Gitte Christensen Dyalog Ltd Managing External Data 3

Gitte ChristensenDyalog Ltd

Managing External Data 3

Relational Algebra

Relational Algebra is :•the formal description of how a relational database operates•the mathematics which underpin SQL operations.

Operators in relational algebra are not necessarily the same as SQL operators, even if they have the same name.

Terminology

• Relation - a set of tuples.• Tuple - a collection of attributes which

describe some real world entity.• Attribute - a real world role played by a

named domain.• Domain - a set of atomic values.• Set - a mathematical definition for a

collection of objects which contains no duplicates.

Operators - Write

• INSERT - provides a list of attribute values for a new tuple in a relation. This operator is the same as SQL.

• DELETE - provides a condition on the attributes of a relation to determine which tuple(s) to remove from the relation. This operator is the same as SQL.

• MODIFY - changes the values of one or more attributes in one or more tuples of a relation, as identified by a condition operating on the attributes of the relation. This is equivalent to SQL UPDATE.

Operators - Retrieval

There are two groups of operations:• Mathematical set theory based relations:

UNION, INTERSECTION, DIFFERENCE, and CARTESIAN PRODUCT.

• Special database operations: SELECT (not the same as SQL SELECT), PROJECT, and JOIN.

Relational SELECT

SELECT is used to obtain a subset of the tuples of a relation that satisfy a select condition.For example, find all employees born after 1st Jan 1950:

SELECT dob > ’01/JAN/1950’ (employee)

Relational PROJECTThe PROJECT operation is used to select a subset of the attributes of a relation by specifying the names of the required attributes.For example, to get a list of all employees surnames and employee numbers:

PROJECT surname,empno (employee)

SELECT and PROJECT

FROM employeeWHERE depno = 1;

SELECT empno

PROJECT empno (SELECT depno = 1 (employee))

Mapping this back to SQL gives:

SELECT and PROJECT can be combined together. For example, to get a list of employee numbers for employees in department number 1:

Set Operations - semanticsConsider two relations R and S.• UNION of R and S

the union of two relations is a relation that includes all the tuples that are either in R or in S or in both R and S. Duplicate tuples are eliminated.

• INTERSECTION of R and Sthe intersection of R and S is a relation that includes all tuples that are both in R and S.

• DIFFERENCE of R and Sthe difference of R and S is the relation that contains all the tuples that are in R but that are not in S.

SET Operations - requirementsFor set operations to function correctly the relations R and S must be union compatible. Two relations are union compatible if

– they have the same number of attributes– the domain of each attribute in column

order is the same in both R and S.

UNION Example

INTERSECTION Example

DIFFERENCE Example

CARTESIAN PRODUCT

The Cartesian Product is also an operator which works on two sets. It is sometimes called the CROSS PRODUCT or CROSS JOIN.

It combines the tuples of one relation with all the tuples of the other relation.

CARTESIAN PRODUCT Example

JOIN OperatorJOIN is used to combine related tuples from two

relations:• In its simplest form the JOIN operator is just the cross

product of the two relations.• As the join becomes more complex, tuples are

removed within the cross product to make the result of the join more meaningful.

• JOIN allows you to evaluate a join condition between the attributes of the relations on which the join is undertaken.

The notation used is R JOIN join condition S

JOIN Example

Natural Join

Invariably the JOIN involves an equality test, and thus is often described as an equi-join. Such joins result in two attributes in the resulting relation having exactly the same value. A ‘natural join’ will remove the duplicate attribute(s).– In most systems a natural join will require that the

attributes have the same name to identify the attribute(s) to be used in the join. This may require a renaming mechanism.

– If you do use natural joins make sure that the relations do not have two attributes with the same name by accident.

OUTER JOINsNotice that much of the data is lost when applying a join to two relations. In some cases this lost data might hold useful information. An outer join retains the information that would have been lost from the tables, replacing missing data with nulls.

There are three forms of the outer join, depending on which data is to be kept.•LEFT OUTER JOIN - keep data from the left-hand table•RIGHT OUTER JOIN - keep data from the right-hand table•FULL OUTER JOIN - keep data from both tables

OUTER JOIN Example 1

OUTER JOIN Example 2

SQL query optimisation

• Optimisation Concept• Implementation of Rel. Algebra

Operations• Oracle Query Execution Plans.• Btree Indexing• Rtree Indexing

SQL query optimisation

SQL query

Parse and TranslateRelational algebra expression

Optimisation

using data statistics

Execution planQuery

result Evaluate

against Database

Optimisation steps• Parse

– check SQL syntax and columns & tables valid

• Translate– SQL into relational algebra expression

• Select most efficient query execution plan – minimisation of the input/output and cpu

requirements

• Evaluate expression– Call required code modules

Example SQL query

S(sno,sname,status,scity) SP(sno,pno,qty) P(pno,pname,colour,weight,pcity)

Get supplier name for suppliers who supply red parts.

SQL select sname

from S,P,SP

where S.sno = SP.sno

and P.pno = Sp.pno and colour = ‘red’;

Example SQL query

• How would you do this?

Query as relational algebra expression and graph(((((( P restrict[ colour = red] ) project[ pno ])

join SP [pno = pno]) project [ sno ]) join S [ sno = sno ]) project [ sname ])

P SP S

project

restrict

project

join

project

join

Relational algebra transformations• The following transformations can be

applied without regard to the actual data values stored in the tables referenced in the SQL query.

• They are being stated to justify some of the common manipulations of the relational algebra during optimisation.

Distributive transformationDistributive Law

f(A o B) = f(A) o f(B)

e.g. restrict(A join B) = restrict (A) join restrict (B)

Restrict distributes over union, intersection, difference and join.

For join the restriction must, at it’s most complex, consist of the AND of two separate conditions one for each operand.

Project distributes over union, intersection, join.

For join the join attributes must be included in the projection.

Thus restricts / projects can often be done before joins.

Commutative transformation

Commutative Law

The operator o is commutative if :-

A o B = B o A for all values of A and B

Union, intersection and join are commutative.

E.g S join SP = SP join S

Thus optimiser can choose best order typically the smallest table is used to drive the join processing.

Associative transformation

Associative Law

Operator o is associative if :-

A o ( B o C ) = ( A o B ) o C for all A,B,C values

e.g. A join ( B join C ) = ( A join B ) join C

Union, intersection and join are associative.

Thus for a 3 table join the optimiser can choose any sequence of pairs of joins.

Semantic transformations

Referential Integrity Based.

select pno from SP,S where SP.sno = S.sno;

If declared S.sno as PK and SP.sno as FK then optimise to:-

select pno from SP;

Application of Negator Functions

where not (status > 20)

this cannot use an index scan so must use table scan but

if transform to status <= 20 - now can use index scan

Assumes that an index on the status column already exists

Will the results be the same?

select sname from s,sp where s,sno = sp.sno and sp.pno = ‘p2’;

select sname from sp,s where sp.pno = ‘p2’ and sp.sno = s.sno;

select sname from s where s.sno in ( select sp.sno from sp

where sp.pno = ‘p2’);

Explain. What is the implication for performance of the three?

Will the results be the same?

Yes - all produce the same result rows• join s to sp can be transformed into join sp to

s as join operation is commutative• restrict can be distributed past the join thus

first two order of operations can be ignored• set membership can be converted to a join• optimisation will convert all 3 queries the

same query so all must have the same performance - user can chose which to use

Break

SQL query optimisation

Algorithm selection• The relational algebra operations typically

have multiple code implementations. It is necessary for the optimiser to select the most appropriate for the current circumstances.

• For example the presence of a direct access path to the data required will be preferred to a complete table scan if only 10 -20% of the table is to be accessed.

Restriction implementationsFull table scan or table scan with early termination

All blocks must be searched unless an early termination condition is found e.g row found and no duplicates or data is ordered and already passed the required data location.

Btree Non Clustered Index Search

Tree search to starting key then linear scan of sorted keys.

Assume to access each record needs a disk block read.

Multiple restrictionsConjunction And’d together

If only some columns are indexed then use most selective index to locate the data record and check out the other restriction conditions in the buffer.

If all columns are indexed then can get rowid’s of all records meeting the restriction conditions and perform an intersection. Use the result set of rowids to locate the data records.

Disjunction Or’d together

If all columns indexed get all rowid’s and perform union. Use the result set of rowids to locate the data records.Else must scan table.

Join implementations

• All give the same final result set of rows as defined by the join operation.

• Algorithm and cost differ.• Nested loop algorithm will function with any

join condition I.e <, <=, =, >=, >• Hash algorithm only possible with =• Relation S 500 rows 50 rows per disk block• Relation R 100000 rows 50 rows per disk

block

Join - Block Nested Loop Inner file pass 1S Outer File Pass.Block Buffers Pass.Block R Inner File

1.1

1.2

1.1

2.1

1.2

2.2

1.3

2.3Join col values equal

Join - Block Nested LoopInner file pass 2S Outer File Pass.Block Buffers Pass.Block R Inner File

1.1

1.2

1.1

2.1

1.2

2.2

1.3

2.3Join col values equal

Join - Block Nested Loop - CostCost

No. Outer Loop Blocks * No. Inner Loop blocks + No. Outer Loop Blocks

Use smallest table as outer loopi.e. (500/50 * 100000/50 ) + 500/50 = 20010

Reduce cost of inner loop by having :-index built on inner loop join column which is also a Primary Key - lookup avoiding full file scan – assume needs 1 disk read per row in outer file

(500 / 50 + 500 ) = 510

Equi Join - Hash example

Relation R 91 1 40 900

Disk Hash Buckets R [ 0] [1]

Disk Hash Buckets S [0] [1]

Relation S 40 101 90 1 1000 91

Value allocated to [0] if even else allocated to [1] if odd

Equi Join - Hash example

Relation R 91 1 40 900

Disk Hash Buckets R [ 0] 40 900 [1] 1 91

Disk Hash Buckets S [0] 40 90 1000 [1] 101 1 91

Relation S 40 101 90 1 1000 91

Value allocated to [0] if even else allocated to [1] if odd

Equi Join - Hash cost• Only need to compare record values from

matching hash buckets to find which records need joining. Assumes mechanism to retrieve required data rows.

• CostRead files in for hashing; Write out to allocated hash bucket; Read in each pair of buckets for the search for matching join values.

3 * ( No Blocks of R + No of blocks of S)3 * ( 2000 + 10 ) = 6030

Join before restrict Example

S(sno,sname,status,scity) 100 rows 20 rows/disk block

SP(sno,pno,qty) 10000 rows 100/disk block 50 with pno = p2

select * from S,SP where pno = ‘p2’ and S.sno = SP.sno;

Approach 1 Cost in Block I/O

restrict write out after restrict 1

join nested loop 5 + (100*5) 505

S SP Total 506

Restrict before join Example

Approach 2 Cost in Block I/O

write out after join 1

join nested loop in RAM

full scan of S 5

restrict full scan of SP 100

S SP Total 106

Three table join query

branch(b_name, b_city, manager, b_address) 100 rows

account(b_name, a_number, balance) 10 000 rows

account_holder(h_name,a_number,h_address) 50 000 rows

select h_name, balance

from account a, account_holder h, branch b

where a.a_number = h.a_number and a.b_name = b.b_name

and b.city = ‘Leicester’ and balance > 10000;

Optimised query plan project(h_name,balance)

join (a_number)

project(a_number,balance)

join (b_name)

project (b_name) project(h_name,a_number)

restrict (b_city = Leicester)

restrict(balance>10000)

branch account account_holder

Three table join query plan justification

Three table join query plan justification• Projects and restricts distributed past

joins• join branch and account as pair of

smallest tables and therefore assume that this will provide the smallest result set.

Optimiser estimation methods• Hard coded values in the optimiser

algorithm– e.g restriction 1% rows on =, 10% on >=

• No consideration of actual data values

• Statistics based on number of different distinct values in column– assumes uniform distribution

• Statistics based on histogram of count of unique values.

Importance of estimates• The use of an index is only efficient if less

than 20% of the data is to be returned. – Each row access may require a separate disk

read. Thus same disk block may be retrieved more than once if buffer space limited. More efficient to execute a full table scan.

• Cost of statistics collection needs to be traded off against improved performance of queries based on more accurate estimates.

Database Implementations

Relational Databases

• The Conceptual model has become the Physical model

• Row based storage• No ordering of data• Row based searches• Very many comparisons is required

to find and manipulate data

Relational Databases/APL• Current APL/K/J databases

– all relational– all Column stores

• kdb• Vstar• flipdb• TakeCare

Relational Databases/APL• The relational model holds• Row stores are implemented with

great engineering skills• Use them if you can

Relational Databases/APL• but

– The physical laws still apply– Some problems might need different

solutions

Semistructured data

Root

cf mh sw

Carrie Fisher

address address

name

city

street

MapleHollywood

LocustMalibu

city

star

starmovie

title year

Star Wars1977

Mark HamillOak

Brentwood

name

street city

street

starInstarOf

starIn

starOf

Multidimensional data