*7.2 See the text in the book.*
**Answer:**

A decomposition {R_{1, }R_{2}}is a lossless-join decomposition
if R_{1} Ç R_{2 }®
R_{1}
or R_{1} Ç R_{2}®
R_{2}.
Let R_{1 }= (A, B, C), R_{2 }= (A, D, E), and R_{1 }Ç
R_{2}
= A. Since A is a candidate key (see exercise 6.15), therefore R_{1}Ç
R_{2
}®
R_{1}.

*7.3 See the text in the book.*
**Answer:**

Following the hint, use the following example of *r*: R_{1
}=
(A, B, C), R_{2 }= (C, D, E)

A | B | C | D | E |

a_{1} |
b_{1} |
c_{1} |
d_{1} |
e_{1} |

a_{2} |
b_{2} |
c_{1} |
d_{2} |
e_{2} |

p_{R1}(r) would be :

A | B | C |

a_{1} |
b_{1} |
c_{1} |

a_{2} |
b_{2} |
c_{1} |

p_{R2}(r) would be :

C | D | E |

c_{1} |
d_{1} |
e_{1} |

c_{1} |
d_{2} |
e_{2} |

p_{R1}(r) np_{R2}(r)
would be:

A | B | C | D | E |

a_{1} |
b_{1} |
c_{1} |
d_{1} |
e_{1} |

a_{1} |
b_{1} |
c_{1} |
d_{2} |
e_{2} |

a_{2} |
b_{2} |
c_{1} |
d_{1} |
e_{1} |

a_{2} |
b_{2} |
c_{1} |
d_{2} |
e_{2} |

Clearly p_{R1}(r) np_{R2}(r)
is not equal to r. Therefore this is a lossy join.

*7.5 See the text in the book.*
**Answer:**

The simple answer: F_{1} contains no dependencies with D on
the right side of the arrow. F_{2} contains no dependencies with
B on the left side of the arrow. Therefore for B ®
D to be preserved there must be an FD B®a
in F_{1}^{+ }and and a®
D in F_{2}^{+}, so B ® D
would follow by transitivity. Since the intersection of the two schemas
is A, a = A. Observe that B ®
A is not in F_{1}^{+} since B^{+} = BD.

The complex answer: The dependency B ®
D is not preserved. F_{1}, the restriction of F to (A, B, C) is
A ® ABC, A ®
AB, A ® AC, A ®
BC, A ® B, A ®
C, A ® A, B ®
B, C ® C, AB ®
AC, AB ® ABC, AB ®
BC, AB ® AB, AB ®
A, AB ® B, AB ®
C, AC (same as AB), BC (same as AB), ABC (same as AB). F_{2}, the
restriction of F to (C, D, E) is A ® ADE,
A ® AD A ®
AE, A ® DE, A ®
A, A ® D, A ®
E, D ® D, E (same as A), AD, AE, DE, ADE
(same as A). (F_{1} È F_{2})^{+
}is
seen not to contain B ® D since the only
FD F_{1} È F_{2} with
B as the left side is B ® B, a trivial FD.
Note that CD ® ABCDE is also not preserved.

*7.8 See the text in the book.*
**Answer:**

From Exercise 6.15 we know that B ®
D is nontrivial and the left hand side is not a super key. By the algorithm
of Figure 7.6. we derive the relations {(A,B,C,E), (B,D)} This is in BCNF.

*7.11 See the text in the book.*
**Answer:**

First we note that the dependencies given in Exercise 7.2 form a canonical
cover. Generating the schema from the algorithm of Figure 7.7 we get:

R' = {(A,B,C), (C, D, E), (B, D), (E, A)}

Schema (A,B,C) contains a candidate key. Therefore R' is a 3NF dependency
preserving lossless-join decomposition.

Note that the original schema R = (A, B, C, D, E) is already in 3NF. Thus, it was not necessary to apply the algorithm as we have done above. The single original schema is trivially a lossless join, dependency preserving decomposition.

*7.16 See the text in the book.*
**Answer:**

A ®® B

A ®® C

C ®® B

C ®® A

*7.23 See the text in the book.*
**Answer:**

The relational schema R = (A, B, C, D, E) and the dependencies of Exercise
7.21 constitutes a BCNF decomposition, however it is not in 4NF. It is
in BCNF because all FDs are trivial.

*7.24 See the text in the book.*
**Answer:**

4NF is more desirable than BCNF because it reduces the repetition of
information. If we consider a BCNF schema not in 4NF we observe that decomposition
into 4NF does not lose information provided that a lossless join decomposition
is used, yet redundancy is reduced.