2016 first half database systems engineering teacher morning test (25-51)

"Five years college entrance examination three years of simulation" equivalent to the entrance examination "Martial arts Cheats" in the "Nine Yin Canon." A large amount of the question bank, the real problem of detailed analysis, the teachers and students sought after. Can be seen, the real problem is to deal with the examination of good information, the following Greek soft Test college for you to organize the first half of 2016 years of database Systems Engineering Test real problem of the morning, help you cultivate a "stunt" For the next year 's Database Systems Engineering Exam .

2016 first half database Systems Engineering exam morning test (25-51)

The Pl.P2.P3.P4 and P5 of the process are shown in the preceding diagram:

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If using the PV operation to control process P1.P2.P3.P4 and P5 concurrent execution process, you need to set 5 semaphore S1.S2.S3.S4 and S5, and the initial value of the semaphore si~s5 is equal to zero. A and B shall be filled out separately (); C and D shall be filled out separately (), and E and F shall be filled out separately ().

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(b) A. V (S1) P (S2) and V (S3)

B P (S1) v (S2) and V (S3)

C V (S1) v (S2) and V (S3)

D P (S1) p (S2) and V (S3)

(A). P (S2) and P (S4)

B P (S2) and V (S4)

C V (S2) and P (S4)

D V (S2) and V (S4)

(b) A. P (S4) and V (S4) v (S5)

B V (S5) and V (S3) P (S5)

C V (S3) and P (S4) p (S5)

D P (S3) and P (S4) p (S5)

In a database system with a three-level schema structure, if you create a clustered index on the table EMP in the database, you should change the database ().

(b) A. Mode

B: Internal mode

C External mode

D User mode

In the design stage of information integrated management system of an enterprise, the employee entity is called "Quality Inspector" in the Quality management subsystem, and it is called "employee" in the Personnel management subsystem, this kind of conflict is called ().

(A). Semantic conflicts

B Naming conflicts

C Attribute conflicts

D. Structural conflicts

For relational mode R (x, Y, z), the following conclusions are wrong ().

(A). If x→y,y→z, then x→z

B If x→z, then xy→z

C If xy→z, then x→z,y→z

D If x→y,x→z, then X→yz

If the relation R1 is calculated by (), the relationship R2 can be obtained.

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a.σ Product name = ' Towel ' V ' pen ' (R1)

b.σ price ≥ ' g ' (R1)

c.π1.2.3. (R1)

d.σ Product number = ' 01020211 ' V ' 02110200 ' (R1)

Relationship normalization is performed at the () stage of the database design.

(b) A. Demand analysis

B Conceptual design

C Logic Design

D Physical design

If the given relationship pattern is good <u,f>,u={a,b,c), f={ab→c,c→b), then the relationship R ().

(A). There are 2 candidate keywords AC and BC, and there are 3 main properties

B There are 2 candidate keywords AC and AB, and there are 3 main properties

C Only 1 candidate keyword AC, and there are 1 non-primary properties and 2 main attributes

D There are only 1 candidate keyword AB, and there are 1 non-main attributes and 2 main attributes

Set the relationship mode R , where U is the attribute set and F is a set of function dependencies on U, then the pseudo-transitive law of the Armstrong Axiom System refers to ().

A. If x→y,y→z is implied by F, then X→z is the implication of F

B If x→y,x→z, then X→yz is the implication of F

C If x→y,wy→z, then xw→z is the implication of F

D. If the x→y is contained in F, and the z& #8838; U, then Xz→yz is the implication of F

Given the relationship R (a,b,c,d) and the relationship s (c,d,e), the natural connection operation r& #9655;& #9665; S attributes are listed as (): With ΣR.B>S.E (r& #9655;& #9665; S) The equivalent algebraic expression of the equation is ().

(A). 4

B 5

C 6

D 7

(a.σ2>7) (RXS)

B π1.2.3.4.7 (σ ' 2 ' > ' 7 ' ^3=5^4=6 (RXS))

C.σ ' 2 ' > ' 7 ' (RXS)

d.π1.2.3.4.7 (Σ2>7^3=5^4=6 (RXS))

The R.S relationship is shown in the following table, and the tuple calculus expression t={t| R (t) ^& #8704; the result of U (S (u) →[3]>u[1] operation is ().

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The function dependency set on the relationship R (A1,A2,A3) F={a1a3→a2,a1a2→a3}, if a decomposition on R is p={(A1,A2), (A1,A3)}, then P () is decomposed.

(A). It's a lossless connection.

B is to keep the function dependent.

C is a lossy join.

D Unable to determine whether to maintain function dependencies

Assuming that the function dependency set on R (A1,A2,A3) is f={a1→a2,a1→a3,a2→a3}, the function depends on ().

(A.A1→A2) is redundant

B A1→A3 is redundant.

C.A2→A3 is redundant.

D.a1→a2,a1→a3,a2→a3 are not redundant.

A Business Sector Relationship Model Dept (Department number, department name, person in charge, time of service), employee relationship Mode EMP (employee number, name, age, monthly salary, department number, telephone, office). The foreign keys for department and Employee Relations are (). Query the following SQL query statements for the highest monthly employee number, name, + department name, and monthly salary for each department:

Select employee number, name, department name, monthly salary

From EMP y,dept

WHERE () and monthly salary = (

SELECT Max (monthly)

From EMP Z

WHERE ())

(+) A. Employee number and department number

B. Person in charge and department number

C Person in charge and employee number

D Department number and employee number

(A) A. Y. Department number =dept. Department number

B.emp. Department number =dept. Department number

C Y. Employee number =dept. Number of persons in charge

D.emp. Department number =dept. Number of persons in charge

(A). Z. Employee number =Y. Employee number

B.Z. Employee number =y. Number of persons in charge

C.z. Department Number = Department number

D.z. Department number =y. Department number

The component relationship mode in a company database is P (component number, component name, supplier, supplier location, inventory), and the function dependency set F is as follows:

f={component number → component name, (component number, supplier) → stock, supplier, supplier location

The primary key for a component relationship is (), which has problems such as redundancy and insertion and deletion exceptions. In order to solve this problem, we need to decompose the component relationship into (), and the decomposed relational pattern can reach ().

(+) A. (Component number, component name)

B (component number, supplier)

C (Component number: supplier location)

D (Supplier, supplier location)

(A). Component 1 (Component number, component name, supplier, vendor location, stock)

B Component L (component number, component name), component 2 (supplier, vendor location, stock)

C Element 1 (Component number, component name), component 2 (component number, supplier, stock)

Component 3 (supplier, supplier location)

D Component L (component number, component name), element 2 (component number, stock)

Component 3 (supplier, supplier location), component 4 (supplier location, stock)

(A.1NF)

B.2nf

C.3nf

D.bcnf

Transactions have multiple properties, "Once a transaction is successfully committed, its update operation to the database will be permanent, even if the database crashes." "This nature belongs to the () nature of the transaction.

(A). Atomic Nature

B Consistency

C Isolation of

D Durability

In the following description of the relationship, it is correct ().

(A). Two lines in an Exchange relationship form a new relationship

B. The values of two columns in a relationship can be taken from the same domain

C Two columns in the interchange relationship form a new relationship

D A column in a relationship can consist of two child columns

A relational database typically contains more than one table, and the association between the table and the table is implemented by (), and the two associated tables are combined into a single information equivalent table by the () operation.

() A. Pointer

B External code

C Index

D View

(A). Choose

B Projection

C Cartesian product

D Natural connection

If the system uses the most frequently used query statement as

select*

From SC

where Sno=x and cno=y;//where x, y is the variable to make the execution of the query statement the most efficient, create ().

(b) A. Index on the Sno

B Index on the CNO

C Index on the Sno,cno

View on D.Sc sc_ V (SNO,CNO)

Grant the EXECUTE permission of the stored procedure P1 to the user U2 the SQL statement.

GRANT () on PROCEDURE Pl to U2;

(Wuyi) A.insert

B.update

C:delete

D.execute

2016 first half database systems engineering teacher morning test (25-51)