Understanding 3NF: plain English please

The best definition I've found for a relation that is in third normal form (3NF) is the following:

A relation schema R is in 3NF if, whenever a function dependency X -> A holds in R, either    (a) X is a superkey of R, or    (b) A is a prime attribute of R.

Now there are three definitions that need clarification, key,superkey, and prime attribute.

For the definitions we will use examples from the R1 relation to describe them:

R1(ABCD)ACD -> B   AC -> D   D -> C   AC -> B

key: A key is the attribute that determines every attribute of the relation. In other words, it is the set of attributes that will give you all the other attributes of the relation that are not in the set. In relation R1 of the above example, the keys are AC and AD. Why is AC a key? Because by knowing attributes A and C you can determine the remaining attributes, B and D. Why is AD a key? The same reason. A and D will ultimately determine B and C.

superkey: A superkey is basically a superset of a key. A superkey will contain the key always and potentially more attributes. In the previous example, AC is a key. Thus AC, ACD, ACB, etc. are superkeys. Note that a key itself is a superkey.

prime attribute: A prime attribute is basically an attribute that is part of a key. Thus A and C are prime attributes as they are part of the key AC. Take note however, the difference between a key and superkey. For the super key ACB, B is not a prime attribute since B is not part of the key. Just think of a prime attribute as a subset of a key.

Now let's look at the four relations:

R1(ABCD)ACD -> B   AC -> D   D -> C   AC -> BR2(ABCD)AB -> C   ABD -> C   ABC -> D   AC -> DR3(ABCD)C -> B   A -> B   CD -> A   BCD -> AR4(ABCD)C -> B   B -> A   AC -> D   AC -> B

For each relation we will write down the keys and the prime attributes. Then we will see if the definition is satisfied.

R1:keys: AC, ADprime attributes: A, C, D

ACD -> B: Left side is a superkey. Satisfies (a).

AC -> D: Left side is a key and thus a superkey. Satisfies (a).

D -> C: Left side is not a superkey. Does not satisfy (a). However, right side is a prime attribute. Satisfies (b).

AC -> B: Left side is a key. Satisfies (a).

Either (a) or (b) is satisfied in all cases. Thus R1 is in 3NF.

R2:keys: ABprime attributes: A, B

AB -> C: Left side is a key and thus a superkey. Satisfies (a).  

ABD -> C: Left side is a superkey. Satisfies (a).

ABC -> D: Left side is a superkey. Satisfies (a).

AC -> D: Left side is not a superkey. Does not satisfy (a). Right side is not a prime attribute. Does not satisfy (b).

Since (a) or (b) is not satisfied in all cases, R2 is not in 3NF.

R3:keys: CD, prime attributes: C, D

C -> B: Left side is not a superkey. Does not satisfy (a). Right side is not a prime attribute. Does not satisfy (b).

Since we have already found a case that does not satisfy either (a) or (b), we can immediately conclude that R3 is not in 3NF.

R4:keys: Cprime attributes: C

C -> B: Left side is a key and thus a superkey. Satisfies (a).  

B -> A: Left side is not a superkey. Does not satisfy (a). Right side is not a prime attribute. Does not satisfy (b).

Again, we can stop here as the second case satisfies neither (a) nor (b). The relation R4 is not in 3NF.

Let me explain in simple words:

For the given relation,R1(ABCD),the functional dependencies are:

ACD -> B

AC -> D

D -> C

AC -> B

Condition to be in 3NF

X->Y here X is a Super key when Y is non-prime attribute else it can be any attribute

Prime attributes are the attributes which belongs to the super key

Non-prime attributes are the attributes which do not belongs to the super key

Let us come back to relation R1,

AC+=ABCD

AD+=ABCD

ACD+=ABCD and

D+=DC

Thus,we get AD,AC,ACD as our Super keys

And A,C,D are prime attributes and B is a non-prime attributes

ACD->B

This Functional dependency is in 3NF as ACD is a super key and B is prime attribute

AC->D

This Functional dependency is also in 3NF as AC is a super key and D is prime attribute

D->C

This Functional dependency is also in 3NF as D is a prime attribute and C is also prime attribute

AC->B

This Functional dependency is also in 3NF as AC is a super key and B is non-prime attribute

  Thus,the relation is not in 3NF only when non-prime attributes  does not depend on super key

Hope,this helps!

In simple english, here are the 3 Normal Forms:

1NF: The existence of "the key" ensures that the table is in 1NF(Key must be there).

2NF:It is required that "each" non-key attributes be dependent on "the whole key" to ensure 2NF.

3NF: further requiring that "each" non-key attributes be dependent on "nothing but the key" ensures 3NF.

Now, for this:

R1(ABCD) ACD -> B   AC -> D   D -> C   AC -> B

Look at these ACD -> B and AC -> B: clearly violates the 2NF condition. Forget 3NF, this relation is not even in 2NF. "the whole key"-->concept does not hold.

I think, you have proved the same using SET.