relational database schemacarlotta/teaching/infs-614-f08/...answer: this relational algebra...

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1 HasPriPhy Patients address name ssn age Doctors exp_year name ssn specialty Pharmacies name address phone PharmCo name phone Makes stDate endDate txt super isContracted date qty Sell prescribes price Drugs tradeName formula

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Page 1: Relational Database Schemacarlotta/teaching/INFS-614-f08/...Answer: This Relational Algebra statement does not return anything because of the sequence of projection operators. Once

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HasPriPhy Patients

address name

ssn age

Doctors

exp_year name

ssn specialty

Pharmacies

name address phone

PharmCo name

phone

Makes

stDate endDate

txt super

isContracted

date qty

Sell

prescribes

price

Drugs tradeName

formula

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Relational Database Schema Doctors(ssn:integer,name:string, specialty:string,year_exp:integer) PK: ssn

Patients(ssn:integer,name:string,age:integer,address:string,ssnDoctor:integer) PK: ssn FK: ssnDoctor references Doctors

Pharmacies(name:string, phone:integer, address:string) PK: name

PharmCo(name:string, phone:integer) PK: name

Drugs(tradeName: string,formula:string,pharmCoName: string) PK: pharmCoName, tradeName FK: pharmCoName references PharmCo Prescribes(ssnPatient:integer, ssnDoctor:integer, pharmCoName: string , tradeName:sring, date:date, qty:integer) PK: FK: ssnPatient references Patients ssnPatient,ssnDoctor,pharCoName,tradeName FK: ssnDoctor references Doctors FK: (pharCoName,tradeName) referencesDrugs

Sells(pharmCoName: string, tradeName: string, pharmacyName: string , price:float) PK: pharmCoName,tradeName,pharmacyName FK: (pharmCoName,tradeName) references Drugs FK: pharmacyName references Pharmacies

Contracts(pharmCoName: string,pharmacyName: string , startDate:date, endDate:date, text:string, supervisor:string)

PK: pharCoName,pharmacyName FK: pharmCoName references PharmCo FK: pharmacyName references Pharmacies

Each faculty is assigned to exactly one office room. An office room is assigned to at least one faculty. Is the following E-R diagram correct? If your answer is NO, indicate what is wrong in this E-R diagram.

Faculty Office AssignedTo

Answer: Correct!

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A faculty teaches at most three classes, but a class is taught by exactly one faculty. Is the following E-R diagram correct? If your answer is NO, indicate what is wrong in this E-R diagram.

Faculty Classes teaches

 Total participation of Faculty w.r.t teaches.  Key Constraint on Faculty w.r.t teaches.  Partial participation of Classes w.r.t teaches.

Answer: NO

Each faculty schedules office hours in exactly one office room at a weekday and time, and this information must be recorded. An office may be assigned to several faculties’ office hours at different days and times. All offices are assigned to faculty’s office hours. In the following E-R Diagram, complete the dashed lines to reflect the

appropriated participation and key constraints.

Faculty Office scheduled

Time

WeekDay

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Each faculty schedules office hours in exactly one office room at a weekday and time, and this information must be recorded. An office may be assigned to several faculties’ office hours at different days and times. All offices are assigned to faculty’s office hours. In the following E-R Diagram, complete the dashed lines to reflect the

appropriated participation and key constraints.

Faculty Office scheduled

Time

WeekDay

Each faculty schedules office hours in exactly one office room at a weekday and time, and this information must be recorded. An office may be assigned to several faculties’ office hours at different days and times. All offices are assigned to faculty’s office hours. In the following E-R Diagram, complete the dashed lines to reflect the

appropriated participation and key constraints.

Faculty Office scheduled

Time

WeekDay

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Given two relations R1 and R2, where R1 contains N1 tuples and R2 contains N2 tuples, with N2>N1>0, give the minimum and maximum possible sizes (in tuples) for the resulting relation produced by each of the following relational algebra expressions. In each case, state any assumptions about the schemas for R1 and R2 needed to make the expression meaningful.

Expression Assumption Min Max

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 Suppose R1 and R2 are multisets of tuples (rows);

 N1 = number of copies of a tuple t in R1;

 N2 = number of copies of a tuple t in R2.

What’s the number of copies of t in

R1 UNION ALL R2 ?

Answer: N1+N2

 Suppose R1 and R2 are multisets of tuples (rows);

 N1 = number of copies of a tuple t in R1;

 N2 = number of copies of a tuple t in R2.

What’s the number of copies of t in

R1 INTERSECT ALL R2 ?

Answer: min{N1,N2}

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 Suppose R1 and R2 are multisets of tuples (rows);

 N1 = number of copies of a tuple t in R1;

 N2 = number of copies of a tuple t in R2.

What’s the number of copies of t in

R1 EXCEPT ALL R2 ?

Answer:

Consider the following schema:

Suppliers(sid:integer, sname:string, address:string)

Parts(pid:integer, pname:string, color:string)

Catalog(sid:integer, pid:integer, cost:real)

State what the following queries compute.

The Catalog relation lists the prices charged for Parts by Suppliers.

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Suppliers(sid:integer, sname:string, address:string)

Parts(pid:integer, pname:string, color:string)

Catalog(sid:integer, pid:integer, cost:real)

Answer: Find the Supplier names of the suppliers who supply a red part that costs less then 100 dollars.

Suppliers(sid:integer, sname:string, address:string)

Parts(pid:integer, pname:string, color:string)

Catalog(sid:integer, pid:integer, cost:real)

Answer: This Relational Algebra statement does not return anything because of the sequence of projection operators. Once the sid is projected, it is the only field in the set. Therefore, projecting on sname will not return anything.

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Suppliers(sid:integer, sname:string, address:string)

Parts(pid:integer, pname:string, color:string)

Catalog(sid:integer, pid:integer, cost:real)

Answer: Find the Supplier names of the suppliers who supply a red part that costs less than 100 dollars and a green part that costs less than 100 dollars.

Suppliers(sid:integer, sname:string, address:string)

Parts(pid:integer, pname:string, color:string)

Catalog(sid:integer, pid:integer, cost:real)

Answer: Find the Supplier ids of the suppliers who supply a red part that costs less than 100 dollars and a green part that costs less than 100 dollars.

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Suppliers(sid:integer, sname:string, address:string)

Parts(pid:integer, pname:string, color:string)

Catalog(sid:integer, pid:integer, cost:real)

Answer: Find the Supplier names of the suppliers who supply a red part that costs less than 100 dollars and a green part that costs less than 100 dollars.

Department (D-code, D-Name, Chair-SSn) Faculty (Ssn, F-Name, D-Code, Rank)

Course (D-code, C-no, Title, Units) Student (Ssn, S-Name, Major, Status)

Prereq (D-code, C-no, P-code, P-no) Enrollment (Class-no, Student-Ssn)

Class (Class-no, D-code, C-no, Instructor-SSn) Transcript (Student-Ssn, D-Code, C-no, Grade)

select F.SSn from Class Co, Faculty F where F.Ssn = Co.Instructor- SSn and not exists (select * from Class C, Faculty F1 where Co.D-code = C.D-code and Co.C-no = C.C-no and C.Instructor-SSn = F1.Ssn and F1.D-code != F.D-Code);

(1) Find the ssn of faculty who are currently teaching courses taught by faculty of several departments. (2) Find the ssn of faculty who are not currently teaching any courses of his/her department. (3) Find the ssn of faculty who are currently teaching courses taught only by faculties of his/her department. (4) Find the ssn of faculty who are currently teaching some course of his/her department. (5) Find the ssn of faculty who are currently teaching exactly one course.

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Flight(flno:integer, from:string, to:string, distance: integer, departs:time, arrives:time) Aircraft(aid:integer, aname:string, cruisingrange:integer) Certified(eid:integer, aid:integer) Employees(eid:integer, ename:string, salary:integer)

 Employees relation describes pilots and other kinds of employees as well;

 Every pilot is certified for some aircraft (otherwise, he/she would not qualify as a pilot);

 Only pilots are certified to fly.

Write the following queries in relational algebra and SQL.

Flight(flno:integer, from:string, to:string, distance: integer, departs:time, arrives:time) Aircraft(aid:integer, aname:string, cruisingrange:integer) Certified(eid:integer, aid:integer) Employees(eid:integer, ename:string, salary:integer)

Find the aids of all aircrafts that can be used on non-stop flights from Bonn to Madrid.

SELECT A.aid FROM Aircraft A, Flight F WHERE F.from=‘Bonn’ AND F.to=‘Madrid’, AND A.cruisingrange >=F.distance

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Flight(flno:integer, from:string, to:string, distance: integer, departs:time, arrives:time) Aircraft(aid:integer, aname:string, cruisingrange:integer) Certified(eid:integer, aid:integer) Employees(eid:integer, ename:string, salary:integer)

Identify the flights (flno) that can be piloted by each pilot whose salary is more than $100,000.

SELECT F.flno FROM Aircraft A, Certified C, Employees E, Flight F WHERE E.eid=C.eid AND C.aid=A.aid AND F.distance<=A.cruisingrange AND E.salary>100,000

Flight(flno:integer, from:string, to:string, distance: integer, departs:time, arrives:time) Aircraft(aid:integer, aname:string, cruisingrange:integer) Certified(eid:integer, aid:integer) Employees(eid:integer, ename:string, salary:integer)

Find the names of pilots who can operate planes with a range greater than 3,000 miles but are not certified on any Boeing aircraft.

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SELECT E.ename FROM Employees E WHERE E.eid IN ((SELECT C1.eid FROM Certified C1, Aircraft A1 WHERE A1.aid=C1.aid AND A1.cruisingrange>3000) EXCEPT (SELECT C2.eid FROM Certified C2, Aircraft A2 WHERE A2.aid=C2.aid AND A2.aname=‘Boeing’))

Find the names of pilots who can operate planes with a range greater than 3,000 miles but are not certified on any Boeing aircraft. Another solution using EXCEPT:

Flight(flno:integer, from:string, to:string, distance: integer, departs:time, arrives:time) Aircraft(aid:integer, aname:string, cruisingrange:integer) Certified(eid:integer, aid:integer) Employees(eid:integer, ename:string, salary:integer)

Find the eids of employees who make the highest salary (RA only). The approach to take is: first find all employees who do not have the highest salary. Subtract these from the original list of employees and what is left is the highest paid employees.

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Flight(flno:integer, from:string, to:string, distance: integer, departs:time, arrives:time) Aircraft(aid:integer, aname:string, cruisingrange:integer) Certified(eid:integer, aid:integer) Employees(eid:integer, ename:string, salary:integer)

Find the eids of employees who make the second highest salary (RA only). The approach is similar to the previous one. First find all employees who do not have the highest salary. Remove these from the original list of employees and what is left is the highest paid employees. Remove the highest paid employees from the original list. What is left is the second highest paid employees together with the rest of the employees. Then find the highest paid employees of this new list. This is the list of the second highest paid employees.

Flight(flno:integer, from:string, to:string, distance: integer, departs:time, arrives:time) Aircraft(aid:integer, aname:string, cruisingrange:integer) Certified(eid:integer, aid:integer) Employees(eid:integer, ename:string, salary:integer)

Find the eids of employees who make the second highest salary (RA only).

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 Sample Midterm (with solutions);

 Solutions of HW1, HW2, HW3;

 You should:

 Review the lectures’ slides and hws;

 Do the readings in the textbook;

 Do exercises at the end of chapters.

Material for you to get ready for Midterm…