# 1 relational algebra and calculus chapter 4. 2 relational query languages query languages: allow...

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Relational Algebra and Calculus

Chapter 4

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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

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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, declarative)

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Example Instances

sid sname rating age

22 dustin 7 45.0

31 lubber 8 55.558 rusty 10 35.0

sid sname rating age28 yuppy 9 35.031 lubber 8 55.544 guppy 5 35.058 rusty 10 35.0

sid bid day

22 101 10/10/9658 103 11/12/96

R1

S1

S2

“Sailors” and “Reserves” relations for our examples

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Relational Algebra Basic operations:

Selection ( ) Selects a subset of rows from relation Projection ( ) Deletes unwanted columns from 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 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”)

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Projectionsname rating

yuppy 9lubber 8guppy 5rusty 10

sname rating

S,

( )2

age

35.055.5

age S( )2

Deletes attributes that are not in projection list

Schema of result contains exactly the fields in the projection list, with the same names that they had in the (only) input relation

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?)

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Selection

rating

S82( )

sid sname rating age28 yuppy 9 35.058 rusty 10 35.0

sname ratingyuppy 9rusty 10

sname rating rating

S,

( ( ))82

Selects rows that satisfy selection condition

No duplicates in result! (Why?)

Schema of result identical to schema of (only) input relation

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

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Union, Intersection, 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 What is the schema of

result?

sid sname rating age

22 dustin 7 45.031 lubber 8 55.558 rusty 10 35.044 guppy 5 35.028 yuppy 9 35.0

sid sname rating age31 lubber 8 55.558 rusty 10 35.0

S S1 2

S S1 2

sid sname rating age

22 dustin 7 45.0

S S1 2

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Cross-Product Each row of S1 is paired with each row of 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

( ( , ), )C sid sid S R1 1 5 2 1 1

(sid) sname rating age (sid) bid day

22 dustin 7 45.0 22 101 10/ 10/ 96

22 dustin 7 45.0 58 103 11/ 12/ 96

31 lubber 8 55.5 22 101 10/ 10/ 96

31 lubber 8 55.5 58 103 11/ 12/ 96

58 rusty 10 35.0 22 101 10/ 10/ 96

58 rusty 10 35.0 58 103 11/ 12/ 96

Renaming operator:

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Joins Condition Join:

Result schema same as that of cross-product Fewer tuples than cross-product, might be

able to compute more efficiently

R c S c R S ( )

(sid) sname rating age (sid) bid day

22 dustin 7 45.0 58 103 11/ 12/ 9631 lubber 8 55.5 58 103 11/ 12/ 96

S RS sid R sid

1 11 1

. .

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Joins

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

Natural Join: Equijoin on all common fields

sid sname rating age bid day

22 dustin 7 45.0 101 10/ 10/ 9658 rusty 10 35.0 103 11/ 12/ 96

S Rsid

1 1

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Division Not supported as a primitive operator, but useful

for expressing queries like: Find sailors who have reserved all boats

Let A have 2 fields, x and y; B have only field y: A/B =

i.e., A/B contains all x tuples (sailors) such that for every y tuple (boat) in B, there is an xy tuple in A

Or: If the set of y values (boats) associated with an x value (sailor) in A contains all y values in B, the x value is in A/B

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Examples of Division A/B

sno pnos1 p1s1 p2s1 p3s1 p4s2 p1s2 p2s3 p2s4 p2s4 p4

pnop2

pnop2p4

pnop1p2p4

snos1s2s3s4

snos1s4

snos1

A

B1B2

B3

A/B1 A/B2 A/B3

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Find names of sailors who’ve reserved boat #103

Solution 1: sname bidserves Sailors(( Re ) )103

Solution 2: ( , Re )Temp servesbid

1103

( , )Temp Temp Sailors2 1

sname Temp( )2

Solution 3: sname bidserves Sailors( (Re ))

103

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Find names of sailors who’ve reserved a red boat

Information about boat color only available in Boats; so need an extra join: sname color red

Boats serves Sailors((' '

) Re )

A more efficient solution:

sname sid bid color redBoats s Sailors( ((

' ') Re ) )

A query optimizer can find this, given the first solution!

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Find names of 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 greenBoats

sname Tempboats serves Sailors( Re )

Can also define Tempboats using union! (How?)

What happens if is replaced by in this query?

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Find names of 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): ( , ((

' ') Re ))Tempred

sid color redBoats serves

sname Tempred Tempgreen Sailors(( ) )

( , ((' '

) Re ))Tempgreensid color green

Boats serves

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Find names of sailors who’ve reserved all boats

Uses division; schemas of the input relations to / must be carefully chosen:

( , (,

Re ) / ( ))Tempsidssid bid

servesbid

Boats

sname Tempsids Sailors( )

To find sailors who’ve reserved all ‘Interlake’ boats:

/ (' '

) bid bname Interlake

Boats.....

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Relational Calculus

Comes in two flavors: Tuple relational calculus (TRC) and Domain relational calculus (DRC)

Calculus has variables, constants, comparison ops, logical connectives and quantifiers TRC: Variables range over (i.e., get bound to) tuples. DRC: Variables range over domain elements (= field

values) Both TRC and DRC are simple subsets of first-order logic

Expressions in the calculus are called formulas An answer tuple is essentially an assignment of

constants to variables that make the formula evaluate to true

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Tuple Relational Calculus Query has the form: {T | p(T)}

p(T) denotes a formula in which tuple variable T appears

Answer is the set of all tuples T for which the formula p(T) evaluates to true.

Formula is recursively defined: start with simple atomic formulas (get tuples

from relations or make comparisons of values) build bigger and better formulas using the

logical connectives

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TRC Formulas An Atomic formula is one of the following:

R Rel R.a op S.bR.a op constant

op is one of A formula can be:

an atomic formula where p and q are

formulas where variable R is a tuple

variable where variable R is a tuple

variable

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Free and Bound Variables

The use of quantifiers and in a formula is said to bind X in the formula A variable that is not bound is free

Let us revisit the definition of a query: {T | p(T)}

• There is an important restriction — the variable T that appears to the left of `|’

must be the only free variable in the formula p(T)

— in other words, all other tuple variables must be bound using a quantifier

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Selection and Projection

Find names and ages of sailors with rating

above 7

{S |S Sailors S.rating > 7}

{S | S1 Sailors(S1.rating > 7 S.sname = S1.sname S.age = S1.age)}

– Modify this query to answer: Find sailors who are older than 18 or have a rating under 9, and are called ‘Bob’

Useful convention: S is a tuple variable of 2 fields (i.e. {S} is a projection of sailors), since only 2 fields are ever mentioned and S is never used to range over any relations in the query

• Find all sailors with rating above 7

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Find sailors rated > 7 who’ve reserved boat #103

Note the use of to find a tuple in Reserves that `joins with’ the Sailors tuple under consideration

{S | SSailors S.rating > 7 R(RReserves R.sid = S.sid R.bid = 103)}

Joins

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Joins (continued)

{S | SSailors S.rating > 7 R(RReserves R.sid = S.sid B(BBoats B.bid = R.bid B.color = ‘red’))}

Find sailors rated > 7 who’ve reserved a red boat

Observe how the parentheses control the scope of each quantifier’s binding. (Similar to SQL!)

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Division (makes more sense here???)

Find all sailors S such that for each tuple B in Boats there is a tuple in Reserves showing that sailor S has reserved it

Find sailors who’ve reserved all boats

(hint, use )

{S | SSailors BBoats (RReserves (S.sid = R.sid B.bid = R.bid))}

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Division – a trickier example…

{S | SSailors B Boats ( B.color = ‘red’ R(RReserves S.sid = R.sid B.bid = R.bid))}

Find sailors who’ve reserved all Red boats

{S | SSailors B Boats ( B.color ‘red’ R(RReserves S.sid = R.sid B.bid = R.bid))}

Alternatively…

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Unsafe Queries, Expressive Power

syntactically correct calculus queries that have an infinite number of answers! Unsafe queries. e.g.,

Solution???? Don’t do that! Expressive Power (Theorem due to Codd):

every query that can be expressed in relational algebra can be expressed as a safe query in DRC / TRC; the converse is also true.

Relational Completeness: Query language (e.g., SQL) can express every query that is expressible in relational algebra (actually, SQL is more powerful, as we will see…)

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Summary

The relational model has rigorously defined query languages that are simple and powerful

Relational algebra is more operational; useful as internal representation for query evaluation plans

Several ways of expressing a given query; a query optimizer should choose the most efficient version

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Summary

Relational calculus is non-operational, and users define queries in terms of what they want, not in terms of how to compute it. (Declarativeness)

Algebra and safe calculus have same expressive power, leading to the notion of relational completeness