CS 540
1.
Assignment 5
Solutions
Let R(A,B,C,D,E,G,H) be a relation schema with set of functional dependencies:
F = {ABC, BD, CDE, CEGH, GA}.
a.
Exhibit a derivation of ABE from F.
in F
Reason
-----------------------------------------------------ABB
reflexivity rule
BD
in F
ABD
transitivity rule
ABC
in F
ABCD
union rule
CDE
in F
ABE
transitivity rule
b.
Exhibit a derivation sequence for BGC using only Armstrong’s Axioms.
in F
Reason
--------------------------------------------------------GA
in F
BGAB
augmentation rule
ABC
in F
BGC
transitivity rule
c.
Find AB+.
AB+ contains AB
AB+ contains ABC
AB+ contains ABCD
AB+ contains ABCDE
AB+ contains ABCDEGH
Hence, AB+ = ABCDEGH
reflexivity
ABC
BD
CDE
CEGH
2.
Let R(A,B,C,D,E,G,H) be a relation schema with set of functional dependencies:
F = {ABC, BD, CDE, CEGH, GA}. Find all keys for R. Show and
explain all work.
The attribute B must appear in every key; but B+ = BD which is not a key.
No key contains H. Since AB+ = ABCDEGH and no proper subset of AB
satisfies this property, AB is a key. Since B is not a key, R does not have a
unique key.
BC+ = ABCDEGH and no proper subset of BC satisfies this property; thus,
BC is a key.
BG+ = ABCDEGH nd no proper subset of BG satisfies this property; thus,
BG is a key.
No other combinations are keys; thus, the only keys are AB, BC and BG.
3.
Let R have arity n. Find an upper bound for the cardinality of F+ in terms of n.
There are 2n – 1 subsets of the attributes of R save the empty set. Hence,
there are (2n – 1)2 possible combinations XY functional dependencies
using the attributes of R.
4.
Show that the closure operator for a set of attributes is monotonic; i.e. if Y  X,
then Y+  X+.
We prove that the closure operator is monotonic. Thus, let Y and X
be two sets of attributes such that Y  X. We show Y+  X+. Let A be an
attribute in the set of attributes Y+. Then, YA derivable from F using the
inference rules. Since Y  X, XY is derivable from F using Armstrong's
Axioms and by transitivity X A is in F+. Thus A is derivable from X and
thus, A is in X+.
5.
A set of attributes Y is said to be closed if Y equals its own closure, i.e. Y = Y+.
Prove that, for a set of attributes X, the smallest closed set containing X is X+.
Prove that, for a set of attributes X, the smallest closed set containing X
is X+. We appeal to contradiction and assume that there is a closed set Z of
attributes such that X  Z  X+. Since the closure operator is monotonic,
we have X+ Z+  (X+)+. But Z is a closed set of attributes; hence, Z+ = Z.
We also have proven that X+ = (X+)+. Therefore, X+ Z  X+ and Z = X+.
6.
Let R be a relation schema with two sets of functional dependencies G and F.
Suppose G  F. Prove that G+  F+, i.e. that the closure operator for functional
dependencies is monotonic.
We prove that G+  F+ if G  F. Let XY  G+. By completeness, XY can
be derived from G using the inference rules. Since G  F, XY can be
derived from F using the inference rules. By soundness, XY  F+.
7.
Let R be a relation schema with set of functional dependencies F. Prove that
F+ = (F+)+.
To show that F+ = (F+)+, we need only show that (F+)+  F+ since F+  (F+)+,
i.e. for any set of functional dependencies G, it is always the case that G  G+.
Thus, let XY  (F+)+. By completeness, XY is derivable from F+ using
the inference rules. By completeness, any functional dependency in F+ is
derivable from F using the inference rules. Hence, XY is derivable from F
using the inference rules. By soundness, XY  F+.
8.
Let R be a relation schema and F a set of functional dependencies for R. Let X, Y,
W and Z be sets of attributes of R. Armstrong’s Axioms are:
A1.
A2.
A3.
if Y  X then XY is in F+.
if XY is in F+, then XZYZ is in F+.
if XY and YZ are in F+, then XZ is in F+.
The following set of inference rules has been introduced:
FD1. XX is in F+.
FD2. if XYZ is in F+, then XY is in F+.
FD3. if XWY and YZ are in F+ then XWYZ is in F+.
Prove the second set of inference rules is equivalent to Armstrong’s Axioms by
proving that:
a.
Armstrong’s Axioms implies the second set of inference rules.
FD1: XX is in F+; follows from A1 since X  X.
FD2: if XYZ is in F+, then XY is in F+; follows from decomposition.
FD3: if XWY and YZ are in F+ then XWYZ is in F+; follows from the
extended transitivity rule.
b.
The second set of inference rules implies Armstrong’s Axioms.
Now assume that FD1, FD2 and FD 3 hold. We prove that Armstrong’s
Axioms may be derived from these three rules.
9.
A1:
Suppose that Y  X.  Z such that X = YZ. Since XX is in F+, by
FD2, XY is in F+.
A2:
XZXZ is in F+ by FD1. XY is in F+ by assumption. By FD3,
XZXYZ is in F+. By FD2, XZXY is in F+.
A3:
Assume XY and YZ are in F+. By FD3, XYZ is in F+. By FD2,
XY is in F+.
We have studied functional dependencies for a relation schema R. We now define
another class of dependencies for a relation schema R. Let X be a set of attributes
of R; we say α(X) holds in an instance r of R if any two records of r that agree
on the X attributes agree on every attribute of R. Let G be a set of these
dependencies, i.e. G = {α (X1), α(X2), …}. An instance r of R is said to bevalid if
every dependency, α(Xi) in G, holds in r.
a.
Find a set of inference rules for this class of dependencies.
Consider the inference rule:
if α(X)  G+ and Y is any set of attributes of R, then α(XY)  G+.
b.
Prove that your inference rules are sound.
Suppose α(X)  G+ and Y is any set of attributes of R. If r is a valid
instance of R and any two records agree on the XY attributes, then
the two records agree on the X attributes. Since α(X)  G+, the two
records agree on all the attributes of R. Hence, the two records agree
on the XY attributes and α(XY)  G+.
c.
Prove that your inference rules are complete.
We need to prove that if α(Z)  G+ for any set of attributes Z of R,
then  α(Xi)  G such that Xi  Z. We prove the contrapositive: if
Xi  Z for any α(Xi)  G, then α(Z)  G+. For each i, select an
attribute Ai  Xi but Ai  Z. Create an instance r of R with two rows
t1 and t2 such that t1 and t2 disagree on each Ai attribute but agree on
all the other attributes. This instance is valid as no two records agree
on any Xi attributes. However, both records agree on all the Z
attributes but are distinct records. Thus, α(Z)  G+
10.
Let R be a relation schema with set of functional dependencies F. If X is a set of
attributes of R, we say X is closed if X = X+. It is obvious that if X is a set of
attributes such that the left hand side of no functional dependency in F is a subset
of X, then X is closed. Prove or disprove the converse of this statement.
The converse is not true. Consider R as a set of attributes R, R is a closed set
of attributes since R+ = R. But any non-trivial functional dependency in F has
a left hand side as a subset of R.