Review Solutions - GMU CS Department

Review for Final Exam
Lecture Week 14
Problems on
Functional Dependencies
and
Normal Forms
Problem 1
R = {Broker (B), Office (O), Investor (I),
Stock (S), Quantity (Q)}
• F = { I -> B, IS -> Q, B -> O}
• a.
What are the keys of the relation?
IS -> ISBQO IS is the only key
• b.
Is the relation in BCNF?
No, because of I->B and B->O
Problem 1 (contd.)
F = { I -> B, IS -> Q, B -> O}
• Produce a lossless-join, BCNF decomposition of the
original relation
R1=(I,B), R2 = (I,S,Q,O) R2 is not BCNF (I->O)
R21= (I,S,Q), R22= (I,O)
• Is your decomposition dependency-preserving?
No, because of B->O is not preserved. You
need a join to enforce it!
Problem 2
Consider the relation R=(A,B,C,D,E) and
F ={ A->BC, CD->E, B-> D, E-> A}
a.
Is this schema in 3NF?
Note: no attributes appear only on left sides or only on right sides. We
start with computing the closure of single attributes:
A+ = ABCDE; B+ = BD; C+ = C; D+ = D; E+ = EABCD. It follows that A
and E are candidate keys.
Now we consider combinations of two attributes:
BC+ = BCDEA, BD+ = BD, CD+ = CDEAB. It follows that BC and CD are
candidate keys. There are no more combinations to be considered.
The candidate keys are: A, E, BC, and CD.
Problem 2 (contd.)
Consider the relation R=(A,B,C,D,E) and
F ={ A->BC, CD->E, B-> D, E-> A}
a.
Is this schema in 3NF?
The candidate keys are: A, E, BC, and CD.
We reduce F in canonical form: F={A->B, A->C, CD->E, B->D, E->A}
A->B is in 3NF since A is a key;
A->C is in 3NF since A is a key;
CD->E is in 3NF since CD is a key;
B->D is in 3NF since D is part of a key;
E->A is in 3NF since E is a key.
It follows that the given schema is in 3NF.
Problem 2 (contd.)
F ={ A->BC, CD->E, B-> D, E-> A}
• Give a lossless-join, decomposition into BCNF
• (The relation is not in BCNF, since B is not a key and B->D)
ABCDE
BD
ABCE
Also
BCNF
The decomposition is lossless-join: R1  R2 = {B}
key of R1=BD. But it is not d.p. CD->E requires a
join!
Problem 2 (contd.)
F ={ A->BC, CD->E, B-> D, E-> A}
Show that the decomposition R1= (A,B,C), R2 =
(A,D,E) is lossless-join, but not dependencypreserving
R1  R2 = A, which is a key of both! So, the
decomposition is lossless-join
But the dependency B->D is not preserved.
Problem 2 (contd.)
F ={ A->BC, CD->E, B-> D, E-> A}
Show that the decomposition R1= (A,B,C), R2 =
(C,D,E) is not lossless-join.
R1  R2 = {C} which is not a key of R1 or R2, so
the decomposition is not lossless-join.
Problem 3
• R(A,B,C,D), F={AB->C, C->D, D->A}
• Is R in BCNF?
We need to check if any of the given functional dependencies
violates the BCNF:
AB+ = ABCD Thus AB->C is in BCNF.
C+ = CDA Thus C->D violates the BCNF.
D+ = DA Thus D->A violates the BCNF.
It follows that R in not in BCNF.
Problem 3 (contd.)
• R(A,B,C,D), F={AB->C, C->D, D->A}
• Decompose R into relations that are in BCNF using a
lossless join decomposition.
From previous analysis: C->D and D->A violate the BCNF.
Using C->D, we decompose R into CD and ABC.
We need to check whether ABC is in BCNF. We observe that C->A
is in F+, and in the projection of F onto ABC. C is not a (super)key
for ABC, thus ABC is not in BCNF. We decompose it further into CA
and BC.
The resulting decomposition CD, CA, and BC is a lossless join
decomposition of ABCD into BCNF relations.
The decomposition is not dependency preserving. E.g., AB->C is
not preserved.
Problem 4
• R(A,B,C,D) F = {B->C, B->D}
• Is R in BCNF? If not, compute a lossless join
decomposition of R into BCNF relations.
Both dependencies in F violate the BCNF condition since
B is not a (super)key.
One possible decomposition is R1=BC and R2=ABD
(using B->C). B->D holds on R2, and B is not a superkey
for R2, so we decompose it further in: R23=BD and
R24=AB. The resulting decomposition is: BC, BD, and AB.
This decomposition preserves dependencies:
(F1  F23)+ = F+
Problem 5
Answer: B and C are consistent with
Country -> Capitol
Final Exam
• Comprehensive, closed books;
• Topics:
– ER Diagrams;
– Relational Database schemas;
– Key constraints, participation constraints, weak
entities, etc;
– Relational algebra;
– SQL (including GROUP BY and HAVING clauses,
aggregate operators, set comparison operators,
nested queries, etc.);
– Functional dependencies;
– Normal forms (BCNF, 3NF) and related
decomposition techniques.
Discussion of
Problems in
Sample Final
Problem 2 (4)
SELECT C.Instructor-SSn,
sum(decode(S.Status,’Grad’,1,0))/count(*) as Percent
FROM
Class C, Enrollment E, Student S
WHERE (C.Class-no=E.Class-no) AND (E.Student-Ssn=S.Ssn)
GROUP BY C.Instructor-SSn
Problem 2 (5)
SELECT C.D-code, C.num-st
FROM (SELECT D.D-code, count(*) as num-st
FROM Department D, Student S
WHERE D.D-code=S.Major
GROUP BY D.D-code) C
WHERE C.num-st <= ALL (SELECT C.num-st
FROM C)

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