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1、华南农业大学期末考试试卷A卷2021-2021 学年第一学期 考试科目： 数据库系统 考试类型：闭卷 考试时间：120 分钟学号 姓名 年级专业 题号一二三四五六总分得分评阅人Question 1: true-false question (15 points) For each of the following statements, indicate whether it is TRUE or FALSE (Using for TRUE and for FALSE). You will get 1 point for each correct answer, -0.5 point for
2、each incorrect answer, and 0 point for each answer left blank. Be sure to write your answer in the answer sheet.1. A primary key is a field (or group of fields) that uniquely describes each record in the database. 2. Data redundancy improves the integrity of a database. 3. SQL is the language used b
3、y relational databases to create objects and to manipulate and retrieve data.4. A relational database management system does not include tools for backing up & restoring databases. 5. An attribute is also known as a row in most databases. 6. An association between entities is known as a relationship
4、. 7. Integrity constraints limit the number of entities that can be placed in a table or database. 8. The Entity-Relationship data model is often used in the physical design phase.9. The concept “relation in relation model is exactly the same as the concept “relationship in ER model.10. Most relatio
5、nship sets in a database system involves two entity sets.11. The closure of an attribute set contains that attribute set.12. Lossless decomposition is necessary in a decomposition algorithm.13. If A B and C D hold, then AC BD also holds.14. It is not necessary that a legal schedule preserves the ord
6、er in which the instructions appear in each individual transaction.15. Update operations in database must be written into log before updating database.Question 2 single-choice question (2 points for each problem, 30 points in total)1. is the only one incorrect description from the followings:A. R=(R
7、-S)(RS) B. R-S=R-(RS)C. RS=S-(R-S)D. RS=S-(S-R)2. Choose the only one correct expression from the followings: _ _.A. ( some) in B. (= all) not in C. exists r r D. X-Y X Y3. of the following four expressions of relational algebra is not equivalent to the other three They are all based on the relation
8、s R(A,B) and S(B,C). A. B. C. D.4. In the following, assume a is an attribute of some character-string type, e.g. CHAR(10), and that it may be NULL.Q1: SELECT * FROM R WHERE a IS NULL;Q2: SELECT * FROM R WHERE a NOT LIKE %;A.Q1 and Q2 produce the same answer. B. The answer to Q1 is always contained
9、in the answer to Q2. C. The answer to Q2 is always contained in the answer to Q1. D. Q1 and Q2 produce different answers.5. The Entity-Relationship data model is A. DBMS dependent B. DBMS independent C. both A and B D. neither A nor B6. In SQL, an UPDATE statement without a WHERE clause:A. Updates e
10、very row in a table. B. Updates no rows in a table.C. Updates every column in a table. D. Results in a Cartesian product.7. If a course can be taught by many teachers, and a teacher can teach only one course, then the mapping cardinality from course to teacher isA. one-to-one B. one-to-many C. many-
11、to-one D. many-to-many8. If there is a many-to-one relationship between entity A and B, then A. there exists a functional dependency from the primary key in B to the primary key in A, i.e., PK(B) PK(A).B. there exists a functional dependency from the primary key in A to the primary key in B, i.e., P
12、K(A) PK(B).C. both A and B. D. neither A nor B9. If a functional dependency ABR holds on relation R(A, B, C), then (A, B) is definitely a _ of R.A. super key B. primary key C. candidate key D. foreign key10. A relational schema R is in _ if the domains of all attributes of R are atomicA. 1NF B. 3NF
13、C. BCNF D. 4NF11. Which one of the following statement is trueA. 3NF is more strict than BCNF B. 4NF is more strict than BCNFC. 1NF is more strict than BCNF D. BCNF is the most strict normal form12. If a transaction Ti has obtain an exclusive lock on data item Q, then transition Tj can _.A. obtain a
14、n exclusive lock on data item Q B. obtain a shared lock on data item Q C. wait for lock granting on data item Q D. read or write Q without a lock13. If both and appear in the log after the nearest checkpoint to system crash, then transaction Ti must be _A. undone B. redone C. deleted D. Neither A or
15、 B14. _ is the final state in a life cycle of a transaction.A. committed B. aborted C. failed D. A or B15. in 2PL protocol, at stage, A transaction may obtain locks, but may not release locks. A. Shrinking phase B. Growing phase C. CommittedD. AbortedQuestion 3 (12 points) Consider the following dat
16、abase requirement:A hospital has properties like ID, name, location, rank, capacity. A doctor can be described by ID, name, age, skill. A patient has properties like ID, name, age, sex, address. The above three entities must satisfy some constraints: Each doctor can be unemployed or employed by one
17、hospital. If a doctor is employed, his salary needs to be recorded in the database. A patient can go to many hospitals.1. Draw ER diagram to illustrate the above database requirement 8 points.2. Translate your ER diagram into relational database schemas, and point out the primary keys and foreign ke
18、ys. You can write your answers in the following format: “R(a1, a2, a3, a4), primary key: a1, foreign key: a4 4 points.Question 4. (24 points) The following five tables are for a company management system:EMPLOYEE (ID, Name, Birthday, Address, Sex, Salary, Dnumber)DEPARTMENT (Dnumber, Dname, MgrID )P
19、ROJECT (Pnumber, Pname, Pcity)WORKS_ON (Pnumber, E-ID, Wdate, Hours)CHILD (E-ID, CHD-ID, CHD_name, Sex, Birthday)1. Based on the giving relations, Specify the following queries using relational algebra (3 points for each).1)List the names of all employees with birthday earlier than 1970-1-1 and sala
20、ry less than $5000. 2)List the names of all employees who have a child. 3)List the cities and the total number of projects which are located on same city. 2.Specify the following operations in SQL(3 points for each). 1)Define the table WORKS_ON, declare Pnumber, E-ID, Wdate as the primary key, Pnumb
21、er as the foreign key referencing the primary key of project, E-ID as the foreign key referencing the primary key of employee, and ensure that the values of Hours are non-negative with default value 8.2)For each employee working on the Network project (Pname), increase his/her salary by 5%. 3)List t
22、he names of all department managers who have no child. 4)Find the names of all employees in department 5 (Dnumber) who have worked on both X project and Y project (Pname). 5)For each project, find the project number, project name and the total hours (by all employees) spent on that project in Oct. 2
23、021. Question 5 (10 points) Consider a relation R(A, B, C, D, E, F) with the set of Functional Dependencies F = A BCD, BC DE, B D, D A 1. Compute the Closures of attribute sets A+, C+, E+ 3 points.2. Give one candidate key of R 2 points.3. Is F equivalent to A BC, BC E, B D, D A 2 points4. What is t
24、he highest normal form of R Explain your reasons 3 points.Question 6 (9 points) There are 3 transactions:Consider the following schedule S on transitions set T1, T2, T3, T4, with R and W denotes read and write operation respectively.S= R1(A) R2(B)R3(A)R2(C)R4(D)W2(B)R1(B)W1(D)R3(B)W3(B)W2(C)1. List
25、all conflict operation pairs in S 3 points.2. S no-conflict operations to see whether it is a serializable schedule 3 points.3. Write all its equivalent serial schedules if S is conflict serializable Or show why if it is not conflict serializable 3 points.华南农业大学期末考试试卷A卷-Answer Sheets2021-2021 学年第1 学
26、期 考试科目： Database system 考试类型：闭卷 考试时间： 120 分钟学号 姓名 年级专业 题号一二三四五六总分得分评阅人 Instructions to candidates:1.Write your name, student number and class on both the question papers and the answer papers.2.DO NOT write your answers on the question papers. Write them ALL ON THE ANSWER PAPERS. 3. Write your answe
27、rs in either Chinese or English. 4.Hand in all papers (both the question papers and the answer papers). Question 1 (15 points)题号1234567891得分Question 2 (30 points)题号1234567891得分CCCDBABBAABCBDB得分Question 3 (12points)hospital(ID, name, location, rank, capacity), primary key: ID,doctor(ID, name, age, sk
28、ill, hospitalID, salary), primary key: ID, foreign key: hospitalID refer to hospital(ID)patient (ID, name, age, sex, address), primary key: ID,livein(patientID, hospitalID) primary key: (hospitalID, PatientID), foreign key hospitalID refer to hospital(ID)，foreign key patientID refer to patient (ID)得
29、分Question 4 24 points1：1)2)3) 2： 1) CREATE TALBE Works_On(Pnumber int,E-ID char(15),Wdate date,Hours int default 8 CHECK (Hours = 0),PRIMARY KEY (Pnumber, E-ID, Wdate), FOREIGN KEY (E-ID) REFERENCES Employee(ID),FOREIGN KEY (Pnumber) REFERENCES Project(Pnumber),); 2) UPDATE Employee SET Salary=Salar
30、y*1.05 WHERE ID IN (SELECT E-ID FROM Project natural join Works_onWHERE Pname=Network); 3) SELECT Name FROM Employee, DepartmentWHERE ID=MrgID AND ID NOT IN (SELECT E-ID FROM Child); 4) SELECT Name FROM Employee WHERE Dnumber=5 AND ID IN(SELECT E-ID FROM Project natural join Works_onWHERE Pname=X)AN
31、D ID IN(SELECT E-ID FROM Project natural join Works_onWHERE Pname=Y); 5)SELECT Pnumber, Pname, SUM(Hours)FROM Project NATURAL JOIN Works_OnWHERE Wdate BETWEEN 2021-10-1 AND 2021-10-31GROUP BY Pnumber, Pname;得分Question 5 10 points1 A+=ABCDE, C+ =C, E+=E2. AF is a candidate key of R3. Yes F equivalent
32、 to A BC, BC E, B D, D A 4. the highest normal form of R is 1NF. The reason lies in that (1) it is not in BCNF, for A is not a super key but A BCD. (2) It is not in 3NF for all candidate keys are: DF, AF, BF, so that primary attributes are ABDF. To A BC, A is not a super key and c is not included in
33、 any candidate key therefore it violate rules of 3NF. (3) all attribute are atomic.得分Question 6 9 points1. conflict operation pairs (1)W2(B)R1(B) (2) W2(B)R3(B) (3)W2(B)W3(B) (4) R1(B)W3(B) (5) R2(B) W3(B) (6) R4(D) W1(D) 2. S= R1(A) R2(B)R3(A)R2(C)R4(D)W2(B)R1(B)W1(D)R3(B)W3(B)W2(C)= R2(B)R1(A) R2(
34、C) R3(A) W2(B) R4(D) R1(B)W1(D)R3(B) W2(C)W3(B)= R2(B) R2(C)R1(A) W2(B) R3(A) R4(D) R1(B)W1(D) W2(C)R3(B) W3(B)= R2(B) R2(C) W2(B) R1(A) R4(D) R3(A) R1(B) W2(C)W1(D) R3(B) W3(B)= R2(B) R2(C) W2(B) R4(D)R1(A) R1(B) R3(A) W2(C) W1(D) R3(B) W3(B)= R2(B) R2(C) W2(B) R4(D) R1(A) R1(B) W2(C) R3(A) W1(D) R
35、3(B) W3(B)= R2(B) R2(C) W2(B) R4(D) R1(A) W2(C)R1(B) W1(D) R3(A) R3(B) W3(B)= R2(B) R2(C) W2(B) R4(D) W2(C)R1(A) R1(B) W1(D) R3(A) R3(B) W3(B)= R2(B) R2(C) W2(B) W2(C)R4(D) R1(B) W1(D) R3(A) R3(B) W3(B)3. it is a conflict serializable schedule. The equivalent serial schedules are:T2, T4, T1, T3 or T4, T2, T1, T3