CS4432: Database Systems II

Query Operator & Algebraic Expressions

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Why SQL

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SQL is a very-high-level language. Say “what to do” rather than “how to do it.” Avoid a lot of data-manipulation details needed in procedural

languages like C++ or Java.

Database management system figures out “best” way to execute query. Called “query optimization.”

Query Processing

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SELECT pNumber, count(*) AS CNT

FROM Student

WHERE sNumber > 1

GROUP BY pNumber;

SQL Query

Query Plans

Query Example

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SELECT B, D FROM R, S WHERE R.A = “c” and S.E = 2 and R.C=S.C

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• How do we execute query?

- Form Cartesian product of all tables in FROM-clause- Select tuples that match WHERE-clause- Project columns that occur in SELECT-clause

One idea

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R X S R.A R.B R.C S.C S.D S.E

a 1 10 10 x 2

a 1 10 20 y 2

. .

C 2 10 10 x 2 . .

Bingo!

Got one...

SELECT B, D FROM R, S WHERE R.A = “c” and S.E = 2 and R.C=S.C

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But ?

Performance would be unacceptable!

We need a better approach for reasoning about queries, their execution orders and their respective costs

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Formal Relational Query Languages

Relational Algebra: More operational, very useful for representing execution plans.

Operators working on relations

Core Relational Algebra (Recap)

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Union, intersection, and difference. Usual set operations, both operands have the same relation schema.

Selection: picking certain rows.

Projection: picking certain columns.

Products and joins: compositions of relations.

Renaming of relations and attributes.

Grouping and Aggregation: Grouping matching tuples

Duplicate Elimination: eliminates identical copies except one

Sorting: Orders tuples based on a given criteria

Relational Algebra Express Query Plans

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B,D

R.A=“c” S.E=2 R.C=S.C

X R S

Recap on Relational Algebra & Operators

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Algebra Behind the Query Language

Relational Algebra

Set of operators that operate on relations

Operator semantics based on Set or Bag theory

Relational algebra form underlying basis (and optimization rules) for SQL

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SELECT pNumber, count(*) AS CNT

FROM Student

WHERE sNumber > 1

GROUP BY pNumber;

Relational Algebra Basic operators

Set Operations (Union: , Intersection: ∪ ∩ ,difference: – ) Select: σ Project: π Cartesian product: x rename: ρ

More advanced operators, e.g., grouping and joins

The operators take one or two relations as inputs and produce a new relation as an output

One input unary operator, two inputs binary operator

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Union over sets: Consider two relations R and S that are

union-compatible (same schema)

A B

1 2

3 4

R

A B

1 2

3 4

5 6

S

A B

1 2

3 4

5 6

R S

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Binary Op.

Difference over sets: – R – S are the tuples that appear in R and not in S R & S must be union-compatible

Defined as: R – S = {t | t ∈R and t ∈ S}

A B

1 2

3 4

R

A B

1 2

5 6

S

A B

3 4

R – S

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Binary Op.

R-S ≠ S-R

Intersection over sets: ∩

Consider two Relations R and S that are union-compatible

A B

1 2

3 4

R

A B

1 2

3 4

5 6

S

A B

1 2

3 4

R ∩ S

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Binary Op.

Selection: σ

Select: σc (R): c is a condition on R’s attributes Select subset of tuples from R that satisfy

selection condition c

A B C

1 2 5

3 4 6

1 2 7

R σ(C ≥ 6) (R)

A B C

3 4 6

1 2 7

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Unary Op.

Selection: Example

Rσ ((A=B) ^ (D>5)) (R)

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σ (D > C) (R)

Project: π πA1, A2, …, An (R), with A1, A2, …, An attributes AR

returns all tuples in R, but only columns A1, A2, …, An

A1, A2, …, An are called Projection List

A B C

1 2 5

3 4 6

1 2 7

1 2 8

R πA, C (R)

A C

1 5

3 6

1 7

1 8

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Unary Op.

Extended Projection: πL (R)Example

π C, VA, X C*3+B (R)

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A B C

1 2 5

3 4 6

1 2 7

1 2 8

R

C V X

5 1 17

6 3 22

7 1 23

8 1 26

Rename column A to VCompute this expression and call it X

Cross Product (Cartesian Product): X

RS

R X S

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Each tuple in R joined with each tuple is S R x S = {t q | t ∈ R and q ∈ S}

Binary Op.

Natural Join: R ⋈ S

R S

R S ⋈

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Implicit condition (R.B = S.B and R.D = S.D)

Binary Op.

An implicit equality condition on the common columns

Theta Join: R ⋈C S

A join based on any arbitrary condition C

It is defined as :

R ⋈C S = (σC (R X S))

A B

1 2

3 2

R

D C

2 3

4 5

4 5

S

R ⋈ R.A>=S.CS

A B D C

3 2 2 3

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Recommendation: Always use Theta join(more explicit and more clear)

Binary Op.

Duplicate Elimination: (R) Delete all duplicate records

Convert a Bag (allows duplicates) to a Set (does not allow duplicates)

RA B

1 2

3 4

1 2

1 2

(R)

A B

1 2

3 4

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Unary Op.

Grouping & Aggregation operator:

Grouing & Aggregate operation in relational algebra g1,g2, …gm, F1(A1), F2(A2), …Fn(An) (R)

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Unary Op.

Group by these columns (can be empty)

Aggregation functions applied over each group

avg: average value

min: minimum value

max: maximum value

sum: sum of values

count: number of values

Grouping & Aggregation Operator: Example

sum(c)(R)

R S

branch_name,sum(balance)(S)

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Assignment Operator: Write query as a sequence of line consisting of:

Series of assignments

Result expression containing the final answer

May use a variable multiple times in subsequent expressions

Example:

R1 (σ ((A=B) ^ (D>5)) (R – S)) ∩ W

R2 R1 ⋈(R.A = T.C) T

Result R1 U R2

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Banking Example branch (branch_name, branch_city, assets)

customer (customer_name, customer_street, customer_city)

account (account_number, branch_name, balance)

loan (loan_number, branch_name, amount)

depositor (customer_name, account_number)

borrower (customer_name, loan_number)

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Example Queries

Find customer names having account balance below 100 or above 10,000

πcustomer_name (depositor ⋈ πaccount_number(σbalance <100 OR balance > 10,000 (account)))

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Example Queries

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Example Queries (Cont’d)

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Example Queries

Find customers’ names who have neither accounts nor loans

πcustomer_name(customer) - (πcustomer_name(borrower) U πcustomer_name(depositer))

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Example Queries

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For branches that gave loans > 100,000 or hold accounts with balances >50,000, report the branch name along whether it is reported because of a loan or an account

R1 πbranch_name, ‘Loan’ As Type (σamount >100,000 (loan))

R2 πbranch_name, ‘Account’ As Type(σbalance > 50,000 (account)))

Result R1 U R2

Example Queries

Find customer names having loans with sum > 20,000

πcustomer_name (σsum > 20,000 (customer_name, sum sum(amount)(loan borrower⋈ )))

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Example Queries

Find the branch name with the largest number of accounts

R1 branch_name, countAccounts count(account_number)(account)

R2 Max max(countAccounts)(R1)

Result πbranch_name(R1 ⋈countAccounts = Max R2)

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Summary of Relational-Algebra Operators

Set operators Union, Intersection, Difference

Selection & Projection & Extended Projection

Joins Natural, Theta, Outer join

Rename & Assignment

Duplicate elimination

Grouping & Aggregation

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