数据库系统原理第三章同步练习
数据库系统原理04735课后习题参考答案
数据库系统原理课后习题第一章. 数据库系统基本概念1.1.名词解释DB——DB是长期存储在计算机内、有组织的、统一管理的相关数据的集合。
DB能为各种用户共享，具有较小冗余度、数据间联系紧密而又有较高的数据独立性等特点。
DBMS——是位于用户与操作系统之间的一层数据管理软件，它为用户或应用程序提供访问DB的方法，包括DB的建立、查询、更新及各种数据控制。
DBS——是实现有组织地、动态地存储大量关联数据、方便多用户访问的计算机硬件、软件和数据资源组成的系统，即它是采用数据库技术的计算机系统。
联系——是实体间的相互关系。
联系的元数——与一个联系有关的实体集个数。
１：１联系——如果实体集E1中每个实体至多和实体集E2中一个实体有联系，反之亦然，那么实体集E1和E2的联系称为“一对一联系”，记为“１：１”。
１：N联系——如果实体集E1中的每个实体可以与实体集E2中的任意个（０个或多个）实体有联系，而E2中的每个实体至多和E1中的一个实体有联系，那么称E1对E2的联系是一对多联系，记作：“１：N ”。
M：N联系——如果实体集E１中的每个实体可以与实体集E2中的任意个（０个或多个）实体有联系，反之亦然，那么称E1和E2的联系是“多对多联系”，记作“M：N”。
数据模型——在数据库技术中，我们用数据模型的概念描述数据库的结构和语义，对现实世界的数据进行抽象。
根据数据抽象级别定义了四种模型：概念数据模型、逻辑数据模型、外部数据模型和内部数据模型。
概念模型——表达用户需求观点的数据全局逻辑结构的模型。
逻辑模型——表达计算机实现观点的DB全局逻辑结构的模型。
主要有层次、网状、关系模型等三种。
外部模型——表达用户使用观点的DB局部逻辑结构的模型。
内部模型——表达DB物理结构的模型。
层次模型——用树型（层次）结构表示实体类型及实体间联系的数据模型。
网状模型——用有向图结构表示实体类型及实体间联系的数据模型。
关系模型——是由若干个关系模式组成的集合。
《数据库系统原理》04735课后习题答案(2018版)
答案仅供参考第一章数据库系统概述选择题B、B、A简答题1.请简述数据,数据库,数据库管理系统,数据库系统的概念。
P27数据是描述事物的记录符号，是指用物理符号记录下来的，可以鉴别的信息。
数据库即存储数据的仓库，严格意义上是指长期存储在计算机中的有组织的、可共享的数据集合。
数据库管理系统是专门用于建立和管理数据库的一套软件，介于应用程序和操作系统之间。
数据库系统是指在计算机中引入数据库技术之后的系统，包括数据库、数据库管理系统及相关实用工具、应用程序、数据库管理员和用户。
2.请简述早数据库管理技术中，与人工管理、文件系统相比，数据库系统的优点。
数据共享性高数据冗余小易于保证数据一致性数据独立性高可以实施统一管理与控制减少了应用程序开发与维护的工作量3.请简述数据库系统的三级模式和两层映像的含义。
P31答：数据库的三级模式是指数据库系统是由模式、外模式和内模式三级工程的，对应了数据的三级抽象。
两层映像是指三级模式之间的映像关系，即外模式/模式映像和模式/内模式映像。
4.请简述关系模型与网状模型、层次模型的区别。
P35使用二维表结构表示实体及实体间的联系建立在严格的数学概念的基础上概念单一，统一用关系表示实体和实体之间的联系，数据结构简单清晰，用户易懂易用存取路径对用户透明，具有更高的数据独立性、更好的安全保密性。
第二章关系数据库选择题C、C、D简答题1.请简述关系数据库的基本特征。
P48答：关系数据库的基本特征是使用关系数据模型组织数据。
2.请简述什么是参照完整性约束。
P55答：参照完整性约束是指：若属性或属性组F是基本关系R的外码，与基本关系S的主码K 相对应，则对于R中每个元组在F上的取值只允许有两种可能，要么是空值，要么与S中某个元组的主码值对应。
3.请简述关系规范化过程。
答：对于存在数据冗余、插入异常、删除异常问题的关系模式，应采取将一个关系模式分解为多个关系模式的方法进行处理。
一个低一级范式的关系模式，通过模式分解可以转换为若干个高一级范式的关系模式，这就是所谓的规范化过程。
1数据库原理习题与答案_第3章数据库系统结构（范文）
1数据库原理习题与答案_第3章数据库系统结构（范文）第一篇：1数据库原理习题与答案_第3章数据库系统结构（范文）简答题1．试述数据库系统三级模式结构，这种结构的优点是什么。
答：数据库系统的三级模式结构由外模式、模式和内模式组成。
外模式，亦称子模式或用户模式，是数据库用户能够看见和使用的局部数据的逻辑结构和特征的描述，是数据库用户的数据视图，是与某一应用有关的数据的逻辑表示。
模式，亦称逻辑模式，是数据库中全体数据的逻辑结构和特征的描述，是所有用户的公共数据视图。
模式描述的是数据的全局逻辑结构，外模式涉及的是数据的局部逻辑结构，通常是模式的子集。
内模式，亦称存储模式，是数据在数据库系统内部的表示，即对数据的物理结构和存储方式的描述。
数据库系统的三级模式是对数据的三个抽象级别，它把数据的具体组织留给DBMS管理，使用户能逻辑抽象地处理数据，而不必关心数据在计算机中的表示和存储。
为了能够在内部实现这三个抽象层次的联系和转换，数据库系统在这三级模式之间提供了两层映像：外模式/模式映像和模式/内模式映像，正是这两层映像保证了数据库系统中的数据能够具有较高的逻辑独立性和物理独立性。
2．定义并解释以下术语：模式、外模式、内模式、DDL、DML。
答：模式，亦称逻辑模式，是数据库中全体数据的逻辑结构和特征的描述，是所有用户的公共数据视图。
外模式，亦称子模式或用户模式，是数据库用户能够看见和使用的局部数据的逻辑结构和特征的描述，是数据库用户的数据视图，是与某一应用有关的数据的逻辑表示。
内模式，亦称存储模式，是数据在数据库系统内部的表示，即对数据的物理结构和存储方式的描述。
DDL：数据定义语言，用来定义数据库模式、外模式、内模式的语言。
DML：数据操纵语言，用来对数据库中的数据进行查询、插入、删除和修改的语句。
3．什么叫数据与程序的物理独立性？什么叫数据与程序的逻辑独立性？为什么数据库系统具有数据与程序的独立性？答：数据与程序的逻辑独立性：当模式改变时，由数据库管理员对各个外模式//模式的映像做相应改变，可以使外模式保持不变。
数据库系统基础教程第三章答案
Exercise 3.1.1Answers for this exercise may vary because of different interpretations.Some possible FDs:Social Security number → nameArea code → stateStreet address, city, state → zipcodePossible keys:{Social Security number, street address, city, state, area code, phone number}Need street address, city, state to uniquely determine location. A person couldhave multiple addresses. The same is true for phones. These days, a person couldhave a landline and a cellular phoneExercise 3.1.2Answers for this exercise may vary because of different interpretationsSome possible FDs:ID → x-position, y-position, z-positionID → x-velocity, y-velocity, z-velocityx-position, y-position, z-position → IDPossible keys:{ID}{x-position, y-position, z-position}The reason why the positions would be a key is no two molecules can occupy the same point.Exercise 3.1.3aThe superkeys are any subset that contains A1. Thus, there are 2(n-1) such subsets, since each of the n-1 attributes A2 through A n may independently be chosen in or out.Exercise 3.1.3bThe superkeys are any subset that contains A1 or A2. There are 2(n-1) such subsets when considering A1 and the n-1 attributes A2 through A n. There are 2(n-2) such subsets when considering A2 and the n-2 attributes A3 through A n. We do not count A1 in these subsets because they are already counted in the first group of subsets. The total number of subsets is 2(n-1) + 2(n-2).Exercise 3.1.3cThe superkeys are any subset that contains {A1,A2} or {A3,A4}. There are 2(n-2) such subsets when considering {A1,A2} and the n-2 attributes A3 through A n. There are 2(n-2) – 2(n-4) such subsets when considering {A3,A4} and attributes A5 through A n along with the individual attributes A1 and A2. We get the 2(n-4) term because we have to discard the subsets that contain the key {A1,A2} to avoid double counting. The total number of subsets is 2(n-2) +2(n-2) – 2(n-4).Exercise 3.1.3dThe superkeys are any subset that contains {A1,A2} or {A1,A3}. There are 2(n-2) such subsets when considering {A1,A2} and the n-2 attributes A3 through A n. There are 2(n-3) such subsets when considering {A1,A3} and the n-3 attributes A4 through A n We do not count A2 in these subsets because they are already counted in the first group of subsets. The total number of subsets is 2(n-2) + 2(n-3).Exercise 3.2.1aWe 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 are:C→A, AB→D, AC→D, BC→A, BC→D, BD→A, BD→C, CD→A, ABC→D, ABD→C, and BCD→A.Exercise 3.2.1bFrom 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 3.2.1cThe superkeys are all those that contain one of those three keys. That is, a superkeythat 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 3.2.2ai) For the single attributes we have {A}+ = ABCD, {B}+ = BCD, {C}+ = C, and {D}+ = D. Thus, the new dependencies are A→C and A→D.Now consider pairs of attributes:{AB}+ = ABCD, {AC}+ = ABCD, {AD}+ = ABCD, {BC}+ = BCD, {BD}+ = BCD, {CD}+ = CD. Thus the new dependencies are AB→C, AB→D, AC→B, AC→D, AD→B, AD→C, BC→D and BD→C.For the triples of attributes, {BCD}+ = BCD, but the closures of the other sets are each ABCD. Thus, we get new dependencies ABC→D, ABD→C, and ACD→B.Since {ABCD}+ = ABCD, we get no new dependencies.The collection of 13 new dependencies mentioned above are:A→C, A→D, AB→C, AB→D, AC→B, AC→D, AD→B, AD→C, BC→D, BD→C, ABC→D, ABD→C and ACD→B.ii) For the single attributes we have {A}+ = A, {B}+ = B, {C}+ = C, and {D}+ = D. Thus, there are no new dependencies.Now consider pairs of attributes:{AB}+ = ABCD, {AC}+ = AC, {AD}+ = ABCD, {BC}+ = ABCD, {BD}+ = BD, {CD}+ = ABCD. Thus the new dependencies are AB→D, AD→C, BC→A and CD→B.For the triples of attributes, all the closures of the sets are each ABCD. Thus, we get new dependencies ABC→D, ABD→C, ACD→B and BCD→A.Since {ABCD}+ = ABCD, we get no new dependencies.The collection of 8 new dependencies mentioned above are:AB→D, AD→C, BC→A, CD→B, ABC→D, ABD→C, ACD→B and BCD→A.iii) For the single attributes we have {A}+ = ABCD, {B}+ = ABCD, {C}+ = ABCD, and {D}+ = ABCD. Thus, the new dependencies are A→C, A→D, B→D, B→A, C→A, C→B, D→B and D→C.Since all the single attributes’ closures are ABCD, any superset of the singleattributes will also lead to a closure of ABCD. Knowing this, we can enumerate the restof the new dependencies.The collection of 24 new dependencies mentioned above are:A→C, A→D, B→D, B→A, C→A, C→B, D→B, D→C, AB→C, AB→D, AC→B, AC→D, AD→B, AD→C, BC→A, BC→D, BD→A, BD→C, CD→A, CD→B, ABC→D, ABD→C, ACD→B and BCD→A.Exercise 3.2.2bi) From the analysis of closures in 3.2.2a(i), we find that the only key is A. All other sets either do not have ABCD as the closure or contain A.ii) From the analysis of closures 3.2.2a(ii), we find that AB, AD, BC, and CD are keys.All other sets either do not have ABCD as the closure or contain one of these four sets.iii) From the analysis of closures 3.2.2a(iii), we find that A, B, C and D are keys. All other sets either do not have ABCD as the closure or contain one of these four sets.Exercise 3.2.2ci) The superkeys are all those sets that contain one of the keys in 3.2.2b(i). The superkeys are AB, AC, AD, ABC, ABD, ACD, BCD and ABCD.ii) The superkeys are all those sets that contain one of the keys in 3.2.2b(ii). The superkeys are ABC, ABD, ACD, BCD and ABCD.iii) The superkeys are all those sets that contain one of the keys in 3.2.2b(iii). The superkeys are AB, AC, AD, BC, BD, CD, ABC, ABD, ACD, BCD and ABCD.Exercise 3.2.3aSince A1A2…A n C contains A1A2…A n, then the closure of A1A2…A n C contains B. Thus it follows that A1A2…A n C→B.Exercise 3.2.3bFrom 3.2.3a, we know that A1A2…A n C→B. Using the concept of trivial dependencies, we can show that A1A2…A n C→C. Thus A1A2…A n C→BC.Exercise 3.2.3cFrom A1A2…A n E1E2…E j, we know that the closure contains B1B2…B m because of the FD A1A2…A n→B1B2…B m. The B1B2…B m and the E1E2…E j combine to form the C1C2…C k. Thus the closure ofA1A2…A n E1E2…E j contains D as well. Thus, A1A2…A n E1E2…E j→D.Exercise 3.2.3dFrom A1A2…A n C1C2…C k, we know that the closure contains B1B2…B m because of the FD A1A2…A n→B1B2…B m. The C1C2…C k also tell us that the closure of A1A2…A n C1C2…C k contains D1D2…D j. Thus, A1A2…A n C1C2…C k→B1B2…B k D1D2…D j.Exercise 3.2.4aIf attribute A represented Social Security Number and B represented a person’s name,then we would assume A→B but B→A would not be valid because there may be many peoplewith the same name and different Social Security Numbers.Exercise 3.2.4bLet attribute A represent Social Security Number, B represent gender and C represent name. Surely Social Security Number and gender can uniquely identify a person’s name . AB→C).A Social Security Number can also uniquely identify a person’s name . A→C). However, gender does not uniquely determine a name . B→C is not valid).Exercise 3.2.4cLet attribute A represent latitude and B represent longitude. Together, both attributes can uniquely determine C, a point on the world map . AB→C). However, neither A nor B can uniquely identify a point . A→C and B→C are not valid).Exercise 3.2.5Given a relation with attributes A1A2…A n, we are told that there are no functional dependencies of the form B1B2…B n-1→C where B1B2…B n-1 is n-1 of the attributes from A1A2…A n and C is the remaining attribute from A1A2…A n. In this case, the set B1B2…B n-1 and any subset do not functionally determine C. Thus the only functional dependencies that we can make are ones where C is on both the left and right hand sides. All of these functional dependencies would be trivial and thus the relation has no nontrivial FD’s.Exercise 3.2.6Let’s prove this by using the contrapositive. We wish to show that if X+ is not a subset of Y+, then it must be that X is not a subset of Y.If X+ is not a subset of Y+, there must be attributes A1A2…A n in X+ that are not in Y+. If any of these attributes were originally in X, then we are done because Y does not contain any of the A1A2…A n. However, if the A1A2…A n were added by the closure, then we must examine the case further. Assume that there was some FD C1C2…C m→A1A2…A j where A1A2…A j is some subset of A1A2…A n. It must be then that C1C2…C m or some subset of C1C2…C m is in X. However, the attributes C1C2…C m cannot be in Y because we assumed that attributes A1A2…A n are only in X+ and are not in Y+. Thus, X is not a subset of Y.By proving the contrapositive, we have also proved if X ⊆ Y, then X+⊆ Y+.Exercise 3.2.7The algorithm to find X+ is outlined on pg. 76. Using that algorithm, we can prove that(X+)+ = X+. We will do this by using a proof by contradiction.Suppose that (X+)+≠ X+. Then for (X+)+, it must be that some FD allowed additional attributes to be added to the original set X+. For example, X+→ A where A is some attribute not in X+. However, if this were the case, then X+ would not be the closure of X. The closure of X would have to include A as well. This contradicts the fact that we were given the closure of X, X+. Therefore, it must be that (X+)+ = X+ or else X+ is not the closure of X.Exercise 3.2.8aIf all sets of attributes are closed, then there cannot be any nontrivial functional dependencies. Suppose A1A2...A n→B is a nontrivial dependency. Then {A1A2...A n}+ contains B and thus A1A2...A n is not closed.Exercise 3.2.8bIf the only closed sets are ø and {A,B,C,D}, then the following FDs hold:A→B A→C A→DB→A B→C B→DC→A C→B C→DD→A D→B D→CAB→C AB→DAC→B AC→DAD→B AD→CBC→A BC→DBD→A BD→CCD→A CD→BABC→DABD→CACD→BBCD→AExercise 3.2.8cIf the only closed sets are ø, {A,B} and {A,B,C,D}, then the following FDs hold:A→BB→AC→A C→B C→DD→A D→B D→CAC→B AC→DAD→B AD→CBC→A BC→DBD→A BD→CCD→A CD→BABC→DABD→CACD→BBCD→AExercise 3.2.9We can think of this problem as a situation where the attributes A,B,C represent cities and the functional dependencies represent one way paths between the cities. The minimal bases are the minimal number of pathways that are needed to connect the cities. We do not want to create another roadway if the two cities are already connected.The systematic way to do this would be to check all possible sets of the pathways. However, we can simplify the situation by noting that it takes more than two pathways to visit the two other cities and come back. Also, if we find a set of pathways that is minimal, adding additional pathways will not create another minimal set.The two sets of minimal bases that were given in example are:{A→B, B→C, C→A}{A→B, B→A, B→C, C→B}The additional sets of minimal bases are:{C→B, B→A, A→C}{A→B, A→C, B→A, C→A}{A→C, B→C, C→A, C→B}Exercise 3.2.10aWe 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}+=ABCDEWe ignore D and E, so a basis for the resulting functional dependencies for ABC is: 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.Exercise 3.2.10bWe need to compute the closures of all subsets of {ABC}, although there is no need tothink about the empty set or the set of all three attributes. Here are the calculationsfor the remaining six sets:{A}+=AD{B}+=B{C}+=C{AB}+=ABDE{AC}+=ABCDE{BC}+=BCWe ignore D and E, so a basis for the resulting functional dependencies for ABC is: AC→B.Exercise 3.2.10cWe need to compute the closures of all subsets of {ABC}, although there is no need tothink about the empty set or the set of all three attributes. Here are the calculationsfor the remaining six sets:{A}+=A{B}+=B{C}+=C{AB}+=ABD{AC}+=ABCDE{BC}+=ABCDEWe ignore D and E, so a basis for the resulting functional dependencies for ABC is: AC→B and BC→A.Exercise 3.2.10dWe need to compute the closures of all subsets of {ABC}, although there is no need tothink about the empty set or the set of all three attributes. Here are the calculationsfor the remaining six sets:{A}+=ABCDE{B}+=ABCDE{C}+=ABCDE{AB}+=ABCDE{AC}+=ABCDE{BC}+=ABCDEWe ignore D and E, so a basis for the resulting functional dependencies for ABC is: A→B, B→C and C→A.Exercise 3.2.11For step one of Algorithm , suppose we have the FD ABC→DE. We want to use Armstrong’s Axioms to show that ABC→D and ABC→E follow. Surely the functional dependencies DE→Dand DE→E hold because they are trivial and follow the reflexivity property. Using the transitivity rule, we can derive the FD ABC→D from the FDs ABC→DE and DE→D. Likewise, we can do the same for ABC→DE and DE→E and derive the FD ABC→E.For steps two through four of Algorithm , suppose we have the initial set of attributesof the closure as ABC. Suppose also that we have FDs C→D and D→E. According toAlgorithm , the closure should become ABCDE. Taking the FD C→D and augmenting both sides with attributes AB we get the FD ABC→ABD. We can use the splitting method in step one to get the FD ABC→D. Since D is not in the closure, we can add attribute D. Taking the FDD→E and augmenting both sides with attributes ABC we get the FD ABCD→ABCDE. Using again the splitting method in step one we get the FD ABCD→E. Since E is not in the closure, we can add attribute E.Given a set of FDs, we can prove that a FD F follows by taking the closure of the left side of FD F. The steps to compute the closure in Algorithm can be mimicked by Armstrong’s axioms and thus we can prove F from the given set of FDs using Armstrong’s axioms.Exercise 3.3.1aIn the solution to Exercise 3.2.1 we found that there are 14 nontrivial dependencies, including the three given ones and eleven derived dependencies. They 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 the violation C→D. Using the above FDs, we get ACD and BC as decomposed relations. BC is surely in BCNF, since any two-attribute relation is. Using Algorithm to discover the projection of FDs on relation ACD, we discover that ACDis not in BCNF since C is its only key. However, D→A is a dependency that holds in ABCD and therefore holds in ACD. We must further decompose ACD into AD and CD. Thus, the three relations of the decomposition are BC, AD, and CD.Exercise 3.3.1bBy computing the closures of all 15 nonempty subsets of ABCD, we can find all thenontrivial FDs. They are B→C, B→D, AB→C, AB→D, BC→D, BD→C, ABC→D and ABD→C. From the closures we can also deduce that the only key is AB. Thus, any dependency above that does not contain AB on the left is a BCNF violation. These are: B→C, B→D, BC→D andBD→C.One choice is to decompose using the violation B→C. Using the above FDs, we get BCD and AB as decomposed relations. AB is surely in BCNF, since any two-attribute relation is. Using Algorithm to discover the projection of FDs on relation BCD, we discover that BCDis in BCNF since B is its only key and the projected FDs all have B on the left side. Thus the two relations of the decomposition are AB and BCD.Exercise 3.3.1cIn the solution to Exercise 3.2.2(ii), we found that there are 12 nontrivial dependencies, including the four given ones and the eight derived ones. They are AB→C, BC→D, CD→A, AD→B, AB→D, AD→C, BC→A, CD→B, ABC→D, ABD→C, ACD→B and BCD→A.We also found out that the keys are AB, AD, BC, and CD. Thus, any dependency above that does not have one of these pairs on the left is a BCNF violation. However, all of the FDs contain a key on the left so there are no BCNF violations.No decomposition is necessary since all the FDs do not violate BCNF.Exercise 3.3.1dIn the solution to Exercise 3.2.2(iii), we found that there are 28 nontrivial dependencies, including the four given ones and the 24 derived ones. They are A→B, B→C, C→D, D→A, A→C, A→D, B→D, B→A, C→A, C→B, D→B, D→C, AB→C, AB→D, AC→B, AC→D,AD→B, AD→C, BC→A, BC→D, BD→A, BD→C, CD→A, CD→B, ABC→D, ABD→C, ACD→B and BCD→A.We also found out that the keys are A,B,C,D. Thus, any dependency above that does nothave one of these attributes on the left is a BCNF violation. However, all of the FDs contain a key on the left so there are no BCNF violations.No decomposition is necessary since all the FDs do not violate BCNF.Exercise 3.3.1eBy computing the closures of all 31 nonempty subsets of ABCDE, we can find all the nontrivial FDs. They are AB→C, DE→C, B→D, AB→D, BC→D, BE→C, BE→D, ABC→D, ABD→C, ABE→C, ABE→D, ADE→C, BCE→D, BDE→C, ABCE→D, and ABDE→C. From the closures we canalso deduce that the only key is ABE. Thus, any dependency above that does not contain ABE on the left is a BCNF violation. These are: AB→C, DE→C, B→D, AB→D, BC→D, BE→C, BE→D, ABC→D, ABD→C, ADE→C, BCE→D and BDE→C.One choice is to decompose using the violation AB→C. Using the above FDs, we get ABCDand ABE as decomposed relations. Using Algorithm to discover the projection of FDs on relation ABCD, we discover that ABCD is not in BCNF since AB is its only key and the FDB→D follows for ABCD. Using violation B→D to further decompose, we get BD and ABC as decomposed relations. BD is in BCNF because it is a two-attribute relation. Using Algorithm again, we discover that ABC is in BCNF since AB is the only key and AB→C isthe only nontrivial FD. Going back to relation ABE, following Algorithm tells us that ABE is in BCNF because there are no keys and no nontrivial FDs. Thus the three relationsof the decomposition are ABC, BD and ABE.Exercise 3.3.1fBy computing the closures of all 31 nonempty subsets of ABCDE, we can find all the nontrivial FDs. They are: C→B, C→D, C→E, D→B, D→E, AB→C, AB→D, AB→E, AC→B, AC→D, AC→E, AD→B, AD→C, AD→E, BC→D, BC→E, BD→E, CD→B, CD→E, CE→B, CE→D, DE→B,ABC→D, ABC→E, ABD→C, ABD→E, ABE→C, ABE→D, ACD→B, ACD→E, ACE→B, ACE→D, ADE→B, ADE→C, BCD→E, BCE→D, CDE→B, ABCD→E, ABCE→D, ABDE→C and ACDE→B. From the closures we can also deduce that the keys are AB, AC and AD. Thus, any dependency above that does not contain one of the above pairs on the left is a BCNF violation. These are: C→B, C→D, C→E, D→B, D→E, BC→D, BC→E, BD→E, CD→B, CD→E, CE→B, CE→D, DE→B, BCD→E, BCE→D and CDE→B.One choice is to decompose using the violation D→B. Using the above FDs, we get BDE and ABC as decomposed relations. Using Algorithm to discover the projection of FDs onrelation BDE, we discover that BDE is in BCNF since D, BD, DE are the only keys and all the projected FDs contain D, BD, or DE in the left side. Going back to relation ABC, following Algorithm tells us that ABC is not in BCNF because since AB and AC are itsonly keys and the FD C→B follows for ABC. Using violation C→B to further decompose, we get BC and AC as decomposed relations. Both BC and AC are in BCNF because they are two-attribute relations. Thus the three relations of the decomposition are BDE, BC and AC.Exercise 3.3.2Yes, we will get the same result. Both A→B and A→BC have A on the left side and part of the process of decomposition involves finding {A}+ to form one decomposed relation and A plus the rest of the attributes not in {A}+ as the second relation. Both cases yield the same decomposed relations.Exercise 3.3.3Yes, we will still get the same result. Both A→B and A→BC have A on the left side and part of the process of decomposition involves finding {A}+ to form one decomposedrelation and A plus the rest of the attributes not in {A}+ as the second relation. Both cases yield the same decomposed relations.Exercise 3.3.4This is taken from Example pg. 95.Suppose that an instance of relation R only contains two tuples.A B C1 2 3425The projections of R onto the relations with schemas {A,B} and {B,C} are:A B1 2 42If we do a natural join on the two projections, we will get:A B C1 2 3 1 2 5 4 2 3 425The result of the natural join is not equal to the original relation R.Exercise 3.4.1aThis is the initial tableau:A B C D Ea b c d 1 e 1 a 1 b c d e 1 ab 1cd 1e→A.A B C D Ea b c d 1 e 1 a b c d e 1 ab 1cd 1eSince there is not an unsubscripted row, the decomposition for R is not lossless for this set of FDs.We can use the final tableau as an instance of R as an example for why the join is not lossless. The projected relations are:B C 2 3 25A B Ca b ca b1 cB C Db c d1b c db1 c d1A C Ea c e1a c eThe joined relation is:A B C D Ea b c d1 e1a b c d e1a b1 c d1 e1a b c d1 ea b c d ea b1 c d1 eThe joined relation has three more tuples than the original tableau. Exercise 3.4.1bThis is the initial tableau:A B C D Ea b c d1 e1a1 b c d e1a b1 c d1 eThis is the final tableau after applying FDs AC→E and BC→DA B C D Ea b c d ea1 b c d e1a b1 c d1 eSince there is an unsubscripted row, the decomposition for R is lossless for this set of FDs.Exercise 3.4.1cThis is the initial tableau:A B C D Ea b c d1 e1a1 b c d e1a b1 c d1 eThis is the final tableau after applying FDs A→D, D→E and B→D.A B C D Ea b c d ea1 b c d ea b1 c d eSince there is an unsubscripted row, the decomposition for R is lossless for this set of FDs.Exercise is the initial tableau:A B C D Ea b c d1 e1a1 b c d e1a b1 c d1 eThis is the final tableau after applying FDs A→D, CD→E and E→DA B C D Ea b c d ea1 b c d ea b1 c d eSince there is an unsubscripted row, the decomposition for R is lossless for this set of FDs.Exercise we decompose a relation into BCNF, we will project the FDs onto the decomposed relations to get new sets of FDs. These dependencies are preserved if the union of these new sets is equivalent to the original set of FDs.For the FDs of the dependencies are not preserved. The union of the new sets of FDs is CE→A. However, the FD B→E is not in the union and cannot be derived. Thus the two sets of FDs are not equivalent.For the FDs of the dependencies are preserved. The union of the new sets of FDs is AC→E and BC→D. This is precisely the same as the original set of FDs and thus the two sets of FDs are equivalent.For the FDs of the dependencies are not preserved. The union of the new sets of FDs isB→D and A→E. The FDs A→D and D→E are not in the union and cannot be derived. Thus the two sets of FDs are not equivalent.For the FDs of the dependencies are not preserved. The union of the new sets of FDs is AC→E. However, the FDs A→D, CD→E and E→D are not in the union and cannot be derived. Thus the two sets of FDs are not equivalent.Exercise the solution to Exercise we found that there are 14 nontrivial dependencies. They 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. Since all the attributes on the right sides of the FDs are prime, there are no 3NF violations.Since there are no 3NF violations, it is not necessary to decompose the relation.Exercise the solution to Exercise we found that there are 8 nontrivial dependencies. They are B→C, B→D, AB→C, AB→D, BC→D, BD→C, ABC→D and ABD→C.We also found out that the only key is AB. FDs where the left side is not a superkey or the attributes on the right are not part of some key are 3NF violations. The 3NF violations are B→C, B→D, BC→D and BD→C.Using algorithm , we can decompose into relations using the minimal basis B→C and B→D. The resulting decomposed relations would be BC and BD. However, none of these two sets of attributes is a superkey. Thus we add relation AB to the result. The final set of decomposed relations is BC, BD and AB.Exercise the solution to Exercise we found that there are 12 nontrivial dependencies. They are AB→C, BC→D, CD→A, AD→B, AB→D, AD→C, BC→A, CD→B, ABC→D, ABD→C, ACD→B and BCD→A.We also found out that the keys are AB, AD, BC, and CD. Since all the attributes on the right sides of the FDs are prime, there are no 3NF violations.Since there are no 3NF violations, it is not necessary to decompose the relation.Exercise the solution to Exercise we found that there are 28 nontrivial dependencies. They are A→B, B→C, C→D, D→A, A→C, A→D, B→D, B→A, C→A, C→B, D→B, D→C, AB→C, AB→D, AC→B, AC→D, AD→B, AD→C, BC→A, BC→D, BD→A, BD→C, CD→A, CD→B, ABC→D,ABD→C, ACD→B and BCD→A.We also found out that the keys are A,B,C,D. Since all the attributes on the right sides of the FDs are prime, there are no 3NF violations.Since there are no 3NF violations, it is not necessary to decompose the relation.Exercise the solution to Exercise we found that there are 16 nontrivial dependencies. They are AB→C, DE→C, B→D, AB→D, BC→D, BE→C, BE→D, ABC→D, ABD→C, ABE→C, ABE→D, ADE→C, BCE→D, BDE→C, ABCE→D, and ABDE→C.We also found out that the only key is ABE. FDs where the left side is not a superkey or the attributes on the right are not part of some key are 3NF violations. The 3NF violations are AB→C, DE→C, B→D, AB→D, BC→D, BE→C, BE→D, ABC→D, ABD→C, ADE→C, BCE→D and BDE→C.Using algorithm , we can decompose into relations using the minimal basis AB→C, DE→C and B→D. The resulting decomposed relations would be ABC, CDE and BD. However, none of these three sets of attributes is a superkey. Thus we add relation ABE to the result. The final set of decomposed relations is ABC, CDE, BD and ABE.Exercise the solution to Exercise we found that there are 41 nontrivial dependencies. They are: C→B, C→D, C→E, D→B, D→E, AB→C, AB→D, AB→E, AC→B, AC→D, AC→E, AD→B, AD→C, AD→E, BC→D, BC→E, BD→E, CD→B, CD→E, CE→B, CE→D, DE→B, ABC→D, ABC→E, ABD→C, ABD→E, ABE→C, ABE→D, ACD→B, ACD→E, ACE→B, ACE→D, ADE→B, ADE→C, BCD→E, BCE→D, CDE→B, ABCD→E, ABCE→D, ABDE→C and ACDE→B.We also found out that the keys are AB, AC and AD. FDs where the left side is not a superkey or the attributes on the right are not part of some key are 3NF violations. The 3NF violations are C→E, D→E, BC→E, BD→E, CD→E and BCD→E.Using algorithm , we can decompose into relations using the minimal basis AB→C, C→D,D→B and D→E. The resulting decomposed relations would be ABC, CD, BD and DE. Since。
数据库第三章习题及答案
第3章关系数据库标准语言SQL一、选择题1、SQL语言是的语言，易学习。
A．过程化 B．非过程化 C．格式化 D．导航式答案：B2、SQL语言是语言。
A．层次数据库 B．网络数据库 C．关系数据库 D．非数据库答案：C3、SQL语言具有的功能。
A．关系规范化、数据操纵、数据控制 B．数据定义、数据操纵、数据控制C．数据定义、关系规范化、数据控制 D．数据定义、关系规范化、数据操纵答案：B4、SQL语言具有两种使用方式，分别称为交互式SQL和。
A．提示式SQL B．多用户SQL C．嵌入式SQL D．解释式SQL 答案：C5、假定学生关系是S(S#，SNAME，SEX，AGE)，课程关系是C(C#，CNAME，TEACHER)，学生选课关系是SC(S#，C#，GRADE)。
要查找选修“COMPUTER”课程的“女”学生姓名，将涉及到关系。
A．S B．SC，C C．S，SC D．S，C，SC 答案：D6、若用如下的SQL语句创建一个student表：CREATE TABLE student(NO C(4) NOT NULL，NAME C(8) NOT NULL，SEX C(2)，AGE N(2))可以插入到student表中的是。
A．(‘1031’，‘曾华’，男，23) B．(‘1031’，‘曾华’，NULL，NULL)C．(NULL，‘曾华’，‘男’，‘23’) D．(‘1031’，NULL，‘男’，23) 答案：B7、当两个子查询的结果时，可以执行并，交，差操作．Ａ.结构完全不一致 B.结构完全一致C.结构部分一致D.主键一致答案：B第8到第10题基于这样的三个表即学生表S、课程表C和学生选课表SC，它们的结构如下：S(S#，SN，SEX，AGE，DEPT)C(C#，CN)SC(S#，C#，GRADE)其中：S#为学号，SN为姓名，SEX为性别，AGE为年龄，DEPT为系别，C#为课程号，CN为课程名，GRADE为成绩。
数据库原理及应用教程第3版课后题答案
第一章习题参考答案一、选择题1. C2. B3. D4. C5. D6. A7. A8. B9. D 10. B11. C 12. D 13. A 14. D 15. B16. C 17. D 18. A 19. D 20. A二、填空题1. 数据库系统阶段2. 关系3. 物理独立性4. 操作系统5. 数据库管理系统（DBMS）6. 一对多7. 独立性8. 完整性控制9. 逻辑独立性10. 关系模型11. 概念结构（逻辑）12. 树有向图二维表嵌套和递归13. 宿主语言（或主语言）14. 数据字典15. 单用户结构主从式结构分布式结构客户/服务器结构浏览器/服务器结构第2章习题参考答案一、选择题1. A2. C3. C4. B5. B6. C7. B8. D9. C 10. A11. B 12. A 13. A 14. D 15. D二、填空题1. 选择（选取）2. 交3. 相容（或是同类关系）4. 并差笛卡尔积选择投影5. 并差交笛卡尔积6. 选择投影连接7. σf(R)8. 关系代数关系演算9. 属性10. 同质11. 参照完整性12. 系编号，系名称，电话办公地点13. 元组关系域关系14. 主键外部关系键15. R和S没有公共的属性第3章习题参考答案一、选择题1. B2. A3. C4. B5. C6. C7. B8. D9. A 10. D二、填空题结构化查询语言（Structured Query Language）数据查询、数据定义、数据操纵、数据控制外模式、模式、内模式数据库、事务日志NULL/NOT NULL、UNIQUE约束、PRIMARY KEY约束、FOREIGN KEY约束、CHECK 约束聚集索引、非聚集索引连接字段行数定义系统权限、对象权限基本表、视图12．（1）INSERT INTO S VALUES('990010','李国栋','男',19)（2）INSERT INTO S(No,Name) VALUES('990011', '王大友')（3）UPDATE S SET Name='陈平' WHERE No='990009'（4）DELETE FROM S WHERE No='990008'（5）DELETE FROM S WHERE Name LIKE '陈%'13．CHAR(8) NOT NULL14．o=o15．ALTER TABLE StudentADDSGrade CHAR(10)第4章习题参考答案一、选择题1. B2. B3. D4. B5. C6. D7. B8. D9. C 10. A二、填空题1. 超键（或超码）2. 正确完备3. 属性集X的闭包X + 函数依赖集F的闭包F +4. 平凡的函数依赖自反性5. {AD→C} φ6. 2NF 3NF BCNF7. 无损连接保持函数依赖8. AB BC BD9. B→φ B→B B→C B→BC10. B→C A→D D→C11. AB 1NF12. AD 3NF第5章习题参考答案一、选择题1. B2. B3. C4. A5. C6. D7. A8. C9. D 10. D11. B 12. B 13. A 14. D 15. A二、填空题安全性控制、完整性控制、并发性控制、数据库恢复数据对象、操作类型授权粒度、授权表中允许的登记项的范围原始数据（或明文）、不可直接识别的格式（或密文）、密文事务、原子性、一致性、隔离性、持久性丢失更新、污读、不可重读封锁、排它型封锁、共享封锁利用数据的冗余登记日志文件、数据转储事务故障、系统故障、介质故障完整性登录账号、用户账号public服务器、数据库第6章习题参考答案一、选择题1. B2. C3. C4. A5. C6. B7. C8. B9. D 10. C11. D 12. B 13. B 14. D二、填空题数据库的结构设计、数据库的行为设计新奥尔良法分析和设计阶段、实现和运行阶段需求分析概念结构设计自顶向下、自底向上属性冲突、命名冲突、结构冲突逻辑结构设计确定物理结构、评价物理结构数据库加载运行和维护物理数据字典需求分析载入第7章习题参考答案一、选择题1. B2.C3.B4.D5.A二、填空题局部变量、全局变量- -、/*……*/DECLARESQL、流程控制AFTER 触发器、INSTEAD OF 触发器插入表、删除表数据库备份、事务日志备份、差异备份、文件和文件组备份简单还原、完全还原、批日志还原硬盘、磁带、管道下面是古文鉴赏，不需要的朋友可以下载后编辑删除！！谢谢！！九歌·湘君屈原朗诵：路英君不行兮夷犹，蹇谁留兮中洲。
数据库系统原理(2018年版)课后习题参考答案解析
第三代数据库系统必须保持或集成第二代数据库系统技术
第三代数据库系统必须对其他系统开放
2.描述数据仓库粒度的概念P182
粒度是指数据仓库数据单位中保存数据的细化或综合程度
3.描述数据挖掘技术的能P183
概念描述
关联分析
分类与预测
聚类
孤立点检测
趋势和演变分析
视图的内容是由存储在数据库中进行查询操作的SQL语句定义的，它的列数据与行数据均来自于定义视图的查询所引用的基本表。
视图不适宜数据集的形式存储在数据库中的，它所对应的数据实际上是存储在视图所引用的基本表中的。
视图是用来查看存储在别处的数据的一种虚拟表，本身不存储数据。
第五章 数据库编程
简答题
1.请简述存储过程的概念P125
答：参照完整性约束是指：若属性或属性组F是基本关系R的外码，与基本关系S的主码K相对应，则对于R中每个元组在F上的取值只允许有两种可能，要么是空值，要么与S中某个元组的主码值对应。
3.请简述关系规范化过程。
答：对于存在数据冗余、插入异常、删除异常问题的关系模式，应采取将一个关系模式分解为多个关系模式的方法进行处理。一个低一级范式的关系模式，通过模式分解可以转换为若干个高一级范式的关系模式，这就是所谓的规范化过程。
数据库系统是指在计算机中引入数据库技术之后的系统，包括数据库、数据库管理系统及相关实用工具、应用程序、数据库管理员和用户。
2.请简述早数据库管理技术中，与人工管理、文件系统相比，数据库系统的优点。
数据共享性高
数据冗余小
易于保证数据一致性
数据独立性高
可以实施统一管理与控制
减少了应用程序开发与维护的工作量
数据库系统原理(2018版)课后习题参考答案
选择题
A、D、D
填空题
数据库系统就是指在计算机中引入数据库技术之后得系统，包括数据库、数据库管理系统及相关实用工具、应用程序、数据库管理员与用户。
2、请简述早数据库管理技术中，与人工管理、文件系统相比，数据库系统得优点。
数据共享性高
数据冗余小
易于保证数据一致性
数据独立性高
可以实施统一管理与控制
减少了应用程序开发与维护得工作量
一对多联系：对于实体集A中得每个实体，实体集B中有N个实体与之联系，反之，对于实体集B中得每个实体，实体集A中最多只有一个实体与之联系。举例：班级与班级成员，每个班级对应多个班级成员，每个班级成员只对应一个班级。
多对多联系：对于实体集A中得每个实体，实体集B中有N个实体与之联系，反之，对于实体集B中得每个实体，实体集A中有M个实体与之联系。举例：授课班级与任课教师，每个班级对应多名任课教师，每个任课教师也对应多个授课班级。
答案仅供参考
第一章 数据库系统概述
选择题
B、B、A
简答题
1、请简述数据,数据库,数据库管理系统,数据库系统得概念。 P27
数据就是描述事物得记录符号，就是指用物理符号记录下来得，可以鉴别得信息。
数据库即存储数据得仓库，严格意义上就是指长期存储在计算机中得有组织得、可共享得数据集合。
数据库管理系统就是专用于建立与管理数据库得一套软件，介于应用程序与操作系统之间。
建立在严格得数学概念得基础上
概念单一，统一用关系表示实体与实体之间得联系，数据结构简单清晰，用户易懂易用
存取路径对用户透明，具有更高得数据独立性、更好得安全保密性。
第二章 关系数据库
选择题
C、C、D
简答题
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性。
8. 消除了非主属性对候选键局部依赖的关系模式， 9. 两个函数依赖集F和G等价的充分必要条件是
10. 消除了每一属性对候选键传递依赖的关系模
式称为 BCNF 模式
11. 一个关系模式属于 4NF ，它必定属于BCNF。
A. 互不相关的
B. 不可分解的
C. 长度可变的
D. 互相关联的
6. 假设关系模式R(A,B)属于3NF，下列说法（ B ）
是正确的
A. 它一定消除了插入和删除异常
B. 仍存在一定的插入和删除异常
C. 一定属于BCNF
D. A和C
7. 设有关系W(工号, 姓名, 工种, 定额), 将其规范
化到第三范式正确的答案是（ C ）
1NF变成了3NF
A. 局部函数依赖和传递函数依赖
B. 完全函数依赖和传递函数依赖
C. 完全函数依赖
D. 局部函数依赖
13. 下述说法正确的是（ D ）
A. 属于BCNF的关系模式不存在存储异常
B. 函数依赖可由属性值决定，不由语义决定
C. 超键就是候选键
D. 键是唯一能决定一个元组的属性或属性组
一、单项选择题
1. 当B属性函数依赖于A属性时，属性A与B的联
系是（B ）
A. 一对多
C. 多对多
B. 多对一
C. 以上都不是
2. 关系模式R中的属性全部是主属性，则R的最高
范式必定是（B ）
A. 2NF
B. 3NF
C. BCNF
D. 4NF
3. 在关系模式R(A,B,C,D)中，有函数依赖集F={
Z=U-X-Y，则 X →→Z
5. 若关系模式R已属于第一范式，且其中的每一
非主属性都
于第二范式
完全函数依赖
于主键，则R属
6. 如果关系R只有一个键，且该键是单属性，那 么R至少属于 2NF 范式
7. “不能从已知的FD集使用推理规则导出的FD不
在F+中”, 这是推理规则的 完备 称为 2NF 模式。
A. W1(工号, 姓名) W2(工种, 定额)
B. W1(工号, 工种, 定额) W2(工号, 姓名)
C. W1(工号, 姓名, 工种) W2(工种, 定额)
D. 以上都不对
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8. 有关系模式学生(学号, 课程号, 名次)，若每一
名学生的每门课程都有一定的名词，每门课
程的每一名词只有一名学生，则以下叙述中
B →C, C →D, D →A}，则R能达到（B）
A. 1NF
B. 2NF
C. 3NF
D. BCNF
4. 下列有关范式的叙述中正确的是（D）
A. 如果关系模式 R ∈ 1NF ，且 R 中主属性完全函
数依赖与候选键，则R是2NF
B. 如果关系模式R∈3NF, X,Y U,若X →Y，则
R是BCNF
错误的是（ B）
A. (学号, 课程号)和(课程号, 名次)都可作为候选 键 B. 只有(学号, 课程号)能作为候选键 C. 关系模式属于第三范式 D. 关系模式属于BCNF
9. 以下叙述中正确的是（A ）
A. 函数依赖 X→Y 的有效性仅决定于两属性集的值；而
多值依赖X→→Y的有效性与属性集的范围有关 B. 函数依赖X→Y与多值依赖X→→Y的有效性都决定于 两属性集的值 C. 多值依赖X→→Y，若在R(U)上成立，则对任何
C. 如果关系模式R∈BCNF，若X → → Y(Y X )
是平凡的多值依赖，则R是4NF D. 一 个 关 系 模 式 如 果 属 于 4NF ， 则 一 定 属 于 BCNF，反之不成立
5. 规范化理论是关系数据库进行逻辑设计的理论
依据。根据这个理论，关系数据库中的关系
必须满足每一属性都是（B ）
二、填空题
1. 如果X→Y和Z
Y成立，则X→Z成立，这条
推理规则称为 分解性
2. 在关系模式R(U)中，X∈U, Y∈U, X→Y,如果
X的某一真子集X’都有X’→Y, 则称为
Y局部函数依赖于X
3. 包含在所有候选键中的属性称为 主属性 所有候选键之外的属性称为 非主属性
4. 在 关 系 模 式 R(U) 中 ， X,Y ∈ U, 若 X→→Y ，
Y' Y都有X Y'
D. 对于函数依赖X→Y在R上成立，不能断言对任何
Y' Y都有X Y'
10. 数据库一般使用（ C）以上的关系
A. 1NF
B. 3NF
C. 4NF
D. 5NF
11. 关系规范化实质是围绕（ B ） 进行的
A. 函数
B. 函数依赖
C. 范式
D. 关系
12. 在关系规范化过程中，消除了（ A）依赖后，