Homework 4 Solution
March 4, 2009
Question 1-7 and 9: 10 points/each. Question 8 a) and b): 5 points/each. Question
8 c) and d): 10 points/each.
1. (a) YES. X+=XYWZ
(b) No. (XW)+=XW, Z is not in there.
2. r satisfies AD โB, C โDE, CD โA, AE โB. But r does not satisfy AโB
since the first two tuples of r have the same values for A, but different values for
B.
3. (a) A+=ABEC
(b) (AE)+=ABEC
(c) (ADE)+=ABCDEI
4. To see that F and G are equivalent, we need to verify that every FD in F is in G+,
and vice versa. We first check if F โG+: (i) A+=ACD, so A โC is in G+. (ii)
AC+
G=ACD, so AC โD is in G+. (iii) (E)+
G=ACDEH, so E โAD is in G+.
Next we verify if G โF+: (i) (A)+
F=ACD, so A โCD is in F+. (ii) (E)+
F=
ACDE, so E โAHE is not in F+.
Thus F and G and NOT equivalent.
5. (a) Let F ={A โBC, B โC, AB โD}. We need to obtain an equivalent set
of FDs that satisfies the three properties of a minimal cover.
โขRight side of each FD in F must ne a single attribute: so we replace F
by F1={A โB, A โC, B โC, AB โD}.
โขNo extraneous attributes on the left side. We first check if A can be
deleted from AB โD. We can do so if B โD follows from F1. Since
(B)+
F1=BC, the answer is NO.
We next check if B can be deleted from AB โD. We can do if A โ
D follows from F1. Since (A)+
F1=ABCD, the answer is YES. Let F2
={A โBC, B โC, A โD}.
โขNo redundant FDs: A โC can be deleted from F2. Minimal cover =
{A โB, B โC, A โD}.
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