CS530
Winter 2002 (Answers)
Question 1 a) Discuss the following terms from Extended-Entity-Relationship (EER) modelling. Show how each of the elements you describe is drawn in an EER model. i. Entity types . What is an entity type? What is the difference between a strong entity type and a weak entity type?
[1 marks]
( knowledge and understanding )
Group of “objects” (real or abstract) with the same properties and having an independent existence. Strong = not dependent on some other entity type for its existence. Weak = dependent on some other entity type for its existence. ii. Relationship type . What is a relationship type? What is the degree of a relationship type? What is a recursive relationship?
[2 marks]
( knowledge and understanding )
A set of meaningful associations among entity types. Degree = number of participating entity types. Recursive = same entity participates more than once in different roles. iii. Attributes . What is an attribute? What is an attribute domain? What is a derived attribute? What is a discriminator attribute? What is the difference between simple attributes and composite attributes and the difference between single-valued attributes and multi-valued attributes?
[3 marks]
( knowledge and understanding )
Property of an entity or relationship type. Domain is set of allowable values. Simple has single independent component, composite made up of multiple independent components. Derived is derivable from value(s) of related attribute(s), not necessarily of same entity. Discriminator is used to distinguish instances of a weak entity type (part of composite primary key). Single-valued has only one value for each instance of entity type, multi-valued may have more than one. iv. Keys . What is the difference between a candidate key, a primary key and a composite key? What considerations might influence the choice of a primary key?

(
[4 marks] knowledge and understanding )
Candidate is minimal set of attributes that uniquely identifies instances of entity. Primary is selected candidate key. Composite is candidate consisting of more than one attribute. b) Figure 1 shows a fragment of an Extended-Entity-Relationship (EER) model describing operations (surgical procedures) in a hospital database. name
Procedure date name name
Patient operation Nurse patient_id nurse_id
Doctor name doctor_id
Figure 1 i. Draw a new entity relationship model in which the operation relation is replaced by a weak entity and which expresses the additional constraints that an operation consists of exactly one surgical procedure being performed on exactly one patient by one or more doctors and one or more nurses. It should also be assumed that a particular procedure can be performed on a particular patient at most once in a single day, but many different procedures could be performed on the same patient in a single day.
[7 marks]
( knowledge and understanding plus intellectual skills ) ii. A Doctor may be either a Consultant or a Registrar, but not both, and there are some Doctors who are neither Consultants nor Registrars. Show how the entity relationship can be extended to express this additional information.
[3 marks]
( knowledge and understanding plus intellectual skills )

name
Patient
1 patient_id name
Procedure
1 on date perform m m operation m m by with name n
Nurse nurse_id n
Doctor name doctor_id d
Consultant Registrar
Question 2 a) Figure 2 shows a fragment of an Extended-Entity-Relationship (EER) model describing products and their distribution.

price description
Product m has-part code supplier-address n
Part weight stock account-id cust-code contact address supplier-name id
Figure 2 i. Map the EER model to a relational schema without performing any normalisation . Clearly indicate all primary keys and foreign keys (also stating into which relations they are keys.
[7 marks]
( knowledge and understanding plus intellectual skills )
Distributor (account-id, contact, address)
Stock (account-id, code, cust-code) account-id FK into Distributor, code FK into Product
Product (code, price, description)
Has-part (code, id) code FK into Product, id FK into Part
Part (id, weight, supplier-name, supplier-address) ii. Given that suppliers have a unique name and exactly one address, say why the Part relation is not in Boyce-Codd normal form (BCNF).
[2 marks]
( knowledge and understanding plus intellectual skills )
Because supplier-address is functionally dependent on suppliername, which is not a candidate key. iii. Show how the Part relation can be decomposed into 2 relations that are both in BCNF.
[2 marks]
( knowledge and understanding plus intellectual skills )

Part (id, weight, supplier-name) supplier-name FK into Supplier
Supplier (supplier-name, supplier-address) iv. Show how the EER model could be modified to directly represent the normalised relations.
[3 marks]
( knowledge and understanding plus intellectual skills ) weight
Part n id has-part supplier-address
1
Supplier supplier-name
Add Supplier entity with name and address attributes related to
Part by 1:m relationship. b) Consider the following relational schema for a database describing accommodation and activities available in national parks:
Park (name, region)
Hotel (id, name, tname)
Town (name, pname)
Activity (pname, activity) i. Write an SQL to create a view called “Climbing” that contains the id, name and tname of all Hotels situated in a park where “climbing” is an available activity. Is this view updatable, and why?
[3 marks]
( knowledge and understanding plus intellectual skills ) tname is a foreign key into Town pname is a foreign key into Park pname is a foreign key into Park
CREATE VIEW Climbing AS
(SELECT h.id, h.name, h.tname
FROM Park p, Hotel h, Town t, Activity a

WHERE h.tname=t.name and t.pname=p.name
and a.pname=p.name and a.activity=`climbing’); ii. Write an SQL query that lists the name of each Town, with the name of the Park it is in and the number of hotels it contains. The list should be ordered first by Park name and then by Town name in ascending alphabetical order.
[3 marks]
( knowledge and understanding plus intellectual skills )
SELECT p.name, t.name, COUNT(*) as hotels
FROM Park p, Hotel h, Town t
WHERE h.tname=t.name and t.pname=p.name
GROUP BY t.name
ORDER BY p.name, t.name;
Question 3
(a) Which of the restrictions below are enforced, and which are not enforced, by the following schema:
[6 marks]
( knowledge and understanding )
CREATE TABLE Stores( storeID INT CHECK(storeID IN (SELECT store FROM Sales)), location CHAR(30),
PRIMARY KEY (storeID) );
CREATE TABLE Items( name CHAR(30) PRIMARY KEY, price REAL CHECK(price >= 0.0) );
CREATE TABLE Sales( store INT, item CHAR(30) REFERENCES Items(name), totalSales REAL,
PRIMARY KEY (store, item) );
CREATE ASSERTION BigSales CHECK(NOT EXISTS(
SELECT * FROM Stores, Sales WHERE Stores.location =
'M21'
AND Stores.storeID = Sales.store
AND Sales.totalSales != 1000000) );
(i) No item has a price of 0.
(ii) Every storeID that appears in the Stores relation must also appear as a store value in the relation sales.
(iii) Every store has total sales of at least 1,000,000 pounds.
(iv) None of the above; i.e., each could be false in some database that satisfies the constraints.
(i) is not enforced; the CHECK only requires the price be nonnegative, which allows 0. (2 marks)

(ii) is not enforced. Remember that attribute-based checks cannot enforce a foreign-key constraint, such as (b), because they are not triggered on a deletion.
Yet a deletion can cause a foreign-key violation. (2 marks)
(iii) is not enforced. The assertion only requires that stores in Chorlton (M21) have sales of at least a million. (2 marks)
(iv) is enforced.
(b) Which of the restrictions below are enforced, and which are not enforced, by the schema above (explain why):
[6 marks]
( knowledge and understanding )
(i) Every name appearing in relation Items also appears in the item column of
Sales.
(ii) Every item appearing in Sales also appears in the name column of Items.
(iii) No two tuples (i.e., rows) of Sales can agree (i.e., have the same value) on either store or item.
(iv) None of the above; i.e., each could be false in some database that satisfies the constraints.
(i) is not enforced, since it is the opposite of the foreign-key constraint. (2 marks)
(ii) is enforced, since this constraint is exactly the foreign-key constraint in Sales.
(2 marks)
(iii) is not enforced, since the primary key is on both attributes. (2 marks)
(c) Suppose we have the following table declarations:
CREATE TABLE Empl (ECode INT PRIMARY KEY);
CREATE TABLE Manager (MCode INT PRIMARY KEY
REFERENCES Empl(ECode) ON DELETE SET NULL;
CREATE TABLE PartTime (PRef INT REFERENCES Empl(ECode));
CREATE TABLE Boss
(MCode INT REFERENCES Manager(MCode) ON DELETE SET
NULL,
ECode INT REFERENCES Empl(ECode) ON UPDATE CASCADE);
Consider the following scripts:
I. DELETE ; DELETE FROM
Manager
; DELETE
FROM
Empl
; DELETE FROM
Boss
;
II. DELETE
PartTime
; DELETE FROM
Boss
; DELETE FROM
Empl
; DELETE FROM B
Manager
III. DELETE FROM
Manager
; DELETE FROM
PartTime
; DELETE
FROM
Boss
; DELETE FROM
Empl
;
Which of the above scripts will empty all four tables, without error (explain why):
(i) III only, (ii) I only, (iii) II and III only, (iv) I and III only.
[8 marks]
( knowledge and understanding )
The answer is (iii). (2 marks)

The foreign-key constraint from
Manager to
Empl doesn't cause problems if we delete from
Empl first, since there is a clause that tells how to handle dangling tuples in
Manager
. (1 mark)
However, the foreign-key constraint from
PartTime to
Empl is a problem, since the default policy is to reject a deletion from
Empl that causes a dangling tuple in
PartTime
. (1 mark)
Thus, in a valid sequence,
PartTime must precede
Empl
. (1 mark)
The foreign-key constraint from
Boss
to
Manager is not a problem, but the one from
Boss
to
Empl is, again because the default policy will cause a rejection of certain deletions from
Empl
. (1 mark)
Thus,
Boss
must also come before
Empl
. (1 mark)
Of the three sequences, only I has
PartTime or
Boss
before
Empl
, so II and
III are OK; I is not. (1 mark)
Question 4
Consider the following relational schema:
LECTURER(LCode, FirstName, LastName)
STUDENT(SCode, FirstName, LastName)
TOPIC(TCode, Description)
LECTURE(TCode, Date, LCode, StudNumber)
Tcode FK into
TOPIC
LCode FK into
LECTURER
(
StudNumber
is the number of students attending the lecture)
ASSESSMENT(Tcode, Date, Scode, Mark)
Tcode, Date FK into
LECTURE
Scode FK into
STUDENT
(Assessments are done during lectures)
Write the SQL expressions for the following queries:
(a) Select the code number, the first name and the last name of the students who where never assessed on a topic with a description “Logic”.
[4 marks]
( knowledge and understanding plus intellectual skill )
(b) Select the code number, of the lecturer who gave lectures on all the topics with description “Logic”.
[4 marks]
( knowledge and understanding plus intellectual skill )
(c) Select the code number of the lecturer who made continuous assessments, i.e. who made at least an assessment for each of her/his lectures.
[4 marks]
( knowledge and understanding plus intellectual skill )
(d) Select, for each topic, the average of the marks assessed on the topic, considering only the students who where assessed at least three times on the topic.
[4 marks]

( knowledge and understanding plus intellectual skill )
(e) Select, for each student, the code number of the lecturer with whom she/he got the greatest number of assessments.
[4 marks]
( knowledge and understanding plus intellectual skill )
(a)
SELECT *
FROM STUDENT
WHERE NOT EXISTS
( SELECT *
FROM ASSESSMENT, TOPIC
WHERE STUDENT.Scode=ASSESSMENT.Scode
AND TOPIC.Tcode=ASSESSMENT.Tcode
AND TOPIC.Description=’Logic’)
(b)
SELECT DISTINCT LCode
FROM LECTURE X
WHERE NOT EXISTS
( SELECT *
FROM TOPIC
WHERE Description=’Logic’
AND NOT EXISTS
( SELECT *
FROM LECTURE Y
WHERE X.LCode=Y.LCode
AND TOPIC.Tcode=Y.Tcode))
(c)
SELECT LX.LCode
FROM LECTURE LX
WHERE NOT EXISTS
( SELECT *
FROM LECTURE LY
WHERE LX.LCode=LY.LCode
AND NOT EXISTS
( SELECT *
FROM ASSESSMENT A
WHERE LY.Date=A.Date
AND LY.Tcode=A.Tcode))
(d)
SELECT X.Tcode, AVG(X.Mark)
FROM ASSESSMENT X
WHERE 3 <= ( SELECT COUNT(*)
FROM ASSESSMENT Y
WHERE Y.Tcode=X.Tcode
AND Y.Scode=X.Scode)
(e)
SELECT AX.Scode, LECTURE.LCode
FROM LECTURE, ASSESSMENT AX
WHERE AX.Tcode=LECTURE.Tcode
AND AX.Date=LECTURE.Date

GROUP BY AX.Scode, LECTURE.LCode
HAVING COUNT(*) >= ALL ( SELECT COUNT(*)
FROM LECTURE L, ASSESSMENT A
WHERE L.Tcode=A.Tcode
AND L.Date=A.Date
AND AX.Scode=A.Scode
GROUP BY A.Scode, L.LCode)