3 16

Find names of sailors who’ve reserved a red boat Reserves R1 Sailors S1 Boats B1 bid bname color 101 Interlake blue 103 Clipper red 17

Find names of sailors who’ve reserved a red boat Information about boat color only available in Boats; so need an extra join: A more efficient solution: A query optimizer can find this given the first solution! 17

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: Can also define Tempboats using union! (How?) What happens if is replaced by in this query? 18

Find sailors who’ve reserved a red and a green boat Previous approach won’t work! Must identify sailors who reserved red boats, sailors who’ve reserved green boats, then find their intersection 19

Find the names of sailors who’ve reserved all boats Uses division; schemas of the input relations to / must be carefully chosen: To find sailors who’ve reserved all ‘Interlake’ boats: ..... 20

Relational Algebra : Some More Operators Beyond Chapter 4

Generalized Projection  F1, F2, … (R)  sname, (rating*2 as myrating) (Sailors) 7

Aggregation operators MIN, MAX, COUNT, SUM, AVG AGGB (R) considers only non-null values of R. SUMB (R) MINB (R) COUNTB (R) R MINB (R) 2 SUMB (R) 9 COUNTB (R) 3 A B 1 2 3 4 null AVGB (R) COUNT* (R) MAXB (R) AVGB (R) 3 COUNT* (R) 4 MAXB (R) 4

Aggregation Operators MIN, MAX, SUM, AVG must be on any 1 attribute. COUNT can be on any 1 attribute or COUNT* (R) An aggregation operator returns a bag, not a single value ! But SQL allows treatment as a single value. σB=MAXB (R) (R) A B 3 4

Grouping Operator: GL, AL (R) GL, AL (R) groups all attributes in GL, and performs the aggregation specified in AL.  starName, MIN (year)→year, COUNT(title) →num (StarsIn) starName year num HF 77 3 KR 94 2 StarsIn title year starName SW1 77 HF Matrix 99 KR 6D&7N 93 SW2 79 Speed 94

Summary The relational model has rigorously defined query languages that are simple and powerful. Relational algebra is operational; useful as internal representation for query evaluation plans. Several ways of expressing a given query; a query optimizer should choose most efficient version.