Decomposition relation model of the 2022 database of tyut Taiyuan University of Technology

Question 4

Decompose relational patterns ： Examination site ： Basic functional dependencies of relational patterns 、 Relationship candidate key 、 Which paradigm does relationship belong to 、

What kind of relationship does it belong to NF、 decompose NF Pattern set （10 branch ）

To master this topic, we must first master many basic concepts , Let's have a look , Just talk about what you want to test and what you will use to do the questions 、 I won't talk about the official words in the book , How to understand how to

One 、 Basic concepts

1、 Function dependency (FD): Functional dependency is the constraint relationship between attributes within the same relationship , That is to say R(A1,A2...) Medium lecture A1 And A2 And so on , For example, set the attribute student number to X、 The name is Y, Because the student number can decide the name , Then there are X decision Y Or say Y Depend on X, Write it down as ：X->Y

2、FD Set （ Functional dependency set ）： Will be multiple FD Put them together ： Such as function dependency set F={A->B,B->C,A->C}, If there is a set, there will be implication , therefore F Logic implies X->Y, Write it down as ：F|=X->Y

3、 Super key ： Set a relationship R The property set of is U,X yes U Subset , If U Some of the properties in X Can decide U, namely X->U, So called X yes R A super key of （ Remove any attribute in the hyperkey ,X->U Still possible ）

4、 Candidate key ： In hyperkeys , If you remove any attribute in the hyperkey ,X->U Will not be established （ That is, there is no redundant attribute in the super key ）, Then call this super key a candidate key

5、L class ：F={A->B,B->C,A->C} All in -> Properties on the left , namely A、B

6、R class ：F={A->B,B->C,A->C} All in -> Properties on the right , namely B、C

7、 Minimum functional dependency ： The difficult words , Pass the examination , But I haven't seen it in the past two years , It's the test site , If you want to learn, you can read by yourself , I won't talk about it here

8、 Decomposition of relational patterns ： We can R(A1,A2,A3,A4) Decompose into R1（A1,A2）,R2（A3,A4）

9、 Completely function dependent ： There are attribute sets X And property sets Y, Functional dependency X->Y, Get rid of X Any attribute in ,X->Y It doesn't work （ That is the X There is no redundant attribute in ）, that Y The complete function depends on X, Otherwise, it is called local dependency

10、 Non primary property ： It refers to the attribute that is not included in any candidate code in the relationship . for example ： In relation —— Student （ Student number , full name , Age , Gender , class ） in , Main attribute 、 The candidate code is “ Student number ”, So the others “ full name ”、“ Age ”、“ Gender ”、“ class ” Can be called non primary property

（ Similar to determining candidate keys , It's all about dependencies L Whether the class has redundant attributes ）

11、 The transitive dependency that will appear in the big question ： There are candidate keys X、 Non primary property Y、A, if X->Y,Y->A,Y!->X,A Do not belong to Y, So called X->A It's delivery dependency （A Delivery depends on X）

Two 、 The paradigm of relationship pattern

There is an inclusive relationship between paradigms ：1NF contain 2NF contain 3NF contain BCNF

1、 First normal form （1NF）

As long as the relationship mode R(A1,A2...) All properties in A1、A2...... Are inseparable attributes , Are atomic properties , Then meet the first paradigm

To be dissatisfied with 1NF Relationship model of R, We're going to R Become satisfied 1NF Relationship model of ： Or decompose its non atomic properties , Or decompose its relationship pattern , You are satisfied with the general questions 1NF

2、 Second normal form （2NF）

If the relational pattern R yes 1NF, And each non primary attribute is completely functionally dependent on candidate keys , be R For the second paradigm （2NF）

Judge whether it is 2NF： Put the candidate key （ The common candidate keys in big questions generally include 2 Attributes ） Determine each non primary attribute in turn , Judge whether the function depends on the right （L class ） Whether there are redundant attributes , If there is no , It's in line with 2NF , On the contrary, it does not meet

To be dissatisfied with 2NF Relationship model of R, We're going to R Become satisfied 2NF Relationship model of ： In fact, it is a simple classification , We put R Decompose into R1、R2....... Non primary attributes that will be redundant with candidate keys 、 And its corresponding candidate keys are placed in R1 in ; Non primary attributes that will not be redundant with candidate keys 、 And its corresponding candidate keys are placed in R2 in . Let's take an example to understand :

example ： Set a relationship R（ Customer number , The number of the item purchased , Number , Name of the manufacturer , Business address ）, Judge R Whether to belong to 2NF, If R Do not belong to 2NF, Please put R Decompose into 2NF Pattern set

Explain ： The candidate key is the customer number , The number of the item purchased , Yes （ Customer number , The number of the item purchased ）->（ Name of the manufacturer , Business address ）, Because of a single attribute “ The number of the item purchased ” You can decide “ Name of the manufacturer , Business address ”, Therefore, there is a case that non primary attributes locally depend on candidate keys ,R Do not belong to 2NF

take R Decompose into 2NF Pattern set thinking （ Don't write on the paper , Write only on scratch paper ）

  take R Decompose into 2NF The pattern set is ：

R1（ The number of the item purchased , Name of the manufacturer , Business address ）

R2（ Customer number , The number of the item purchased , Number ）

3、 Third normal form （3NF）

Brush up on ：11、 The transitive dependency that will appear in the big question ： There are candidate keys X、 Non primary property Y、A, if X->Y,Y->A,Y!->X,A Do not belong to Y, So called X->A It's delivery dependency （A Delivery depends on X）

If the relational pattern R yes 2NF, And each non primary attribute does not rely on candidate keys , be R For the third paradigm （3NF）

Judge whether it is 3NF： By definition , See if there is a phenomenon that non primary attribute transmission depends on candidate keys in the relational pattern .

To be dissatisfied with 3NF Relationship model of R, We're going to R Become satisfied 3NF Relationship model of ：

If in R（X,Y,A,B,C） in , The candidate key is X、 Non primary attribute is Y、A、B、C, There is X->Y,Y->A,Y!->X,A Do not belong to Y（ namely X->A,A Delivery depends on X）, Then just put R Decompose into R1（X,Y,B,C）,R2（Y,A） To satisfy 3NF, It can be understood as R1 Only the Y,R2 Only two non primary attributes that are transitively dependent on candidate keys are retained Y,A

Look at an example to understand

example ： There's a relational model ： readers （ Reader number , full name , Company , Address ）, Each unit has only one address , Please normalize it into 3NF

Explain ：

From the meaning of the question ： Reader number -> Company -> Address

readers 1（ Reader number , full name , Company ） Only the address at the end of the delivery position is removed

Company （ Company , Address ） Only two non primary attributes that are transitively dependent on candidate keys are retained ： Company , Address

The real question ：