Lecture 4: Relational algebra www.cl.cam.ac.uk/Teaching/current/Databases/

Today’s lecture What’s the (core) relational algebra? How can we write queries using the relational algebra? How powerful is the relational algebra?

Relational query languages Query languages allow the manipulation and retrieval of data from a database The relational model supports simple, powerful query languages Strong formal foundation Allows for much (provably correct) optimisation NOTE: Query languages are not (necessarily) programming languages

Formal relational query languages Two formal query languages Relational Algebra Simple ‘operational’ model, useful for expressing execution plans Relational Calculus Logical model (‘declarative’), useful for theoretical results Both languages were introduced by Codd in a series of papers They have equivalent expressive power They are the 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 Schema of relations are fixed (cf. types) The query will then execute over any valid instance The schema of the result can also be determined

Example relation instances A database of boats, sailors, and reservations S2 sid sname rating age 10 Myleene 6 23 22 Tim 8 26 99 Julia 100 20 88 Gavin 21 R1 sid bid day 22 101 101001 99 103 111201 S1 sid sname rating age 11 Sue 7 26 22 Tim 8 33 Bob 9 28 55 Kim 10 B1 bid colour 101 red 102 blue 103 green

Core relational algebra Five basic operator classes: Selection Selects a subset of rows Projection Picking certain columns Renaming Renaming attributes Set theoretic operations The familiar operations: union, intersection, difference, … Products and joins Combining relations in useful ways

Selection Selects rows that satisfy a condition, written R1 = c(R2) where c is a condition involving the attributes of R2, e.g. rating>8(S2) returns the relation instance sid sname rating age 99 Julia 100 20 88 Gavin 21

sname=“Julia”(rating>8(S2)) Selection cont. Note: The schema of the result is exactly the same as the schema of the input relation instance There are no duplicates in the resulting relation instance (why?) The resulting relation instance can be used as the input for another relational algebra operator, e.g. sname=“Julia”(rating>8(S2))

Projection Deletes fields that are not in the projection list R1=A(R2) where A is a list of attributes from the schema of R2, e.g. sname,rating(S2) returns the relation instance sname rating Myleene 6 Tim 8 Julia 100 Gavin

Projection cont. Note: Projection operator has to eliminate duplicates (why?) Aside: Real systems don’t normally perform duplicate elimination unless the user explicitly asks for it (why not?)

Renaming R1= A:=B(R2) Returns a relation instance identical to R2 except that field A is renamed B For example, sname:=nom(S1) sid nom rating age 11 Sue 7 26 22 Tim 8 33 Bob 9 28 55 Kim 10

Familiar set operations We have the familiar set-theoretic operators, e.g. , , - There is a restriction on their input relation instances: they must be union compatible Same number of fields Same field names and domains E.g. S1S2 is valid, but S1R1 is not

Cartesian products AB Concatenate every row of A with every row of B What do we do if A and B have some field names in common? Several choices, but we’ll simply assume that the resulting duplicate field names will have the suffix 1 and 2

Example S1R1 Note! sid.1 sname rating age sid.2 bid day 11 Sue 7 26 22 101 101001 99 103 111201 Tim 8 33 Bob 9 28 55 Kim 10

Theta join Theoretically, it is a derived operator R1 Vc R2 @ c(R1R2) E.g., S1 Vsid.1<=sid.2R1 sid.1 sname rating age sid.2 bid day 11 Sue 7 26 22 101 101001 99 103 111201 Tim 8 33 Bob 9 28 55 Kim 10

Theta join cont. The result schema is the same as for a cross-product Sometimes this operator is called a conditional join Most commonly the condition is an equality on field names, e.g. S1 Vsid.1=sid.2R1

Equi- and natural join Equi-join is a special case of theta join where the condition is equality of field names, e.g. S1 Vsid R1 Natural join is an equi-join on all common fields where the duplicate fields are removed. It is written simply A V B sid.1 sname rating age sid.2 bid day 22 Tim 8 26 101 101001

Natural join cont. Note that the common fields appear only once in the resulting relation instance This operator appears very frequently in real-life queries It is always implemented directly by the query engine (why?)

Find sailors who have reserved all the boats Division Not a primitive operator, but useful to express queries such as Find sailors who have reserved all the boats Consider the simple case, where relation A has fields x and y, and relation B has field y A/B is the set of xs (sailors) such that for every y (boat) in B, there is a row (x,y) in A

Division cont. Can you code this up in the relational algebra?

Division cont. Can you code this up in the relational algebra? x’s that are disqualified: x((x(A)  B) – A) Thus: x(A)-x((x(A)  B) – A)

Find names of sailors who’ve reserved boat 103 Example 1 Find names of sailors who’ve reserved boat 103 Solution 1: sname(bid=103(Reserves) V Sailors) Solution 2: sname(bid=103(Reserves V Sailors)) Which is more efficient? Query optimisation

Find names of sailors who’ve reserved a red boat Example 2 Find names of sailors who’ve reserved a red boat DO THIS ON SLIDE MYSELF

Find names of sailors who’ve reserved a red boat Example 2 Find names of sailors who’ve reserved a red boat sname(colour=“red”(Boats) V Reserves V Sailors) Better: sname(sid(bid(colour=“red”(Boats)) V Reserves) V Sailors) DO THIS ON SLIDE MYSELF

Find sailors who’ve reserved a red or a green boat Example 3 Find sailors who’ve reserved a red or a green boat MARS Bar challenge!!! Need to know that SQL has logical AND and OR operators

Find sailors who’ve reserved a red or a green boat Example 3 Find sailors who’ve reserved a red or a green boat let T = colour=“red”colour=“green”(Boats) in sname(T V Reserves V Sailors) MARS Bar challenge!!! Need to know that SQL has logical AND and OR operators

Find sailors who’ve reserved a red and a green boat Example 4 Find sailors who’ve reserved a red and a green boat Mars Bar challenge

Find sailors who’ve reserved a red and a green boat Example 4 Find sailors who’ve reserved a red and a green boat NOTE: Can’t just trivially modify last solution! let T1 = sid (colour=“red”(Boats) V Reserves) T2 = sid (colour=“green”(Boats) V Reserves) in sname((T1  T2) V Sailors) Mars Bar challenge

Find the names of sailors who’ve reserved at least two boats Example 5 Find the names of sailors who’ve reserved at least two boats let T = sid.1:=sid (sid.1,sname,bid (Sailors V Reserves)) in sname.1 (sid.1=sid.2bid.1bid.2(T  T)) Mars bar challenge

Find the names of sailors who’ve reserved all boats Example 6 Find the names of sailors who’ve reserved all boats let T = sid,bid (Reserves) / bid (Boats) in sname(T V Sailors) But this is pretty much given on the division slides

Computational limitations Suppose we have a relation SequelOf of movies and their immediate sequels We want to compute the relation ‘isFollowedBy’ … movie sequel Naked Gun Naked Gun 2½ Naked Gun 33 1/3 Rocky Rocky II Rocky III Rocky IV Rocky V

Computational limitations We could compute fst,thd(movie:=fst,sequel:=snd(SequelOf) V movie:=snd,sequel:=thd(SequelOf)) This provides us with sequels-of-sequels We could write three joins to get sequels-of- sequels-of-sequels and union the results What about Friday the 13th (9 sequels)?  In general we need to be able to write an arbitrarily large union… The relational algebra needs to be extended to handle these sorts of queries

Summary You should now understand: The core relational algebra Operations and semantics Union compatibility Computational limitations of the relational algebra Next lecture: Relational calculus