Normalization theory exercises
Dr. Shiyong Lu
Department of Computer Science
Wayne State University
[email protected]
©copyright 2007, all rights reserved
Database schema for the
exercises
Consider schema(R, F), where
R=ABCDE, F={E→AC, A→B,
BD→E, CE→A}
Exercise 1
Consider schema(R, F), where
R=ABCDE, F={E→AC, A→B,
BD→E, CE→A}
Find all the keys of the relation.
Solution
Consider schema(R, F), where
R=ABCDE, F={E→AC, A→B,
BD→E, CE→A}
Step 1: for X to be a key, X+F =ABCDE
take the union of the right hand side, we
got ABCE, therefore D must be part of a
key since otherwise, D X+F ?
Solution
Consider schema(R, F), where
R=ABCDE, F={E→AC, A→B,
BD→E, CE→A}
Step 2: is D already a key?
We have D+F=D, so no.
Try to calculate the attribute closure of
DA, DB, DC, DE.
Solution
Consider schema(R, F), where
R=ABCDE, F={E→AC, A→B,
BD→E, CE→A}
Step 3:
DA+F= DABEC, DB+F= DBEAC
DC+F = DC, DE+F = DEACB, so DA,
DB, DE are keys, what about DC?
Continue to try DCA, DCB, DCE?
Exercise 2
Consider schema(R, F), where
R=ABCDE, F={E→AC, A→B,
BD→E, CE→A}
Decompose the schema into a set of
BCNF schemas.
Solution
R=ABCDE, F={E→AC, A→B, BD→E, CE→A}
Step 1: Pick up a dependency that violates BCNF
requirements. (DA, DB, DE are keys). Pick up
E→AC. Decompose:
R1 = EAC, F1 = {E→AC, CE→A}
R2=BDE, F2 = {BD→E}
Both of them are now in BCNF! Why?
Exercise 3
Consider schema(R, F), where
R=ABCDE, F={E→AC, A→B,
BD→E, CE→A}
Is this decomposition lossless? Is it
dependency-preserving?
Solution
R=ABCDE, F={E→AC, A→B, BD→E, CE→A}
Yes, it is lossless since the intersection of the
two relations is a superkey of one of the
relations.
Not dependency preserving, because F1 
F2 ={E→AC, BD→E, CE→A} and
A→B  (F1  F2)+!
May I drop the above +?
Exercise 4
Consider schema(R, F), where
R=ABCDE, F={E→AC, A→B,
BD→E, CE→A}
What is the attribute closure of
ACD?
Solution
R=ABCDE, F={E→AC, A→B, BD→E, CE→A}
[ACD+F ]1 = ACD
[ACD+F ]2 = ACDB, after applying A→B
[ACD+F ]3 = ACDBE, after applying BD→E
Done! [ACD+F ] = ABCDE, ACD is a
superkey?
Can we conclude ACD is a key?
Exercise 5
Consider schema(R, F), where
R=ABCDE, F={E→AC, A→B,
BD→E, CE→A}
What is the minimal cover of F?
Solution
R=ABCDE, F={E→AC, A→B, BD→E, CE→A}
Step 1: minimize the right hand side
F1={E→A, E → C, A→B, BD→E, CE→A}
Solution
R=ABCDE, F={E→AC, A→B, BD→E, CE→A}
Step 2: minimize the left hand side of F1
F1={E→A, E → C, A→B, BD→E, CE→A}
B+F1 = B, D+F1 = D, C+F1 = D, E+F1 = EAC, so
the last dependency can be minimized to
E→A.
We got F2 = {E→A, E → C, A→B, BD→E}
Solution
R=ABCDE, F={E→AC, A→B, BD→E, CE→A}
Step 3: F2 = {E→A, E → C, A→B, BD→E},
can we eliminate any dependency from F2?
Consider eliminating E→A, see if E→A 
(F2- E→A)+. Since E+F2-f =EC, we cannot
eliminate E→A.
How about other dependencies? None. So F2
is the minimal cover!
Exercise 6
Consider schema(R, F), where
R=ABCDE, F={E→AC, A→B,
BD→E, CE→A}
Decompose the schema into a set
of 3NF schemas.
Solution
R=ABCDE, F={E→AC, A→B, BD→E, CE→A}
Step 1: we got the minimal cover as F2 =
{E→A, E → C, A→B, BD→E},
Step 2: grouping by left hands
U1 = {E→A, E → C}, U2={A→B}, U3 ={
BD→E}
Step 3: R1=(EAC, E→A, E → C), R2 = (AB,
A→B), R3=(BDE, BD→E), do we need step
Exercise 7
Consider schema(R, F), where
R=ABCDE, F={E→AC, A→B,
BD→E, CE→A}
Is the decomposition lossless? Is it
functional dependency preserving,
why?