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Model Answers to Assignment 4
Since this assignment involves making a lot of diagrams, handwritten submissions while be allowed,
though diagrams made with the builtin drawing facilities of Word or other software is preferred.
Each exercise is worth 10 points (50 points total).
CISC 437 students
1.
Exercise 3.5.2.
SOLUTION
(a)
Since neither H nor S appears on the right hand side of any of the functional dependencies, the only
way that they can be in the closure of a key is if they are in the key to begin with. Thus, H and S must
be in any key of the relation.
Computing {H, S}
+
, we get all the attributes of the relation, so HS is the
only key for the relation.
(b)
If we take each FD in turn, remove it from the set and compute the closure of the left side of the
FD using the set, we do not get the right side of the FD in the closure.
(For example, choosing FD HT
→
C, computing the closure of {H, T} using the FD set {C
→
T, HT
→
R, HS
→
R, CS
→
G}, we see
that C is not in the closure, so HT
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 Fall '08
 Schwartz,L

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