Relational Algebra

Lecture Notes: Relational Algebra

What? Why?

• Similar to normal algebra (as in 2+3*x-y), except we use relations as values instead of numbers, and the operations and operators are different.
• Not used as a query language in actual DBMSs. (SQL instead.)
• The inner, lower-level operations of a relational DBMS are, or are similar to, relational algebra operations. We need to know about relational algebra to understand query execution and optimization in a relational DBMS.
• Some advanced SQL queries requires explicit relational algebra operations, most commonly outer join.
• Relations are seen as sets of tuples, which means that no duplicates are allowed. SQL behaves differently in some cases. Remember the SQL keyword distinct.
• SQL is declarative, which means that you tell the DBMS what you want, but not how it is to be calculated. A C++ or Java program is procedural, which means that you have to state, step by step, exactly how the result should be calculated. Relational algebra is (more) procedural than SQL. (Actually, relational algebra is mathematical expressions.)

Set operations

Relations in relational algebra are seen as sets of tuples, so we can use basic set operations.

Review of concepts and operations from set theory

• set
• element
• no duplicate elements (but: multiset = bag)
• no order among the elements (but: ordered set)
• subset
• proper subset (with fewer elements)
• superset
• union
• intersection
• set difference
• cartesian product

Projection

Example: The table E (for EMPLOYEE)

nr	name	salary
1	John	100
5	Sarah	300
7	Tom	100

SQL	Result	Relational algebra
select salary from E	salary 100 300	PROJECTsalary(E)
select nr, salary from E	nr salary 1 100 5 300 7 100	PROJECTnr, salary(E)

Note that there are no duplicate rows in the result.

Selection

The same table E (for EMPLOYEE) as above.

SQL	Result	Relational algebra
select * from E where salary <>	nr name salary 1 John 100 7 Tom 100	SELECTsalary <>(E)
select * from E where salary <>= 7	nr name salary 7 Tom 100	SELECTsalary <>= 7(E)

Note that the select operation in relational algebra has nothing to do with the SQL keyword select. Selection in relational algebra returns those tuples in a relation that fulfil a condition, while the SQL keyword select means "here comes an SQL statement".

Relational algebra expressions

SQL	Result	Relational algebra
select name, salary from E where salary <>	name salary John 100 Tom 100	PROJECTname, salary (SELECTsalary <>(E)) or, step by step, using an intermediate result Temp <- SELECTsalary <>(E) Result <- PROJECTname, salary(Temp)

Notation

The operations have their own symbols. The symbols are hard to write in HTML that works with all browsers, so I'm writing PROJECT etc here. The real symbols:

Operation My HTML Symbol Projection PROJECT Selection SELECT Renaming RENAME Union UNION Intersection INTERSECTION Assignment <-

Operation My HTML Symbol Cartesian product X Join JOIN Left outer join LEFT OUTER JOIN Right outer join RIGHT OUTER JOIN Full outer join FULL OUTER JOIN Semijoin SEMIJOIN

Example: The relational algebra expression which I would here write as

PROJECT_Namn ( SELECT_{Medlemsnummer <> ( Medlem ) )}

_{should actually be written}

_{Cartesian product}

_{The cartesian product of two tables combines each row in one table with each row in the other table.}

_{Example: The table E (for EMPLOYEE)}

enr	ename	dept
1	Bill	A
2	Sarah	C
3	John	A

_{Example: The table D (for DEPARTMENT)}

dnr	dname
A	Marketing
B	Sales
C	Legal

SQL	Result	Relational algebra
select * from E, D	enr ename dept dnr dname 1 Bill A A Marketing 1 Bill A B Sales 1 Bill A C Legal 2 Sarah C A Marketing 2 Sarah C B Sales 2 Sarah C C Legal 3 John A A Marketing 3 John A B Sales 3 John A C Legal	E X D

• _{Seldom useful in practice.}
• _{Usually an error.}
• _{Can give a huge result.}

_{Join (sometimes called "inner join")}

_{The cartesian product example above combined each employee with each department. If we only keep those lines where the dept attribute for the employee is equal to the dnr (the department number) of the department, we get a nice list of the employees, and the department that each employee works for:}

SQL	Result	Relational algebra
select * from E, D where dept = dnr	enr ename dept dnr dname 1 Bill A A Marketing 2 Sarah C C Legal 3 John A A Marketing	SELECTdept = dnr (E X D) or, using the equivalent join operation E JOINdept = dnr D

• _{A very common and useful operation.}
• _{Equivalent to a cartesian product followed by a select.}
• _{Inside a relational DBMS, it is usually much more efficient to calculate a join directly, instead of calculating a cartesian product and then throwing away most of the lines.}
• _{Note that the same SQL query can be translated to several different relational algebra expressions, which all give the same result.}
• _{If we assume that these relational algebra expressions are executed, inside a relational DBMS which uses relational algebra operations as its lower-level internal operations, different relational algebra expressions can take very different time (and memory) to execute.}

_{Natural join}

_{A normal inner join, but using the join condition that columns with the same names should be equal. Duplicate columns are removed.}

_{Renaming tables and columns}

_{Example: The table E (for EMPLOYEE)}

nr	name	dept
1	Bill	A
2	Sarah	C
3	John	A

_{Example: The table D (for DEPARTMENT)}

nr	name
A	Marketing
B	Sales
C	Legal

_{We want to join these tables, but:}

• _{Several columns in the result will have the same name (nr and name).}
• _{How do we express the join condition, when there are two columns called nr?}

_Solutions:

• _{Rename the attributes, using the rename operator.}
• _{Keep the names, and prefix them with the table name, as is done in SQL. (This is somewhat unorthodox.)}

SQL	Result	Relational algebra
select * from E as E(enr, ename, dept), D as D(dnr, dname) where dept = dnr	enr ename dept dnr dname 1 Bill A A Marketing 2 Sarah C C Legal 3 John A A Marketing	(RENAME(enr, ename, dept)(E)) JOINdept = dnr (RENAME(dnr, dname)(D))
select * from E, D where dept = D.nr	nr name dept nr name 1 Bill A A Marketing 2 Sarah C C Legal 3 John A A Marketing	E JOINdept = D.nr D

_{You can use another variant of the renaming operator to change the name of a table, for example to change the name of E to R. This is necessary when joining a table with itself (see below).}

_RENAMER(E)

_{A third variant lets you rename both the table and the columns:}

_{RENAMER(enr, ename, dept)(E)}

_{Aggregate functions}

_{Example: The table E (for EMPLOYEE)}

nr	name	salary	dept
1	John	100	A
5	Sarah	300	C
7	Tom	100	A
12	Anne	null	C

SQL	Result	Relational algebra
select sum(salary) from E	sum 500	Fsum(salary)(E)

_Note:

• _{Duplicates are not eliminated.}
• _{Null values are ignored.}

SQL	Result	Relational algebra
select count(salary) from E	Result: count 3	Fcount(salary)(E)
select count(distinct salary) from E	Result: count 2	Fcount(salary)(PROJECTsalary(E))

_{You can calculate aggregates "grouped by" something:}

SQL	Result	Relational algebra
select sum(salary) from E group by dept	dept sum A 200 C 300	deptFsum(salary)(E)

_{Several aggregates simultaneously:}

SQL	Result	Relational algebra
select sum(salary), count(*) from E group by dept	dept sum count A 200 2 C 300 1	deptFsum(salary), count(*)(E)

_{Standard aggregate functions: sum, count, avg, min, max}

_Hierarchies

_{Example: The table E (for EMPLOYEE)}

nr	name	mgr
1	Gretchen	null
2	Bob	1
5	Anne	2
6	John	2
3	Hulda	1
4	Hjalmar	1
7	Usama	4

_{Going up in the hierarchy one level: What's the name of John's boss?}

SQL	Result	Relational algebra
select b.name from E p, E b where p.mgr = b.nr and p.name = "John"	name Bob	PROJECTbname ([SELECTpname = "John"(RENAMEP(pnr, pname, pmgr)(E))] JOINpmgr = bnr [RENAMEB(bnr, bname, bmgr)(E)]) or, in a less wide-spread notation PROJECTb.name ([SELECTname = "John"(RENAMEP(E))] JOINp.mgr = b.nr [RENAMEB(E)]) or, step by step P <- RENAMEP(pnr, pname, pmgr)(E) B <- RENAMEB(bnr, bname, bmgr)(E) J <- SELECTname = "John"(P) C <- J JOINpmgr = bnr B R <- PROJECTbname(C)

_{Notes about renaming:}

• _{We are joining E with itself, both in the SQL query and in the relational algebra expression: it's like joining two tables with the same name and the same attribute names.}
• _{Therefore, some renaming is required.}
• _{RENAMEP(E) JOIN... RENAMEB(E) is a start, but then we still have the same attribute names.}

_{Going up in the hierarchy two levels: What's the name of John's boss' boss?}

SQL	Result	Relational algebra
select ob.name from E p, E b, E ob where b.mgr = ob.nr where p.mgr = b.nr and p.name = "John"	name Gretchen	PROJECTob.name (([SELECTname = "John"(RENAMEP(E))] JOINp.mgr = b.nr [RENAMEB(E)]) JOINb.mgr = ob.nr [RENAMEOB(E)]) or, step by step P <- RENAMEP(pnr, pname, pmgr)(E) B <- RENAMEB(bnr, bname, bmgr)(E) OB <- RENAMEOB(obnr, obname, obmgr)(E) J <- SELECTname = "John"(P) C1 <- J JOINpmgr = bnr B C2 <- C1 JOINbmgr = bbnr OB R <- PROJECTobname(C2)

_{Recursive closure}

_{Both one and two levels up: What's the name of John's boss, and of John's boss' boss?}

SQL	Result	Relational algebra
(select b.name ...) union (select ob.name ...)	name Bob Gretchen	(...) UNION (...)

_{Recursively: What's the name of all John's bosses? (One, two, three, four or more levels.)}

• _{Not possible in (conventional) relational algebra, but a special operation called transitive closure has been proposed.}
• _{Not possible in (standard) SQL (SQL2), but in SQL3, and using SQL + a host language with loops or recursion.}

_{Outer join}

_{Example: The table E (for EMPLOYEE)}

enr	ename	dept
1	Bill	A
2	Sarah	C
3	John	A

_{Example: The table D (for DEPARTMENT)}

dnr	dname
A	Marketing
B	Sales
C	Legal

_{List each employee together with the department he or she works at:}

SQL	Result	Relational algebra
select * from E, D where edept = dnr or, using an explicit join select * from (E join D on edept = dnr)	enr ename dept dnr dname 1 Bill A A Marketing 2 Sarah C C Legal 3 John A A Marketing	E JOINedept = dnr D

_{No employee works at department B, Sales, so it is not present in the result. This is probably not a problem in this case. But what if we want to know the number of employees at each department?}

SQL	Result	Relational algebra
select dnr, dname, count(*) from E, D where edept = dnr group by dnr, dname or, using an explicit join select dnr, dname, count(*) from (E join D on edept = dnr) group by dnr, dname	dnr dname count A Marketing 2 C Legal 1	dnr, dnameFcount(*)(E JOINedept = dnr D)

_{No employee works at department B, Sales, so it is not present in the result. It disappeared already in the join, so the aggregate function never sees it. But what if we want it in the result, with the right number of employees (zero)?}

_{Use a right outer join, which keeps all the rows from the right table. If a row can't be connected to any of the rows from the left table according to the join condition, null values are used:}

SQL	Result	Relational algebra
select * from (E right outer join D on edept = dnr)	enr ename dept dnr dname 1 Bill A A Marketing 2 Sarah C C Legal 3 John A A Marketing null null null B Sales	E RIGHT OUTER JOINedept = dnr D
select dnr, dname, count(*) from (E right outer join D on edept = dnr) group by dnr, dname	dnr dname count A Marketing 2 B Sales 1 C Legal 1	dnr, dnameFcount(*)(E RIGHT OUTER JOINedept = dnr D)
select dnr, dname, count(enr) from (E right outer join D on edept = dnr) group by dnr, dname	dnr dname count A Marketing 2 B Sales 0 C Legal 1	dnr, dnameFcount(enr)(E RIGHT OUTER JOINedept = dnr D)

_{Join types:}

• _{JOIN = "normal" join = inner join}
• _{LEFT OUTER JOIN = left outer join}
• _{RIGHT OUTER JOIN = right outer join}
• _{FULL OUTER JOIN = full outer join}

_{Outer union}

_{Outer union can be used to calculate the union of two relations that are partially union compatible. Not very common.}

_{Example: The table R}

A	B
1	2
3	4

_{Example: The table S}

B	C
4	5
6	7

_{The result of an outer union between R and S:}

A	B	C
1	2	null
3	4	5
null	6	7

_Division

_{Who works on (at least) all the projects that Bob works on?}

_Semijoin

_{A join where the result only contains the columns from one of the joined tables. Useful in distributed databases, so we don't have to send as much data over the network.}

_Update

_{To update a named relation, just give the variable a new value. To add all the rows in relation N to the relation R:}

_{R <- R UNION N}

_{Taken from:}

http://www.databasteknik.se/webbkursen/relalg-lecture/index.html