数据库系统基础教程(第二版)课后习题答案资源-CSDN文库

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在《数据库系统基础教程(第二版)》中，学习数据库理论和实践操作是至关重要的。本教程涵盖了数据库设计的基础，包括实体-关系模型（E/R模型）、键的概念以及关系的性质。以下是对部分章节习题答案的详细解析：在第2.1节中，主要讨论了如何构建E/R图来表示现实世界的实体和它们之间的关系。练习2.1.1要求绘制E/R图。E/R图是数据库设计的初步阶段，用于可视化地表示实体、属性和联系。在练习2.1.8(a)中，讨论了银行客户和地址的关系。如果每个客户只有一个地址，那么地址应该作为客户实体的一个属性。如果有多个地址，那么需要创建一个“Addresses”实体集，并用“Lives-at”表示多对多关系。同时，Acct-Sets实体集被指出是冗余的，因为它仅仅表示了客户与其账户集合的关系，而这个信息可以通过直接建立客户与账户的多对多关系来表达。第2.2节集中于实体集和它们之间的联系。在练习2.2.1中，讨论了"Addresses"实体集和"Customers"实体集的关系。如果银行记录每个客户的多个地址，那么"Addresses"实体集是有意义的。但是，如果每个客户只有一个地址，将地址作为"Customers"实体集的一个属性更为合适。对于"Acct-Sets"，由于每个客户有唯一的账户集合，可以简化模型，直接建立客户与账户的多对多联系。第2.3节涉及键和关系的基数约束。练习2.3.1(a)中，"ssNo"被识别为"Customers"实体的适当键，"number"为"Accounts"实体的键。关于"Owns"关系，考虑到一个账户不能与零个客户相关联，所以应将其与"Customers"连接的边设置为非空约束。另一方面，允许客户拥有0个账户是合理的，例如，客户可能只是借款人，因此"Owns"到"Accounts"的连接没有基数约束。这在E/R图中通过下划线表示的关键属性和基数约束得以体现。练习2.3.2(b)探讨了多对一关系的键属性。如果关系R是从E1到E2的多对一，那么两个R的关系元组(e1, e2)和(f1, f2)必须相同，如果它们在E1的关键属性上一致。这是因为e1和f1必须相同，而由于R是从E1到E2的多对一，所以e2和f2也必须相同，从而证明了这两个元组是相同的。第2.4节介绍了实体集间的联系和属性。在练习2.4.1中，构建了一个E/R图，显示了"Students"和"Courses"实体的关键属性，以及"Enrollments"实体的键，它由"Students"的"studID"和"Courses"的"dept"和"number"组成。属性"grade"不是"Enrollments"的键。在练习2.4.4b中，再次构建了E/R图，但省略了关系名称和其他非关键属性。通过这些习题，读者可以深化对数据库设计基础的理解，包括如何正确选择键、构建E/R模型以及理解和应用基数约束。这些知识对于数据库设计和管理至关重要，能够帮助开发者创建高效、稳定且适应性强的数据存储解决方案。

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Database Systems: The Complete Book

Solutions for Chapter 2

Solutions for Section 2.1

Exercise 2.1.1

The E/R Diagram.

Exercise 2.1.8(a)

The E/R Diagram

Solutions for Section 2.2

Exercise 2.2.1

The Addresses entity set is nothing but a single address, so we would prefer to make address an

attribute of Customers. Were the bank to record several addresses for a customer, then it might

make sense to have an Addresses entity set and make Lives-at a many-many relationship.

The Acct-Sets entity set is useless. Each customer has a unique account set containing his or her

accounts. However, relating customers directly to their accounts in a many-many relationship

conveys the same information and eliminates the account-set concept altogether.

Solutions for Section 2.3

Exercise 2.3.1(a)

Keys ssNo and number are appropriate for Customers and Accounts, respectively. Also, we think

it does not make sense for an account to be related to zero customers, so we should round the edge

connecting Owns to Customers. It does not seem inappropriate to have a customer with 0

accounts; they might be a borrower, for example, so we put no constraint on the connection from

Owns to Accounts. Here is the The E/R Diagram，

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.

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.

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.

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,

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