[easy to understand] relational schema paradigm decomposition tutorial 3NF and BCNF formula! Xiaobai can also understand "suggestions collection"

Hello everyone , I meet you again , I'm your friend, Quan Jun .

It was originally to review the database final exam , As a result, I found no better explanation after looking around , Through consulting materials and summarizing , Provide you with an easy to understand solution , You can hear it ！ And equipped with Shorthand formula ！ The new version of the ship you haven't played includes minimum dependency set algorithm, candidate code algorithm

Before schema decomposition , First of all, for 1NF,2NF,3NF,BCNF Make a brief introduction .

1NF Each column of a database table is an indivisible basic data item , That is, an attribute in an entity cannot have multiple values or duplicate attributes .

2NF The attribute is required to be completely dependent on the primary key , There cannot be a property that depends only on part of the primary keyword .

3NF Each non primary attribute is required to be neither partially dependent on the code nor transitively dependent on the code .

BCNF It eliminates the dependence of the main attribute on the candidate code and the transfer function .

notes ：1. be relative to BCNF,3NF It is allowed to have the transitive dependency and partial dependency of the primary attribute on the candidate code .

2.BCNF More abstract , A little explanation ： In the student information form , Student number is a candidate code , The student number can determine the student's name ;( class , The student's name ） It is also a group of candidate codes , Yes ( class , The student's name ）-> Student number , Therefore, a transitive dependency is formed between the main attributes .

3. If the concept is not clear , About code 、 Candidate code 、 Main attribute 、 Please refer to ：

https://blog.csdn.net/sumaliqinghua/article/details/85872446#commentBox

Our focus is on paradigm decomposition ：

One 、3NF decompose

It is divided into maintaining dependency and lossless connection

To illustrate that the solution remains dependent , First, we need to know how to find the minimum dependency set

（1） Minimum dependency set method ：

formula ： First split the order on the right , Dependencies are deleted in turn .

Restore and delete , Then dismantle the left non single .

Explain the pithy formula by finding the following minimum dependency set ,

（2）3NF decompose ：

formula ：

Letter of guarantee depends on decomposition , Let's find the minimum dependency set first .

Dependence on both sides does not appear , Divide it into subsets and put it aside , Residual dependent variable set .

To connect without damage , Add candidates for subsets .

Here are some examples to explain the pithy formula ：

example 1. It is known that R(ABCDE), F={A ->D,E->D,D->B,BC->D,DC->A} To keep the function dependent 3NF decompose , And with lossless connectivity and functional dependency 3NF decompose

First step ： Letter of guarantee depends on decomposition , Let's find the minimum dependency set first . Find out first R The minimum set of dependencies , Available F={A ->D,E->D,D->B,BC->D,DC->A}

The second step ： Dependence on both sides does not appear , Divide it into subsets and put it aside . First of all, we can find that there are no elements that do not appear on both sides, and there is no need to separate a subset ,“ Residual dependent variable set ” Then we divide each dependency into subsets to get ：{AD} {ED} {DB} {BCD} {DCA}, That is to keep the function dependent 3NF decompose

The third step ： To connect without damage , Add candidates for subsets .

(1) Solution of candidate code ： The so-called candidate code can determine the whole relationship , We look for Does not appear on the right of dependency And Elements that do not appear on either side You can find ,

(2) You can find C E Not on the right , So the candidate code is {CE}. Therefore, the method has lossless connectivity and function dependence 3NF Decompose into {AD} {ED} {DB} {BCD} {DCA} {CE}

example 2. Relationship model R, Yes U={A,B,C,D,E,G},F={B->G,CE->B,C->A,CE->G,B->D,C->D}, Break down the relationship pattern into 3NF And keep the function dependent

Break down the relationship pattern into 3NF And keep the function dependent ：

First step ： Letter of guarantee depends on decomposition , Let's find the minimum dependency set first . Find out first R The minimum set of dependencies ,

hypothesis B->G redundancy , be (B)+=BD, No, G So it's not redundant .

hypothesis CE->B redundancy , be （CE）+=CEGDA, No, B So it's not redundant .

hypothesis C->A redundancy , be （C）+=CD, So it's not redundant .

We can get the minimum function dependency set at one time Fm={B->G,CE->B,C->A,B->D,C->D}

The second step ： Dependence on both sides does not appear , Divide it into subsets and put it aside , Residual dependent variable set . First of all, it can be found that there are no elements that do not appear on both sides , Then we divide the dependencies into subsets and get {BG} {CEB} {CA} {BD} {CD}, That is to keep the function dependent 3NF decompose

The third step ： To connect without damage , Add candidates for subsets . find R A candidate code for is {CE}. Therefore, the method has lossless connectivity and function dependence 3NF Decompose into {BG} {CEB} {CA} {BD} {CD} {CE} ( notes ： Paradigm decomposition is not unique , Just be right )

Two 、BCNF decompose ：

Model relationships R<U,F> Break down into one BCNF The basic step is

1） Let's find the minimum dependency set first , Waiting codes are not subsets

3） The remaining left side is all waiting , complete BCNF topic .

example . Relationship model R, Yes U={A,B,C,D,E,G},F={B->G,CE->B,C->A,CE->G,B->D,C->D}, Break down the relationship pattern into 3NF And keep the function dependent

Break down the relationship pattern into 3NF And keep the function dependent ：

First step ： Let's find the minimum dependency set first . You can find CE->G redundant , Therefore, the minimum dependency set is F={B->G,CE->B,C->A,B->D,C->D}.

The second step ： Waiting codes are not subsets . Because the candidate code is (CE) So it will CE->B Divide into subsets （BCE）, and B->G,B->D There are no primary attributes on the left (C、E) Any one of them is divided into (BG),(BD）

The third step ： At this time, the remaining dependencies F={C->A,C->D} The remaining elements {A,C,D} Check and find that the left side of the function dependency is all candidate codes, that is, complete BCNF decompose , If not, continue to decompose the rest .

therefore BCNF The final result of the decomposition is {（BG）,(BD),(ACD),(BCE)}.

If you have any questions, please leave a message in the comment area , If you are helpful, please point a like in the upper right corner ~~ Crab

3、 ... and 、 summary

1. Closure

2. Candidate code

3. Minimum dependency set

4.3NF decompose

5.BCNF decompose

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