Database Systems: The Complete Book: Solutions for Chapter 3

Database Systems: The Complete Book

Solutions for Chapter 3

Solutions for Section 3.1
Solutions for Section 3.2
Solutions for Section 3.3
Solutions for Section 3.4
Solutions for Section 3.5
Solutions for Section 3.6
Solutions for Section 3.7

Solutions for Section 3.1

Exercise 3.1.2(a)

We can order the three tuples in any of 3! = 6 ways. Also, the columns can be ordered in any of 3! = 6 ways. Thus, the number of presentations is 6*6 = 36.

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Solutions for Section 3.2

Exercise 3.2.1

     Customers(ssNo, name, address, phone)
     Flights(number, day, aircraft)
     Bookings(ssNo, number, day, row, seat)

Being a weak entity set, Bookings' relation has the keys for Customers and Flights and Bookings' own attributes.

Notice that the relations obtained from the toCust and toFlt relationships are unnecessary. They are:

     toCust(ssNo, ssNo1, number, day)
     toFlt(ssNo, number, day, number1, day1)

That is, for toCust, the key of Customers is paired with the key for Bookings. Since both include ssNo, this attribute is repeated with two different names, ssNo and ssNo1. A similar situation exists for toFlt.

Exercise 3.2.3

     Ships(name, yearLaunched)
     SisterOf(name, sisterName)

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Solutions for Section 3.3

Exercise 3.3.1

Since Courses is weak, its key is number and the name of its department. We do not have a relation for GivenBy. In part (a), there is a relation for Courses and a relation for LabCourses that has only the key and the computer-allocation attribute. It looks like:

     Depts(name, chair)
     Courses(number, deptName, room)
     LabCourses(number, deptName, allocation)

For part (b), LabCourses gets all the attributes of Courses, as:

     Depts(name, chair)
     Courses(number, deptName, room)
     LabCourses(number, deptName, room, allocation)

And for (c), Courses and LabCourses are combined, as:

     Depts(name, chair)
     Courses(number, deptName, room, allocation)

Exercise 3.3.4(a)

There is one relation for each entity set, so the number of relations is e. The relation for the root entity set has a attributes, while the other relations, which must include the key attributes, have a+k attributes.

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Solutions for Section 3.4

Exercise 3.4.2

Surely ID is a key by itself. However, we think that the attributes x, y, and z together form another key. The reason is that at no time can two molecules occupy the same point.

Exercise 3.4.4

The key attributes are indicated by capitalization in the schema below:

     Customers(SSNO, name, address, phone)
     Flights(NUMBER, DAY, aircraft)
     Bookings(SSNO, NUMBER, DAY, row, seat)

Exercise 3.4.6(a)

The superkeys are any subset that contains A1. Thus, there are 2^{n-1} such subsets, since each of the n-1 attributes A2 through An may independently be chosen in or out.

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Solutions for Section 3.5

Exercise 3.5.1(a)

We could try inference rules to deduce new dependencies until we are satisfied we have them all. A more systematic way is to consider the closures of all 15 nonempty sets of attributes.

For the single attributes we have A+ = A, B+ = B, C+ = ACD, and D+ = AD. Thus, the only new dependency we get with a single attribute on the left is C->A.

Now consider pairs of attributes:

AB+ = ABCD, so we get new dependency AB->D. AC+ = ACD, and AC->D is nontrivial. AD+ = AD, so nothing new. BC+ = ABCD, so we get BC->A, and BC->D. BD+ = ABCD, giving us BD->A and BD->C. CD+ = ACD, giving CD->A.

For the triples of attributes, ACD+ = ACD, but the closures of the other sets are each ABCD. Thus, we get new dependencies ABC->D, ABD->C, and BCD->A.

Since ABCD+ = ABCD, we get no new dependencies.

The collection of 11 new dependencies mentioned above is: C->A, AB->D, AC->D, BC->A, BC->D, BD->A, BD->C, CD->A, ABC->D, ABD->C, and BCD->A.

Exercise 3.5.1(b)

From the analysis of closures above, we find that AB, BC, and BD are keys. All other sets either do not have ABCD as the closure or contain one of these three sets.

Exercise 3.5.1(c)

The superkeys are all those that contain one of those three keys. That is, a superkey that is not a key must contain B and more than one of A, C, and D. Thus, the (proper) superkeys are ABC, ABD, BCD, and ABCD.

Exercise 3.5.3(a)

We must compute the closure of A1A2...AnC. Since A1A2...An->B is a dependency, surely B is in this set, proving A1A2...AnC->B.

Exercise 3.5.4(a)

Consider the relation

A	B
0	2
1	2

This relation satisfies A->B but does not satisfy B->A.

Exercise 3.5.8(a)

If all sets of attributes are closed, then there cannot be any nontrivial functional dependencies. For suppose A1A2...An->B is a nontrivial dependency. Then A1A2...An+ contains B and thus A1A2...An is not closed.

Exercise 3.5.10(a)

We need to compute the closures of all subsets of {ABC}, although there is no need to think about the empty set or the set of all three attributes. Here are the calculations for the remaining six sets:

A+ = A
B+ = B
C+ = ACE
AB+ = ABCDE
AC+ = ACE
BC+ = ABCDE

We ignore D and E, so a basis for the resulting functional dependencies for ABC are: C->A and AB->C. Note that BC->A is true, but follows logically from C->A, and therefore may be omitted from our list.

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Solutions for Section 3.6

Exercise 3.6.1(a)

In the solution to Exercise 3.5.1 we found that there are 14 nontrivial dependencies, including the three given ones and 11 derived dependencies. These are: C->A, C->D, D->A, AB->D, AB-> C, AC->D, BC->A, BC->D, BD->A, BD->C, CD->A, ABC->D, ABD->C, and BCD->A.

We also learned that the three keys were AB, BC, and BD. Thus, any dependency above that does not have one of these pairs on the left is a BCNF violation. These are: C->A, C->D, D->A, AC->D, and CD->A.

One choice is to decompose using C->D. That gives us ABC and CD as decomposed relations. CD is surely in BCNF, since any two-attribute relation is. ABC is not in BCNF, since AB and BC are its only keys, but C->A is a dependency that holds in ABCD and therefore holds in ABC. We must further decompose ABC into AC and BC. Thus, the three relations of the decomposition are AC, BC, and CD.

Since all attributes are in at least one key of ABCD, that relation is already in 3NF, and no decomposition is necessary.

Exercise 3.6.1(b)

(Revised 1/19/02) The only key is AB. Thus, B->C and B->D are both BCNF violations. The derived FD's BD->C and BC->D are also BCNF violations. However, any other nontrivial, derived FD will have A and B on the left, and therefore will contain a key.

One possible BCNF decomposition is AB and BCD. It is obtained starting with any of the four violations mentioned above. AB is the only key for AB, and B is the only key for BCD.

Since there is only one key for ABCD, the 3NF violations are the same, and so is the decomposition.

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Solutions for Section 3.7

Exercise 3.7.1

Since A->->B, and all the tuples have the same value for attribute A, we can pair the B-value from any tuple with the value of the remaining attribute C from any other tuple. Thus, we know that R must have at least the nine tuples of the form (a,b,c), where b is any of b1, b2, or b3, and c is any of c1, c2, or c3. That is, we can derive, using the definition of a multivalued dependency, that each of the tuples (a,b1,c2), (a,b1,c3), (a,b2,c1), (a,b2,c3), (a,b3,c1), and (a,b3,c2) are also in R.

Exercise 3.7.2(a)

First, people have unique Social Security numbers and unique birthdates. Thus, we expect the functional dependencies ssNo->name and ssNo->birthdate hold. The same applies to children, so we expect childSSNo->childname and childSSNo->childBirthdate. Finally, an automobile has a unique brand, so we expect autoSerialNo->autoMake.

There are two multivalued dependencies that do not follow from these functional dependencies. First, the information about one child of a person is independent of other information about that person. That is, if a person with social security number s has a tuple with cn,cs,cb, then if there is any other tuple t for the same person, there will also be another tuple that agrees with t except that it has cn,cs,cb in its components for the child name, Social Security number, and birthdate. That is the multivalued dependency

ssNo->->childSSNo childName childBirthdate

Similarly, an automobile serial number and make are independent of any of the other attributes, so we expect the multivalued dependency

ssNo->->autoSerialNo autoMake The dependencies are summarized below:

     ssNo -> name birthdate
     childSSNo -> childName childBirthdate
     autoSerialNo -> autoMake
     ssNo ->-> childSSNo childName childBirthdate
     ssNo ->-> autoSerialNo autoMake

Exercise 3.7.2(b)

We suggest the relation schemas

     {ssNo, name, birthdate}
     {ssNo, childSSNo}
     {childSSNo, childName childBirthdate}
     {ssNo, autoSerialNo}
     {autoSerialNo, autoMake}

An initial decomposition based on the two multivalued dependencies would give us

     {ssNo, name, birthDate}
     {ssNo, childSSNo, childName, childBirthdate}
     {ssNo, autoSerialNo, autoMake}

Functional dependencies force us to decompose the second and third of these.

Exercise 3.7.3(a)

Since there are no functional dependencies, the only key is all four attributes, ABCD. Thus, each of the nontrvial multivalued dependencies A->->B and A->->C violate 4NF. We must separate out the attributes of these dependencies, first decomposing into AB and ACD, and then decomposing the latter into AC and AD because A->->C is still a 4NF violation for ACD. The final set of relations are AB, AC, and AD.

Exercise 3.7.7(a)

Let W be the set of attributes not in X, Y, or Z. Consider two tuples xyzw and xy'z'w' in the relation R in question. Because X ->-> Y, we can swap the y's, so xy'zw and xyz'w' are in R. Because X ->-> Z, we can take the pair of tuples xyzw and xyz'w' and swap Z's to get xyz'w and xyzw'. Similarly, we can take the pair xy'z'w' and xy'zw and swap Z's to get xy'zw' and xy'z'w.

In conclusion, we started with tuples xyzw and xy'z'w' and showed that xyzw' and xy'z'w must also be in the relation. That is exactly the statement of the MVD X ->-> Y-union-Z. Note that the above statements all make sense even if there are attributes in common among X, Y, and Z.

Exercise 3.7.8(a)

Consider a relation R with schema ABCD and the instance with four tuples abcd, abcd', ab'c'd, and ab'c'd'. This instance satisfies the MVD A ->-> BC. However, it does not satisfy A ->-> B. For example, if it did satisfy A ->-> B, then because the instance contains the tuples abcd and ab'c'd, we would expect it to contain abc'd and ab'cd, neither of which is in the instance.

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