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How do normal forms decompose?
If a relation is not in a desired normal form, it can be decomposed into multiple relations that each are in that normal form. Every attribute of R appears as an attribute of at least one of the new relations.
How do you convert a table to 3 NF?
The normalization of 2NF relations to 3NF involves the removal of transitive dependencies. If a transitive dependency exists, we remove the transitively dependent attribute(s) from the relation by placing the attribute(s) in a new relation along with a copy of the determinant. Consider the examples given below.
What relation is 2NF?
A relation is in 2NF if it has No Partial Dependency, i.e., no non-prime attribute (attributes which are not part of any candidate key) is dependent on any proper subset of any candidate key of the table.
How do you decompose a 3NF?
How to decompose this relation into 3NF relations?
- Find a minimal basis of F, say G.
- For each FD X → A in G, use {X, A} as the schema of one of the relations in the decomposition.
- If none of the sets of relations from Step2 is a superkey for R, add another relation whose schema is a key for R.
How to normalize a relation into a 3NF relation?
First of all, note that the original relation is already in Third Normal Form, since each attribute is prime (each attribute is a key, actually), so that the definition of 3NF is respected. Then, note that the algorithm is incomplete. The steps are: Delete all the relations whose attributes are contained in another relation.
Which is correct the original relation or the 3NF relation?
My question is: which one is correct? First of all, note that the original relation is already in Third Normal Form, since each attribute is prime (each attribute is a key, actually), so that the definition of 3NF is respected. Then, note that the algorithm is incomplete. The steps are:
What is the relation between 2NF and 3NF?
D and E depend only on A, F depends only on B, G, H, I and J don’t depend on the key (directly) at all. R0, R1, and R2 contain no partial dependencies (or repeating groups) so they are 2NF. However R1 and R2 are still an issue, because they contain transitive dependencies. I and J depend on D, not on the key of R1.
Can you check if a relation is in 3NF?
Of course, if one could check at the beginning if the relation is already in 3NF, than he can avoid to apply the algorithm. But this in general cannot be done, since the check requires the exponential computation of all the keys, to find the prime attributes of the relation.