3NF Normal form

I have a question about 3NF normal form:

Normalize, with respect to 3NF, the relational scheme E(A, B, C, D, E, F) 
by assuming that (A, B, C) is the unique candidate key and that the following additional functional dependencies hold: 
    A,B -> D 
    C,D -> E 
    E -> F 

My understanding is that if I apply the 3NF which says that a schema is 3NF if all attributes

non-prime do not transitively depend on any key candidate , the result should be:

E'=(A,B,C,E,F), E''= (B,D) , E''开发者_Python百科'= A,B,C,D,F) , E''''=(D,E) , E''''''= (A,B,C,D,E),

E''''''= (E,F)

but I do think I'm wrong...

Can someone help understand the issue?

Thanks

(Reformatted for readability)

My understanding is that if I apply the 3NF which says that a schema is 3NF if all attributes non-prime do not transitively depend on any key candidate , the result should be:

• E¹= {A,B,C,E,F}
• E²= {B,D}
• E³= {A,B,C,D,F}
• E⁴= {D,E}
• E⁵= {A,B,C,D,E}
• E⁶= {E,F}

3NF means that a) the relation is in 2NF, and b) every non-prime attribute is directly dependent (that is, not transitively dependent) on every candidate key.

In turn, 2NF means that a) the relation is in 1NF, and b) every non-prime attribute is dependent on the whole of every candidate key, not just on part of any candidate key.

Given {ABC} is a candidate key, and given {AB->D}, you can see that D depends on part of a candidate key. So

• E⁰ = {A,B,C,D,E,F}

is not in 2NF. You fix that by moving that dependent attribute to a new relation, and you copy the attributes that determine it to the same relation.

• R⁰ = {ABC DEF} This relation—which we started with, and which is not in 2NF—goes away, to be replaced with

• R¹ = {ABC EF}

• R² = {AB D}

You want to continue from here?

When it comes to getting normalization right, there is no substitute for understanding the formal definitions. If you're still working on building that understanding, there's a cute little mnemonic that people use to help remember the essence of 3NF and to judge whether a table that they're looking at is 3NF or not.

"The key, the whole key, and nothing but the key, so help me Codd."

How do you apply it? Every attribute of the relation must depend on the key. It must depend on the whole key. I must not depend on anything that isn't the key. When you look at your example, clearly there are problems and you need to normalize. You need to get to a point where every non-key column which violates 3NF is out of your original relation. Each of the non-key columns, D, E, and F all violate 3NF.

Note that your additional functional dependencies cover all of the non-key columns in your original relation. Each of these additional functional dependencies is going to result in a relation:

{ A B D } - This solves 3NF for attribute D
{ C D E } - This solves 3NF for attribute E
{ E F } - This solves 3NF for attribute F

What is left to cover from your original relation? Nothing except the candidate key:

{ A B C }