This preview shows pages 1–2. Sign up to view the full content.
Tutorial 3
Problem 1: BCNF form
Consider a relation with schema R(A,B,C,D,E) and FD’s AB
→
C, DE
→
C, and B
→
D. Is this relation in
BCNF? Why or why not? If it's not, decompose the relation into collections of relations that are in BCNF.
Make sure to indicate which dependency you apply to each decomposition
.
Solution
Definition: A relation R is in BCNF if and only if every determinant (the left hand side of the FD) is a
candidate key.
R is not in BCNF. In fact, every FD is a violation. We decompose in steps:
Step 1: AB
→
C is a violate FD, (AB)+ is (ABCD) (recall that (AB)+ means the closure of AB).
Therefore, we decompose R to R1(ABE) and R2(ABCD). We now examine R1 and R2.
Step 2: R1 has no FDs, so it's in BCNF.
Step 3: R2 has the FDs AB
→
CD and B
→
D. AB is a key, but B is not. Therefore we decompose R2 to
R3(ABC) and R4(BD).
Step 4: R3's FD is AB
→
C and AB is a key, so it's in BCNF.
Step 5: R4's FD is B
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '07
 Winslett

Click to edit the document details