DATABASE SYSTEMS IADMIN STUFF

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Mid-term exam Tuesday, Oct 19 @ 2:30pm Room 3005 (usual room) Closed book No cheating, blah blah

No class on Oct 21

The amount of time we spent on each topic in class is a good indication of importance & likelihood of exam question

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Assignment 4 solution posted (with algebra)

Assignment 6 posted For procedure question, sample on next page

might help

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DATABASE SYSTEMS IWEEK 6: REVIEW

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REVIEW A transaction has the following properties:

Atomicity: all-or-nothing propertyConsistency: must leave the DB in a

consistent state if DB is consistent when the transaction begins

Isolation: transaction is performed as if only one transaction at a time (serial processing)

Durability: effects of completed transactions are permanent

What is a transaction?

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ENTITY-RELATIONSHIP DIAGRAMS

An Entity-Relationship diagram (ER diagram) is a graph with nodes representing entity sets, attributes and relationship sets. Entity sets denoted by rectangles. Attributes denoted by ovals. Relationship sets denoted by diamonds. Edges (lines) connect entity sets to their attributes and

relationship sets to their entity sets.

lot

dname

budgetdid

sincename

Works_In DepartmentsEmployees

ssn

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RELATIONAL DATABASE Rows are called tuples (or records), columns called

attributes (or fields). Attributes are referenced not by column number,

but by name. Order of attributes does not matter

Attribute types are called domains. Domains consist of atomic values such as integers or strings.

No structured values such as lists or sets The order of tuples does not matter, a relation is a

set of tuples. The order of tuples resulting from a relational query is undefined.

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RELATIONSHIP SETS

ER Diagram

Relational

CREATE TABLE Works_In( ssn CHAR(11), did INTEGER, since DATE, PRIMARY KEY (ssn, did), FOREIGN KEY (ssn) REFERENCES Employees, FOREIGN KEY (did) REFERENCES Departments);

SQL

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FORMAL QUERY LANGUAGES

Two mathematical query languages form the basis for “real” languages (e.g. SQL), and for implementation: Relational Algebra (RA)

More procedural, very useful for representing execution plans, relatively close to SQL.

Composed of a collection of operators A step-by-step procedure for computing the answer

Relational Calculus (RC) Lets users describe what they want, rather than how to

compute it. (Non-procedural, declarative.) Describes the answer, not the steps.

Understanding these formal query languages is important for understanding SQL and query processing.

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PRELIMINARIES A query is applied to relation instances, and

the result of a query is also a relation instance. Inputs and Outputs of Queries are relations Query evaluated on instances of input relations

Different instance (DB?) as input = different answer Schemas of input relations for a query are fixed

(but query will run regardless of instance!) The schema for the result of a given query is also

fixed! Determined by definition of input relations and query language constructs.

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RELATIONAL ALGEBRA OPERATIONS Basic operations

Selection ( ) Selects a subset of rows from

relation. Projection ( )

Deletes unwanted columns from relation.

Cartesian product ( ) Combine two relations.

Set-difference ( ) Tuples in relation 1, but not in

relation 2. Union ( )

Tuples in relation 1 or in relation 2.

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Renames relations / attributes, without changing the relation instance.

relation R is renamed to S, attributes are renamed A1, . . ., An

Rename only some attributes

using the positional notation to reference attributes

No renaming of attributes, just the relation

RENAMING

)(),...,2,1( RAnAAS

)(),...,11( RAkkAS

)(RS

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PROJECTION Similar in concept to VIEWs

sname rating

yuppy 9lubber 8guppy 5rusty 10

sname rating

S,

( )2

age

35.055.5

age S( )2

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

S2

Other fields are projected out

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SELECTION

rating

S8

2( )sid sname rating age28 yuppy 9 35.058 rusty 10 35.0

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

S2

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UNION

Concatenates S1 and S2 Result contains ALL tuples that occur in either S1 or S2

Schemas must be identical If they have the same number of fields Fields have same domains

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

S S1 2

sid sname rating age

22 dustin 7 45.0

31 lubber 8 55.558 rusty 10 35.0

S1sid sname rating age28 yuppy 9 35.031 lubber 8 55.544 guppy 5 35.058 rusty 10 35.0

S2

SELECT * FROM S1

UNION

SELECT * FROM S2

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INTERSECTION

Result contains ALL tuples that occur in both S1 or S2 Schemas must be identical

21 SS

sid sname rating age

22 dustin 7 45.0

31 lubber 8 55.558 rusty 10 35.0

S1sid sname rating age28 yuppy 9 35.031 lubber 8 55.544 guppy 5 35.058 rusty 10 35.0

S2

SELECT * FROM S1

INTERSECT

SELECT * FROM S2

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SET-DIFFERENCE

Result contains ALL tuples that occur in S1 but not in S2 Schemas must be identical

21 SS

sid sname rating age

22 dustin 7 45.0

31 lubber 8 55.558 rusty 10 35.0

S1sid sname rating age28 yuppy 9 35.031 lubber 8 55.544 guppy 5 35.058 rusty 10 35.0

S2

SELECT * FROM S1

MINUS

SELECT * FROM S2

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CARTESIAN PRODUCT Field names in conflict become unnamed

(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

sid sname rating age

22 dustin 7 45.0

31 lubber 8 55.558 rusty 10 35.0

sid bid day

22 101 10/10/96 58 103 11/12/96

R1

S1

21 SS

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JOIN Condition Join

Each tuple of the one relation is paired with each tuple of the other relation if the two tuples satisfy the join condition.

Condition c refers to attributes of both R and S.

)( SRcScR

(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

11:Example.1.1RS

sidRsidS

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Find names of sailors who’ve reservedboat #103.

Solution 1:

Solution 2:

Solution 3:

Which is most efficient? Why?

))Re((103

Sailorsservesbidsname

)Re(103

1 servesbid

Temp

)1(2 SailorsTempTemp

sname bidserves Sailors( (Re ))

103

EXAMPLE QUERIES

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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:

A more efficient solution:

A query optimizer can find the second solution given the first one.

sname color redBoats serves Sailors((

' ') Re )

sname sid bid color redBoats s Sailors( ((

' ') Re ) )

EXAMPLE QUERIES

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How to find number of sailors?

EXAMPLE QUERIES

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How to find oldest sailor?

EXAMPLE QUERIES