Title: Relational Algebra and Relational Calculus
1Lecture 13 (10/03/2005)
- Relational Algebra and Relational Calculus
2Aggregation over Nulls and Duplicates
- Aggregate functions ignore nulls
- If all aggregated values are null? return null
- SUM, AVERAGE, MIN and MAX include duplicates
- COUNT is not clear
- COUNT ? includes duplicates
- COUNT_DISTINCT ? excludes duplicates
- Grouping over null attributes
- ? a new group
3Binary Relational Operations
- Theta-JOIN Operation
- R ltjoin conditiongtS where R(A1, A2, , An) and
S(B1, B2, , Bm) - The result of a join operation is a relation
Q(A1, A2, , An, B1, B2, , Bm) and in this order
- with nm attributes
- Contains one tuple for each combination of tuples
(one from R and one from S) whenever the
combination satisfies the ltjoin conditiongt - Theta is any operator
- Difference from Cartesian product?
- For every employee, find any other employee that
earns more than s/he does.
4Binary Relational Operations
- EQUIJOIN Operation
- The most common use of join involves join
conditions with equality comparisons only - In the result of an EQUIJOIN we always have one
or more pairs of attributes (whose names need not
be identical) that have identical values in
every tuple - Theta is the operator
- NATURAL JOIN Operation (additional attribute)
- Because one of each pair of attributes with
identical values is extra, a new operation called
natural joindenoted by was created to get rid
of the second (superfluous) attribute in an
EQUIJOIN condition - The standard definition of natural join requires
that the two join attributes, or each pair of
corresponding join attributes, have theĀ same name
in both relations - If this is not the case, a renaming operation
isĀ applied first
5Additional Relational Operations
- The OUTER JOIN Operation
- In regular JOIN tuples without a matching (or
related) tuple are eliminated - Inner join
- Tuples with null in the join attributes are also
eliminated - This amounts to loss of information
- Outer joins, can be used when we want to keep all
the tuples in R, or all those in S, or all those
in both relations in the result of the join - Regardless of whether or not they have matching
tuples
6Additional Relational Operations
- The left outer join operation
- Keeps every tuple in the first or left relation R
in R S if no matching tuple is found in
S, then the attributes of S in the join result
are filled or padded with null values - A similar operation, right outer join, keeps
every tuple in the second or right relation S in
the result of R S - A third operation, full outer join, denoted by
keeps all tuples in both the left and the
right relations when no matching tuples are
found, padding them with null values as needed
7Additional Relational Operations
- For every employee, find any other employee that
earns more than s/he does. - Display all EMPLOYEEs including managers (with
their managed DEPARTMENTs) - Next slide
- Display all DEPARTMENTs names with their
controlled PROJECTs (display names) including
projects with no controlling DEPARTMENTS - Assume new PROJECTs are not assigned DEPARTMENTs
initially - Display all DEPARTMENTs names with their
controlled PROJECTs (display names) including
projects with no controlling DEPARTMENTS - Assume new PROJECTs are not assigned DEPARTMENTs
initially - Assume Accounting Dept has no controlled projects
8(No Transcript)
9Additional Relational Operations
- Recursive Closure Operations
- Another type of operation that, in general,
cannot be specified in the basic relational
algebra is recursive closure - This operation is applied to a recursive
relationship - Retrieve all SUPERVISEES of an EMPLOYEE e at all
levels - That is, all employees e directly supervised by
e all employees e directly supervised by each
employee e all employees e directly
supervised by each employee e and so on - Although it is possible to retrieve employees at
each level and then take their union, we cannot,
in general, specify a query such as retrieve the
supervisees of James Borg at all levels
without utilizing a looping mechanism - The SQL3 standard includes syntax for recursive
closure
10SSN
SUPERSSN
11Reading Assignment
- Read about relational calculus
- Tuple and Domain
- Assignment 2
- Tuesday 10/11
- Exam 1
- Wednesday 10/19
12Sailor Schema
- SAILOR (Sid, Sname, Rating, Age)
- BOAT (Bid, Bname, Color)
- RESERVES (Sid, Bid, Date)
- 1- Find names of sailors whove reserved boat
103 - 2- Find names of sailors whove reserved a red
boat - 3- Find sailors whove reserved a red or a green
boat
13Sailor Schema
- SAILOR (Sid, Sname, Rating, Age)
- BOAT (Bid, Bname, Color)
- RESERVES (Sid, Bid, Date)
- 4- Find sailors whove reserved a red and a green
boat - 5- Find sailors whove made at least two
reservations - 6- Find sailors whove reserved at least two
different boats - 7- Find the names of sailors whove reserved all
boats - 8- Find the names of sailors whove reserved all
Interlake boats - 9- Find the names of sailors over 20 who have not
reserved a red Boat
14Find names of sailors whove reserved boat 103
- Reserved103 ? ?Sid(? Bid103 (Reserves))
- Result ? Reserved103 SidSid Sailor
- Or, Result ? Reserved103 Sailor (Natural Join)
15Find names of sailors whove reserved a red boat
- Information about boat color only available in
Boats so need an extra join - RedBoats ? ?Sid(? ColorRed (Boats))
- ReservedRed ? ?Sid(Reserves BidBid RedBoats)
- Result ? ?name(ReservedRed SidSid Sailor)
16Find sailors whove reserved a red or a green boat
- Can identify all red or green boats, then find
sailors whove reserved one of these boats - RGBoats ? ?Sid(? ColorRed or Green (Boats))
- ReservedRG ? ?Sid(Reserves BidBid RGBoats)
- Result ? ?name(ReservedRG SidSid Sailor)
- Using Union?
17Find sailors whove reserved a red and a green
boat
- Must identify sailors whove reserved red boats,
sailors whove reserved green boats, then find
the intersection - RedBoats ? ?Sid(? ColorRed (Boats))
- ReservedRed ? ?Sid(Reserves BidBid RedBoats)
- GreenBoats ? ?Sid(? ColorGreen (Boats))
- ReservedGreen ? ?Sid(Reserves BidBid
GreenBoats) - ReservedRG ? ReservedRed ? ReservedGreen)
- Result ? ?name(ReservedRG SidSid Sailor)
18Find sailors whove made at least two reservations
- Must identify sailors whove reserved red boats,
sailors whove reserved green boats, then find
the intersection - CountReservations ? SIDFCount(BID) (Reserves)
- TwoRes ? ?Sid(? Count_BIDgt2 (CountReservations))
- Result ? ?name (TwoRes SidSid Sailor)
19Find sailors whove reserved at least two
different boats
- Res1(Sid1,Bid1) ? ?Sid, Bid(Reserves)
- Res2(Sid2,Bid2) ? ?Sid, Bid(Reserves)
- Reserved2 ? Res1 Sid1Sid2 Res2)
- SailorsReserved2 ? ?Sid1(? Bid1ltgt Bid1
(Reserved2)) - Result ? ?name(SailorsReserved2 Sid1Sid
Sailor)
20Find the names of sailors whove reserved all
boats
- Reserved ? ?Sid, Bid(Reserves)
- AllBoats ??Bid(Boat)
- SailorsReservedAll ? (Resreved AllBoats)
- Result ? ?name(SailorsReserved2All SidSid
Sailor)
21Find the names of sailors whove reserved all
Interlake boats
- Reserved ? ?Sid, Bid(Reserves)
- AllBoats ??Bid(? nameInterlake Boat)
- SailorsReservedAll ? (Resreved AllBoats)
- Result ? ?name(SailorsReserved2All SidSid
Sailor)
22Find the names of sailors over 20 who have not
reserved a Red Boat
- SailorsOver20 ? ?Sid (? agegt20 Sailor)
- RedBoats ??Bid(? colorRed Boat)
- SailorsReservedRed ? ?Sid(Reserves BidBid
RedBoats) - SailorsOver20NoRed ? SailorsOver20 -
SailorsReservedRed - Result ? ?name(SailorsOver20NoRed SidSid
Sailor)