The database has a relational schema R A, B, C, D and a dependency F A B, B C BC paradigm problem

Decide B, B decides C, so A decides C

But nothing decides d

So. a,d

It's the key, or the candidate code and the main code, whatever you want to call it.

2.Partially dependent on a->b, because b only needs a, d

A can be decided.

It is obvious that A decides B, B decides C, and A decides C to pass through B, so A->C passes the dependency and can separate another table.

b,c｝3.It does not belong to the third normal form because there is a transitive dependency, which translates into 2 tables {adb} and {b

c} is now satisfied with both the third normal and the bc paradigm.

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The candidate key is either a or b.

a->c, a->b, a->bc, bc->d, a->d, so a is the key

b->a, so b is also key

It is the first paradigm, because it is necessary to satisfy each component and cannot be redivided;

is the second normal because of the non-principal property c

d is completely dependent on the key;

It's the third paradigm.

Because of the non-primary attribute c

d has no transfer function dependence on the main attributes a and b;

is the BC paradigm, because every determinant must contain a or b;

It's the fourth paradigm.

Because there is no such thing as a non-trivial and non-function-dependent multi-valued dependency.

Two multi-value dependencies, both with primary attributes).

Therefore, the highest is the fourth normal form.

Because baia b, b c, and therefore a bc, the relationship model is a candidate

The code is ad, that is, ad bc, and dao because of d bc, so there is a partial dependence of the non-main answer attribute on the code. Therefore, this relational model is the first paradigm.

If the relationship conforms to 1nf, and for each function depends on x y, x must contain a candidate key, or every set of determining attributes in the relationship is a candidate key, then the relationship meets the requirements of bcnf.

Because a b, b c, and therefore a bc, the candidate code for this relationship pattern is ad, that is, ad bc, and because d bc, there is a partial dependence of non-primary attributes on the code. Therefore, this relational model is the first paradigm.

It must be copied from the first paradigm, because from.

On the point that the b, bc function determines a and d, it is obvious that both dub and bc may be zhi

Master code. If b is the main code, if you look closely, you will find that there is no one in f to determine c, which obviously does not make sense, (because c will at least be determined by the main code function b); If BC is the primary code, then there is a partial dependence of the non-primary attribute on the candidate code in F, which does not meet the requirements of the second normal form, so it is the first normal code.

It must be the first normal form, because from the point where b and bc functions determine a and d, it is obvious that b and bc are both likely to be the master code. If B is the main code, if you look closely, you will find that there is no one in F who has the right to determine C, which obviously does not make sense, (because C will at least be determined by the main code function of B); If BC is the primary code, then there is a partial dependence of the non-primary attribute on the candidate code in F, which does not meet the requirements of the second normal form, so it is the first normal code.

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Summary. 1.All candidate keywords in the relational pattern R are .

Because both a and b are hyperkeys, and there are no other properties that can be removed. 2.This relational model satisfies the third normal form at most.

Because all non-primary properties are fully functionally dependent on candidate keywords, there are no transitive dependencies. 3.The process of decomposing the relational pattern so that it finally satisfies the third normal form is as follows:

According to the function dependency set, two relational patterns r1(a, d) and r2(c, e, b) are obtained. Both -r1 and r2 satisfy the third normal because they have no transitive dependencies. - The final decomposed relationship patterns are r1(a, d) and r2(c, e, b).

When decomposed in this way, there is only one candidate keyword for each attribute in the relationship pattern, satisfying the third normal formula.

Three normals (keeping functions dependent).

There is a relational pattern r (a, b, c, d, e) in which the relation.

There is a set of function dependencies on the schema:, please:

1.Try to find out all the candidate keywords in the relationship pattern only.

2.What is the highest paradigm that this relational model can satisfy?

3.Try to break down the relationship pattern so that it finally satisfies the first step.

There is a relational pattern r (a, b, c, d, e) in which the relation.

Three normals (keeping functions dependent).

3.Try to break down the relationship pattern so that it finally satisfies the first step.

Thank you. I see.

Thank you. There is a set of function dependencies on the schema:, please:

There is a set of function dependencies on the schema:, please:

There is a set of function dependencies on the schema:, please:

There is a set of function dependencies on the schema:, please:

There is a set of function dependencies on the schema:, please:

There is a set of function dependencies on the schema:, please:

There is a set of function dependencies on the schema:, please:

There is a set of function dependencies on the schema:, please:

There is a set of function dependencies on the schema: {a->d.). c->e.(a,b)-> has a relational pattern r (a, b, c, d, e) in which relation.

Candidate Code: (

1）a；（bai2）cd；（3）e；These are DU candidate codes:

Therefore, the main attributes are a, (c, d), zhie; The non-primary attributes are: daob;

If: Within. There is no non-primary attribute of the transfer function dependency so it belongs to 3nf, how to normalize it to bcnf, bcnf if it is.

To investigate whether each nontrivial function depends on the x--y determinant including codes, the following schema decomposition is made:

a，b，c）；capacity (c, d, e); This makes it possible to include code for every nontrivial function dependency. i.e. a--b, c; cd--e;Each non-functional dependency includes code.

(1)(e,c) is the only candidate key The simplest way to understand it is that the closure of (e,c) is the entire set (a,b,c,d,e,f), i.e. only these two elements are needed to determine all the elements.

2) The highest belongs to the first normal form, because there is a pass-through dependency of c b, b a, and the second normal form stipulates that all elements that are not principal attributes must be directly related to the principal attribute, but element a is passed from b, so it does not conform to the second normal form, but conforms to the first normal form and has a unique primary key (e, c) to identify.

3) Decompose into r1(b,c,d,e,f) r2(b,a).

r, for example, relational patterns.

Student (student number, name, gender, department).

The student is the relationship name r, the four fields in parentheses are the attribute set u, f is the function dependency set, one of them is the student number - > name is a group, and a student number can determine a student, right? That student number can determine a student's n attributes, compared to determining the student's name, age, or other attributes.

Student number and age (or name, department) are a subset of U, and the relationship can be regarded as a table, does any row in the table also contain these attributes such as student number and name?

Does this line also have a function dependency on student number - > name and so on, then this line satisfies the function dependency f, and also contains a subset of the property u

Is it possible to think of either line as a relational model?

r is a line in r and a small relational pattern in r.